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https://www.gams.com/latest/docs/S_BARON.html | BARON
Date
5 June 2015
Introduction
The Branch-And-Reduce Optimization Navigator (BARON) is a GAMS solver for the global solution of nonlinear (NLP) and mixed-integer nonlinear programs (MINLP).
While traditional NLP and MINLP algorithms are guaranteed to converge only under certain convexity assumptions, BARON implements deterministic global optimization algorithms of the branch-and-bound type that are guaranteed to provide global optima under fairly general assumptions. These include the existence of finite lower and upper bounds on nonlinear expressions in the NLP or MINLP to be solved.
BARON implements algorithms of the branch-and-bound type enhanced with a variety of constraint propagation, interval analysis, and duality techniques for reducing ranges of variables in the course of the algorithm. Rigorous relaxations are constructed by enlarging the feasible region and/or underestimating the objective function.
Parts of the BARON software were created at the University of Illinois at Urbana-Champaign. The algorithms implemented in the software, the theory behind them, and some related applications are (partly) described in [208, 73, 209, 215, 123, 160, 108, 239, 220, 240, 212, 107, 210, 230, 231, 232, 235, 236, 214, 211, 216, 10, 233, 213, 234, 56, 25, 199, 26, 140, 141, 262, 263, 142, 264, 27] .
Licensing and software requirements
In order to use GAMS/BARON, users will need to have a GAMS/BARON license. BARON comes with several embedded LP/MIP and NLP solvers (CBC; IPOPT, FilterSD, FilterSQP). Additionally, GAMS/BARON users can expedite convergence by accessing CPLEX and XPRESS to solve BARON's LP/MIP subproblems and MINOS, SNOPT, and any GAMS NLP solver, such as CONOPT, to solve BARON's NLP subproblems. These solvers need to be licensed separately.
By default, GAMS/BARON will attempt to use CPLEX as the LP solver and select an NLP solver automatically. The user can use the options LPSol and NLPSol to specify the LP or NLP solver. If the user does not have a license for the default or user-specified LP solver, BARON will switch to CLP. If the user does not have a license for the user-specified NLP solver, BARON will automatically select a licensed NLP solver and may default to IPOPT if no other NLP solver is available. BARON can be used without a local NLP solver by setting DoLocal and and NumLoc to 0.
Running GAMS/BARON
BARON is capable of solving models of the following types: LP, MIP, RMIP, NLP, DNLP, RMINLP, and MINLP. If BARON is not specified as the default solver for these models, it can be invoked by issuing the following command before the solve statement:
option <modeltype>=baron;
where <modeltype> stands for LP, MIP, RMIP, QCP, MIQCP, RMIQCP, CNS, NLP, DNLP, MINLP, or RMINLP.
Model requirements
In order to achieve convergence to global optimality, additional model constraints may be required. The additional constraints may speed up solver time and increase the probability of success.
Variable and expression bounds
All nonlinear expressions in the mathematical program to be solved must be bounded below and/or above. It is important that finite lower and upper bounds be provided by the user on all problem variables. Note that providing finite bounds on variables alone may not suffice to guarantee finite bounds on nonlinear expressions arising in the model.
For example, consider the term 1/x for x[0,1], which has finite variable bounds, but is unbounded. It is important to provide bounds for problem variables that guarantee that the problem functions are finitely-valued. If the user model does not include variable bounds that guarantee that all nonlinear expressions are finitely-valued, BARON will attempt to infer appropriate bounds from problem constraints. If this step fails, global optimality of the solutions provided is not guaranteed. Occasionally, because of the lack of bounds no numerically stable lower bounding problems can be constructed, in which case BARON may terminate.
See section Some BARON features on how to specify variable bounds.
Allowable nonlinear functions
In addition to multiplication and division, GAMS/BARON can handle nonlinear functions that involve exp(x), ln(x), xα for real α, βx for real β, xy, and |x|. Currently, there is no support for other functions, including the trigonometric functions sin(x), cos(x), etc.
BARON output
BARON log output
The log output below is obtained for the MINLP model gear.gms from the GAMS model library using absolute and relative optimality tolerances of 1e-6.
===========================================================================
BARON version 15.6.5. Built: LNX-64 Fri Jun 5 08:34:09 EDT 2015
If you use this software, please cite:
Tawarmalani, M. and N. V. Sahinidis, A polyhedral
branch-and-cut approach to global optimization,
Mathematical Programming, 103(2), 225-249, 2005.
BARON is a product of The Optimization Firm, LLC. http://www.minlp.com/
Parts of the BARON software were created at the
University of Illinois at Urbana-Champaign.
===========================================================================
This BARON run may utilize the following subsolver(s)
For LP/MIP: ILOG CPLEX
For NLP: MINOS, SNOPT, GAMS external NLP, COIN IPOPT with MUMPS and METIS, FILTERSD
===========================================================================
Starting solution is feasible with a value of 36.1767610000
Doing local search
Solving bounding LP
Starting multi-start local search
Preprocessing found feasible solution with value 4.23791612465
Done with local search
===========================================================================
Iteration Open nodes Time (s) Lower bound Upper bound
1 1 0.02 1.00000 4.23792
* 1+ 1 0.02 1.00000 3.29321
* 1+ 1 0.02 1.00000 2.20487
1 1 0.02 1.00000 2.20487
* 2 2 0.02 1.00000 1.06987
* 3+ 1 0.03 1.00000 1.01273
* 4+ 2 0.03 1.00000 1.00117
* 4+ 2 0.03 1.00000 1.00018
* 4 2 0.03 1.00000 1.00004
* 14 8 0.04 1.00000 1.00001
* 30 0 0.05 1.00000 1.00000
30 0 0.05 1.00000 1.00000
Cleaning up
*** Normal completion ***
Wall clock time: 0.05
Total CPU time used: 0.05
Total no. of BaR iterations: 30
Best solution found at node: 30
Max. no. of nodes in memory: 11
All done
===========================================================================
The solver first tests feasibility of the user-supplied starting point. This point is found to be feasible with an objective function value of 36.1767610000. BARON subsequently performs a randomized local search procedure and, eventually, finds a feasible solution with an objective of 4.23791612465. Then, the iteration log provides information every 1,000,000 branch-and-bound iterations and every 30 seconds. Additionally, information is printed at the end of the root node, whenever the value of the incumbent is improved by at least 10-5, and at the end of the search. A star (*) in the first position of a line indicates that a better feasible solution was found that improves the value of previous best known solution by at least 10-5. The log fields include the iteration number, number of open branch-and-bound nodes, the CPU time taken thus far, the lower bound, and the upper bound for the problem. The log output fields are summarized below:
Field Description
Itn. no. The number of the current iteration. A plus (+) following the iteration number denotes reporting while solving a probing (as opposed to a relaxation) subproblem of the corresponding node.
Open Nodes Number of open nodes in branch-and-reduce tree.
Time Current computational time in seconds. CPU time is reported for single-threaded jobs and wall clock time is reported for multi-threaded jobs.
Lower BoundCurrent lower bound on the model.
Upper BoundCurrent upper bound on the model.
Once the branch-and-reduce tree is searched, the best solution is isolated and a corresponding dual solution is calculated. Finally, the total number of branch-and-reduce iterations (number of search tree nodes) is reported, followed by the node where the best solution was identified (a -1 indicates preprocessing as explained in the next section on termination messages).
Termination messages, model and solver statuses
Upon normal termination, BARON will report the node where the optimal solution was found. We refer to this node as nodeopt. Associated with this node is a return code indicating the status of the solution found at nodeopt. The return code is given in the log line:
Best solution found at node: (return code)
The return codes have the following interpretation:
-3 : no feasible solution found
-2 : the best solution found was the user-supplied
-1 : the best solution was found during preprocessing
i : the best solution was found in the i-th node of the tree
In addition to reporting nodeopt, upon termination, BARON will issue one of the following statements:
*** Normal completion ***
This is the most desirable termination status. The problem has been solved within tolerances in this case. If BARON returns a code of -3, then no feasible solution exists.
*** Heuristic termination ***
While global optimality is not guaranteed in this case, BARON will terminate with this message when (a) a feasible solution has been found and (b) the progress of lower/upper bounds satisfies the heuristic termination criterion set by the user through BARON's DeltaTerm option.
*** User did not provide appropriate variable bounds ***
The user will need to read the BARON output (in file sum.dat in the scratch directory, use GAMS parameter keep=1 to prevent the automatic removal of this directory) for pointers to variables and expressions with missing bounds. The model should be modified in order to provide bounds for variables and intermediate expressions that make it possible for BARON to construct reliable relaxations. Even though relaxation bounds are printed on the screen to give the user a feeling for convergence, these bounds may not be valid for the problem at hand. This message is followed by one of the following two messages:
***Infeasibility is therefore not guaranteed ***
This indicates that, because of missing bounds, no feasible solution was found, but model infeasibility was not proven.
*** Globality is therefore not guaranteed ***
This indicates that, because of missing bounds, a feasible solution was found, but global optimality was not proven.
*** Max. allowable nodes in memory reached ***
The user will need to increase the physical memory of the computer or change algorithmic options, such as branching and node selection rules, to reduce the size of the search tree and memory required for storage.
*** Max. allowable BaR iterations reached ***
The user will need to increase the maximum number of allowable iterations. The BARON option is MaxIter. To specify this in GAMS, one can use the NodLim option. We remark that the BARON option MaxIter overrides NodLim.
*** Max. allowable CPU time exceeded ***
The user will need to increase the maximum of allowable CPU time. The BARON option is MaxTime. To specify this in GAMS, one can use the ResLim option. We remark that the BARON option MaxTime overrides ResLim.
*** Problem is numerically sensitive ***
BARON is designed to automatically handle problems with numerical difficulties. However, for certain problems, the global optimum is numerically sensitive. This occurs, for instance, when the objective function value varies significantly over small neighborhoods of points that are strictly outside the feasible region but nonetheless feasible within numerical tolerances. When BARON returns this message, the "Best possible" reported on the objective is likely correct.
*** Search interrupted by user ***
The run was interrupted by the user (Ctrl-C).
*** Insufficient memory for data structures ***
More memory is needed to set up the problem data structures. The user will need to increase the physical memory available on the computer in order to accommodate problems of this size.
*** A potentially catastrophic access violation just took place ***
In the unlikely event of a memory access violation, BARON will terminate the search and return the best known solution. Please report problems that lead to this condition to Nick Sahinidis (niksa.nosp@m.h@mi.nosp@m.nlp.c.nosp@m.om).
Some BARON features
The features described in this section rely on options that are further detailed in the next section. For details of the algorithmic implementations, the user may wish to consult publications cited at the end of this document.
No starting point is required
In contrast to many NLP algorithms that require a feasible starting point, a starting point is not required for BARON. A user may optionally provide a starting point for all or even some of the problem variables. BARON will judiciously initialize any variables that are not initialized by the user. Even when the problem functions cannot be evaluated at a user-provided starting point, BARON is still capable of carrying out its global search.
For problems for which GAMS compilation is aborted because the nonlinear functions cannot be evaluated at the starting point, the user can use the following commands before the SOLVE statement:
MaxExecError = 100000;
option sys12 = 1;
The first command asks GAMS to continue compilation for as many as MaxExecError execution errors. The sys12 option will pass the model to the BARON despite the execution errors. Even though the starting point is bad in this case, BARON is capable of carrying out its global search.
Finding a few of the best or all feasible solutions
BARON offers a facility, through its NumSol option, to find the best few, or even all feasible, solutions to a model. The development of this facility was motivated by combinatorial optimization problems but the facility is applicable to continuous problems as well. Even for combinatorial problems, BARON does not rely on integer cuts to find multiple solutions. Instead, it utilizes a single search tree, thus providing a computationally efficient method for finding multiple solutions. Furthermore, because BARON's approach applies to integer as well as continuous programs, it can be used to find all feasible solutions to a system of nonlinear equality and inequality constraints.
Once a model is solved by BARON with the NumSol option, the solutions found can be recovered using the GAMS GDX facility. An example is provided below.
$eolcom !$Ontext
Purpose: demonstrate use of BARON option 'numsol' to obtain the best
numsol solutions of an optimization problem in a single branch-and-bound
search tree.
The model solved here is a linear general integer problem with 18 feasible
solutions. BARON is run with a request to find up to 20 solutions. The
model solved is the same as the one solved in gamslib/icut.gms.
$Offtext set i index of integer variables / 1 * 4 / variables x(i) variables z objective variable integer variable x; x.lo(i) = 2; x.up(i) = 4; x.fx('2') = 3; ! fix one variable x.up('4') = 3; ! only two values equation obj obj definition; * pick an objective function which will order the solutions obj .. z =e= sum(i, power(10,card(i)-ord(i))*x(i)); model enum / all /; * instruct BARON to return numsol solutions$onecho > baron.opt
numsol 20
gdxout multsol
\$offecho
enum.optfile=1; option mip=baron, limrow=0, limcol=0, optca=1e-5,
optcr=1e-5; solve enum minimizing z using mip;
* recover BARON solutions through GDX
set sol /multsol1*multsol100/; variables xsol(sol,i), zsol(sol);
execute 'gdxmerge multsol*.gdx > %gams.scrdir%merge.%gams.scrext%';
option decimals=8;
display xsol.l, zsol.l;
Using BARON as a multi-start heuristic solver
To gain insight into the difficulty of a nonlinear program, especially with regard to existence of multiple local solutions, modelers often make use of multiple local searches from randomly generated starting points. This can be easily done with BARON's NumLoc option, which determines the number of local searches to be done by BARON's preprocessor. BARON can be forced to terminate after preprocessing by setting the number of iterations to 0 through the MaxIter option. In addition to local search, BARON's preprocessor performs extensive reduction of variable ranges. To sample the search space for local minima without range reduction, one would have to set to 0 the range reduction options TDo, MDo, LBTTDo, and OBTTDo. On the other hand, leaving these options to their default values increases the likelihood of finding high quality local optima during preprocessing. If NumLoc is set to -1, local searches in preprocessing will be done from randomly generated starting points until global optimality is proved or MaxTime seconds have elapsed.
Systematic treatment of unbounded problems
If BARON declares a problem as unbounded, it will search for and may report a vertex and direction of an unbounded ray. In addition, BARON will report the best solution found. This will be a feasible point that is as far as possible along an unbounded ray while avoiding numerical errors due to floating point arithmetic.
Systematic treatment of infeasible problems
If BARON declares a problem as infeasible, it has the capability to identify a subset of the constraints that are infeasible and become feasible once any one of them is eliminated. This, so-called, irreducibly inconsistent system (IIS) can be obtained by BARON for all types of problems handled by BARON, including linear and nonlinear, continuous and integer, convex and nonconvex, and problems with complementarity constraints. BARON's CompIIS option can be used to identify an IIS.
As an example, consider the problem of minimizing the nonconvex function $$x_1 x_3$$ over the following nonconvex constrained set:
\begin{eqnarray*} & e_1: & 85 + 0.006 x_2 x_5 + 0.0006 x_1 x_4 - 0.002 x_3 x_5 <= 92 \\ & e_2: & 0.8 x_2 x_5 + 0.003 x_1 x_2 + 0.002 x_3^2 = 110 \\ & e_3: & 9 + 0.005 x_3 x_5 + 0.001 x_1 x_3 + 0.002 x_3 x_4 <= 25 \\ & & 78 \leq x_1 \leq 102 \\ & & 33 \leq x_2 \leq 45 \\ & & 27 \leq x_i \leq 45, \quad i=3,\ldots,5 \end{eqnarray*}
When this problem is solved with CompIIS equal to 1, BARON provides the following result:
IIS contains 1 row and 3 columns as follows:
e2 Upper
x1 Lower
x2 Lower
x5 Lower
The IIS consists of the lower bounds of variables $$x_1$$, $$x_2$$, and $$x_5$$, along with the $$\le$$ part of the equality constraint $$e_2$$. This suggests that constraint $$e_2$$ and the entire model can be made feasible by lowering the lower bound of any of the three variables that are part of the IIS, whereas modifying the bounds of $$x_3$$ would not make the model feasible.
If a problem is known to be infeasible and the user desires to identify an IIS, it may be beneficial to set BARON's NumLoc option to zero. Doing so will deactivate BARON's initial upper bounding search, which involves multiple local searches. On the other hand, DoLocal should be nonzero in order to permit local search during the solution of certain subproblems that BARON solves while searching for an IIS.
Handling of complementarity constraints
Complementarity relationships of the type f(x)g(x) = 0 are automatically recognized and exploited algorithmically by BARON. The functions f and g may be univariate or multivariate, linear or nonlinear, convex or nonconvex, in terms of continuous and/or integer variables, and may be subject to additional constraints in the model. These complementarity relationships can be inferred by BARON even when implied by problem constraints and variable bounds. As a result, BARON can solve general mathematical programs with equilibrium constraints (MPECs). This class of problems includes the classical linear complementarity problem
(LCP): Find $$z\geq 0$$ and $$q$$ such that
\begin{align*} Mz + q & \geq 0, \\ z^t(Mz + q) & = 0 \end{align*}
as well as the more general mixed complementarity problem
(MCP): Given a function $$f: \mathbb{R}^n \rightarrow \mathbb{R}^n$$ and bounds $$l, u \in \mathbb{R}^n$$ with $$\overline{\mathbb{R}} = \mathbb{R}\cup \{-\infty,+\infty\}$$, find $$z \in \mathbb{R}^n$$ and $$w, v\in \mathbb{R}^n_+$$ such that
\begin{align*} f(z)&=w-v,\\ l \le z &\le u, \\ (z-l)^t w &= 0,\\ (u-z)^t v &= 0. \end{align*}
Both problems are automatically recognized and exploited by BARON without the user having to mark complementarities in any special way. In GAMS, all these problems can be solved by BARON when declared as NLP or MINLP models.
Parallel capabilities
For difficult problems with integer variables, most of BARON's time is spent on solving MIP relaxations. Hence, considerable speedups may be obtained via parallel solution of the MIP subproblems. For this purpose, the option threads may be used to specify the number of cores that BARON's MIP subsolver is allowed to use. By default, this option has the value of 1, meaning that a single core will be utilized.
The BARON options
The BARON options allow the user to control termination tolerances, branching and relaxation strategies, heuristic local search options, and output options as detailed in this section.
Many options can also be set in the GAMS model. The most relevant GAMS options are ResLim, NodLim, OptCA, OptCR, OptFile, and CutOff. The IterLim option is not implemented. specify BARON iterations, the user can set the MaxIter option, which is equivalent to the GAMS option NodLim.
Additionally, a BARON Options file can be provided. See section The Solver Options File for general use of solver option files.
For branching, users can specify separate branching priorities for any discrete and continuous variables using Dot Options. To specify variable branching priorities, one specifies
(variable).prior(value)
in the BARON options file, where (value) can be any non-negative real value. The lower the value, the higher the priority for branching, see also Setting Priorities for Branching. Specifying maxdouble for (value) translates to passing 0 as branching priority for (variable).
Termination options
Option Description Default
AbsConFeasTol Absolute constraint feasibility tolerance
This tolerance is used for general constraints and variable bounds. AbsConFeasTol must be ≥ 1e-12. A point is considered feasible for a constraint/bound if the absolute or relative constraint feasibility tolerance is satisfied.
Range: [1e-12, ∞]
1e-5
AbsIntFeasTol Absolute integer feasibility tolerance
All integer variable values must satisfy this tolerance. AbsIntFeasTol must be ≥ 1e-12. A point is considered integer feasible for a variable if integrality is satisfied using the absolute or relative integer feasibility tolerance.
Range: [1e-12, ∞]
1e-5
BoxTol Box elimination tolerance.
Boxes will be eliminated if smaller than this tolerance. BoxTol must be ≥ 1e-12.
1e-8
CutOff Eliminate solutions that are no better than this value
Can also be used with the GAMS model attribute option CutOff.
GAMS CutOff
DeltaA Absolute improvement for insufficient progress termination
DeltaAa) must be ≥ 1e-12.
Range: [1e-12, ∞]
∞
DeltaR Relative improvement for insufficient progress termination
DeltaRr) must be ≥ 1e-12.
Range: [1e-12, ∞]
1
DeltaT Time interval for insufficient progress termination
If DeltaTerm is set to 1, BARON will terminate if insufficient progress is made over DeltaTt) consecutive seconds. If δt is set to a non-positive quantity, BARON will automatically set δt equal to -δt times the CPU time taken till the end of root node processing. DeltaT can take any real value.
Range: [-∞, ∞]
-100
DeltaTerm Indicates whether insufficient progress termination is on or off
Users have the option to request BARON to terminate if insufficient progress is made over DeltaTt) consecutive seconds. Progress is measured using the absolute and relative improvement thresholds DeltaAa) and DeltaRr). Termination will occur if, over a period of δt consecutive seconds, the value of the best solution found by BARON is not improved by at least an absolute amount δa or an amount equal to δr times the value of the incumbent at time tt. This termination condition is enforced after processing the root node and only after a feasible solution has been obtained. Because it relies on CPU time measurements, which may depend on machine load, this option may result in nondeterministic behavior.
0: do not enforce this termination condition
1: terminate if progress is insufficient
0
EpsA Absolute termination tolerance.
BARON terminates if |U-L|EpsA, where U and L are the upper and lower bound, respectively, on the optimal value of the optimization problem at the current iteration. EpsA must be ≥ 1e-12.
GAMS OptCA
EpsR Relative termination tolerance.
BARON terminates if |U-L|EpsR|L|, where U and L are the upper and lower bound, respectively, on the optimal value of the optimization problem at the current iteration. EpsR must be ≥ 0.
GAMS OptCR
FirstFeas Changes the search for first numsol solutions
If set to 1, BARON will terminate once it finds NumSol feasible solutions, irrespective of solution quality.
0: search for the best NumSol feasible solutions
1: find NumSol solutions irrespective of solution quality
0
FirstLoc Terminate the search as soon as a local optimum is found 0
ISolTol Solution Distance
Separation distance between solutions. This option is used in conjunction with NumSol. For combinatorial optimization problems, feasible solutions are isolated. For continuous problems, feasible solution points within an l distance that does not exceed IsolTol will be treated as identical by BARON. IsolTol must be ≥ 1e-12.
1e-4
MaxIter Maximum number of branch-and-reduce iterations allowed
Setting MaxIter to 0 will force BARON to terminate after root node preprocessing. Setting MaxIter to 1 will result in termination after the solution of the root node. MaxIter must be ≥ -1, where -1 implies unlimited.
Range: [-1, ∞]
GAMS NodeLim
MaxTime Maximum time allowed (sec)
MaxTime must be -1 or > 0, where -1 implies unlimited. For single-threaded jobs, i.e., when threads equals 1, this limit is enforced on CPU time consumed by the job. For multi-threaded jobs, the limit is enforced on wall clock time.
GAMS ResLim
NumSol Number of feasible solutions to be found
Solutions found will be listed in the results file. As long as NumSol ≠ -1, these solutions will be sorted from best to worse. If NumSol is set to -1, BARON will search for all feasible solutions to the given model and print them, in the order in which they are found, in the results file. NumSol must be ≥ -1.
Range: [-1, ∞]
1
RelConFeasTol Relative constraint feasibility tolerance
This tolerance is used for general constraints and variable bounds. RelConFeasTol must be ≥ 0. A point is considered feasible for a constraint/bound if the absolute or relative constraint feasibility tolerance is satisfied.
Range: [0, 0.1]
0
RelIntFeasTol Relative integer feasibility tolerance
All integer variable values must satisfy this tolerance. RelIntFeasTol must be ≥ 0. A point is considered integer feasible for a variable if integrality is satisfied using the absolute or relative integer feasibility tolerance.
Range: [0, 0.1]
0
Relaxation options
Option Description Default
Nouter1 Number of outer approximators of convex univariant functions
NOuter1 must be ≥ 0.
4
NoutIter Number of rounds of cutting plane generation at node relaxation
NOutIter must be a ≥ 0.
4
NoutPerVar Number of outer approximations per variable for convex multivariate functions
NOutPerVar must be ≥ 0.
4
OutGrid Number of grid points per variable for convex multivariate approximators of BARON's CONVEX_EQUATIONS.
OutGrid must be a ≥ 0.
20
Threads Number of cores used for solution of MIP subproblems
The value of this option is passed to CBC, CPLEX, and XPRESS.
Range: [1, ∞]
max(1,GAMS Threads)
Range reduction options
Option Description Default
LBTTDo Linear-feasibility-based range reduction option (poor man's LPs)
0: no range reduction based on feasibility.
1: range reduction done based on feasibility.
1
MDo Marginals-based reduction option
0: no range reduction based on marginals.
1: range reduction done based on marginals.
1
OBTTDo Optimality based tightening option
0: no range reduction based on optimality.
1: range reduction done based on optimality.
1
PDo Number of probing problems allowed
-2: automatically decided by BARON.
-1: probing on all variables.
0: no range reduction by probing.
n: probing on n variables.
-2
TDo Nonlinear-feasibility-based range reduction option (poor man's NLPs)
0: no bounds tightening is performed.
1: bounds tightening is performed.
1
Tree management options
Option Description Default
BrPtStra Branching point selection strategy
0: BARONs dynamic strategy
1: w-branching
2: bisection-branching
3: convex combination of 1 and 2
0
BrVarStra Branching variable selection strategy
0: BARONs dynamic strategy
1: largest violation
2: longest edge
0
NodeSel Specifies the node selection rule to be used for exploring the search tree
Specifies the node selection rule to be used for exploring the search tree.
0: BARONs mixed selection scheme
1: best bound
2: last in first out [LIFO]
3: minimum infeasibility
0
Local search options
Option Description Default
DoLocal Local search option for upper bounding
0: no local search is done during upper bounding
1: BARON's dynamic local search decision rule
1
NumLoc Number of local searches done in preprocessing
The first local search begins with the user-specified starting point. Subsequent local searches are done from judiciously chosen starting points. If NumLoc is set to -1, local searches in preprocessing will be done until proof of globality or MaxTime is reached. If NumLoc is set to -2, BARON decides the number of local searches in preprocessing based on problem and NLP solver characteristics. NumLoc must be ≥ -2.
Range: [-2, ∞]
-2
Output Options
Option Description Default
LocRes Option to control output from local search
0: no local search output
1: detailed results from local search will be printed to res.dat file
0
prfreq Log output frequency in number of nodes 1000000
prlevel Defines the level of log output printed.
0: all log output is suppressed
1: print log output
1
prtimefreq Log output frequency in number of seconds 30
Subsolver Options
Option Description Default
AllowExternal Indicator for use of External NLP solver with automatic NLP solver selection
In case of automatic NLP solver selection, this option can be used to selectively permit or disallow the use of external GAMS NLP solver as an NLP subsolver.
0: do not use the GAMS external NLP solver for local search
1: consider the GAMS external NLP solver for local search
1
AllowFilterSD Indicator for use of FILTERSD with automatic NLP solver selection
In case of automatic NLP solver selection, this option can be used to selectively permit or disallow the use of FILTERSD as an NLP subsolver.
0: do not use FILTERSD for local search
1: consider FILTERSD for local search
1
AllowFilterSQP Indicator for use of FILTERSQP with automatic NLP solver selection
In case of automatic NLP solver selection, this option can be used to selectively permit or disallow the use of FILTERSQP as an NLP subsolver.
0: do not use FILTERSQP for local search
1: consider FILTERSQP for local search
1
AllowIpopt Indicator for use of IPOPT with automatic NLP solver selection
In case of automatic NLP solver selection, this option can be used to selectively permit or disallow the use of IPOPT as an NLP subsolver. Currently, this option defaults to 0 on Mac OS X.
0: do not use IPOPT for local search
1: consider IPOPT for local search
1
AllowMinos Indicator for use of MINOS with automatic NLP solver selection
In case of automatic NLP solver selection, this option can be used to selectively permit or disallow the use of MINOS as an NLP subsolver.
0: do not use MINOS for local search
1: consider MINOS for local search
1
AllowSnopt Indicator for use of SNOPT with automatic NLP solver selection
In case of automatic NLP solver selection, this option can be used to selectively permit or disallow the use of SNOPT as an NLP subsolver.
0: do not use SNOPT for local search
1: consider SNOPT for local search
1
ExtNLPsolver External GAMS NLP solver and option file (e.g. conopt.1)
Specifies the GAMS NLP solver to be used when NLPSol is set to 6. All GAMS NLP solvers are available through this option. If a non-existing solver is specified or the solver chosen cannot solve NLPs, NLPSol will be reset to its default. A GAMS solver options file can be specified for the GAMS NLP solver by adding a dot followed by the options file number to the solver name, e.g., setting ExtNLPSolver to CONOPT.42 would instruct GAMS/CONOPT to use options file conopt.o42.
conopt
LPAlg Specifies the LP algorithm to be used (available only with CPLEX as the LP solver)
0: automatic selection of LP algorithm
1: primal simplex
2: dual simplex
3: barrier
0
LPSol Specifies the LP/MIP Solver to be used
3: CPLEX
7: XPRESS
8: CLP
3
NLPSol Specifies the NLP solver to be used
By default, BARON will select the NLP solver and may switch between different NLP solvers during the search, based on problem characteristics and solver performance. Any combination of licensed NLP solvers may be used in that case. A single specific NLP solver can be specified by setting this option to a value other than the default. If the specified solver is not licensed, BARON will default to automatic solver selection.
-1: Automatic NLP solver selection and switching strategy
0: Local search based on function evaluations alone with no calls to local solvers
2: MINOS
4: SNOPT
6: GAMS NLP solver (see ExtNLPsolver)
9: IPOPT
10: FILTERSD
14: FILTERSQP
-1
Other Options
Option Description Default
CompIIS Request the computation of an Irreducible Inconsistent Set (IIS)
In case of an infeasible problem, this option can be used to search for an IIS. Setting this option 1, works very well for most problems.
0: do not search for an IIS
1: the search for an IIS is based on a fast heuristic
2: an IIS is obtained using a deletion filtering algorithm
3: an IIS is obtained using an addition filtering algorithm
4: an IIS is obtained using an addition-deletion filtering algorithm
5: an IIS is obtained using a depth-first search algorithm
0
IISInt Indicates whether general integers should be considered as potential members of the IIS
When search for an IIS is requested through CompIIS, BARON assumes that the model is unlikely to include an error in terms of binaries, i.e., the binary definitions are assumed correct and the IIS output should be interpreted with respect to binary definitions. General integer bounds may be assumed as correct or can be questioned using the option IISint. Integrality is enforced in both cases.
0: do not consider general integers as part of an IIS, assume them to be correct
1: consider general integers (but not binaries) as part of an IIS
0
IISOrder Order in which constraints are considered in the search for an IIS
-1: auto set to aim for a small IIS depending on the value of CompIIS
1: arrange constraints in problem order
2: arrange constraints in ascending order of degree
3: arrange constraints in descending order of degree
>3: random order using IISorder as seed
-1
InfBnd infinity value to be used on bounds
If set to 0, then no bounds are used.
0
WantDual whether to try hard to provide dual solution values
0: use an inexpensive way to solve the KKT system to provide dual values
1: make a final call to the NLP solver to compute dual values if the inexpensive way of solving the KKT system failed
1 for LP/RMIP, 0 otherwise
Interface and Conversion
Option Description Default
ClockType Type of clock to use when reporting solving time back to GAMS
wall: report time according to "clock on the wall" (as used by most GAMS solver links)
cpu: report time used by CPU (summed up over all cores)
baron: report same time as used by BARON ("cpu" if one thread, "wall" if multiple threads)
wall
.EquClass Equation Classification
Specifies nature of named constraint in the users model. This is a Dot Option. Slices like supply.EquClass("new-york") 1 are allowed.
0: Regular constraint.
1: Relaxation-only constraint. These constraints are provided to BARON as RELAXATION_ONLY_EQUATIONS and used to help strengthen the relaxation bound but are not considered as part of the user model and thus not used for feasibility testing of solutions or local search. Adding, for instance, the first-order optimality conditions as relaxation-only constraints often expedites convergence.
2: Convex constraint. These constraints are provided to BARON as CONVEX_EQUATIONS and used to generate cutting planes from the set of outer approximating supporting hyperplanes of the convex constraint set.
3: Convex constraint that is relaxation-only.
0
GDXOut Prefix for GDX file names for multiple solutions if NumSol > 1. | 2020-01-26 05:07:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.631123423576355, "perplexity": 2579.4979414289746}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579251687725.76/warc/CC-MAIN-20200126043644-20200126073644-00148.warc.gz"} |
https://zbmath.org/?q=an:1206.11122 | ## Sums of two relatively prime $$k$$-th powers.(English)Zbl 1206.11122
Let $$k\geq 3$$ be a natural number. Let $$V_k(x)$$ be the number of solutions $$(u,v)$$ in $${\mathbb Z}^2$$ of $|u|^k+|v|^k\leq x,\quad (u,v)=1$ and let $E_k(x)=V_k(x)-c_kx^{\frac{2}{k}},\quad c_k=\frac{3}{\pi^2}\frac{\Gamma^2(\frac{1}{k})}{\Gamma(\frac{2}{k})}.$ The author proves:
Theorem 1. If $$\zeta(s)$$ has no zero with real part greater than $$\frac{123\theta_3-30}{90\theta_3-20}$$, $$\theta_3=\frac{9581}{36864}$$ then for every $$\varepsilon>0$$ $E_3(x)=O(x^{\theta_3+\varepsilon}).$
Theorem 2. If $$\zeta(s)$$ has no zero with real part greater than $$\frac{32\theta_4-5}{16\theta_4-1}$$, $$\theta_4=\frac{7801}{37616}$$ then for every $$\epsilon>0$$ $E_4(x)=O(x^{\theta_4+\varepsilon}).$ The mean value of $$E_k(x)$$ is also considered, and here an asymptotic formula is provided under the assumption of a zero-free strip of width $$>\frac{1}{k}$$.
### MSC:
11P21 Lattice points in specified regions 11D75 Diophantine inequalities 11J25 Diophantine inequalities 11D45 Counting solutions of Diophantine equations
### Keywords:
sums of two $$k$$-th powers
Full Text: | 2023-03-28 15:54:24 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9207763671875, "perplexity": 327.8941148554719}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948867.32/warc/CC-MAIN-20230328135732-20230328165732-00165.warc.gz"} |
https://codegolf.meta.stackexchange.com/questions/2140/sandbox-for-proposed-challenges/1938 | What is the Sandbox?
This "Sandbox" is a place where Code Golf users can get feedback on prospective challenges they wish to post to the main page. This is useful because writing a clear and fully specified challenge on the first try can be difficult. There is a much better chance of your challenge being well received if you post it in the Sandbox first.
To post to the Sandbox, scroll to the bottom of this page or click on the "Add Proposal" link below, and click "Answer This Question". Click "OK" when it asks if you really want to add another answer. Write your challenge just as you would when actually posting it. You may also add some notes about specific things you would like to clarify before posting it. Other users will help you improve your challenge by rating and discussing it. When you think your challenge is ready for the public, go ahead and post it, replace the post here with a link to the challenge and delete it.
See the Sandbox FAQ for more information on how to use the Sandbox.
Get the Sandbox Viewer to view the sandbox more easily
To add an inline tag to a proposal use shortcut link syntax with a prefix: [tag:king-of-the-hill]
The social network code-golf
On my social network, two users are "friends" if their name share a common letter. For exemple, bob and bill are friends, as they share the letter b.
Given a list of user names:
• display a falsy value if there exist in the list two distinct users x and y that cannot be related through a friendship chain;
• else, display a truthy value.
Examples
abc cde efg ghi should return true, as abc is friend with cde, which is friend with efg, which is friend with ghi : all users are related.
abc cde fgh hij should return false, as for example abc and fgh cannot be related through a friendship chain.
abc should return true, as we cannot find in that list two unrelated users.
Input
• You can read the name list in any convenient format for your language.
• You can assume all the names are lowercase and use only the characters a-z.
• You don't need to handle the empty list, any result (true, false, program crash) is acceptable for it.
• We've had transitive closure questions before (1, 2). This may be different enough to run, though (especially because the format increases the chance of a regex solution doing well), although it's particularly close to my second link there. I'd recommend the use of the graph-theory tag, though, as it's clearly heavily related to the other transitive closure questions. – user62131 Dec 31 '16 at 23:11
• @ais523 yes that's almost same than codegolf.stackexchange.com/questions/8647/…... will not post then – Arnaud Jan 2 '17 at 3:10
Smallest integer divisible by 2..n
Given an integer n, output the smallest integer divisible by 2,3,4,...,n inclusive.
Example
2520 is divisible by every integer from 2 to 10.
Scoring
Shortest code in bytes wins.
Sandbox
• Dup?
• Better Wording?
• Restrictions/Rules?
• So just lcm(2..n)? – FlipTack Dec 31 '16 at 12:18
• you're right. would be marked as dup I guess :D – Seims Dec 31 '16 at 13:09
• Actually, I don't think there's been challenges exactly like this before. I wouldn't call it a dupe. – FlipTack Dec 31 '16 at 14:41
Pseudoku Cops and Robbers King of the Hill
(I know that another user, @NathanMerrill, is proposing a similar contest. I started playing with the idea for this type of contest independently yesterday, but have since chatted with in The Nineteenth Byte. He is currently undecided on the type of puzzle to use and has some different ideas on how to evaluate participants' performance, so I feel comfortable proposing my idea as a separate challenge.)
Sudoku is a well-known logic puzzle. It is a puzzle of four nines: nine rows of cells, nine columns of cells, nine 3x3 adjacent and distinct blocks of cells, and nine values that any cell can have. A valid Sudoku arrangement or solution is one in which every row, cell, and block has all nine values exactly one time. For example, consider the following valid solution:
+-----+-----+-----+
|4 6 1|5 7 3|2 8 9|
|5 7 8|2 1 9|4 6 3|
|3 2 9|6 8 4|1 7 5|
+-----+-----+-----+
|9 8 4|7 6 2|3 5 1|
|7 5 6|3 4 1|9 2 8|
|2 1 3|9 5 8|7 4 6|
+-----+-----+-----+
|8 3 5|1 2 7|6 9 4|
|6 9 7|4 3 5|8 1 2|
|1 4 2|8 9 6|5 3 7|
+-----+-----+-----+
These are turned into puzzles by removing many of the values in the arrangement in such a way that all blanks are mirrored horizontally and vertically across the puzzle and so there is only one valid way to fill in the blanks to get a valid Sudoku solution. For the above puzzle, this might look like this:
+-----+-----+-----+
|4 | |2 8 |
| 7 | 1 | 3|
| 9| 4|1 |
+-----+-----+-----+
| |7 6 |3 5 1|
| | | |
|2 1 3| 5 8| |
+-----+-----+-----+
| 5|1 |6 |
|6 | 3 | 1 |
| 4 2| | 7|
+-----+-----+-----+
Someone who wished to play this Sudoku puzzle would then use the information provided to find the original solution.
Sudoku has some interesting properties that allow it to be generalized to similar puzzles with different rules that are sometimes called "Pseudoku" (which is pronounced the same way as the actual puzzle, SOO-DOE-KOO, so please stop saying SOO-DOO-KOO). For our purposes, we will make two differences. First, it may be possible to generate harder puzzles by removing the restriction for symmetric removals. The following is a valid puzzle by Sudoku rules, so why not allow it?
+-----+-----+-----+
|4 1| 7 |2 |
|5 |2 9| |
|3 | 8 | |
+-----+-----+-----+
| 8 4|7 2| 1|
| |3 1| 2 |
| | | |
+-----+-----+-----+
| | |6 |
| 9 7| | |
|1 | |5 3 |
+-----+-----+-----+
Second, Sudoku's properties allow us to define games with different sizes. You can define a Pseudoku game with a parameter N where the resulting board has N^2 rows, N^2 columns, N^2 blocks of size NxN, and N^2 values for each cell. Standard Sudoku would be a Pseudoku variant with N=3. So the following would be a valid Pseudoku(2) game:
+---+---+
|1 | |
| | 3|
+---+---+
| 1|4 |
|4 | |
+---+---+
and an example Pseudoku(4) game:
+-----------+-----------+-----------+-----------+
|11 | 10 4| 1 9 | 2 16|
| 6 5 | 15 1 | 3 2|12 8 |
|10 13 14| 12| 5 15| 4 7 |
| 2 3 | 6 13| 8 11| 5 10|
+-----------+-----------+-----------+-----------+
| 7 11 12 9|14 2|16 1 4 | |
| | 10 4 | 3|13 16 11|
| 4 10| 16 15| 12 | 6|
| 1 16 2 |11 3 | 10 8| |
+-----------+-----------+-----------+-----------+
| 3 2| 6 |13 | 5 14 1|
| 7 | 5| | |
| 13 14 4|12 16 | | 8 9 |
| 5 16 |13 9 | 4 2 1| |
+-----------+-----------+-----------+-----------+
|14 12 7| | 15 4|11 6 |
| 9 | 6 |11 16 | 3 |
| 11 | 8 13 1| 3 14 | 7 |
|13 8 | 7 11 | 2 | 15 |
+-----------+-----------+-----------+-----------+
Since Sudoku is NP-complete, so is Pseudoku. That means that it gets more difficult to solve a Pseudoku puzzle the larger N gets. However, it can take more time to generate Pseudoku puzzles than it does to solve them, since the naive algorithm for generating a puzzle requires solving the puzzle each time a value is removed! Solving Pseudoku puzzles is fun, but if it takes longer to generate them than it does to solve them, it becomes more work than play.
So help me out! I propose a Cops and Robbers style King of the Hill. The Cops will compete by writing programs to generate lots of Pseudoku puzzles to consume as much time as possible for solving, while the Robbers will compete by writing programs to solve Pseudoku puzzles to consume as little as time possible solving these puzzles.
I need some help ironing out the format, but here is what I have so far:
1. I will provide a Java framework for running the contest. This framework will connect to clients by TCP/IP so contestants can choose whatever language they want to write their Cops and Robbers (so long as I can run them on my system). I will also provide a basic Cop and Robber for these users to try out to see what sorts of times they take. I will publish the times they generate on my system so contestants can estimate how their entries will run on my system.
2. I will give each Cop ten minutes to generate as many Pseudoku(N>= 3) puzzles as they can, but they should be able to generate at least Pseudoku(N=4) puzzles. They can choose what sizes they want the puzzles to be, but they have to be valid with exactly one solution. My server will naively check each one to guarantee their validity; any Cop that generates an invalid puzzle is disqualified. I recommend configuring the Cop programs to be parameterized externally so that Robbers can test their code against basic Cop configurations, but then the Cops can send me secret, optimized configurations before the contest completes for their actual execution. I will provide a couple days after the deadline ends for conferring with the Cop programmers if their settings do not work as expected on my system. Cops should generate different puzzles every time with reasonable expectations; that means no spamming with the same puzzle repeatedly or reading pregenerated puzzles from a file system, Internet source, or internal cache. In addition, I don't want to see a Cop that uses the same removal pattern for every puzzle (that may not guarantee valid puzzles, anyway).
3. Each Robber will be tested against each puzzle generated by the Cops. The Robber will have to generate the correct solution for each puzzle as quickly as possible. I will probably need to see some timings before I make a final decision, but each Robber will be capped with some amount of time to solve a puzzle (maybe an hour?) before the time-to-completion defaults to twice that cap. These Robbers will be permitted to use any technique for solving the puzzles that my system supports except for packet sniffing. I am on the fence as to whether the Robbers will be on an honor code to not study Cop code since I plan to have secret parameterizations anyway.
4. All the times for all the puzzles will be sorted from least to greatest and then assigned an index as one would in a Mann-Whitney U test. Each Cop and Robber will be scored using the sum of the indices of their contributions: Cops for the times the Robbers spent solving their puzzzles, and Robbers for the times they spent solving puzzles. The winning Cop will have the highest sum and the winning Robber the lowest. Cop ties will be broken first by the average time required to solve one of its problems (more is better), then by the number of puzzles generated (more is better), then by the name I deem cooler (here's hoping that doesn't happen). Robber ties will be broken first by the average time spent solving puzzles, then by the sum of the time, then by the standard deviation, then by the cooler name.
5. This scoring scheme poses an interesting challenge to Cops: balancing the size of the problems (and the likely amount of time needed to solve them) against the number of problems generated. A Cop that generates only one puzzle that no Robber can solve in the time limit is likely to lose to another Cop that generates many moderate problems. Similarly, a Cop that spams many small problems is likely to be beaten by another Cop that generates fewer problems of larger sizes. Since the official contest configurations should be kept secret until the contest starts, other Cops can study the other programs to try to determine what their opponents are likely to do and plan accordingly.
• I pronounce Pseudoku as SOO-do-ku and Sudoku as soo-DO-ku, to align with the pronunciation of Pseudo. – Pavel Jan 6 '17 at 0:22
• I don't know if you'll get a lot of submissions. Sudoku is a bit difficult to program. Also, TCP-IP is not something people are used to using for their submissions. – mbomb007 Jan 6 '17 at 16:25
• It is? Well, part of what I am looking for is whether people would participate. I would need at a minimum two Cops and two Robbers or there is no point. Could people comment saying whether they would play and whether they would play as a Cop, a Robber, or both? – sadakatsu Jan 6 '17 at 17:17
• The scoring system you've chosen adds a large incentive to submit a huge number of programs that are almost identical to your own submission but slightly worse. This means that if some opposing programs generate some puzzles that are harder than yours and some puzzles that are easier (which is likely), you'll push the easy ones right down the leaderboard, making your programs look better in comparison. – user62131 Jan 6 '17 at 17:47
• Good point. This can be resolved by a "one-submission-per-category" rule. – sadakatsu Jan 6 '17 at 18:07
• If you want, I have some code that can help you communicate with submissions (over standard in/out). However, I think the best solution is to run the cops' submission and kill it after 10 minutes. They should write each sudoku puzzle as a file, which you would then read.. I also wouldn't worry about automated checking to see if the puzzles are valid. I generally assume good faith in these types of challenges, unless it becomes an issue. – Nathan Merrill Jan 6 '17 at 18:33
• Re: validation... there's a problem here. For it to make any sense, I would need to have at least as good a solver as the best Robber entry, in which case someone would just copy mine in a faster language than Java. I think I would still require a unit testing protocol where I pass solutions to the Cops, the Cops return problems, and then I validate that they sent unique puzzles. – sadakatsu Jan 6 '17 at 18:45
• Re: standard input/output versus sockets... I don't get the aversion to network programming. I used TCP in my Speed Clue contest a couple years ago, and it worked great (though I admittedly had few entries). So long as I guarantee an environment for the contest (probably Linux), even C/C++ developers can write code with platform-specific libraries if they wish. Using networking also allows a good method for timing responses: I start the clock once I get the ACK after sending a command, and I stop the clock after I ACK that I received a response. File dumping makes timing Cops difficult. – sadakatsu Jan 6 '17 at 18:50
• 1. I don't think you've taken account of how badly things scale. It's easy to spit out valid pseudokus for N=100; validating them in a reasonable time requires supporting every rule which the cop knows. 2. The stuff about secret parameterisations doesn't really make sense to me. Taken to extremes, that could mean that we make the actual generation code the "parameterisation" and the cop "program" is just an eval. 3. The cop/robber setup means there's inherently a submission deadline. That's generally a bad thing, but even more so with something which can get extremely complicated. – Peter Taylor Jan 10 '17 at 15:49
broken keyboard workaround
|nspired by BASTA´s song and memories from earlier work:
Your keyboard is broken but there is some urgent work you have to complete; you have no back^up hardware - and the shops are closed so you can´t buy a new keyboard!
All you have left to work with is your mouse.
6iven two texts as input (the one you have and the one you want to have), create a program or function that tells you the cut, copy&paste actions that will turn the one text into the other.
Using the mouse is strenuous, so you don´t want too many cut/copy/paste actions. Keep your output as short as possible.
Remember: Your keyboard is broken = you can´t use any characters in your code that you don't have ~ you must get alon9 with those that are provided in th1s {["te%t"]}. For7unately your keyboard 7ook qu17e a wh1le 7o bre4k down c0mpletely 4nd y0u used numb3rs 4nd sp3c*4l ch@r$t0 r3pl@c3 br0k3n l3773r$; $0 y0u$h0uld h@v3 m0$7 0f 7h3m @v@*l@bl3. Also, you don´t want to do too much C&P to cre8 your code, so keep that as short as possible, too. • You can assume that the second text contains no characters that are not present in the first text. • You can pick any input format and method that is convenient for you; but the output format should match that. (e.g. if you take input from files, output should also go to a file). NOTES Note that the challenge description contains all letters and digits except j and z. If you absolutely need them: they are hidden in the YouTube link. (I didn´t check for upper/lower case though.) Curlys, brackets, braces, single and double quotes and all operators I could think of are there, so the challenge should be fine for most languages that use printable ASCII. Still trying to find a more fluent way to include curlys, brackets, double quotes and circumflex, though. I thought there was a tag [string-manipulation]; but couldn´t find it in the list. I think about dropping the "output method should match input method" restriction. • Try [tag:code-golf] and [tag:restricted-source] – TrojanByAccident Jan 9 '17 at 16:49 • The part of your question with weird characters in it is a bit hard to read. Maybe tone it down a bit. Also, some examples would be helpful for understanding the challenge. – mbomb007 Feb 1 '17 at 4:51 Find B1nar0 Solutions B1nar0 puzzle is a paper and pencil game with 0 and 1. The goal is to fill the grid accoring to 3 rules : 1. No more than 2 consecutive 0s or 1s 2. Each row/column has half 0s and half 1s 3. No identical row/columns Example : [ • A is 0 according to rule 1 • B is 1 according to rule 1 • C is 0 according to rule 2 • D is 1 according to rule 2 • etc. Edit : Grids are square grids of even size (4, 6, 8, 10, 12 or 16 are usual sizes). Input : Any binary grid (array or string) with 0,1 and any other character you want for empty cells. Output : Same format as input but filled with a correct grid. Test case (see GIF) 0 11 0 0 1 1 • Are the grid dimensions always 4x4 ? – Arnaud Jan 17 '17 at 8:14 • Non, any even number should fit, generally 4-6-8-10-12 or 16 games. – Crypto Jan 17 '17 at 8:16 • Nice challenge, but I suggest you add some test cases for the larger grid sizes too. Also, may we assume that we will get input that makes it possible to solve the puzzle? For instance not: [[1,1,1, ],[ , , , ],[ , , , ],[ , , , ]]. Can we assume that there will only be one valid solution? For instance, not an empty grid. – Stewie Griffin Mar 9 '17 at 14:22 • Also, the 6x6 test case is a lot harder than the 4x4. The 4x4 can be solved going through the matrix checking the different rules one after the other. To solve the 6x6 grid you need an algorithm that's a lot more sophisticated. Do you require that the program should be able to solve any input, regardless of the difficulty. Even if it requires brute-forcing the solution (which may take a loong time for a 16x16 matrix – Stewie Griffin Mar 9 '17 at 14:22 • Then you should add a few 10x10, 12x12 and 16x16 test cases (with solutions). Requiring that submissions can solve all possible boards regardless of difficulty makes this a really hard challenge. You should also impose a time limit. Otherwise I can just write a script that checks all possible combinations and claim that it will eventually find the right solution – Stewie Griffin Mar 9 '17 at 14:22 • I added the comments I had on the post when it was on main. Some of them are a bit out of context now, but I guess you remember what they're about. :) – Stewie Griffin Mar 9 '17 at 14:23 • Nice challenge, but I suggest you add some test cases for the larger grid sizes too. Also, may we assume that we will get input that makes it possible to solve the puzzle? For instance not: [[1,1,1, ],[ , , , ],[ , , , ],[ , , , ]]. Can we assume that there will only be one valid solution? For instance, not an empty grid – Stewie Griffin Mar 9 '17 at 14:23 Enthusiastically Russianify a String Greetings Comrades, Many of you may have interacted with people from Russia on the internet at some point, and a subset of you may have noticed the slightly odd method they have of expressing themselves. e.g. деинсталляция игра нуб))) - (forgive the google translate) where the ))) are added for emphasis on the previous statement, I have been working on a theory that the ratio of )'s to the rest of the string is directly proportional to the amount of implied emphasis, however I oftentimes find it difficult to compute the ratio on the fly, as I am also trying to cope with a slew of abuse, so I would like the shortest possible code to help me calculate what the resulting string should be, for a value of enthusiasm between 0 and 500%, given the original, unenthusiastic string, this will aid my research greatly as i will not have to type out bulky scripts every time I wish to test my hypothesis. so, the challenge: write a full program or function, which, provided two arguments, a string of unknown length, and a number, in either integer format (between 0 and 500) or in decimal format (between 0 and 5, with 2 points of accuracy) will • return the original string, suffixed with a number of )'s • the number will be the calculated as a ratio of the input number to the string length. • so if the number 200, or 2.00 was provided, 200% of the string must be suffixed as )'s • the number of brackets rounded to in decimal situations does not matter. • script is required to support Printable ASCII characters. • only has to support one input number format, of your choice. examples: "codegolf" 125 = codegolf)))))))) "codegolf" 75 = codegolf)))))) "noob team omg" 0.5 = noob team omg)))))) "hi" 4.99 = hi!))))))))))))))) example code (powershell) (with decimal input): Function Get-RussianString ([string]$InputStr,[decimal]$Ratio){$StrLen = $InputStr.Length$SuffixCount = $StrLen *$Ratio
$Suffix = [string]::New(")",$SuffixCount)
return $InputStr +$Suffix
}
Get-RussianString "codegolf" 0.5
codegolf))))
this is so shortest code wins!
This is my first challenge, any feedback is greatly appreciated.
• Privyet tovarisch, but challenges on PPCG need an objective winning criterion (eg code-golf for shortest code) – TuxCrafting Jan 25 '17 at 16:27
• @TùxCräftîñg - apologies this is code-golf, I included a mention of it in the 'background' block shortest possible code i'll include the tag now though. – colsw Jan 25 '17 at 16:49
• @AdmBorkBork the minimum character set would be that, full Cyrillic alphabet support would be ideal, but I decided to simplify that aspect as much as possible, I could change it to the full ASCII set or otherwise if you believe it would be of benefit? - i'll include space as a default charachter, and remove the ! in the examples for now though. – colsw Jan 25 '17 at 16:52
• Restricting the input to "Printable ASCII" would probably be sufficient. – AdmBorkBork Jan 25 '17 at 16:58
• If anything that's actually more understandable - i'll edit that in now, thanks! – colsw Jan 25 '17 at 17:00
• Please edit the answer down to a hyperlink to the posted answer on the main site and delete it now that it is posted. – mbomb007 Feb 1 '17 at 4:49
Animate the text in your terminal
The goal
The goal is to "animate" the string "Hello world" in your output so that each character gets capitalised after each other.
Your program can exit after each letter has been capitalised.
For example;
# Iteration 1
Hello world
# Iteration 2
hEllo world
# Iteration 3
heLlo world
# Iteration 4
helLo world
# Iteration 5
hellO world
# Iteration 5
hello world
# Iteration 6
hello World
# Iteration 7
hello wOrld
# Iteration 8
hello woRld
# Iteration 9
hello worLd
# Iteration 10
hello worlD
Input
No input is required, but "Hello world" must be the string that is "animated".
Output
The string "Hello world" must be animated. The output must be as 1 line to create a sort of wave animation. Example gif;
I saw this on a metasploit youtube video and thought the effect was pretty cool, which is where I recorded the gif from, so it's a little laggy, but I hope it illustrates the output fine
This is code-golf, lowest byte-count will be deemed the winner.
• I think that you should make it that you take input and animate that. – caird coinheringaahing Jun 9 '17 at 15:01
Write a program or function that takes two arbitrarily long, nonnegative binary integers and adds them.
Rules
• The point of this challenge is to do the addition in binary; therefore, you may not use any base-conversion builtins. Writing your own base conversion code, while not banned, is highly discouraged.
• Your algorithm must work in theory for arbitrarily long inputs. It may fail because your language's storage method isn't large enough to hold the result. However, it must work for all of the test cases below (the last of which has a 71-bit result).
• These are nonnegative integers: no need to worry about one's/two's complement, fractions, etc.
• You may use big-endian or little-endian order for your input and output. That is, the numbers' most significant bit may be either on the left end or the right end of the input/output.
• If one of the numbers is shorter than the other, you may, optionally, pad it with zeros to the same length.
• Don't use loopholes.
Input and output
Input and output are flexible to accommodate as many languages as possible. You may use any of the default I/O methods. If your language has an unusual I/O method, leave a comment and I may allow it.
You may take input and produce output in the form of strings ("1001"), lists of numbers / strings / characters / booleans ([1,0,0,1], ["1","0","0","1"], [true,false,false,true]), or integers whose digits are all 1's and 0's (1001). You may not use decimal integers (9).
You may use other characters besides 0 and 1, as long as you pick two printable ASCII characters (or single digits/single-digit numbers) and use them consistently. You may, if you prefer, take input as a single string, with a one-character delimiter between the two numbers.
Test cases
0, 0 -> 0
1, 0 -> 1
1, 1 -> 10
11, 10 -> 101
111, 10 -> 1001
0, 1000101 -> 1000101
1, 1111111111111111111 -> 10000000000000000000
1100011001001101111011111010101010011010000111100001010111101000110111, 1100010011110101001101110111010000000110101001010111000010110001101110 -> 11000101101000011001001110001111010100000110000111000011010011010100101
0, 0 -> 0
1, 0 -> 1
1, 1 -> 01
11, 01 -> 101
111, 010 -> 1001
0000000, 1010001 -> 1010001
1000000000000000000, 1111111111111111111 -> 00000000000000000001
1110110001011110101000011110000101100101010101111101111011001001100011, 0111011000110100001110101001010110000000101110111011001010111100100011 -> 10100101011001011000011100001100000101011110001110010011000010110100011
This is . I will not be accepting an answer; the shortest code in each language wins.
Some related challenges, all of which ask for decimal integers:
• related? – steenbergh Feb 21 '17 at 9:44
• @steenbergh Yeah, I just found that one. Is it similar enough to count as a dupe? – DLosc Feb 21 '17 at 10:28
• 'fraid so. Your challenge specifically defines arbitrarily large in-/output, that might be the only difference. But adding the numbers is only a subset of what the other challenge asks us to do. – steenbergh Feb 21 '17 at 11:57
• @steenbergh Would it perhaps not be a dupe if I disallowed padding with zeros? The other challenge guarantees the inputs will be the same length, which allows for some algorithmic shortcuts. (Also, just a note: the accepted answer on the other challenge does not, in fact, work with arbitrarily sized input.) – DLosc Feb 22 '17 at 3:50
• I don't know - this looks too close to me. On the other hand, my latest challenge could've been seen as a dupe, yet it ran fine on main. You could always post on main, getting it closed is about the worst that can happen... – steenbergh Feb 22 '17 at 6:41
Introduction
You've calculated which of the first n numbers are prime, and want to save your achievement for all future generations. Unfortunately, you're broke, and want to minimize storage costs (you'll be paying them forever, after all.)
You need to determine the best way to pack all of the primes <=n and still be able to answer the question "is p prime?" in O(1) time.
Challenge
A submission to this challenge must include both a compress algorithm and an isPrime algorithm.
compress
Input: n -- the number that you have checked prime-hood through.
Output: Bytes to feed into your isPrime algorithm.
isPrime
Input: The output of your compression algorithm, and an integer i >= 0. i is guaranteed to be <= n.
Output: True if i is prime, otherwise False.
This algorithm must run in O(1).
The winner of this challenge is the (compression, isPrime) pair that is
• Correct
• Has the best compression ratio, as determined by the average compression ratio for
n in {10^3, 10^4, 10^5, 10^6, 10^7, 10^8, 10^9}
as compared to the naive solution below.
Consider the following solution in Python:
def compress(n):
# simple sieve of Eratosthenes. Note: this is not a
# prime generation challenge; a list of the first
# billion numbers will be provided in this format.
primes = [1] * (n + 1)
primes[0] = 0
primes[1] = 0
upper_bound = int(math.sqrt(n)) + 1
for i in range(2, upper_bound):
factor = i
if not primes[factor]:
continue
factor += i
while factor <= n:
primes[factor] = 0
factor += i
primePackStr = ''.join(str(i) for i in primes)
return primePackStr
def isPrime(compressed, i):
return compressed[i] == '1'
Example Input and Output
Input to compress:
20
Output:
"001101010001010001010"
Input to isPrime:
("001101010001010001010", 13)
Output:
True
Notes
• This is not a prime generation challenge. The compress executable can assume that there is a file called primes.txt in the same directory that contains the first billion numbers in the format s[i] = 1 if i is prime, 0 otherwise. (Zero-indexed)
• Naturally, the isPrime executable cannot make use of this file.
• The isPrime executable must not hardcode any primes.
• Please provide instructions on how to compile/run your code on either OSX 10.12 or Ubuntu 16.04, if it's not obvious.
• This is not a code golf challenge. Any length of code is fine, as long as isPrime doesn't attempt to cheat.
Notes for sandbox
• Any thoughts on a better restriction than "The isPrime executable must not hardcode any primes?"
• Should I test on random values of n instead?
• thanks!
• This is an interesting idea, and I hope it can be made to work, but it does have a big problem in the subtlety of what you mean by saying that isPrime must run in O(1) time. Interpreted with maximum pedantry, it's impossible because O(1) time isn't sufficient to read i from the input, even assuming random access to the input (which some key models of computation don't, and many interpreters won't give you). – Peter Taylor Mar 8 '17 at 9:12
• If you instead restrict answers to accessing a fixed (independent of n and of the length of the compressed text) number of bytes of the compressed text and doing a fixed (independent of n and i) amount of processing on them, you're pretty much killing the challenge because the only feasible compression will be bit-packing with a wheel and the competition will just be how big to make the wheel. – Peter Taylor Mar 8 '17 at 9:12
• In particular, a wheel size of 10^9 would trivialise the challenge. – Peter Taylor Mar 8 '17 at 12:38
• As far as your first comment goes, I could clarify to say that isPrime can assume that the entire output of compress is already in memory - or that isPrime may be called many times, with different i but the same compressed and it only has to be amortized O(1). Unfortunately, you're totally right about the prime wheel - though the idea is that the algorithm would work for arbitrary values of n, not just up to 10^9. – vroomfondel Mar 8 '17 at 14:34
• Maybe I could entirely remove the isPrime in O(1) restriction and simply make this a challenge about the most efficient compression algorithm for prime numbers. (Allowing arbitrary compression.) @PeterTaylor – vroomfondel Mar 8 '17 at 14:34
• If you do that then everyone will compress the list to 0 bytes unless you fix the decompression. A variant which might work is to ditch isPrime and say that the output of compress will be passed through zcat | sort -n so that the challenge is finding a good ordering of the primes which exploits Lempel-Ziv behaviour. – Peter Taylor Mar 8 '17 at 14:45
• That might be interesting, though I'd need to add some sort of polynomial time restriction - you could theoretically test all O((n/(log n))!) orderings of primes <= n otherwise. I'm going to abandon this for now, but may come back eventually if I have an epiphany. Thanks for your help! – vroomfondel Mar 8 '17 at 14:56
• This is a really interesting idea, and I hope you can come up with a way to make it work successfully. – AdmBorkBork Mar 8 '17 at 21:00
Challenge about loudly interjecting in a courtroom
One of the most important things about being a courtroom lawyer is loudly interjecting before you make your point. In this challenge, we're going to edit a typical courtroom transcript to include these interjections.
Any lawyer (and in fact, any character at all in the transcript), uses these rules to interject:
1. Use an interjection when the character who is speaking changes to you.
Take the following example:
SAHWIT: I remember the time I found the body exactly.
SAHWIT: It was 1 P.M.
PHOENIX: Frankly, I find that hard to believe!
There is one change in speaker, so an interjection will be added in at that point like this:
SAHWIT: I remember the time I found the body exactly.
SAHWIT: It was 1 P.M.
PHOENIX: Hold it! Frankly, I find that hard to believe!
1. Use Hold it! if the previous statement ends with a single full-stop or exclamation mark, Take that! if the previous statement ends with an elipses (...), and Objection! if the previous statement ends with a question mark.
For instance:
JUDGE: What evidence proves the clock is running slow?
PHOENIX: The victim had just returned from abroad the day before the murder.
PHOENIX: The time difference between here and Paris is 9 hours!
PAYNE: But modern day clocks automatically adjust for time zones...
PHOENIX: This is an antique!
Becomes:
JUDGE: What evidence proves the clock is running slow?
PHOENIX: Objection! The victim had just returned from abroad the day before the murder.
PHOENIX: The time difference between here and Paris is 9 hours!
PAYNE: Hold it! But modern day clocks automatically adjust for time zones...
PHOENIX: Take that! This is an antique!
That's about it. I'll write some longer test cases a bit later. This challenge is probably Retina-bait to be honest.
• This challenge might work better if the interjections made more sense in context. For example, "Objection!" likely works best after questions (as most objections in an actual court case are to try to invalidate a question that fails to follow the rules). – user62131 Mar 14 '17 at 4:35
Garbled Phone Numbers
(de)
You know how you get a voicemail message and the person's connection wasn't great, and you're trying to figure out how to call them back, but you're not sure if that was a "5" or an "8" they said?
That's this challenge.
The good news is that the caller read off their number twice, but it's garbled in both places.
Your program should take input like this:
5551231234 / 5551231234
Where the first seven digits are the first time the phone number is said in the voice mail and the second set are the second time it's said. Only...it'll look more like this:
555?AAA1_36? / 55?522_1?234
A digit followed by a question mark means that that's the best-guess for that digit (e.g. "5?" means "probably a 5, compare with repeat"). An underscore indicates a known missing digit, something too fuzzed by static to be deciphered at all. Letters are just that: letters. Treat them as their respective digits (ABC -> 2, DEF ->3, HIJ -> 4, etc).
You can safely assume the following judgement calls:
5? / _ -> 5 //5 is the best guess we have, use it
5? / 4? -> ? //conflict
5 / 4 -> ? //conflict
5? / 4 -> 4 //solid information overrides possible value
5 / 4? -> 5 //solid information overrides possible value
_ / _ -> ? //no information available
Additionally you can assume that all inputs will contain ten-digit phone numbers, not including the question marks. Inputs that aren't ten digits (e.g. 1234567 / 1234567) can either be treated as unsolvable (falsey output) or throw an error.
Output option A: Output a truthy value indicating whether or not a given input can be resolved to a single valid ten-digit phone number.
Output option B: If it can be parsed to a single valid ten-digit phone number, output the phone number. Otherwise output some form of error indication (e.g. -1, false, empty line).
Shortest wins, as per usual.
[Sample inputs]
• I'm not sure what your intended meaning for letters is. If it's just A=1,B=2,C=3... then they're a bit pointless and weird in this context. You should also probably choose only one between option A and option B before posting (I vote for B). – Leo Apr 9 '17 at 16:39
• @Leo Letters as they appear on a dial pad: A,B,C = 1, DEF = 2, GHI = 3, etc. – Draco18s no longer trusts SE Apr 9 '17 at 21:05
• You need an explicit mapping for letter→number. Most phones I've seen map A/B/C to 2 (apparently they follow this international standard). – user62131 Apr 9 '17 at 22:07
• @ais523 Whoops, that's what I get for posting late at night just before bed, then making the comment gia tablet at a rest stop somewhere in western Pennsylvania, 6 hours from home. – Draco18s no longer trusts SE Apr 10 '17 at 4:43
• I think you should omit 'output option A' and just keep B; B includes A pretty much. – officialaimm Apr 10 '17 at 9:36
• @officialaimm I think that's the beret idea, yes. There were a mess of ideas running around in my head, such as scoring based on a given input list, but never congealed well enough to make it to paper. A and B were the only two that did. – Draco18s no longer trusts SE Apr 10 '17 at 12:18
• Any other comments before I start generating some inputs and posting it? – Draco18s no longer trusts SE Apr 17 '17 at 19:46
Write a "21" game in exactly 21 characters code-bowling
Challenge
You must write a program which implements the following algorithm:
Let x = 0
Let y = truthy value
while (y is not falsy AND x <= 21) do:
Let x = x + a uniform random number from {1,2,3,...,11}
Output the value of x
Input a value of y from the user (you may assume input is valid)
Output the value of x
(You do not have to follow the pseudocode exactly. For example, if your language happens to initialise variables to a truthy value automatically, you don't have to include the y:=TRUE line. Similarly, you don't have to use a while loop. The important thing is that it repeatedly takes user input until either x exceeds 21 or the user chooses to stop, and it outputs the current value of x after each user input.)
Score
Let n be the length of the shortest program which meets the spec which can be obtained by deleting 0 or more characters from your code. Then your score is:
- 500 if n > 21
- 1 + (n-21)^2 if n < 21
The winner in each language is the program with the lowest score.
Questions
• Is this a resonable idea? I can't find similar challenges, so maybe there is a problem with ones like this? (Trivial solutions etc.)
• Is the specification too complicated (maybe more languages could enter if it was a simpler algorithm, for example just taking user input once?)
• This victory condition is code-shuffleboard. I don't think it adds much over just doing golf, though; it's normally fairly easy to pad out a program in a way that can't be fooled via simple character deletion. (Also, I suspect 21 characters isn't enough in most languages, although golfing languages should be able to beat that; it'll be interesting to see whether some of the terser practical languages can.) – user62131 Apr 11 '17 at 12:31
Tetris Programming
Your program or function should take as input one character from the set IJLOSTZ, which represents one of the seven tetrominos as usual:
I J L O S T Z
# # # ## ## ### ##
# # # ## ## # ##
# ## ##
#
You should output the number of shapes which are equivalent to this tetromino up to rotation. For example, the I only has two arrangements, horizontal and vertical, whereas the J has four different orientations. The O looks the same no matter how you rotate it, so there's only one shape. Here all possible input/output pairs:
I 2
J 4
L 4
O 1
S 2
T 4
Z 2
The Source Code
The main part of this challenge is the source code restriction:
• You may choose either linefeeds (LF, 0x0A), carriage returns (CR, 0x0D) or LR-CR pairs to represent newlines in your code (consistently). These split the source code into a 2D grid of lines (which aren't necessarily the same length).
• This 2D grid must be completely made up of tetrominos where all 4 characters in each tetromino must be the same. For example, this would be a valid program:
aaa
bba
cbddd
cbd
c
c
eeee
Note that the individual characters don't necessarily need to be distinct, so there may be larger connected regions of the same character, as long as this region can be segmented into non-overlapping tetrominos. Also note that this restriction also applies to spaces, so the following is not a valid program, because the two spaces don't form a tetromino.:
x
xxx
The Score
For each of the seven tetrominos count how often it appears in your source code. Your score is the maximum of these seven values.
That means you don't want to make up your code entirely of Is but instead try to use about the same number of each of them to keep the maximum of the seven values down.
The Small Print
You may either write a programs or a functions and use any of the standard methods of receiving input and providing output, as usual. Note that these loopholes are forbidden by default.
• It strikes me that most solutions are going to put most of the code in a comment, and it wouldn't surprise me if a score of 1 were achievable in some languages. Do you consider that reasonable? – user62131 May 5 '17 at 12:25
• @ais523 I'm not sure about the task yet. It's hard to find something that isn't too hard for the restriction to become a pain but that is intricate enough to allow people to use several parts of a tetromino. – Martin Ender May 5 '17 at 14:25
So I've been puzzling over the best way to present this idea I had, so this will probably need a lot of help. I am open to completely reworking the challenge, but this is the best polish I've managed to figure out so far.
At work I have to secure my laptop with one of those 4-digit cable locks and it occurred to me that there was a puzzle here: figuring out the combination by looking at the typical behavior of setting the lock: never allowing any given wheel to "rest" on its unlocked value. e.g. if the combination is 1234 then never walking away with a 1 in the first position, a 2 in the second, and so on (e.g. 1111 would not be considered locked, but 2111 would be). Or possibly by not letting any digit of the unlocked combination be visible (so even 2111 would be "bad" but 6789 would be ok...unless a transpose was also considered to be insufficiently random, however such choices are often up to the user of the lock). I also subsequently changed my behavior (not that I have any real risk of my laptop being stolen).
A standard challenge of "write some code that examines a series of locked values to determine the unlocked value, scoring by number of entries needed" is non-viable, as the sequence list would need to be carefully chosen such that there is a strictly known optimal solution (i.e. a minimum number of locked values), as finding a shortcut in that specific sequence might be possible, but invalid on another sequence.
Then the other night it occurred to me that it might be possible to do this as : one side has to randomize their locks (albeit following a set of rules that allows exploitation), the other side has to break them open. The downside being that it will be a nightmare to validate scores as there will be no easy way to pipe input and output back and forth between two programs running arbitrary languages.
I'm also not sure if there's enough room for freedom in designing the lock randomization code (i.e. interesting for the cop) for it to be plausibly crackable without resorting to brute force (an uninteresting challenge for the robber). Ostensibly the robber half is brute force, but it's guided in some manner towards a determinable value ("ah, I see, the first spindle is never set to 1 when locked, ergo the first digit in the code must be a 1) rather than indeterminate ("ah, I see, no spindle is ever set to 1, ergo there is a 1 somewhere in the code" -> 4 digits ^ 4 spindles -> 256 plausible values with further attempts gaining no new information).
Combination Locks (Cops)
Your goal is to write a program that produces a 4 digit random number as a combination lock entry code. Your program needs to keep this value a secret, but must produce output that is the result of the lock being locked and its tumblers spun, the value printed being the digits shown along the set row (8585 in the above image).
Your program will then take input of a 4 digit code that is an attempt to unlock the lock. If it is the correct value, output the number of attempts made and the seed value, otherwise print another randomized lock value. Repeat until successfully unlocked.
Rules:
• Your program must have some way of setting the combination (for scoring), eg. providing a seed value for the random number generator (inputting the correct combination is allowed).
• All locked combinations must be considered random. However:
• The nature of "random" is what is to be exploited here. Obviously you wouldn't want your random lock to actually remain unlocked after shuffling the dials!
• You may chose any rules by which to keep the lock locked, provided that it can be exploited. No outputting 0000 every iteration or cycling between predetermined sets (1234,4567,7890,1234). You're trying to emulate what appears to be smart behavior of a human being, not create an unbreakable lock.
• Every digit from 0-9 should be possible with some degree of uniformity. That is, if the correct combination is 1234 you are allowed to prevent 1 from showing up as the fist digit, but you may not prevent 1 from showing up in other positions.
• Blanket removal of all four digits of the combination from all four columns reduces the problem to brute forcing 256 possible combinations.
• Similarly, allowing a ban on a digit for up to three columns reduces it to brute force against 3136 possible combinations (banning only the combination digits from 3 columns is 81 possible combinations). None of these are interesting challenges.
• Entries shown to devolve to a brute force guessing will score based on the worst-case lucky guess (i.e. the number of attempts needed to identify the brute-force point, +1).
• If your language does not have a way to "wait for input" then....??? (requirements for fixed-seed randomness across multiple attempts, e.g. for a given combination and the same number of attempts made, the next output should be the same)
• Your program should store no data about attempts to break the lock or prior output values, the only data that may be stored are the Random instance (if needed), the correct combination, and number of attempts made. Outputting an attempt value back out (intentionally) would be underhanded.
Scoring
The ratio of your code's byte-length to the best (lowest) number of attempts needed by any robber against your lock.
Combination Locks (Robbers)
Your goal is to exploit the non-pure-random nature of locked 4 digit combination lock. After all, no one leaves a portion of the correct code in the lock after they shuffle the wheels!
You are to write a program that attempts to deduce the correct combination for a given lock, given only a series of locked (incorrect) combinations. Your program will read a single 4 digit number as the current state of the lock and produce a 4 digit number output as an attempt to unlock the lock. If additional input is given, the attempt was unsuccessful. Your program need not self-terminate (i.e. there is no requirement to take input telling your program that it was successful; ctrl-C interrupt is succificient).
As you are an accomplished thief, you know exactly how each lock gets randomized. You are to exploit the built-in rules to bypass the lock in the fewest number of attempts by looking for patterns in the lock's "output" and narrowing down the list of possible correct combinations.
• Locks will have a way to predetermine their combination (e.g. random seed or specific 4 digit combination). Your program may not know these values, they are used for scoring only. Remember the standard loopholes: hardcoding the output is disallowed.
• If a human is unable to find the solution with the data known at that the point of solving, the number of attempts will not count for scoring as it can be considered a lucky guess (arbitrary threshold: 10 or fewer attempts will be automatically assumed as such). This should be treated like a logic puzzle, not a slot machine.
• If your language of choice doesn't do "programs" it is acceptable to write a function taking in an array of inputs [XXXX, AAAA, XXXX, BBBB, XXXX] (where XXXX represents the combinations displayed on the lock, and AAAA/BBBB represents the prior attempts made) or similar. Note that there will be one more value from the lock than values from attempts, as your function would be producing the paired half as its output.
• Supplementary output to support ease of alternative input methods acceptable (e.g. a newline followed by the input array for the next iteration to be copy-pasted).
Scoring
The ratio of your code's byte-length to the best (lowest) number of attempts needed top open any lock.
• @Ilikemydog I browsed other cops and robbers questions while writing, scoring did not seem unusual. This one has scoring, so does this one, and this one calls for short code (typical scoring method), while this one scores based on number of different cops entries cracked. However, that's not my main concern, "I'm also not sure if there's enough room for freedom in designing a lock.." – Draco18s no longer trusts SE May 28 '17 at 18:06
• never mind then. I'll delete the comment. I'd like to give you some more feedback but I didn't follow some of it (mfny) and all I can say is have a +1 – caird coinheringaahing May 28 '17 at 18:11
• No feedback, except I've had this exact thought that a code could be guessed by tracking where the lock was left set to when it was locked over time. Similar idea, but you actually fleshed it out. – BLT Mar 11 '18 at 19:48
• @BLT That was pretty much where I'd been approaching from. Still not sure there's a good code challenge here, though. :\ Thanks for the look over! – Draco18s no longer trusts SE Mar 12 '18 at 3:15
Wasn't able to find anything in my searches, but please let me know if this or something very similar has been done before. Appreciate any feedback, first post in sandbox.
Can I leave yet?
I'm bored at work, and want to know how close I am to being able to go home. To represent this, I wish to know what percentage of work I have completed for the day.
Inputs
None - Current local/computer time shall be used
Outputs
Percentage of work completed for this day
• Formatted as either a percentage value or a decimal value: 0.57, .57, 57%, 57.0%
• Output should be accurate to at least +/-0.5%. Additional accuracy/digits are allowed.
• The work day is a total of 8 hours.
• Work starts at 08:00, and ends at 17:00.
• Lunch is between 12:00 and 13:00. Working during lunch is forbidden, and thus should not count towards the percentage of work completed.
• Output should be correct during any time of day, including before work starts (0%) and after the work day ends (100%).
The response for a full 24 hour day is shown below:
Valid Test Cases
Time Ran Output
03:55 0%
04:31 0.000
08:00 0
09:00 12.5%
11:31 44%
12:37 .5
16:30 0.94
21:08 100%
Incorrect Test Cases
09:00 12.5 (Interpreted as 1250%
11:31 43% (Error of 0.9%)
12:37 .58 (Did not account for lunch)
16:30 0.9 (Error of 3.8%)
Notes
• I work 7 days a week 365 days a year; you do not need to check if it's a weekend, holiday, etc.
• I live in an area with no Daylight Savings Time, Leap Seconds, or any other confusing time-changing events.
This is , so lowest byte-count score wins.
• Does no feedback mean it's perfect the way it is and I should post it? Or it's so bad that nobody wants to touch it? – qoou May 24 '17 at 16:08
• I think it's pretty solid. – CalculatorFeline Jun 20 '17 at 14:57
This is mainly an idea for something I could potentially host on my KOTH server.
Everybody knows that bitcoins are the next big thing. It's just a question of when they are going to take off. Right now, they are worth $250 each, but who knows, maybe someday they will be worth over$1000! The growth trend is phenomenal.
You are a tech-savvy investor who wants to get in on this action.
The Challenge
Your goal is to write a bot that can predict the market and tell you how you should invest your money, given hourly updates of the Bitcoin price.
Keeping Balance
To overcome the fastest-gun-in-the-west effect, wherein early answers have more time to make more money, this challenge will not keep track of any absolute balances. Instead, the assets of each entrant will be scaled up/down between each round.
Each entrant will be given a single float in the range 0 to 1 representing the percent of total assets are currently invested in Bitcoin. This is calculated by (BTC_cur_rate*BTC_owned)/(USD_owned + BTC_cur_rate*BTC_owned).
A value of 0 means that you currently have nothing invested in Bitcoin, while a value of 1 means that you have everything invested in Bitcoin. An input of 0.3 means that 30% of your total value is in Bitcoin, while the other 70% of your value is in dollars.
Examples
input -> assets as portion of your total value
BTC % USD %
0.0 -> 0.0 0% 1.0 100%
0.3 -> 0.3 30% 0.7 70%
0.6 -> 0.6 60% 0.4 40%
Price Data
Players will also have access to a file history.txt which will contain the BTC price history, measured in cents, over the duration of the competition. Each time a player is called, they are presented with a fresh copy of history.txt, with one line appended each turn. Do not attempt to modify this file.
Example File
This could be the history.txt file after 3 hours of competition. The most recent price is $247.49. 24694 24724 24749 There will be a trailing newline at the end of the file. Making a Trade (ouput) The output of your program should be another float in the range of 0 to 1, representing the new portion of your assets that you want invested in Bitcoin. The difference between your input number and output number represent the amount of value being exchanged. Examples input -> BTC USD | output -> BTC USD | trade being made 0.3 -> 0.3 0.7 | 0.2 -> 0.2 0.8 | 0.1 in BTC -> 0.1 in USD 0.3 -> 0.3 0.7 | 0.6 -> 0.6 0.4 | 0.3 in USD -> 0.3 in BTC Calculating Score for a Round Your score for a round is based on your change of value for that round. You start every round with a total value of 1, but your ending value is influenced by two things: • A 0.2% commission on your trade taken by the controller • The change in Bitcoin value over the next hour Taking Commission Commission is taken whenever you buy or sell bitcoins. Whenever you convert a certain amount of value from one currency to the other, you will receive 0.2% less of the new currency than what you actually ordered. Examples input | output | trade | commission | result after commission 0.3 | 0.6 | 0.3 -> BTC | 0.0006 BTC | 0.5994 BTC & 0.4 USD 0.75 | 0.05 | 0.7 -> USD | 0.0014 USD | 0.05 BTC & 0.9486 USD Adjusting BTC Value After taking commission, your value of BTC is multiplied by the price percent change in BTC over the next hour. The amount of value you have in USD will stay constant. Examples BTC after commission | prices in cents | % change | new BTC value 0.5 | 30000 -> 29850 | -0.5% | 0.4975 0.236 | 20000 -> 30447 | +2.0% | 0.24072 Overall Process of a Round Below is an example showing all of the steps in a single round. BTC USD .3 .7 = 1.0 input to entrant is 0.3 .6 .4 = 1.0 output of entrant is 0.6 .0006 .0 0.2% commission of the trade .5994 .4 = 0.9994 result after commission +0.3% percent change in bitcoin price over 1 hour .6012 .4 = 1.0012 result after the flow of time = score for that round .60048 .39952 = 1 input for the next round is 0.60048 after scaling Determining the Winner For a given round, your score is your new total value. This is after taking the 0.2% commission and calculating the change in Bitcoin value. For the above example, the score was about 1.0012. At any given time, the aggregate score for an entrant shall be the product of the scores for its most recent (up to) 50 rounds. At any given time, the current winner is the player with the highest aggregate score. For example, a bot could get these scores for its first 5 rounds: 1.001 1.002 0.998 0.999 1.003. The total score of the bot is about 1.00299. The Controller I haven't written the controller yet, but I think it's going to be written in Perl with support for entrants in a variety of other languages (Java/Python/Ruby/C++). I plan to use this API for bitcoin price data. The controller will probably run all of the entrants in parallel, each with their own thread. This simply allows it to put a stop to any infinite looping that may occur. I hope it will work if all of the programs are reading the same history file at once. Additional Rules Since this is a PvE competition and not a PvP competition, and takes place on a server, there are some slight differences in rules. • There's no set restriction on submitting multiple bots, since you can't make a team. • Similar to always, you can't call other programs, like the controller or other bots, during your turn. • The time limit is loose. A long as a single round with all of the bots doesn't take up most of an hour to perform, it'll be fine. It really shouldn't take more than a couple minutes for each bot to make a move. • You may create a single file, with the filename [botname]-data.txt, in the current directory. This file will persist, even across updates of your bot or the controller. • 1. What determines whether the commission is taken in bitcoin, dollars, or some mixture of the two? 2. What is the score of a round? (It's mentioned in the example, but it needs to be more prominent). 3. Is there any input other than the balance? Or are entries allowed to access external data sources? Or is it pure uninformed guessing? – Peter Taylor Oct 11 '15 at 20:09 • @PeterTaylor I've added some details. I'm not completely decided on how much data the entrants will be able to access/store. Right now it is just the price history. – PhiNotPi Oct 11 '15 at 23:41 • Do you have access to Mathematica? Also, how precise are the input and output? – LegionMammal978 Oct 16 '15 at 1:04 • How much of your assets are in BC to start with? – LegionMammal978 Oct 21 '15 at 23:59 • IT'S 2016 AND PHINOTPHI STILL HASN'T... /s – noɥʇʎԀʎzɐɹƆ Dec 12 '16 at 2:29 Build the Chain Quine This is an puzzle. Each person will write a program that is not a true Quine but does output its source when given the source of the last program as input. If anything else is input your program may do whatever you wish (undefined behavior) as long as it does not print the source code. The first program will be a true Quine. Rules • Standard rules apply • You may not write a submission in a language that has already been used • You may not answer twice in a row Goal The goal is to have as many valid links in the chain as possible. Sandbox This is a little sparse because I am still in the brain-storming phase of development. I just wanted to write this down so I wont forget it and, of course, to get feedback. I am not even really set on a winning criterion yet. If you have any ideas/suggestions I am really excited to hear them (thats why I put it in the sandbox). • This is semi similar to my answer chaining quine. So I wish you good luck – Christopher Feb 27 '17 at 16:36 • yeah sure thing! meta.codegolf.stackexchange.com/posts/11615 – Christopher Feb 27 '17 at 16:38 • – Christopher Feb 27 '17 at 16:40 • As with so many quine challenges, this falls afoul of universal quine constructors (and is also likely to get longer and longer over time, due to the need for each program to be able to reconstruct the previous entry, and thus implicitly all previous entries). – user62131 May 23 '17 at 8:51 Quine without a character Write a program in any language that takes as input any character, and outputs a quine in the same language, that does not contain that character. For instance (for some made up language): Input: b Output: s(fg;fg) Run "s(fg;fg)" Output: s(fg;fg) Input: ( Output: s[fg;fg] Run "s[fg;fg]" Output: s[fg;fg] Your program must handle as input every character within the range of characters that are valid in the source code of the language you're using, including new lines, punctuation, etc. Scoring This is with penalties. Your score is L + 1000xC where: • L is the length of your program in bytes • C is the number of characters it fails to meet the requirements on. So, if you produce a 50-character program that passes every character except for ( and ), your score is 2050. Standard loopholes are forbidden, and standard methods of input/output are ok. Outputs must be a proper quine, whatever that is. • Basically a duplicate. The two challenges aren't quite the same (this one is more lenient), but the solution technique will be the same in both cases. – user62131 May 23 '17 at 6:17 • Ah, I didn't see that one. I think that one is much harder, because your program has to output the whole quine factory. In my version, it's just a quine factory that outputs quines to order. – Steve Bennett May 23 '17 at 7:35 The alphabet - my way Your pointy-haired boss gives you a list of words and tells you to sort them. So you give him back a sorted list. "Wrong!" he says. "I want them sorted according to MY alphabet..." Challenge Given a new ordering of the alphabet and a list of words, sort the words according to that new alphabet ordering. The new ordering is given as a 26-character string, guaranteed to contain all letter of the alphabet exactly once, in lower case. All words in the list of words will be made up of lower case letters only -- no capital letters or punctuation. There will be no repeat words. If there is a word in the list that is the prefix of another word, then the shorter (prefix) word should appear first in the sorting. For example, "golf" should appear before "golfing". Examples Example 1 Input: qwertyuiopasdfghjklzxcvbnm apple banana currant dragonfruit elderberry fejoia Output: elderberry apple dragonfruit fejoia currant banana Example 2 Input: qwertyuiopasdfghjklzxcvbnm uranium plutonium uranus pluto polonium Output: uranus uranium polonium pluto plutonium Example 3 Input: abcdefghijklmnopqrstuvwxyz four score and seven years ago Output: ago and four score seven years • Shouldn't pluto/plutonium be before polonium – ASCII-only May 29 '17 at 11:26 • No, because 'o' comes before 'l' in that alphabet. Just like when you alphabetise using the standard alphabet, ties are broken by the alphabetical rank of subsequent letters. – Computronium May 29 '17 at 11:42 • Oh wait read polonium wrong sorry – ASCII-only May 29 '17 at 11:43 Title Challenge In your language of choice, write 25 programs, functions, or snippets that output or return the integers 1 through 25, inclusive. However, the goal is to simultaneously minimize the number of distinct chars used and the length of the code. Scoring This is a variation on : If your 25 entries have N distinct characters and a total length of L, your score is N × (L + N). The submission with the lowest score wins. Example Say the challenge only went up to 10, and your ten snippets were: 1 2 3 4 5 6 7 8 9 10 That's 11 bytes total, and 10 distinct chars; therefore, your final score would be 10 × (11 + 10) = 210. Now, if your snippets were: 1 1+1 1+1+1 1+1+1+1 1+1+1+1+1 1+1+1+1+1+1 1+1+1+1+1+1+1 1+1+1+1+1+1+1+1 1+1+1+1+1+1+1+1+1 1+1+1+1+1+1+1+1+1+1 That's 100 bytes total, and 2 distinct characters (+ and 1); thus, the final score is 2 × (100 + 2) = 204, a small improvement over the literal numbers. One last example. If your snippets were: 1 -~1 1-~1 -~1-~1 1-~1-~1 -~1-~1-~1 11-1-1-1-1 11-1-1-1 11-1-1 11-1 That's 58 bytes total, and 3 distinct chars (-, ~, 1); therefore, your final score would be 3 × (58 + 3) = 183, an improvement over both. Rules • Each output may be a string of digits rather than a literal number. • Each output may have trailing decimals, as long as they are all 0s (e.g. 1.000 is allowed, but 1.000001 is not). Meta • Is 25 a good number? I originally had it at 100, but that seems a little tedious. • Will the scoring system work well enough? • Is this even a good idea? • Title and tag suggestions? • Unary/Lenguege/Glypho would automatically win, – Okx Jun 5 '17 at 17:37 • @Okx The length still factors into the score, so no, they wouldn't. – ETHproductions Jun 5 '17 at 17:41 • Ah, sorry. My mistake :P – Okx Jun 5 '17 at 17:49 • The "make a lot of snippets for the first few natural numbers" bandwagon already seems overloaded this week. – Peter Taylor Jun 5 '17 at 22:27 • @PeterTaylor Yeah, I'll wait a while before considering posting this to main... – ETHproductions Jun 6 '17 at 11:11 Compare string repetitiveness code-challengedecision-problem Given two strings, decide which one is more repetitive. This is the string whose most common character appears more often in it. If these are equal, then tiebreak by counting their respective second most common characters, and so on. Once a string's distinct characters have been exhausted, all further counts are zero. Give one consistent output if the first string is more repetitive, and a different consistent output if the second one is. You will never be given a complete tie. You may assume input strings will be non-empty and use only ASCII characters. Scoring: Your code's score is its repetitiveness, with comparing lower being better. Put in your header the counts of the top 3 most common characters, and the full frequency list in your body. Test cases: TODO • additional tie break is probably just timing (of the post, not the programs computations) – Destructible Lemon Jan 3 '17 at 8:17 • @DestructibleWatermelon I was thinking alphabetical order (in the code page). Maybe that's interesting to optimize? If not, yes, I'll make it earlier post as standard. – xnor Jan 3 '17 at 8:19 • A somewhat meatier task might be good, otherwise golfing languages might easily score 1 without even trying. Otherwise, I do think this scoring mechanism could be interesting. – Martin Ender Jan 3 '17 at 9:14 • Maybe you could use order-0 entropy. calculate it and the score is your own entropy? – Christoph Jan 3 '17 at 11:42 • I can't edit, so: "alternative is too have" -> "alternative is to have". To make the task harder, how about, when counting a specific letter, you start with the original string, but then each time it's found, the string you're going through is shifted by one through the Caesar cipher? – 0WJYxW9FMN Jan 3 '17 at 13:27 • I think I prefer it less-meaty. It presents it as an interesting scoring mechanism to main, and since languages don't compete against each other, I'd be rather interested in how well non-golfing languages fair. – Nathan Merrill Jan 4 '17 at 16:37 • I think I like this version better, as it isn't quite so easy for golfing languages while not being out of reach for some esoteric languages. – FryAmTheEggman Jun 7 '17 at 20:06 Range without Range Builtin Challenge Given an integer y where y > 0, output a list of in some reasonable format that contains every integer in increasing order up to but not including y, without using any builtin that generates a range of any sort. If your language has a feature whose specification uses a range, that is not allowed (for example, you cannot use the map quick in Jelly on a single integer because that maps over the range). You can assume that y will not exceed your program's capacity for integers, but it must be able to theoretically work on any integer given no memory, time, or otherwise language constraints. A format is reasonable for a list a = [a0, a1, ..., an] if and only if there exists a string x, a string y, and a non-empty string z such that the output is x + z.join(a) + y. • This seems extremely simple :P for(i=0;i<y;i++)printf("%d ",i); – MD XF May 26 '17 at 19:44 • @MDXF Yes, but there's a 4-byte solution in Jelly. :P – user42649 May 26 '17 at 19:48 • Which you have disallowed.... – MD XF May 26 '17 at 19:48 • @MDXF No, there's a 1-byte solution that I disallowed. There's also a 2-byte solution which I disallowed with the rule about the specification backend. Hint: Iterating through a list or iterating like you did is permitted. – user42649 May 26 '17 at 19:52 • Ah, now I see. This would probably get 50 submissions in 10 minutes, have fun with the clogged inbox :P – MD XF May 26 '17 at 19:53 • @MDXF I don't mind :P – user42649 May 26 '17 at 19:53 • I don't think this question would be closed, but I have to voice my dislike for this challenge. I am not a fan of banning builtins in the first place, I feel it has a number of issues. My main problem is it is hard to enforce/judge when someone is using a builtin. This challenge is going to take that and push it to its limits. I feel that this for this question to work you would need a very solid definition of what a built in is and I don't think such a thing can be made. – Ad Hoc Garf Hunter May 27 '17 at 0:24 • @WheatWizard Thank you for your feedback. I will try to get a fully objective way to determine what a builtin is and if I can find one I will mention you again in a comment for your review, and if I can't then this challenge will probably die and be buried by the rest of the sandbox posts because I don't want bad challenges :P – user42649 May 27 '17 at 0:34 • I'd recommend posting this after adding a few rules - in what order do the integers need to be printed? What's the max value y can be? "In a list of some reasonable format" is a bit too broad IMO. – MD XF Jun 7 '17 at 20:45 • @MDXF I will clarify in the post, thanks. – user42649 Jun 7 '17 at 22:07 • @MDXF "in increasing order" I think I clarified. – user42649 Jun 7 '17 at 22:07 • Looks great! My last suggestion is to make it a bit more readable: The # Challenge right at the top doesn't really add anything. And I'd recommend splitting the paragraph in half, i.e. making If your language has a feature whose specification uses a range... its own paragraph. – MD XF Jun 7 '17 at 22:09 • @MDXF Yes, that makes it much more readable. Thanks! – user42649 Jun 7 '17 at 22:10 • Also, the # Challenge is the header of the actual question; the title won't be there in the actual post. :) – user42649 Jun 7 '17 at 22:10 Fence Matrix Given a positive integer n, output the 2n+1 x 2n+1 "fence"-matrix 0 1 0 1 ... 0 1 2 1 2 ... 1 0 1 0 1 ... 0 1 2 1 2 ... 1 ⋮ ⋮ ⋮ ⋮ ⋮ 0 1 0 1 ... 0 Alternatively you can also return a nested array or print a string (even with other entry delimiters than spaces or none at all) or output a raster image where each entry is represented by one pixel. Examples n = 1 0 1 0 1 2 1 0 1 0 n = 2 0 1 0 1 0 1 2 1 2 1 0 1 0 1 0 1 2 1 2 1 0 1 0 1 0 • related, but not at all close to a dupe – Destructible Lemon Jun 7 '17 at 23:13 • 1. The challenge should specify better about what constitutes a fence matrix. I suggest stating "A fence matrix is a representation of a square matrix where the 0-indexed element at index a of 0-indexed line b has the value a%2+b%2" – fireflame241 Jun 7 '17 at 23:25 • @fireflame241 This suggests using your particular solution, but there are many more to achieve that. – flawr Jun 8 '17 at 8:54 • @DestructibleLemon Ha, I would never have seen the connection if you did not point it out :) – flawr Jun 8 '17 at 8:54 Obsfucation: Use Uncommon Chars Note that I do not have the SE lookup skills to set this up. An automated query would take this challenge a low way. If someone would like this to happen, I would greatly appreciate some help setting this up :) Basic search (currently searches for Jelly, anywhere) First, choose your language. It must have at least 100 answers on PPCG before the posting of this challenge, and at least 25 chars that are not no-ops. /* Click this query and insert your language's name to see if it is valid. If it is, this query will give you legal chars, and their point value */. Use those chars the complete the challenge. The lowest point value wins. /*This Query //Can we do challenges for SE queries? If so this would be a decent one :) What I would like for this query to be: Gets all chars from the first codeblock following the language name in a heading. Accumulates all chars into a frequency table. Make sure that there are at least 25 relevant chars - we can human inspect this to insure no-ops are not polluting the data. It will then return the bottom 20% (frequency) of chars, rounded down, along with point values from 1-length for each of them. The least common chars will receive lower scores. Ties will have the same point value (do it like tournament rankings - 1,1,3, not 1,1,2.) Challenge answerers will only be able to use the chars provided. /Should whitespace be excluded (and allowed to be used, with a penalty) because of how SE treats whitespace?/ RE 25 char min: I really would like to get rid of this, but I don't know how else to prevent languages like BF from having an inherent advantage. Even if I restrict BF to 2 chars, it will score really low because they will have point scores 1 and 2.*/ /*The Challenge I have not yet decided what the challenge should be - snippets to solve as many challenges, with point value as tiebreaker, might work, or a more difficult challenge. Input requested :) */ Anyone who would like may post this challenge to main. Just give credit to @Lordofdark. How long will I sleep You need to go to bed, but what you need more is to know how long you will sleep until your alarm rings. Your task Write a program or function that takes a time (hours and minutes) as input, and outputs the number of hours and minutes until the next occurence of this time. Rules • In this challenge every clock in 24h format. • You must always get the current time for the same timezone; you can assume the input is in this timezone Input The input time must be in hours and minutes in any convenient 24h format for your language. Hours and minutes must always be separated by at least one character Valid inputs for 8h30: "8h30" 8H30M 8,30 8 30 [8,30] ... Invalid inputs for 8h30: 8.5 830 510min Ouptut The output is the difference between current time and the next occurrence of the input time (it can only be today or tomorrow). The same formatting rules apply : hours and minutes separated by at least one character an in 24h format. Note that the output will always be between 0h00 and 23h59 Examples : If it is currently 20h10 : 7h30 -> 11h20m 20h -> 23h50m 21h -> 0h50m Challenge This is , so the shortest answer in bytes wins. Standard loopholes are prohibited • This is my first challenge so I'm not sure about the I/O rules – Fabich Jan 31 '17 at 14:51 • Hello and welcome to PPCG! :) Your challenge seems fine, though I recommend specifying that you should use a 24h clock earlier in the post. Aside from that, the "minutes are always two figures" part is a bit odd. Does that mean that if I used the format [8, 30] I would then have to return [8, 05]? Personally I would recommend just saying that hours and minutes have to be separate, as it is simpler and would require less space to show. – FryAmTheEggman Jan 31 '17 at 17:07 • What timezone is the input and the current time? Do we assume UTC? What happens if a language can't get the current time? – Artyer Jan 31 '17 at 22:16 • @FryAmTheEggman You are right I removed the 2 figures condition, and I specified at the beginning the 24h format. – Fabich Jan 31 '17 at 22:58 • @Artyer I guess I should add a condition about input and current being in the same timezone. Something like "You must always get the current time for the same timezone; you can assume the input is in this timezone". What do you think I could do for language without access to current time ? – Fabich Jan 31 '17 at 23:02 • @Lordofdark in general, it's good to not exclude challenges for arbitrary reasons, but in this case there's a very good reason for certain languages to not be allowed. If it can't get the time, it can't compete. – Pavel Feb 1 '17 at 0:37 • Are multi-character separators allowed in the input format (e.g. 8 hours 30 minutes)? They probably shouldn't be, or people may well figure out a way to put their entire program in the separators and thus get a score of 0 (or however many bytes eval is in their language). – user62131 Feb 8 '17 at 5:36 • @Lordofdark can I adopt this abandoned challenge? – programmer5000 Jun 9 '17 at 12:03 • @programmer5000 yes you can. Sorry I totally forgot about this – Fabich Jun 9 '17 at 13:27 This message is open for anyone to adopt and post to main. For more details, see the chat room or meta post. Visualized Tree of 3n+1 Conjecture Originally by @KeyuGan. Thanks for letting me use this! Introduction Probably you are already familiar with 3n+1 conjecture (aka Collatz conjecture). As is stated in this golfing problem: • Start with an integer n > 1. • Repeat the following steps: • If n is even, divide it by 2. • If n is odd, multiply it by 3 and add 1. And it is proven that for all positive integers up to 5 * 260, or about 5764000000000000000, n will eventually reach 1. It is easy to draw a chain of the whole process for an integer (e.g. for 5, the chain is 1<-2<-4<-8<-16<-5). Task Description You are asked to print a string of a visualized 3n+1 tree of all chains resulting from positive integers from 2 to n, containing new lines if necessary. Input and Output There is only one input n, which can be read from stdin, be a function parameter or from any external sources. You can safely assume input is valid and does not exceed your language's processing ability. However, your code should be able to deal with inputs of 2 - 446. Under such circumstances, the biggest number involved is 13120. [Sandbox note: Is 446 a proper minimum ? It turns out to be the largest number in which the biggest number involves is less than 32767] The output is flexible, as long as: • It is a textfile, or a string, or an array of characters, or an array of lines. • It contains only 0-9, -, |, <, >, ^, v, spaces or new lines, where <, >, ^, v are for arrows, -, | are for lines. • Not hardcoded • Correctly visualized and in proper directions (for instance, 1->2->4->8->16->5 and 1-2-4-8-16-5 are not accepted.) • All numbers included in the output occur exactly once. • All leaves of the visualized tree should lie in the range of 2 ~ n, that is, all numbers in the output must be necessary for the result. • the destination of every chain is 1 Besides, the output should meet the following formatting criterion: • A number should be arranged horizontally and connected. • There should not be horizontally-adjancent digits from different numbers. For instance, in the following example, 17<34 23<46 is OK, while 17<3423<46 is not. Space(s) should be put between the two numbers under this circumstance. • There should not be vertically-adjancent digits as well. • There should not be zero(es) before a number (such as 0016). • There should be only one arrow for a line. • The line between two numbers must be straight. • As is demonstrated in output, - and | can be omitted if not neccessary. • Lines should not be crossed. A solution without crossed lines is proved to be available. A simple explaination is: Thinking in reverse, you can start from integer x, and draw 2x and (x-1)/3 (if result is odd) following x, and repeat the process for every new number. Stop when you have all required integers from 2 ~ n in the graph and erase all unneccessary numbers. • You can only draw a line onto and from a number directly, that is, the arrow of the line must be pointing at a digit. e.g. |<--, ^<--, |-- and ^-- are not accepted. • The direction of arrows and lines must be correct. e.g. ^- and <| are not accepted. • There should not be spaces between arrow and number. • There should not be spaces between arrows and lines, neither in lines. [Sandbox note: Tell me plz if you come up with other loopholes.] Output is assumed to be printed in a monospaced font (all characters have same width). Sample Input 15 Possible Output 1 1<2<4<8<16 ^ 5<10<20<40<80<160 ^ ^ ^ 3<6<12| 53<106 | ^ 13<26<52 35<70 ^ ^ 17<34 23<46 ^ ^ 11<22 15 ^ 7<14<28 ^ 9 Possible Output 2 15 v 46 v 23 v 70 v 35 v 106 v 53 v 160 v 80 v 1<2<4<8<16<5<10<20<40<13<26<52<17<34<11<22<7<14<28<9 ^ 3 ^ 6 ^ 12 Possible Output 3 1<2<4<8<16<5 80<160<53<106<35<70<23<46<15 ^ v 12>6>3>10<20<40<13<26<52<17<34<11<22<7<14<28<9 Scoring Your answer should include verifiable output of input 42, without a violation to output requirements. And you should verify your answers with different answers on this page: TBD [Sandbox note: I will provide a js checker on my site to validate an output.] Among all accepted codes, shortest code wins. • Smallest output can be a useful winning criterion in some cases, where there is the possibility of continuously finding smaller outputs with little chance of finding an optimal solution. However, in this case the sequence will always be the same, so the winning criterion is how short an output format can be made before being judged unreadable. This has two problems: 1. This will force output formats towards the subjective boundary between readable and unreadable, making judging validity difficult. 2. An output format does not require programming skill. – trichoplax Oct 22 '16 at 19:47 • You might want to consider taking out the "readable" requirement and just keeping the objective description about adjacent numbers and spacing, as that cannot lead to ambiguity. Then you can use a different winning criterion (such as code golf), and people can be flexible in the output format they choose depending on what allows for the shortest code in their language. – trichoplax Oct 22 '16 at 19:56 • Looks like I was commenting on the version before your edit - apologies if some of this no longer applies... – trichoplax Oct 22 '16 at 19:57 • @trichoplax thanks. I believe your words have convinced me that subjective 'readable' judgement is not that good for this challenge. – Keyu Gan Oct 22 '16 at 20:04 • @trichoplax I have modified the problem a little bit to take out that requirement – Keyu Gan Oct 22 '16 at 20:15 • One way to test that your requirements are objective is to write a validator program that takes the output of a submission as input and indicates whether it is valid. If you can write this program then the requirements are objective, and it will also ensure everyone is working with the same definition. Any problems you run into while writing it will also help to identify any ambiguities in the requirements. – trichoplax Oct 23 '16 at 10:51 • Can I post this abandoned proposal? – programmer5000 Jun 9 '17 at 12:50 • @programmer5000 Sure. Mention me if it doesn't bother you. XD – Keyu Gan Jun 9 '17 at 13:26 Perfect Hash Generator Given N words you are to generate a perfect hash function (ala gperf). A perfect hash function for a set of strings is a hash function with no collisions. An additional condition is that the range of the generated hash function must be [0...O(N)] (i.e. at most a constant times larger than N). You can use any language for the generated function. Can we get some feedback on this old post? I'm wondering if it is possible to avoid the obvious loophole of a cat program. • Sounds good at first blush. Do you foresee this as a [code-golf] or some more extensive challenge? If the later what metric would be used to judge it? I think that evaluation of results for compliance is easy enough if the resulting hashes are composed into programs---in unix: entry < testfile > hash_program && hash_program < testfile | sort -u | wc -l and compare to wc -l testfile---but less obvious if the submitters don't provide a scaffold (and if they do should it be counted toward length in the event that this is a [code-golf]?). – dmckee --- ex-moderator kitten Jun 22 '11 at 1:24 • Perl solution, 2 bytes (1, plus 1 for -pE instead of -E): ; Or, wait, did you mean that our program has to print another program that generates a hash? Then say";" I suppose, at 6 bytes. – msh210 Jun 16 '16 at 14:52 This message is open for anyone to adopt and post to main. For more details, see the chat room or meta post. Find the mines! You are a mine remover. Your job is to find all mines on a field, without a mine explodes. So, you write an application that can find the mines carefully. The input The input can either be provided through command line arguments or through STDIN (tell what you use in your submission). The input items are separated by commas. The input looks like this: <current step (zero-based)>,<mine count>,<field width>,<field height>,<field data> The field data is like a Minesweeper field. Rows in the field data are separated by semicolons, columns are separated by nothing, as each column is just one character. Here are the characters you can get: • X This means that you don't yet know what's there, the real data is still hidden. At the start, you get a field full of Xs. • / This means that there is nothing on that location. • <number> Specifies the count of mines around the location of the number. • F This is marked by a flag by you. • ? This got a question mark from you. There might be a mine on it, but you are not sure. This is just used as a reminder for you, it doesn't mean something specific to the controller. Example input: 2,1,3,3,XX1;X1X;XXX That input means that it's currently your third step, there is one mine, the field is 3x3, and the field looks like: X X 1 X 1 X X X X The output The output consists of 4 parts: the X of which you want to see the data (like a click in Minesweeper), the location on which you want to put a flag mark, the location on which you want to put a question mark and a sign, used to let the controller know whether you are finished or not (0 for not finished, 1 for finished). Locations are written as X;Y, zero-based. If there is something you don't want to do, output -1. You can also remove flags/question marks using the same way. Example output: 4;3,-1,3;3 Specifications • If your first output data is the location of a mine, you hit the mine and you die, but you'll still get a score. • If you select a X which hides an empty location (/), all adjacent empty fields (and their borders, which are numbers) will be revealed. • For every step, your program is executed again with the updated arguments. • When looking for mines, you are allowed to have more flags than the amount of mines. Only if you finish, the amount of flags must not be more than the amount of mines. If the amount of flags is greater than the amount of mines, your submission is disqualified (for every test!) and excluded from the scoreboard. Testing When I test your submission, I'll run 100 tests on every submission, with randomly generated fields, which I created using a program that I'll write after I got some feedback. Every submission gets the same test fields, so it's fair. Test fields look like this: • 10 tests with a 10x10 field and 10 mines. • 10 tests with a 10x10 field and 12 mines. • 10 tests with a 12x12 field and 14 mines. • 10 tests with a 15x10 field and 16 mines. • 10 tests with a 15x15 field and 35 mines. • 10 tests with a 20x20 field and 40 mines. • 10 tests with a 25x25 field and 50 mines. • 10 tests with a 25x25 field and 60 mines. • 10 tests with a 50x50 field and 100 mines. • 10 tests with a 50x50 field and 125 mines. Scoring You get 10 points for every mine you find, you lose 5 points if you think there is a mine somewhere when there is none and you lose 2 points for every mine you missed. The scoring is always the same, it doesn't matter whether you finish or die. The highest score wins. In case of a tie, the count of steps is a tie breaker. Controller I'll start working on this after I got some feedback. • You say "like Minesweeper" a couple of times, but on a cursory read I didn't see anything which differentiates it from Minesweeper. Why is this not a dupe of codegolf.stackexchange.com/q/24118/194 ? – Peter Taylor Jul 24 '14 at 14:58 • @PeterTaylor You are right, only the winning criterion is different. As there is many discussion going on about these dupes with only a different winning criterion, I'll wait for some more opinions about whether it is different enough or not. – ProgramFOX Jul 24 '14 at 15:08 • Maybe you could distinguish it by more than just the winning criterion. How about something crazy like a 3d grid of cubes where you can only access cubes that can be reached from the outside, so you slowly clear it from the outside in. – trichoplax Jul 28 '14 at 22:36 • @githubphagocyte I'm not sure what you mean by "where you can only access cubes that can be reached from the outside". – ProgramFOX Jul 29 '14 at 6:49 • That bit isn't essential - a 3d grid would work without that restriction. What I mean is restricting the cubes that can be uncovered or marked to just those on the outside of the big cube at first. Imagine it like breaking blocks to get through to blocks behind them. – trichoplax Jul 29 '14 at 8:58 • The equivalent in the 2d minefield would be treating the 2d playing area as an actual field which you have to walk across, so you can't walk to a square you want to test without testing the squares on a path to it first. – trichoplax Jul 29 '14 at 9:00 • 3d was just my 1st idea - but you could make it different in other ways. You could stick with the integer grid of squares to uncover, but let the mines beneath the grid take on floating point positions. The number in an uncovered square would be floating point because each of the eight squares adjacent to it may contain only part of a mine (which would explode if any of the squares it is overlapping were uncovered). If mines are squares the same size as the grid squares, then it may take 1, 2 or 4 flagged squares per mine, and each flagged square may contain overlap with 1, 2 or 4 mines. – trichoplax Jul 29 '14 at 9:03 • A simpler change would be to keep everything integer but let the mines be 2x2 squares. – trichoplax Jul 29 '14 at 9:09 • @githubphagocyte Thanks for your comments! What about just changing it into a 3d grid, but keep the 'normal' rules? Doing what you said about only accessing blocks if you broke the block that hides it looks complicated to implement. Unfortunately, there will still be one problem left: if everyone would post an optimal solution, then the scoreboard will boil down to luck. – ProgramFOX Jul 29 '14 at 9:33 • Yes I think with the normal rules there will be a clear optimal solution. I guess even working in from the outside there would still be only one objectively best move at each step. – trichoplax Jul 29 '14 at 10:03 • If you want to avoid the possibility of an optimal solution, there are 2 possibilities. 1.Make it a king of the hill and somehow have bots competing against each other in the same minefield. That way an optimal solution against one bot will be sub-optimal against another. 2.Make a change to the game that makes the search space too large for an optimal solution to be found. Then answers will consist of interesting heuristics and there will be the possibility of continually finding better solutions over a long period of time. – trichoplax Jul 29 '14 at 10:05 • @githubphagocyte Your KotH suggestion is a good idea, thanks! I'll think of a good way to do this. – ProgramFOX Jul 29 '14 at 10:13 • I guess just taking turns would work. You'd just need to decide the winning criteria: survivor when the opponent hits a mine / player who identifies the most mine / player who uncovers the most safe squares / ... – trichoplax Jul 29 '14 at 10:54 • Hello! This looks like a good but abandoned meta post, would you be willing to offer it for adoption? (If you want to, you can still post to main.) Due to community guidelines, if you don't respond to this comment in 7 days I have permission to adopt this. – programmer5000 Jun 9 '17 at 16:54 • Hey @programmer5000, feel free to adopt it. – ProgramFOX Jun 9 '17 at 16:55 Write a brute-forcer for the 3-byte input 'emoticon numbers' challenge The Emoticon numbers! challenge asks you to identify the 3-byte snippet which evaluates to the highest numeric value in your language, and which also has the bytes in the form ABA (where the outer two are identical and the middle one is different), and which generates an output that is only digits. I trust your claim that you have identified the best possible one is honestly intended, but as a casual scripter I'm not totally convinced, and can't be reassured by unfamiliar language specification references - since there are only 256^2 possible values, can you convince me with brute force instead? Write a program or function which: 1. takes no input 2. generates all the possible 3-byte sequences matching the pattern ABA and evaluates them in the same language. (No using one language to generate the best pattern for a different language). 3. Processes all the ones which evaluate to digits-only (output text matches the regex ^[0-9]+$, with or without trailing newline).
4. Outputs just the ABA sequence which evaluated to the highest value, and an optional trailing newline. No errors or stderr output from failed evaluations.
Clarifications:
• There's no limit on runtime, but your program must at least plausibly finish if run for long enough. Particularly, if evaluating one of the byte sequences would get stuck prompting for keyboard input, or go into an infinite loop, or quit the interpreter, you must avoid or handle that.
• If you are able to usefully reduce the search space (and explain why it's valid for your language) to avoid searching 256^2 options, that's OK. Especially if you need to do so to get past an infinite loop, etc.
Show off your brute-force strength by forcing your brute-forcer into the smallest possible space. Fewest bytes wins.
Tag:
• The whole challenge seems really brute-force :)) – Mr. Xcoder Jun 9 '17 at 17:17
• I actually did that to prove that 9E9 is the optimal JS solution. – programmer5000 Jun 9 '17 at 17:33
• -100 points if your brute forcer fits ABA :D – CalculatorFeline Jun 9 '17 at 18:47
• Now I'm wondering if there are languages for which some program of the form ABA neither halts nor obviously doesn't halt. It's likely too short, but who knows. – user62131 Jun 10 '17 at 3:55
• [With no downvotes or critique it's not too badly formed, but with only +1 upvotes it's not popular either so I am not going to post it] – TessellatingHeckler Jun 28 '17 at 0:31
Inspired by lifecompetes.com
Multiplayer Game of Life (GOL)
There are n players that play the Game of Life (standard rules) on an 50 x 50 grid. (Size, border conditions? Toroidal, Absorbing, Mirroring?)
Before first GOL-Step
When the game starts, each player has 12 cells that he can place anywhere he want as long as they do not overlap. Before the first GOL-step occures every player has to place 6 cells. If two players place their cell on the same spot no cell wil be placed an thei cells will remain in each players bank.
During game
In each GOL-Step, each player can place as many new cell on the grid as he has in his bank. If two or more players want to place a cell on the same spot, no cell will be placed there (the cells will remain in each players bank). Every six GOL-steps all the players who have less that 12 cells in their bank will get a new cell in their bank.
Goal
The goal is achieving the maximum number of cells on the grid during 1000 GOL-steps.
How to participate
Each participant has to write a javascript function of the following form (multiple return statements allowed.)
function my_bots_name(field, bank, golstep){
return p;
}
Where p is a 2d array of points [[x1,y1],[x2,y2],[x3,y3],...] that the player wants to set. field contains a 2d array of the GOL grid, bank indicates the number of cells in the player's bank, and golstep is the index of the current GOL step (golstep == 0 before the first GOL-step occurs). The function may not the global variables and cannot access Math.random() or Date(). In field` the empty cells will be set to 0, the own cells will be set to 2 and the other player's cells will be set to 1. (You will not be able to distinguish between various other players.)
The winner will be determined with a game that contains all valid submissions after one week after the first submission.
TODO
The exact environment will be provided so everyone can test the own function before the official runs.
• What size of the grid is appropriate? (dependent on number of players?)
• How many steps should be computed?
• What border conditions should be chosen?
• Is the restriction of Math.random() and Date() apropriate? (The idea was that the games will be the same no matter of who/when they will be run. (Deterministic) )
• – Martin Ender Aug 6 '14 at 12:20
• It looks like it was abandoned, since the user has not been here for more than a month. I was not able to read everything yet - is there something important that should be considered or is it a challenge that shouln't even be started? – flawr Aug 6 '14 at 13:03
• I just thought you might want to have a look to get some inspiration from a previous spec. – Martin Ender Aug 6 '14 at 13:09
• Ok thank you, I wil read them later. – flawr Aug 6 '14 at 13:21
• What do you mean by "standard rules"? The standard rules for Life have binary cells, and this doesn't. – Peter Taylor Aug 7 '14 at 22:07
• Well each cell of the grid can have two states: occupied by a live cell or not. For executing a GOL-Step it does not matter which cell is of which player, they are all treated the same. (As it is in lifecompetes.com) – flawr Aug 7 '14 at 22:15
• What happens to an empty cell that has 3 neighbours of different players? Are new cells only born if they have 3 neighbours of the same player? – trichoplax Aug 10 '14 at 20:23
• Thank you for pointing this out, I did never think about this special case but I just checked livecompetes and they handle this as follows: A new cell is born if two or three of the neighbours are from the same player. If there are three different players invovled the space remains empty. – flawr Aug 10 '14 at 20:40
• why n players instead of 2 players? – Sparr Aug 18 '14 at 4:31
• Can you provide some thoughts why only two players would make a better game? If you could start it with n players at once you can let all submissions compete against each other as you do in the original lifecompetes.com – flawr Aug 18 '14 at 8:11
• @programmer5000 Yes, feel free to adopt! Just make sure work out the details in the sandbox before posting. If you want me to help in one way or another, just ping me! – flawr Jun 9 '17 at 17:43
• Can you add support for other languages? – CalculatorFeline Jun 9 '17 at 18:46
• @programmer5000 I thought you wanted to adopt it?!? – flawr Jun 9 '17 at 19:32
• Please next time say that you want to list it for adoption rather than adopting it yourself! – flawr Jun 9 '17 at 19:39 | 2020-07-05 23:47:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.39106929302215576, "perplexity": 1446.615057754615}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655889877.72/warc/CC-MAIN-20200705215728-20200706005728-00366.warc.gz"} |
https://pendulumedu.com/blog/qotd-quantitative-aptitude-hcf-lcm-26-09-2019 | # QOTD - Quantitative Aptitude - HCF & LCM Problem
2019-09-26 | Team PendulumEdu
The HCF of $$\frac{2}{25}$$,$$\frac{6}{5}$$, and $$\frac{8}{15}$$ will be
Options:
Solution:
HCF of any fraction is given as
HCF of two or more numbers is the greatest number which divides each of them exactly.
Thus, numbers which can exactly divide all the numbers 2, 6 and 8 will be either 1 or 2
Greatest number which divides each of them exactly will be 2
Thus, HCF of (2, 6 and 8) will be 2
LCM is the smallest number which is the multiple of all the numbers.
Thus, LCM of the numbers 25, 5, and 15 will be
Therefore,
Hence, (D) is the correct answer.
Such type of question is asked in Quantitative Aptitude/Numerical Ability section of various exams like SSC CGL, SSC MTS, SSC CPO, SSC CHSL, RRB JE, RRB NTPC, RRB GROUP D, RRB Office Assistant, IBPS Clerk, RBI Assistant, etc. Try and attempt free mock tests at PendulumEdu.
RRB NTPC Quizzes
Attempt Now
Share Blog | 2020-09-19 10:05:55 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5335214138031006, "perplexity": 5386.92192041832}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400191160.14/warc/CC-MAIN-20200919075646-20200919105646-00013.warc.gz"} |
https://www.rdocumentation.org/packages/igraph/versions/0.5.2-2 | # igraph v0.5.2-2
0
0th
Percentile
## Routines for simple graphs, network analysis.
Routines for simple graphs and network analysis. igraph can handle large graphs very well and provides functions for generating random and regular graphs, graph visualization, centrality indices and much more. | 2020-01-26 20:14:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.1952866017818451, "perplexity": 2861.456439383375}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579251690379.95/warc/CC-MAIN-20200126195918-20200126225918-00478.warc.gz"} |
http://www.physicsforums.com/showpost.php?p=3139571&postcount=14 | View Single Post
Quote by rahuld.exe $$m_{SR}=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$$ according to this formula... anything that travels with speed of light will have infinite mass... but light also travels like a matter.... and so light(matter) obviously travels at the speed of light.... doesnt it imply that light(as a matter) has infinite mass?
$$E^{2}=m_{0}c^{4}+p^{2}c^{2}$$ | 2014-07-30 01:00:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7013634443283081, "perplexity": 3032.8749018521903}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510268363.15/warc/CC-MAIN-20140728011748-00182-ip-10-146-231-18.ec2.internal.warc.gz"} |
https://www.nature.com/articles/s41539-018-0028-7?error=cookies_not_supported&code=9a683c2e-93fe-4d71-a298-4221f75092c6 | ## Introduction
Adult gender differences in science, technology, engineering, and math (STEM) career representation sometimes are thought to originate from inborn differences between the sexes in aptitude for STEM fields.1,2,3,4,5 Gender differences could be biological differences that are present at birth, or they might emerge over time with maturation.4 In this study, we focus on gender differences in early childhood. Although adult STEM talent is derived from a large suite of cognitive abilities and unlikely to be traceable to a single domain or skill, if intrinsic differences between the sexes are indeed a root cause for the under-representation of women in STEM, one expectation is that gender differences in quantitative cognition will emerge early in human development.
Understanding the nature of gender differences in mathematics has been a focus of research for many years. However, differences in measurements, analyses, and participant samples have led to a variety of findings. For one, differences can emerge in mean performance on mathematical tasks,6,7,8 and small differences in favor of boys have been reported in a range of numerical skills by the end of kindergarten.9 Although most studies of school-aged children that find gender differences report higher performance in boys, some studies have only found advantages for boys when tasks involve more reasoning or are more spatial in nature.2,10 In contrast, elementary school girls sometimes show an advantage on computational tasks and when performance is assessed using school grades.11 Other studies find no differences, trivial differences, or differences in older children, but not younger children.10,12,13,14 Group differences can sometimes be attributed to cohort effects. For instance, some studies show that differences between US and Chinese children in mathematics depend on generation or school,15,16 and a recent study showed that the strength of any advantage in mathematics for boys vs. girls varies by country.17 Gender differences may also emerge in the variability of mathematical performance across boys and girls. When these gender differences in cognition are observed, boys tend to show greater variability than girls, resulting in more boys than girls at the high-performing and low-performing ends of distributions.6,7,8,17,18 This may cause gender differences in mean performance to be absent at the group level12,14 but detectable at the high-performing and low-performing ends of the distributions.18
Another major obstacle in assessing such gender differences on school-based mathematics metrics is that sociocultural influences, such as stereotype threat and the influence of teachers and parents, make it difficult to tease apart gender differences in experience from differences in intrinsic abilities.19,20,21,22,23,24,25,26,27,28,29 For example, school-aged children could show gender differences in mathematics abilities because girls are given less or different exposure to mathematics than boys or are told that “math is not for girls.” Therefore, it is unclear whether differences in mathematics abilities are rooted in intrinsic differences in numerical reasoning in early childhood or whether gender differences emerge as a result of differences in cultural exposure to mathematical concepts. Understanding the sources of any gender differences is crucial for optimizing early childhood math and science curricula.
Previous research30 described evidence against the existence of gender differences in visuospatial reasoning in early childhood. Across six tasks, boys and girls performed similarly on measures of object tracking (the ability to follow multiple, independent moving objects), early numerical processing, and core geometric abilities (Fig. 1). Those data revealed no gender differences in some basic cognitive abilities of children aged 3–10 years. However, that research leaves open key areas for investigating gender differences in core numerical processing, including patterns of looking at quantitative information during infancy, early discrimination acuity during quantity processing, and formal mathematics learning.
We examined children’s early mathematical cognition during infancy and early childhood to provide insight into whether gender differences are evident in early childhood. With the exception of the infant data, these data were collected as part of standard testing batteries measuring numerical processing skills. While we acknowledge that there are other ways to measure mathematical thinking in this age range, we combined published data31,32,33,34,35 with unpublished data from our longitudinal records that measured children’s performance in three key areas of numerical processing from our standard testing battery of early childhood numerical cognition. First, we assessed numerosity perception and acuity in infancy and childhood. Numerosity perception allows us to estimate the quantity of a set without knowing exactly how many items are in the set—we measured children’s acuity to detect differences in numerosity. Next, we examined two aspects of verbal counting acquisition during preschool, which is the earliest emerging exact understanding of quantities. Finally, we evaluated school-based mathematics during the first few years of schooling when children learn to manipulate numbers. School-based mathematics refers to comprehensive, standardized testing of a variety of mathematical skills including counting proficiency, numeral knowledge, concrete set comparison and transformation, word problems with numerical comparisons and basic arithmetic transformations, and part-whole concepts. Because the tests are age-based, the tasks completed by each child varied. These data are largely unpublished but were combined with published data31,32,33,34,35 in order to examine gender differences in over 500 children.
We conducted several analyses to test for statistical differences and statistical equivalence in performance, the emergence or disappearance of differences with age, and statistical differences in variability between groups. Similarities and differences between boys’ and girls’ performance were assessed using independent-samples t tests to identify statistical differences in mean performance and Schuirmann’s two one-sided tests of equivalence36 to identify statistical equivalence in mean performance (similarity within ½ standard deviation (s.d.) of the group data; implementation of this test for SAT-Math scores.37) Testing for both statistical differences and statistical equivalence is important. Non-significant t tests only allow us to conclude that there is not enough evidence to reject the assumption that performance is equivalent between groups. However, this does not necessarily mean that the groups are statistically equivalent. By including tests of equivalence, we can determine whether the lack of a significant difference between groups reflects statistically equivalent distributions of scores between groups. To date, tests of equivalence have not been conducted on data on mathematical abilities in early childhood, but these tests are especially important for informing the “Gender Similarities Hypothesis.”38,39 To determine whether the results of the t test were consistent across age, we also conducted simultaneous linear regressions with age, gender, and their interaction entered as predictors. A main effect of gender would suggest that there is a difference between boys and girls when controlling for age and an interaction would suggest that differences may emerge only at one end of the age range. In addition to assessing children’s mean performance, we determined whether boys and girls showed equal variance in performance using Levene’s test. Testing for equality of variance is particularly important in light of previous work that suggests that there are more high-performing and low-performing males than females because males show greater variability in measures of quantitative processing.4 For thoroughness, tests of statistical equivalence and differences in variability on scores controlled for age are reported in Supplement 1 (statistical differences in age-controlled scores should be evident in the regression analyses). Finally, for visualization purposes, we calculated growth curves at the group level following previous work.40 Because these curves were calculated at the group level, we do not statistically test for differences between boys’ and girls’ growth rates and simply provide these curves as a way to visualize changes in performance with age.
Across all three aspects of early mathematical cognition assessed here, we would expect that if boys and girls truly differ in their capacities for numerical processing, we should find evidence of statistical differences in mean performance (independent-samples t tests), and we should see that this effect is consistent across age (main effect of gender in the linear regressions) or driven by one end of the age range (interaction between gender and age in the linear regression). However, the cross-sectional analysis indicates that there are no robust gender differences in early numerical processing including preverbal numerosity perception, counting acquisition, and school-based mathematics ability.
### Core numerosity perception
Humans have the ability to perceptually estimate the numerical magnitude of a set of objects without counting. For example, without counting, people can rapidly determine that a set of 20 objects is numerically greater than a set of 10. Because numerosity representations are only noisy estimations of number, discrimination between quantities depends on the numerical ratio of the sets based on Weber’s law.41 For example, using estimation it is equally easy for people to choose the larger quantity of 10 vs. 5 as 20 vs. 10, because they have the same ratio (2:1 ratio)—quantities with finer ratios like 7 vs. 5 or 15 vs. 10 will be more difficult to discriminate. Research has shown that this ability to represent and discriminate numerosities emerges within the first year of life42,43,44,45 and that it is evident in nonhuman animals,46,47,48,49,50,51 suggesting an evolutionarily primitive origin. At 6 months, human infants can discriminate quantities that differ by a ratio of 2:1 (e.g., 16 vs. 8 dots),44,45 but by 9 months, infants can discriminate quantities at a 3:2 ratio.45 Numerosity representations become more refined with age such that 4-year-old children can discriminate at a 4:3 ratio and adults can discriminate at a ratio of 10:9.52 The visuospatial nature of numerosity perception makes it an important ability to investigate in children because gender differences in mathematics have sometimes been attributed to fundamental visuospatial skills, such as mental rotation.53 Moreover, because the acuity of these representations has been shown to relate to math ability54,55,56 (but note opposing views57), understanding whether there are gender differences in early numerical processing is essential to understanding the fundamental nature of gender differences in math achievement. Here we examined data from infants, preschool children, and early school-aged children.
To test for gender differences in numerosity representations in infancy, we analyzed previously published data from 80 6-month-old infants35 (range = 5 months 13 days–6 months 17 days, 38 girls, 42 boys). The precision of infants’ numerosity representations was assessed using a preferential looking paradigm in which infants were presented with two image streams: one in which numerosities alternated between images and one in which numerosity was constant (see Fig. 2a). Infants preferred to look to the numerically alternating image stream if numerosities differed by at least a 2:1 ratio, and there were individual differences in infants’ preferences.35 An independent-samples t test and Schuirmann’s test of equivalence revealed no gender differences (t test: t(78) = 0.14, p = 0.89, difference = 0.41%, 95% confidence interval (CI) = −6 to 7; equivalence test: t1(78) = 2.36, p = 0.01; t2(78) = −2.08, p = 0.026), with boys showing a mean preference for the numerically alternating image stream of 7.5% (±14.6) and girls showing a mean preference of 7.09% (±14.3). Levene’s test of Equality of Variances revealed no significant differences in variance between girls and boys (F(1, 78) < 0.01, p = .94, boys’ s.d. = 14.65, girls’ s.d. = 14.27). This is consistent with the previous work showing no overall differences between boys’ and girls’ sensitivity to numerosity in infancy.58,59,60
We also tested for gender differences in numerosity perception in the earliest years of formal education. Two hundred forty-one scores were collected from 3- to 7-year-old children (mean age = 5.48 years, 125 girls, 116 boys; data from 68 children have been previously reported31,32). All children completed a computerized numerical comparison task. In this task, children were shown two side-by-side dot arrays and were asked to choose the side that had more dots. The numerical ratio between dot arrays varied between 4:1 and 10:9. This type of numerical discrimination task permits a psychophysical evaluation of numerosity representation and is consistent with previous literature using this task in adults and children.52,61,62,63,64,65 Furthermore, performance on this type of task has been shown to be similar to neural measures of numerosity encoding,31,63 indicating that this is a fundamental aspect of numerical cognition. Although previous work found that women and girls performed better than men and boys,52 sample sizes were small (n = 16 per age group), so it is unclear whether these differences are representative of the general population.
To assess the acuity of boys’ and girls’ numerosity representations, Weber fractions (w) were calculated for each child.66 The w score represents the acuity of numerosity representations such that a smaller w indicates greater acuity. An independent-samples t test and Schuirmann’s equivalence test revealed that boys and girls showed equal acuity of numerosity representations in early childhood (Fig. 3; t test: t(239) = 0.23, p = 0.82, boys’ mean = 0.56, girls’ mean = 0.58, difference = 0.02, 95% CI = −0.12 to 0.15; equivalence test: t1(239) = 4.10, p = 0.00003; t2(239) = −3.64, p = 0.0002). A simultaneous regression further revealed that while acuity improves with age, the effect of age on acuity of numerosity representations does not differ between boys and girls (F(3, 237) = 32.03, p < 0.0001, R2 = 0.29; Gender: b = 0.10, t(237) = 0.33, p = 0.74; Age: b = −0.27, t(237) = −7.33, p < 0.0001; Age × Gender: b = 0.02, t(237) = 0.30, p = 0.77). Finally, Levene’s test of Equality of Variances did not reveal a difference in variance between boys and girls (F(1, 239) = 0.09, p = 0.76; boys’ s.d. = 0.57, girls’ s.d. = 0.49).
Taken together, we find that from infancy into early childhood, boys and girls do not differ in their earliest numerosity perceptions. Boys and girls are equally capable of discriminating numerosities.
### Culturally trained counting
Verbal counting is the first culturally trained symbolic mathematics concept to develop in children. Knowledge of the verbal counting routine emerges gradually between the ages of 2 and 5 years. First, children learn to rote recite the count list (2–2.5 years). Over the next 6 to 12 months, children begin to acquire the meanings of the number words one at a time: they learn that the number word “one” corresponds to exactly one item, then that the word “two” corresponds to exactly two items, then “three,” and finally “four.” Around 3.5 years, children seemingly suddenly become cardinal-principle knowers in that they learn that each number word refers to a specific quantity and that a number word can be used to label the size of a set as determined by counting.67,68,69,70,71 We tested children’s knowledge of the rote-memorized counting sequence with the “How High?” task, and we tested their cardinal knowledge of number and counting principle knowledge with the “Give-N” task.69,70 Although there are other ways to assess counting skills and knowledge of the cardinal principles,9,69,70,71,72 these tasks are commonly used and standardized across the literature. These two measures of culturally trained counting allowed us to determine whether boys or girls show a general advantage for early number word learning or whether there are different patterns of gender differences in memorizing the counting sequence (“How High” task) vs. learning the meanings of number words (“Give-N” task). A general advantage for early number word learning would be supported by differences in favor of one gender on both measures of early number word knowledge. An advantage on only one test would suggest the advantage is isolated to a specific skill.
For the “How High?” task, children were asked to count as high as they could until they reached 100. One hundred forty-three children aged 2–5.5 years old were tested (mean age = 4.10 years, 71 girls, 72 boys). An independent-samples t test revealed that boys and girls did not show a difference in their ability to memorize the verbal counting sequence (t(141) = 1.48, p = 0.14, boys’ mean = 30 girls mean = 23, difference = 7, 95% CI = −2 to 16), and Schuirmann’s tests of equivalence found marginal statistical equivalence (t1(141) = 4.48, p = 0.00001; t2(141) = −1.52, p = 0.06). A simultaneous regression confirmed that although children’s ability to recite the count list improves across age, differences do not emerge when controlling for age or at one end of the age range (Fig. 4a for scatterplot of data by age; F(3, 139) = 20.96, p < 0.0001, R2 = 0.31; Gender: b = 12.75, t(139) = 0.58, p = 0.56; Age: b = 18.26, t(139) = 5.10, p < 0.0001; Age × Gender: b = 4.20, t(139) = 0.80, p = 0.43). Furthermore, Levene’s test of Equality of Variances revealed no difference in variability (F(1, 141) = 1.41, p = 0.24; boys’ s.d. = 30, girls’ s.d. = 25). Taken together, this suggests that from 2 to 5.5 years of age, boys and girls show equal proficiency in memorizing and reciting the count list.
Performance on the “How High?” task only represents verbal learning of the culturally trained, rote-memorized list of count terms and is not an index of children’s quantitative or logical reasoning during counting. To test children’s understanding of the counting procedure, we tested children on the “Give-N” task. In the “Give-N” task,69,70 children were asked to count in order to produce sets of 1 to 10 objects. One hundred and twenty-three children aged 2.98–5.47 years completed the tasks (mean age = 3.87 years, 65 girls, 58 boys). Children were scored by the highest set size that they could correctly produce. An independent-samples t test revealed no statistical difference between boys and girls, but Schuirmann’s tests of equivalence test failed to find statistical equivalence (t test: t(121) = 1.67, p = 0.097, boys’ mean = 6.38, girls’ mean = 5.26, difference = 1.12, 95% CI = −0.2 to 2.44; equivalence tests: t1(121) = 4.46, p = 0.00001; t2(121) = −1.12, p = 0.13). The simultaneous regression revealed a main effect of age, but no effect of gender or interaction between gender and age (Fig. 4b for scatterplot of data by age. F(3, 119) = 31.63, p < 0.0001, R2 = 0.44; Gender: b = 1.67, t(119) = 0.57, p = 0.57; Age: b = 3.58, t(119) = 7.45, p < 0.0001; Age × Gender: b = 0.23, t(119) = 0.30, p = 0.76). In addition, we did not detect differences in variance between boys and girls (F(1, 121) = 0, p = 0.99; boys’ s.d. = 3.61, girls’ s.d. = 3.78). Overall, there are no strong differences between boys and girls in their ability to use counting to produce sets.
Thus, boys and girls do not significantly differ in their cardinal and logical knowledge of the counting sequence during early childhood. The lack of a difference between boys and girls is consistent with the findings depicted in Fig. 1 that tested 194 3-year-old children on similar counting tasks.30
In sum, we find that boys and girls show equal proficiency in memorizing and reciting the count list, and comparable abilities to learn the logic of the counting sequence. We conclude that there is no true gender difference in children’s early counting.
### Formal and informal early elementary mathematics
Children begin to learn school-based numerical and mathematical concepts shortly after acquiring the counting principles. To test for early gender differences in the foundations of school-based mathematical concepts, we administered the Test of Early Mathematics Ability Third Edition (TEMA-373) to 275 children aged 3.07–7.92 years (mean age = 5.45 years, 133 boys, 142 girls; data from 77 children have been previously reported32,33,34). The TEMA-3 is a comprehensive test of school-based mathematical knowledge for children aged 3–9 years. Items are categorized as “formal” and “informal”: Formal items tap into knowledge that is formally taught such as numeral names, numeral writing, and arithmetic facts. Informal items tap into children’s abilities to count and reason about quantitative relations and transformations that draw on acquired knowledge but are not explicitly trained or memorized. Although some test items overlap with the skills measured in the previous section on verbal counting acquisition, the TEMA-3 represents math achievement at a broader level. Importantly, the achievement scores that result from the TEMA-3 reflect knowledge on a wide range of mathematical skills including, but not limited to, counting ability. We compared boys’ and girls’ performance on the TEMA-3 overall and on items tapping into formal vs. informal math achievement separately.
Boys and girls did not differ in overall math achievement, suggesting that children show equal understanding of math concepts in early childhood (Fig. 5; t test: t(273) = 1.11, p = 0.27, boys’ mean = 32.32, girls’ mean = 30.04, difference = 2.28, 95% CI = −1.76 to 6.31; equivalence test: t1(273) = 5.25, p < 0.001; t2(212) = −3.04, p = 0.001; test of equality of variances: F(1, 273) = 0.002, p = 0.99; boys’ s.d. = 16.96 girls’ s.d. 17.02). This pattern was consistent across age suggesting that during early childhood boys and girls show equal competency for math concepts (regression: F(3, 271) = 224.3, p < 0.00001, R2 = 0.71; Gender: b = 3.81, t(271) = 0.70, p = 0.49; Age: b = 12.72, t(271) = 19.00, p < 0.0001; Gender × Age: b = 0.19, t(271) = 0.19, p = 0.85).
To look at differences in boys’ and girls’ performance by question type, we compared formal vs. informal math scores. We conducted a 2 (Formal/Informal) × 2 (Boys/Girls) repeated-measures analysis of variance (ANOVA) on a subset of the data from children who answered at least four formal questions and at least four informal questions (for a similar approach74). We found no interaction between gender and question type nor did we find a main effect of gender (Fig. 6; Gender: F(1, 207) = 0.56, p = 0.46; Question Type: F(1, 207) = 235.98, p < 0.0001; Gender × Question Type: F(1, 207) = 0.30, p = 0.58). Furthermore, we found statistical equivalence between boys’ scores and girls’ scores and no differences in variances for both formal and informal questions (Formal Questions: equivalence tests: t1(207) = 3.77, p = 0.0001; t2(107) = −3.44, p = 0.0003, boys’ mean = 0.46, girls’ mean = 0.46; variance test: F(1, 207) = 0.24, p = 0.63, boys’ s.d. = 0.17, girls’ s.d. = 0.17; Informal Questions: equivalence tests: t1(207) = 4.57, p < 0.00001; t2(207) = −2.65, p = 0.004, boys’ mean = 0.71, girls’ mean = 0.69; variance test: F(1, 207) = 0.32, p = 0.57; boys’ s.d. = 0.15, girls’ s.d. = 0.16). In addition, differences did not emerge for either question type when controlling for age or testing for interactions between gender and age (Formal Questions: F(3, 205) = 8.25, p = 0.00003, R2 = 0.12; Gender: b = 0.05, t(205) = 0.14, p = 0.74; Age: b = 0.06, t(205) = 3.70, p = 0.0003; Gender × Age: b = −0.01, t(205) = −0.25, p = 0.81; Informal Questions: F(3, 205) = 19.62, p < 0.0001, R2 = 0.22; Gender: b = 0.16, t(205) = 1.29, p = 0.20; Age: b = 0.09, t(205) = 6.09, p < 0.00001; Gender × Age: b = −0.02, t(205) = −1.05, p = 0.29).
In sum, we did not find any robust performance differences in early childhood math ability between boys and girls. Differences did not emerge with age or by question type. This suggests that boys and girls show equal competency forming mathematics concepts in early childhood.
## Discussion
Recent public discussions surrounding the under-representation of women in STEM fields have suggested that differences in career choices between men and women could be due to intrinsic differences in aptitude in STEM domains. This claim would predict that gender differences should be evident from early on in childhood. Our data, compiled across studies from over 500 infants and children, provide a comprehensive analysis of the effect of gender on early mathematical cognition, and show that in fact, there are no substantive gender differences in mathematical thinking skills during infancy or early childhood. Boys and girls perform equivalently on numerosity perception, counting acquisition, and early school-based math concepts. Our results are consistent with those of a previous study of nearly 200 children who were tested on knowledge of the counting procedure using the “Give-N” task and found no evidence of a statistical difference between boys and girls.30 Furthermore, early school-based mathematical concepts that build upon knowledge of the logical principles of counting did not show any gender-based differences, suggesting that boys and girls learn mathematics similarly even beyond counting acquisition, into early schooling. This interpretation is consistent with a prior analysis of three million elementary school children showing that school test performance differences in mathematics between boys and girls are non-existent or trivial during elementary school, but steadily increase through high school and college.10,13 Thus, boys and girls begin education with equivalent early mathematical thinking skills.
Although these results are consistent with some previous work in this age range,10,12,30 these results contrast with other work in this age range. For example, a small advantage for boys in a variety of numerical skills by the end of kindergarten has been previously reported.9 However, the growth curve trajectories they fit for each test suggest that these differences were not consistent across every timepoint assessed during kindergarten. For some tests, such as numerical estimation and counting skills, boys and girls were indistinguishable at the initial timepoint. For other tests, such as patterns, number recognition, and number combinations, boys and girls had overlapping scores in the middle timepoints. This shows that even when gender differences are detected, they are inconsistent and highlights the importance of future work that measures gender differences using a longitudinal approach. In contrast, their work found consistent differences in math ability based on socioeconomic status.9 Although gender differences between socioeconomic statuses could not be assessed in the present study, it is important to take this into consideration in future work. Comparing the present study to previous work also emphasizes the reality that there are many ways to measure mathematical thinking in early childhood and group differences could vary across tasks, cohorts, and age.
The absence of statistical differences across the major developmental milestones of early mathematical cognition are unlikely to be due to sample size. Power analyses suggest that given the sizes of the samples analyzed here, we should have been able to detect small to medium effect sizes ranging from Cohen’s d = 0.34 to 0.65 (80% power, p = 0.05; Infant Numerosity Comparison (looking time): d = 0.65; Early Childhood Numerosity Comparison (w): d = 0.37; Recitation of Count List (“How High?” task): d = 0.47; Counting Principles (“Give-N” task): d = 0.52; Math Concepts (TEMA): d = 0.34, (Formal/Informal Math Scores): d = 0.40). Importantly, even if smaller effects do exist, they are unlikely to reliably, meaningfully, or consistently manifest in children. Caution should be taken when interpreting any small effects in large sample to ensure that their importance is not over-exaggerated.13,75
The origin of adult gender differences in science, technology, engineering, and mathematics likely has a complex sociological explanation2,4 and cannot be easily reduced to intrinsic differences in aptitude in early childhood. Women have been discouraged from participating in mathematics and science, and there is a long legacy of sexism in academics. Stereotype threat has been shown to have deleterious effects on girls’ and women’s mathematics performance19,20 (but see Ganley et al.76), and the strength of implicit stereotypes associating men over women with science predicted gender differences in 8th grade math achievement.77 Prior studies have found that science and mathematics teachers are more likely to encourage boys to ask and answer questions, explain concepts to boys, praise boys, and spend more time interacting with boys.22,23,24,25 Another source for gender differences includes parental perceptions of children’s abilities.26 Parents who believed that men are superior at math gave significantly higher math-ability estimates to their sons than to their daughters even when controlling for the children’s actual scores.27,28 Several studies have also found that parental expectations for children’s abilities and success are correlated with their children’s self-concepts of their own abilities and later performance.26,29 In fact, parental perceptions of children’s abilities may influence children’s beliefs in their abilities more than grades.78 In addition, teachers’ perceptions of students’ math ability have been shown to predict later math achievement scores when earlier measures of ability are controlled.21 Taken together, there is a strong cultural influence on math achievement throughout childhood. Expelling the stereotype that boys have an intrinsic advantage for mathematics in early childhood may lead to increased mathematics exposure and improved parental and societal perceptions, resulting in improved success in mathematics for girls.
The findings presented here provide strong evidence that boys and girls have comparable cognitive faculties for reasoning about mathematics during early childhood. Although it remains possible that gender differences in STEM involvement emerge later in development from maturation,4 in other cognitive skills,79,80 or from interactions between cultural stereotypes, training, and sexually dimorphic behaviors,4,81,82 there is compelling evidence that males’ and females’ abilities are shaped by different cultural experiences that affect their self-image, treatment, and opportunities, and little evidence to support claims of intrinsic or biological gender differences in early mathematical cognition.
## Methods
### Participants
Five hundred and seven children (256 girls, 251 boys, range = 5 months 13 days to 7.92) contributed 868 measures of quantitative reasoning. Informed consent was obtained from children’s parents and assent was obtained from children aged 7 years and older. Children were rewarded with small toys and stickers, and their parents were compensated for travel expenses.
### Sites
Children were tested across three testing sites: the University of Rochester in Rochester, NY, the University of Pittsburgh in Pittsburgh, PA, and Duke University in Durham, NC. Protocols were approved by the local Institutional Review Board at each location and parents of all children provided written consent.
### Testing procedures
#### Preferential looking paradigm for numerosity perception
Eighty children (38 girls, 42 boys, boys’ mean age = 6 months 1 day, girls’ mean age = 6 months and 2 days, range = 5 months 13 days to 6 months 17 days) participated in one of five conditions (16 infants per condition). For full details on additional infants who were tested but excluded from the analyses see previously reported work.35
Infants were seated on a parent’s lap or in a high chair approximately 105 cm away from the middle of three 17-inch computer screens. The experimenter began the trials when the infant looked at the attractor on the middle screen (see Fig. 2). Infants completed four 60-s trials. During each trial, infants were simultaneously presented with image streams on each of the two outer computer screens. One stream continuously alternated between two different numbers of dots (numerosities), while the other stream contained images with a constant number of dots. The numerosities on the alternating stream differed by one of five ratios. Infants were randomly assigned to one of these five conditions: 24 vs. 6 (4:1 ratio), 18 vs. 6 (3:1 ratio), 20 vs. 10 (2:1 ratio), 16 vs. 8 (2:1 ratio), and 18 vs. 12 (3:2 ratio). Each image was presented for 500 ms followed by 300 ms of blank screen. Every other image was the same between the two streams, and identical images were interspersed with images that differed in numerosity. One-third of the images that differed between the two streams were matched on either total surface area, individual element size, or total perimeter of the dots. Because half of the images differing in numerosity were matched on density, the two streams could not be differentiated using element size, cumulative surface area, cumulative perimeter, or density. The side of the changing image stream alternated between trials, and the order was counterbalanced between participants. Half of the participants in each condition saw a non-changing image stream containing the larger numerosity, while the other half saw the smaller numerosity.
Looking behavior was recorded digitally and was later coded by an experienced observer using a custom-made coding program written in RealBasic.83 A second observer coded more than one-fourth of all participants. Reliability between the two observers was very high (r = 0.99). For each stream, we calculated the proportion of time each infant spent looking at the changing and non-changing image streams as a function of total looking behavior to both screens for each infant. The analyses presented here were then conducted on preference scores, which were calculated by subtracting the average percent looking time to the non-changing stream from the percent looking time to the changing stream across all four trials, such that a positive score indicates a preference for changing over non-changing streams.
Two hundred and fifty children (129 girls, 121 boys, girls’ mean age = 5.43 years, boys’ mean age = 5.46 years, range = 3.07–7.92 years) completed a computerized numerical discrimination task to measure the acuity of their numerosity perception. During the task, children were shown two side-by-side dot arrays too brief to count and were asked to indicate which array contained more dots. Dots varied in location from trial to trial and correct answers (i.e., the larger quantities) were equally presented on the left and right sides of the screen. Children completed one of four versions of this task:
### Version A
Dot arrays consisted of 1 to 30 dots. Comparisons were defined as having a small, medium, or large number of dots and were made across five different ratios (24 trials per ratio): 4.0 (e.g., 16 dots vs. 4 dots), 3.0, 2.0, 1.43, and 1.11. To ensure that participants used numerical information to make their decision, on half of the trials, the dots were the same size in both arrays, and on the other half of the trials. the dots varied in size between the arrays such that cumulative surface area was the same in both arrays. Dot densities (average interitem distance) varied equally between a large density and a small density.
### Version B
Dot arrays contained between 12 and 36 dots, and dot size varied within single arrays (average dot diameter = 36 pixels; allowed variation = 20%). There were 18 trials for each of four ratio categories: 2.0, 1.5, 1.17, and 1.14. To ensure that participants used numerical information instead of other perceptual cues to determine the correct response, three trial types were included: Congruent (i.e., the array with the larger number had the larger cumulative area), Incongruent (i.e., the array with the smaller number had the larger cumulative area but both arrays had equal cumulative perimeter), and Neutral (i.e., the arrays had equal cumulative area).
### Version C
Dot arrays contained 3 to 31 dots and dot size varied within single arrays. There were three ratio categories: 2.0 (easy trial), 1.2533 (medium trial), and 1.11 or less (difficult trial). Children who were successful on the first ten trials (accuracy >80%) completed more medium or difficult trials in the remaining trials to ensure children stayed motivated across the task. Children completed between 30 and 48 trials. In this version, after children made a decision, they placed a bet on how sure they were of their answers. Correct answers were reward with virtual tokens and incorrect answers resulted in taking away tokens.
### Version D
Dot arrays contained between 3 and 20 dots. Comparisons were made across three ratios (24 trials per ratio): 4.0, 2.0, and 1.25. On ¼ of the trials, dot size was held constant between the two arrays. On another ¼ of the trials, cumulative surface area was constant between the arrays. On ½ the trials, the cumulative surface area varied at a ratio of 2.5 such that on half of those the larger cumulative surface area was congruent with the correct answer and on the other half the larger surface area was incongruent.
To assess the acuity of their numerosity perception, Weber fractions were calculated following Pica et al.66 Children were excluded from analyses if their performance on the task was at or below chance (w > 74, based on simulated 50% accuracy data; n = 1 girl, 2 boys) or if their weber fraction was >2 s.d. from the remaining mean (w > 3.99, n = 3 girls, 3 boys). The final sample consisted of data from 241 children (125 girls, 116 boys, girls’ mean age = 5.47, boys’ mean age = 5.49). Of these children, data from 32 children who completed version A and 36 children who completed version C had been previously reported, respectively31,32 (sample sizes in previous work differ due to differences in age ranges examined in previous vs. present analyses or because some children had also completed the task at an earlier date—we only report the earliest test date in the present work).
One hundred and forty-three children (71 girls, 72 boys, girls’ mean age = 4.04 years, boys’ mean age = 4.15 years, range = 2.98–5.46 years) completed a rote memorization counting task where they were asked to count out loud as high as they could up to 100. Performance was measured as the highest number counted with no errors.
One hundred and twenty-three children (65 girls, 58 boys, girls’ mean age = 3.83 years, boys’ mean age = 3.92 years, range = 2.98–5.47 years) completed a test of knowledge of number word meanings.69,70 In our version of this task, children were asked to remove a specified number of gold coins from a treasure chest, place them on the table, and count them out loud one at a time. Children were allowed to correct their mistakes until they verbally confirmed that the requested number of coins was on the table. The trial was scored as correct or incorrect based on the final number of coins produced. Children were tested on 1 to 10 items, starting with 1 item and continuing until they failed to correctly produced a requested quantity on 2 out of 3 trials of that quantity. If the correct amount was produced, on the next trial, the child was asked to produce a set size of one more than the previous set. If the amount produced was incorrect, then the child was asked for a set size of one less than the previous set. If the child reached 10 items, trials alternated between 9 and 10 until three trials of each set size were administered. Performance was measured as the highest number that was correctly produced on at least two out of three trials.
### Analyses
Statistical tests were performed using R (version 3.3.1) and R-Studio (version 1.0.44). Independent-samples t tests were conducted using the “ttest” function assuming equal variance. Tests of equivalence were conducted using the “TOSTtwo.raw” function from the TOSTER package (upper and lower bounds set to ±0.5 × standard deviation of the entire group; α = 0.05). This function returns two t values (t1 and t2). For statistical equivalence, both t values must be statistically significant. Statistical equivalence is rejected if either t1 or t2 does not reach significance. Regressions were conducted using the “lm” function. Levene’s test of equivalence was carried out using the “leveneTest” function in the car package. ANOVAs were conducted using the “ezANOVA” function (type = 3) from the ez package.
Growth curves were calculated at the group level.40 Specifically, Eq. 1 was fit to the data using the nls function (“port” algorithm) in R. Equation 1 fits two free parameters, alpha (α), which indicates the rate of growth, and lambda (λ), which indicates the age at which children are halfway to the maximum possible score. In Eq. 1 β0 and β1 represent the lower and upper score limits and are set to the lowest and highest possible scores for a given measure before the free parameters are fit to the data.
$${\mathrm{Quantitative}}\,{\mathrm{Reasoning}}\,{\mathrm{Ability}} = \beta _0 + \frac{{\beta _1 - \beta _0}}{{1 + {\mathrm{e}}^{ - \alpha \left( {{\mathrm{Age}} - \lambda } \right)}}}.$$
(1)
The growth curves displayed in Figs. 35 were calculated across the group data, and the standard deviation (shaded areas) at each age was calculated across 10,000 growth curves fit by bootstrapping the data. Because these curves were fit at the group level rather than the individual level, no statistical tests of gender-specific growth rates were conducted. Instead, these growth curves are meant to simply provide a visualization of boys’ and girls’ performance within the age range of the children in our sample.
### Data availability
The datasets analyzed during the current study are available from the corresponding author on reasonable request. | 2022-10-06 11:57:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.35135412216186523, "perplexity": 2191.387172228127}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337803.86/warc/CC-MAIN-20221006092601-20221006122601-00047.warc.gz"} |
https://ineffectivetheory.com/ | Archive
# Ineffective Theory
Over-incarceration as a market failure. It seems like much more could be written on this perspective.
Deepmind declares victory! Mohammed AlQuraishi outlines the method, and discusses the implications for protein structure prediction and related fields. And here is a video about the result from a machine learning researcher.
Gowers issues a challenge: create a computer-checkable proof of a foundational result in “condensed mathematics”. The challenge is issued as a result of his own fear that his proof is flawed (after spending about a year working on it).
Socrates as the bed guy.
Bryan Caplan doesn’t think much of elite universities.
Another claim of quantum supremacy appears! As with all these early claims, a key question is whether or not a classical computer could perform the same computation. In this case, Gil Kalai thinks so.
The essence of economics, in a parody of he-who-must-not-be-named. Nearly half applies to physics without modification.
Microsoft’s report on the SolarWinds hack is worth a read. One can draw many parallels between public health (particularly pandemic preparedness) and cyber security. One way in which they’re not similar: we have a pretty good idea what “the worst” pandemic looks like. We have no idea what a worst-case cybersec incident looks like.
# Lies That (Probably) Don't Matter
Here are a few commonly repeated, casually believed statements (in or adjacent to physics), which are unambiguously false but extraordinarily useful. Most of the time, the fact that they’re false doesn’t matter at all; hence the “commonly repeated/extraordinarily useful” part. But once in a while, they might matter. At the very least, it’s fun to keep them in mind.
The universe has no privileged reference frame. Yes, it does (and it’s not Lorentz-invariant). Relatedly, this is not a picture of the CMB. The Cosmic Microwave Background has an enormous dipole moment, because the Earth is at motion with respect to the universe’s (locally) privileged reference frame. Photons on one side of the sky are mostly coming toward us, and so are higher energy than photons on the other side of the sky. That picture is a picture of the CMB with the dipole moment subtracted.
If the universe didn’t have a privileged reference frame, then either there would be no CMB, or we’d be immediately ripped apart by infinitely blueshifted cosmic rays.
Cosmology also breaks time-translation invariance, but that’s much more obvious.
Anything about the infinite volume limit. Commonly in the theory of phase transitions, for instance, there’s a step where the thermodynamic limit is taken. Similarly, computational complexity is founded on the nearly exclusive study of asymptotic behavior. Of course these are only (very very good) approximations to reality. The conflict between computational complexity and physics (which isn’t just about a finite universe) sometimes gets attention; I’m not aware of any reason to care about the conflict between thermodynamics and cosmology. Maybe at very late times, the number of particles within your event horizon is below any reasonably thermodynamic limit? Seems pretty uninteresting.
I suppose that black hole formation matters for thermodynamics too. Because a “small” system is only approximately thermodynamic, and black holes prevent systems from getting too large, there’s some sort of fundamental limit on how accurate the thermodynamic approximation can be before a black hole forms. Again, probably consequence-free.
General relativity is generally a great killer of general beliefs. Energy conservation gets mauled, for instance.
Church-Turing thesis. Thanks to the finite volume/lifespan of the visible universe, most computable functions can’t actually be computed.
Worrying about cosmology is getting boring. Even ignoring that, I should mention the extended Church-Turing thesis. (That is, the claim that Turing machines can efficiently perform any task the universe can efficiently perform.) Generally believed to be false now thanks to quantum computers, but still “mostly true” for natural problems. If you take anthropic arguments too seriously, this principle runs into some trouble there, too.
The classification of phase transitions. This is a complete mess, and can’t really be summarized here. To start with, there is no “the” classification of phrase transitions, because there are several conflicting conventions — go read the wikipedia article if you want to cry. More importantly: the point of a classification is to say “objects of type A all have properties A1, A2, and A3, whereas objects of type B have properties B1 and B2 instead”. If you can do this, then the categories don’t change much when you tweak the definition. With phase transitions, this just cannot be done. The most prominent example is the BKT transition. It’s common to say (and you may have heard in class) that divergent correlation lengths, and divergent susceptibilities, occur in the same systems. With the most common (and most insane) naming scheme, either property can be taken as the definition of “second-order transition”. But, the BKT transition has a divergent correlation length without a divergent susceptibility; in fact, the partition function is smooth (but non-analytic).
Some other trivia about the difficulty of classifying transitions. Simple discontinuities (as opposed to smooth divergences) can occur in any derivative of the partition function: it needn’t be a first derivative. First-order transitions need not have metastable phases (possible example). No type of phase transition is necessarily associated to an order parameter. Both sides of a first-order transition can be “in the same phase”, in the sense that there’s a smooth change of parameters to get from one side to the other, without crossing the transition.
A bit more niche, but the phrase “low-energy Hilbert space” is problematic.
And, of course, I must mention Tsirelson’s problem, now apparently answered in the negative — giving lie to pretty much everybody’s understanding of quantum mechanics. But that’s a topic for its own post…
# Harmonic Oscillator Weirdness
The phrase “low-energy Hilbert space” is massively deceptive.
Some background: working with infinite-dimensional vector spaces is hard. We like to truncate them, to obtain a finite-dimensional space, so that operators can be treated as matrices on a computer. The standard truncation is to take some well-understood Hamiltonian, diagonalize it, and consider only the (vector space spanned by the) lowest $N$ eigenstates. As $N$ is increased, the approximation is improved, and the properties of the true (infinite-dimensional) system are recovered in the limit.
Physically, the idea is that high-energy physics can be ignored when you know that the energy of your system is, well, not that high. Since we took only the low-energy states, it seems natural to call the resulting vector space the “low-energy Hilbert space”.
So far, seems fine. Now consider the (not normalized) state of the harmonic oscillator (hamiltonian $H = x^2 + p^2 - 1$): $$\Psi(x) = \frac{1}{1 + x^2} \text.$$ This is, of course, just the glorious Cauchy-Lorentz distribution. The expectation value $\langle x^2\rangle$ is some finite number. Same for $\langle p^2\rangle$. Therefore, the energy of this state is finite! The expectation value $\langle x^4 \rangle$, though, is infinite: $$\langle x^4\rangle = \int dx\; \Psi(x)^* x^4 \Psi(x) = \int dx\; \frac{x^4}{x^4 + O(x^2)} “=” \infty \text.$$
But wait! The low-energy Hilbert space (for any finite cutoff) is spanned by states with finite $\langle x^4\rangle$ (and indeed finite expectations of every polynomial in $x$ and $p$). And here we have a low-energy state with an infinite expectation value. What gives?
Well, the state $\Psi$ is not in any low-energy Hilbert space. It has amplitudes (decaying algebraically) for every excited state of the harmonic oscillator. For certain observables (like $x^4$), its physics is dominated by those high-lying states. It can be approximated arbitrarily well within the cutoff, but some qualitative features can never be recovered.
This isn’t just about divergent integrals, either. If $|E\rangle$ is the harmonic oscillator eigenstate with energy $E$, then by considering the state $|0\rangle + \epsilon^2|\frac 1 \epsilon\rangle$, we see that there are arbitrarily low-energy states that don’t lie within any low-energy Hilbert space.
In short: the “low-energy Hilbert space” does not contain all (or most) low-energy states. Conversely, the space of low-energy states is not a vector space, since it fails to be closed under linear combination. Moreover, if you ignore the fact that it’s not a vector space, it appears to be infinite-dimensional; more evidence that it really looks nothing like the space constructed from a truncation of the eigenbasis.
Of course, this is not actually about the harmonic oscillator. The same logic should apply to every quantum system, including field theories. | 2021-01-19 12:35:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7841271758079529, "perplexity": 709.2785263205969}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703518240.40/warc/CC-MAIN-20210119103923-20210119133923-00709.warc.gz"} |
https://brilliant.org/problems/dave-the-mischievous-astronomer/ | # Dave the mischievous astronomer
Calculus Level 3
Dave is very mischievous. He presents Carl a problem:
Take the function $f(n)=\sum^n_{k=0}\log\left(\frac{k+2}{k+3}\right)$ What is $$f(-2)$$?
Carl immediately studies the problem carefully. After some thought he tells Dave that it cannot be done. "You cannot have a smaller upper limit on the summation! This is ludicrous!".
Dave replies with a smile "Ah! But to the eyes of an astronomer, the problem can be solved!".
Can you help Carl solve Dave's problem?
× | 2018-06-22 23:07:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9103698134422302, "perplexity": 5183.434147900835}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267864822.44/warc/CC-MAIN-20180622220911-20180623000911-00451.warc.gz"} |
https://solvedlib.com/n/question-4-15-points-using-the-curve-sketching-guidelines,19851998 | Question 4: (15 points)Using the curve sketching guidelines (and showing all the supporting work, as was done in class and
Question:
Question 4: (15 points) Using the curve sketching guidelines (and showing all the supporting work, as was done in class and in the text) , sketch the graph of this function: 5r2 b(x) = e
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Question 6 (20 points). Determine whether the following series converge or diverge, and by which test: o arctan(k) k-0 oo ok k! k-0 Question 6 (20 points). Determine whether the following series...
Question 6 (20 points). Determine whether the following series converge or diverge, and by which test: o arctan(k) k-0 oo ok k! k-0 Question 6 (20 points). Determine whether the following series converge or diverge, and by which test: o arctan(k) k-0 oo ok k! k-0...
2. Exercise 5-6 Rachel Sells is unable to reconcile the bank balance at January 31. Rachel's reconciliation is shown as follows. Cash balance per bank Add: NSF check Less: Bank service charge $3,687.20 460.00 33.00 Adjusted balance per bank$4,114.20 Cash balance per books Less: Deposits in tra... | 2023-03-26 18:28:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5966783761978149, "perplexity": 4372.843764326768}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296946445.46/warc/CC-MAIN-20230326173112-20230326203112-00711.warc.gz"} |
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# Simple Oscillation
Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24
## The Position Equation
This section shows how to form the equation describing the position of a mass on a spring.
For a simple oscillator consisting of a mass m attached to one end of a spring with a spring constant s, the restoring force, f, can be expressed by the equation
${\displaystyle f=-sx\,}$
where x is the displacement of the mass from its rest position. Substituting the expression for f into the linear momentum equation,
${\displaystyle f=ma=m{d^{2}x \over dt^{2}}\,}$
where a is the acceleration of the mass, we can get
${\displaystyle m{\frac {d^{2}x}{dt^{2}}}=-sx}$
or,
${\displaystyle {\frac {d^{2}x}{dt^{2}}}+{\frac {s}{m}}x=0}$
Note that the frequency of oscillation ${\displaystyle \omega _{0}}$ is given by
${\displaystyle \omega _{0}^{2}={s \over m}\,}$
To solve the equation, we can assume
${\displaystyle x(t)=Ae^{\lambda t}\,}$
The force equation then becomes
${\displaystyle (\lambda ^{2}+\omega _{0}^{2})Ae^{\lambda t}=0,}$
Giving the equation
${\displaystyle \lambda ^{2}+\omega _{0}^{2}=0,}$
Solving for ${\displaystyle \lambda }$
${\displaystyle \lambda =\pm j\omega _{0}\,}$
This gives the equation of x to be
${\displaystyle x=C_{1}e^{j\omega _{0}t}+C_{2}e^{-j\omega _{0}t}\,}$
Note that
${\displaystyle j=(-1)^{1/2}\,}$
and that C1 and C2 are constants given by the initial conditions of the system
If the position of the mass at t = 0 is denoted as x0, then
${\displaystyle C_{1}+C_{2}=x_{0}\,}$
and if the velocity of the mass at t = 0 is denoted as u0, then
${\displaystyle -j(u_{0}/\omega _{0})=C_{1}-C_{2}\,}$
Solving the two boundary condition equations gives
${\displaystyle C_{1}={\frac {1}{2}}(x_{0}-j(u_{0}/\omega _{0}))}$
${\displaystyle C_{2}={\frac {1}{2}}(x_{0}+j(u_{0}/\omega _{0}))}$
The position is then given by
${\displaystyle x(t)=x_{0}cos(\omega _{0}t)+(u_{0}/\omega _{0})sin(\omega _{0}t)\,}$
This equation can also be found by assuming that x is of the form
${\displaystyle x(t)=A_{1}cos(\omega _{0}t)+A_{2}sin(\omega _{0}t)\,}$
And by applying the same initial conditions,
${\displaystyle A_{1}=x_{0}\,}$
${\displaystyle A_{2}={\frac {u_{0}}{\omega _{0}}}\,}$
This gives rise to the same position equation
${\displaystyle x(t)=x_{0}cos(\omega _{0}t)+(u_{0}/\omega _{0})sin(\omega _{0}t)\,}$
## Alternate Position Equation Forms
If A1 and A2 are of the form
${\displaystyle A_{1}=Acos(\phi )\,}$
${\displaystyle A_{2}=Asin(\phi )\,}$
Then the position equation can be written
${\displaystyle x(t)=Acos(\omega _{0}t-\phi )\,}$
By applying the initial conditions (x(0)=x0, u(0)=u0) it is found that
${\displaystyle x_{0}=Acos(\phi )\,}$
${\displaystyle {\frac {u_{0}}{\omega _{0}}}=Asin(\phi )\,}$
If these two equations are squared and summed, then it is found that
${\displaystyle A={\sqrt {x_{0}^{2}+({\frac {u_{0}}{\omega _{0}}})^{2}}}\,}$
And if the difference of the same two equations is found, the result is that
${\displaystyle \phi =tan^{-1}({\frac {u_{0}}{x_{0}\omega _{0}}})\,}$
The position equation can also be written as the Real part of the imaginary position equation
${\displaystyle \mathbf {Re} [x(t)]=x(t)=Acos(\omega _{0}t-\phi )\,}$
Due to euler's rule (e = cosφ + jsinφ), x(t) is of the form
${\displaystyle x(t)=Ae^{j(\omega _{0}t-\phi )}\,}$
Example 1.1
GIVEN: Two springs of stiffness, ${\displaystyle s}$, and two bodies of mass, ${\displaystyle M}$
FIND: The natural frequencies of the systems sketched below
${\displaystyle s_{TOTAL}=s+s{\text{ (springs are in parallel)}}}$
${\displaystyle \omega _{0}={\sqrt {\frac {s_{TOTAL}}{m_{TOTAL}}}}={\sqrt {\frac {2s}{M}}}}$
${\displaystyle \mathbf {f_{0}} ={\frac {\omega _{0}}{2\pi }}=\mathbf {{\frac {1}{2\pi }}{\sqrt {\frac {2s}{M}}}} }$
${\displaystyle \omega _{0}={\sqrt {\frac {s_{TOTAL}}{m_{TOTAL}}}}={\sqrt {\frac {s}{2M}}}}$
${\displaystyle \mathbf {f_{0}} ={\frac {\omega _{0}}{2\pi }}=\mathbf {{\frac {1}{2\pi }}{\sqrt {\frac {s}{2M}}}} }$
${\displaystyle \mathbf {1.} {\text{ }}s(x_{1}-x_{2})=sx_{2}}$
${\displaystyle \mathbf {2.} {\text{ }}-s(x_{1}-x_{2})=m{\frac {d^{2}x}{dt^{2}}}}$
${\displaystyle {\frac {d^{2}x_{1}}{dt^{2}}}+{\frac {s}{2m}}x_{1}=0}$
${\displaystyle \omega _{0}={\sqrt {\frac {s}{2m}}}}$
${\displaystyle \mathbf {f_{0}={\frac {1}{2\pi }}{\sqrt {\frac {s}{2m}}}} }$
${\displaystyle \omega _{0}={\sqrt {\frac {2s}{m}}}}$
${\displaystyle \mathbf {f_{0}={\frac {1}{2\pi }}{\sqrt {\frac {2s}{m}}}} }$
# Forced Oscillations(Simple Spring-Mass System)
Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24
## Recap of Section 1.3
In the previous section, we discussed how adding a damping component (e. g. a dashpot) to an unforced, simple spring-mass system would affect the response of the system. In particular, we learned that adding the dashpot to the system changed the natural frequency of the system from to a new damped natural frequency , and how this change made the response of the system change from a constant sinusoidal response to an exponentially-decaying sinusoid in which the system either had an under-damped, over-damped, or critically-damped response.
In this section, we will digress a bit by going back to the simple (undamped) oscillator system of the previous section, but this time, a constant force will be applied to this system, and we will investigate this system’s performance at low and high frequencies as well as at resonance. In particular, this section will start by introducing the characteristics of the spring and mass elements of a spring-mass system, introduce electrical analogs for both the spring and mass elements, learn how these elements combine to form the mechanical impedance system, and reveal how the impedance can describe the mechanical system’s overall response characteristics. Next, power dissipation of the forced, simple spring-mass system will be discussed in order to corroborate our use of electrical circuit analogs for the forced, simple spring-mass system. Finally, the characteristic responses of this system will be discussed, and a parameter called the amplification ratio (AR) will be introduced that will help in plotting the resonance of the forced, simple spring-mass system.
## Forced Spring Element
Taking note of Figs. 1, we see that the equation of motion for a spring that has some constant, external force being exerted on it is...
${\displaystyle {\hat {F}}=s_{M}\Delta {\hat {x}}\qquad (1.4.1)\,}$
where ${\displaystyle s_{M}\,}$ is the mechanical stiffness of the spring.
Note that in Fig. 1(c), force ${\displaystyle {\hat {F}}}$ flows constantly (i.e. without decreasing) throughout a spring, but the velocity ${\displaystyle {\hat {u}}}$ of the spring decrease from ${\displaystyle {\hat {u_{1}}}}$ to ${\displaystyle {\hat {u_{2}}}}$ as the force flows through the spring. This concept is important to know because it will be used in subsequent sections.
In practice, the stiffness of the spring ${\displaystyle s_{M}\,}$, also called the spring constant, is usually expressed as ${\displaystyle C_{M}={\frac {1}{s_{M}}}\,}$ , or the mechanical compliance of the spring. Therefore, the spring is very stiff if ${\displaystyle s_{M}\,}$ is large ${\displaystyle \Rightarrow \;C_{M}}$ is small. Similarly, the spring is very loose or “bouncy” if ${\displaystyle s_{M}\,}$ is small ${\displaystyle \Rightarrow \;C_{M}}$ is large. Noting that force and velocity are analogous to voltage and current, respectively, in electrical systems, it turns out that the characteristics of a spring are analogous to the characteristics of a capacitor in relation to, and, so we can model the “reactiveness” of a spring similar to the reactance of a capacitor if we let ${\displaystyle C=C_{M}\,}$ as shown in Fig. 2 below.
${\displaystyle Reactance\ of\ Capacitor:\ X_{C}={\frac {1}{j\omega C}}\qquad (1.4.2a)}$
${\displaystyle Reactance\ of\ Spring:\ X_{MS}={\frac {1}{j\omega C_{M}}}\qquad (1.4.2b)}$
## Forced Mass Element
Taking note of Fig. 3, the equation for a mass that has constant, external force being exerted on it is...
${\displaystyle {\hat {F}}=M_{M}{\hat {a}}=M_{M}{\hat {\dot {u}}}=M_{M}{\hat {\ddot {x}}}\qquad (1.4.3)}$
If the mass ${\displaystyle M_{M}\,}$ can vary its value and is oscillating in a mechanical system at max amplitude ${\displaystyle A_{M}\,}$ such that the input the system receives is constant at frequency ${\displaystyle \omega \,}$, as ${\displaystyle M_{M}\,}$ increases, the harder it will be for the system to move the mass at ${\displaystyle \omega \,}$ at ${\displaystyle A_{M}\,}$ until, eventually, the mass doesn’t oscillate at all . Another equivalently way to look at it is to let ${\displaystyle \omega \,}$ vary and hold ${\displaystyle M_{M}\,}$ constant. Similarly, as ${\displaystyle \omega \,}$ increases, the harder it will be to get ${\displaystyle M_{M}\,}$ to oscillate at ${\displaystyle \omega \,}$ and keep the same amplitude ${\displaystyle A_{M}\,}$ until, eventually, the mass doesn’t oscillate at all. Therefore, as ${\displaystyle \omega \,}$ increases, the “reactiveness” of mass ${\displaystyle M_{M}\,}$ decreases (i.e. ${\displaystyle M_{M}\,}$ starts to move less and less). Recalling the analogous relationship of force/voltage and velocity/current, it turns out that the characteristics of a mass are analogous to an inductor. Therefore, we can model the “reactiveness” of a mass similar to the reactance of an inductor if we let ${\displaystyle L=M_{M}\,}$ as shown in Fig. 4.
${\displaystyle Reactance\ of\ Inductor:\ X_{L}=j\omega L\qquad (1.4.4a)}$
${\displaystyle Reactance\ of\ Mass:\ X_{MM}=j\omega L_{M}\qquad (1.4.4b)}$
## Mechanical Impedance of Spring-Mass System
As mentioned twice before, force is analogous to voltage and velocity is analogous to current. Because of these relationships, this implies that the mechanical impedance for the forced, simple spring-mass system can be expressed as follows:
${\displaystyle {\hat {Z_{M}}}={\frac {\hat {F}}{\hat {u}}}\qquad (1.4.5)\,}$
In general, an undamped, spring-mass system can either be “spring-like” or “mass-like”. “Spring-like” systems can be characterized as being “bouncy” and they tend to grossly overshoot their target operating level(s) when an input is introduced to the system. These type of systems relatively take a long time to reach steady-state status. Conversely, “mass-like” can be characterized as being “lethargic” and they tend to not reach their desired operating level(s) for a given input to the system...even at steady-state! In terms of complex force and velocity, we say that “ force LEADS velocity” in mass-like systems and “velocity LEADS force” in spring-like systems (or equivalently “ force LAGS velocity” in mass-like systems and “velocity LAGS force” in spring-like systems). Figs. 5 shows this relationship graphically.
## Power Transfer of a Simple Spring-Mass System
From electrical circuit theory, the average complex power ${\displaystyle P_{E}\,}$ dissipated in a system is expressed as ...
${\displaystyle P_{E}={\frac {1}{2}}\mathbf {Re} \left\{{\hat {V}}{\hat {I^{*}}}\right\}\qquad (1.4.6)\,}$
where ${\displaystyle {\hat {V}}\,}$ and ${\displaystyle {\hat {I^{*}}}\;}$ represent the (time-invariant) complex voltage and complex conjugate current, respectively. Analogously, we can express the net power dissipation of the mechanical system ${\displaystyle {\hat {P}}_{E}\,}$ in general along with the power dissipation of a spring-like system ${\displaystyle {\hat {P}}_{MS}\,}$ or mass-like system ${\displaystyle {\hat {P}}_{MM}\,}$ as...
${\displaystyle {\hat {P}}_{E}={\frac {1}{2}}\mathbf {Re} \left\{{\hat {F}}{\hat {u^{*}}}\right\}\qquad \qquad \qquad (1.4.7a)\,}$
${\displaystyle {\hat {P}}_{MS}={\frac {1}{2}}\mathbf {Re} \left\{{\hat {F}}\left({\frac {j{\hat {F}}\omega }{s_{M}}}\right)^{*}\right\}\qquad (1.4.7b)}$
${\displaystyle {\hat {P}}_{MM}={\frac {1}{2}}\mathbf {Re} \left\{{\hat {F}}\left({\frac {\hat {F}}{j\omega M_{M}}}\right)^{*}\right\}\qquad (1.4.7c)}$
In equations 1.4.7, we see that the product of complex force and velocity are purely imaginary. Since reactive elements, or commonly called, lossless elements, cannot dissipate energy, this implies that the net power dissipation of the system is zero. This means that in our simple spring-mass system, power can only be (fully) transferred back and forth between the spring and the mass. But this is precisely what a simple spring-mass system does. Therefore, by evaluating the power dissipation, this corroborates the notion of using electrical circuit elements to model mechanical elements in our spring-mass system.
## Responses For Forced, Simple Spring-Mass System
Fig. 6 below illustrates a simple spring-mass system with a force exerted on the mass.
This system has response characteristics similar to that of the undamped oscillator system, with the only difference being that at steady-state, the system oscillates at the constant force magnitude and frequency versus exponentially decaying to zero in the unforced case. Recalling equations 1.4.2b and 1.4.4b, letting be the natural (resonant) frequency of the spring-mass system, and letting ${\displaystyle \omega _{n}\,}$ be frequency of the input received by the system, the characteristic responses of the forced spring-mass systems are presented graphically in Figs. 7 below.
${\displaystyle \mathbf {Figs.7} \,}$
## Amplification Ratio
The amplification ratio is a useful parameter that allows us to plot the frequency of the spring-mass system with the purports of revealing the resonant freq of the system solely based on the force experienced by each, the spring and mass elements of the system. In particular, AR is the magnitude of the ratio of the complex force experienced by the spring and the complex force experienced by the mass, i.e.
${\displaystyle \mathbf {AR} =\left|{\frac {s_{M}{\hat {x}}}{M_{M}{\hat {a}}}}\right|=\left|{\frac {s_{M}{\hat {x}}}{M_{M}{\hat {\dot {u}}}}}\right|=\left|{\frac {s_{M}{\hat {x}}}{M_{M}{\hat {\ddot {x}}}}}\right|\qquad (1.4.8)\,}$
If we let ${\displaystyle \zeta ={\frac {\omega }{\omega _{n}}}}$, be the frequency ratio, it turns out that AR can also be expressed as...
${\displaystyle \mathbf {AR} ={\frac {1}{1-\zeta ^{2}}}\qquad (1.4.9)\,}$
.
AR will be at its maximum when ${\displaystyle \left|X_{MS}\right|=\left|X_{MM}\right|\,}$. This happens precisely when ${\displaystyle \zeta ^{2}=1\,}$ . An example of an AR plot is shown below in Fig 8.
# Mechanical Resistance
Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24
## Mechanical Resistance
For most systems, a simple oscillator is not a very accurate model. While a simple oscillator involves a continuous transfer of energy between kinetic and potential form, with the sum of the two remaining constant, real systems involve a loss, or dissipation, of some of this energy, which is never recovered into kinetic nor potential energy. The mechanisms that cause this dissipation are varied and depend on many factors. Some of these mechanisms include drag on bodies moving through the air, thermal losses, and friction, but there are many others. Often, these mechanisms are either difficult or impossible to model, and most are non-linear. However, a simple, linear model that attempts to account for all of these losses in a system has been developed.
## Dashpots
The most common way of representing mechanical resistance in a damped system is through the use of a dashpot. A dashpot acts like a shock absorber in a car. It produces resistance to the system's motion that is proportional to the system's velocity. The faster the motion of the system, the more mechanical resistance is produced.
As seen in the graph above, a linear relationship is assumed between the force of the dashpot and the velocity at which it is moving. The constant that relates these two quantities is ${\displaystyle R_{M}}$, the mechanical resistance of the dashpot. This relationship, known as the viscous damping law, can be written as:
${\displaystyle F=R\cdot u}$
Also note that the force produced by the dashpot is always in phase with the velocity.
The power dissipated by the dashpot can be derived by looking at the work done as the dashpot resists the motion of the system:
${\displaystyle P_{D}={\frac {1}{2}}\Re \left[{\hat {F}}\cdot {\hat {u^{*}}}\right]={\frac {|{\hat {F}}|^{2}}{2R_{M}}}}$
## Modeling the Damped Oscillator
In order to incorporate the mechanical resistance (or damping) into the forced oscillator model, a dashpot is placed next to the spring. It is connected to the mass (${\displaystyle M_{M}}$) on one end and attached to the ground on the other end. A new equation describing the forces must be developed:
${\displaystyle F-S_{M}x-R_{M}u=M_{M}a\rightarrow F=S_{M}x+R_{M}{\dot {x}}+M_{M}{\ddot {x}}}$
It's phasor form is given by the following:
${\displaystyle {\hat {F}}e^{j\omega t}={\hat {x}}e^{j\omega t}\left[S_{M}+j\omega R_{M}+\left(-\omega ^{2}\right)M_{M}\right]}$
## Mechanical Impedance for Damped Oscillator
Previously, the impedance for a simple oscillator was defined as ${\displaystyle \mathbf {\frac {F}{u}} }$. Using the above equations, the impedance of a damped oscillator can be calculated:
${\displaystyle {\hat {Z_{M}}}={\frac {\hat {F}}{\hat {u}}}=R_{M}+j\left(\omega M_{M}-{\frac {S_{M}}{\omega }}\right)=|{\hat {Z_{M}}}|e^{j\Phi _{Z}}}$
For very low frequencies, the spring term dominates because of the ${\displaystyle {\frac {1}{\omega }}}$ relationship. Thus, the phase of the impedance approaches ${\displaystyle {\frac {-\pi }{2}}}$ for very low frequencies. This phase causes the velocity to "lag" the force for low frequencies. As the frequency increases, the phase difference increases toward zero. At resonance, the imaginary part of the impedance vanishes, and the phase is zero. The impedance is purely resistive at this point. For very high frequencies, the mass term dominates. Thus, the phase of the impedance approaches ${\displaystyle {\frac {\pi }{2}}}$ and the velocity "leads" the force for high frequencies.
Based on the previous equations for dissipated power, we can see that the real part of the impedance is indeed ${\displaystyle R_{M}}$. The real part of the impedance can also be defined as the cosine of the phase times its magnitude. Thus, the following equations for the power can be obtained.
${\displaystyle W_{R}={\frac {1}{2}}\Re \left[{\hat {F}}{\hat {u^{*}}}\right]={\frac {1}{2}}R_{M}|{\hat {u}}|^{2}={\frac {1}{2}}{\frac {|{\hat {F}}|^{2}}{|{\hat {Z_{M}}}|^{2}}}R_{M}={\frac {1}{2}}{\frac {|{\hat {F}}|^{2}}{|{\hat {Z_{M}}}|}}cos(\Phi _{Z})}$
# Characterizing Damped Mechanical Systems
Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24
## Characterizing Damped Mechanical Systems
Characterizing the response of Damped Mechanical Oscillating system can be easily quantified using two parameters. The system parameters are the resonance frequency (${\displaystyle '''wresonance'''}$ and the damping of the system ${\displaystyle '''Q(qualityfactor)orB(TemporalAbsorption''')}$. In practice, finding these parameters would allow for quantification of unknown systems and allow you to derive other parameters within the system.
Using the mechanical impedance in the following equation, notice that the imaginary part will equal zero at resonance.
(${\displaystyle Z_{m}=F/u=R_{m}+j(w*M_{m}-s/w)}$)
Resonance case:(${\displaystyle w*M_{m}=s/w}$)
## Calculating the Mechanical Resistance
The decay time of the system is related to 1 / B where B is the Temporal Absorption. B is related to the mechancial resistance and to the mass of the system by the following equation.
${\displaystyle B=Rm/2*Mm}$
The mechanical resistance can be derived from the equation by knowing the mass and the temporal absorption.
## Critical Damping
The system is said to be critically damped when:
${\displaystyle Rc=2*M*sqrt(s/Mm)=2*sqrt(s*Mm)=2*Mm*wn}$
A critically damped system is one in which an entire cycle is never completed. The absorption coefficient in this type of system equals the natural frequency. The system will begin to oscillate, however the amplitude will decay exponentially to zero within the first oscillation.
## Damping Ratio
${\displaystyle DampingRatio=Rm/Rc}$
The damping ratio is a comparison of the mechanical resistance of a system to the resistance value required for critical damping. Rc is the value of Rm for which the absorption coefficient equals the natural frequency (critical damping). A damping ratio equal to 1 therefore is critically damped, because the mechanical resistance value Rm is equal to the value required for critical damping Rc. A damping ratio greater than 1 will be overdamped, and a ratio less than 1 will be underdamped.
## Quality Factor
The Quality Factor (Q) is way to quickly characterize the shape of the peak in the response. It gives a quantitative representation of power dissipation in an oscillation.
${\displaystyle Q=wresonance/(wu-wl)}$
Wu and Wl are called the half power points. When looking at the response of a system, the two places on either side of the peak where the point equals half the power of the peak power defines Wu and Wl. The distance in between the two is called the half-power bandwidth. So, the resonant frequency divided by the half-power bandwidth gives you the quality factor. Mathematically, it takes Q/pi oscillations for the vibration to decay to a factor of 1/e of its original amplitude.
# Electro-Mechanical Analogies
Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24
## Why Circuit Analogs?
Acoustic devices are often combinations of mechanical and electrical elements. A common example of this would be a loudspeaker connected to a power source. It is useful in engineering applications to model the entire system with one method. This is the reason for using a circuit analogy in a vibrating mechanical system. The same analytic method can be applied to Electro-Acoustic Analogies.
## How Electro-Mechanical Analogies Work
An electrical circuit is described in terms of its potential (voltage) and flux (current). To construct a circuit analog of a mechanical system we define flux and potential for the system. This leads to two separate analog systems. The Impedance Analog denotes the force acting on an element as the potential and the velocity of the element as the flux. The Mobility Analog equates flux with the force and velocity with potential.
Mechanical Electrical Equivalent
Impedance Analog
Potential: Force Voltage
Flux: Velocity Current
Mobility Analog
Potential: Velocity Voltage
Flux: Force Current
For many, the mobility analog is considered easier for a mechanical system. It is more intuitive for force to flow as a current and for objects oscillating the same frequency to be wired in parallel. However, either method will yield equivalent results and can also be translated using the dual (dot) method.
## The Basic Elements of an Oscillating Mechanical System
The Mechanical Spring:
The ideal spring is considered to be operating within its elastic limit, so the behavior can be modeled with Hooke's Law. It is also assumed to be massless and have no damping effects.
${\displaystyle F=-cx,\ }$
The Mechanical Mass
In a vibrating system, a mass element opposes acceleration. From Newton's Second Law:
${\displaystyle F=mx^{\prime \prime }=ma=m{\frac {du}{dt}}}$
${\displaystyle F=K\int \,udt}$
The Mechanical Resistance
The dashpot is an ideal viscous damper which opposes velocity.
${\displaystyle F=Ru\displaystyle }$
Ideal Generators
The two ideal generators which can drive any system are an ideal velocity and ideal force generator. The ideal velocity generator can be denoted by a drawing of a crank or simply by declaring ${\displaystyle u(t)=f(t)}$, and the ideal force generator can be drawn with an arrow or by declaring ${\displaystyle F(t)=f(t)}$
Simple Damped Mechanical Oscillators
In the following sections we will consider this simple mechanical system as a mobility and impedance analog. It can be driven either by an ideal force or an ideal velocity generator, and we will consider simple harmonic motion. The m in the subscript denotes a mechanical system, which is currently redundant, but can be useful when combining mechanical and acoustic systems.
## The Impedance Analog
The Mechanical Spring
In a spring, force is related to the displacement from equilibrium. By Hooke's Law,
${\displaystyle F(t)=c_{m}\Delta x=c_{m}\int _{0}^{t}u(\tau )d\tau }$
The equivalent behaviour in a circuit is a capacitor:
${\displaystyle V(t)={\frac {1}{C}}\int _{0}^{t}\,i(\tau )d\tau }$
The Mechanical Mass
The force on a mass is related to the acceleration (change in velocity). The behaviour, by Newton's Second Law, is:
${\displaystyle F(t)=m_{m}a=m_{m}{\frac {d}{dt}}u(t)}$
The equivalent behaviour in a circuit is an inductor:
${\displaystyle V(t)=L{\frac {d}{dt}}i(t)}$
The Mechanical Resistance
For a viscous damper, the force is directly related to the velocity
${\displaystyle F=R_{m}u\displaystyle }$
The equivalent is a simple resistor of value ${\displaystyle R_{m}\displaystyle }$
${\displaystyle V=Ri\displaystyle }$
Example:
Thus the simple mechanical oscillator in the previous section becomes a series RCL Circuit:
The current through all three elements is equal (they are at the same velocity) and that the sum of the potential drops across each element will equal the potential at the generator (the driving force). The ideal voltage generator depicted here would be equivalent to an ideal force generator.
IMPORTANT NOTE: The velocity measured for the spring and dashpot is the relative velocity ( velocity of one end minus the velocity of the other end). The velocity of the mass, however, is the absolute velocity.
Impedances:
Element Impedance
Spring Capacitor ${\displaystyle Z_{c}={\frac {V_{c}}{I_{c}}}={\frac {c_{m}}{j\omega }}}$
Mass Inductor ${\displaystyle Z_{m}={\frac {V_{m}}{I_{m}}}=j\omega m_{m}}$
Dashpot Resistor ${\displaystyle Z_{d}={\frac {V_{m}}{I_{m}}}=R_{m}}$
## The Mobility Analog
Like the Impedance Analog above, the equivalent elements can be found by comparing their fundamental equations with the equations of circuit elements. However, since circuit equations usually define voltage in terms of current, in this case the analogy would be an expression of velocity in terms of force, which is the opposite of convention. However, this can be solved with simple algebraic manipulation.
The Mechanical Spring
${\displaystyle F(t)=c_{m}\int u(t)dt}$
The equivalent behavior for this circuit is the behavior of an inductor.
${\displaystyle \int Vdt=\int L{\frac {d}{dt}}i(t)dt}$
${\displaystyle i={\frac {1}{L}}\int \,Vdt}$
The Mechanical Mass
${\displaystyle F=m_{m}a=m_{m}{\frac {d}{dt}}u(t)}$
Similar to the spring element, if we take the general equation for a capacitor and differentiate,
${\displaystyle {\frac {d}{dt}}V(t)={\frac {d}{dt}}{\frac {1}{C}}\int \,i(t)dt}$
${\displaystyle i(t)=C{\frac {d}{dt}}V(t)}$
The Mechanical Resistance
Since the relation between force and velocity is proportionate, the only difference is that the mechanical resistance becomes inverted:
${\displaystyle F={\frac {1}{r_{m}}}u=R_{m}u}$
${\displaystyle i={\frac {1}{R}}V}$
Example:
The simple mechanical oscillator drawn above would become a parallel RLC Circuit. The potential across each element is the same because they are each operating at the same velocity. This is often the more intuitive of the two analogy methods to use, because you can visualize force "flowing" like a flux through your system. The ideal voltage generator in this drawing would correspond to an ideal velocity generator.
IMPORTANT NOTE: Since the measure of the velocity of a mass is absolute, a capacitor in this analogy must always have one terminal grounded. A capacitor with both terminals at a potential other than ground may be realized physically as an inverter, which completes all elements of this analogy.
Impedances:
Element Impedance
Spring Inductor ${\displaystyle Z_{c}={\frac {V_{m}}{I_{m}}}={\frac {j\omega }{c_{m}}}}$
Mass Capacitor ${\displaystyle Z_{m}={\frac {V_{c}}{I_{c}}}={\frac {1}{j\omega m_{m}}}}$
Dashpot Resistor ${\displaystyle Z_{d}={\frac {V_{m}}{I_{m}}}=r_{m}={\frac {1}{R_{m}}}}$
# Methods for checking Electro-Mechanical Analogies
Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24
After drawing the electro-mechanical analogy of a mechanical system, it is always safe to check the circuit. There are two methods to accomplish this:
## Review of Circuit Solving Methods
Kirchkoff's Voltage law
"The sum of the potential drops around a loop must equal zero."
${\displaystyle v_{1}+v_{2}+v_{3}+v_{4}=0\displaystyle }$
Kirchkoff's Current Law
"The Sum of the currents at a node (junction of more than two elements) must be zero"
${\displaystyle -i_{1}+i_{2}+i_{3}-i_{4}=0\displaystyle }$
Hints for solving circuits:
Remember that certain elements can be combined to simplify the circuit (the combination of like elements in series and parallel)
If solving a ciruit that involves steady-state sources, use impedances. Any circuit can eventually be combined into a single impedance using the following identities:
Impedances in series: ${\displaystyle Z_{\mathrm {eq} }=Z_{1}+Z_{2}+\,\cdots \,+Z_{n}.}$
Impedances in parallel: ${\displaystyle {\frac {1}{Z_{\mathrm {eq} }}}={\frac {1}{Z_{1}}}+{\frac {1}{Z_{2}}}+\,\cdots \,+{\frac {1}{Z_{n}}}.}$
## Dot Method: (Valid only for planar network)
This method helps obtain the dual analog (one analog is the dual of the other). The steps for the dot product are as follows:
1) Place one dot within each loop and one outside all the loops.
2) Connect the dots. Make sure that there is only one line through each element and that no lines cross more than one element.
3) Draw in each line that crosses an element its dual element, including the source.
4) The circuit obtained should have an equivalent behavior as the dual analog of the original electro-mechanical circuit.
Example:
The parallel RLC Circuit above is equivalent to a series RLC driven by an ideal current source
## Low-Frequency Limits
This method looks at the behavior of the system for very large or very small values of the parameters and compares them with the expected behavior of the mechanical system. For instance, you can compare the mobility circuit behavior of a near-infinite inductance with the mechanical system behavior of a near-infinite stiffness spring.
Very High Value Very Low Value
Capacitor Short Circuit Open Circuit
Inductor Open Circuit Closed Circuit
Resistor Open Circuit Short Circuit
# Additional Resources for solving linear circuits
Thomas & Rosa, "The Analysis and Design of Linear Circuits", Wiley, 2001
Hayt, Kemmerly & Durbin, "Engineering Circuit Analysis", 6th ed., McGraw Hill, 2002
# Examples of Electro-Mechanical Analogies
Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24
# Example System
Note: The crank indicates an ideal velocity generator, with an amplitude of ${\displaystyle u_{0}}$ rotating at ${\displaystyle \omega }$ rad/s.
# Primary variables of interest
Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24
## Basic Assumptions
Consider a piston moving in a tube. The piston starts moving at time t=0 with a velocity u=${\displaystyle u_{p}}$. The piston fits inside the tube smoothly without any friction or gap. The motion of the piston creates a planar sound wave or acoustic disturbance traveling down the tube at a constant speed c>>${\displaystyle u_{p}}$. In a case where the tube is very small, one can neglect the time it takes for acoustic disturbance to travel from the piston to the end of the tube. Hence, one can assume that the acoustic disturbance is uniform throughout the tube domain.
File:Acousticplanewave1.gif
### Assumptions
1. Although sound can exist in solids or fluid, we will first consider the medium to be a fluid at rest. The ambient, undisturbed state of the fluid will be designated using subscript zero. Recall that a fluid is a substance that deforms continuously under the application of any shear (tangential) stress.
2. Disturbance is a compressional one (as opposed to transverse).
3. Fluid is a continuum: infinitely divisible substance. Each fluid property assumed to have definite value at each point.
4. The disturbance created by the motion of the piston travels at a constant speed. It is a function of the properties of the ambient fluid. Since the properties are assumed to be uniform (the same at every location in the tube) then the speed of the disturbance has to be constant. The speed of the disturbance is the speed of sound, denoted by letter ${\displaystyle c_{0}}$ with subscript zero to denote ambient property.
5. The piston is perfectly flat, and there is no leakage flow between the piston and the tube inner wall. Both the piston and the tube walls are perfectly rigid. Tube is infinitely long, and has a constant area of cross section, A.
6. The disturbance is uniform. All deviations in fluid properties are the same across the tube for any location x. Therefore the instantaneous fluid properties are only a function of the Cartesian coordinate x (see sketch). Deviations from the ambient will be denoted by primed variables.
## Variables of interest
### Pressure (force / unit area)
Pressure is defined as the normal force per unit area acting on any control surface within the fluid.
File:Acousticcontrolsurface.gif
${\displaystyle p={\frac {{\tilde {F}}.{\tilde {n}}}{dS}}}$
For the present case,inside a tube filled with a working fluid, pressure is the ratio of the surface force acting onto the fluid in the control region and the tube area. The pressure is decomposed into two components - a constant equilibrium component, ${\displaystyle p_{0}}$, superimposed with a varying disturbance ${\displaystyle p^{'}(x)}$. The deviation ${\displaystyle p^{'}}$is also called the acoustic pressure. Note that ${\displaystyle p^{'}}$ can be positive or negative. Unit: ${\displaystyle kg/ms^{2}}$. Acoustical pressure can be measured using a microphone.
File:Acousticpressure1.gif
### Density
Density is mass of fluid per unit volume. The density, ρ, is also decomposed into the sum of ambient value (usually around ρ0= 1.15 kg/m3) and a disturbance ρ’(x). The disturbance can be positive or negative, as for the pressure. Unit: ${\displaystyle kg/m^{3}}$
### Acoustic volume velocity
Rate of change of fluid particles position as a function of time. Its the well known fluid mechanics term, flow rate.
${\displaystyle U=\int _{s}{\tilde {u}}.{\tilde {n}}\,dS}$
In most cases, the velocity is assumed constant over the entire cross section (plug flow), which gives acoustic volume velocity as a product of fluid velocity ${\displaystyle {\tilde {u}}}$ and cross section S.
${\displaystyle U={\tilde {u}}.S}$
# Electro-acoustic analogies
Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24
## Electro-acoustical Analogies
### Acoustical Mass
Consider a rigid tube-piston system as following figure.
Piston is moving back and forth sinusoidally with frequency of f. Assuming ${\displaystyle f<<{\frac {c}{l\ or\ {\sqrt {S}}}}}$ (where c is sound velocity ${\displaystyle c={\sqrt {\gamma RT_{0}}}}$), volume of fluid in tube is,
${\displaystyle \Pi _{v}=S\ l}$
Then mass (mechanical mass) of fluid in tube is given as,
${\displaystyle M_{M}=\Pi _{v}\rho _{0}=\rho _{0}S\ l}$
For sinusoidal motion of piston, fluid move as rigid body at same velocity as piston. Namely, every point in tube moves with the same velocity.
Applying the Newton's second law to the following free body diagram,
${\displaystyle SP'=(\rho _{0}Sl){\frac {du}{dt}}}$
${\displaystyle {\hat {P}}=\rho _{0}l(j\omega ){\hat {u}}=j\omega ({\frac {\rho _{0}l}{S}}){\hat {U}}}$
Where, plug flow assumption is used.
"Plug flow" assumption:
Frequently in acoustics, the velocity distribution along the normal surface of
fluid flow is assumed uniform. Under this assumption, the acoustic volume velocity U is
simply product of velocity and entire surface. ${\displaystyle U=Su}$
#### Acoustical Impedance
Recalling mechanical impedance,
${\displaystyle {\hat {Z}}_{M}={\frac {\hat {F}}{\hat {u}}}=j\omega (\rho _{0}Sl)}$
acoustical impedance (often termed an acoustic ohm) is defined as,
${\displaystyle {\hat {Z}}_{A}={\frac {\hat {P}}{\hat {U}}}={\frac {Z_{M}}{S^{2}}}=j\omega ({\frac {\rho _{0}l}{S}})\quad \left[{\frac {Ns}{m^{5}}}\right]}$
where, acoustical mass is defined.
${\displaystyle M_{A}={\frac {\rho _{0}l}{S}}}$
#### Acoustical Mobility
Acoustical mobility is defined as,
${\displaystyle {\hat {\xi }}_{A}={\frac {1}{{\hat {Z}}_{A}}}={\frac {\hat {U}}{\hat {P}}}}$
#### Acoustical Resistance
Acoustical resistance models loss due to viscous effects (friction) and flow resistance (represented by a screen).
File:Ra analogs.png rA is the reciprocal of RA and is referred to as responsiveness.
### Acoustical Generators
The acoustical generator components are pressure, P and volume velocity, U, which are analogus to force, F and velocity, u of electro-mechanical analogy respectively. Namely, for impedance analog, pressure is analogus to voltage and volume velocity is analogus to current, and vice versa for mobility analog. These are arranged in the following table.
Impedance and Mobility analogs for acoustical generators of constant pressure and constant volume velocity are as follows:
File:Acoustic gen.png
### Acoustical Compliance
Consider a piston in an enclosure.
File:Enclosed Piston.png
When the piston moves, it displaces the fluid inside the enclosure. Acoustic compliance is the measurement of how "easy" it is to displace the fluid.
Here the volume of the enclosure should be assumed to be small enough that the fluid pressure remains uniform.
Assume no heat exchange 1.adiabatic 2.gas compressed uniformly , p prime in cavity everywhere the same.
from thermo equitation File:Equ1.jpg it is easy to get the relation between disturbing pressure and displacement of the piston File:Equ3.gif where U is volume rate, P is pressure according to the definition of the impendance and mobility, we can getFile:Equ4.gif
Mobility Analog VS Impedance Analog
File:Comp.gif
### Examples of Electro-Acoustical Analogies
Example 1: Helmholtz Resonator
Assumptions - (1) Completely sealed cavity with no leaks. (2) Cavity acts like a rigid body inducing no vibrations.
Solution:
- Impedance Analog -
File:Example2holm1sol.JPG
Example 2: Combination of Side-Branch Cavities
File:Exam2prob.JPG
Solution:
- Impedance Analog -
File:Exam2sol.JPG
# Transducers - Loudspeaker
Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24
# Acoustic transducer
The purpose of an acoustic transducer is to convert electrical energy into acoustic energy. Many variations of acoustic transducers exist, such as electrostatic, balanced armature and moving-coil loudspeakers. This article focuses on moving-coil loudspeakers since they are the most commonly used type of acoustic transducer. First, the physical construction and principle of a typical moving coil transducer are discussed briefly. Second, electro-mechano-acoustical modeling of each element composing the loudspeaker is presented in a tutorial way to reinforce and supplement the theory on electro-mechanical analogies and electro-acoustic analogies previously seen in other sections. Third, the equivalent circuit is analyzed to introduce the theory behind Thiele-Small parameters, which are very useful when designing loudspeaker enclosures. A method to experimentally determine Thiele-Small parameters is also included.
## Moving-coil loudspeaker construction and principle
The classic moving-coil loudspeaker driver can be divided into three key components:
1) The magnet motor drive system, comprising the permanent magnet, the center pole and the voice coil acting together to produce a mechanical force on the diaphragm from an electrical current.
2) The loudspeaker cone system, comprising the diaphragm and dust cap, permitting mechanical force to be translated into acoustic pressure;
3) The loudspeaker suspension, comprising the spider and surround, preventing the diaphragm from breaking due to over excursion, allowing only translational movement and tending to bring the diaphragm back to its rest position.
The following illustration shows a cut-away view of a typical moving coil-permanent magnet loudspeaker. A coil is mechanically coupled to a diaphragm, also called cone, and rests in a fixed magnetic field produced by a magnet. When an electrical current flows through the coil, a corresponding magnetic field is emitted, interacting with the fixed field of the magnet and thus applying a force to the coil, pushing it away or towards the magnet. Since the cone is mechanically coupled to the coil, it will push or pull the air it is facing, causing pressure changes and emitting a sound wave.
Figure 1: A cross-sectional view of a typical moving-coil loudspeaker
An equivalent circuit can be obtained to model the loudspeaker as a lumped system. This circuit can be used to drive the design of a complete loudspeaker system, including an enclosure and sometimes even an amplifier that is matched to the properties of the driver. The following section shows how such an equivalent circuit can be obtained.
## Electro-mechano-acoustical equivalent circuit
Electro-mechanico-acoustical systems such as loudspeakers can be modeled as equivalent electrical circuits as long as each element moves as a whole. This is usually the case at low frequencies or at frequencies where the dimensions of the system are small compared to the wavelength of interest. To obtain a complete model of the loudspeaker, the interactions and properties of electrical, mechanical, and acoustical subsystems composing the loudspeaker driver must each be modeled. The following sections detail how the circuit may be obtained starting with the amplifier and ending with the acoustical load presented by air. A similar development can be found in [1] or [2].
### Electrical subsystem
The electrical part of the system is composed of a driving amplifier and a voice coil. Most amplifiers can be approximated as a perfect voltage source in series with the amplifier output impedance. The voice coil exhibits an inductance and a resistance that may be directly modeled as a circuit.
Figure 2: The amplifier and loudspeaker electrical elements modeled as a circuit
### Electrical to mechanical subsystem
When the loudspeaker is fed an electrical signal, the voice coil and magnet convert current to force. Similarly, voltage is related to the velocity. This relationship between the electrical side and the mechanical side can be modeled by a transformer.
${\displaystyle {\tilde {f_{c}}}=Bl{\tilde {i}}}$; ${\displaystyle {\tilde {u_{c}}}={\dfrac {\tilde {e}}{Bl}}}$
Figure 3: A transformer modeling transduction from the electrical impedance to mechanical mobility analogy
### Mechanical subsystem
In a first approximation, a moving coil loudspeaker may be thought of as a mass-spring system where the diaphragm and the voice coil constitute the mass and the spider and surround constitute the spring element. Losses in the suspension can be modeled as a resistor.
Figure 4: Mass spring system and associated circuit analogies of the impedance and mobility type.
The equation of motion gives us :
${\displaystyle {\tilde {f_{c}}}=R_{m}{\tilde {u_{c}}}+{\dfrac {\tilde {u_{c}}}{j\omega C_{MS}}}+j\omega M_{MD}{\tilde {u_{c}}}}$
${\displaystyle {\dfrac {\tilde {f_{c}}}{\tilde {u_{c}}}}=R_{m}+{\dfrac {1}{j\omega C_{MS}}}+j\omega M_{MD}}$
Which yields the mechanical impedance type analogy in the form of a series RLC circuit. A parallel RLC circuit may also be obtained to get the mobility analog following mathematical manipulation:
${\displaystyle {\dfrac {\tilde {u_{c}}}{\tilde {f_{c}}}}={\dfrac {1}{R_{m}+{\dfrac {1}{j\omega C_{MS}}}+j\omega M_{MD}}}}$
${\displaystyle {\dfrac {\tilde {u_{c}}}{\tilde {f_{c}}}}={\dfrac {1}{{\dfrac {1}{G_{m}}}+{\dfrac {1}{j\omega C_{MS}}}+{\dfrac {1}{\dfrac {1}{j\omega M_{MD}}}}}}}$
Which expresses the mechanical mobility type analogy in the form of a parallel RLC circuit where the denominator elements are respectively a parallel conductance, inductance, and compliance.
### Mechanical to acoustical subsystem
A loudspeaker’s diaphragm may be thought of as a piston that pushes and pulls on the air facing it, converting mechanical force and velocity into acoustic pressure and volume velocity. The equations are as follow:
${\displaystyle {\tilde {P_{d}}}={\dfrac {\tilde {f_{c}}}{\tilde {S_{D}}}}}$; ${\displaystyle {\tilde {U_{c}}}={\tilde {u_{c}}}{S_{D}}}$
These equations can be modeled by a transformer.
Figure 5: A transformer modeling the transduction from mechanical mobility to acoustical mobility analogy performed by a loudspeaker's diaphragm
### Acoustical subsystem
The impedance presented by the air load on the loudspeaker's diaphragm is both resistive due to sound radiation and reactive due to the air mass that is being pushed radially but does not contribute to sound radiation to the far field. The air load on the diaphragm can be modeled as an impedance or an admittance. Specific values and approximations can be found in [1], [2] or [3]. Note that the air load depends on the mounting conditions of the loudspeaker. If the loudspeaker is mounted in a baffle, the air load will be the same on each side of the diaphragm. Then, if the air load on one side is ${\displaystyle Y_{AR}}$ in the admittance analogy, then the total air load is ${\displaystyle Y_{AR}/2}$ as both loads are in parallel.
### Complete electro-mechano-acoustical equivalent circuit
Using electrical impedance, mechanical mobility and acoustical admittance yield the following equivalent circuit, modeling the entire loudspeaker drive unit.
Figure 6: A complete electro-mechano-acoustical equivalent circuit of a loudspeaker drive unit
This circuit can be reduced by substituting the transformers and connected loads by an equivalent loading that would present the same impedance as the loaded transformer. An example of this is shown on figure 7, where acoustical and electrical loads and sources have been "brought over" to the mechanical side.
Figure 7: Mechanical equivalent circuit modeling of a loudspeaker drive unit
The advantage of doing such manipulations is that we can then directly relate electrical measurements with elements in the circuit. This will later allow us to obtain values for the different components of the model and match this model to real loudspeaker drivers. We can further simplify this circuit by using Norton's theorem and converting the series electrical components and voltage source into an equivalent current source and parallel electrical components. Then, using a technique called the Dot method, presented in section Solution Methods: Electro-Mechanical Analogies, we can obtain a single loop series circuit which is the dual of the parallel circuit previously obtained with Norton's theorem. If we are mainly interested in the low frequency behavior of the loudspeaker, as should be the case when using lumped element modeling, we can neglect the effect of the voice coil inductance, which has an effect only at high frequencies. Furthermore, the air load impedance at low frequencies is mass-like and can be modeled by a simple inductance ${\displaystyle M_{M1}}$. This results in a simplified low frequency model equivalent circuit, shown of figure 8, which is easier to manipulate than the circuit of figure 7. Note that the analogy used for this circuit is of the impedance type.
Figure 8: Low frequency approximation mechanical equivalent circuit of a loudspeaker drive unit
Where ${\displaystyle M_{M1}=2.67a^{3}\rho }$ if ${\displaystyle a}$ is the radius of the loudspeaker and ${\displaystyle \rho }$, the density of air. Mass elements, in this case the mass of the diaphragm and voice coil ${\displaystyle M_{MS}}$ and the air mass loading the diaphragm ${\displaystyle 2M_{M1}}$ can be regrouped in a single element:
${\displaystyle M_{MS}=M_{MD}+2M_{M1}}$
## Thiele-Small Parameters
### Theory
The complete low frequency behavior of a loudspeaker drive unit can be modeled with just six parameters, called Thiele-Small parameters. Most of these parameters result from algebraic manipulation of the equations of the circuit of figure 8. Loudspeaker driver manufacturers seldom provide electro-mechano-acoustical parameters directly and rather provide Thiele-Small parameters in datasheets, but conversion from one to the other is quite simple. The Thiele-Small parameters are as follow:
1. ${\displaystyle R_{e}}$, the voice coil DC resistance;
2. ${\displaystyle Q_{ES}}$, the electrical Q factor;
3. ${\displaystyle Q_{MS}}$, the mechanical Q factor;
4. ${\displaystyle f_{s}}$, the loudspeaker resonance frequency;
5. ${\displaystyle S_{D}}$, the effective surface area of the diaphragm;
6. ${\displaystyle V_{AS}}$, the equivalent suspension volume: the volume of air that has the same acoustic compliance as the suspension of the loudspeaker driver.
These parameters can be related directly from the low frequency approximation circuit of figure 8, with ${\displaystyle R_{e}}$ and ${\displaystyle S_{D}}$ being explicit.
${\displaystyle Q_{MS}={\dfrac {1}{R_{MS}}}{\sqrt {\dfrac {M_{MS}}{C_{MS}}}}}$; ${\displaystyle Q_{ES}={\dfrac {R_{g}+R_{e}}{(Bl)^{2}}}{\sqrt {\dfrac {M_{MS}}{C_{MS}}}}}$ ; ${\displaystyle f_{s}={\dfrac {1}{2\pi {\sqrt {M_{MS}C_{MS}}}}}}$; ${\displaystyle V_{AS}=C_{MS}S_{D}^{2}\rho c^{2}}$
Where ${\displaystyle \rho c^{2}}$ is the Bulk modulus of air. It follows that, if given Thiele-Small parameters, one can extract the values of each component of the circuit of figure 8 using the following equations :
${\displaystyle C_{MS}={\dfrac {V_{AS}}{S_{D}^{2}\rho c^{2}}}}$; ${\displaystyle M_{MS}={\dfrac {1}{(2\pi f_{s})^{2}C_{MS}}}}$; ${\displaystyle R_{MS}={\dfrac {1}{Q_{MS}}}{\sqrt {\dfrac {M_{MS}}{C_{MS}}}}}$; ${\displaystyle Bl={\sqrt {\dfrac {R_{e}}{2\pi f_{s}Q_{ES}C_{MS}}}}}$; ${\displaystyle M_{MD}=M_{MS}-2M_{M1}}$;
### Measurement
Many methods can be used to measure Thiele-Small parameters of drivers. Measurement of Thiele-Small parameters is sometimes necessary if a manufacturer does not provide them. Also, the actual Thiele-Small parameters of a given loudspeaker can differ from nominal values significantly. The method described in this section comes from [2]. Note that for this method, the loudspeaker is considered to be mounted in an infinite baffle. In practice, a baffle with a diameter of four times that of the loudspeaker is sufficient. Measurements without a baffle are also possible: the air mass loading will simply be halved and can be easily accounted for. The setup for this method includes an FFT analyzer or a mean to obtain an impedance curve. A signal generator of variable frequency and an AC meter can also be used.
Figure 9: Simple experimental setup to measure the impedance of a loudspeaker drive unit
${\displaystyle Z_{spk}=R{\dfrac {V_{spk}}{V_{s}\left(1-{\dfrac {V_{spk}}{V_{s}}}\right)}}}$
Figure 10: A typical loudspeaker drive unit impedance curve
Once the impedance curve of the loudspeaker is measured, ${\displaystyle R_{e}}$ and ${\displaystyle f_{s}}$ can be directly identified by looking at the low frequency asymptote of the impedance value and the center frequency of the resonance peak. If the frequencies where ${\displaystyle Z_{spk}={\sqrt {R_{e}R_{c}}}}$ are identified as ${\displaystyle f_{l}}$ and ${\displaystyle f_{h}}$, Q factors can be calculated.
${\displaystyle Q_{MS}={\dfrac {f_{s}}{f_{h}-f_{l}}}{\sqrt {\dfrac {R_{c}}{R_{e}}}}}$
${\displaystyle Q_{ES}={\dfrac {Q_{MS}}{{\dfrac {R_{c}}{R_{e}}}-1}}}$
${\displaystyle S_{D}}$ can simply be approximated by ${\displaystyle \pi a^{2}}$, where ${\displaystyle a}$ is the radius of the loudspeaker driver. The last remaining Thiele-Small parameter, ${\displaystyle V_{AS}}$ is slightly trickier to measure. The idea is to either increase mass or reduce compliance of the loudspeaker drive unit and note the shift in resonance frequency. If a known mass ${\displaystyle M_{x}}$ is added to the loudspeaker diaphragm, the new resonance frequency will be:
${\displaystyle f_{s}^{'}={\dfrac {1}{2\pi {\sqrt {(M_{MS}+M_{x})C_{MS}}}}}}$
And the equivalent suspension volume may be obtained with:
${\displaystyle V_{AS}=\left(1-{\dfrac {f_{s}^{'2}}{f_{s}^{2}}}\right){\dfrac {S_{D}^{2}\rho c^{2}}{(2\pi f_{s}^{'})^{2}M_{x}}}}$
Hence, all Thiele-Small parameters modeling the low frequency behavior of the loudspeaker drive unit can be obtained from a fairly simple setup. These parameters are of tremendous help in loudspeaker enclosure design.
### Numerical example
This section presents a numerical example of obtaining Thiele-Small parameters from impedance curves. The impedance curves presented in this section have been obtained from simulations using nominal Thiele-Small parameters of a real woofer loudspeaker. Firsy, these Thiele-Small parameters have been transformed into an electro-mechano-acoustical circuit using the equation presented before. Second, the circuit was treated as a black box and the method to extract Thiele-Small parameters was used. The purpose of this simulation is to present the method, step by step, using realistic values so that the reader can get more familiar with the process, the magnitude of the values and with what to expect when performing such measurements.
For this simulation, a loudspeaker of radius ${\displaystyle a=6.55cm}$ is mounted on a baffle sufficiently large to act as an infinite baffle. Its impedance is obtained and plotted in figure 11, where important cursors have already been placed.
Figure 11: Simulated measurement of an impedance curve for a woofer loudspeaker
The low frequency asymptote is immediately identified as ${\displaystyle Re=6.6\Omega }$. The resonance is clear and centered at ${\displaystyle fs=33Hz}$. The value of the impedance at this frequency is about ${\displaystyle 66\Omega }$. This yields ${\displaystyle {\sqrt {R_{e}R_{c}}}=20.8\Omega }$, which occurs at ${\displaystyle f_{l}=19.5Hz}$ and ${\displaystyle f_{h}=52.5Hz}$. With this information, we can compute some of the Thiele-Small parameters.
${\displaystyle Q_{MS}={\dfrac {f_{s}}{f_{h}-f_{l}}}{\sqrt {\dfrac {R_{c}}{R_{e}}}}={\dfrac {33}{52.5-19.5}}*{\sqrt {\dfrac {66}{6.6}}}=3.1}$
${\displaystyle Q_{ES}={\dfrac {Q_{MS}}{{\dfrac {R_{c}}{R_{e}}}-1}}={\dfrac {3.1}{{\dfrac {66}{6.6}}-1}}=0.35}$
As a next step, a mass of ${\displaystyle M_{x}=10g}$ is fixed to the loudspeaker diaphragm. This shifts the resonance frequency and yields a new impedance curve, as shown on figure 12.
Figure 12: Simulated measurement of an impedance curve for a woofer loudspeaker
${\displaystyle S_{D}=\pi a^{2}=0.0135m^{2}}$
${\displaystyle V_{AS}=\left(1-{\dfrac {27.5^{2}}{33^{2}}}\right){\dfrac {0.0135^{2}*1.18*344^{2}}{(2\pi 27.5)^{2}*0.01}}=0.0272m^{3}}$
Once all six Thiele-Small parameters have been obtained, it is possible to calculate values for the electro-mechano-acoustical circuit modeling elements of figure 6 or 7. From then, the design of an enclosure can start. This is discussed in application sections Sealed box subwoofer design and Bass reflex enclosure design.
# References
[1] Kleiner, Mendel. Electroacoustics. CRC Press, 2013.
[2] Beranek, Leo L., and Tim Mellow. Acoustics: sound fields and transducers. Academic Press, 2012.
[3] Kinsler, Lawrence E., et al. Fundamentals of Acoustics, 4th Edition. Wiley-VCH, 1999.
[4] Small, Richard H. "Direct radiator loudspeaker system analysis." Journal of the Audio Engineering Society 20.5 (1972): 383-395.
# Moving Resonators
Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24
## Moving Resonators
Consider the situation shown in the figure below. We have a typical Helmholtz resonator driven by a massless piston which generates a sinusoidal pressure ${\displaystyle P_{G}}$, however the cavity is not fixed in this case. Rather, it is supported above the ground by a spring with compliance ${\displaystyle C_{M}}$. Assume the cavity has a mass ${\displaystyle M_{M}}$.
Recall the Helmholtz resonator (see Module #9). The difference in this case is that the pressure in the cavity exerts a force on the bottom of the cavity, which is now not fixed as in the original Helmholtz resonator. This pressure causes a force that acts upon the cavity bottom. If the surface area of the cavity bottom is ${\displaystyle S_{C}}$, then Newton's Laws applied to the cavity bottom give
${\displaystyle \sum {F}=p_{C}S_{C}-{\frac {x}{C_{M}}}=M_{M}{\ddot {x}}\Rightarrow p_{C}S_{C}=\left[{\frac {1}{j\omega C_{M}}}+j\omega M_{M}\right]u}$
In order to develop the equivalent circuit, we observe that we simply need to use the pressure (potential across ${\displaystyle C_{A}}$) in the cavity to generate a force in the mechanical circuit. The above equation shows that the mass of the cavity and the spring compliance should be placed in series in the mechanical circuit. In order to convert the pressure to a force, the transformer is used with a ratio of ${\displaystyle 1:S_{C}}$.
## Example
A practical example of a moving resonator is a marimba. A marimba is a similar to a xylophone but has larger resonators that produce deeper and richer tones. The resonators (seen in the picture as long, hollow pipes) are mounted under an array of wooden bars which are struck to create tones. Since these resonators are not fixed, but are connected to the ground through a stiffness (the stand), it can be modeled as a moving resonator. Marimbas are not tunable instruments like flutes or even pianos. It would be interesting to see how the tone of the marimba changes as a result of changing the stiffness of the mount.
# Transverse vibrations of strings
Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24
## Introduction
This section deals with the wave nature of vibrations constrained to one dimension. Examples of this type of wave motion are found in objects such a pipes and tubes with a small diameter (no transverse motion of fluid) or in a string stretched on a musical instrument.
Stretched strings can be used to produce sound (e.g. music instruments like guitars). The stretched string constitutes a mechanical system that will be studied in this chapter. Later, the characteristics of this system will be used to help to understand by analogies acoustical systems.
## What is a wave equation?
There are various types of waves (i.e. electromagnetic, mechanical, etc.) that act all around us. It is important to use wave equations to describe the time-space behavior of the variables of interest in such waves. Wave equations solve the fundamental equations of motion in a way that eliminates all variables but one. Waves can propagate longitudinal or parallel to the propagation direction or perpendicular (transverse) to the direction of propagation. To visualize the motion of such waves click here (Acoustics animations provided by Dr. Dan Russell,Kettering University)
## One dimensional Case
Assumptions :
- the string is uniform in size and density
- stiffness of string is negligible for small deformations
- effects of gravity neglected
- no dissipative forces like frictions
- string deforms in a plane
- motion of the string can be described by using one single spatial coordinate
Spatial representation of the string in vibration:
The following is the free-body diagram of a string in motion in a spatial coordinate system:
From the diagram above, it can be observed that the tensions in each side of the string will be the same as follows:
Using Taylor series to expand we obtain:
## Characterization of the mechanical system
A one dimensional wave can be described by the following equation (called the wave equation):
${\displaystyle \left({\frac {\partial ^{2}y}{\partial x^{2}}}\right)=\left({\frac {1}{c^{2}}}\right)\left({\frac {\partial ^{2}y}{\partial t^{2}}}\right)}$
where,
${\displaystyle y(x,t)=f(\xi )+g(\eta )\,}$ is a solution,
With ${\displaystyle \xi =ct-x\,}$ and ${\displaystyle \eta =ct+x\,}$
Another way to solve this equation is the Method of separation of variables. This is useful for modal analysis. This assumes the solution is of the form:
${\displaystyle y(x,t)=f(x)g(t)\ }$
The result is the same as above, but in a form that is more convenient for modal analysis.
For more information on this approach see: Eric W. Weisstein et al. "Separation of Variables." From MathWorld—A Wolfram Web Resource. [2]
Please see Wave Properties for information on variable c, along with other important properties.
For more information on wave equations see: Eric W. Weisstein. "Wave Equation." From MathWorld—A Wolfram Web Resource. [3]
Example with the function ${\displaystyle f(\xi )}$ :
Example: Java String simulation
This show a simple simulation of a plucked string with fixed ends.
# Time-Domain Solutions
Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24
## d'Alembert Solutions
In 1747, Jean Le Rond d'Alembert published a solution to the one-dimensional wave equation.
The general solution, now known as the d'Alembert method, can be found by introducing two new variables:
${\displaystyle \xi =ct-x\,}$ and ${\displaystyle \eta =ct+x\,}$
and then applying the chain rule to the general form of the wave equation.
From this, the solution can be written in the form:
${\displaystyle y(\xi ,\eta )=f(\xi )+g(\eta )\,=f(x+ct)+g(x-ct)}$
where f and g are arbitrary functions, that represent two waves traveling in opposing directions.
A more detailed look into the proof of the d'Alembert solution can be found here.
## Example of Time Domain Solution
If f(ct-x) is plotted vs. x for two instants in time, the two waves are the same shape but the second displaced by a distance of c(t2-t1) to the right.
The two arbitrary functions could be determined from initial conditions or boundary values.
# Boundary Conditions and Forced Vibrations
Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24
## Boundary Conditions
The functions representing the solutions to the wave equation previously discussed,
i.e. ${\displaystyle y(x,t)=f(\xi )+g(\eta )\,}$ with ${\displaystyle \xi =ct-x\,}$ and ${\displaystyle \eta =ct+x\,}$
are dependent upon the boundary and initial conditions. If it is assumed that the wave is propagating through a string, the initial conditions are related to the specific disturbance in the string at t=0. These specific disturbances are determined by location and type of contact and can be anything from simple oscillations to violent impulses. The effects of boundary conditions are less subtle.
The most simple boundary conditions are the Fixed Support and Free End. In practice, the Free End boundary condition is rarely encountered since it is assumed there are no transverse forces holding the string (e.g. the string is simply floating).
### For a Fixed Support
The overall displacement of the waves travelling in the string, at the support, must be zero. Denoting x=0 at the support, This requires:
${\displaystyle y(0,t)=f(ct-0)+g(ct+0)=0\,}$
Therefore, the total transverse displacement at x=0 is zero.
The sequence of wave reflection for incident, reflected and combined waves are illustrated below. Please note that the wave is traveling to the left (negative x direction) at the beginning. The reflected wave is ,of course, traveling to the right (positive x direction).
t=0
t=t1
t=t2
t=t3
### For a Free Support
Unlike the Fixed Support boundary condition, the transverse displacement at the support does not need to be zero, but must require the sum of transverse forces to cancel. If it is assumed that the angle of displacement is small,
${\displaystyle \sin(\theta )\approx \theta =\left({\frac {\partial y}{\partial x}}\right)\,}$
and so,
${\displaystyle \sum F_{y}=T\sin(\theta )\approx T\left({\frac {\partial y}{\partial x}}\right)=0\,}$
But of course, the tension in the string, or T, will not be zero and this requires the slope at x=0 to be zero:
i.e. ${\displaystyle \left({\frac {\partial y}{\partial x}}\right)=0\,}$
Again for free boundary, the sequence of wave reflection for incident, reflected and combined waves are illustrated below:
t=0
t=t1
t=t2
t=t3
### Other Boundary Conditions
There are many other types of boundary conditions that do not fall into our simplified categories. As one would expect though, it isn't difficult to relate the characteristics of numerous "complex" systems to the basic boundary conditions. Typical or realistic boundary conditions include mass-loaded, resistance-loaded, damping-loaded, and impedance-loaded strings. For further information, see Kinsler, Fundamentals of Acoustics, pp 54–58.
Here is a website with nice movies of wave reflection at different BC's: Wave Reflection
## Wave Properties
To begin with, a few definitions of useful variables will be discussed. These include; the wave number, phase speed, and wavelength characteristics of wave travelling through a string.
The speed that a wave propagates through a string is given in terms of the phase speed, typically in m/s, given by:
${\displaystyle c={\sqrt {T/\rho _{L}}}\,}$ where ${\displaystyle \rho _{L}\,}$ is the density per unit length of the string.
The wave number is used to reduce the transverse displacement equation to a simpler form and for simple harmonic motion, is multiplied by the lateral position. It is given by:
${\displaystyle k=\left({\frac {\omega }{c}}\right)\,}$ where ${\displaystyle \omega =2\pi f\,}$
Lastly, the wavelength is defined as:
${\displaystyle \lambda =\left({\frac {2\pi }{k}}\right)=\left({\frac {c}{f}}\right)\,}$
and is defined as the distance between two points, usually peaks, of a periodic waveform.
These "wave properties" are of practical importance when calculating the solution of the wave equation for a number of different cases. As will be seen later, the wave number is used extensively to describe wave phenomenon graphically and quantitatively.
For further information: Wave Properties
## Forced Vibrations
1.forced vibrations of infinite string suppose there is a string very long , at x=0 there is a force exerted on it.
F(t)=Fcos(wt)=Real{Fexp(jwt)}
use the boundary condition at x=0,
neglect the reflected wave
it is easy to get the wave form
where w is the angular velocity, k is the wave number.
according to the impedance definition
it represents the characteristic impedance of the string. obviously, it is purely resistive, which is like the resistance in the mechanical system.
The dissipated power
Note: along the string, all the variables propagate at same speed.
link title a useful link to show the time-space property of the wave.
Some interesting animations of the wave at different boundary conditions.
1.hard boundary (which is like a fixed end)
2.soft boundary (which is like a free end)
3.from low density to high density string
4.from high density to low density string
# Room Acoustics and Concert Halls
Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24
# Room Acoustics and Concert Halls
## Introduction
From performing on many different rooms and stages all over the United States, I thought it would be nice to have a better understanding and source about the room acoustics. This Wikibook page is intended to help the user with basic technical questions/answers about room acoustics. Main topics that will be covered are: what really makes a room sound good or bad, alive or dead. This will lead into absorption and transmission coefficients, decay of sound in the room, and reverberation. Different use of materials in rooms will be mentioned also. There is no intention of taking work from another. This page is a switchboard source to help the user find information about room acoustics.
## Sound Fields
Two types of sound fields are involved in room acoustics: Direct Sound and Reverberant Sound.
### Direct Sound
The component of the sound field in a room that involves only a direct path between the source and the receiver, before any reflections off walls and other surfaces.
### Reverberant Sound
The component of the sound field in a room that involves the direct path and the path after it reflects off of walls or any other surfaces. How the waves deflect off of the mediums all depends on the absorption and transmission coefficients.
Good example pictures are shown at Crutchfield Advisor, a Physics Site from MTSU, and Voiceteacher.com
## Room Coefficients
In a perfect world, if there is a sound shot right at a wall, the sound should come right back. But because sounds hit different materials types of walls, the sound does not have perfect reflection. From 1, these are explained as follows:
### Absorption & Transmission Coefficients
The best way to explain how sound reacts to different mediums is through acoustical energy. When sound impacts on a wall, acoustical energy will be reflected, absorbed, or transmitted through the wall.
Absorption Coefficient: NB: this chemical structure is unrelated to the acoustics being discussed.
Transmission Coefficient:
If all of the acoustic energy hits the wall and none is reflected, the alpha would equal 1. The energy had zero reflection and was absorbed or transmitted. This would be an example of a dead or soft wall because it takes in everything and doesn't reflect anything back. Rooms that are like this are called Anechoic Rooms which looks like this from Axiomaudio.
If all of the acoustic energy hits the wall and all reflects back, the alpha would equal 0. This would be an example of a live or hard wall because the sound bounces right back and does not go through the wall. Rooms that are like this are called Reverberant Rooms like this McIntosh room. Look how the walls have nothing attached to them. More room for the sound waves to bounce off the walls.
### Room Averaged Sound Absorption Coefficient
Not all rooms have the same walls on all sides. The room averaged sound absorption coefficient can be used to have different types of materials and areas of walls averaged together.
RASAC:
### Absorption Coefficients for Specific Materials
Basic sound absorption Coefficients are shown here at Acoustical Surfaces.
Brick, unglazed, painted alpha ~ .01 - .03 -> Sound reflects back
An open door alpha equals 1 -> Sound goes through
Units are in Sabins.
## Sound Decay and Reverberation Time
In a large reverberant room, a sound can still propagate after the sound source has been turned off. This time when the sound intensity level has decay 60 dB is called the reverberation time of the room.
## References
[1] Lord, Gatley, Evensen; Noise Control for Engineers, Krieger Publishing, 435 pgs
Back to Engineering Acoustics
Created by Kevin Baldwin
# Bass Reflex Enclosure Design
Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24
## Introduction
Bass-reflex enclosures improve the low-frequency response of loudspeaker systems. Bass-reflex enclosures are also called "vented-box design" or "ported-cabinet design". A bass-reflex enclosure includes a vent or port between the cabinet and the ambient environment. This type of design, as one may observe by looking at contemporary loudspeaker products, is still widely used today. Although the construction of bass-reflex enclosures is fairly simple, their design is not simple, and requires proper tuning. This reference focuses on the technical details of bass-reflex design. General loudspeaker information can be found here.
## Effects of the Port on the Enclosure Response
Before discussing the bass-reflex enclosure, it is important to be familiar with the simpler sealed enclosure system performance. As the name suggests, the sealed enclosure system attaches the loudspeaker to a sealed enclosure (except for a small air leak included to equalize the ambient pressure inside). Ideally, the enclosure would act as an acoustical compliance element, as the air inside the enclosure is compressed and rarified. Often, however, an acoustic material is added inside the box to reduce standing waves, dissipate heat, and other reasons. This adds a resistive element to the acoustical lumped-element model. A non-ideal model of the effect of the enclosure actually adds an acoustical mass element to complete a series lumped-element circuit given in Figure 1. For more on sealed enclosure design, see the Sealed Box Subwoofer Design page.
Figure 1. Sealed enclosure acoustic circuit.
In the case of a bass-reflex enclosure, a port is added to the construction. Typically, the port is cylindrical and is flanged on the end pointing outside the enclosure. In a bass-reflex enclosure, the amount of acoustic material used is usually much less than in the sealed enclosure case, often none at all. This allows air to flow freely through the port. Instead, the larger losses come from the air leakage in the enclosure. With this setup, a lumped-element acoustical circuit has the following form.
Figure 2. Bass-reflex enclosure acoustic circuit.
In this figure, ${\displaystyle Z_{RAD}}$ represents the radiation impedance of the outside environment on the loudspeaker diaphragm. The loading on the rear of the diaphragm has changed when compared to the sealed enclosure case. If one visualizes the movement of air within the enclosure, some of the air is compressed and rarified by the compliance of the enclosure, some leaks out of the enclosure, and some flows out of the port. This explains the parallel combination of ${\displaystyle M_{AP}}$, ${\displaystyle C_{AB}}$, and ${\displaystyle R_{AL}}$. A truly realistic model would incorporate a radiation impedance of the port in series with ${\displaystyle M_{AP}}$, but for now it is ignored. Finally, ${\displaystyle M_{AB}}$, the acoustical mass of the enclosure, is included as discussed in the sealed enclosure case. The formulas which calculate the enclosure parameters are listed in Appendix B.
It is important to note the parallel combination of ${\displaystyle M_{AP}}$ and ${\displaystyle C_{AB}}$. This forms a Helmholtz resonator (click here for more information). Physically, the port functions as the “neck” of the resonator and the enclosure functions as the “cavity.” In this case, the resonator is driven from the piston directly on the cavity instead of the typical Helmholtz case where it is driven at the “neck.” However, the same resonant behavior still occurs at the enclosure resonance frequency, ${\displaystyle f_{B}}$. At this frequency, the impedance seen by the loudspeaker diaphragm is large (see Figure 3 below). Thus, the load on the loudspeaker reduces the velocity flowing through its mechanical parameters, causing an anti-resonance condition where the displacement of the diaphragm is a minimum. Instead, the majority of the volume velocity is actually emitted by the port itself instead of the loudspeaker. When this impedance is reflected to the electrical circuit, it is proportional to ${\displaystyle 1/Z}$, thus a minimum in the impedance seen by the voice coil is small. Figure 3 shows a plot of the impedance seen at the terminals of the loudspeaker. In this example, ${\displaystyle f_{B}}$ was found to be about 40 Hz, which corresponds to the null in the voice-coil impedance.
Figure 3. Impedances seen by the loudspeaker diaphragm and voice coil.
## Quantitative Analysis of Port on Enclosure
The performance of the loudspeaker is first measured by its velocity response, which can be found directly from the equivalent circuit of the system. As the goal of most loudspeaker designs is to improve the bass response (leaving high-frequency production to a tweeter), low frequency approximations will be made as much as possible to simplify the analysis. First, the inductance of the voice coil, ${\displaystyle {\it {L_{E}}}}$, can be ignored as long as ${\displaystyle \omega \ll R_{E}/L_{E}}$. In a typical loudspeaker, ${\displaystyle {\it {L_{E}}}}$ is of the order of 1 mH, while ${\displaystyle {\it {R_{E}}}}$ is typically 8${\displaystyle \Omega }$, thus an upper frequency limit is approximately 1 kHz for this approximation, which is certainly high enough for the frequency range of interest.
Another approximation involves the radiation impedance, ${\displaystyle {\it {Z_{RAD}}}}$. It can be shown [1] that this value is given by the following equation (in acoustical ohms):
${\displaystyle Z_{RAD}={\frac {\rho _{0}c}{\pi a^{2}}}\left[\left(1-{\frac {J_{1}(2ka)}{ka}}\right)+j{\frac {H_{1}(2ka)}{ka}}\right]}$
Where ${\displaystyle J_{1}(x)}$ and ${\displaystyle H_{1}(x)}$ are types of Bessel functions. For small values of ka,
${\displaystyle J_{1}(2ka)\approx ka}$ and ${\displaystyle H_{1}(2ka)\approx {\frac {8(ka)^{2}}{3\pi }}}$ ${\displaystyle \Rightarrow Z_{RAD}\approx j{\frac {8\rho _{0}\omega }{3\pi ^{2}a}}=jM_{A1}}$
Hence, the low-frequency impedance on the loudspeaker is represented with an acoustic mass ${\displaystyle M_{A1}}$ [1]. For a simple analysis, ${\displaystyle R_{E}}$, ${\displaystyle M_{MD}}$, ${\displaystyle C_{MS}}$, and ${\displaystyle R_{MS}}$ (the transducer parameters, or Thiele-Small parameters) are converted to their acoustical equivalents. All conversions for all parameters are given in Appendix A. Then, the series masses, ${\displaystyle M_{AD}}$, ${\displaystyle M_{A1}}$, and ${\displaystyle M_{AB}}$, are lumped together to create ${\displaystyle M_{AC}}$. This new circuit is shown below.
Figure 4. Low-Frequency Equivalent Acoustic Circuit
Unlike sealed enclosure analysis, there are multiple sources of volume velocity that radiate to the outside environment. Hence, the diaphragm volume velocity, ${\displaystyle U_{D}}$, is not analyzed but rather ${\displaystyle U_{0}=U_{D}+U_{P}+U_{L}}$. This essentially draws a “bubble” around the enclosure and treats the system as a source with volume velocity ${\displaystyle U_{0}}$. This “lumped” approach will only be valid for low frequencies, but previous approximations have already limited the analysis to such frequencies anyway. It can be seen from the circuit that the volume velocity flowing into the enclosure, ${\displaystyle U_{B}=-U_{0}}$, compresses the air inside the enclosure. Thus, the circuit model of Figure 3 is valid and the relationship relating input voltage, ${\displaystyle V_{IN}}$ to ${\displaystyle U_{0}}$ may be computed.
In order to make the equations easier to understand, several parameters are combined to form other parameter names. First, ${\displaystyle \omega _{B}}$ and ${\displaystyle \omega _{S}}$, the enclosure and loudspeaker resonance frequencies, respectively, are:
${\displaystyle \omega _{B}={\frac {1}{\sqrt {M_{AP}C_{AB}}}}}$ ${\displaystyle \omega _{S}={\frac {1}{\sqrt {M_{AC}C_{AS}}}}}$
Based on the nature of the derivation, it is convenient to define the parameters ${\displaystyle \omega _{0}}$ and h, the Helmholtz tuning ratio:
${\displaystyle \omega _{0}={\sqrt {\omega _{B}\omega _{S}}}}$ ${\displaystyle h={\frac {\omega _{B}}{\omega _{S}}}}$
A parameter known as the compliance ratio or volume ratio, ${\displaystyle \alpha }$, is given by:
${\displaystyle \alpha ={\frac {C_{AS}}{C_{AB}}}={\frac {V_{AS}}{V_{AB}}}}$
Other parameters are combined to form what are known as quality factors:
${\displaystyle Q_{L}=R_{AL}{\sqrt {\frac {C_{AB}}{M_{AP}}}}}$ ${\displaystyle Q_{TS}={\frac {1}{R_{AE}+R_{AS}}}{\sqrt {\frac {M_{AC}}{C_{AS}}}}}$
This notation allows for a simpler expression for the resulting transfer function [1]:
${\displaystyle {\frac {U_{0}}{V_{IN}}}=G(s)={\frac {(s^{3}/\omega _{0}^{4})}{(s/\omega _{0})^{4}+a_{3}(s/\omega _{0})^{3}+a_{2}(s/\omega _{0})^{2}+a_{1}(s/\omega _{0})+1}}}$
where
${\displaystyle a_{1}={\frac {1}{Q_{L}{\sqrt {h}}}}+{\frac {\sqrt {h}}{Q_{TS}}}}$ ${\displaystyle a_{2}={\frac {\alpha +1}{h}}+h+{\frac {1}{Q_{L}Q_{TS}}}}$ ${\displaystyle a_{3}={\frac {1}{Q_{TS}{\sqrt {h}}}}+{\frac {\sqrt {h}}{Q_{L}}}}$
## Development of Low-Frequency Pressure Response
It can be shown [2] that for ${\displaystyle ka<1/2}$, a loudspeaker behaves as a spherical source. Here, a represents the radius of the loudspeaker. For a 15” diameter loudspeaker in air, this low frequency limit is about 150 Hz. For smaller loudspeakers, this limit increases. This limit dominates the limit which ignores ${\displaystyle L_{E}}$, and is consistent with the limit that models ${\displaystyle Z_{RAD}}$ by ${\displaystyle M_{A1}}$.
Within this limit, the loudspeaker emits a volume velocity ${\displaystyle U_{0}}$, as determined in the previous section. For a simple spherical source with volume velocity ${\displaystyle U_{0}}$, the far-field pressure is given by [1]:
${\displaystyle p(r)\simeq j\omega \rho _{0}U_{0}{\frac {e^{-jkr}}{4\pi r}}}$
It is possible to simply let ${\displaystyle r=1}$ for this analysis without loss of generality because distance is only a function of the surroundings, not the loudspeaker. Also, because the transfer function magnitude is of primary interest, the exponential term, which has a unity magnitude, is omitted. Hence, the pressure response of the system is given by [1]:
${\displaystyle {\frac {p}{V_{IN}}}={\frac {\rho _{0}s}{4\pi }}{\frac {U_{0}}{V_{IN}}}={\frac {\rho _{0}Bl}{4\pi S_{D}R_{E}M_{A}S}}H(s)}$
Where ${\displaystyle H(s)=sG(s)}$. In the following sections, design methods will focus on ${\displaystyle |H(s)|^{2}}$ rather than ${\displaystyle H(s)}$, which is given by:
${\displaystyle |H(s)|^{2}={\frac {\Omega ^{8}}{\Omega ^{8}+\left(a_{3}^{2}-2a_{2}\right)\Omega ^{6}+\left(a_{2}^{2}+2-2a_{1}a_{3}\right)\Omega ^{4}+\left(a_{1}^{2}-2a_{2}\right)\Omega ^{2}+1}}}$ ${\displaystyle \Omega ={\frac {\omega }{\omega _{0}}}}$
This also implicitly ignores the constants in front of ${\displaystyle |H(s)|}$ since they simply scale the response and do not affect the shape of the frequency response curve.
## Alignments
A popular way to determine the ideal parameters has been through the use of alignments. The concept of alignments is based upon well investigated electrical filter theory. Filter development is a method of selecting the poles (and possibly zeros) of a transfer function to meet a particular design criterion. The criteria are the desired properties of a magnitude-squared transfer function, which in this case is ${\displaystyle |H(s)|^{2}}$. From any of the design criteria, the poles (and possibly zeros) of ${\displaystyle |H(s)|^{2}}$ are found, which can then be used to calculate the numerator and denominator. This is the “optimal” transfer function, which has coefficients that are matched to the parameters of ${\displaystyle |H(s)|^{2}}$ to compute the appropriate values that will yield a design that meets the criteria.
There are many different types of filter designs, each which have trade-offs associated with them. However, this design approach is limited because of the structure of ${\displaystyle |H(s)|^{2}}$. In particular, it has the structure of a fourth-order high-pass filter with all zeros at s = 0. Therefore, only those filter design methods which produce a low-pass filter with only poles will be acceptable methods to use. From the traditional set of algorithms, only Butterworth and Chebyshev low-pass filters have only poles. In addition, another type of filter called a quasi-Butterworth filter can also be used, which has similar properties to a Butterworth filter. These three algorithms are fairly simple, thus they are the most popular. When these low-pass filters are converted to high-pass filters, the ${\displaystyle s\rightarrow 1/s}$ transformation produces ${\displaystyle s^{8}}$ in the numerator.
More details regarding filter theory and these relationships can be found in numerous resources, including [5].
## Butterworth Alignment
The Butterworth algorithm is designed to have a maximally flat pass band. Since the slope of a function corresponds to its derivatives, a flat function will have derivatives equal to zero. Since as flat of a pass band as possible is optimal, the ideal function will have as many derivatives equal to zero as possible at s = 0. Of course, if all derivatives were equal to zero, then the function would be a constant, which performs no filtering.
Often, it is better to examine what is called the loss function. Loss is the reciprocal of gain, thus
${\displaystyle |{\hat {H}}(s)|^{2}={\frac {1}{|H(s)|^{2}}}}$
The loss function can be used to achieve the desired properties, then the desired gain function is recovered from the loss function.
Now, applying the desired Butterworth property of maximal pass-band flatness, the loss function is simply a polynomial with derivatives equal to zero at s = 0. At the same time, the original polynomial must be of degree eight (yielding a fourth-order function). However, derivatives one through seven can be equal to zero if [3]
${\displaystyle |{\hat {H}}(\Omega )|^{2}=1+\Omega ^{8}\Rightarrow |H(\Omega )|^{2}={\frac {1}{1+\Omega ^{8}}}}$
With the high-pass transformation ${\displaystyle \Omega \rightarrow 1/\Omega }$,
${\displaystyle |H(\Omega )|^{2}={\frac {\Omega ^{8}}{\Omega ^{8}+1}}}$
It is convenient to define ${\displaystyle \Omega =\omega /\omega _{3dB}}$, since ${\displaystyle \Omega =1\Rightarrow |H(s)|^{2}=0.5}$ or -3 dB. This definition allows the matching of coefficients for the ${\displaystyle |H(s)|^{2}}$ describing the loudspeaker response when ${\displaystyle \omega _{3dB}=\omega _{0}}$. From this matching, the following design equations are obtained [1]:
${\displaystyle a_{1}=a_{3}={\sqrt {4+2{\sqrt {2}}}}}$ ${\displaystyle a_{2}=2+{\sqrt {2}}}$
## Quasi-Butterworth Alignment
The quasi-Butterworth alignments do not have as well-defined of an algorithm when compared to the Butterworth alignment. The name “quasi-Butterworth” comes from the fact that the transfer functions for these responses appear similar to the Butterworth ones, with (in general) the addition of terms in the denominator. This will be illustrated below. While there are many types of quasi-Butterworth alignments, the simplest and most popular is the 3rd order alignment (QB3). The comparison of the QB3 magnitude-squared response against the 4th order Butterworth is shown below.
${\displaystyle \left|H_{QB3}(\omega )\right|^{2}={\frac {(\omega /\omega _{3dB})^{8}}{(\omega /\omega _{3dB})^{8}+B^{2}(\omega /\omega _{3dB})^{2}+1}}}$ ${\displaystyle \left|H_{B4}(\omega )\right|^{2}={\frac {(\omega /\omega _{3dB})^{8}}{(\omega /\omega _{3dB})^{8}+1}}}$
Notice that the case ${\displaystyle B=0}$ is the Butterworth alignment. The reason that this QB alignment is called 3rd order is due to the fact that as B increases, the slope approaches 3 dec/dec instead of 4 dec/dec, as in 4th order Butterworth. This phenomenon can be seen in Figure 5.
Figure 5: 3rd-Order Quasi-Butterworth Response for ${\displaystyle 0.1\leq B\leq 3}$
Equating the system response ${\displaystyle |H(s)|^{2}}$ with ${\displaystyle |H_{QB3}(s)|^{2}}$, the equations guiding the design can be found [1]:
${\displaystyle B^{2}=a_{1}^{2}-2a_{2}}$ ${\displaystyle a_{2}^{2}+2=2a_{1}a_{3}}$ ${\displaystyle a_{3}={\sqrt {2a_{2}}}}$ ${\displaystyle a_{2}>2+{\sqrt {2}}}$
## Chebyshev Alignment
The Chebyshev algorithm is an alternative to the Butterworth algorithm. For the Chebyshev response, the maximally-flat passband restriction is abandoned. Now, a ripple, or fluctuation is allowed in the pass band. This allows a steeper transition or roll-off to occur. In this type of application, the low-frequency response of the loudspeaker can be extended beyond what can be achieved by Butterworth-type filters. An example plot of a Chebyshev high-pass response with 0.5 dB of ripple against a Butterworth high-pass response for the same ${\displaystyle \omega _{3dB}}$ is shown below.
Figure 6: Chebyshev vs. Butterworth High-Pass Response.
The Chebyshev response is defined by [4]:
${\displaystyle |{\hat {H}}(j\Omega )|^{2}=1+\epsilon ^{2}C_{n}^{2}(\Omega )}$
${\displaystyle C_{n}(\Omega )}$ is called the Chebyshev polynomial and is defined by [4]:
${\displaystyle C_{n}(\Omega )={\big \lbrace }}$ ${\displaystyle {\rm {{cos}[{\it {{n}{\rm {{cos}^{-1}(\Omega )]}}}}}}}$ ${\displaystyle |\Omega |<1}$ ${\displaystyle {\rm {{cosh}[{\it {{n}{\rm {{cosh}^{-1}(\Omega )]}}}}}}}$ ${\displaystyle |\Omega |>1}$
Fortunately, Chebyshev polynomials satisfy a simple recursion formula [4]:
${\displaystyle C_{0}(x)=1}$ ${\displaystyle C_{1}(x)=x}$ ${\displaystyle C_{n}(x)=2xC_{n-1}-C_{n-2}}$
For more information on Chebyshev polynomials, see the Wolfram Mathworld: Chebyshev Polynomials page.
When applying the high-pass transformation to the 4th order form of ${\displaystyle |{\hat {H}}(j\Omega )|^{2}}$, the desired response has the form [1]:
${\displaystyle |H(j\Omega )|^{2}={\frac {1+\epsilon ^{2}}{1+\epsilon ^{2}C_{4}^{2}(1/\Omega )}}}$
The parameter ${\displaystyle \epsilon }$ determines the ripple. In particular, the magnitude of the ripple is ${\displaystyle 10{\rm {{log}[1+\epsilon ^{2}]}}}$ dB and can be chosen by the designer, similar to B in the quasi-Butterworth case. Using the recursion formula for ${\displaystyle C_{n}(x)}$,
${\displaystyle C_{4}\left({\frac {1}{\Omega }}\right)=8\left({\frac {1}{\Omega }}\right)^{4}-8\left({\frac {1}{\Omega }}\right)^{2}+1}$
Applying this equation to ${\displaystyle |H(j\Omega )|^{2}}$ [1],
${\displaystyle \Rightarrow |H(\Omega )|^{2}={\frac {{\frac {1+\epsilon ^{2}}{64\epsilon ^{2}}}\Omega ^{8}}{{\frac {1+\epsilon ^{2}}{64\epsilon ^{2}}}\Omega ^{8}+{\frac {1}{4}}\Omega ^{6}+{\frac {5}{4}}\Omega ^{4}-2\Omega ^{2}+1}}}$ ${\displaystyle \Omega ={\frac {\omega }{\omega _{n}}}}$ ${\displaystyle \omega _{n}={\frac {\omega _{3dB}}{2}}{\sqrt {2+{\sqrt {2+2{\sqrt {2+{\frac {1}{\epsilon ^{2}}}}}}}}}}$
Thus, the design equations become [1]:
${\displaystyle \omega _{0}=\omega _{n}{\sqrt[{8}]{\frac {64\epsilon ^{2}}{1+\epsilon ^{2}}}}}$ ${\displaystyle k={\rm {{tanh}\left[{\frac {1}{4}}{\rm {{sinh}^{-1}\left({\frac {1}{\epsilon }}\right)}}\right]}}}$ ${\displaystyle D={\frac {k^{4}+6k^{2}+1}{8}}}$ ${\displaystyle a_{1}={\frac {k{\sqrt {4+2{\sqrt {2}}}}}{\sqrt[{4}]{D}}},}$ ${\displaystyle a_{2}={\frac {1+k^{2}(1+{\sqrt {2}})}{\sqrt {D}}}}$ ${\displaystyle a_{3}={\frac {a_{1}}{\sqrt {D}}}\left[1-{\frac {1-k^{2}}{2{\sqrt {2}}}}\right]}$
## Choosing the Correct Alignment
With all the equations that have already been presented, the question naturally arises, “Which one should I choose?” Notice that the coefficients ${\displaystyle a_{1}}$, ${\displaystyle a_{2}}$, and ${\displaystyle a_{3}}$ are not simply related to the parameters of the system response. Certain combinations of parameters may indeed invalidate one or more of the alignments because they cannot realize the necessary coefficients. With this in mind, general guidelines have been developed to guide the selection of the appropriate alignment. This is very useful if one is designing an enclosure to suit a particular transducer that cannot be changed.
The general guideline for the Butterworth alignment focuses on ${\displaystyle Q_{L}}$ and ${\displaystyle Q_{TS}}$. Since the three coefficients ${\displaystyle a_{1}}$, ${\displaystyle a_{2}}$, and ${\displaystyle a_{3}}$ are a function of ${\displaystyle Q_{L}}$, ${\displaystyle Q_{TS}}$, h, and ${\displaystyle \alpha }$, fixing one of these parameters yields three equations that uniquely determine the other three. In the case where a particular transducer is already given, ${\displaystyle Q_{TS}}$ is essentially fixed. If the desired parameters of the enclosure are already known, then ${\displaystyle Q_{L}}$ is a better starting point.
In the case that the rigid requirements of the Butterworth alignment cannot be satisfied, the quasi-Butterworth alignment is often applied when ${\displaystyle Q_{TS}}$ is not large enough.. The addition of another parameter, B, allows more flexibility in the design.
For ${\displaystyle Q_{TS}}$ values that are too large for the Butterworth alignment, the Chebyshev alignment is typically chosen. However, the steep transition of the Chebyshev alignment may also be utilized to attempt to extend the bass response of the loudspeaker in the case where the transducer properties can be changed.
In addition to these three popular alignments, research continues in the area of developing new algorithms that can manipulate the low-frequency response of the bass-reflex enclosure. For example, a 5th order quasi-Butterworth alignment has been developed [6]; its advantages include improved low frequency extension, and much reduced driver excursion at low frequencies and typically bi-amping or tri-amping, while its disadvatages include somewhat difficult mathematics and electronic complication (electronic crossovers are typically required). Another example [7] applies root-locus techniques to achieve results. In the modern age of high-powered computing, other researchers have focused their efforts in creating computerized optimization algorithms that can be modified to achieve a flatter response with sharp roll-off or introduce quasi-ripples which provide a boost in sub-bass frequencies [8].
## References
[1] Leach, W. Marshall, Jr. Introduction to Electroacoustics and Audio Amplifier Design. 2nd ed. Kendall/Hunt, Dubuque, IA. 2001.
[2] Beranek, L. L. Acoustics. 2nd ed. Acoustical Society of America, Woodbridge, NY. 1993.
[3] DeCarlo, Raymond A. “The Butterworth Approximation.” Notes from ECE 445. Purdue University. 2004.
[4] DeCarlo, Raymond A. “The Chebyshev Approximation.” Notes from ECE 445. Purdue University. 2004.
[5] VanValkenburg, M. E. Analog Filter Design. Holt, Rinehart and Winston, Inc. Chicago, IL. 1982.
[6] Kreutz, Joseph and Panzer, Joerg. "Derivation of the Quasi-Butterworth 5 Alignments." Journal of the Audio Engineering Society. Vol. 42, No. 5, May 1994.
[7] Rutt, Thomas E. "Root-Locus Technique for Vented-Box Loudspeaker Design." Journal of the Audio Engineering Society. Vol. 33, No. 9, September 1985.
[8] Simeonov, Lubomir B. and Shopova-Simeonova, Elena. "Passive-Radiator Loudspeaker System Design Software Including Optimization Algorithm." Journal of the Audio Engineering Society. Vol. 47, No. 4, April 1999.
## Appendix A: Equivalent Circuit Parameters
Name Electrical Equivalent Mechanical Equivalent Acoustical Equivalent
Voice-Coil Resistance ${\displaystyle R_{E}}$ ${\displaystyle R_{ME}={\frac {(Bl)^{2}}{R_{E}}}}$ ${\displaystyle R_{AE}={\frac {(Bl)^{2}}{R_{E}S_{D}^{2}}}}$
Driver (Speaker) Mass See ${\displaystyle C_{MEC}}$ ${\displaystyle M_{MD}}$ ${\displaystyle M_{AD}={\frac {M_{MD}}{S_{D}^{2}}}}$
Driver (Speaker) Suspension Compliance ${\displaystyle L_{CES}=(Bl)^{2}C_{MS}}$ ${\displaystyle C_{MS}}$ ${\displaystyle C_{AS}=S_{D}^{2}C_{MS}}$
Driver (Speaker) Suspension Resistance ${\displaystyle R_{ES}={\frac {(Bl)^{2}}{R_{MS}}}}$ ${\displaystyle R_{MS}}$ ${\displaystyle R_{AS}={\frac {R_{MS}}{S_{D}^{2}}}}$
Enclosure Compliance ${\displaystyle L_{CEB}={\frac {(Bl)^{2}C_{AB}}{S_{D}^{2}}}}$ ${\displaystyle C_{MB}={\frac {C_{AB}}{S_{D}^{2}}}}$ ${\displaystyle C_{AB}}$
Enclosure Air-Leak Losses ${\displaystyle R_{EL}={\frac {(Bl)^{2}}{S_{D}^{2}R_{AL}}}}$ ${\displaystyle R_{ML}=S_{D}^{2}R_{AL}}$ ${\displaystyle R_{AL}}$
Acoustic Mass of Port ${\displaystyle C_{MEP}={\frac {S_{D}^{2}M_{AP}}{(Bl)^{2}}}}$ ${\displaystyle M_{MP}=S_{D}^{2}M_{AP}}$ ${\displaystyle M_{AP}}$
Enclosure Mass Load See ${\displaystyle C_{MEC}}$ See ${\displaystyle M_{MC}}$ ${\displaystyle M_{AB}}$
Low-Frequency Radiation Mass Load See ${\displaystyle C_{MEC}}$ See ${\displaystyle M_{MC}}$ ${\displaystyle M_{A1}}$
Combination Mass Load ${\displaystyle C_{MEC}={\frac {S_{D}^{2}M_{AC}}{(Bl)^{2}}}}$
${\displaystyle ={\frac {S_{D}^{2}(M_{AB}+M_{A1})+M_{MD}}{(Bl)^{2}}}}$
${\displaystyle M_{MC}=S_{D}^{2}(M_{AB}+M_{A1})+M_{MD}}$ ${\displaystyle M_{AC}=M_{AD}+M_{AB}+M_{A1}}$
${\displaystyle ={\frac {M_{MD}}{S_{D}^{2}}}+M_{AB}+M_{A1}}$
## Appendix B: Enclosure Parameter Formulas
Figure 7: Important dimensions of bass-reflex enclosure.
Based on these dimensions [1],
${\displaystyle C_{AB}={\frac {V_{AB}}{\rho _{0}c_{0}^{2}}}}$ ${\displaystyle M_{AB}={\frac {B\rho _{eff}}{\pi a}}}$ ${\displaystyle B={\frac {d}{3}}\left({\frac {S_{D}}{S_{B}}}\right)^{2}{\sqrt {\frac {\pi }{S_{D}}}}+{\frac {8}{3\pi }}\left[1-{\frac {S_{D}}{S_{B}}}\right]}$ ${\displaystyle \rho _{0}\leq \rho _{eff}\leq \rho _{0}\left(1-{\frac {V_{fill}}{V_{B}}}\right)+\rho _{fill}{\frac {V_{fill}}{V_{B}}}}$ ${\displaystyle V_{AB}=V_{B}\left[1-{\frac {V_{fill}}{V_{B}}}\right]\left[1+{\frac {\gamma -1}{1+\gamma \left({\frac {V_{B}}{V_{fill}}}-1\right){\frac {\rho _{0}c_{air}}{\rho _{fill}c_{fill}}}}}\right]}$ ${\displaystyle V_{B}=hwd}$ (inside enclosure gross volume) ${\displaystyle S_{B}=wh}$ (baffle area of the side the speaker is mounted on) ${\displaystyle c_{air}=}$specific heat of air at constant isovolumetric process (about ${\displaystyle 0.718{\frac {\rm {kJ}}{\rm {kg.K}}}}$ at 300 K) ${\displaystyle c_{fill}=}$specific heat of filling at constant volume (${\displaystyle V_{filling}}$) ${\displaystyle \rho _{0}=}$mean density of air (about ${\displaystyle 1.3{\frac {\rm {kg}}{\rm {m^{3}}}}}$ at 300 K) ${\displaystyle \rho _{fill}=}$ density of filling ${\displaystyle \gamma =}$ ratio of specific heats (Isobaric/Isovolumetric processes) for air (about 1.4 at 300 K) ${\displaystyle c_{0}=}$ speed of sound in air (about 344 m/s) ${\displaystyle \rho _{eff}}$ = effective density of enclosure. If little or no filling (acceptable assumption in a bass-reflex system but not for sealed enclosures), ${\displaystyle \rho _{eff}\approx \rho _{0}}$
# New Acoustic Filter For Ultrasonics Media
Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24
## Introduction
Acoustic filters are used in many devices such as mufflers, noise control materials (absorptive and reactive), and loudspeaker systems to name a few. Although the waves in simple (single-medium) acoustic filters usually travel in gases such as air and carbon-monoxide (in the case of automobile mufflers) or in materials such as fiberglass, polyvinylidene fluoride (PVDF) film, or polyethylene (Saran Wrap), there are also filters that couple two or three distinct media together to achieve a desired acoustic response. General information about basic acoustic filter design can be perused at the following wikibook page Acoustic Filter Design & Implementation. The focus of this article will be on acoustic filters that use multilayer air/polymer film-coupled media as its acoustic medium for sound waves to propagate through; concluding with an example of how these filters can be used to detect and extrapolate audio frequency information in high-frequency "carrier" waves that carry an audio signal. However, before getting into these specific type of acoustic filters, we need to briefly discuss how sound waves interact with the medium(media) in which it travels and how these factors can play a role when designing acoustic filters.
## Changes in Media Properties Due to Sound Wave Characteristics
As with any system being designed, the filter response characteristics of an acoustic filter are tailored based on the frequency spectrum of the input signal and the desired output. The input signal may be infrasonic (frequencies below human hearing), sonic (frequencies within human hearing range), or ultrasonic (frequencies above human hearing range). In addition to the frequency content of the input signal, the density, and, thus, the characteristic impedance of the medium (media) being used in the acoustic filter must also be taken into account. In general, the characteristic impedance ${\displaystyle Z_{0}\,}$ for a particular medium is expressed as...
${\displaystyle Z_{0}=\pm \rho _{0}c\,}$ ${\displaystyle (Pa\cdot s/m)}$
where
${\displaystyle \pm \rho _{0}\,}$ = (equilibrium) density of medium ${\displaystyle (kg/m^{3})\,}$
${\displaystyle c\,}$ = speed of sound in medium ${\displaystyle (m/s)\,}$
The characteristic impedance is important because this value simultaneously gives an idea of how fast or slow particles will travel as well as how much mass is "weighting down" the particles in the medium (per unit area or volume) when they are excited by a sound source. The speed in which sound travels in the medium needs to be taken into consideration because this factor can ultimately affect the time response of the filter (i.e. the output of the filter may not radiate or attentuate sound fast or slow enough if not designed properly). The intensity ${\displaystyle I_{A}\,}$ of a sound wave is expressed as...
${\displaystyle I_{A}={\frac {1}{T}}\int _{0}^{T}pu\quad dt=\pm {\frac {P^{2}}{2\rho _{0}c}}\,}$ ${\displaystyle (W/m^{2})\,}$.
${\displaystyle I_{A}\,}$ is interpreted as the (time-averaged) rate of energy transmission of a sound wave through a unit area normal to the direction of propagation, and this parameter is also an important factor in acoustic filter design because the characteristic properties of the given medium can change relative to intensity of the sound wave traveling through it. In other words, the reaction of the particles (atoms or molecules) that make up the medium will respond differently when the intensity of the sound wave is very high or very small relative to the size of the control area (i.e. dimensions of the filter, in this case). Other properties such as the elasticity and mean propagation velocity (of a sound wave) can change in the acoustic medium as well, but focusing on frequency, impedance, and/or intensity in the design process usually takes care of these other parameters because most of them will inevitably be dependent on the aforementioned properties of the medium.
## Why Coupled Acoustic Media in Acoustic Filters?
In acoustic transducers, media coupling is employed in acoustic transducers to either increase or decrease the impedance of the transducer, and, thus, control the intensity and speed of the signal acting on the transducer while converting the incident wave, or initial excitation sound wave, from one form of energy to another (e.g. converting acoustic energy to electrical energy). Specifically, the impedance of the transducer is augmented by inserting a solid structure (not necessarily rigid) between the transducer and the initial propagation medium (e.g. air). The reflective properties of the inserted medium is exploited to either increase or decrease the intensity and propagation speed of the incident sound wave. It is the ability to alter, and to some extent, control, the impedance of a propagation medium by (periodically) inserting (a) solid structure(s) such as thin, flexible films in the original medium (air) and its ability to concomitantly alter the frequency response of the original medium that makes use of multilayer media in acoustic filters attractive. The reflection factor and transmission factor ${\displaystyle {\hat {R}}\,}$ and ${\displaystyle {\hat {T}}\,}$, respectively, between two media, expressed as...
${\displaystyle {\hat {R}}={\frac {pressure\ of\ reflected\ portion\ of\ incident\ wave}{pressure\ of\ incident\ wave}}={\frac {\rho c-Z_{in}}{\rho c+Z_{in}}}\,}$
and
${\displaystyle {\hat {T}}={\frac {pressure\ of\ transmitted\ portion\ of\ incident\ wave}{pressure\ of\ incident\ wave}}=1+{\hat {R}}\,}$,
are the tangible values that tell how much of the incident wave is being reflected from and transmitted through the junction where the media meet. Note that ${\displaystyle Z_{in}\,}$ is the (total) input impedance seen by the incident sound wave upon just entering an air-solid acoustic media layer. In the case of multiple air-columns as shown in Fig. 2, ${\displaystyle Z_{in}\,}$ is the aggregate impedance of each air-column layer seen by the incident wave at the input. Below in Fig. 1, a simple illustration explains what happens when an incident sound wave propagating in medium (1) and comes in contact with medium (2) at the junction of the both media (x=0), where the sound waves are represented by vectors.
As mentioned above, an example of three such successive air-solid acoustic media layers is shown in Fig. 2 and the electroacoustic equivalent circuit for Fig. 2 is shown in Fig. 3 where ${\displaystyle L=\rho _{s}h_{s}\,}$ = (density of solid material)(thickness of solid material) = unit-area (or volume) mass, ${\displaystyle Z=\rho c=\,}$ characteristic acoustic impedance of medium, and ${\displaystyle \beta =k=\omega /c=\,}$ wavenumber. Note that in the case of a multilayer, coupled acoustic medium in an acoustic filter, the impedance of each air-solid section is calculated by using the following general purpose impedance ratio equation (also referred to as transfer matrices)...
${\displaystyle {\frac {Z_{a}}{Z_{0}}}={\frac {\left({\frac {Z_{b}}{Z_{0}}}\right)+j\ \tan(kd)}{1+j\ \left({\frac {Z_{b}}{Z_{0}}}\right)\tan(kd)}}\,}$
where ${\displaystyle Z_{b}\,}$ is the (known) impedance at the edge of the solid of an air-solid layer (on the right) and ${\displaystyle Z_{a}\,}$ is the (unknown) impedance at the edge of the air column of an air-solid layer.
## Effects of High-Intensity, Ultrasonic Waves in Acoustic Media in Audio Frequency Spectrum
When an ultrasonic wave is used as a carrier to transmit audio frequencies, three audio effects are associated with extrapolating the audio frequency information from the carrier wave: (a) beating effects, (b) parametric array effects, and (c) radiation pressure.
Beating occurs when two ultrasonic waves with distinct frequencies ${\displaystyle f_{1}\,}$ and ${\displaystyle f_{2}\,}$ propagate in the same direction, resulting in amplitude variations which consequently make the audio signal information go in and out of phase, or “beat”, at a frequency of ${\displaystyle f_{1}-f_{2}\,}$.
Parametric array effects occur when the intensity of an ultrasonic wave is so high in a particular medium that the high displacements of particles (atoms) per wave cycle changes properties of that medium so that it influences parameters like elasticity, density, propagation velocity, etc. in a non-linear fashion. The results of parametric array effects on modulated, high-intensity, ultrasonic waves in a particular medium (or coupled media) is the generation and propagation of audio frequency waves (not necessarily present in the original audio information) that are generated in a manner similar to the nonlinear process of amplitude demodulation commonly inherent in diode circuits (when diodes are forward biased).
Another audio effect that arises from high-intensity ultrasonic beams of sound is a static (DC) pressure called radiation pressure. Radiation pressure is similar to parametric array effects in that amplitude variations in the signal give rise to audible frequencies via amplitude demodulation. However, unlike parametric array effects, radiation pressure fluctuations that generate audible signals from amplitude demodulation can occur due to any low-frequency modulation and not just from pressure fluctuations occurring at the modulation frequency ${\displaystyle \omega _{M}\,}$ or beating frequency ${\displaystyle f_{1}-f_{2}\,}$.
## References
[1] Minoru Todo, "New Type of Acoustic Filter Using Periodic Polymer Layers for Measuring Audio Signal Components Excited by Amplitude-Modulated High-Intensity Ultrasonic Waves," Journal of Audio Engineering Society, Vol. 53, pp. 930–41 (2005 October)
[2] Fundamentals of Acoustics; Kinsler et al., John Wiley & Sons, 2000
[3] ME 513 Course Notes, Dr. Luc Mongeau, Purdue University
# Noise in Hydraulic Systems
Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24
## Noise in Hydraulic Systems
Hydraulic systems are the most preferred source of power transmission in most of the industrial and mobile equipments due to their advantages in power density, compactness, flexibility, fast response and efficiency. The field hydraulics and pneumatics is also known as 'Fluid Power Technology'. Fluid power systems have a wide range of applications which include industrial, off-road vehicles, automotive system and aircrafts. In spite of these advantages,there are also some disadvantages. One of the main drawbacks with the hydraulic fluid power systems is the vibration and noise generated by them. The health and safety issues relating to noise, vibration and harshness (NVH) have been recognized for many years and legislation is now placing clear demands on manufacturers to reduce noise levels [1]. Hence, a lot of attention has been paid on reducing the noise of hydraulic fluid power systems both from the industrial and academic researchers. A good understanding of the noise generation, transmission and propagation is very important in order to improve the NVH performance of hydraulic fluid power systems.
## Sound in fluids
The speed of sound in fluids can be determined using the following relation.
${\displaystyle c={\sqrt {\frac {K}{\rho }}}}$ where K - fluid bulk modulus, ${\displaystyle \rho }$- fluid density, c - velocity of sound
Typical value of bulk modulus range from 2e9 to 2.5e9 N/m2. For a particular oil, with a density of 889 kg/m3,
speed of sound ${\displaystyle c={\sqrt {\frac {2e9}{889}}}=1499.9m/s}$
## Source of Noise
The main source of noise in hydraulic systems is the pump which supplies the flow. Most of the pumps used are positive displacement pumps. Of the positive displacement pumps, axial piston swash plate type is mostly preferred due to their controllability and efficiency.
The noise generation in an axial piston pump can be classifeid under two categories:
(i) fluidborne noise and
(ii) Structureborne noise
## Fluidborne Noise (FBN)
Among the positive displacement pumps, highest levels of FBN are generated by axial piston pumps and lowest levels by screw pumps and in between these lie the external gear pump and vane pump [1]. The discussion in this page is mainly focused on axial piston swash plate type pumps. An axial piston pump has a fixed number of displacement chambers arranged in a circular pattern separated from each other by an angular pitch equal to ${\displaystyle \phi ={\frac {360}{n}}}$ where n is the number of displacement chambers. As each chamber discharges a specific volume of fluid, the discharge at the pump outlet is sum of all the discharge from the individual chambers. The discontinuity in flow between adjacent chambers results in a kinematic flow ripple. The amplitude of the kinematic ripple can be theoretically determined given the size of the pump and the number of displament chambers. The kinematic ripple is the main cause of the fluidborne noise. The kinematic ripples is a theoretical value. The actual flow ripple at the pump outlet is much larger than the theoretical value because the kinematic ripple is combined with a compressibility component which is due to the fluid compressibility. These ripples (also referred as flow pulsations) generated at the pump are transmitted through the pipe or flexible hose connected to the pump and travel to all parts of the hydraulic circuit.
The pump is considered an ideal flow source. The pressure in the system will be decided by resistance to the flow or otherwise known as system load. The flow pulsations result in pressure pulsations. The pressure pulsations are superimposed on the mean system pressure. Both the flow and pressure pulsations easily travel to all part of the circuit and affect the performance of the components like control valve and actuators in the system and make the component vibrate, sometimes even resonate. This vibration of system components adds to the noise generated by the flow pulsations. The transmission of FBN in the circuit is discussed under transmission below.
A typical axial piston pump with 9 pistons running at 1000 rpm can produce a sound pressure level of more than 70 dBs.
## Structureborne Noise (SBN)
In swash plate type pumps, the main sources of the structureborne noise are the fluctuating forces and moments of the swas plate. These fluctuating forces arise as a result of the varying pressure inside the displacement chamber. As the displacing elements move from suction stroke to discharge stroke, the pressure varies accordingly from few bars to few hundred bars. This pressure changes are reflected on the displacement elements (in this case, pistons) as forces and these force are exerted on the swash plate causing the swash plate to vibrate. This vibration of the swash plate is the main cause of structureborne noise. There are other components in the system which also vibrate and lead to structureborne noise, but the swash is the major contributor.
Fig. 1 shows an exploded view of axial piston pump. Also the flow pulsations and the oscillating forces on the swash plate, which cause FBN and SBN respectively are shown for one revolution of the pump.
## Transmission
### FBN
The transmission of FBN is a complex phenomenon. Over the past few decades, considerable amount of research had gone into mathematical modeling of pressure and flow transient in the circuit. This involves the solution of wave equations, with piping treated as a distributed parameter system known as a transmission line [1] & [3].
Lets consider a simple pump-pipe-loading valve circuit as shown in Fig. 2. The pressure and flow ripple at ay location in the pipe can be described by the relations:
${\displaystyle {\frac {}{}}P=Ae^{-kx}+Be^{-kx}}$ .........(1)
${\displaystyle Q={\frac {1}{Z_{0}}}(Ae^{-kx}-Be^{-kx})}$.....(2)
where ${\displaystyle {\frac {}{}}A}$ and ${\displaystyle {\frac {}{}}B}$ are frequency dependent complex coefficients which are directly proportional to pump (source) flow ripple, but also functions of the source impedance ${\displaystyle {\frac {}{}}Z_{s}}$, characteristic impedance of the pipe ${\displaystyle {\frac {}{}}Z_{0}}$ and the termination impedance ${\displaystyle {\frac {}{}}Z_{t}}$. These impedances ,usually vary as the system operating pressure and flow rate changes, can be determined experimentally.
Fig.2 Schematic of a pump connected to a hydraulic line
For complex systems with several system compenents, the pressure and flow ripples are estimated using the transformation matrix approach. For this, the system compenents can be treated as lumped impedances (a throttle valve or accumulator), or distributed impedances (flexible hose or silencer). Various software packages are available today to predict the pressure pulsations.
### SBN
The transmission of SBN follows the classic source-path-noise model. The vibrations of the swash plate, the main cause of SBN, is transferred to the pump casing which encloses all the rotating group in the pump including displacement chambers (also known as cylinder block), pistons and the swash plate. The pump case, apart from vibrating itself, transfers the vibration down to the mount on which the pump is mounted. The mount then passes the vibrations down to the main mounted structure or the vehicle. Thus the SBN is transferred from the swash plate to the main strucuture or vehicle via pumpcasing and mount.
Some of the machine structures, along the path of transmission, are good at transmitting this vribational energy and they even resonate and reinforce it. By converting only a fraction of 1% of the pump structureborne noise into sound, a member in the transmission path could radiate more ABN than the pump itself [4].
## Airborne noise (ABN)
Both FBN and SBN , impart high fatigue loads on the system components and make them vibrate. All of these vibrations are radiated as airborne noise and can be heard by a human operator. Also, the flow and pressure pulsations make the system components such as a control valve to resonate. This vibration of the particular component again radiates airborne noise.
## Noise reduction
The reduction of the noise radiated from the hydraulic system can be approached in two ways.
(i) Reduction at Source - which is the reduction of noise at the pump. A large amount of open literature are availabale on the reduction techniques with some techniques focusing on reducing FBN at source and others focusing on SBN. Reduction in FBN and SBN at the source has a large influence on the ABN that is radiated. Even though, a lot of progress had been made in reducing the FBN and SBN separately, the problem of noise in hydraulic systems is not fully solved and lot need to be done. The reason is that the FBN and SBN are interlated, in a sense that, if one tried to reduce the FBN at the pump, it tends to affect the SBN characteristics. Currently, one of the main researches in noise reduction in pumps, is a systematic approach in understanding the coupling between FBN and SBN and targeting them simultaneously instead of treating them as two separte sources. Such an unified approach, demands not only well trained researchers but also sophisticated computer based mathematical model of the pump which can accurately output the necessary results for optimization of pump design. The amplitude of fluid pulsations can be reduced, at the source, with the use of an hydraulic attenuator(5).
(ii) Reduction at Component level - which focuses on the reduction of noise from individual component like hose, control valve, pump mounts and fixtures. This can be accomplished by a suitable design modification of the component so that it radiates least amount of noise. Optimization using computer based models can be one of the ways.
## Hydraulic System noise
Fig.3 Domain of hydraulic system noise generation and transmission (Figure recreated from [1])
## References
1. Designing Quieter Hydraulic Systems - Some Recent Developements and Contributions, Kevin Edge, 1999, Fluid Power: Forth JHPS International Symposium.
2. Fundamentals of Acoustics, L.E. Kinsler, A.R. Frey, A.B.Coppens, J.V. Sanders. Fourth Edition. John Wiley & Sons Inc.
3. Reduction of Axial Piston Pump Pressure Ripple, A.M. Harrison. PhD thesis, University of Bath. 1997
4. Noise Control of Hydraulic Machinery, Stan Skaistis, 1988. MARCEL DEKKER , INC.
5. Hydraulic Power System Analysis, A. Akers, M. Gassman, & R. Smith, Taylor & Francis, New York, 2006, ISBN 0-8247-9956-9
6. Experimental studies of the vibro-acoustic characteristics of an axial piston pump under run-up and steady-state operating conditions, Shaogan Ye et al., 2018, Measurement, 133.
7. Sound quality evaluation and prediction for the emitted noise of axial piston pumps, Junhui Zhang, Shiqi Xia, Shaogan Ye et al., 2018, Applied Acoustics 145:27-40.
# Basic Acoustics of the Marimba
Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24
## Introduction
Marimba Band "La Gloria Antigueña", Antigua Guatemala, 1979
Like a xylophone, a marimba has octaves of wooden bars that are struck with mallets to produce tones. Unlike the harsh sound of a xylophone, a marimba produces a deep, rich tone. Marimbas are not uncommon and are played in most high school bands.
Now, while all the trumpet and flute and clarinet players are busy tuning up their instruments, the marimba player is back in the percussion section with her feet up just relaxing. This is a bit surprising, however, since the marimba is a melodic instrument that needs to be in tune to sound good. So what gives? Why is the marimba never tuned? How would you even go about tuning a marimba? To answer these questions, the acoustics behind (or within) a marimba must be understood.
## Components of Sound
What gives the marimba its unique sound? It can be boiled down to two components: the bars and the resonators. Typically, the bars are made of rosewood (or some synthetic version of wood). They are cut to size depending on what note is desired, then the tuning is refined by shaving wood from the underside of the bar.
### Example: Rosewood bar, middle C, 1 cm thick
The equation that relates the length of the bar with the desired frequency comes from the theory of modeling a bar that is free at both ends. This theory yields the following equation: ${\displaystyle Length={\sqrt {\frac {3.011^{2}\cdot \pi \cdot t\cdot v}{8\cdot {\sqrt {12}}\cdot f}}}}$ where t is the thickness of the bar, v is the speed of sound in the bar, and f is the frequency of the note. For rosewood, v = 5217 m/s. For middle C, f=262 Hz. Therefore, to make a middle C key for a rosewood marimba, cut the bar to be: ${\displaystyle Length={\sqrt {\frac {3.011^{2}\cdot \pi \cdot .01\cdot 5217}{8\cdot {\sqrt {12}}\cdot 262}}}=.45m=45cm}$
The resonators are made from metal (usually aluminum) and their lengths also differ depending on the desired note. It is important to know that each resonator is open at the top but closed by a stopper at the bottom end.
### Example: Aluminum resonator, middle C
The equation that relates the length of the resonator with the desired frequency comes from modeling the resonator as a pipe that is driven at one end and closed at the other end. A "driven" pipe is one that has a source of excitation (in this case, the vibrating key) at one end. This model yields the following: ${\displaystyle Length={\frac {c}{4\cdot f}}}$ where c is the speed of sound in air and f is the frequency of the note. For air, c = 343 m/s. For middle C, f = 262 Hz. Therefore, to make a resonator for the middle C key, the resonator length should be: ${\displaystyle Length={\frac {343}{4\cdot 262}}=.327m=32.7cm}$
### Resonator Shape
The shape of the resonator is an important factor in determining the quality of sound that can be produced. The ideal shape is a sphere. This is modeled by the Helmholtz resonator. (For more see Helmholtz Resonator page) However, mounting big, round, beach ball-like resonators under the keys is typically impractical. The worst choices for resonators are square or oval tubes. These shapes amplify the non-harmonic pitches sometimes referred to as “junk pitches”. The round tube is typically chosen because it does the best job (aside from the sphere) at amplifying the desired harmonic and not much else.
As mentioned in the second example above, the resonator on a marimba can be modeled by a closed pipe. This model can be used to predict what type of sound (full and rich vs dull) the marimba will produce. Each pipe is a "quarter wave resonator" that amplifies the sound waves produced by of the bar. This means that in order to produce a full, rich sound, the length of the resonator must exactly match one-quarter of the wavelength. If the length is off, the marimba will produce a dull or off-key sound for that note.
## Why would the marimba need tuning?
In the theoretical world where it is always 72 degrees with low humidity, a marimba would not need tuning. But, since weather can be a factor (especially for the marching band) marimbas do not always perform the same way. Hot and cold weather can wreak havoc on all kinds of percussion instruments, and the marimba is no exception. On hot days, the marimba tends to be sharp and for cold days it tends to be flat. This is the exact opposite of what happens to string instruments. Why? The tone of a string instrument depends mainly on the tension in the string, which decreases as the string expands with heat. The decrease in tension leads to a flat note. Marimbas on the other hand produce sound by moving air through the resonators. The speed at which this air is moved is the speed of sound, which varies proportionately with temperature! So, as the temperature increases, so does the speed of sound. From the equation given in example 2 from above, you can see that an increase in the speed of sound (c) means a longer pipe is needed to resonate the same note. If the length of the resonator is not increased, the note will sound sharp. Now, the heat can also cause the wooden bars to expand, but the effect of this expansion is insignificant compared to the effect of the change in the speed of sound.
## Tuning Myths
It is a common myth among percussionists that the marimba can be tuned by simply moving the resonators up or down (while the bars remain in the same position.) The thought behind this is that by moving the resonators down, for example, you are in effect lengthening them. While this may sound like sound reasoning, it actually does not hold true in practice. Judging by how the marimba is constructed (cutting bars and resonators to specific lengths), it seems that there are really two options to consider when looking to tune a marimba: shave some wood off the underside of the bars, or change the length of the resonator. For obvious reasons, shaving wood off the keys every time the weather changes is not a practical solution. Therefore, the only option left is to change the length of the resonator. As mentioned above, each resonator is plugged by a stopper at the bottom end. So, by simply shoving the stopper farther up the pipe, you can shorten the resonator and sharpen the note. Conversely, pushing the stopper down the pipe can flatten the note. Most marimbas do not come with tunable resonators, so this process can be a little challenging. (Broomsticks and hammers are common tools of the trade.)
### Example: Middle C Resonator lengthened by 1 cm
For ideal conditions, the length of the middle C (262 Hz) resonator should be 32.7 cm as shown in example 2. Therefore, the change in frequency for this resonator due to a change in length is given by: ${\displaystyle \Delta Frequency=262Hz-{\frac {c}{4\cdot (.327+\Delta L)}}}$ If the length is increased by 1 cm, the change in frequency will be: ${\displaystyle \Delta Frequency={\frac {343}{4\cdot (.327+.01)}}-262Hz=7.5Hz}$
The acoustics behind the tuning a marimba go back to the design that each resonator is to be ¼ of the total wavelength of the desired note. When marimbas get out of tune, this length is no longer exactly equal to ¼ the wavelength due to the lengthening or shortening of the resonator as described above. Because the length has changed, resonance is no longer achieved, and the tone can become muffled or off-key.
## Conclusions
Some marimba builders are now changing their designs to include tunable resonators. Since any leak in the end-seal will cause major loss of volume and richness of the tone, this is proving to be a very difficult task. At least now, though, armed with the acoustic background of their instruments, percussionists everywhere will now have something to do when the conductor says, “tune up!”
# How an Acoustic Guitar works
Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24
## Introduction
There are three main parts of the guitar that contribute to sound production.
First of all, there are the strings. Any string that is under tension will vibrate at a certain frequency. The tension and gauge in the string determine the frequency at which it vibrates. The guitar controls the length and tension of six differently weighted strings to cover a very wide range of frequencies.
Second of all, there is the body of the guitar. The guitar body is connected directly to one end of each of the strings. The body receives the vibrations of the strings and transmits them to the air around the body. It is the body’s large surface area that allows it to “push” a lot more air than a string.
Finally, there is the air inside the body. This is very important for the lower frequencies of the guitar. The air mass just inside the sound hole oscillates, compressing and decompressing the compliant air inside the body. In practice this concept is called a Helmholtz resonator. Without this, it would difficult to produce the wonderful timbre of the guitar.
## The Strings
The strings of the guitar vary in linear density, length, and tension. This gives the guitar a wide range of attainable frequencies. The larger the linear density is, the slower the string vibrates. The same goes for the length; the longer the string is the slower it vibrates. This causes a low frequency. Inversely, if the strings are less dense and/or shorter they create a higher frequency. The lowest resonance frequencies of each string can be calculated by
${\displaystyle f_{1}={\frac {1}{2L}}{\sqrt {\frac {T}{\rho _{1}}}}}$ where ${\displaystyle T}$= string tension, ${\displaystyle \rho _{1}}$=linear density, ${\displaystyle L}$ = string length
The string length, L, in the equation is what changes when a player presses on a string at a certain fret. This will shorten the string which in turn increases the frequency it produces when plucked. The spacing of these frets is important. The length from the nut to bridge determines how much space goes between each fret. If the length is 25 inches, then the position of the first fret should be located (25/17.817) inches from the nut. Then the second fret should be located (25-(25/17.817))/17.817 inches from the first fret. This results in the equation
When a string is plucked, a disturbance is formed and travels in both directions away from point where the string was plucked. These "waves" travel at a speed that is related to the tension and linear density and can be calculated by
The waves travel until they reach the boundaries on each end where they are reflected back. The link below displays how the waves propagate in a string.
The strings themselves do not produce very much sound because they are so thin. This is why they are connected to the top plate of the guitar body. They need to transfer the frequencies they are producing to a large surface area which can create more intense pressure disturbances.
## The Body
The body of the guitar transfers the vibrations of the bridge to the air that surrounds it. The top plate contributes to most of the pressure disturbances, because the player dampens the back plate and the sides are relatively stiff. This is why it is important to make the top plate out of a light springy wood, like spruce. The more the top plate can vibrate, the louder the sound it produces will be. It is also important to keep the top plate flat, so a series of braces are located on the inside to strengthen it. Without these braces the top plate would bend and crack under the large stress created by the tension in the strings. This would also affect the magnitude of the sound being transmitted. The warped plate would not be able to "push" air very efficiently. A good experiment to try, in order to see how important this part of the guitar is in the amplification process, is as follows:
2. Stretch the rubber band and pluck it a few times to get a good sense for how loud it is.
3. Stretch the plastic wrap over the bowl to form a sort of drum.
4. Tape down one end of the rubber band to the plastic wrap.
5. Stretch the rubber band and pluck it a few times.
6. The sound should be much louder than before.
## The Air
The final part of the guitar is the air inside the body. This is very important for the lower range of the instrument. The air just inside the sound hole oscillates, compressing and expanding the air inside the body. This is just like blowing across the top of a bottle and listening to the tone it produces. This forms what is called a Helmholtz resonator. For more information on Helmholtz resonators go to Helmholtz Resonance. This link also shows the correlation to acoustic guitars in great detail. The acoustic guitar makers often tune these resonators to have a resonance frequency between F#2 and A2 (92.5 to 110.0 Hz)(Hz stands for Hertz). Having such a low resonance frequency is what aids the amplification of the lower frequency strings. To demonstrate the importance of the air in the cavity, simply play an open A on the guitar (the second string). Now, as the string is vibrating, place a piece of cardboard over the sound hole. The sound level is reduced dramatically. This is because you've stopped the vibration of the air mass just inside the sound hole, causing only the top plate to vibrate. Although the top plate still vibrates and transmits sound, it isn't as effective at transmitting lower frequency waves, thus the need for the Helmholtz resonator.
# Specific application-automobile muffler
Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24
## Introduction
A muffler is a part of the exhaust system on an automobile that plays a vital role. It needs to have modes that are located away from the frequencies that the engine operates at, whether the engine be idling or running at the maximum amount of revolutions per second.A muffler that affects an automobile in a negative way is one that causes noise or discomfort while the car engine is running.Inside a muffler, you'll find a deceptively simple set of tubes with some holes in them. These tubes and chambers are actually as finely tuned as a musical instrument. They are designed to reflect the sound waves produced by the engine in such a way that they partially cancel themselves out.( cited from www.howstuffworks.com )
It is very important to have it on the automobile. The legal limit for exhaust noise in the state of California is 95 dB (A) - CA. V.C. 27151 .Without a muffler the typical car exhaust noise would exceed 110 dB.A conventional car muffler is capable of limiting noise to about 90 dB. The active-noise canceling muffler enables cancellation of exhaust noise to a wide range of frequencies.
## How Does automobile muffler function?
### General Concept
The simple and main part of designing the automobile muffler is to use the low-pass filter. It typically makes use of the change of the cross section area which can be made as a chamber to filter or reduce the sound wave which the engine produced.
### Low-Pass Filter
A low-pass filter is a circuit that passes low frequency signals but stops the high frequency signals. Once the low pass filter is set by the user at a specific cutoff frequency, all frequencies lower than that will be passed through the filter, while higher frequencies will be attenuated in amplitude. This circuit made up of passive components (resistor, capacitors and inductors) capable of accomplishing this objective. File:Inductive law pass filter.jpg
the formula to be used:
### Human ear sound reaction feature
When these pressure pulses reach your ear, the eardrum vibrates back and forth. Your brain interprets this motion as sound. Two main characteristics of the wave determine how we perceive the sound:
1.sound wave frequency. 2.air wave pressure amplitude.
It turns out that it is possible to add two or more sound waves together and get less sound.
### Benefits of an Active Noise-Canceling Muffler
1.By using an active muffler the exhaust noise can be easily tuned, amplified, or nearly eliminated.
2.The backpressure of a conventional muffler can be essentially eliminated, thus increasing engine performance and efficiency.
3.By increasing engine efficiency and performance, less fuel will be used and the emissions will be reduced.
## Absorptive muffler
### Lined ducts
It can be regarded as simplest form of absorptive muffler. Attach absorptive material to the bare walls of the duct.( in car that is the exhaustion tube) The attenuation performance improves with the thickness of absorptive material.
The attenuation curves like a skewed bell. Increase the thickness of the wall will get the lower maximum attenuation frequency. For higher frequency though, thinner absorbent layers are effective, but the large gap allows noise to pass directly along. Thin layers and narrow passages are therefore more effective at high frequencies. For good absorption over the widest frequency range, thick absorbent layers and narrow passages are best.
### Parallel and block-line-of-sight baffles
Divide the duct into several channels or turn the flow channels so that there is no direct line-of-sight through the baffles. Frequently the materials line on the channels. Attenuation improves with the thickness of absorptive material and length of the baffle. Lined bends can be used to provide a greater attenuation and attenuate best at high frequency. Comparatively, at low frequency attenuation can be increased by adding thicker lining.
### Plenum chambers
They are relatively large volume chambers, usually fabricated from sheet metal, which interconnect two ducts. The interior of the chamber is lined with absorbing material to attenuate noise in the duct. Protective facing material may also be necessary if the temperature and velocity conditions of the gas stream are too severe.
The performance of a plenum chamber can be improved by: 1.increase the thickness of the absorbing lining 2.blocking the direct line of sight from the chamber inlet to the outlet. 3.increase the cross-sectional area of the chamber.
# Bessel Functions and the Kettledrum
Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24
# Abstract
In class, we have begun to discuss the solutions of multidimensional wave equations. A particularly interesting aspect of these multidimensional solutions are those of Bessel functions for circular boundary conditions. The practical application of these solutions is the kettledrum. This page will explore in qualitative and quantitative terms how the of the kettledrum works. More specifically, the kettledrum will be introduced as a circular membrane and its solution will be discussed in visual (e.g. visualization of Bessel functions, video of kettledrums and audio forms (wav files of kettledrums playing. In addition, links to more information about this material, including references, will be included.
# What is a kettledrum
A kettledrum is a percussion instrument with a circular drumhead mounted on a "kettle-like" enclosure. When one strikes the drumhead with a mallet, it vibrates which produces its sound. The pitch of this sound is determined by the tension of the drumhead, which is precisely tuned before playing. The sound of the kettldrum (called the Timpani in classical music) is present in many forms of music from many difference places of the world.
# The math behind the kettledrum:the brief version
When one looks at how a kettledrum produces sound, one should look no farther than the drumhead. The vibration of this circular membrane (and the air in the drum enclosure) is what produces the sound in this instrument. The mathematics behind this vibrating drum are relatively simple. If one looks at a small element of the drum head, it looks exactly like the situation for the vibrating string (see:). The only difference is that there are two dimensions where there are forces on the element, the two dimensions that are planar to the drum. As this is the same situation, we have the same equation, except with another spatial term in the other planar dimension. This allows us to model the drumhead using a helmholtz equation. The next step (solved in detail below) is to assume that the displacement of the drumhead (in polar coordinates) is a product of two separate functions for theta and r. This allows us to turn the PDE into two ODES which are readily solved and applied to the situation of the kettledrum head. For more info, see below.
# The math behind the kettledrum:the derivation
So starting with the trusty general Helmholtz equation:
${\displaystyle \nabla ^{2}\Psi +k^{2}\Psi =0}$
Where k is the wave number, the frequency of the forced oscillations divided by the speed of sound in the membrane.
Since we are dealing with a circular object, it make sense to work in polar coordinates (in terms of radius and angle) instead of rectangular coordinates. For polar coordinates the Laplacian term of the Helmholtz relation (${\displaystyle \nabla ^{2}}$) becomes ${\displaystyle \partial ^{2}\Psi /\partial r^{2}+1/r\partial ^{2}\Psi /\partial r+1/r^{2}\partial ^{2}\Psi /\partial \theta ^{2}}$
Now lets assume that:${\displaystyle \Psi (r,\theta )=R(r)\Theta (\theta )}$
This assumption follows the method of separation of variables. (see Reference 3 for more info) Substituting this result back into our trusty Helmholtz equation gives the following:
${\displaystyle r^{2}/R(d^{2}R/dr^{2}+1/rdR/dr)+k^{2}r^{2}=-1/\Theta d^{2}\Theta /d\theta ^{2}}$
Since we separated the variables of the solution into two one-dimensional functions, the partial derivatives become ordinary derivatives. Both sides of this result must equal the same constant. For simplicity, I will use ${\displaystyle \lambda }$ as this constant. This results in the following two equations:
${\displaystyle d^{2}\Theta /d\theta ^{2}=-\lambda ^{2}\Theta }$
${\displaystyle d^{2}R/dr^{2}+1/rdR/dr+(k^{2}-\lambda ^{2}/r^{2})R=0}$
The first of these equations readily seen as the standard second order ordinary differential equation which has a harmonic solution of sines and cosines with the frequency based on ${\displaystyle \lambda }$ | 2019-12-13 06:07:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 509, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6341553926467896, "perplexity": 675.1079581411404}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540548544.83/warc/CC-MAIN-20191213043650-20191213071650-00208.warc.gz"} |
https://www.semanticscholar.org/paper/Chapter-%CE%B1-Engineering-Complex-Systems-Norman-Kuras/40faf511fec354018e82b9e7d168e5ce34692c0e?p2df | • Corpus ID: 16846877
Chapter α Engineering Complex Systems
@inproceedings{Norman2004ChapterE,
title={Chapter $\alpha$ Engineering Complex Systems},
author={Douglas O. Norman and Michael L. Kuras},
year={2004}
}
• Published 2004
Using the current instantiation of the Air and Space Operations Center (AOC), and the desired evolution of it, the AOC is shown to be best thought of as a complex system. Complex Systems are alive and constantly changing. They respond and interact with their environments – each causing impact on (and inspiring change in) the other. We make the case that a traditional systems engineering (TSE) approach does not scale to the AOC; consequently, we don’t believe TSE scales to the “enterprise.”
1 Citations
References
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When systems engineering fails-toward complex systems engineering
• Y. Bar-Yam
SMC'03 Conference Proceedings. 2003 IEEE International Conference on Systems, Man and Cybernetics. Conference Theme - System Security and Assurance (Cat. No.03CH37483)
• 2003
It is suggested that there are two effective strategies for overcoming problems with systems engineering: restricting the conventional systems engineering process to not-too-complex projects and adopting an evolutionary paradigm for complex systems engineering that involves rapid parallel exploration and a context designed to promote change.
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https://www.physicsforums.com/insights/frequently-made-errors-mechanics-forces/ | # Frequently Made Errors in Mechanics: Forces
Notation: On this page, a circumflex signifies an average.
## 1. Forces as vectors
A force is a vector, i.e. has magnitude and direction. But it also has a point of application, which vectors, in general, do not have.
A force can be thought of as a vector pair, one vector specifying its magnitude and direction, the other specifying a point of application.
For a force applied to a rigid body, the precise point of application does not matter; instead, the line of action suffices. This is the straight-line through the point of application, in the direction of the force. The force applied at any point in that line would have the same effect.
## 2. Tension and Compression
X “If a light string is pulled with a force of F at each end, the tension force in the string is 2F”
Tension is not exactly force; it is more like a pair of opposing forces. Similarly compression.
If you pick a point in a string and consider the portions on either side of that point as separate bodies, the tension at that point is the magnitude of each of the pair of forces, action, and reaction, with which each portion pulls the other.
“If a light string is pulled with a force of F at each end, the tension in the string is F”
Note that this includes e.g. a string pulled with force F at one end and tied to a wall at the other. The wall pulls on the string with force F.
## 3. Average force
“Work = force × distance,
X Therefore average force can be calculated as work done divided by distance”
What is meant by “average force”? Average acceleration is the change in velocity divided by the elapsed time:
##\hat {\vec a} = \Delta \vec v/\Delta t##
A sensible definition of average force is such that ##\hat {\vec F} = m\hat {\vec a}##:
##\hat {\vec F} = m\hat {\vec a} = m\Delta \vec v/\Delta t = \Delta \vec p/\Delta t##
where ##\vec p## is momentum.
If the force happens to be constant then this will produce the same numerical answer as work/distance, but only in that case.
It is possible to define averages in different ways, provided it is made clear. One could specify “force averaged over distance”. But there is another reason for rejecting the work/distance calculation. The completely correct form of Work = force × distance involves vectors:
##\Delta E = \int \vec F.\vec{ds}##, ##\vec{\Delta s} = \int \vec{ds}##
Turning that into a division results in dividing by a vector, which is not a defined operation:
X ##\hat {\vec F} = \Delta E/\vec{\Delta s}##
SHM example
“A mass m collides elastically at speed v with a rubber pad of spring constant k. What is the average force bringing the mass to stationary?”
The distance x the pad is compressed is given by
##\frac 12 m v^2 = \frac 12 k x^2##
##x = v\sqrt{\frac mk}##
If we calculate the average force from this we get
##\hat {\vec F_{dist}}=\frac 12 v\sqrt{mk}##
The time to maximum compression is ##T=\frac{\pi}2\sqrt{\frac mk}## . Dividing this into the momentum change mv gives
##\hat{\vec F} = \frac 2{\pi}v\sqrt{mk}##
## 4. Available force equations
In two dimensions there are three force equations available. These are usually taken as linear force sums,##\Sigma F_x=m a_x##, in (any) two orthogonal directions, plus a moment sum about some reference point; but there are other valid choices:
The two linear equations do not need to be orthogonal; they can be based on any two directions that are not parallel.
One linear equation and moments about two points also works, provided the line through the two points is not orthogonal to the linear choice.
Equations for any other linear direction or moments about any other point can be deduced from the three chosen, so provide no further information
A linear force can be thought of as a moment about a point at infinity (the point being infinitely far away in a direction orthogonal to the linear direction taken).
## 5. Sign Conventions and Gravity
A common practice is to take ‘up’ as positive. Taking down as positive is fine, but it is advisable to be consistent, using the same convention for displacements, velocities and accelerations. Defining some variables one way and some the other is valid but tends to confuse.
X “If up is positive then the force due to gravity is -mg”
g” is just a variable name, commonly reserved in mechanics for gravitational acceleration at Earth’s surface. Writing the force as -mg is valid, but in effect adopts an inverted convention for the variable g, i.e. makes g positive downwards. It is more consistent to take the gravitational force as +mg and, when plugging in numbers, provide a negative value for g.
“If up is positive then the force due to gravity is mg, unless g is defined positive down”
That said, it is extremely common to take g positive down but all else positive up.
## 6. Force and Work
X “The work done by a force on a body is the magnitude of the force multiplied by the distance the body moves.”
That is only true if the force is constant and the distance the body moves is in the same direction as the force.
In general,
Work = ##\int \vec F. \vec{ds}##
If the force is constant, this simplifies to ##\vec F.\vec{\Delta s}##.
If the distance the body moves is entirely in the direction of the constant force, this further simplifies to ##F \Delta s##, where ##F## and ##\Delta s## are signed scalars.
## 6.1 Work done by the normal force
X “The normal force does no work”
That is true if the surface is not moving along its perpendicular, but consider a standing person mass m riding up an escalator through a height h. The vertical displacement is h and the normal force is -mg, so the work done is -mgh (g being negative).
Example: Riding an escalator
It gets a little more complicated if the person is also walking up the escalator.
X “If a person is walking up the escalator at a constant speed then the normal force exceeds mg”
If the normal force were to exceed mg then the person must accelerate.
Suppose the step height is s, and in the time the person moves up one step the escalator rises height r. Consider the person as consisting of a point mass m atop an extensible massless leg.
Since the mass is not accelerating, the compression in the leg is |mg|. The escalator rises height r, doing work –mgr. In the same time, the leg extends by length s, doing work –mgs.
## 6.2 Infinitesimal work
“What is the minimum work needed to move a box weight W on the floor at point A to a point B on the floor distance s away? The floor is level and has a coefficient of kinetic friction ##\mu_k##.”
The minimum force needed to drag the box is when it is applied at angle ##\tan^{-1}\mu_k##,
X so the minimum work done is ##\frac{Ws\mu_k}{1+{\mu_k}^2}##
The minimum work does not necessarily correspond to the minimum force. The work can be lower if a greater force is applied but the force advances a much shorter distance. In the extreme case, a force advancing an infinitesimal distance does infinitesimal work.
In the problem above, a force exceeding W by a tiny amount, acting over a tiny distance, is sufficient to lift the box clear of the floor and hold it clear. A second force acting horizontally over a tiny distance accelerates the box to a tiny speed. No further work, just adequate time, is required to reach point B. In principle, even these small quantities of work might then be recoverable in bringing the box to rest and lowering it to the floor.
The work required to move the box is arbitrarily small
Tags:
35 replies
1. haruspex says:
[QUOTE=”haruspex, post: 5101720, member: 334404″]Point 3 I have covered in another post under development.
Points 1 and 2 belong in a much more general FME. I’ll add them to my list![/QUOTE]
Correction:
Point 2 I have already covered in an imminent post on Friction.
2. robphy says:
In my experience,
the errors I listed are much more frequent than any of the ones you listed.
3. haruspex says:
[QUOTE=”robphy, post: 5101707, member: 9587″]Here are other common errors…
1) “the magnitude of the normal force is always mg” (because of a formula they saw).
2) “the magnitude of the static friction force is always $mu_k N$” (because of a formula they saw)
3) “the centripetal force is an additional force drawn on a free-body diagram”[/QUOTE]
Point 3 I have covered in another post under development.
Points 1 and 2 belong in a much more general FME. I’ll add them to my list!
4. haruspex says:
[QUOTE=”robphy, post: 5100610, member: 9587″]Probably the [B]biggest error[/B] that students make with vectors is
the incorrect thinking that “vectors add like numbers”…
in particular, that magnitude of a sum of vectors is the sum of their magnitudes.[/QUOTE]
Thinking about this some more, it belongs in a separate FME (“Frequently Made Errors”) on vectors. Care to write one? If not, I’ll get to it eventually. Should be able to link to it from the forces paragraph.
5. haruspex says:
[QUOTE=”Ben Niehoff, post: 5101020, member: 99109″]I have to object to the first item. A “line of action” is insufficient to describe a force except in the special case of a rigid body. In full generality, a force is a vector that is attached to a point, which I guess you can call the “point of application”. The point should be thought of as described by some suitable coordinates; not as a “second vector”, which is not really the appropriate way to think about it. The “line of action” is merely the line passing through the point of application such that it is collinear with the force vector.
From here it is easy to generalize to fields of force density, as in continuum mechanics.[/QUOTE]
I can certainly add the rigid body rider (as done already in this thread).
A vector to describe the point of application is perfectly reasonable. It would be an appropriate vector for determining the moment about the origin.
6. rumborak says:
Lol, I had the same argument the last two pages :D
7. haruspex says:
[QUOTE=”robphy, post: 5100610, member: 9587″]Probably the [B]biggest error[/B] that students make with vectors is
the incorrect thinking that “vectors add like numbers”…
in particular, that magnitude of a sum of vectors is the sum of the their magnitudes.[/QUOTE]
True. If I can find out how to edit the post now that it has been published (anyone know?) I will add something on that.
[QUOTE=”rumborak, post: 5100604, member: 536216”]What? I’ve been saying that the whole time. It is my very point that a force can be described with 2 3-dimensional vectors. You seem to want to add another entity to it, or at the very least consider it “a frequently made error” to not consider that line of action.[/QUOTE]
Then I misunderstood your objection. It seems all we are arguing about is whether a force has magnitude, direction and point of action, or magnitude, direction and line of action. Line of action is enough to define the force, but you find it less confusing to refer to point of action. I’ll try to reword it a little.
8. haruspex says:
[QUOTE=”robphy, post: 5100610, member: 9587″]Probably the [B]biggest error[/B] that students make with vectors is
the incorrect thinking that “vectors add like numbers”…
in particular, that magnitude of a sum of vectors is the sum of the their magnitudes.[/QUOTE]
True. If I can find out how to edit the post now that it has been published (anyone know?) I will add something on that.
9. rumborak says:
What? I’ve been saying that the whole time. It is my very point that a force can be described with 2 3-dimensional vectors. You seem to want to add another entity to it, or at the very least consider it “a frequently made error” to not consider that line of action.
10. haruspex says:
[QUOTE=”rumborak, post: 5100597, member: 536216″]the vector+point description uniquely and sufficient describes it[/QUOTE]
So you’ve shifted your position to “a force is a vector plus a point of application”?
11. rumborak says:
But either way, I find your “Insight” article to be wrong in the sense that it makes the claim that people are mistaken in not considering the line of action. It is redundant, because the vector+point description uniquely and sufficient describes it. A person using the latter definition is in now way wrong, and in fact is better equipped to use the regular formulas in physics.
Not only that, the term “line of action” produces almost no results on the internet, which means its use will likely be of no help to anybody studying physics.
12. haruspex says:
[QUOTE=”rumborak, post: 5100591, member: 536216″]The problem is, while it may make some certain sense in mechanics of rigid bodies, once the student moves on to things like Lorentz force, “line of action” makes no sense at all anymore.[/QUOTE]
Sure it does. The basic definition is for a point particle, so the line of action is clear. In principle, though, it also applies to a larger rigid body. The Lorentz force on the body as a whole will be the net of the forces on the particles, and it will have a line of action.
13. rumborak says:
The problem is, while it may make some certain sense in mechanics of rigid bodies, once the student moves on to things like Lorentz force, “line of action” makes no sense at all anymore. A student asking “so, what’s the line of action on an electron” will be met with the answer “yeah, forget about that thing here. Force is something different”.
BTW, a line may look like it needs only two values, but it also requires a defined x+y offset to a coordinate system. You actually end up with more that way.
14. haruspex says:
[QUOTE=”rumborak, post: 5100544, member: 536216″]Maybe this is just my personal Occam’s Razor here, but between a point in space, and a line in space, if the point describes it fully, the point should be preferred.
[/QUOTE]
The point requires three coordinates. Since we know the direction, the line of action only requires two.
Is there any problem in the mechanics of rigid bodies where the line of action is inadequate and the point of application needs to be known?
15. rumborak says:
Maybe this is just my personal Occam’s Razor here, but between a point in space, and a line in space, if the point describes it fully, the point should be preferred. For example, what meaning do the infinite other points on that “line of action” have? The line of action extends indefinitely away from the body it applies to; what physical reality does this correspond to?
TLDR, I personally find these additional redundant concepts will wreak more havoc on a student’s understanding than the simple, and minimal, “vector + point in space” definition of a force. Especially when the math in physics nowhere mentions any lines, at all.
16. haruspex says:
[QUOTE=”rumborak, post: 5100535, member: 536216″]I think this is more than semantics. The boeingconsult site says “a force has magnitude, line of action, direction, and point of application”.
So, that means, it’s
– magnitude and direction => 3-dimensional vector
– line of action => 3-dimensional vector (slope and intercept essentially)
– point of application => 3-dimensional point
A force vector needs 9 scalars to be described? I very much doubt so.[/QUOTE]
The boeingconsult statement is clearly excessive. The line of action can be deduced from point of application and direction. The only question is whether to discard line of action or point of application. All the other references I’ve found agree with me that it’s the line of action that matters.
17. rumborak says:
I think this is more than semantics. The boeingconsult site says “a force has magnitude, line of action, direction, and point of application”.
So, that means, it’s
– magnitude and direction => 3-dimensional vector
– line of action => 3-dimensional vector (slope and intercept essentially)
– point of application => 3-dimensional point
A force vector needs 9 scalars to be described? I very much doubt so. In my book, a force is fully described by its 3-dimensional vector of direction, and the 3-dimensional vector of where it is applied.
When I look at things like Lorentz force, I also don’t see the “line of action” anywhere.
18. haruspex says:
[QUOTE=”rumborak, post: 5100506, member: 536216″]Yes, but that is due to the moment. The force is still just the force, simply acting on a different spot. Don’t conflate what it *does* to a body (whose center of mass might be anywhere) with the clean concept of a force.[/QUOTE]
You yourself wrote that a force “has a point of application”. That is, the point of application is an attribute of the force. When you say “clean concept of a force”, you are really referring to the clean concept of a vector. What you call a force, I would call the force vector.
Seems to me we are arguing about semantics. Is a force a disembodied vector, and the application of a force (for want of a better term) the combination of a force and a line of action? Or is a force what we think of in the real world as a force, having attributes of magnitude, direction and line of action?:
[I][URL]http://en.wikipedia.org/wiki/Statics[/URL] “[B]Force[/B] is the action of one body on another.”[/I]
[URL]https://books.google.com.au/books?isbn=812190952X[/URL] “[I]A force[/I] is completely specified by [I]its Vector[/I] and [I]its point of application[/I].”
[URL=”http://www.boeingconsult.com/tafe/structures/struct1/”]www.boeingconsult.com/tafe/structures/struct1/[/URL][B]forces[/B]/[B]forces[/B].htm “a force has magnitude, line of action, direction, and point of application”
If we take the pure vector view, it makes no sense to ask what moment a force has about a point. We must instead ask what moment a particular application of the force has about that point.
19. rumborak says:
Yes, but that is due to the moment. The force is still just the force, simply acting on a different spot. Don’t conflate what it *does* to a body (whose center of mass might be anywhere) with the clean concept of a force.
20. haruspex says:
[QUOTE=”rumborak, post: 5100496, member: 536216″]Sorry, yeah, I meant moment.
But, please explain this to me. You say “it matters greatly whether its line of action is through the mass centre”. So, you have a force vector, applying on the center of the mass. And then you have *another* vector? What does this additional vector refer to?[/QUOTE]
No, I have one force that maybe is not acting through the mass centre. If it is not, it will cause the body to rotate, thereby doing more work on it.
21. rumborak says:
Sorry, yeah, I meant moment.
But, please explain this to me. You say “it matters greatly whether its line of action is through the mass centre”. So, you have a force vector, applying on the center of the mass. And then you have *another* vector? What does this additional vector refer to?
22. haruspex says:
[QUOTE=”rumborak, post: 5100465, member: 536216″]So, I went to college in Europe, and I can say I have never heard of the concept of “line of action”, nor, after various Google searches, do I understand the point of it. Force has a vector to it that indicates the magnitude and direction, and obviously there is a point that it applies to.
I looked at [URL]http://web.mit.edu/4.441/1_lectures/1_lecture4/1_lecture4.html[/URL] , and it seems to make a mangle of momentum and other things. Momentum is a separate thing, don’t conflate it with force.[/QUOTE]
[URL]http://en.wikipedia.org/wiki/Line_of_action[/URL] states:
[INDENT][I]”The concept is essential, for instance, for understanding the net effect of multiple forces applied to a body.”[/I][/INDENT]
But I would go further. With just one force applied to a body it matters greatly whether its line of action is through the mass centre.
The point of application will obviously tell you the line of action, but of the two it’s the line of action that matters.
I’m not conflating momentum with force. Perhaps you mean moments?
23. rumborak says:
[QUOTE=”haruspex, post: 5100324, member: 334404″]
I’m unsure how important it is, but I notice that most seem to think of a force as being completely described by its vector. The vector does not tell you the line of action, which as we all know is crucial when it comes to moments. A force really is more like a pair of vectors, but it’s never described that way.
[/QUOTE]
So, I went to college in Europe, and I can say I have never heard of the concept of “line of action”, nor, after various Google searches, do I understand the point of it. Force has a vector to it that indicates the magnitude and direction, and obviously there is a point that it applies to.
I looked at [URL]http://web.mit.edu/4.441/1_lectures/1_lecture4/1_lecture4.html[/URL] , and it seems to make a mangle of momentum and other things. Momentum is a separate thing, don’t conflate it with force.
24. haruspex says:
[QUOTE=”CWatters, post: 5100209, member: 423469″]There are plenty of other common errors. For example the friction force is frequently taken to be μN in situations where that’s actually the maximum force before slipping occurs.[/QUOTE]
I have separate one on friction in the works. Just wanted to get some feedback on this simple one first. Others to follow…. moments etc.
[QUOTE=”rumborak, post: 5100098, member: 536216″]Yeah, number 1, in my opinion, makes it more confusing to say the least. I for one think I have a reasonable understanding of what a force is, but I have no idea what a “line of action” is supposed to be in this context.[/QUOTE]
I’m unsure how important it is, but I notice that most seem to think of a force as being completely described by its vector. The vector does not tell you the line of action, which as we all know is crucial when it comes to moments. A force really is more like a pair of vectors, but it’s never described that way.
I’ll expand it, mentioning point of application.
[QUOTE=”PeroK, post: 5100014, member: 493650″]
In point 2, you could add that a string pulled at both ends (with a force F) is equivalent to a string pulled at one end with a force F and attached to a wall at the other. That seems often to be missed by those who think the tension should be 2F.[/QUOTE]
Good idea.
25. CWatters says:
There are plenty of other common errors. For example the friction force is frequently taken to be μN in situations where that’s actually the maximum force before slipping occurs.
26. rumborak says:
Yeah, number 1, in my opinion, makes it more confusing to say the least. I for one think I have a reasonable understanding of what a force is, but I have no idea what a “line of action” is supposed to be in this context.
27. alva says:
“Tension is not exactly force; it is more like a pair of opposing forces. Similarly compression.” Yes, force is always tension or compression at least in Mechanics. Newton’s third law.
• haruspex says:
No, a force is neither a tension nor a compression. A force is something one body exerts on another. Tension and compression usually refer to extensive states within a body. You could describe an action/reaction pair as a compression or a tension, but not the two individual forces.
28. robphy says:
Here are other common errors…1) "the magnitude of the normal force is always mg" (because of a formula they saw).2) "the magnitude of the static friction force is always ##\mu_kN##" (because of a formula they saw)3) "the centripetal force is an additional force drawn on a free-body diagram"
29. Ben Niehoff says:
I have to object to the first item. A "line of action" is insufficient to describe a force except in the special case of a rigid body. In full generality, a force is a vector that is attached to a point, which I guess you can call the "point of application". The point should be thought of as described by some suitable coordinates; not as a "second vector", which is not really the appropriate way to think about it. The "line of action" is merely the line passing through the point of application such that it is collinear with the force vector.From here it is easy to generalize to fields of force density, as in continuum mechanics.
30. AlephNumbers says:
My favorite insights article so far. I would love to see more articles on common mistakes in other areas of mechanics.
31. robphy says:
Probably the biggest error that students make with vectors is the incorrect thinking that "vectors add like numbers"… in particular, that magnitude of a sum of vectors is the sum of the their magnitudes.
32. PeroK says:
Point 1, perhaps, needs an illustration. And, maybe, something about two equal and opposite vectors with different lines of action not canceling out to a nett zero force?In point 2, you could add that a string pulled at both ends (with a force F) is equivalent to a string pulled at one end with a force F and attached to a wall at the other. That seems often to be missed by those who think the tension should be 2F.
33. Greg Bernhardt says:
Great first post haruspex! Nice resource! | 2022-11-28 22:37:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7500450015068054, "perplexity": 927.2431016794084}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710662.60/warc/CC-MAIN-20221128203656-20221128233656-00359.warc.gz"} |
http://marcofrasca.wordpress.com/tag/atlas/ | Waiting for EPS HEP 2013: Some thoughts
13/07/2013
On 18th July the first summer HEP Conference will start in Stockholm. We do not expect great announcements from CMS and ATLAS as most of the main results from 2011-2012 data were just unraveled. The conclusions is that the particle announced on 4th July last year is a Higgs boson. It decays in all the modes foreseen by the Standard Model and important hints favor spin 0. No other resonance is seen at higher energies behaving this way. It is a single yet. There are a lot of reasons to be happy: We have likely seen the guilty for the breaking of the symmetry in the Standard Model and, absolutely for the first time, we have a fundamental particle behaving like a scalar. Both of these properties were looked upon for a long time and now this search is finally ended. On the bad side, no hint of new physics is seen anywhere and probably we will have to wait the restart of LHC on 2015. The long sought SUSY is at large yet.
Notwithstanding this hopeless situation for theoretical physics, my personal view is that there is something that gives important clues to great novelties that possibly will transmute into something of concrete at the restart. It is important to note that there seem to exist some differences between CMS and ATLAS and this small disagreement can hide interesting news for the future. I cannot say if, due to the different conception of this two detectors, something different should be seen but is there. Anyway, they should agree in the end of the story and possibly this will happen in the near future.
The first essential point, that is often overlooked due to the overall figure, is the decay of the Higgs particle in a couple of W or Z. WW decay has a significantly large number of events and what CMS claims is indeed worth some deepening. This number is significantly below one. There is a strange situation here because CMS gives $0.76\pm 0.21$ and in the overall picture just write $0.68\pm 0.20$ and so, I cannot say what is the right one. But they are consistent each other so not a real problem here. Similarly, ZZ decay yields $0.91^{+0.30}_{-0.24}$. ATLAS, on the other side, yields for WW decay $0.99^{+0.31}_{-0.28}$ and for ZZ decay $1.43^{+0.40}_{-0.35}$. Error bars are large yet and fluctuations can change these values. The interesting point here, but this has the value of a clue as these data agree with Standard Model at $2\sigma$, is that the lower values for the WW decay can be an indication that this Higgs particle could be a conformal one. This would mean room for new physics. For ZZ decay apparently ATLAS seems to have a lower number of events as this figure is somewhat larger and the error bar as well. Anyway, a steady decrease has been seen for the WW decay as a larger dataset was considered. This decrease, if confirmed at the restart, would mean a major finding after the discovery of the Higgs particle. It should be said that ATLAS already published updated results with the full dataset (see here). I would like to emphasize that a conformal Standard Model can imply SUSY.
The second point is a bump found by CMS in the $\gamma\gamma$ channel (see here). This is what they see
but ATLAS sees nothing there and this is possibly a fluke. Anyway, this is about $3\sigma$ and so CMS reported about on a publication of them.
Finally, it is also possible that heavier Higgs particles could have depressed production rates and so are very rare. This also would be consistent with a conformal Standard Model. My personal view is that all hopes to see new physics at LHC are essentially untouched and maybe this delay to unveil it is just due to the unlucky start of the LHC on 2008. Meantime, we have to use the main virtue of a theoretical physicist: keeping calm and being patient.
Update: Here is the press release from CERN.
ATLAS Collaboration (2013). Measurements of Higgs boson production and couplings in diboson final
states with the ATLAS detector at the LHC arXiv arXiv: 1307.1427v1
Higgs and beyond
06/06/2013
I am writing these few lines while the conference “Higgs and beyond” is still going on at Tohoku University (Sendai) in Japan. Talks can be found here. Both ATLAS and CMS presented a lot of results about Higgs particle and the most relevant of them is the combination of the data from the two experiments (see here). I am following the excellent recount by Richard Ruiz on twitter (@bravelittlemuon) that also takes care of CERN’s blog. Some interesting point is that there seems to be a bump in $Z\gamma$ channel that is persistent also in other channels. About decay rates, improvements confirm yet nearly Standard Model behavior of the Higgs particle but with the rates of WW and ZZ going down with a too large error bars yet (see my preceding post). Hopes are that CMS and ATLAS could combine also these data reducing error bars. No other Standard Model heavy Higgs particle is seen. Both CMS and ATLAS are looking for evidence of more Higgs particles to no avail yet. Of course, my view is that these excitations should be searched with somewhat different rates from Standard Model expectations. In any case, Standard Model confirms itself as one of the most successful theories in the history of physics. As said by one of ATLAS speakers: “There is overwhelming evidence for a new boson; there is overwhelming evidence for nothing else.” Both experiments plan to complete the analysis of data at 8 TeV for the summer conferences. My personal expectations are that just improvements in the precision of the measurements of the decay rates could eventually give hints of new physics. To fulfill other hopes, we need LHC upgrade that will restart operations on the spring of 2015, hopefully.
CMS harbors new physics beyond the Standard Model
17/05/2013
In these days is ongoing LHCP 2013 (First Large Hadron Collider Physics Conference) and CMS data seem to point significantly toward new physics. Their measurements on the production modes for WW and ZZ are agreeing with my recent computations (see here) and overall are deviating slightly from Standard Model expectations giving
$\frac{\sigma}{\sigma_SM}=0.80\pm 0.14$
Note that Standard Model is alive and kicking yet but looking at the production mode of WW you will read
$\frac{\sigma_{WW}}{\sigma_{WW\ SM}}=0.68\pm 0.20$
in close agreement with results given in my paper and improved respect to Moriond that was $0.71\pm 0.21$. The reason could be that: Higgs model is a conformal one. Data from ZZ yield
$\frac{\sigma_{ZZ}}{\sigma_{ZZ\ SM}}=0.92\pm 0.28$
that is consistent with the result for WW mode, though. I give here the full table from the talk
For the sake of completeness I give here also the same results from ATLAS at the same conference that, instead, seems to go the other way round obtaining overall $1.30\pm 0.20$ and this is already an interesting matter.
At CMS, new physics beyond the Standard Model is peeping out and, more inteestingly, the Higgs model tends to be a conformal one. If this is true, supersymmetry is an inescapable consequence (see here). I would like to conclude citing the papers of other people working on this model and that will be largely cited in the foreseeable future (see here and here).
Marco Frasca (2013). Revisiting the Higgs sector of the Standard Model arXiv arXiv: 1303.3158v1
Marco Frasca (2010). Mass generation and supersymmetry arXiv arXiv: 1007.5275v2
T. G. Steele, & Zhi-Wei Wang (2013). Is Radiative Electroweak Symmetry Breaking Consistent with a 125 GeV
Higgs Mass? Physical Review Letters 110, 151601 arXiv: 1209.5416v3
Krzysztof A. Meissner, & Hermann Nicolai (2006). Conformal Symmetry and the Standard Model Phys.Lett.B648:312-317,2007 arXiv: hep-th/0612165v4
Conformal Standard Model is consistent with the observed Higgs particle
12/04/2013
Robert Garisto is an Editor of Physical Review Letters, the flagship journal of American Physical Society and the one with the highest impact factor in physics. I follow him on twitter (@RobertGaristo) and he points out interesting papers that appear in the journal he works in. This time I read the following
and turned immediately my attention to the linked paper: This one (if you have not a subscription you can find it at arxiv) by Tom Steele and Zhi-Wei Wang showing, with the technique of Padè approximants and an average method how to compute the exact mass of Higgs particle from Coleman-Weinberg mechanism arriving to estimate the ninth order contribution. This is so beacuse they need a stronger coupling with respect to the original Higgs mechanism. They reach an upper bound of 141 GeV for the mass and 0.352 for the self-coupling while they get the mass of 124 GeV for a self-coupling of 0.23. This shows unequivocally that the quadratic term, the one generating the hierarchy problem, is absolutely not needed and the Standard Model, in its conformal formulation, is able to predict the mass of the Higgs particle. Besides, the production rates are identical to the original model but differ for the production of Higgs pairs and this is where one could tell which way nature has chosen. This implies that, at the moment, one has no way to be sure this is the right solution but we have to wait till 2015 after LHC upgrade. So, once again, the precise measurements of these decay rates are essential to tell if we are coping with the original Higgs mechanism or something different or if we need two more years to answer this question. In any case, it is possible that Nobel committee has to wait yet before to take a decision. However, in the sixties that formulation was the only possible and any other solution would have been impossible to discover for the lack of knowledge. They did a great job even if we will prove a different mechanism at work as they provided credibility to the Standard Model and people could trust it.
Finally, I would like to note how the value of the coupling is consistent with my recent estimation where I get 0.36 for the self-interaction. I get different production rates and I would be just curious to see how pictures from ATLAS and CMS would change comparing differently from the Standard Model in order to claim no other Higgs-like particle is seen.
What we can conclude is that the conformal Standard Model is in even more better shape than before and just a single Higgs particle would be needed. An astonishing result.
Steele, T., & Wang, Z. (2013). Is Radiative Electroweak Symmetry Breaking Consistent with a 125 GeV Higgs Mass? Physical Review Letters, 110 (15) DOI: 10.1103/PhysRevLett.110.151601
Marco Frasca (2013). Revisiting the Higgs sector of the Standard Model arXiv arXiv: 1303.3158v1
Much closer to the Standard Model
18/03/2013
Today, the daily from arxiv yields a contribution from John Ellis and Tevong You analyzing new data presented at Aspen and Moriond the last two weeks by CMS and ATLAS about Higgs particle (see here). Their result can be summarized in the following figure
that is really impressive. This means that the updated data coming out from LHC constraints even more the Higgs particle found so far to be the Standard Model one. Another impressive conclusion they are able to draw is that the couplings appear to be proportional to the masses as it should be expected from a well-behaved Higgs particle. But they emphasize that this is “a” Higgs particle and the scenario is well consistent with supersymmetry. Citing them:
The data now impose severe constraints on composite alternatives to the elementary Higgs boson of the Standard Model. However, they do not yet challenge the predictions of supersymmetric models, which typically make predictions much closer to the Standard Model values. We therefore infer that the Higgs coupling measurements, as well as its mass, provide circumstantial support to supersymmetry as opposed to these minimal composite alternatives, though this inference is not conclusive.
They say that further progress on the understanding of this particle could be granted after the upgraded LHC will run and, indeed, nobody is expecting some dramatic change into this scenario from the data at hand.
John Ellis, & Tevong You (2013). Updated Global Analysis of Higgs Couplings arXiv arXiv: 1303.3879v1
A Higgs particle but which one?
14/03/2013
After Moriond conference last week, and while Moriond QCD and Aspen conferences are running yet, an important conclusion can be drawn and it is the one given in this CERN press release. The particle announced on 4th July last year is for certain a Higgs particle as it has spin 0, positive parity and couples almost like the Standard Model Higgs particle to all others. The agreement with Standard Model is embarrassingly increasing as cumulated data since last year are analyzed. Today, CMS will also update their results for the decay $H\rightarrow\gamma\gamma$ and we will know if the small deviation observed by ATLAS will be confirmed. It is true that they see such a deviation with a larger dataset but, rather to increase, it has slightly diminished and this is not really encouraging.
So far, no other particle has been seen and no new physics beyond the Standard Model is seen at the horizon. There is some people pushing for a conclusive assignment of the nature of this boson to the vanilla Higgs particle postulated in the sixties. But it is really too early yet to draw such a conclusion and I have explained why in a paper of mine appeared today on arxiv (see here). Indeed, a formulation of the Higgs field is possible such that, at the tree level, coincides with the original Higgs field (a Higgs impostor). This is due to the existence of exact solutions of the equations of motion of such a field (see here). The relevant point to tell which one is realized in nature is through the decay rate in WW and ZZ and, with the current data, there is agreement for both yet. But, being amplitudes exponentially damped, higher excited states of the Higgs boson cannot be easily seen presently and their eventual observation appears as a statistical fluctuation yet. This can be evaluated quantitatively. It is important because the ZZ decay is sensible to higher masses and displays some peaks that reveal themselves as statistical fluctuations. Increasing the number of events could turn these peaks into real observations.
The interesting point here is that we are moving form the discovery moment to the study phase with a lot of room for improving measurements on this Higgs particle. But the analysis for the existence of higher excited states, Higgs’ brothers, is just at its infancy.
Update: This the analogous figure from ATLAS while the figure for $H\rightarrow\gamma\gamma$ from CMS agrees quite well with the Standard Model: $0.8\pm 0.3$.
Marco Frasca (2013). Revisiting the Higgs sector of the Standard Model arXiv arXiv: 1303.3158v1
Marco Frasca (2009). Exact solutions of classical scalar field equations J.Nonlin.Math.Phys.18:291-297,2011 arXiv: 0907.4053v2
Fabiola Gianotti at Accademia dei Lincei
11/01/2013
On November 7th last year, Fabiola Gianotti, spokesperson of ATLAS experiment at CERN and one of the discoverers of the Higgs-like boson, has been nominated fellow of the Accademia dei Lincei. This is one of the oldest and most prestigious scientific societies that held fellows like Galileo Galilei and Enrico Fermi. Today, she held a public conference with fellows both of moral and scientific classes about “The Higgs boson and our life”. Of course, I was there to see and listen to her personally. As I entered the room, I asked “excuse me” to three people blocking my passage to the chair. When I sat, I looked at them and I realized who were: Carlo di Castro, Francesco de Martini and Giovanni Jona-Lasinio. They were all my former professors. Also Giorgio Parisi was there and later Luciano Maiani entered the audience. Undoubtedly, the audience was truly remarkable.
Lamberto Maffei, president of the Accademia, introduced Gianotti through her main achievements and awards. I would like to remember that she gave the money of the Fundamental Physics Prize for student grants.
The aim of this conference was to convey to all fellows of the Accademia and public at large what was behind the discovery of the Higgs-like particle announced on July 4th last year. For me has been a good chance to hear, from one of the persons mastering this matter, a talk addressed to everybody without the use of technical jargon and using several nice images. Gianotti has shown a very fine gift for this. I would like to reassure my readers that she used comic sans.
By my side, I was proud to hear that 1400 scientists working at CERN are Italians and that an Italian company, Ansaldo Superconduttori Genova, is responsible for one third of the realization of the superconductors at LHC and are also installed in ATLAS detector. At CERN it is working a great majority of young people. Gianotti said that it does not matter if you are a graduate student just entered the team. If your idea is good it is taken and applied. This is what makes scientific enterprise quite different from other realities and renders it so effective. Ideas count more than any authority.
Gianotti pointed out how difficult the situation is for Italy as we have a lot of young people leaving the country for academic positions at foreign universities while there are very few students coming in Italy to do research. Also, reduced budgets from our government with nonsensical cuts can produce a gap between generations of a line that produced excellent people. Recovering would be difficult then.
Turning attention to the discovery, I would like to emphasize that Gianotti repeated more and more times that the only certainty is that Standard Model, a beautiful theory, is verified with very high precision without no hint of breaking so far. But she warned the audience that we know that it must be overcome motivating this mostly from evidence of dark matter. The new particle, she said “Higgs-like”, resemble more and more the one originally postulated by Peter Higgs et al. but they have a lot of data to analyse yet and cannot be certain it is that one yet. They hope to clarify this matter with these other data (Moriond?). She used an interesting image to describe the Higgs field to common people and then turned to the technical one to recover with respect to the formidable physicists were present there. Who speaks Italian can appreciate this video: Gianotti, Tonelli and Bertolucci explain Higgs field with children on similar lines.
The reason why she referred to our life is that most people generally ask “Why?”. Why all this effort to catch such a particle? She gave the beautiful example of J. J. Thompson and the discovery of the electron. When this happened both Thompson’s life and that of his neighbourhood did not change at all. But with the discovery of electronics and its application we all know now what all that has meant. For the Higgs particle can happen the same. From the discovery to its possible applications can pass some time and we need fundamental physics as a priori we cannot foresee the consequences but when they appear can be devastating and change our life definitely and forever for better. Gianotti said that without fundamental research, applied research dries up and eventually dies causing serious troubles to the economy of a country. I completely share her view. She also showed how hadron therapy and pet imaging were by-products of such endeavour.
Questions took more time than expected as the talk was really exciting and several people asked questions. She took this chance to recognize her debt with Ettore Fiorini, in the audience, that introduced her to particle physics and taught her a lot about it. Also Giorgio Salvini was present and asked for beyond LHC. Gianotti said that they hope to have LHC running for more than twenty years as also happened to other accelerator facilities. Salvini participated to most of the history of particle physics since Fermi’s time. He was in the experiment at CERN that produced W and Z particles for the first time with Carlo Rubbia. Francesco de Martini asked a technical question: Has Higgs particle cosmological implications? He was referring to a paper by Lee Smolin that claims that, due to this field, geometry should change from a Riemann to a Weyl one. Gianotti answered immediately that the cosmological implications for the Higgs particle are enormous. The reason is that this is the first scalar particle ever discovered and inflation, the main mechanism in the Standard Model of cosmology to solve the problem of the homogeneity of the universe, has as a basic ingredient a scalar field. CERN discovery shows once again that the idea of inflation is in the right direction. de Martini was not satisfied with the answer turning back to the Smolin’s paper. Then Gianotti asked support to Giorgio Parisi, Parisi is one of the greatest Italian theoretical physicist, that confirmed Gianotti’s answer and said that, even if he is not an expert in the field of general relativity, people working in this research area have devised everything but the kitchen sink and so he would not be surprised if something like this was conceived.
In the end, a very beautiful talk from a great physicist. I would like to paraphrase what Gianotti said about Higgs and its light mass: Thanks Nature for giving us Gianotti! | 2014-08-29 20:12:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 17, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5477834343910217, "perplexity": 853.7973220385652}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500833115.44/warc/CC-MAIN-20140820021353-00247-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://itecnotes.com/electrical/electronic-why-does-more-bandwidth-mean-higher-bit-rate-in-digital-transmission/ | # Electronic – Why does more bandwidth mean higher bit rate in digital transmission
digital-logichigh frequencysignaltransmissiontransmission line
I understand that similar questions like this one have been asked before on this site, listed below. However, I am confused about the answers. If I explain what I think I understand, can somebody please point out where i'm wrong?
I'll start with what I do know:
Shannon Law gives the theoretical upper limit
$$C_{noisy}=B*log_{2}(1+\frac{S}{N})$$
if S = N, then C = B
As N→∞, C→0
As N→0, C→∞
Nyquist Formula says approximately how many levels are needed to achieve this limit
$$C_{noiseless}=2*B*log_{2}M$$
(If you do not use enough logic levels you can not approach the shannon limit, but by using more and more levels you will not exceed the shannon limit)
My problem is that I'm having a hard time understanding why bandwidth relates to bit rate at all. To me it seems like the upper limit of the frequency that can be sent down the channel is the important factor.
Here's a very simplified example: No noise at all, 2 logic levels (0V and 5V), no modulation, and a bandwidth of 300 Hz (30 Hz – 330 Hz). It will have a Shannon Limit of ∞, and a Nyquist Limit of 600bps. Also assume that the channel is a perfect filter so anything outside of the bandwidth is completely dissipated. As I double the bandwidth, I double the bit rate etc.
But why is this? For two level digital transmission With a bandwidth of 300 Hz (30 Hz – 330 Hz), the digital signal of "0V's" and "5V's" will be a (roughly) square wave. This square wave will have the harmonics below 30 Hz and above 330 Hz dissipated, so it will not be perfectly square. If it has a fundamental frequency at the minimum 30 Hz, (so the "0V's" and "5V's" are switching 30 times a second), then there will be a good amount of harmonics and a nice square wave. If it has a fundamental frequency at the max 330 Hz, the signal will be a pure sine wave as there are no higher order harmonics to make it square. However, as there is no noise the receiver will still be able to discriminate the zeros from the ones. In the first case the bit rate will be 60 bps, as the "0V's" and "5V's" are switching 30 times a second. In the second case the bit rate will be a maximum of 660bps, (if the threshold switching voltage of the receiver is exactly 2.5V), and slightly less if the threshold voltage is different.
However this differs from the expected answer of 600 bps for the upper limit. In my explanation it is the upper limit of the channel frequency that matters, not the difference between the upper and lower limit (bandwidth). Can somebody please explain what have I misunderstood?
Also when my logic is applied to the same example but using FSK modulation (frequency shift keying), I get the same problem.
If a zero is expressed as a 30 Hz carrier frequency, a one is expressed as a 330 Hz carrier frequency, and the modulation signal is 330 Hz, then the max bit rate is 660 bps.
Again, can somebody please clear up my misunderstanding?
Also why use a square wave in the first place? Why cant we just send sine waves and design the receivers to have a switching threshold voltage exactly in the middle between the max and min value of the sin wave? This way the signal would take up much less bandwidth.
Thanks for reading!
#### Best Answer
It's a subtle point, but your thinking is going astray when you think of a 330-Hz tone as somehow conveying 660 bits/second of information. It doesn't — and in fact, a pure tone conveys no information at all other than its presence or absence.
In order transmit information through a channel, you need to be able to specify an arbitrary sequence of signaling states that are to be transmitted, and — this is the key point — be able to distinguish those states at the other end.
With your 30-330 Hz channel, you can specify 660 states per second, but it will turn out that 9% of those state sequences will violate the bandwidth limitations of the channel and will be indistinguishable from other state sequences at the far end, so you can't use them. This is why the information bandwidth turns out to be 600 b/s. | 2023-04-01 19:23:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7807847857475281, "perplexity": 455.83334199488417}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296950247.65/warc/CC-MAIN-20230401191131-20230401221131-00162.warc.gz"} |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-10th-edition/chapter-10-analytic-geometry-chapter-review-cumulative-review-page-701/8 | ## Precalculus (10th Edition)
$\theta=30$ degrees $-\infty\lt r\lt \infty$
A polar equation of the line containing the origin and making an angle $\theta_0=30$ degrees with the positive $x$-axis is: $\theta=\theta_0$ $\theta=30$ degrees $-\infty\lt r\lt \infty$ | 2021-10-16 05:26:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.952745795249939, "perplexity": 576.1703777973819}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323583423.96/warc/CC-MAIN-20211016043926-20211016073926-00261.warc.gz"} |
http://math.stackexchange.com/questions/158987/parametric-equation-of-a-cone | # Parametric equation of a cone.
I have a cone with vertex (a, b, c) and base circumference with center $(x_0,y_0)$ and radius R. I can't understand what is the parametric representation of three dimensional space inside the cone. Any suggestions please?
-
The parametric equation of the circle is:
$$\gamma(u) = (x_0 + R\cos u, y_0 + R\sin u, 0)$$
Each point on the cone lies on a line that passes through $p(a, b, c)$ and a point on the circle. Therefore, the direction vector of such a line is:
$$\gamma(u) - p = (x_0 + R\cos u, y_0 + R\sin u, 0) - (a, b, c) = (x_0 - a + R\cos u, y_0 - b + R\sin u, -c)$$
And since the line passes through $p(a, b, c)$, the parametric equation of the line is:
$$p + v\left(\gamma(u) - p\right) = (a, b, c) + v \left((x_0 - a + R\cos u, y_0 - b + R\sin u, -c)\right)$$
Simplify to get the parametric equation of the cone:
$$\sigma(u, v) = \left(a(1-v) + v(x_0 + R\cos u), b(1-v) + v(y_0 + R\sin u), c(1 - v)\right)$$
Here is a plot of the cone for $p(0, 0, 2)$, $(x_0, y_0) = (-1, -1)$ and $R = 1$, when $u$ scans $[0, 2\pi]$ and $v$ scans $[0, 1]$:
-
Begin with a parametric representation of the base, using polar coordinates (with the center shifted to $(x_0,y_0)$. For any given point inside the cone, draw a straight line from the vertex through the point to the base of the cone. Let two parameters specify the latter point. Use $z$ as the third parameter, or perhaps a variable $t$ that varies linearly with $z$ so that $t=0$ at the vertex and $t=1$ at the base of the cone. Many minor variations over this theme are possible. | 2014-03-08 21:28:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8665115833282471, "perplexity": 128.65978367086393}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1393999664120/warc/CC-MAIN-20140305060744-00018-ip-10-183-142-35.ec2.internal.warc.gz"} |
http://commons.apache.org/proper/commons-math/apidocs/org/apache/commons/math3/analysis/MultivariateFunction.html | org.apache.commons.math3.analysis
## Interface MultivariateFunction
• ### Method Summary
Methods
Modifier and Type Method and Description
double value(double[] point)
Compute the value for the function at the given point.
• ### Method Detail
• #### value
double value(double[] point)
Compute the value for the function at the given point.
Parameters:
point - Point at which the function must be evaluated.
Returns:
the function value for the given point.
Throws:
DimensionMismatchException - if the parameter's dimension is wrong for the function being evaluated.
MathIllegalArgumentException - when the activated method itself can ascertain that preconditions, specified in the API expressed at the level of the activated method, have been violated. In the vast majority of cases where Commons Math throws this exception, it is the result of argument checking of actual parameters immediately passed to a method. | 2016-02-08 12:27:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.1987483948469162, "perplexity": 1584.880627761748}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701153323.32/warc/CC-MAIN-20160205193913-00225-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://topfuturepoint.com/1-3-as-a-percentage/ | # 1/3 as a percentage
Let us know about 1/3 as a percentage.
example value
Also, what does 1/3 have as a decimal?
Answer: 1/3 is expressed as 0.3333 in its decimal form.
Here, how do you calculate 1/3 of the total?
Therefore, it is one third of a sum. Counting thirds 3. is done by dividing by . For example: One third of 24 = 3/24 of 1 = 24/3 = 8.
Also to know what is 1/3 as a decimal and as a percentage? Some common decimals and fractions
What is 3/4 as a percentage?
Answer: 3/4 is expressed in terms of percentage up to 75% .
#### What is 3 and 1/3 as a decimal?
So the answer is that 3 1/3 as a decimal is 3.3333333333333 .
#### What is 3/3 rounded off to 1 decimal place as a decimal?
Most people would write it as 0.33 , 0.333, 0.3333, etc. Use 13 as 0.333 or 0.33 depending on the level of accuracy required in the exercise.
#### What is 1/3 called?
are called fractions . 1/3 represents 3 parts out of 1 equal parts.
#### 1 2 or 1 3 Which fraction is greater?
Half is more than a third. Because two fractions, 1/3 and 1/2, have the same numerator (remember, the numerator is the number at the top), it’s easy to compare them. If two fractions have the same numerator, the fraction with the smaller denominator is the larger fraction. Therefore, 1/2 is greater than 1/3.
#### What is 1 and 3/4 as a decimal?
Method 1: Write 1 3/4 in decimal using the division method. To convert a fraction to a decimal form, we need to divide its numerator by the denominator. It gives the answer 1.75 . So, the decimal to 1 3/4 is 1.75.
#### What is 3/2 rounded off to 1 decimal place as a decimal?
Most people would write it as 0.33 , 0.333, 0.3333 , etc. Use 13 as 0.333 or 0.33 depending on the level of accuracy required in the exercise. 0.333,0.3333 is precise and therefore accurate.
#### What is 3/4 as a decimal number?
Answer: 3/4 is expressed as 0.75 in decimal form.
#### What is 3/4 percent as a decimal?
3/4 as a decimal is 0.75 .
#### How do you simplify 4 3?
Subtract 4/3 in lowest terms
1. Find the GCD (or HCF) of the numerator and denominator. The GCD of 4 and 3 is 1.
2. 4 13 1.
3. Reduced fraction: 43. Therefore, 4/3 is 4/3 from the simplified to the lowest terms.
#### What is 3 and 3/4 as a decimal?
So the decimal section for this is 0.75 .
#### What is 1/3 in calculator?
1 / 3 = 13 0.3333333 .
#### What is 2 plus 3 as a decimal?
Answer: 3/2 is expressed as a decimal 1.5 .
#### How do you write 0.08 3 as a fraction?
0.083 equals 112 as a fraction.
#### Which fraction is greater than 1/4 or 1 3?
0.333 is greater than 0.25. Therefore, 1/3 is greater than 1/4 and the answer to the question “Is 1/3 greater than 1/4?” Yes it is. Note: When comparing fractions like 1/3 and 1/4, you can also convert the fractions (if necessary) so that they have the same denominator and then compare which fraction is greater.
#### What is 3 out of 30?
1/3 x 100 = 33.33
Which means our answer of 33.33 is 100 percent of 33.33.
#### What is greater 1/4 or 2 3?
The numerator of the first fraction 8 is greater than the numerator of the second fraction 3, which means that the first fraction 812 is greater than the second fraction 312 and that 23 is greater than 14 .
#### What is smaller than half?
2 7 is only a fraction smaller than a half, so it is the smallest. 1 2 is the next fraction because all other fractions are greater than one half. 3 5 and 3 4 are greater than one half but less than one. Fourth is greater than fifth, so three fourths is greater than three fifths.
#### Is 2 thirds more than half?
“On a measuring cup, the line of two-thirds is above the line of one-half,” Ramon said. “It’s like halfway through a full cup after half a cup.” … “If two-thirds were the same as one-half, then two would have to be half of three. But it is greater, therefore two-thirds must be greater “.
#### What is 1 and 2/3 as a decimal?
So the answer is that 1 2/3 in the decimal form is 1.6666666666667 .
#### How do you convert 3/4 to decimal?
To convert a fraction to a decimal form, we just need to divide the numerator by the denominator. The fraction here is 3/4 which means we need to divide by: 3 4. Therefore, 3/4 = 0.75 . | 2023-04-02 12:59:56 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8548421263694763, "perplexity": 807.4107317606478}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296950528.96/warc/CC-MAIN-20230402105054-20230402135054-00088.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/282542-logarithmic-functions-their-graphs.html | # Thread: logarithmic functions and their graphs
1. ## logarithmic functions and their graphs
The problem: Choose 2 numbers and add their common logarithms. Then find the common logarithm of the product of those 2 numbers. What do you observe? Does it work for 3 numbers?
What I know: common logarithms use base-10.
What I think I know: a logarithm is an exponent.
I don't know how to parlay my limited knowledge into a solution, please help
2. ## Re: logarithmic functions and their graphs
I guess what they are after is that $\log(abc) = \log(a)+\log(b)+\log(c)$
it's a standard property of logarithms
3. ## Re: logarithmic functions and their graphs
Thank u, appreciate the help | 2019-06-24 15:50:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5019199252128601, "perplexity": 1325.8388601831334}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999615.68/warc/CC-MAIN-20190624150939-20190624172939-00536.warc.gz"} |
http://clay6.com/qa/3137/the-area-of-the-region-bounded-by-the-curve-y-x-1-and-the-lines-x-2-and-x-3 | Browse Questions
# The area of the region bounded by the curve $y=x+1$ and the lines $x=2$ and $x=3$ is $(A)\frac{7}{2} sq.units\quad(B)\;\frac{9}{2}sq. units\quad(C)\frac{11} {2}sq.units\quad(D)\frac{13}{2} sq.units$
Toolbox:
• The area enclosed by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$ is given by $\int_a^b ydx.$
• $\int x^ndx=\large\frac{x^{n+1}}{n+1}$$+c$
Step 1:
Given curve is $y=x+1$ which is a straight line bounded between the lines $x=2$ and $x=3$
The required area is the shaded portion as shown in the fig.
Area of the shaded portion is $A=\int_2^3(x+1)dx.$
Step 2:
On integrating we get,
$\begin{bmatrix}\large\frac{x^2}{2}\normalsize +x\end{bmatrix}_2^3$
On applying limits we get,
$\begin{bmatrix}\large\frac{1}{2}\normalsize(3^2-2^2)+(3-2)\end{bmatrix}=\begin{bmatrix}\large\frac{1}{2}\normalsize(9-4)+(3-2)\end{bmatrix}$
$\qquad\qquad\qquad\qquad\qquad=\begin{bmatrix}\large\frac{5}{2}\normalsize +1\end{bmatrix}=\large\frac{7}{2}$sq.units.
Hence the correct option is A. | 2017-02-22 01:45:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9804207682609558, "perplexity": 259.7177834532147}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501170875.20/warc/CC-MAIN-20170219104610-00148-ip-10-171-10-108.ec2.internal.warc.gz"} |
http://www.maths.usyd.edu.au/u/About/prizes2016.html | # School of Mathematics and Statistics: Prizes and Scholarships 2016–2017
[Photos are available: see below.]
The School's annual Prizes and Scholarships presentation ceremony was held in the Forum Restaurant, in the Darlington Centre, on the evening of May 2nd. The presentations were made by Professor Duncan Ivison (Deputy Vice Chancellor (Research)), and Professor Trevor Hambley (Dean of the Faculty of Science).
The guest speakers were Mr Harry Jack and Dr Sabrina Rodrigues.
Harry Jack completed his Honours in applied mathematics at the University of Sydney in 2013, and went on to do a graduate diploma of meteorology at the Bureau of Meteorology, where he has remained. His roles have included writing weather forecasts, chatting on the radio about weather forecasting, and most recently working as a data scientist to automate the production of weather forecasts.
Sabrina Rodrigues completed a mathematics degree with a specialization in teaching at the University of Minho in Portugal. From there she went to University of Melbourne, and enrolled in a MSc with a thesis component in mathematical logic, but developed an interest in bioinformatics, an area that combines biology, computer science, statistics and mathematics. In 2011 Sabrina she completed her MSc in Mathematics and Statistics with a thesis on bioinformatics, and then enrolled in a PhD in Statistics. In 2016, while finishing her PhD, she started working at the CBA as an Associate Data Scientist. In June 2017 Sabrina will be joining the Department of Epidemiology and Biostatistics at Imperial College London as a postdoctoral research associate.
## The Joye Prize
Established in 2002 by donation from Mr Ian Edward Joye LLB and Mr Christopher Ronald Edward Joye BCom, the Joye Prize is awarded annually to the most outstanding student completing fourth year Honours in Applied Mathematics, Pure Mathematics or Mathematical Statistics (provided the work is of sufficient merit).
Recipient: Eric Hester
## The T. G. Room Medal
Established in 1987 by a donation from Mrs Jessica Room and her family in memory of Professor Thomas Gerald Room, who was Professor of Pure Mathematics from 1935 to 1968. Awarded for a PhD thesis in Pure Mathematics of outstanding merit.
Recipient: Ross Ogilvie
for his thesis: "Deformations of harmonic tori in $$S^3$$".
## The David Jackson Prize
Awarded in memory of the late Dr David GA Jackson, who studied and taught in the School of Mathematics and Statistics from 1989 to 1998. In remembrance of David's particular talents the prize is awarded to an undergraduate for creativity and originality in Pure Mathematics.
Recipient: Kane Townsend and Yee Yau.
## Prizes for Honours Students
The KE Bullen Memorial Prize was established in 1989 by public subscription in memory of Professor K E Bullen who was Professor of Applied Mathematics from 1946 to 1971. Awarded for proficiency in Applied Mathematics Honours.
Joint recipients: Eric Hester and Adarsh Kumbhari.
The MJ and M Ashby Prize for Mathematics in Science was established in 1994 through a donation from Mrs M Ashby. Awarded for the best honours essay in the School of Mathematics and Statistics submitted by a student in the Faculty of Science.
Recipient: Sheng Xu.
The Norbert Quirk prize was founded in 1886 by a gift from the subscribers to a memorial for the Reverend John Aloysius Norbert Quirk, Principal of Lyndhurst College, Glebe. Three separate prizes are now awarded.
The Norbert Quirk Prize number 4 is awarded for the best essay by an Honours student.
Joint recipients: Samuel Jelbart and Yee Yau.
The Chris Cannon Prize was established in memory of Chris Cannon, a former staff member in Applied Mathematics. It is awarded for the best Applied Mathematics Honours seminar presentation.
Recipient: Eric Hester.
The Rolf Adams Prizes were established in 2011, following a bequest by Mrs Joan Adams. The prizes are in memory of her grandson, Rolf Adams, who graduated with first class honours and the University medal in Pure Mathematics in 1986. Rolf went on to graduate studies at University of California, Berkeley and then returned to work in image analysis at CSIRO. He died tragically at the age of 26.
Three prizes have been established to encourage outstanding communication and presentation skills in mathematics at the undergraduate and honours levels.
The Rolf Adams Prize No. 1 is awarded for the best Pure Mathematics Honours seminar presentation.
Recipient: Yee Yau.
The Veronica Thomas Prize was established in 1993 by Alice and Richard Thomas in memory of their daughter Veronica, who graduated with first class honours in Mathematical Statistics. Veronica shared the Tim Brown prize for Mathematical Statistics 2, the Statistical Society of Australia Prize and the Australian Federation of University Women (NSW) Prize.
The Veronica Thomas Prize is awarded for the best Honours seminar presentation in Mathematical Statistics.
Joint recipients: Ming Qiu. and Sheng Xu
The Australian Federation of Graduate Women (NSW) Prize in Mathematics is awarded to the most distinguished woman candidate graduating with first class honours in the School.
Recipient: Ming Qiu.
The Barker Scholarship fund was established in 1853 by a gift of £1000 by Thomas Barker, for the encouragement of mathematical science. It funds the Barker Scholarships (see below) and the Barker Prize.
The Barker Prize (established in 1935) is awarded for proficiency in Honours Pure Mathematics, Applied Mathematics or Mathematical Statistics.
Joint recipients: Edwin Spark and Sheng Xu.
## Prizes for Senior Students
The Norbert Quirk Prize number 3 is awarded for the best entry by a third year student in the Norbert Quirk Mathematical Essay Competition.
Recipient: Travis McKenna.
The Applied Probability Trust Prize was established in 2001 by a donation from the Applied Probability Trust. It is awarded to the Senior student in the advanced Stochastic Processes unit of study who demonstrates the greatest proficiency.
Recipient: Gabriel Gregory.
The Tim Brown prizes were established in 1966 by a gift from Mr and Mrs T H Brown to found a prize in memory of their son, T. A. I. Brown. The fund was later increased by further gifts from the family. Tim gained first class honours in 1963 and began a PhD in the Department of Mathematical Statistics. A paper with Professor H O Lancaster was published posthumously in 1965.
The Tim Brown Prize number 2 is awarded for proficiency in Senior Statistics.
Recipients: Gabriel Gregory.
## Prizes for Intermediate Students
The John Spark Memorial Prize was established in 2012 by a donation from Mrs Una Spark in memory of her husband, John Spark, who graduated from the University of Sydney in 1929 with majors in Mathematics, Latin and Psychology.
The John Spark Memorial Prize is awarded for the best academic performance in 18 credit points of Intermediate Mathematics.
Recipient: Ashwin Singh.
The Tim Brown Prize number 1 is awarded for proficiency in Intermediate Statistics.
Recipient: Vaishnavi Calisa.
The Rolf Adams Prize number 2 is awarded for the best Special Studies Program Talk among the students enrolled in MATH2916.
Recipient: Andy Tran.
The Rolf Adams Prize number 3 is awarded for the best Special Studies Program Talk among the students enrolled in MATH2917.
Recipient: Stephen Huang.
## Prizes for Junior Students
The Norbert Quirk Prize number 1 is awarded for the best entry by a first year student in the Norbert Quirk Mathematical Essay Competition.
Recipient: Gordon Li.
The Cengage Learning Prizes were established in 1981 by the offer of Wadsworth Publishing Company (Australasia) to establish a prize for Junior Mathematics.
The Cengage Learning Prize I is awarded for proficiency in junior advanced level mathematics.
Joint recipients: Hazel Browne and Maggie Tong.
The Cengage Learning Prize II is awarded for proficiency in junior mainstream level mathematics.
Recipient: Siyu Mao.
## The George Allen Scholarships
The George Allen Scholarships were founded in 1877 by a bequest from the Honourable George Allen. Three scholarships are now awarded, one for each of Applied Mathematics, Pure Mathematics and Mathematical Statistics.
The George Allen Scholarship for Applied Mathematics is awarded to a student in Applied Mathematics Honours on the basis of performance in Senior Mathematics and Statistics.
Recipient: Angus Leck.
The George Allen Scholarship for Pure Mathematics is awarded to a student in Pure Mathematics Honours on the basis of performance in Senior Mathematics and Statistics.
Joint recipients: Oliver Alexander and Timothy Collier..
The George Allen Scholarship for Mathematical Statistics is awarded to a student in Mathematical Statistics Honours on the basis of performance in Senior Mathematics and Statistics.
Recipient: Chunxi Jiao.
## The KE Bullen Scholarships
Recognizing the work of Professor K E Bullen, these scholarships were established in 1982 by public subscription, and supplemented by further donations. The scholarships are to encourage study in mathematics with the particular emphasis of excellence in Applied Mathematics.
The KE Bullen Scholarship I is awarded to the student who has shown greatest proficiency in Senior units of study and who enrols full-time in Applied Mathematics Honours.
Recipient: Timothy Roberts.
The KE Bullen Scholarship II is awarded to a student who has shown proficiency in Senior units of study and who enrols full-time in Applied Mathematics Honours.
Recipient: Mitchell Curran
The KE Bullen Scholarship III is awarded to the woman student who has shown greatest proficiency in Senior units of study and who enrols full-time in Applied Mathematics Honours.
Joint recipients: Madeleine Cartwright and Cecilia Li.
## Barker Scholarship No. 1
Esstablished in 1853. Awarded for proficiency in Intermediate Mathematics.
Recipient: Leo Jiang.
## Barker Scholarship No. 2
Established in 1881. Awarded for proficiency in Junior Mathematics.
Recipient: Hazel Browne.
## Barker Scholarship No. 4
Awarded for proficiency in Senior units of study in the School of Mathematics and Statistics.
Joint recipients: Timothy Gibson and Nelson Ma.
## Mrs Elva Rae Talented Mathematics Student Award
Awarded to encourage and support talented female mathematics students to reach their full potential in mathematics.
Recipient: Hazel Browne. | 2019-08-20 17:27:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17599886655807495, "perplexity": 5910.431286042893}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027315551.61/warc/CC-MAIN-20190820154633-20190820180633-00544.warc.gz"} |
https://www.physicsforums.com/threads/equivalent-expressions.890975/ | Equivalent Expressions
1. Oct 27, 2016
Veronica_Oles
1. The problem statement, all variables and given/known data
State an equivalent expression in terms of the related acute angle?
(A) cot(-π/4)
2. Relevant equations
3. The attempt at a solution
(A) I made the the unit circle and since its negative I went clockwise and made it into the fourth quadrant. I ended up getting -cotπ/4. However the answer is not the one that the text book has given. Im a little confused.
The text book has given the answer cot3π/4, but I thought we were supposed to go clockwise because it is negative?
2. Oct 27, 2016
PeroK
An acute angle is one that is less than $\pi /2$. I've never seen this used for negative angles, so I'm not sure whether $-\pi /4$ would be considered acute. In any case, $3\pi /4$ is not an acute angle. It's actually a "reflex" angle (I had to look that one up).
I thought your answer was correct. $\pi /4$ is definitely an acute angle.
Last edited: Oct 27, 2016
3. Oct 27, 2016
Staff: Mentor
I agree with @PeroK -- $3\pi/4$ is not an acute angle.
I'll bet what you actually looked up was "reflex".
4. Oct 27, 2016
Veronica_Oles
I just asked my teacher, I got the angle correct. Thx. | 2017-08-22 04:29:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6423775553703308, "perplexity": 863.4911744687134}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886109893.47/warc/CC-MAIN-20170822031111-20170822051111-00257.warc.gz"} |
https://zbmath.org/?q=an:0915.58059 | # zbMATH — the first resource for mathematics
Calculation of Fuchsian groups associated to billiards in a rational triangle. (English) Zbl 0915.58059
Summary: We define, following Veech, the Fuchsian group $$\Gamma(P)$$ of a rational polygon $$P$$. If $$P$$ is simply-connected, then ‘rational’ is equivalent to the condition that all interior angles of $$P$$ be rational multiples of $$\pi$$. Should it happen that $$\Gamma(P)$$ has finite covolume in $$\text{PSL}(2,\mathbb{R})$$ (and is thus a lattice), then a theorem of Veech states that every billiard path in $$P$$ is either finite or uniformly distributed in $$P$$.
We consider the Fuchsian groups of various rational triangles. First, we calculate explicitly the Fuchsian groups of a new sequence of triangles, and discover they are lattices. Interestingly, the lattices found are not commensurable with those previously known. We then demonstrate a class of triangles whose Fuchsian groups are not lattices. These are the first examples of such triangles. Finally, we end by showing how one may specify algebraically, i.e., by an explicit polynomial in two variables, the Riemann surfaces and holomorphic one-forms that are associated to a simply-connected rational polygon. Previously, these surfaces were known by their geometric description. As an example, we show a connection between the billiard in a regular polygon and the well-known Fermat curves of the algebraic equation $$x^n+ y^n= 1$$.
##### MSC:
37A99 Ergodic theory 37G05 Normal forms for dynamical systems
Full Text: | 2021-05-16 00:25:58 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.674271285533905, "perplexity": 384.15866517097845}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991488.53/warc/CC-MAIN-20210515223209-20210516013209-00363.warc.gz"} |
http://www.physicsforums.com/showthread.php?t=490842 | # AC Stark shift
by Niles
Tags: shift, stark
P: 1,820 1. The problem statement, all variables and given/known data Hi In my book, they do calculations on a 2-level atom. After some approximations, they reach the following equation for the coefficient for the first (lower) level: $$c_1(t)\approx c_1(0)e^{-i|\Omega|^2t/\delta}$$ where Omega is the Rabi frequency and delta is the detuning. They say that this equations is basically the AC Stark shift, since the amplitude evolves as if the state energy was shifted |Omega|2/delta. My questions are 1) How do we see that the state energy is shifted by |Omega|2/delta? 2) Is there a specific reason why we call it AC Stark shift? Any help will be greatly appreciated. Best, Niles.
Related Discussions General Physics 1 Atomic, Solid State, Comp. Physics 9 Career Guidance 9 Advanced Physics Homework 2 Advanced Physics Homework 16 | 2013-12-12 08:00:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5812666416168213, "perplexity": 969.6344842581277}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386164571932/warc/CC-MAIN-20131204134251-00079-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/dimensional-analysis-exam.835379/ | Dimensional Analysis exam
1. Sep 30, 2015
1. The problem statement, all variables and given/known data
In a desperate attempt to come up with an equation to use during an examination, a student tries the following equations. Use dimensional analysis to determine which of these equations could be correct. Here x, v, and a, are the position, velocity and acceleration.
1. v2=ax
2. a=v2/x
3. x=v2/a
4. a=xv2
5. x=av2
6. v=ax
7. v=a/x
8. x=av
9. a=xv
10. v=a/t
11. v=at
12. v=t/a
13. a=v/t
14. t=av
15. t=v/a
16. t=a/v
17. a=t/v
2. Relevant equations
a=d/t^2
v=d/t
3. The attempt at a solution
This dimensional analysis thing is really killing me, I understand that the units have to match, thats about it I got out of my textbook and some youtube videos. How about to I apply that to solve this problem (T/F)
2. Sep 30, 2015
andrewkirk
Well, you've got two equations that express a and v as dimensions (ie in terms of d and t). All you need is the third that expresses position x as dimensions, which is of course x = d.
So now use those three equations to substitute everywhere for x, a and v into the list of 17 equations. The ones that balance (ie when both sides are the same when simplified) are possible. The others are not.
3. Sep 30, 2015
Staff: Mentor
Welcome to the PF.
The dimension of position is meters [m]. The dimension of velocity is meters per second [m/s]. The dimension of acceleration is meters per second squared [m/s^2].
The dimensions of the lefthand side (LHS) and the RHS must match. Multiply and divide dimensions the same way you multiply and divide numbers. If you end up with [m]/[m], those units cancel. If you end up with [m/s]/[m], you cancel the [m] units in the numerator and denominator, and are left with units of 1/[seconds], which is the same as Hz (Hertz, which is the unit of frequency, 1/[seconds]).
Does that help? Can you now show us the dimensions for the LHS and RHS for each of the questions in your post?
4. Sep 30, 2015
Staff: Mentor
The idea is to replace each variable with its fundamental units and see if the resulting equation balances. Fundamental units include:
[L] = Length
[M] = Mass
[T] = Time
So, for example, your first equation becomes:
$v^2 = a x \rightarrow ([L] [T]^{-1})^2 = [L][T]^{-2} [L]$
Do the algebra and see if it balances. Then see if you can do the rest.
5. Sep 30, 2015
Staff: Mentor
Piling on!
Now it's up to the OP...
6. Sep 30, 2015
So i just sub in units correct? Also two things, how can we assume d=x?
And the whole idea of why this works, what makes it so special if they are equal (rs=ls) in dimension value?
7. Sep 30, 2015
YEs thank you!
8. Oct 1, 2015
haruspex
To clarify, this L, M, T notation (and likewise Q for charge etc.) represents dimensions. Fundamental units are the base set of standard units laid down for the different dimensions according to some convention. Thus in modern SI units (MKS), the metre is the fundamental unit of length dimension, etc.
The square bracket notation does not appear to be completely standardised. A common usage is to put square brackets around a variable to signify the dimension of the variable (not around the dimensions themselves). Thus one may write for a velocity variable v: [v]=LT-1 | 2017-10-17 00:41:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8260085582733154, "perplexity": 1205.76873351237}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187820487.5/warc/CC-MAIN-20171016233304-20171017013304-00410.warc.gz"} |
https://rviews.rstudio.com/2018/07/09/solver-interfaces-in-cvxr/?utm_campaign=News&utm_medium=Community&utm_source=DataCamp.com | # Solver Interfaces in CVXR
by Anqi Fu and Balasubramanian Narasimhan
## Introduction
In our previous blog post, we introduced CVXR, an R package for disciplined convex optimization. The package allows one to describe an optimization problem with Disciplined Convex Programming rules using high level mathematical syntax. Passing this problem definition along (with a list of constraints, if any) to the solve function transforms it into a form that can be handed off to a solver. The default installation of CVXR comes with two (imported) open source solvers:
• ECOS and its mixed integer cousin ECOS_BB via the CRAN package ECOSolveR
• SCS via the CRAN package scs.
CVXR (version 0.99) can also make use of several other open source solvers implemented in R packages:
The real work of finding a solution is done by solvers, and writing good solvers is hard work. Furthermore, some solvers work particularly well for certain types of problems (linear programs, quadratic programs, etc.). Not surprisingly, there are commercial vendors who have solvers that are designed for performance and scale. Two well-known solvers are MOSEK and GUROBI. R packages for these solvers are also provided, but they require the problem data to be constructed in a specific form. This necessitates a bit of work in the current version of CVXR and is certainly something we plan to include in future versions. However, it is also true that these commercial solvers expose a much richer API to Python programmers than to R programmers. How, then, do we interface such solvers with R as quickly as possible, at least in the short term?
## Reticulate to the Rescue
The current version of CVXR exploits the reticulate package for commercial solvers such as MOSEK and GUROBI. We took the Python solver interfaces in CVXPY version 0.4.11, edited them suitably to make them self-contained, and hooked them up to reticulate.
This means that one needs two prerequisites to use these commercial solvers in the current version of CVXR:
## Installing MOSEK/GUROBI
Both MOSEK and GUROBI provide academic versions (registration required) free of charge. For example, Anaconda users can install MOSEK with the command:
conda install -c mosek mosek
Others can use the pip command:
pip install -f https://download.mosek.com/stable/wheel/index.html Mosek
GUROBI is handled in a similar fashion. The solvers must be activated using a license provided by the vendor.
Once activated, one can check that CVXR recognizes the solver; installed_solvers() should list them.
> installed_solvers()
[1] "ECOS" "ECOS_BB" "SCS" "MOSEK" "LPSOLVE" "GLPK" "GUROBI"
## Further information
More information on these solvers, along with a number of tutorial examples are available on the CVXR site. If you are attending useR! 2018, you can catch Anqi’s CVXR talk on Friday, July 13. | 2019-06-25 15:33:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.33246752619743347, "perplexity": 2714.694369328676}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999853.94/warc/CC-MAIN-20190625152739-20190625174739-00492.warc.gz"} |
https://github.com/ivanov/vim-ipython/issues/20 | # Windows Issues#20
Closed
opened this Issue Jan 25, 2012 · 28 comments
### 11 participants
commented Jan 25, 2012
Hi,
I've been trying to get GVIM and IPython to play together, without success so far on Windows 7.
Here are the steps I've taken:
1. I launch IPython qtconsole successfully. I'm using version 0.12, what came with the EPD Python 2.7.2.
2. :echo has('python') returns 1 for me.
3. I source ipy.vim successfully in vim.
Things go bad for me when I attempt to run ":IPython" in vim.
VIM reports the following:
Traceback (most recent call last):
File "", line 1, in
File "", line 39, in km_from_string
File "C:\Python27\lib\site-packages\IPython\zmq__init__.py", line 39, in
check_for_zmq('2.1.4')
File "C:\Python27\lib\site-packages\IPython\zmq__init__.py", line 17, in check_for_zmq
import zmq
File "C:\Python27\lib\site-packages\zmq__init__.py", line 32, in
File "C:\Python27\Lib\ctypes__init__.py", line 431, in LoadLibrary
return self.dlltype(name)
File "C:\Python27\Lib\ctypes__init_
.py", line 353, in init
self._handle = _dlopen(self._name, mode)
Any idea what my issue is? From IPython i can import ctypes and zmq. I'm pretty new to Python and IPython, which is probably the biggest problem!
Thanks for any help!
-Patrick
Owner
commented Jan 25, 2012
I think I found the problem, I am pretty sure the version of vim I have was not compiled with the +python libraries. I faked myself out by adding the python.vim to the plugin directory, which is why it was returning 1! Now to get a version compiled with python!
I'll let you know if I get it going.
Thanks!
commented Jan 25, 2012
Well I reinstalled gvim, and the python libraries are there this time! Not sure that the issue was the first go around. In any case, :include shows +python. This is gvim 7.3.46, which is supposedly compiled with Python 2.7, the flavor I'm using.
Checking
$vim -c ':py import os; print os.file' -c ':q'$ python -c 'import os; print os.file'
Vim says C:\Python27\Lib\os.pyc
Python says C:\Python27\lib\os.pyc
The case shouldn't matter on Windows right? The actual folder is Lib.
I get the same error,
Traceback (most recent call last): File "<string>", line 1, in <module> File "<string>", line 39, in km_from_string File "C:\Python27\lib\site-packages\IPython\zmq\__init__.py", line 39, in <module> check_for_zmq('2.1.4') File "C:\Python27\lib\site-packages\IPython\zmq\__init__.py", line 17, in check_for_zmq import zmq File "C:\Python27\lib\site-packages\zmq\__init__.py", line 32, in <module> ctypes.cdll.LoadLibrary(libzmq) File "C:\Python27\Lib\ctypes\__init__.py", line 431, in LoadLibrary return self._dlltype(name) File "C:\Python27\Lib\ctypes\__init__.py", line 353, in __init__ self._handle = _dlopen(self._name, mode)
The same as before! Any more thoughts?
Thanks!
commented Jan 25, 2012
Yeah, here is some output:
import zmq
zmq.zmq_version()
'2.1.11'
Importing ctypes also returns no errors. It is version 1.1.0.
Owner
commented Jan 27, 2012
@youngpm: you can also take a look at #21 to see if you're running into a similar issue, and if the solution proposed in this pyodbc bug solves it for you
commented Feb 6, 2012
Sorry I've been out of the office on jury duty! In any case, I'm going to try what you suggest. edouardp says
"I did this with all of the *.pyd files and now I've got gvim <-> ipython 0.12 interaction going great!"
It is unclear to me which *.pyd files he is referring to.
Thanks!
commented Feb 15, 2012
I think he is referring to all the *.pyd files in the pyzmq package:
zmq\core\constants.pyd
zmq\core\context.pyd
zmq\core\device.pyd
zmq\core\error.pyd
zmq\core\message.pyd
zmq\core\poll.pyd
zmq\core\socket.pyd
zmq\core\stopwatch.pyd
zmq\core\version.pyd
zmq\devices\monitoredqueue.pyd
zmq\utils\rebuffer.pyd
I had the same problem as in #21 and applying the fix referred by edouardp to these files fixed it for me!
HTH
commented May 30, 2012
I still get the error the first time I type :IPython, but everything seems to function right after that, so I can confirm this still works!
commented Jun 17, 2012
Here are the commands I used to make it run and their outputs.
Windows xp, Python 2.7.1 (r271:86832, Nov 27 2010, 18:30:46) [MSC v.1500 32 bit (Intel)], zmq 2.2.0
Note that dependencywalker is missing ieshims.dll and wer.dll, but they don't seem to matter
C:\Python27\Lib\site-packages\pyzmq-2.2.0-py2.7-win32.egg\zmq\utils>dir
Directory of C:\Python27\Lib\site-packages\pyzmq-2.2.0-py2.7-win32.egg\zmq\utils
06/17/2012 09:06 AM
.
06/17/2012 09:06 AM ..
06/16/2012 02:39 PM 642 allocate.h
06/16/2012 02:39 PM 9,753 buffers.pxd
06/16/2012 02:39 PM 2,407 jsonapi.py
06/16/2012 02:39 PM 2,492 jsonapi.pyc
06/12/2008 03:58 PM 739,376 mt.exe
06/16/2012 02:39 PM 1,324 pyversion_compat.h
06/16/2012 02:39 PM 286 rebuffer.py
06/16/2012 02:39 PM 609 rebuffer.pyc
06/17/2012 09:06 AM 25,600 rebuffer.pyd
06/16/2012 02:39 PM 980 strtypes.py
06/16/2012 02:39 PM 859 strtypes.pyc
06/16/2012 02:39 PM 3,514 zmq_compat.h
06/16/2012 02:39 PM 0 init.py
06/16/2012 02:39 PM 166 init.pyc
Microsoft (R) Manifest Tool version 5.2.3790.2076
C:\Python27\Lib\site-packages\pyzmq-2.2.0-py2.7-win32.egg\zmq\utils>mt.exe -inputresource:c:\windows\system32\python27.dll;#2 -outputresource:rebuffer.pyd;#2
Microsoft (R) Manifest Tool version 5.2.3790.2076
C:\Python27\Lib\site-packages\pyzmq-2.2.0-py2.7-win32.egg\zmq\utils>cd. .
C:\Python27\Lib\site-packages\pyzmq-2.2.0-py2.7-win32.egg\zmq\utils>
C:\Python27\Lib\site-packages\pyzmq-2.2.0-py2.7-win32.egg\zmq\utils>mt.exe -inputresource:c:\windows\system32\python27.dll;#2 -outputresource:..\core\constants.
pyd;#2
Microsoft (R) Manifest Tool version 5.2.3790.2076
C:\Python27\Lib\site-packages\pyzmq-2.2.0-py2.7-win32.egg\zmq\utils>mt.exe -inputresource:c:\windows\system32\python27.dll;#2 -outputresource:..\core\context.py
d;#2
Microsoft (R) Manifest Tool version 5.2.3790.2076
C:\Python27\Lib\site-packages\pyzmq-2.2.0-py2.7-win32.egg\zmq\utils>mt.exe -inputresource:c:\windows\system32\python27.dll;#2 -outputresource:..\core\device.pyd
;#2
Microsoft (R) Manifest Tool version 5.2.3790.2076
C:\Python27\Lib\site-packages\pyzmq-2.2.0-py2.7-win32.egg\zmq\utils>mt.exe -inputresource:c:\windows\system32\python27.dll;#2 -outputresource:..\core\error.pyd;
#2
Microsoft (R) Manifest Tool version 5.2.3790.2076
C:\Python27\Lib\site-packages\pyzmq-2.2.0-py2.7-win32.egg\zmq\utils>mt.exe -inputresource:c:\windows\system32\python27.dll;#2 -outputresource:..\core\message.py
d;#2
Microsoft (R) Manifest Tool version 5.2.3790.2076
C:\Python27\Lib\site-packages\pyzmq-2.2.0-py2.7-win32.egg\zmq\utils>mt.exe -inputresource:c:\windows\system32\python27.dll;#2 -outputresource:..\core\stopwatch.
pyd;#2
Microsoft (R) Manifest Tool version 5.2.3790.2076
C:\Python27\Lib\site-packages\pyzmq-2.2.0-py2.7-win32.egg\zmq\utils>mt.exe -inputresource:c:\windows\system32\python27.dll;#2 -outputresource:..\core\socket.pyd
;#2
Microsoft (R) Manifest Tool version 5.2.3790.2076
C:\Python27\Lib\site-packages\pyzmq-2.2.0-py2.7-win32.egg\zmq\utils>mt.exe -inputresource:c:\windows\system32\python27.dll;#2 -outputresource:..\core_poll.pyd;
#2
Microsoft (R) Manifest Tool version 5.2.3790.2076
C:\Python27\Lib\site-packages\pyzmq-2.2.0-py2.7-win32.egg\zmq\utils>mt.exe -inputresource:c:\windows\system32\python27.dll;#2 -outputresource:..\core_version.p
yd;#2
Microsoft (R) Manifest Tool version 5.2.3790.2076
C:\Python27\Lib\site-packages\pyzmq-2.2.0-py2.7-win32.egg\zmq\utils>mt.exe -inputresource:c:\windows\system32\python27.dll;#2 -outputresource:..\devices\monitor
edqueue.pyd;#2
Microsoft (R) Manifest Tool version 5.2.3790.2076
I have been trying to get this to work for a few days now; I am having the exact same problems.
A pop-up says:
Runtime Error!
Program: C:\opt\vim\vim73\gvim.exe
R6034
An application has made an attempt to load the C runtime library incorrectly.
I check the versions of python in vim, and it is correct. I also checked that the python_path is the same. I am trying to follow the instructions in here, but haven't gotten it to work yet.
I am on windows 64bit,
Python 2.7.3
newest version of vim (7.3. something)
IPython 0.12
commented Jun 21, 2012
I never got a popup, just an import error, pyzmq version not ok, import error, initthreads import error etc. Maybe you have a different error,
Did you run the dependencywalker on the dlls to see if anything is missing, and then try to fix the manifest with mt?
It works very nicely for me now.
commented Jun 21, 2012
I was getting the exact behavior johndgiese is seeing - with the popup error. Running mt as explained above on all of the pyd files fixed it for me. Note that some of my files didn't match the exact ones from the pydoc thread, i had to run it on ALL the pyd files in the pyzmq directory.
commented Jul 26, 2012
after i run all the mt commands, the pop up window is still there with error R6034 and zmq import error in vim output when i put :IPython first time, but when i put second time, then it seems the vim is communicating with ipython but ipython does not fire output, say, i run :py run_command('a=1234') and there is nothing happend in ipython, but when i type print a in ipython console, the a is 1234.
run enviroment: python 2.7.2, ipython 0.12.1, any one has any idea?
When you run a command in vim-ipython it sends the command to the ipython server; this is a separate process than the python that vim has internally for scripting; the namespaces are completely separate. Running :py commands will always run in vim's python process, while writing commands in the ipython console will send the command via zmq to the ipython process you started yourself, it will then execute the command, and send the output back to vim via zmq, at which point vim will update the vim-ipython prompt.
commented Jul 26, 2012
thanks for explaining, after reading your comments several times, i think i got some idea, so, here is the problem i encounter:
1 start ipython qtconsole
2. start gvim, open foo.py file (which only has lines: 'a= 1234; print a')
3 run :IPython in vim, window pop up giving runtime library loading error R6034,
4. run :IPython again in vim, no error shows up
5. press F5 in vim, the minibuffer shows:
In[1]: In[]: run -i 'Path\to\foo.py'
press enter or type command to continue
6. switch to ipython qtconsole,found no output, but type a, it prints 1234, which means a is there with value from foo.py
So, i know vim and ipython are talking to each other, but this is not so useful if i can not get output from ipython.. and also the error R 6034 shows the first time ? actually if i skip step 4 above, the phenomena is the same...
i just wonder if there is a solution to this...
thanks again!
P.S. in step 5, the number '1' in the 'In[1]' matches the prompt in ipython, that means if you run several command in ipython qtconsole, and then run a script in vim by F5, the number will change to current ipython input prompt number
This is a bit complicated, but basically there is an ipython kernel, and the console (and vim) are clients that talk to the kernel. Currently the output resulting from commands from one client (e.g. vim) are not displayed in other clients (the IPython kernel).
I actually have my own version of vim-ipython that I have heavily modified from ivanov's. If you want to give it a try you can. I am setting up a repository for you, in case you want to.
commented Jul 26, 2012
i actually saw the same issue here: ipython/ipython#1873
did you really solve the problem? that would be nice and greatly appreciated if you can share your script,
i am working on windows 7 btw,
thx
I still get the R6034 error, even after running the above mt script, but this time not on zmq but IPython itself, e.g. when running
:py import IPython
Turns out I was having issues with import uuid. I've now fixed this by merging my python manifest into my vim/gvim's manifest, like so:
mt -inputresource:C:\windows\syswow64\python27.dll;#2 -out:%TEMP%\python.manifest
mt -inputresource:"C:\Program Files (x86)\vim\vim73\gvim.exe" -out:%TEMP%\gvim.manifest
mt -manifest %TEMP%\python.manifest %TEMP%\gvim.manifest -outputresource:"C:\Program Files (x86)\vim\vim73\gvim.exe"
mt -manifest %TEMP%\python.manifest -outputresource:"C:\Program Files (x86)\vim\vim73\vim.exe"
(The above is from memory, make sure to backup your vim/gvim first)
commented Dec 8, 2012
Hi all, i had exactly the same issues . My environment is win7 , python27 , pyzmq 2.2.0.1 , ipython 13.1 .Gvim 7.3.754 build with gcc 4.6.2 and python27
I resolved them following these steps after testing with Dep.Walker that there were errors with MSVCR90.DLL loading
when opened *.PYD and *.DLL (1) files from pyzmq foder.
1) went to the folder pyzmq with VS command line tools 2012 x86 (my os) and searched for all pyd files.
2) i went to every subfolder where these .pyd existed and executed:
mt.exe -inputresource:c:\windows\system32\python27.dll;#2 -outputresource:(XXX).pyd;#2
where (XXX) was every pyd (around 10) in every subfolder..The console didn't show any mistake in the procedure
3) i did the same with the .dll file AND VOILA!
Everything is working like charm! I must confess that DANBE'S POST above was the real key ...
I just wanted to post this in order to assure that this works..
Owner
commented Mar 26, 2013
closing this issue
closed this Mar 26, 2013
referenced this issue Nov 7, 2013
Closed
### Runtime error when starting up #75
commented Apr 27, 2014
I came across this problem yesterday. XP + python27 + vim7.3 (+python)
I fixed it by executing mt.exe -inputresource:"d:\Python27\python.exe" -outputresource:"d:\Program Files\Vim\vim73\gvim.exe"
This command actually replaced the manifest embedded in gvim.exe with the one in python.exe.
But it DO fix the R6034 error when executing IPython command in gvim.
Hope this helps.
Closed | 2016-04-28 20:16:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.37558653950691223, "perplexity": 8971.538155042743}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860106452.21/warc/CC-MAIN-20160428161506-00139-ip-10-239-7-51.ec2.internal.warc.gz"} |
http://tug.org/pipermail/texhax/2009-February/011868.html | # [texhax] problems with circumflex
Uwe Lück uwe.lueck at web.de
Sat Feb 28 22:51:36 CET 2009
At 09:14 28.02.09, Benjamin Sambale wrote:
>I wonder about the difference between $^x \operatorname{N}$ and
>$\operatorname{N}^x$. Here, I use \operatorname only to write it
>upright. Apart from that it doesn't play a role. The problem is that the
>space between ^x and N is larger than the space between N and ^x.
>Another problem is that ^x in front of N lies lower than ^x after N. I
>want that both ^x look exactly the same. If I use
>$^x{\operatorname{N}}$, then the space reduces a bit, but it is still to
>large. The space in $^x\!\operatorname{N}$ looks also better, but I
>think this is not the best solution.
At 14:56 28.02.09, Benjamin Sambale wrote:
>The problem is not to write it upright. But you're right,
>$\operatorname{N}^x$ and $\mathrm{N}^x$ are different. In fact I want to
>use N as an mathematical operator. But even if you use $^xN$ and $N^x$,
>the spaces between ^x and N are different.
The TeXbook explains horizontal spacing in math formulas on, say, pp.
154ff., 158f., 168ff. ...
[I think to have learnt:]
The circumflex attaches a superscript to something *previous*.
"left superscript" (as in non-standard analysis or ...!?) is not supported
by TeX (from scratch, tried tricks, don't remember). "Symmetry" between
left and right superscripts is not intended and occurs "by chance" if it does.
^x \operatorname ...' yields a space because there is a space between an
operator and something previous (to which the superscript is attributed).
With \mathrm{N}, there is the symmetry! ("by chance")
With $^xN$ and $N^x$ there is some symmetry again, but the "left
superscript" is somewhat unable to account for the italics shift of the
N', while a "right superscript" does account for the italcs shift. There
is no space between the "left supercript" and the *box* enclosing N'.
There is only some (apparent) space between the "left superscript" and the
upper part of the N'.
Before working at "superscript symmetry", one might reason whether it
really is needed and worth it.
-- Uwe. | 2017-12-18 03:10:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8370383977890015, "perplexity": 5834.993157018019}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948604248.93/warc/CC-MAIN-20171218025050-20171218051050-00233.warc.gz"} |
https://giswqs.github.io/whiteboxR/index.html | WhiteboxTools R Frontend.
This repository is related to the whitebox R package for geospatial analysis, which is an R frontend of a stand-alone executable command-line program called WhiteboxTools.
Contents
## Description
The whitebox R package is built on WhiteboxTools, an advanced geospatial data analysis platform developed by Prof. John Lindsay (webpage; jblindsay) at the University of Guelph’s Geomorphometry and Hydrogeomatics Research Group. WhiteboxTools can be used to perform common geographical information systems (GIS) analysis operations, such as cost-distance analysis, distance buffering, and raster reclassification. Remote sensing and image processing tasks include image enhancement (e.g. panchromatic sharpening, contrast adjustments), image mosaicing, numerous filtering operations, simple classification (k-means), and common image transformations. WhiteboxTools also contains advanced tooling for spatial hydrological analysis (e.g. flow-accumulation, watershed delineation, stream network analysis, sink removal), terrain analysis (e.g. common terrain indices such as slope, curvatures, wetness index, hillshading; hypsometric analysis; multi-scale topographic position analysis), and LiDAR data processing. LiDAR point clouds can be interrogated (LidarInfo, LidarHistogram), segmented, tiled and joined, analyzed for outliers, interpolated to rasters (DEMs, intensity images), and ground-points can be classified or filtered. WhiteboxTools is not a cartographic or spatial data visualization package; instead it is meant to serve as an analytical backend for other data visualization software, mainly GIS. Suggested citation: Lindsay, J. B. (2016). Whitebox GAT: A case study in geomorphometric analysis. Computers & Geosciences, 95, 75-84. doi: http://dx.doi.org/10.1016/j.cageo.2016.07.003
## Installation
There are three ways to install the whitebox R package.
### 1. CRAN
whitebox is now available on CRAN, so you can install it with:
install.packages("whitebox")
### 2. R-Forge
whitebox is also available on R-Forge, so you can install it with:
install.packages("whitebox", repos="http://R-Forge.R-project.org")
### 3. GitHub
You can alternatively install the development version of whitebox from GitHub as follows:
if (!require("remotes")) install.packages('remotes')
remotes::install_github("giswqs/whiteboxR", build = FALSE)
## Usage
The function wbt_init() checks the path to the WhiteboxTools binary.
wbt_init()
It (invisibly) returns a logical (TRUE/FALSE) value reflecting whether a file is found at one of the default paths, including those specified by package options. See ?whitebox::whitebox.
If you have WhiteboxTools installed already but in a non-standard location run wbt_init(exe_path=...) to set up your options for your current R session.
wbt_init(exe_path = 'C:/home/user/path/to/whitebox_tools.exe')
Additional arguments to wbt_init() can set other package options, such as whether tools print their standard console output with cat().
If you want to turn this off, set verbose = FALSE, for instance:
wbt_init(exe_path = 'C:/home/user/path/to/whitebox_tools.exe', verbose = FALSE)
## Documentation
For whitebox package documentation in R, ask for help:
??whitebox
For list of functions, try the wbt_ prefix in your search:
??wbt_
A complete list of functions available in the whitebox R package can be found HERE.
Check out this demo for examples.
## Installing WhiteboxTools
The quickest way to get started if you are on 64-bit Windows, Linux or MacOS architectures is to download and install the WhiteboxTools binary. A method install_whitebox() is provided to download a version of the binaries that corresponds to the wrapper functions available in the package.
whitebox::install_whitebox()
By default this will install to your whitebox R package installation directory (e.g. in your R package library), subdirectory “WBT”.
## whitebox
How do I run tools?
Tool names in the whitebox R package can be called by corresponding function using wbt_snake_case() naming convention (e.g. wbt_lidar_info() R function for "LidarInfo" WhiteboxTools tool name).
First we load the library and perform any necessary setup.
library(whitebox)
# set up as needed
wbt_init()
Many WhiteboxTools will take GeoTIFF files as input. There is a sample Digital Elevation Model (DEM) included in the whitebox package. You can get path as follows for this demo:
# Set input raster DEM file
dem <- system.file("extdata", "DEM.tif", package = "whitebox")
We will use this DEM in and perform a sequence of processing routines.
1. Applies feature-preserving smoothing (FeaturePreservingSmoothing)
2. Breaches depressions in a DEM (BreachDepressions)
3. Calculates D-Infinity flow accumulation (DInfFlowAccumulation)
Run tools passing file paths for input and output grids.
In this case we will deal with GeoTIFF input, but WhiteboxTools supports several grid input formats.
The filter argument sets the size of the filter kernel.
## 1. FeaturePreservingSmoothing
wbt_feature_preserving_smoothing(
dem = dem,
output = "./smoothed.tif",
filter = 9
)
## 2. BreachDepressions
wbt_breach_depressions(dem = "./smoothed.tif", output = "./breached.tif")
## 3. DInfFlowAccumulation
wbt_d_inf_flow_accumulation(input = dem, output = "./flow_accum.tif")
if (requireNamespace('raster')) {
raster::plot(raster::raster("./flow_accum.tif"))
}
#> Loading required namespace: raster
About WhiteboxTools
library(whitebox)
# cat() output in non-interactive mode
wbt_verbose(TRUE)
# Prints the whitebox-tools help...a listing of available commands
wbt_help()
#> WhiteboxTools Help
#>
#> The following commands are recognized:
#> --cd, --wd Changes the working directory; used in conjunction with --run flag.
#> -h, --help Prints help information.
#> -l, --license Prints the whitebox-tools license. Tool names may also be used, --license="Slope"
#> --listtools Lists all available tools. Keywords may also be used, --listtools slope.
#> -r, --run Runs a tool; used in conjunction with --wd flag; -r="LidarInfo".
#> --toolbox Prints the toolbox associated with a tool; --toolbox=Slope.
#> --toolhelp Prints the help associated with a tool; --toolhelp="LidarInfo".
#> --toolparameters Prints the parameters (in json form) for a specific tool; --toolparameters="LidarInfo".
#> -v Verbose mode. Without this flag, tool outputs will not be printed.
#> --viewcode Opens the source code of a tool in a web browser; --viewcode="LidarInfo".
#> --version Prints the version information.
#>
#> Example Usage:
#> >> ./whitebox_tools -r=lidar_info --cd="/path/to/data/" -i=input.las --vlr --geokeys
# Prints the whitebox-tools license
wbt_license()
#> WhiteboxTools License
#> Copyright 2017-2021 John Lindsay
#>
#> Permission is hereby granted, free of charge, to any person obtaining a copy of this software and
#> associated documentation files (the "Software"), to deal in the Software without restriction,
#> including without limitation the rights to use, copy, modify, merge, publish, distribute, sublicense,
#> and/or sell copies of the Software, and to permit persons to whom the Software is furnished to do so,
#> subject to the following conditions:
#>
#> The above copyright notice and this permission notice shall be included in all copies or substantial
#> portions of the Software.
#>
#> THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT
#> NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND
#> NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES
#> OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN
#> CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.
# Prints the whitebox-tools version
wbt_version()
#> WhiteboxTools v2.1.0 by Dr. John B. Lindsay (c) 2017-2021
#>
#> WhiteboxTools is an advanced geospatial data analysis platform developed at
#> the University of Guelph's Geomorphometry and Hydrogeomatics Research
#> Group (GHRG). See www.whiteboxgeo.com for more details.
# Prints the toolbox for a specific tool.
wbt_toolbox("AsciiToLas")
#> LiDAR Tools
# Lists tools with 'lidar' in tool name or description.
wbt_list_tools("lidar")
#> All 48 Tools containing keywords:
#> AsciiToLas: Converts one or more ASCII files containing LiDAR points into LAS files.
#> ClassifyBuildingsInLidar: Reclassifies a LiDAR points that lie within vector building footprints.
#> ClassifyOverlapPoints: Classifies or filters LAS points in regions of overlapping flight lines.
#> ClipLidarToPolygon: Clips a LiDAR point cloud to a vector polygon or polygons.
#> ErasePolygonFromLidar: Erases (cuts out) a vector polygon or polygons from a LiDAR point cloud.
#> FilterLidarClasses: Removes points in a LAS file with certain specified class values.
#> FilterLidarScanAngles: Removes points in a LAS file with scan angles greater than a threshold.
#> FindFlightlineEdgePoints: Identifies points along a flightline's edge in a LAS file.
#> FlightlineOverlap: Reads a LiDAR (LAS) point file and outputs a raster containing the number of overlapping flight lines in each grid cell.
#> HeightAboveGround: Normalizes a LiDAR point cloud, providing the height above the nearest ground-classified point.
#> LasToAscii: Converts one or more LAS files into ASCII text files.
#> LasToMultipointShapefile: Converts one or more LAS files into MultipointZ vector Shapefiles. When the input parameter is not specified, the tool grids all LAS files contained within the working directory.
#> LasToShapefile: Converts one or more LAS files into a vector Shapefile of POINT ShapeType.
#> LasToZlidar: Converts one or more LAS files into the zlidar compressed LiDAR data format.
#> LidarBlockMaximum: Creates a block-maximum raster from an input LAS file. When the input/output parameters are not specified, the tool grids all LAS files contained within the working directory.
#> LidarBlockMinimum: Creates a block-minimum raster from an input LAS file. When the input/output parameters are not specified, the tool grids all LAS files contained within the working directory.
#> LidarClassifySubset: Classifies the values in one LiDAR point cloud that correspond with points in a subset cloud.
#> LidarColourize: Adds the red-green-blue colour fields of a LiDAR (LAS) file based on an input image.
#> LidarDigitalSurfaceModel: Creates a top-surface digital surface model (DSM) from a LiDAR point cloud.
#> LidarElevationSlice: Outputs all of the points within a LiDAR (LAS) point file that lie between a specified elevation range.
#> LidarGroundPointFilter: Identifies ground points within LiDAR dataset using a slope-based method.
#> LidarHexBinning: Hex-bins a set of LiDAR points.
#> LidarHillshade: Calculates a hillshade value for points within a LAS file and stores these data in the RGB field.
#> LidarHistogram: Creates a histogram of LiDAR data.
#> LidarIdwInterpolation: Interpolates LAS files using an inverse-distance weighted (IDW) scheme. When the input/output parameters are not specified, the tool interpolates all LAS files contained within the working directory.
#> LidarInfo: Prints information about a LiDAR (LAS) dataset, including header, point return frequency, and classification data and information about the variable length records (VLRs) and geokeys.
#> LidarJoin: Joins multiple LiDAR (LAS) files into a single LAS file.
#> LidarKappaIndex: Performs a kappa index of agreement (KIA) analysis on the classifications of two LAS files.
#> LidarNearestNeighbourGridding: Grids LiDAR files using nearest-neighbour scheme. When the input/output parameters are not specified, the tool grids all LAS files contained within the working directory.
#> LidarPointDensity: Calculates the spatial pattern of point density for a LiDAR data set. When the input/output parameters are not specified, the tool grids all LAS files contained within the working directory.
#> LidarPointStats: Creates several rasters summarizing the distribution of LAS point data. When the input/output parameters are not specified, the tool works on all LAS files contained within the working directory.
#> LidarRansacPlanes: Performs a RANSAC analysis to identify points within a LiDAR point cloud that belong to linear planes.
#> LidarRbfInterpolation: Interpolates LAS files using a radial basis function (RBF) scheme. When the input/output parameters are not specified, the tool interpolates all LAS files contained within the working directory.
#> LidarRemoveDuplicates: Removes duplicate points from a LiDAR data set.
#> LidarRemoveOutliers: Removes outliers (high and low points) in a LiDAR point cloud.
#> LidarRooftopAnalysis: Identifies roof segments in a LiDAR point cloud.
#> LidarSegmentation: Segments a LiDAR point cloud based on differences in the orientation of fitted planar surfaces and point proximity.
#> LidarSegmentationBasedFilter: Identifies ground points within LiDAR point clouds using a segmentation based approach.
#> LidarTINGridding: Creates a raster grid based on a Delaunay triangular irregular network (TIN) fitted to LiDAR points.
#> LidarThin: Thins a LiDAR point cloud, reducing point density.
#> LidarThinHighDensity: Thins points from high density areas within a LiDAR point cloud.
#> LidarTile: Tiles a LiDAR LAS file into multiple LAS files.
#> LidarTileFootprint: Creates a vector polygon of the convex hull of a LiDAR point cloud. When the input/output parameters are not specified, the tool works with all LAS files contained within the working directory.
#> LidarTophatTransform: Performs a white top-hat transform on a Lidar dataset; as an estimate of height above ground, this is useful for modelling the vegetation canopy.
#> NormalVectors: Calculates normal vectors for points within a LAS file and stores these data (XYZ vector components) in the RGB field.
#> SelectTilesByPolygon: Copies LiDAR tiles overlapping with a polygon into an output directory.
#> ZlidarToLas: Converts one or more zlidar files into the LAS data format.
#> LidarShift: Shifts the x,y,z coordinates of a LiDAR file.
# Prints the help for a specific tool.
wbt_tool_help("lidar_info")
#> LidarInfo
#> Description:
#> Prints information about a LiDAR (LAS) dataset, including header, point return frequency, and classification data and information about the variable length records (VLRs) and geokeys.
#> Toolbox: LiDAR Tools
#> Parameters:
#>
#> Flag Description
#> ----------------- -----------
#> -i, --input Input LiDAR file.
#> -o, --output Output HTML file for summary report.
#> --vlr Flag indicating whether or not to print the variable length records (VLRs).
#> --geokeys Flag indicating whether or not to print the geokeys.
#>
#>
#> Example usage:
#> >>./whitebox_tools -r=LidarInfo -v --wd="/path/to/data/" -i=file.las --vlr --geokeys"
#> ./whitebox_tools -r=LidarInfo --wd="/path/to/data/" -i=file.las
# Retrieves the tool parameter descriptions for a specific tool.
wbt_tool_parameters("slope")
#> {"parameters": [{"name":"Input DEM File","flags":["-i","--dem"],"description":"Input raster DEM file.","parameter_type":{"ExistingFile":"Raster"},"default_value":null,"optional":false},{"name":"Output File","flags":["-o","--output"],"description":"Output raster file.","parameter_type":{"NewFile":"Raster"},"default_value":null,"optional":false},{"name":"Z Conversion Factor","flags":["--zfactor"],"description":"Optional multiplier for when the vertical and horizontal units are not the same.","parameter_type":"Float","default_value":null,"optional":true},{"name":"Units","flags":["--units"],"description":"Units of output raster; options include 'degrees', 'radians', 'percent'","parameter_type":{"OptionList":["degrees","radians","percent"]},"default_value":"degrees","optional":true}]}
# View the source code for a specific tool on the source code repository.
wbt_view_code("breach_depressions")
#> https://github.com/jblindsay/whitebox-tools/blob/master/whitebox-tools-app/src/tools/hydro_analysis/breach_depressions.rs
## Available Tools
The WhiteboxTools library currently contains more than 518 tools, which are each grouped based on their main function into one of the following categories: Data Tools, GIS Analysis, Hydrological Analysis, Image Analysis, LiDAR Analysis, Mathematical and Statistical Analysis, Stream Network Analysis, and Terrain Analysis. For a listing of available tools, complete with documentation and usage details, please see the WhiteboxTools User Manual.
If you are interested in using the WhiteboxTools command-line program, check WhiteboxTools Usage
## Contributing
If you would like to contribute to the project as a developer, follow these instructions to get started:
1. Fork the whiteboxR repository (https://github.com/giswqs/whiteboxR)
2. Create your feature branch (git checkout -b my-new-feature)
3. Commit your changes (git commit -am ‘Add some feature’)
4. Push to the branch (git push origin my-new-feature)
5. Create a new Pull Request
Unless explicitly stated otherwise, any contribution intentionally submitted for inclusion in the work shall be licensed as the MIT license without any additional terms or conditions.
## License
The whitebox R package is distributed under the MIT license, a permissive open-source (free software) license.
## Reporting Bugs
whitebox is distributed as is and without warranty of suitability for application. If you encounter flaws with the software (i.e. bugs) please report the issue. Providing a detailed description of the conditions under which the bug occurred will help to identify the bug. Use the Issues tracker on GitHub to report issues with the software and to request feature enchancements. Please do not email Dr. Lindsay directly with bugs. | 2022-06-27 18:53:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.19448800384998322, "perplexity": 13109.998053242298}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103337962.22/warc/CC-MAIN-20220627164834-20220627194834-00218.warc.gz"} |
https://covid19-data.nist.gov/pid/rest/local/paper/comix_comparing_mixing_patterns_in_the_belgian_population_during_and_after_lockdown | ## comix comparing mixing patterns in the belgian population during and after lockdown CORD-Papers-2022-06-02 (Version 1)
Title: CoMix: comparing mixing patterns in the Belgian population during and after lockdown The COVID-19 pandemic has shown how a newly emergent communicable disease can lay considerable burden on public health. To avoid system collapse governments have resorted to several social distancing measures. In Belgium this included a lockdown and a following period of phased re-opening. A representative sample of Belgian adults was asked about their contact behaviour from mid-April to the beginning of August during different stages of the intervention measures in Belgium. Use of personal protection equipment (face masks) and compliance to hygienic measures was also reported. We estimated the expected reproduction number computing the ratio of [Formula: see text] with respect to pre-pandemic data. During the first two waves (the first month) of the survey the reduction in the average number of contacts was around 80% and was quite consistent across all age-classes. The average number of contacts increased over time particularly for the younger age classes still remaining significantly lower than pre-pandemic values. From the end of May to the end of July the estimated reproduction number has a median value larger than one although with a wide dispersion. Estimated [Formula: see text] fell below one again at the beginning of August. We have shown how a rapidly deployed survey can measure compliance to social distancing and assess its impact on COVID-19 spread. Monitoring the effectiveness of social distancing recommendations is of paramount importance to avoid further waves of COVID-19. 2020-12-14 Sci Rep 10.1038/s41598-020-78540-7 http://doi.org/10.1038/s41598-020-78540-7 Coletti Pietro https://covid19-data.nist.gov/pid/rest/local/author/coletti_pietro Wambua James https://covid19-data.nist.gov/pid/rest/local/author/wambua_james Gimma Amy https://covid19-data.nist.gov/pid/rest/local/author/gimma_amy Willem Lander https://covid19-data.nist.gov/pid/rest/local/author/willem_lander Vercruysse Sarah https://covid19-data.nist.gov/pid/rest/local/author/vercruysse_sarah Vanhoutte Bieke https://covid19-data.nist.gov/pid/rest/local/author/vanhoutte_bieke Jarvis Christopher I https://covid19-data.nist.gov/pid/rest/local/author/jarvis_christopher_i Van Zandvoort Kevin https://covid19-data.nist.gov/pid/rest/local/author/van_zandvoort_kevin Edmunds John https://covid19-data.nist.gov/pid/rest/local/author/edmunds_john Beutels Philippe https://covid19-data.nist.gov/pid/rest/local/author/beutels_philippe Hens Niel https://covid19-data.nist.gov/pid/rest/local/author/hens_niel 3d35638e8ce7a35203afd21d4187048dcb8b299d cc-by https://creativecommons.org/licenses/by/4.0/ Medline; PMC https://www.medline.com/https://www.ncbi.nlm.nih.gov/pubmed/ 33318521 https://www.ncbi.nlm.nih.gov/pubmed/33318521 PMC7736856 https://www.ncbi.nlm.nih.gov/pmc/articles/PMC7736856 https://doi.org/10.1038/s41598-020-78540-7 https://www.ncbi.nlm.nih.gov/pubmed/33318521/ TRUE lockdown COVID-19 Shanghai 5 face-masks post-lockdown direct/ Zenodo 31 COVID-19 children line Brussels and Netherlands 7 EC UZA 20/13/147 lockdown globe contacts US 9 UK matrix Shanghai and participants 5-17 phone/tablet Face coronavirus UK 6 15,26 close-contact Wuhan households social contacts France S15 Wuhan 5 multi-wave SARS-CoV-2 www.nature.com/scientificreports/ supermarkets/shops contact matrix www.nature.com/scientificreports/ Figure 3 (= people children contacts SocialMixr age-values ses/by/4.0/. line Innovations Programme-Project EpiPose http://creat iveco mmons COVID-19's Public Health First 5000 Characters:The COVID-19 pandemic has shown how a newly emergent communicable disease can lay considerable burden on public health. To avoid system collapse, governments have resorted to several social distancing measures. In Belgium, this included a lockdown and a following period of phased re-opening. A representative sample of Belgian adults was asked about their contact behaviour from mid-April to the beginning of August, during different stages of the intervention measures in Belgium. Use of personal protection equipment (face masks) and compliance to hygienic measures was also reported. We estimated the expected reproduction number computing the ratio of R 0 with respect to pre-pandemic data. During the first two waves (the first month) of the survey, the reduction in the average number of contacts was around 80% and was quite consistent across all age-classes. The average number of contacts increased over time, particularly for the younger age classes, still remaining significantly lower than pre-pandemic values. From the end of May to the end of July , the estimated reproduction number has a median value larger than one, although with a wide dispersion. Estimated R 0 fell below one again at the beginning of August. We have shown how a rapidly deployed survey can measure compliance to social distancing and assess its impact on COVID-19 spread. Monitoring the effectiveness of social distancing recommendations is of paramount importance to avoid further waves of COVID-19. OPEN The COVID-19 pandemic due to the novel coronavirus (SARS-CoV-2) has shown how newly emerging infectious diseases can lay considerable burden on public health and social economic welfare of the society. Since its emergence, over million confirmed cases and deaths have been recorded as of 2020 1 . In the absence of established pharmaceutical interventions, many countries across the globe have resorted to non-pharmaceutical interventions, advocacy of proper hygienic measures (hand washing, sanitizing), as well as promotion of wide-spread usage of masks to help combat the spread of this disease. However, sustainability of some of the imposed measures is infeasible in the long term, due to an urgent need to returning back to normal social life as well as rekindling the economy. Thus governments have been prompted to lift some of the measures in a phased manner whilst enforcing new/existing rules such as wearing masks in designated places such as in public transport, hospitals, schools, workplaces and other places that attract large crowds and gatherings. As COVID-19 is primarily transmitted through close-contact interaction with infected individuals 2 , data on social contacts is indispensable in informing mathematical modeling studies being employed to explore the evolution of this disease. The last decade of research in infectious disease modeling has shown how quantifying contact patterns is crucial to capture disease dynamics 3 . However, social contact data capturing behavioral changes in the population during and across different stages of an epidemic is mostly lacking and mathematical models need to rely on various assumptions, which might be unverifiable. This raises validity concerns on their appropriateness in guiding decision making. Thus, as many governments are carefully monitoring the situation to avoid further waves of COVID-19, continual data collection is vitally important to closely monitor changes in social mixing. This can provide insights Scientific Reports | (2020) 10:21885 | https://doi.org/10.1038/s41598-020-78540-7 www.nature.com/scientificreports/ on the impact of different intervention measures as well as help in real-time management of the COVID-19 crisis, together with other insights from social and behavioral sciences 4 . Studies comparing social contact patterns before and during the COVID-19 pandemic have been reported for Wuhan and Shanghai 5 , the UK 6 , the Netherlands 7 , Luxembourg 8 , the US 9 and in multiple countries (Belgium, France, Germany, Italy, the Netherlands, Spain, the UK, and the US) 10 . The overall reduction in the total number of contacts made by individuals ranged from 48% to 85%, stressing once again the importance of quantifying the impact of social distancing separately for each country. Also, although little variations in the number of contacts over time were measured 10 up to mid-April, this may change as countries relieve stricter measures and social interactions need to adjust to the new post-lockdown reality 8 . In this paper, we present results from a longitudinal survey of the adult population in Belgium, representative by age, gender and region of residence. The survey involves multiple waves of data collection, and is part of a wider study to look at changes in contact patterns across European countries (see e.g. UK 6 ). Here, we present results for eight waves (= 16 weeks). We quantify the changes in social contact patterns comparing pre-pandemic, lockdown and post-lockdown period \documentclass[12pt]{minimal} UK SARS-CoV-2 sciences4 US)10 time25 initiative6 EC UZA 20/13/147 contact matrix Shanghai measured10 globe households people children21 adults19 S15 US9 lockdown \documentclass[12pt]{minimal children contacts mandatory23 multi-wave number15 place22 survey22 face-masks France phone/tablet contacts July14 Netherlands7 Zenodo31 bias3,24 Wuhan line children Wuhan5 's Brussels and reality8 Luxembourg8 close-contact participants age-values hypothesis"15,26 in29 19–65 individuals2 0.02813 SocialMixr matrix supermarkets/shops restaurants conversational contacts individuals UK6 Shanghai5 face Fig. 7 post-lockdown coronavirus COVID-19 First 5000 Characters:The COVID-19 pandemic due to the novel coronavirus (SARS-CoV-2) has shown how newly emerging infectious diseases can lay considerable burden on public health and social economic welfare of the society. Since its emergence, over million confirmed cases and deaths have been recorded as of 20201. In the absence of established pharmaceutical interventions, many countries across the globe have resorted to non-pharmaceutical interventions, advocacy of proper hygienic measures (hand washing, sanitizing), as well as promotion of wide-spread usage of masks to help combat the spread of this disease. However, sustainability of some of the imposed measures is infeasible in the long term, due to an urgent need to returning back to normal social life as well as rekindling the economy. Thus governments have been prompted to lift some of the measures in a phased manner whilst enforcing new/existing rules such as wearing masks in designated places such as in public transport, hospitals, schools, workplaces and other places that attract large crowds and gatherings. As COVID-19 is primarily transmitted through close-contact interaction with infected individuals2, data on social contacts is indispensable in informing mathematical modeling studies being employed to explore the evolution of this disease. The last decade of research in infectious disease modeling has shown how quantifying contact patterns is crucial to capture disease dynamics3. However, social contact data capturing behavioral changes in the population during and across different stages of an epidemic is mostly lacking and mathematical models need to rely on various assumptions, which might be unverifiable. This raises validity concerns on their appropriateness in guiding decision making. Thus, as many governments are carefully monitoring the situation to avoid further waves of COVID-19, continual data collection is vitally important to closely monitor changes in social mixing. This can provide insights on the impact of different intervention measures as well as help in real-time management of the COVID-19 crisis, together with other insights from social and behavioral sciences4. Studies comparing social contact patterns before and during the COVID-19 pandemic have been reported for Wuhan and Shanghai5, the UK6, the Netherlands7 , Luxembourg8, the US9 and in multiple countries (Belgium, France, Germany, Italy, the Netherlands, Spain, the UK, and the US)10. The overall reduction in the total number of contacts made by individuals ranged from 48% to 85%, stressing once again the importance of quantifying the impact of social distancing separately for each country. Also, although little variations in the number of contacts over time were measured10 up to mid-April, this may change as countries relieve stricter measures and social interactions need to adjust to the new post-lockdown reality8. In this paper, we present results from a longitudinal survey of the adult population in Belgium, representative by age, gender and region of residence. The survey involves multiple waves of data collection, and is part of a wider study to look at changes in contact patterns across European countries (see e.g. UK6). Here, we present results for eight waves (= 16 weeks). We quantify the changes in social contact patterns comparing pre-pandemic, lockdown and post-lockdown periods and its impact on the transmission dynamics of COVID-19 based on the changes in the basic reproduction number relying on the next generation principle. We use a published survey of the Flemish region (Belgium) conducted in 201011,12 as reference for the pre-pandemic social mixing. Also, we assess the uptake of face mask wearing and adherence to hygienic measures in the population over time. During the first wave, 1542 participants took part in the survey, divided among 732 males (47.5%) and 810 females (52.5%) (Table 1). Table S2 presents information on participation rates for each wave. The average participant's age was 48.4 years (\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\hbox {standard deviation (sd)} =16.3\hbox { years}$$\end{document}standard deviation (sd)=16.3years), with a median age of 50 years, and an inter-quartile range (IQR) of [35–65]. The average household size was 2.8 (\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\hbox {sd} = 1.4$$\end{document}sd=1.4), IQR [2–4] with a maximum household size of 10. In total, data on 4290 household members, including the participants, was collected. Nearly half of the participants were living with children (51. document_parses/pdf_json/3d35638e8ce7a35203afd21d4187048dcb8b299d.json document_parses/pmc_json/PMC7736856.xml.json comix_comparing_mixing_patterns_in_the_belgian_population_during_and_after_lockdown | 2022-08-15 19:30:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4411126971244812, "perplexity": 4918.498042379544}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882572198.93/warc/CC-MAIN-20220815175725-20220815205725-00485.warc.gz"} |
http://www.scholarpedia.org/article/An_introduction_to_Lie_algebra_cohomology/Lecture_8 | # User:Jan A. Sanders/An introduction to Lie algebra cohomology/Lecture 8
## The trace and Killing form
Let $$R$$ be $$\mathbb{C}$$ and $$\dim_\mathbb{C}\mathfrak{a}<\infty\ .$$ Then define $$K_\mathfrak{a}\in C^2(\mathfrak{g},\mathbb{C})$$ by $K_\mathfrak{a}(x,y)=\mathrm{tr}(d_1(x) d_1(y))$ In the case $$\mathfrak{a}=\mathfrak{g}$$ and $$d_1=\mathrm{ad}\ ,$$ this is called the Killing form. In general, one calls $$K_\mathfrak{a}$$ the trace form.
### example - of a trace form
Let $$\mathfrak{g}=\mathfrak{sl}_2$$ and $$\mathfrak{a}=\R^2\ ,$$ with the standard representation (see Lecture 1). Then $K_{\R^2}(M,M)=0, \quad K_{\R^2}(M,N)=1,\quad K_{\R^2}(M,H)=0,$ $K_{\R^2}(N,M)=1, \quad K_{\R^2}(N,N)=0,\quad K_{\R^2}(N,H)=0,$ $K_{\R^2}(H,M)=0, \quad K_{\R^2}(H,N)=0,\quad K_{\R^2}(H,H)=2.$
### proposition - trace form symmetric
$$K_\mathfrak{a}$$ is symmetric.
### proof
This follows from $$\mathrm{tr}(AB)=\mathrm{tr}(BA)$$.$$\square$$
### proposition - trace form invariant
$$K_\mathfrak{a}$$ is $$\mathfrak{g}$$-invariant, that is, $$K_\mathfrak{a}\in C^2(\mathfrak{g},\mathbb{C})^\mathfrak{g}\ .$$
### proof
Given the trivial action of $$\mathfrak{g}$$ on $$\mathbb{C}\ ,$$ one has $d_1^{2}(x)K_\mathfrak{a}(y,z)=-K_\mathfrak{a}([x,y],z)-K_\mathfrak{a}(y,[x,z])\ :$ $=-\mathrm{tr}(d_1([x,y]) d_1(z))-\mathrm{tr}(d_1(y) d_1([x,z]))\ :$ $=-\mathrm{tr}(d_1(x) d_1(y) d_1(z))+\mathrm{tr}(d_1(y) d_1(x) d_1(z)) -\mathrm{tr}(d_1(y) d_1(x)d_1(z))+\mathrm{tr}(d_1(y) d_1(z)d_1(x))\ :$ $=0$
### proposition - $$d^2 K_\mathfrak{a}$$ antisymmetric
$d^2 K_\mathfrak{a}\in C_{\wedge}^3(\mathfrak{g},\mathbb{C})$
### proof
From the $$\mathfrak{g}$$-invariance it follows that $d^2 K_\mathfrak{a}(x,y,z)=K_\mathfrak{a}(x,[y,z])$ Furthermore, $K_\mathfrak{a}(x,[z,y])=-K_\mathfrak{a}(x,[y,z])$ and $K_\mathfrak{a}(z,[x,y])=-K_\mathfrak{a}(z,[y,x])=K_\mathfrak{a}([y,z],x)=K_\mathfrak{a}(x,[y,z])$$$\square$$
### corollary - nontrivial third cohomology
Let $$\mathfrak{g}$$ be a Lie algebra. Then $[d^2 K_\mathfrak{a}]\in H_{\wedge}^3(\mathfrak{g},\mathbb{C})$ Observe that this class is not trivial, since $$K_\mathfrak{a}$$ is symmetric, not antisymmetric.
### musical maps
Let $$\mathfrak{g}^\star=C^1(\mathfrak{g},\mathbb{C})$$ and define $$\flat: \mathfrak{g}\rightarrow \mathfrak{g}^\star$$ by $\flat(x)(y)=K_\mathfrak{a}(x,y)$
### proposition
$\flat\in Hom_\mathfrak{g}(\mathfrak{g},\mathfrak{g}^\star)$
### proof
$\flat([x,y])(z)=K_\mathfrak{a}([x,y],z)\ :$ $=-K_\mathfrak{a}(y,[x,z])\ :$ $=-\flat(y)([x,z])\ :$ $=d_1^{1}(x)\flat(y)(z)$ or $$\flat([x,y])=d_1^{1}(x)\flat(y)\quad \square\ .$$
Define $\sharp:\mathfrak{g}^\star\rightarrow \mathfrak{g}$ by $K_\mathfrak{a}(\sharp(c_1),y)=c_1(y)$ Then $K_\mathfrak{a}(x,y)=\flat(x)(y)=K_\mathfrak{a}(\sharp(\flat(x)),y)\ ,$ or $$x-\sharp(\flat(x))\in \ker K_\mathfrak{a}\ .$$
### proposition
$$\ker K_\mathfrak{a}$$ is an ideal.
### proof
Let $$y\in\ker K_\mathfrak{a}\ ,$$ that is $$K_\mathfrak{a}(x,y)=0$$ for all $$x\in\mathfrak{g}\ .$$
Then it follows from the invariance of $$K_\mathfrak{a}$$ that $K_\mathfrak{a}([y,x],z)+K_\mathfrak{a}(y,[x,z])=0$ and therefore $$K_\mathfrak{a}([y,x],z)=0$$ for all $$z\in\mathfrak{g}\ .$$
This shows that $$[\mathfrak{g},\ker K_\mathfrak{a}]\subset \ker K_\mathfrak{a}\ .$$
The statement that $$[\ker K_\mathfrak{a},\mathfrak{g}]\subset \ker K_\mathfrak{a}$$ follows by a symmetry argument.
### definition
A Lie algebra $$\mathfrak{g}$$ is called simple if $$[\mathfrak{g},\mathfrak{g}]\neq 0$$ and $$\mathfrak{g}$$ contains no ideals besides $$0$$ and itself.
### proposition - simple Lie algebra
If $$\mathfrak{g}$$ is simple, then $$\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]\ .$$
### proof
$$[\mathfrak{g},\mathfrak{g}]\neq 0$$ is an ideal, so it must equal $$[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}\ .$$
### proposition
If $$K_\mathfrak{a}$$ is nonzero, and $$\mathfrak{g}$$ is simple, then $$\flat$$ is injective.
### proof
Let $$x\in\ker\flat\ .$$
Then $$0=\flat(x)(y)=K_\mathfrak{a}(x,y)$$ for all $$y\in\mathfrak{g}\ ,$$ that is, $$x\in \ker K_\mathfrak{a}\ .$$
But $$\ker K_\mathfrak{a}$$ must be zero, so $$x=0\ .$$
### proposition
Let $$\mathfrak{h}$$ be an ideal in $$\mathfrak{g}\ .$$ Define $\mathfrak{h}^\perp=\{x\in\mathfrak{g}|K_\mathfrak{g}(x,y)=0 \quad \forall y\in \mathfrak{h}\}$ Then $$\mathfrak{h}^\perp$$ is an ideal in $$\mathfrak{g}\ .$$
### proof
Let $$g\in\mathfrak{g}\ ,$$ $$h\in\mathfrak{h}$$ and $$k\in\mathfrak{h}^\perp\ .$$ Then $K_\mathfrak{g}([g,k],h)=-K_\mathfrak{g}(k,[g,h])=0$ This shows that $$[\mathfrak{g},\mathfrak{h}^\perp]\subset \mathfrak{h}^\perp$$ and similarly $$[\mathfrak{h}^\perp,\mathfrak{g}]\subset \mathfrak{h}^\perp\ .$$
### definition - derived series, solvable
One defines a series of ideals of $$\mathfrak{g}\ ,$$ the derived series, as follows. $\mathfrak{g}^{(0)}=\mathfrak{g}$ $\mathfrak{g}^{(i+1)}=[\mathfrak{g}^{(i)},\mathfrak{g}^{(i)}]$ If, for some $$n\in\mathbb{N}\ ,$$ $$\mathfrak{g}^{(n)}=0$$ then $$\mathfrak{g}$$ is called solvable.
### well defined
$$\mathfrak{g}^{(0)}$$ is an ideal in $$\mathfrak{g}\ .$$ Suppose that $$\mathfrak{g}^{(i)}$$ is an ideal for $$i=0,\dots,n\ .$$ Then $[\mathfrak{g},\mathfrak{g}^{(n+1)}]=[\mathfrak{g},[\mathfrak{g}^{(n)},\mathfrak{g}^{(n)}]]\ :$ $\subset [[\mathfrak{g},\mathfrak{g}^{(n)}],\mathfrak{g}^{(n)}]+[\mathfrak{g}^{(n)},[\mathfrak{g},\mathfrak{g}^{(n)}]]\ :$ $\subset [\mathfrak{g}^{(n)},\mathfrak{g}^{(n)}]=\mathfrak{g}^{(n+1)}$ The inclusion $$[\mathfrak{g}^{(n+1)},\mathfrak{g}]\subset \mathfrak{g}^{(n+1)}$$ follows in a similar way. By induction it follows that all the $$g^{(i)}$$'s are ideals in $$\mathfrak{g}$$
### corollary
For $$i\leq j \ ,$$ $$\mathfrak{g}^{(j)}$$ is an ideal in $$\mathfrak{g}^{(i)}\ .$$
### remark
If $$\mathfrak{g}$$ is solvable (that is, $$\mathfrak{g}^{(n)}=0$$ for some $$n$$), then it contains an abelian ideal (namely $$\mathfrak{g}^{(n-1)}$$).
### proposition - solvable
If $$\mathfrak{g}$$ is solvable, then all its subalgebras and homomorphic images are.
### proof
Let $$\mathfrak{h}$$ be a subalgebra.
Then $$\mathfrak{h}^{(0)}\subset\mathfrak{g}^{(0)}\ .$$
Assume $$\mathfrak{h}^{(i)}\subset\mathfrak{g}^{(i)}\ .$$
Then $\mathfrak{h}^{(i+1)}=[\mathfrak{h}^{(i)},\mathfrak{h}^{(i)}]\subset [\mathfrak{g}^{(i)},\mathfrak{g}^{(i)}]=\mathfrak{g}^{(i+1)}$ and the statement is proved by induction.
Similarly, let $$\phi:\mathfrak{g}\rightarrow \mathfrak{h}$$ be surjective, and assume $$\phi:\mathfrak{g}^{(i)}\rightarrow \mathfrak{h}^{(i)}$$ to be surjective.
Then $\phi(\mathfrak{g}^{(i+1)})=\phi([\mathfrak{g}^{(i)},\mathfrak{g}^{(i)}])=[\phi(\mathfrak{g}^{(i)}),\phi(\mathfrak{g}^{(i)})]= [\mathfrak{h}^{(i)},\mathfrak{h}^{(i)}]=\mathfrak{h}^{(i+1)}$
### proposition - solvable quotient
If $$\mathfrak{h}$$ is a solvable ideal such that $$\mathfrak{g}/\mathfrak{h}$$ is solvable, then $$\mathfrak{g}$$ is solvable.
### proof
Say $$(\mathfrak{g}/\mathfrak{h})^{(n)}=0\ .$$
Let $$\pi:\mathfrak{g}\rightarrow \mathfrak{g}/\mathfrak{h}$$ be the canonical projection.
Then $$\pi(\mathfrak{g}^{(n)})=(\mathfrak{g}/\mathfrak{h})^{(n)}=0$$ or $$\mathfrak{g}^{(n)}\subset \mathfrak{h}\ .$$
Since $$\mathfrak{h}^{(m)}=0\ ,$$ $$\mathfrak{g}^{(n+m)}=(\mathfrak{g}^{(n)})^{(m)}\subset \mathfrak{h}^{(m)}=0\ ,$$ implying the statement.
### proposition
If $$\mathfrak{h}, \mathfrak{k}$$ are solvable ideals of $$\mathfrak{g}\ ,$$ then so is $$\mathfrak{h}+\mathfrak{k}\ .$$
### proof
One has $(\mathfrak{h}+ \mathfrak{k})/\mathfrak{k}\equiv \mathfrak{h}/(\mathfrak{h}\cap \mathfrak{k})$ Since $$\mathfrak{h}$$ is solvable, so is $$\mathfrak{h}/(\mathfrak{h}\cap \mathfrak{k}) \ .$$ But this implies that $$\mathfrak{h}+\mathfrak{k}\ ,$$ since $$\mathfrak{k}$$ is solvable.
If $$\dim \mathfrak{g}<\infty\ ,$$ there exists a unique maximal solvable ideal in $$\mathfrak{g}\ ,$$ the radical of $$\mathfrak{g}\ ,$$ denoted by $$\mathrm{Rad\ }\mathfrak{g}\ .$$
### proof
Let $$\mathfrak{s}$$ be a maximal solvable ideal in $$\mathfrak{g}\ .$$
Suppose $$\mathfrak{h}$$ is another solvable ideal.
Then $$\mathfrak{s}+\mathfrak{h}\supset \mathfrak{s}$$ is solvable, and by the maximality, $$\mathfrak{s}+\mathfrak{h}= \mathfrak{s}\ ,$$ that is, $$\mathfrak{h}\subset\mathfrak{s}\ .$$
### definition - semisimple
A Lie algebra $$\mathfrak{g}$$ is called semisimple if $$\mathrm{Rad\ }\mathfrak{g}=0\ .$$
### proposition - simple implies semisimple
If $$\mathfrak{g}$$ is simple, it is semisimple
### proof
For a simple Leibniz algebra the derived series is stationary, that is, $$\mathfrak{g}^{(i)}=\mathfrak{g}$$ for all $$i\in\mathbb{N}\ .$$
The only other possible ideal is $$0\ ,$$ so this must be $$\mathrm{Rad\ }\mathfrak{g}\ .$$
### proposition
$$\mathfrak{g}/\mathrm{Rad\ }\mathfrak{g}$$ is semisimple.
### proof
Let $$[\mathfrak{h}]$$ be a nonzero solvable ideal in $$\mathfrak{g}/\mathrm{Rad\ }\mathfrak{g}\ .$$
Then $$\mathfrak{h}+\mathrm{Rad\ }\mathfrak{g}$$ strictly contains $$\mathrm{Rad\ }\mathfrak{g}\ ,$$ which is in contradiction with its maximality.
Thus $$\mathfrak{h}\subset \mathrm{Rad\ }\mathfrak{g}\ ,$$ that is, $$[\mathfrak{h}]$$ is the zero ideal in $$\mathfrak{g}/\mathrm{Rad\ }\mathfrak{g}\ .$$
### proposition
If $$\ker K_\mathfrak{g}=0\ ,$$ then $$\mathfrak{g}$$ is semisimple.
### proof
Let $$\mathfrak{h}$$ be an abelian ideal of $$\mathfrak{g}\ .$$ Take $$h\in\mathfrak{h}, g\in\mathfrak{g}\ .$$ Then $$ad(h)ad(g)$$ maps $$\mathfrak{g}$$ to $$\mathfrak{h}\ .$$ Thus $$(ad(h)ad(g))^2=0\ .$$ This implies that $K_\mathfrak{g}(h,g)=\mathrm{tr}(ad(h)ad(g))=0$ In other words, $$\mathfrak{h}\subset\ker K_\mathfrak{g}=0\ .$$ If there are no abelian ideals, then there are no solvable ideals besides $$0\ ,$$ that is, $$\mathfrak{g}$$ is semisimple.
### theorem - common eigenvector
Let $$\mathfrak{g}$$ be a solvable subalgebra of $$\mathfrak{gl}(\mathfrak{a})\ ,$$ $$\dim\mathfrak{a}<\infty\ .$$ If $$\mathfrak{a}\neq 0\ ,$$ then $$\mathfrak{a}$$ contains a common eigenvector for all endomorphisms in $$\mathfrak{g}\ .$$
### proof
Induction on $$\dim\mathfrak{g}\ .$$ Since $$\mathfrak{g}$$ is solvable, it properly contains $$\mathfrak{g}^{(1)}=[\mathfrak{g},\mathfrak{g}]\ ,$$ otherwise $$\mathfrak{g}^{(i)}=\mathfrak{g}$$ for $$i\in\mathbb{N}\ .$$
Since $$\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]$$ is abelian, subspaces are ideals.
Take a subspace of codimension one.
Then the inverse image $$\mathfrak{h}$$ in $$\mathfrak{g}$$ is an ideal of codimension one which includes $$[\mathfrak{g},\mathfrak{g}]\ .$$
$$\mathfrak{h}$$ is solvable, and by the induction assumption there exists a vector $$a\in\mathfrak{a}$$ such that $$a$$ is an eigenvector for each $$h\in\mathfrak{h}\ ,$$ that is, $h a=\lambda(h)a,\quad\lambda\in C^1(\mathfrak{h},\mathbb{C})$ (the exceptional case here is when $$\dim \mathfrak{h}=0\ .$$
In that case, $$\mathfrak{g}$$ onedimensional and abelian, so one takes an eigenvector of a generator of $$\mathfrak{g}$$). Let $\mathcal{W}=\{a\in\mathfrak{a}|x a=\lambda(x)a \quad \forall x\in \mathfrak{h}\}$ Now for $$x\in\mathfrak{g}$$ and $$y\in\mathfrak{h}$$ one finds $y x w=x y w-[x,y] w=\lambda(y) x w-\lambda([x,y])w\ .$ If one can prove that $$\lambda([x,y])=0$$ then $$\mathcal{W}$$ is invariant under the action of $$\mathfrak{g}\ .$$
Fix $$x\in \mathfrak{g}\ ,$$ $$w\in\mathcal{W}\ .$$
Let $$n>0$$ be the smallest integer such that $$w, xw, \dots, x^n w$$ are linearly dependent.
Let $$\mathcal{W}_0=0$$ and $$\mathcal{W}_i$$ be the subspace of $$\mathfrak{a}$$ spanned by $$w, xw,\dots, x^{i-1} w\ .$$
It follows that $$\dim\mathcal{W}_n=n$$ and $$W_{n+i}=W_n, i\geq 0\ .$$
Each $$\mathcal{W}_i$$ is invariant under $$y\in\mathfrak{h}\ .$$
The matrix of $$y$$ is upper triangular with eigenvalue $$\lambda(y)$$ on the diagonal.
This implies $$\mathrm{tr}_{\mathcal{W}_i}(y)=i\lambda(y)\ .$$
Since $$[x,y]\in\mathfrak{h}\ ,$$ one also has $\mathrm{tr}_{\mathcal{W}_n}([x,y])=i\lambda([x,y])$ Both $$x$$ and $$y$$ leave $$\mathcal{W}_n$$ invariant, so the trace of $$[x,y]$$ must be zero.
Thus $$n\lambda([x,y])=0\ .$$
This shows that $$\mathcal{W}$$ is invariant under the action of $$\mathfrak{g}\ .$$
Write $$\mathfrak{g}=\mathfrak{h}+\mathbb{C} z\ .$$ Let $$w_0 \in\mathcal{W}$$ be an eigenvector of $$z$$ (acting on $$\mathcal{W}$$).
Then $$w_0$$ is a common eigenvector of $$\mathfrak{g}\ .$$
### definition - flag
Let $$\mathfrak{a}$$ be a finite dimensional vectorspace ($$\dim\mathfrak{a}=n$$). A flag is a chain of subspaces $0=\mathfrak{a}_0\subset\mathfrak{a}_1\subset\dots\subset\mathfrak{a}_n=\mathfrak{a},\quad \dim\mathfrak{a}_i=i$ If $$x\in\mathrm{End}(\mathfrak{a})\ ,$$ one says that $$x$$ leaves the flag invariant if $$x \mathfrak{a}_i\subset \mathfrak{a}_i$$ for $$i=1,\dots,n\ .$$
### theorem (Lie)
Let $$\mathfrak{g}$$ be a solvable subalgebra of $$\mathfrak{gl}(\mathfrak{a}), \dim\mathfrak{a}=n<\infty\ .$$ Then $$\mathfrak{g}$$ leaves a flag in $$\mathfrak{a}$$ invariant.
### proof
It follows from the proof above that there exists a codimension one $$\mathfrak{g}$$-invariant subspace.
Let that be $$\mathfrak{a}_{n-1}\ .$$
Repeat the argument starting with $$\mathfrak{a}_{n-1}$$ instead of $$\mathfrak{a}_{n}$$ and use induction.
### lemma - flag of ideals
Let $$\mathfrak{g}$$ be solvable. Then there exists a flag of ideals $0=\mathfrak{g}_0\subset\mathfrak{g}_1\subset\dots\subset\mathfrak{g}_n=\mathfrak{g},\quad \dim\mathfrak{g}_i=i$
### proof
Let $$d_1:\mathfrak{g}\rightarrow \mathfrak{gl}(\mathfrak{a})$$ be a finite dimensional representation of $$\mathfrak{g}\ .$$
Then $$d_1(\mathfrak{g})$$ is solvable, and stabilizes a flag in $$\mathfrak{a}\ .$$
Take $$\mathfrak{a}=\mathfrak{g}$$ and $$d_1=\mathrm{ad}\ ,$$ then the $$\mathfrak{g}_i$$ are ideals (since they are $$\mathfrak{g}$$-invariant) and they obey the flag condition.
### lemma
Let $$\mathfrak{g}$$ be solvable. Then $$x\in\mathfrak{g}^{(1)}$$ implies that $$\mathrm{ad}_\mathfrak{g}(x)$$ is nilpotent.
### proof
From the flag of ideals construct a basis. relative to this basis the matrix of $$\mathrm{ad}_\mathfrak{g}(y), y\in \mathfrak{g}\ ,$$ is upper triangular.
Thus the matrix of $$\mathrm{ad}_\mathfrak{g}(x), x\in \mathfrak{g}^{(1)}$$ is strictly upper triangular, and therefore nilpotent.
### remark
In the next lecture it is shown that this implies that $$\mathfrak{g}^{(1)}$$ is nilpotent (to be defined). | 2019-11-23 00:06:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9743638634681702, "perplexity": 254.36255403978402}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496672170.93/warc/CC-MAIN-20191122222322-20191123011322-00280.warc.gz"} |
https://zbmath.org/?q=an%3A0991.35059 | ## A Kato type theorem on zero viscosity limit of Navier-Stokes flows.(English)Zbl 0991.35059
The author presents a necessary and sufficient condition for the convergence of a solution $$u^{\nu}$$ of the incompressible Navier-Stokes equations in a spatial domain $$\Omega$$ to a solution $$u^0$$ of the Euler equations at vanishing viscosity. He proves that $$u^{\nu} \rightarrow u^0$$ as $$\nu\rightarrow 0$$ in $$C([0,T], L^2(\Omega))$$ iff there exist $$\delta (\nu)$$ such that $$\lim_{\nu\rightarrow 0} \nu/\delta (\nu)=0$$ and $$\lim_{\nu\rightarrow 0} \int_0^T \int_{\Gamma_{\delta}} |\nabla_t u^{\nu}_n|^2 dx dt=0$$, where $$\Gamma_{\delta}$$ is a $$\delta$$-neighborhood of the wall $$\partial \Omega$$, $$\nabla_t$$ denotes tangential derivatives, and $$u^{\nu}_n$$ denotes the normal component of the velocity.
### MSC:
35Q30 Navier-Stokes equations 76D05 Navier-Stokes equations for incompressible viscous fluids 35Q05 Euler-Poisson-Darboux equations
Full Text: | 2022-08-16 18:38:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.895318329334259, "perplexity": 209.4373802318338}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882572515.15/warc/CC-MAIN-20220816181215-20220816211215-00665.warc.gz"} |
https://artofproblemsolving.com/wiki/index.php?title=Ring_of_integers&diff=prev&oldid=20924 | # Difference between revisions of "Ring of integers"
Let $K$ be a finite algebraic field extension of $\mathbb{Q}$. Then the integral closure of ${\mathbb{Z}}$ in $K$, which we denote by $\mathfrak{o}_K$, is called the ring of integers of $K$. Rings of integers are always Dedekind domains with finite class numbers. | 2021-01-15 18:28:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9574864506721497, "perplexity": 79.7178517662498}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703495936.3/warc/CC-MAIN-20210115164417-20210115194417-00369.warc.gz"} |
https://docs.tigergraph.com/graph-ml/3.7/community-algorithms/strongly-connected-components | Strongly Connected Components
Supported Graph Characteristics
Unweighted edges Directed edges Undirected edges Homogeneous vertex types Heterogeneous vertex types
A strongly connected component (SCC) is a subgraph such that there is a path from any vertex to every other vertex. A graph can contain more than one separate SCC. This SCC algorithm finds the maximal SCCs within a graph.
Our implementation is based on the Divide-and-Conquer Strong Components (DCSC) algorithm:
• In each iteration, pick a pivot vertex `v` randomly.
• Find its descendant and predecessor sets:
• Descendant set `D_v` is the set of vertices reachable from `v` using regular edges.
• Predecessor set `P_v` is the set of vertices reachable from `v` using reverse edges.
• The intersection of these two sets is a strongly connected component `SCC_v`.
The graph can be partitioned into 4 sets:
• `SCC_v`
• Descendants `D_v` excluding `SCC_v`
• Predecessors `P_v` excluding `SCC_v`
• Remainders `R_v`.
It is proved that any SCC is a subset of one of the 4 sets.
Thus, we can divide the graph into different subsets and detect the SCCs independently and iteratively.
Notes
The implementation of this algorithm requires reverse edges for all directed edges considered in the graph.
The problem of this algorithm is unbalanced load and slow convergence when there are a lot of small SCCs, which is often the case in real-world use cases.
We added two trimming stages to improve the performance: Size-1 SCC Trimming and Weakly Connected Components.
References
DCSC Algorithm: Fleischer, Lisa K., Bruce Hendrickson, and Ali Pınar. "On identifying strongly connected components in parallel." International Parallel and Distributed Processing Symposium. Springer, Berlin, Heidelberg, 2000.
Size-1 SCC Trimming: Mclendon III, William, et al. "Finding strongly connected components in distributed graphs." Journal of Parallel and Distributed Computing 65.8 (2005): 901-910.
Weakly Connected Components: Hong, Sungpack, Nicole C. Rodia, and Kunle Olukotun. "On fast parallel detection of strongly connected components (SCC) in small-world graphs." Proceedings of the International Conference on High Performance Computing, Networking, Storage and Analysis. ACM, 2013.
Specifications
``````tg_scc (SET<STRING> v_type, SET<STRING> e_type, SET<STRING> rev_e_type,
INT top_k_dist, INT output_limit, INT max_iter = 500, INT iter_wcc = 5,
BOOL print_accum = TRUE, STRING attr= "", STRING file_path="")``````
Parameters
Parameter Description Default Value
`SET<STRING> v_type`
The vertex types to use
(empty set of strings)
`SET<STRING> e_type`
The edge types to use
(empty set of strings)
`SET<STRING> rev_e_type`
The reverse edge types to use
(empty set of strings)
`INT top_k_dist`
The top k results in the SCC distribution
N/A
`INT output_limit`
The maximum number of vertices to output in JSON format.
N/A
`INT max_iter`
The maximum number of iterations of the algorithm.
500
`INT iter_wcc`
Find weakly connected components for the active vertices in this iteration, since the largest SCCs are already found after several iterations. Usually a small number (3 to 10).
5
`BOOL print_accum`
If true, print output in JSON format to the standard output.
True
`STRING result_attr`
If not empty, store community values in `INT` format to this vertex attribute
(empty string)
`STRING file_path`
If not empty, write output to this file.
(empty string)
Output
Assigns a component id (INT) to each vertex, such that members of the same component have the same ID value.
Time complexity
This algorithm has a time complexity of O(k * d), where k is equal to the number of iterations and d is equal to the maximum component diameter.
Example
Consider a graph with Person vertices and Friend and Coworker edges. The following illustration shows a group of these vertices and edges. Not shown are a number of other vertices that have no connection to other vertices.
We run the SCC query with default values, making sure to include both types of edges in the query.
A portion of the JSON result is shown below.
``````[
{
"@@cluster_dist_heap": [
{
"csize": 9,
"num": 1
},
{
"csize": 1,
"num": 17
}
]
}
]``````
The `@@.cluster_dist_heap` object reports on the size distribution of SCCs.
There is one SCC with 9 vertices, and 17 SCCs with only 1 vertex in the graph. | 2022-12-03 05:57:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3164821267127991, "perplexity": 2728.6930610072313}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710924.83/warc/CC-MAIN-20221203043643-20221203073643-00022.warc.gz"} |
https://www.vandyke.com/products/vshell/docs/windows/Using_the_Windows_Command_S.htm | Using the Windows Command Shell
When performing remote administration of a Windows system, it is normal to think of using a graphical user interface. However, trying to force a bandwidth-intensive interface over a LAN, WAN or other network connection can lead to extremely slow or unreliable behavior of the interface.
Fortunately, the Windows command line allows you to perform almost all of the Windows tasks that you need to administer the system. The syntax in most cases is straightforward and easy to learn, and command-line administration is much more efficient remotely, because only a few characters are transmitted over the connection. The Windows command shell, which resembles MS-DOS, is a powerful environment and contains a strong complement of Win32 console utilities to support it including net config and net start/stop.
Many of the Windows integrated administrative facilities are grouped under the net command. Typing net at the command line will print out a list of these commands. Details about the use of any net command are available by typing:
net <command> /?
For example, to get a list of the commands and options available with the net config command, type:
net config /?
You can find additional information on command-line administration in the Windows Resource Guide available from Microsoft Press.
To help in the administration of your VShell server , VanDyke Software also ships two utilities with VShell. These utilities, Who.exe and Whoami.exe, provide you with information about who is connected to your system.
The Who.exe utility displays the following information about VShell connections:
User names
Connection IDs
Date and time the connection was made
IP Address from which the connection was made (using the -h option with the Who command will display the hostname instead of the IP address)
Below is an example of Who.exe output.
Bob 00093 Nov 13 20:55 192.168.0.54
Alice 00130 Nov 14 13:26 192.168.0.112
Dave 00099 Nov 14 08:08 192.168.0.32
Carol 00100 Nov 14 08:14 192.168.0.28
VShell also supports a who "--kill" command option which terminates the specified session. Only users with administrator privileges are allowed to use the --kill option. Below are three usage examples (the latter two examples can be useful if the session-id to be terminated occurs in both SSH2 and FTPS):
who --kill session-id
who --kill ssh2:session-id
who --kill ftps:session-id
The Whoami.exe utility displays the username with which you are currently connected.
Silent Installation of VShell
If you are installing VShell on a number of machines, it can be time-consuming to run the installation wizard on each machine. Or, if you are upgrading your software, you may want a quick way to install the software and keep all your current settings.
A silent installation allows you to bypass the wizard panels and run the installer non-interactively from the command line. When run silently, the installer accepts all default settings, or keeps all of your existing settings, if you are upgrading.
Note: The following example of how to silently install VShell 3.0 (or later) assumes that you have a copy of the installer already downloaded to a local c:\Installers directory. In the command, substitute the correct name of the executable (.exe) file that you are installing.
To silently install VShell, follow these steps:
1. Create an answer file. Silently installing VShell 3.0 (or later) requires the use of an answer file that will allow you to override the installer's default settings and control the reboot behavior of the installer. Here is an example answer file.
2. If you do not want the installer to reboot your machine at the end of the installation, add the following line to your answer file:
AUTO_REBOOT=NO
To allow the installer to reboot your machine, add the following line:
AUTO_REBOOT=YES
Note: Since Microsoft Windows only loads authentication modules during startup, the machine must be rebooted in order for the public-key authentication module (Vdspka10.dll) to be loaded and used by the Windows operating system, allowing public-key only authentication with VShell. Before reboot, you can still use combination of password and public-key authentication.
If you are upgrading, and the Vdspka10.dll file has not been changed by VanDyke Software since your last installation or upgrade, a reboot is not required. The Vdspka10.dll file does not change very often. If Vdspka10.dll has changed, and you do not reboot, public-key authentication may fail. Rebooting may solve this problem.
3. Save the answer file and copy it to each machine where you want to install VShell 3.0 (or later).
4. At a command prompt, type:
Exporting a VShell for Windows Configuration for Use with an Answer File
If you want to install the same configuration settings on a number of machines, follow these steps:
1. Configure VShell on one machine. Run Regedit and then export the following registry key:
HKEY_LOCAL_MACHINE\Software\VanDyke\VShell\Server
2. Copy the resulting .reg file (for example, “VShellconfig.reg") onto the other machines on which you will install VShell, or place the .reg file on a network drive to which every machine has access.
3. Modify the value of the REGISTRY_CONFIGURATION_FILE option in the VShell answer file to reference the path to the .reg file that exists either on the target machine or in a shared folder on the network. For example:
REGISTRY_CONFIGURATION_FILE=C:\vshellconfig.reg
4. You can now install VShell on all machines using the installation wizard or silently from the command line, for example: | 2018-01-19 19:28:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5252585411071777, "perplexity": 4282.343016139974}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084888113.39/warc/CC-MAIN-20180119184632-20180119204632-00072.warc.gz"} |
https://www.splessons.com/lesson/sebi-assistant-manager-phase-i-practice-test/ | # SEBI Assistant Manager Phase I Practice Test
5 Steps - 3 Clicks
# SEBI Assistant Manager Phase I Practice Test
### Introduction
Phase I Examination is important to qualify for the Phase II Examination. SEBI Assistant Manager Phase I Practice & Mock Test section allows candidates to practice the learned material to ensure the candidate understands the pattern of the exam and the expected questions that would appear in the actual test.
### Pattern
Stream Tests No. of
Questions
Maximum Marks Version Duration
General, Legal,
Information
Technology and
Engineering
General Awareness 40 40 Bilingual i.e.
Hindi and
English except
English Language.
120 minutes
English Language 40 40
Quantitative Aptitude 40 40
Test of Reasoning 40 40
Securities Market
40 40
Total 200 200
### Samples
1. India has invited Pakistan to visit sites of the Pakal Dul and Lower Kalnal hydro-electric projects on which river?
A. Yamuna
B. Chenab
C. Jhelum
D. None of the above
2. The 115th meeting of the India-Pakistan Permanent Indus Commission (PIC) was held where?
A. Lahore
C. Istanbul
D. None of the above
3. National Trust Centre would be created for certifying devices and applications for machine-to-machine communication by which commission?
A. Telephone Commission
B. Maritime Commission
C. Military Commission
D. None of the above
4. SBI Hikes Benchmark Lending Rate By how much?
A. 0.2 %
B. 0.4%
C. 0.3%
D. None of the above
5. GDP growth soars to how much in first quarter?
A. 8.2 %
B. 8.5%
C. 7.68%
D. None of the above
6. Who wrote the famous book – ‘We the people’?
A. T.N.Kaul
B. J.R.D. Tata
C. Khushwant Singh
D. Nani Palkhivala
7. Who is the author of the book ‘Forbidden Verses’?
A. Salman RushDie
B. Abu Nuwas
C. Ms. Taslima Nasrin
D. D.H. Lawrence
8. Which of the following books has been written by Vikram Seth?
A. My God Died Young
B. Islamic Bomb
C. Look Back in Anger
D. A Suitable Boy
9. Who wrote the line: ‘ A thing of beauty is a joy forever’?
A. John Keats
B. Robert Browing
C. P.B.Shelley
D. William Wordsworth
10. The Battle of Plassey was fought in
A. 1757
B. 1782
C. 1748
D. 1764
11. The territory of Porus who offered strong resistance to Alexander was situated between the rivers of
A. Sutlej and Beas
B. Jhelum and Chenab
C. Ravi and Chenab
D. Ganga and Yamuna
12. Tripitakas are sacred books of
A. Buddhists
B. Hindus
C. Jains
D. None of the above
13. The trident-shaped symbol of Buddhism does not represent
A. Nirvana
B. Sangha
C. Buddha
D. Dhamma
14. The Paithan (Jayakwadi) Hydro-electric project, completed with the help of Japan, is on the river
A. Ganga
B. Cauvery
D. Godavari
15. The percentage of irrigated land in India is about
A. 45
B. 65
C. 35
D. 25
16. The southernmost point of peninsular India, that is, Kanyakumari, is
A. north of Tropic of Cancer
B. south of the Equator
C. south of the Capricorn
D. north of the Equator
17. The pass located at the southern end of the Nilgiri Hills in south India is called
A. the Palghat gap
B. the Bhorghat pass
C. the Thalgat pass
D. the Bolan pass
18. The Yarlung Zangbo river, in India, is known as
A. Ganga
B. Indus
C. Brahmaputra
19. The Salal Project is on the river
A. Chenab
B. Jhelum
C. Ravi
D. Sutlej
20. The only zone in the country that produces gold is also rich in iron is
A. North-eastern zone
B. North-western zone
C. Southern zone
D. None of the above
21. The Parliament of India cannot be regarded as a sovereign body because
A. it can legislate only on subjects entrusted to the Centre by the Constitution
B. it has to operate within the limits prescribed by the Constitution
C. the Supreme Court can declare laws passed by parliament as unconstitutional if they contravene the provisions of the Constitution
D. All of the above
22. The name of the Laccadive, Minicoy and Amindivi islands was changed to Lakshadweep by an Act of Parliament in
A. 1970
B. 1971
C. 1972
D. 1973
23. The members of the Rajya Sabha are elected by
A. the people
B. Lok Sabha
C. elected members of the legislative assembly
D. elected members of the legislative council
24. The members of the panchayat are
A. nominated by the district officer
B. the electorates of the respective territorial constituencies
C. nominated by local self-government minister of the state
D. nominated by the block development organization
25. The power to decide an election petition is vested in the
A. Parliament
B. Supreme Court
C. High courts
D. Election Commission
26. The present Lok Sabha is the
A. 14th Lok Sabha
B. 15th Lok Sabha
C. 16th Lok Sabha
D. 17th Lok Sabha
27. Development expenditure of the Central government does not include
A. defence expenditure
B. expenditure on economic services
C. expenditure on social and community services
D. grant to states
28. ICICI is the name of a
A. chemical industry
B. bureau
C. corporation
D. financial institution
29. Gilt-edged market means
A. bullion market
B. market of government securities
C. market of guns
D. market of pure metals
30. In the last one decade, which one among the following sectors has attracted the highest foreign direct investment inflows into India?
A. Chemicals other than fertilizers
B. Services sector
C. Food processing
D. Telecommunication
31. The ratio of width of our National flag to its length is
A. 3:5
B. 2:3
C. 2:4
D. 2:4
32. Rabindranath Tagore’s ‘Jana Gana Mana’ has been adopted as India’s National Anthem. How many stanzas of the said song were adopted?
A. Only the first stanza
B. The whole song
C. Third and Fourth stanza
D. First and Second stanza
33. ‘Natya – Shastra’ the main source of India’s classical dances was written by
A. Nara Muni
B. Bharat Muni
C. Abhinav Gupt
D. Tandu Muni
34. ‘Dandia’ is a popular dance of
A. Punjab
B. Gujarat
D. Maharashtra
35. The words ‘Satyameva Jayate’ inscribed below the base plate of the emblem of India are taken from
A. Rigveda
B. Satpath Brahmana
D. Ramayana
36. Who was the first Indian to win the World Amateur Billiards title?
A. Geet Sethi
B. Wilson Jones
C. Michael Ferreira
D. Manoj Kothari
37. Former Australian captain Mark Taylor has had several nicknames over his playing career. Which of the following was NOT one of them?
A. Tubby
B. Stodge
C. Helium Bat
D. Stumpy
38. Which was the 1st non Test playing country to beat India in an international match?
B. Sri Lanka
C. Zimbabwe
D. East Africa
39. Track and field star Carl Lewis won how many gold medals at the 1984 Olympic games?
A. Two
B. Three
C. Four
D. Eight
40. Who is the first Indian woman to win an Asian Games gold in 400m run?
A. M.L.Valsamma
B. P.T.Usha
C. Kamaljit Sandhu
D. K.Malleshwari
Direction(1-10):So let us pose the question differently — who suffers in the absence of a Uniform Civil Code? Is it Muslim women, victims of polygamy and triple talaq, as Hindutvavadi wisdom has it? But for decades, feminist legal practice has successfully used both the Protection of Women from Domestic Violence Act, 2005 — that is available to all Indian citizens regardless of religious identity — as well as the Muslim Women (Protection of Rights on Divorce) Act, 1986, to deal with polygamy and triple talaq, and to obtain maintenance, child custody and rights to matrimonial home for countless Muslim women. In addition, feminist legal activists have used the landmark Shamim Ara v. State of U.P. (2002) ruling to buttress their claim that arbitrary triple talaq is invalid.
Moreover, polygamy is not exclusive to Muslims. Hindu men are polygamous too, except that because polygamy is legally banned in Hindu law, subsequent wives have no legal standing and no protection under the law. Under Sharia law, on the contrary, subsequent wives have rights and husbands have obligations towards them. If gender justice is the value we espouse, rather than monogamy per se, we would be thinking about how to protect “wives” in the patriarchal institution of marriage. “Wives” are produced through the institution of compulsory heterosexual marriage, the basis of which is the sexual division of labour. This institution is sustained by the productive and reproductive labour of women, and almost all women are exclusively trained to be wives alone.
Thus, when a marriage fails to fulfil its patriarchal promise of security in return for that labour, all that most women are left with is the capacity for unskilled labour. Or they remain trapped in marriage with children to provide for, while men marry again, legally or otherwise, producing still more dependent, exploited wives and children for whom they take no responsibility. If gender justice is the point of legal reforms, the centrality and power of the compulsory heterosexual, patriarchal marriage, and the damage it can do to women, is what must be mitigated. This would mean recognising the reality of multiple “wives” as a common practice across communities, and the protection of the rights of all women in such relationships.In this sense, recent Supreme Court rulings that have granted rights to second wives in Hindumarriages dilute the legal standing of monogamy for Hindus but empower women.
A survey conducted by the Bharatiya Muslim Mahila Andolan, a significant voice in the debate, found that more than 90 per cent of Muslim women in India want a ban on “triple talaq” and polygamy in Muslim Personal Law. That is, the demand is made within the framework of codifying Muslim Personal Law, not in favour of a Uniform Civil Code, partly because there is no clarity on what a uniform code would look like, but also because the demand comes from clearly Hindutvavadi quarters which have shown that both women and minorities are expendable for them.
1. What according to the passage founded in a survey conducted by the Bharatiya Muslim Mahila Andolan?
A. polygamy is not only exclusive to Muslims
B. “Wives” are produced through the institution of compulsory heterosexual marriage, the basis
C. more than 90 per cent of Muslim women in India want a ban on “triple talaq” and polygamy
D. Both A and B
Explanation – more than 90 per cent of Muslim women in India want a ban on “triple talaq” and polygamy
in Muslim Personal Law
2. Which among the following is TRUE according to the passage given above?
A. feminist legal activists have used the landmark Shamim Ara v. State of U.P. (2012) ruling to buttress their claim that arbitrary triple talaq is invalid
B. Domestic Violence Act, 2006 that is available to all Indian citizens regardless of religious
identity
C. polygamy is legally banned in Hindu law
D. Both A and B
Explanation – polygamy is legally banned in Hindu law
3. What is the recent Supreme Court rule?
A. Under Sharia law, on the contrary, subsequent wives have rights and husbands have
obligations towards them
B. granted rights to second wives in Hindu marriages dilute the legal standing of monogamy for Hindus but empower women
C. when a marriage fails to fulfil its patriarchal promise of security in return for that labour, all that most women are left with is the capacity for unskilled labour
D. Both A and C.
Explanation – granted rights to second wives in Hindu marriages dilute the legal standing of monogamy for Hindus but empower women
4. Which of the following is/are true regarding Muslim Women (Protection of Rights on Divorce) Act, 1986?
A. it deal with polygamy and triple talaq
B. wives have no legal standing and no protection under the law
C. and to obtain maintenance child custody and rights to matrimonial home for countless
Muslim women
D. both A and C
Explanation – both A and C
5. Which of the following would be a suitable title of the passage?
A. The real issue of gender justice
B. Lessons from the Goa experience
C. A stick to beat Muslims with
Explanation – The real issue of gender justice
6. Which among the following is MOST SIMILAR in meaning to the word “subsequent”?
A. previous
B. former
C. earlier
D. consequent
Explanation – consequent – following as a result or effect.
7. Which among the following is MOST OPPOSITE in meaning to the word “buttress”?
A. stanchion
B. shore
C. weaken
D. underpinning
Explanation – weaken – make or become weaker in power, resolve, or physical strength.
8. Which among the following is MOST SIMILAR in meaning to the word “arbitrary”?
A. reasonable
B. approximate
C. reasoned
D. circumspect
Explanation – circumspect
9. Which among the following is MOST OPPOSITE in meaning to the word “exploited”?
A. easily
B. Apply
C. Employ
D. noncommercial
Explanation – noncommercial – not having a commercial objective; not intended to make a profit.
10. Which among the following is MOST SIMILAR in meaning to the word “dilute”?
A. deliquesce
B. concentrate
C. strengthen
D. thicken
Explanation – deliquesce – (of organic matter) become liquid, typically during decomposition.
11. I. Students soon grow _______ of listening to a parade of historical facts
II. He has laid out all his strength and is _______.
III. To attempt to perform the double act is instead to __________ between the two.
A. weary , oscillate
B. distracted , reeling
Explanation – Weary – feeling or showing extreme tiredness
Oscillate- move or swing back and forth in a regular rhythm
12. I. The athlete will have one final ______ before retirement
II. He had a / an _______ with his neighbour’s wife
III. Our horses plodded down the muddy _____.
A. comical , freaky
B. witty , screwy
C. droll , goofy
D. fling , track
Explanation – Fling-throw or hurl forcefully
13. The Andaman and Nicobar islands ……………………………… terra nullius, or empty space wherein government would not inscribe their authority.
A. has been historically treated as
B. was historically been treated as
C. have historically been treated as
D. ad historically being treated as
Explanation – have historically been treated as is the best suited option
14. After falling consistently against the US Dollar for most of the year, the Rupee ……………………………. at the year-end.
A. would have managed to gain some ground
B. had managed to gaining some ground
C. has been managing for gaining some ground
D. has managed to gain some ground
Explanation – has managed to gain some ground is the best suited alternative
15. It is a known fact that Indian states with a better Human Development Index ………………………………… voter turnouts.
A. were tending to have
B. tends to have
C. has been tends to have
D. tend to have
Explanation – tend to have is the best suited alternative
16. A well-functioning bond market ………………………………………. expectations of all bond market participants are incorporated into the bond price.
A. had led to efficient pricing of credit risk when
B. can lead to efficient pricing of credit risk as
C. has been leading to efficient pricing of credit risk when
D. would lead to efficient pricing of credit risk while
Explanation – can lead to efficient pricing of credit risk as is the best suited alternative
17.Nature
A. eternal
B. debase
C. aspect
D. accord
Explanation – Nature – the basic or inherent features, character, or qualities of something
Direction (18-22): Rearrange the following sentences in the proper sequence to form a meaningful paragraph then answer the following questions.
A. Because the story isn’t dying.
B. Chanda Kochhar’s current term as ICICI Bank CEO runs till March 31, 2019.
C. That means a bluechip business brand, a stock market-listed major bank will have to ride out 300-plus daily news cycles when investigators and investors will both be asking questions.
D. That the ICICI board initially cleared Kochhar and issued public statements of stout support for the CEO are seeming less relevant every day.
E. It’s evolving into new life forms every 2-3 days.
F. So, there are 300 plus daily news cycles between now and Kochhar’s regulation term end.
G. Is that really a smart strategy?
18. Which is the Fourth sentence after Rearrangement?
A. E
B. F
C. C
D. G
19. Which is the First sentence after Rearrangement?
A. F
B. A
C. E
D. B
20. Which is the Fifth sentence after Rearrangement?
A. C
B. A
C. D
D. F
21. Which is the Second sentence after Rearrangement?
A. F
B. D
C. B
D. A
22. Which is the Third sentence after Rearrangement?
A. D
B. C
C. A
D. B
23. With borrowing costs set of rise and global trade tensions adding / to uncertainties in India’s exporters who are yet to capitalise / on the rupee weakness, policymakers will need / to eschew populism and stick at policy prudence if the tenuous momentum / is to be sustained.
A. With borrowing costs set of rise and global trade tensions adding
B. to uncertainties in India’s exporters who are yet to capitalise
C. on the rupee weakness, policymakers will need
D. to eschew populism and stick at policy prudence if the tenuous momentum
Explanation – In 1 , replace ‘of’ with ‘off’
In 2 , replace ‘in’ with ‘for’
In 4 , replace ‘at’ with ‘to’
24. It is ironical that in spite for using toxic chemicals, / producers in the conventional non-organic / sector do not had to spend to / guarantee compliance with foods / safety regulatory requirements.
A. It is ironical that in spite for using toxic chemicals,
B. producers in the conventional non-organic
C. sector do not had to spend to
D. guarantee compliance with foods
Explanation – In 1 , replace ‘for’ with ‘of’
In 3 , replace ‘had’ with ‘have’
In 4 , replace ‘foods’ with ‘food’
25. Couplet : sonnet : limerick
A. prose
B. songs
C. lyrics
D. poetry
26. Kerosene : petrol : diesel
A. firewood
B. engine
C. fuel
D. coal
27. Fly : bee : ant
A. cockroach
B. spider
C. termite
D. insect
28. Which of the following five words is wrongly spelt?
A. neonatal
C. predominant
D. mainstrem
Explanation – The correct spelling of the word is ‘Mainstream’. It means the ideas, attitudes, or activities that are shared by most people and regarded as normal or conventional.
29. Which of the following five words is wrongly spelt?
A. divergience
B. cumbersome
C. cumbrous
D. enormous
Explanation – The correct spelling of the word is ‘Divergence’. It means the process or state of diverging
30. Which of the following five words is wrongly spelt?
A. braced
B. Plaguied
C. invigorating
D. rejuvenated
Explanation – The correct spelling of the word is ‘Plagued’. It means cause continual trouble or distress to.
31. A. venom
B. abysmal
C. bottomless
D. antipathy
A. B – A
B. C – A
C. B – C
D. B – D
Explanation – B – C both are synonyms.
32. A. Candid
B. cunning
C. surge
D. patron
A. B – A
B. C – A
C. C – B
D. B – D
Explanation – B – A both are Antonyms.
Candid – truthful and straightforward; frank.
Cunning – having or showing skill in achieving one’s ends by deceit or evasion.
33. A. laconic
B. concise
C. obscene
D. indecent
A. B – A
B. C – A
C. C – B
D. B – D
Explanation – A – B are synonyms.
laconic – (of a person, speech, or style of writing) using very few words.
Concise – giving a lot of information clearly and in a few words; brief but comprehensive.
Directions (34-37): Five statements given below, labeled as a,b,c,d and e. Among these, four statements are in logical error and form coherent paragraph/passage. From the given options choose the option that doesn’t fit into the theme of the passage.
34. A. Midnight has an eerie and slightly occult hold on our imaginations.
B. It is the hour when magic is supposed to take place, when the old falls away and a new day arrives with one twitch of a clock’s minute-hand.
C. It is not feasible for India to progress through the 21st century with 19th-century administrative systems.
D. Midnight also has a sinister ring to it – it is the ‘witching hour’ when bad things happen, or when the knock on the door heralds the assassin or the secret police or the gang of abductors.
E. Or, as Eliot wrote, midnight’s lunar incantations “Dissolve the floors of memory/And all its clear relations,/Its divisions and precisions”.
Explanation – Only sentence C. is different from other sentences which talks about the Administrative reforms: Rebolting the steel frame hence, correct answer will be option third.
35. A. Not only is there no one-to-one relationship between interest rates and investment, demonetisation has already done what a rate cut could hope to achieve.
B. Indeed, it has done what repeated signals from RBI have failed to do.
C. Add to this the fact that the US Fed is poised to raise rates at least once, if not twice, this year, liquidity is still surplus,
D. there are signs of asset price bubbles in some asset markets, and the case for a rate cut becomes much weaker.
E. But because it is using economic diplomacy to make more friends or at least transactional acquaintances.
36. A. Despite the positive propaganda, a lot of misinformation about GST is swirling in markets and homes.
B. Shops are trying to cash in with pre-GST clearance sales, although they can claim ‘transition credits’ on tax they have already paid on their inventories when GST kicks in.
C. Registered businesses are living in fear of having to file three GST returns a month, but Jaitley says after the first lot are loaded on the GST Network, the technology platform, two of the three returns will be ‘self-populating’ and require little work.
D. It is also a means to break walls between different government agencies and enable best practices being shared.
E. Most of all, there is fear that GST will raise prices across the board.
Explanation – only forth statement is different than others which talks about Administrative reforms: Rebolting the steel frame
37. A. Of course, the government loses no opportunity to paint things in political colours, rather than with the broad brush of public good. Now they say there are reforms, but no job creation.
B. At a long Q&A on Tuesday, Jaitley complained that critics will not cease cavilling. “I think some people like to create controversies,” he said.
C. “In 2014 they complained that our reforms are incremental, not big-bang. That argument has been demolished by big-bang reforms like GST, demonetization and the insolvency and bankruptcy law.
D. The third reform has been the creation of an immense sense of competitive spirit among states and districts by ranking them, and increasingly selecting them for projects through a challenge method.
E. ” He said a lot of jobs are being created outside the formal sector, as in self-employment.
Explanation – only forth statement is different than others which talks about competitive spirit among states and districts while other statements talk about GST.
38. This surface feels smooth.
A. This surface is felt smooth
B. This surface is smooth when it is felt
C. This surface when felt is smooth
D. This surface is smooth as felt
A. We completed our task before sunset.
B. We have completed our task before sunset
C. We complete our task before sunset.
40. Why do you tell a lie?
A. Why a lie told by you?
B. Why is a lie be told by you?
C. Why is a lie told by you?
D. Why is a lite being told you?
1. Simplify the given question and identify what value should come in the place of question mark (?) in the questions given below? 40% of ($$\frac {20}{4}$$× ?) = 48
A. 20
B. 24
C. 28
D. 32
2. A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?
A. 45%
B. 45$$\frac {5}{11}$$%
C.54$$\frac {6}{11}$$%
D.55%
Explanation –Number of runs made by running = 110 – (3 x 4 + 8 x 6)
= 110 – (60)
= 50.
Required percentage = $$\frac {5}{110$$X100 % = 45$$\frac {5}{11}$$%
3. Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are:
A. 39, 30
B. 41, 32
C. 42, 33
D. 43, 34
Explanation – Let their marks be (x + 9) and x.
Then, x + 9 = $$\frac {56}{100}$$ (x + 9 + x)
25(x + 9) = 14(2x + 9)
=> 3x = 99
=> x = 33
So, their marks are 42 and 33.
4. What percentage of numbers from 1 to 70 have 1 or 9 in the unit’s digit?
A. 1
B. 14
C. 20
D. 21
Explanation – Clearly, the numbers which have 1 or 9 in the unit’s digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.
Number of such number =14
Required percentage = $$\frac {14}{70}$$ X 100% = 20%.
5. A man buys a cycle for Rs. 1400 and sells it at a loss of 15%. What is the selling price of the cycle?
A. Rs. 1090
B. Rs. 1160
C. Rs. 1190
D. Rs. 1202
Explanation – S.P. = 85% of Rs. 1400 = Rs $$\frac {85}{100}$$x 1400% = Rs. 1190
6. In a mixture 60 litres, the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2, then the quantity of water to be further added is:
A. 20 litres
B. 30 litres
C. 40 litres
D. 60 litres
Explanation – Quantity of milk = 60 X $$\frac {2}{3}$$litres = 40 litres
Quantity of water in it = (60- 40) litres = 20 litres.
New ratio = 1 : 2
Let quantity of water to be added further be x litres.
Then, milk : water = $$\frac {40}{20 + X}$$
Now; $$\frac {40}{20 + x}$$ = $$\frac {1}{2}$$
20 + x = 80
=> x = 60.
=> Quantity of water to be added = 60 litres.
7. If Rs. 782 be divided into three parts, proportional to $$\frac {1}{2}$$$$\frac {2}{3}$$$$\frac {3}{4}$$then the first part is:
A. Rs. 182
B. Rs. 190
C. Rs. 196
D. Rs. 204
Explanation – Given ratio = $$\frac {1}{2}$$$$\frac {2}{3}$$$$\frac {3}{4}$$= 6 : 8 : 9.
1st part = Rs.782.X $$\frac {6}{23}$$ = Rs. 204
8. A, B and C jointly thought of engaging themselves in a business venture. It was agreed that A would invest Rs. 6500 for 6 months, B, Rs. 8400 for 5 months and C, Rs. 10,000 for 3 months. A wants to be the working member for which, he was to receive 5% of the profits. The profit earned was Rs. 7400. Calculate the share of B in the profit.
A. Rs. 1900
B. Rs. 2660
C. Rs. 2800
D. Rs. 2840
Explanation – For managing, A received = 5% of Rs. 7400 = Rs. 370.
Balance = Rs. (7400 – 370) = Rs. 7030.
Ratio of their investments = (6500 x 6) : (8400 x 5) : (10000 x 3)
= 39000 : 42000 : 30000
= 13 : 14 : 10
B’s share = Rs. 7030 X $$\frac {14}{37}$$= Rs. 2660.
9. Ravi, Gagan and Nitin are running a business firm in partnership. What is Gagan’s share in the profit earned by them?
I. Ravi, Gagan and Nitin invested the amounts in the ratio of 2 : 4 : 7.
II. Nitin’s share in the profit is Rs. 8750.
A. I alone sufficient while II alone not sufficient to answer
B. Both I and II are necessary to answer
C. Both I and II are not sufficient to answer
D. Either I or II alone sufficient to answer
Explanation – Let us name Ravi, Gagan and Nitin by R, G and N respectively.
I. R : G : N = 2 : 4 : 7.
II. N = 8750..
From I and II, we get:
When N = 7, then G = 4.
When N = 8750, then G = $$\frac {4}{7}$$ X 8750 = 5000.7
Thus, both I and II are needed to get the answer.
10. In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs?
A. 6.25
B. 6.5
C. 6.75
D. 7
Explanation – Required run rate = $$\frac {282 – (3.2 x 10)}{40}$$ = $$\frac {250}{40}$$ = 6.25
11. A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500?
A. Rs. 4991
B. Rs. 5992
C. Rs. 9855
D. 7854
12. Veena has to pay Rs. 2460 to Sita, 5 Months later at 6% SI per annum, and Gita has to pay Sita the same amount at 7.5% SI per annum after certain months. If both took the same amount of loan from Sita then Gita paid loan after how many months?
A. 3 Months
B. 4 Months
C. 6 Months
D. 12 Months
Explanation – 2460= p + p x 6 x $$\frac{5}{12}$$x 100
p = 2400
Now Gita
2460 = 2400+2400 x 7.5 x $$\frac{x}{12}$$ x 100
x = 4
13. If 2994 ÷ 14.5 = 172, then 29.94 ÷ 1.45 = ?
A. 0.172
B. 4325
C. 4475
D. 4675
Explanation – $$\frac{29.94}{1.45}$$ = $$\frac{299.4}{14.5}$$ 299.4
= ($$\frac{29.94}{1.45}$$ x $$\frac{1}{10}$$)[ Here, Substitute 172 in the place of 2994/14.5$$\frac{2994}{14.5}$$ ]
= $$\frac{172}{10}$$
= 17.2
14. Show That$$\frac{.009}{?}$$ ?
A. .0009
B. .09
C. .9
D. 9
Explanation – Let $$\frac{.009}{X}$$ = .01;Then x = $$\frac{.009}{.01}$$ = $$\frac{.9}{1}$$ = .9
15. 3889 + 12.952 – ? = 3854.002
A. 47.095
B. 47.752
C. 47.932
D. 47.95
Explanation –
Let 3889 + 12.952 – x = 3854.002.
Then x = (3889 + 12.952) – 3854.002
= 3901.952 – 3854.002
= 47.95.
16. 0.002 x 0.5 = ?
A. 0.0001
B. 0.001
C. 0.01
D. 0.1
Explanation – 2 x 5 = 10.
Sum of decimal places = 4
0.002 x 0.5 = 0.001
17. The six digit number 54321A is divisible by 9 where A is a single digit whole number. Find A.
A. 0
B. 2
C. 4
D. 3
Explanation – A number is divisible by 9, when the sum of its digits is divisible by 9. Here, 5 + 4 + 3 + 2 + 1 + A = 15 + A should be divisible by 9. Therefore, A = 3 gives 15+ 3 = 18 as the sum of digits, which is divisible by 9. So, answer is option D.
18. The product of 40 odd numbers is
A. even
B. odd
C. 625
D. Can’t say
Explanation –The product of 40 odd numbers will give an odd number. So answer is option B.
19. Find the greatest three number which is multiple of 7
A. 993
B. 994
C. 995
D. None of these
Explanation –Greatest three digits number = 999. When 999 is divided by 7, the remainder will be 5. Required number= 999 – 5 = 994.
20. Simplify the expression using BODMAS rule (3/7) of (4/5) of 20 ($${25}^{2}$$ – $${24}^{2}$$)
A. 336
B. 168
C. 84
D. None of these
Explanation –(3/7) × (4/5) of 20 (625 – 576)⇒ (3/7) × (4/5) × 20 × 49 =336
21. It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?
A. Sunday
B. Saturday
C. Friday
D. Wednesday
Explanation – On 31st December, 2005 it was Saturday.
Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.
=> On 31st December 2009, it was Thursday.
Thus, on 1st Jan, 2010 it is Friday.
22. Today is Monday. After 61 days, it will be:
A. Wednesday
B. Saturday
C. Tuesday
D. Thursday
Explanation – Each day of the week is repeated after 7 days.
So, after 63 days, it will be Monday.
After 61 days, it will be Saturday.
23. Which of the following is not a leap year?
A. 700
B. 800
C. 1200
D. 2000
Explanation – The century divisible by 400 is a leap year.
The year 700 is not a leap year.
24. January 1, 2008 is Tuesday. What day of the week lies on Jan 1, 2009?
A. Monday
B. Wednesday
C. Thursday
D. Sunday
Explanation – The year 2008 is a leap year. So, it has 2 odd days.
1st day of the year 2008 is Tuesday (Given)
So, 1st day of the year 2009 is 2 days beyond Tuesday.
Hence, it will be Thursday.
25. A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500?
A. Rs. 4991
B. Rs. 5991
C. Rs. 6001
D. Rs. 6001
Explanation – Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009.
Required sale = Rs. [ (6500 x 6) – 34009 ]
=> Rs. (39000 – 34009)
=> Rs. 4991.
26. The average weight of 8 person’s increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person?
A. 76 kg
B. 76.5 kg
C. 85 kg
D. None of these
Explanation – Total weight increased = (8 x 2.5) kg = 20 kg.
Weight of new person = (65 + 20) kg = 85 kg.
27. The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average age of the team?
A. 23 years
B. 24 years
C. 25 years
D. None of these
Explanation – Let the average age of the whole team by x years.
=> 11x – (26 + 29) = 9(x -1)
=> 11x – 9x = 46
=> 2x = 46
=> x = 23.
So, average age of the team is 23 years.
28. Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?
A. 9
B. 10
C. 12
D. 20
Explanation – Due to stoppages, it covers 9 km less.
Time taken to cover 9 km = $$\frac {9}{54}$$ X 60min = 10 min
29. In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less than Sameer. Abhay’s speed is:
A. 5 kmph
B. 6 kmph
C. 6.25 kmph
D. 7.5 kmph
Explanation – Let Abhay’s speed be x km/hr.
Then, $$\frac {30}{X}$$ – $$\frac {30}{2X}$$ = 3
=> 6x = 30
=> x = 5 km/hr.
30. The ratio between the speeds of two trains is 7 : 8. If the second train runs 400 km in 4 hours, then the speed of the first train is:
A. 70 km/hr
B. 75 km/hr
C. 84 km/hr
D. 87.5 km/hr
Explanation – Let the speed of two trains be 7x and 8x km/hr.
Then, 8x = $$\frac {400}{4}$$ = 100
=> X = $$\frac {100}{8}$$ = 12.5
Speed of first train = (7 x 12.5) km/hr = 87.5 km/hr.
31. A man has Rs.480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ?
A. 45
B. 60
C. 75
D. 90
Explanation – Let number of notes of each denomination be x.
Then x + 5x + 10x = 480
=> 16x = 480
=> x = 30.
Hence, total number of notes = 3x = 90.
32. If a – b = 3 and $${a}^{2}$$ + $${b}^{2}$$ = 29, find the value of ab.
A. 10
B. 12
C. 15
D. 18
Explanation – 2ab = $${(a}^{2}$$ + $${b}^{2)}$$ – $${(a-b)}^{2}$$
=> 29 – 9 = 20
=> ab = 10.
33. The sum of first five prime numbers is:
A. 11
B. 18
C. 26
D. 28
Explanation – Required sum = (2 + 3 + 5 + 7 + 11) = 28.
Note: 1 is not a prime number.
34. The difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and the 15 as remainder. What is the smaller number ?
A. 240
B. 270
C. 295
D. 360
Explanation – Let the smaller number be x. Then larger number = (x + 1365).
=> x + 1365 = 6x + 15
=> 5x = 1350
=> x = 270
Smaller number = 270.
Directions (35 – 40):The following pie chart shows the amount of subscriptions generated for India Bonds from different categories of investors.
35. In the corporate sector, approximately how many degrees should be there in the central angle ?
A. 121
B. 120
C. 122
D. 124
Explanation – 34 x 3.6 = 122.4 (since 1% = 3.6 degrees)
36. If the investment by NRI’s are Rs 4,000 crore, then the investments by corporate houses and FII’s together is:
A. 24,000 crore
B. 24,363 crore
C. 25,423 crore
D. 25,643 crore
Explanation – (67/11) x 4000 = 24 363.6364
37. What percentage of the total investment is coming from FII’s and NRI’s ?
A. 33%
B. 11%
C. 44%
D. 22%
Explanation – (33 + 11) = 44
38. If the total investment other than by FII and corporate houses is Rs 335,000 crore, then the investment by NRI’s and Offshore funds will be (approximately) ?
A. 274,100
B. 285,600
C. 293,000
D. Cannot be determined
Explanation – Investment other than NRI and corporate houses is 33% = 335000. Also, investment by offshore funds and NRI’s is equal to 27%.
Hence, 27 x 335,000/33 = 274 090.909
39. If the total investment flows from FII’s were to be doubled in the next year and the investment flows from all other sources had remained constant at their existing levels for this year, then what would be the proportion of FII investment in the total investment into India Bonds next year (in US \$ millions) ?
A. 40 %
B. 50 %
C. 60 %
D. 70 %
Explanation – FII’s currently account for 33 out of 100.
If their value is doubled and all other investments are kept constant then their new value would be 66 out of 133 = approximately equal to 50%
40. A clock is started at noon. By 10 minutes past 5, the hour hand has turned through:
A.145°
B.150°
C.155°
D. 160°
Explanation – Angle traced by hour hand in 12 hrs = 360°.
Angle traced by hour hand in 5 hrs 10 min. i.e.$$\frac {31}{6}$$ Hrs = $$\frac {360}{12}$$ x
$$\frac {31}{6}$$ = 155°
1. It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?
A. Sunday
B. Saturday
C. Friday
D. Friday
Explanation –
On 31st December, 2005 it was Saturday.
Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.
In 31st December 2009, it was Thursday.
Thus, on 1st Jan 2010 it is Friday.
2. Today is Monday. After 61 days, it will be:
A. Wednesday
B. Saturday
C. Tuesday
D. Thursday
Explanation –
Each day of the week is repeated after 7 days.
So, after 63 days, it will be Monday.
After 61 days, it will be Saturday.
3. The calendar for the year 2007 will be the same for the year:
A. 2014
B. 2016
C. 2017
D. 2018
Explanation –
Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.
Year : 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017
Odd day : 1 2 1 1 1 2 1 1 1 2 1
Sum = 14 odd days 0 odd days.
Calendar for the year 2018 will be the same as for the year 2007.
4. Amit Khanna, born on 5th june, 1973 has done his post-graduation in Marketing Management with first class. He has secured 50 % marks in the written Test. He has been working in an organisation as a Marketing Officer for the last four years.
A. be a graduate with at least 50 % marks.
B. have secured atleast 40 % marks in the written test
C. not be less than 24 years and more than 29 years as on 10th October, 1997.
D. should have work experience of at least two years as an officer.
5. Rohit Verma, has been working in an organisation as officer for the last ten years. His date of birth is 17th February, 1964. He has secured 60% marks in the degree examination and 40% marks in the written test.
A. be a graduate with at least 50 % marks.
B. have secured atleast 40 % marks in the written test.
C. not be less than 24 years and more than 29 years as on 10th October, 1997.
D. should have work experience of at least two years as an officer.
6. Manju sharma is a first class graduate and has done a diploma in Marketing Management. She has secured 50% marks in the written test. She was 23 years old as on 5th September, 1996.
A. be a graduate with at least 50 % marks.
B. have secured atleast 40 % marks in the written test.
C. have secured atleast 40 % marks in the written test.
D. should have work experience of at least two years as an officer.
7. A’s mother is the only daughter of B’s father. How is B’s husband related to A?
A. Uncle
B. Brother
C. Father
D. Grandfather
8. A is brother of B, B is the son of C, D is C’s father then what is A of D?
A. Brother
B. Son
C. Grandson
D. Grandfather
9. A is the father of C and D is son of B. E is brother of A. If C is sister of D, how is B related to E?
A. Brother
B. Sister
C. Brother-in-law
D. Sister-in-law
10. A and B are brothers. C and D are sisters. A’s son is D’s brother. How is B related to C?
A. Father
B. Brother
C. Grandfather
D. Uncle
11. Choose the word which is least like the others word in a group
A. Chemistry
B. Geography
C. Zoology
D. Botany
12. Choose the word which is least like the others word in a group?
A. Mechanic
B. Engineer
C. Mason
D. Blacksmith E. Architect
13. Choose the word which is least like the others word in a group?
A. Zinc
B. Aluminium
C. Copper
D. Mercury E. Iron
14. A woman introduces a man as the son of the brother of her mother. How is the man related to the woman?
A. Nephew
B. Uncle
C. Son
D. Cousin
15. A man said to a lady, “Your mother’s husband’s sister is my aunt”. How is that lady related to that man?
A. Daughter
B. Grand-daughter
C. Mother
D. Sister
16. If X is brother of the son of Y’s son, how is X related to Y?
A. Son
B. Brother
C. Cousin
D. Grand-Son
17. If B says that his mother is the only daughter of A’s mother, how is A related to B?
A. Son
B. Grandfather
C. Uncle
D. Brother
18. If FISH is written as EHRG in a certain code, how would JUNGLE be written in that code?
A. ITMFKD
B. ITNFKD
C. KVOHMF
D. TIMFKD
19. In a certain code, TWINKLE is written as SVHOJKD, then how would FILTERS be written in the same code?
A. EHKSDQR
B. EHKUDQR
C. EGKUDQR
D. GJMSFST
20. How many such 5s are there in the following number sequence each of which is immediately preceded by 3 or 4 but not immediately followed by 8 or 9?
3 5 9 5 4 5 5 3 5 8 4 5 6 7 3 5 7 5 5 4 5 2 3 5 1 0
A. None
B. Three
C. Four
D. Five
21. Rohan ranks seventh from the top and twenty-sixth from the bottom in a class. How many students are there in the class?
A. 31
B. 32
C. 33
D. 34
22. Look at this series: 7, 10, 8, 11, 9, 12, … What number should come next?
A. 7
B. 10
C. 12
D. 13
Explanation –
This is a simple alternating addition and subtraction series. In the first pattern, 3 is added; in the second, 2 is subtracted.
23. Look at this series: 36, 34, 30, 28, 24, … What number should come next?
A. 20
B. 22
C. 23
D. 26
24. Look at this series: 22, 21, 23, 22, 24, 23, … What number should come next?
A. 22
B. 24
C. 25
D. 26
25. One morning Udai and Vishal were talking to each other face to face at a crossing. If Vishal’s shadow was exactly to the left of Udai, which direction was Udai facing?
A. East
B. West
C. North
D. South
Explanation –
26. Rahul put his timepiece on the table in such a way that at 6 P.M. hour hand points to North. In which direction the minute hand will point at 9.15 P.M. ?
A. South-East
B. South
C. North
D. West
Explanation –
27. Choose the word which is different from the rest.
A. Chicken
B. Snake
C. Swan
D. Crocodile
Explanation – All except Chicken can live in water.
28. hoose the word which is different from the rest.
A. Kiwi
B. Eagle
C. Emu
D. Ostrich
Explanation – All except Eagle are flightless birds.
29. Choose the word which is different from the rest.
A. Rigveda
B. Yajurveda
C. Atharvaveda
D. Ayurveda
Explanation – All except Ayurveda are names of holy scriptures, the four Vedas. Ayurveda is a branch of medicine.
30. Choose the word which is different from the rest.
A. Curd
B. Butter
C. Oil
D. Cheese
Explanation – All except Oil are products obtained from milk.
31. A, P, R, X, S and Z are sitting in a row. S and Z are in the centre. A and P are at the ends. R is sitting to the left of A. Who is to the right of P ?
A. A
B. X
C. S
D. Z
Explanation – The seating arrangement is as follows:
32. A, B, C, D and E are sitting on a bench. A is sitting next to B, C is sitting next to D, D is not sitting with E who is on the left end of the bench. C is on the second position from the right. A is to the right of B and E. A and C are sitting together. In which position A is sitting ?
A. Between B and D
B. Between B and C
C. Between E and D
D. Between C and E
Explanation –
33. An institute organised a fete and 1/5 of the girls and 1/8 of the boys participated in the same. What fraction of the total number of students took part in the fete ?
A. 2/13
B. 13/40
D. None of these
34. A, B, C, D and E play a game of cards. A says to B, “If you give me three cards, you will have as many as E has and if I give you three cards, you will have as many as D has.” A and B together have 10 cards more than what D and E together have. If B has two cards more than what C has and the total number of cards be 133, how many cards does B have ?
A. 22
B. 23
C. 25
D. 35
Explanation – Clearly, we have :
B-3 = E …(i)
B + 3 = D …(ii)
A+B = D + E+10 …(iii)
B = C + 2 …(iv)
A+B + C + D + E= 133 …(v)
From (i) and (ii), we have : 2 B = D + E …(vi)
From (iii) and (vi), we have : A = B + 10 …(vii)
Using (iv), (vi) and (vii) in (v), we get:
(B + 10) + B + (B – 2) + 2B = 133 5B = 125 B = 25.
35. I have a few sweets to be distributed. If I keep 2, 3 or 4 in a pack, I am left with one sweet. If I keep 5 in a pack, I am left with none. What is the minimum number of sweets I have to pack and distribute ?
A. 25
B. 35
C. 45
D. 33
Explanation – Clearly, the required number would be such that it leaves a remainder of 1 when divided by 2, 3 or 4 and no remainder when divided by 5. Such a number is 25.
36. 120, 99, 80, 63, 48, ?
A. 35
B. 45
C. 33
D. 54
Explanation – The pattern is – 21, – 19, – 17, – 15,…..
So, missing term = 48 – 13 = 35.
37. Pointing to a photograph of a boy Suresh said, “He is the son of the only son of my mother.” How is Suresh related to that boy?
A. Brother
B. Uncle
C. Cousin
D. Father
Explanation – The boy in the photograph is the only son of the son of Suresh’s mother i.e., the son of Suresh. Hence, Suresh is the father of boy.
38. Introducing a boy, a girl said, “He is the son of the daughter of the father of my uncle.” How is the boy related to the girl?
A. Brother
B. Nephew
C. Uncle
D. Son-in-law
Explanation – The father of the boy’s uncle → the grandfather of the boy and daughter of the grandfather → sister of father.
39. Find the number of triangles in the given figure.
A.
B.
C.
D.
Explanation – The figure may be labelled as shown.
The simplest triangles are AHG, AIG, AIB, JFE, CJE and CED i.e. 6 in number.
The triangles composed of two components each are ABG, CFE, ACJ and EGI i.e. 4 in number.
The triangles composed of three components each are ACE, AGE and CFD i.e. 3 in number.
There is only one triangle i.e. AHE composed of four components.
Therefore, There are 6 + 4 + 3 + 1 = 14 triangles in the given figure.
40. _ a _ b _ abaa _ bab _ abb
A. aaabb
B. babab
C. ababb
D. babba
Explanation –The series is baa/bba/baa/bba/baa/bb. Thus, the pattern baa/bba is repeated.
1. Which if the following is not true of about Money Market
A. It is a market for short term loans
B. The maturity period is anywhere between a day and a year
C. Treasury Bills are not a component of Money Market
D. Central Banks take part in Money Market
2. FEMA stands for
A. Foreign Exchange Money Act
B. Future Exchange Money Act
C. Future Exchange Management Act
D. Foreign Exchange Management Act
3. The value of Indian Forex Reserves as of September 2018 was
A. 399.2Billion
B. 400.1Billion
C.402.2Billion
D. 403.2Billion
4. The Finance Commission of India is established as per which Article of the Indian Constitution
A. Art 124
B. Art 356
C. Art 324
D. Art 280
5. Euronext is the Stock Exchange of
A. The United Kingdom
B. France
C. European Union
D. None of the Above
6. SEBI stands for –
A. Securities and E-commerce Board of India
B. Stock Exchange Board of India
C. Securities and Exchange Board of India
D. None of the Above
7. Which company among the following would be buying a share in the IDBI Bank?
A. LIC
B. ONGC
C. SBI
D. NTPC
8. Currently which country in talks with the IMF for an economic bailout –
A. India
B. Pakistan
C. North Korea
D. Sri Lanka
9. The recently recommended merger of Bank of Baroda, Dena Bank and Vijaya Bank if done will make it the –
A. d Largest Bank in India
B. Largest Bank in India
C. 3rd Largest Bank in India
D. None of the Above
10. The BSE-SENSEX comprises of the Stock of how many companies
A. 30
B. 20
C. 10
D. 40
11. What is the Tenure of the Chairman of SEBI
A. 2 years
B. 3 years
C. 5 years
D. Not Fixed
12. As per the Time Value of Money, the Value of Money in Hand is ________ the value of Money in Future
A. Lesser than
B. Greater than
C. Same as
D. Not related to
13. In which type of Bonds does the issuer has the option of reducing the tenor of the security?
A. Puttable Bonds
B. Callable Bonds
C. Zero Coupon Bonds
D. Floating Rate Bonds
14. The IMF currently has
A. 189 Members
B. 200 Members
C. 180 Members
D. 280 Menbers
15. What is the full form of NPS ?
16. NBFC deals in?
17. Who is the author of the book My years with Rajiv and Sonia ?
18. The Mutual funds in India follow accounting standards laid by ?
19. Interest on savings account is calculated on which basis?
20. What is the full form of IFSC ?
Answer – Indian Financial System code
21. SBI launched minor a/c for childern above 10 years named as ?
22. Mutual funds regulated by ?
Answer – Securities and Exchange Board of india
23. Full form of GSLV ?
Answer – Geo – Synochorous Launch vehicle
24. HDFC erdo is which type of company ?
20. SEBI is which kind of body?
21. There are how many digits in MICR ?
Answer – Nine Character or Digits
22. BSBDA stands for ?
Answer – Basic Savings Bank deposit account
23. RBI established and nationalized in ?
Answer – Established 1935,.Natonalized in 1949
24. Cheque is a ___ instrument ?
25. Full Form of ATM ?
26. IFSC code consist of _____ character ?
27. CRISIL is a ?
29. The extended date for pre-2005 bank notes ?
30. Who is the president of BRICS Bank ?
31. Banking Ombudsman is appointed by:
32. Name of Union Territory which as recently presented a tax free budget of Rs. 6,100 crore for the union territory:
33. Maximum age for retirement for MD/CEO of all private banks is:
34. ‘Financial Exclusion’ is:
35. Vijay Mallya as ‘Willful Defaulter’ is declared by:
Answer – United Bank of India
36. Mutual Funds are regulated by:
Answer – Securities and Exchange Board of India
37. To solve the problems of Balance of Payments of member countries is function of:
38. Which term is not associated with banking operations:
39. U.T.I. officially changed into:
40. Regulatory Authority for Regional Rural Bank is of:
### Phase 2
Securities Market
1. Which among the following is the oldest Public Sector Bank of India?
A. Punjab National Bank
B. Central Bank of India
D. Bank of Baroda
2. When was SEBI constituted?
A. April, 1988
B. March, 1982
C.July, 1992
D. Dec. 1974
Explanation – SEBI was established on the April 12, 1988. It was constituted as a non statutory body through a resolution of the government of India.
3. Which of the following statement is NOT correct about the SEBI?
A. At present it is a non statutory body
B. At present it is a statutory body
C. It got statutory powers by an ordinance in 1992
D. SEBI is managed by 6 members
Explanation – At present it is a statutory body; which got statutory status and powers by an ordinance in Jan. 1992
4. Who is the current chairman of the SEBI?r
A. U.K. Sinha
B. Ajay Tyagi
C. T.S.Vijayan
D. A. K. Mathur
Explanation – Ajay Tyagi appointed new chairman of SEBI; to take charge on March 1, 2017.He is 9th chairman of the SEBI.
5. Who of the following never became chairman of the SEBI?
A. C. B. Bhave
B. M. Damodaran
C. D. R. Mehta
D. Ashok Ganguly
Explanation – Ashok Ganguly never appointed as the chairman of the SEBI while rest of the persons mentioned in the options have been appointed as the chairman.
6. Which of the following words does not belong to the stock exchange?
A. NAV
B. NSE
C. IPO
D. KPO
Explanation – The KPO’s full form is Knowledge Process Outsourcing; while rests of the words are related to the security market.
7. Where is the headquarter of the SEBI?
A. Delhi
B. Bengaluru
C. Chennai
D. Chennai
Explanation – The headquarter of the SEBI is situated in the Mumbai, Maharashtra.
8. Chairman of the SEBI is appointed for ………
A. For maximum 3 years
B. For maximum 6 years
C. For maximum 5 years
D. Tenure not fixed
Explanation – The chairman of the SEBI is made for a period not exceeding five years, or till the age of 65 years or until further orders.
9. How many companies are included in the SENSEX?
A. 30
B. 45
C. 65
D. 32
Explanation – The S&P BSE SENSEX (S&P Bombay Stock Exchange Sensitive Index) also called the BSE 30 or simply the SENSEX. These 30 companies are big and renowned companies of their sector.
10. Which of the following is responsible for the fluctuations in the SENSEX?
A. Rain
B. Monetary policy
C. Political instability
D. All of the above
Explanation – SENSEX fluctuates due to very small reasons. That is why it is known as sensitive index as well. SENSEX fluctuates due to forecasting of good weather, any political decision by the central government and any change in the government policy.
11. The Finance Commission of India is established as per which Article of the Indian Constitution
A. Art 124
B. Art 356
C. Art 324
D. Art 280
12. Euronext is the Stock Exchange of
A. The United Kingdom
B. France
C. European Union
D. None of the Above
<strong13. SEBI stands for –
A. Securities and E-commerce Board of India
B. Stock Exchange Board of India
C. Securities and Exchange Board of India
D. None of the Above
14. Which company among the following would be buying a share in the IDBI Bank?
A. LIC
B. ONGC
C. SBI
D. NTPC
<strong15. Currently which country in talks with the IMF for an economic bailout –
A. India
B. Pakistan
C. North Korea
D. Sri Lanka
16. The recently recommended merger of Bank of Baroda, Dena Bank and Vijaya Bank if done will make it the –
A. d Largest Bank in India
B. Largest Bank in India
C. 3rd Largest Bank in India
D. None of the Above
17. The BSE-SENSEX comprises of the Stock of how many companies
A. 30
B. 20
C. 10
D. 40
18. What is the Tenure of the Chairman of SEBI
A. 2 years
B. 3 years
C. 5 years
D. Not Fixed
19. Which of the following statements is correct?
A. FTSE-100 is a stock exchange of London, which monitors European market activities
B. Nikkei is related to Singapore Stock Exchange
C. MIDDEX belongs to Japan
D. BSE does not belong to SENSEX
Explanation – FTSE-100 is a stock exchange of London, which monitors the activities of the European security market.
20. The transaction cost of trading of financial instruments in centralized market is classified as
A. flexible costs
B. low transaction costs
C. high transaction costs
D. constant costs
21. In primary markets, the property of shares which made it easy to sell newly issued security is considered as
A. increased liquidity
B. decreased liquidity
C. money flow
D. large funds
22. The money market where debt and stocks are traded and maturity period is more than a year is classified as
A. shorter term markets
B. capital markets
C. counter markets
D. long-term markets
23. The type of market in which securities with less than one year maturity are traded, is classified as
A. money market
B. capital market
C. transaction market
D. global market
24. In capital markets, the major suppliers of trading instruments are
A. government and corporations
B. liquid corporations
C. instrumental corporations
D. manufacturing corporations
25. The rate of return on a bond held to maturity is known as its:
A. duration.
B. interest yield.
C. redemption yield.
D. present value.
26. The MRTP commission was replaced by which commission ?
A. Competition Commission of India
B. Finance Commission
C. Nanavati-Shah commission
D. .National Statistical Commission
Explanation – The Monopolies and Restrictive Trade Practices Act, 1969 [MRTP Act] repealed and is replaced by the Competition Act, 2002, with effect from 01st September, 2009
27. MRTP Act was enacted in which year ?
A. 1997
B. 1969
C. 1996
D. 1970
Explanation – On the basis of recommendation of Dutt Committee, MRTP Act was enacted in 1969 to ensure that concentration of economic power in hands of few rich. The act was there to prohibit monopolistic and restrictive trade practices.
28. The minimum number of people involved in the exchange marketing is _____________
A. 1
B. 2
C. 3
D. 5
Explanation – A marketing exchange is what happens any time two or more people trade goods or services. In marketing theory, every exchange is supposed to produce “utility,”
29. Which of the following center is also known as ‘decision-making units’ (DMUs) ?
A. Marketing Center
B. Shopping Center
C. Sales Center
Explanation – A buying center brings together “all those members of an organization who become involved in the buying process for a particular product or service”.Buying centers are also known as ‘decision-making units’ (DMUs)
30. In which type of exchange, only two parties are involved ?
A. Restricted Exchanges
B. Generalized Exchanges
C. Complex Exchanges
D. Complex Exchanges
Explanation – Simple or “restricted” exchanges are those in which there are only two parties to the exchange. Restricted exchanges are one-on-one relationships, so both parties must receive approximately equal utility if the exchange is to be repeated.
31. Market segmentation as the subdivision of a market into homogeneous subjects of customer’ said by whom ?
A. Dirk Morschett
B. Thomas Rudolph
C. Philip Kotler
D. American Marketing Association
Explanation –Marketing Guru Philip Kotler defines market segmentation as the subdivision of a market into homogeneous subjects of customers
32. MRP stands for
A. Manufacturing Resource Planning
B. Manufacturing Resource Process
C. Making Retail Product
D. Material Resource Planning
Explanation – Manufacturing resource planning (MRP II) is defined as a method for the effective planning of all resources of a manufacturing company.
33. Money market instruments are generally used for Liquidity adjustments, considering this
statement to be true. Who are the users mostly?
A. Banks
B. Government
C. Hospital
D. Dealers
Explanation –Money market instruments are generally used for financing short term liquidity requirements, the
same is used by all scheduled commercial and private banks. A dealer is an intermediary between
the counterparts, called a dealer, announces a bid and an offer rate with the difference between
the two representing a spread, or the dealer’s income
34. Which of the following comprises the domestic debt market in India?
I. Government Securities
II. Private corporate debt
III. PSU bonds
IV. DFIs bonds
A. I and III
B. I, II and III
C. I, II, III and IV
D. None of these
Explanation – The domestic debt market comprises of two main segments, viz., the Government securities and other (mainly corporate) securities comprising private corporate debt, PSU bonds and DFIs bonds.
The government securities market is pre-dominant, while the other segment is not very deep and
liquid
35. Who among the following are the main participants in the securities market?
I. Issuer of securities
II. Investors in securities
III. Intermediaries
IV. The state governments
A. Only IV
B. Both I and II
C. I, II and III
D. I, II, III and IV
Explanation – The securities market essentially has three categories of participants, namely,
the issuers of securities, investors in securities and the intermediaries, such as
merchant bankers, brokers etc. While the corporates and the government raise
resources from the securities market to meet their obligations, it is households
that invest their savings in the securities market
36. Which of the following is the clearing agency for G-secs in India?
A. CCIL
B. RBI
C. SBI
D. Fema
Explanation – The Clearing corporation of India Ltd (CCIL) is the clearing agency for G-Secs. It acts as a Central
Counter Party (CCP) for all transactions in G-Secs by interposing itself between two counterparties.
In effect, during settlement, the CCP becomes the seller to the buyer and buyer to the seller of
the actual transaction. All outright trades undertaken in the OTC market and on the NDS-OM
platform are cleared through the CCIL.
37. There are 2 types of derivatives, futures and options. Which of the following is correct about
Futures contracts?
A. Futures contracts grant buyers rights and no obligations
B. In Futures contracts both parties face a lot of risk
C. It permits the buyer an option to buy call or put
D. Both 1 and 2
38. Commercial Paper is a note in evidence of the _____________of the issuer.
A. Equity obligation
B. Debt Obligation
C. Market Obligation
D. None of the above
Explanation – Commercial Paper is a note in evidence of the debt obligation of the issuer. On issuing commercial paper the debt is transformed into an instrument.
CP is an unsecured promissory note privately placed with investors at a discount rate of face value
determined by market forces.
39. Treasury Bills are short-term (up to one year) borrowing instruments issued by __________
A. State Government
B. Union Goverment
C. President
D. Governer
Explanation – T Bills are issued by the union government. It’s a promise by the Government to pay the stated sum after the expiry of the stated period from the date of issue (less than one year). They are
issued at a discount off the face value and on maturity, the face value is paid to the holder.
40. Which of the following is a power of SEBI?
A. Regulating the business in stock exchanges only
B. Regulating the business in other securities markets
C. Regulating the flow of funds to the various organisations registered with SEBI
D. Both 1 and 2
Explanation – SEBI has no power to regulate the flow of funds to the various organisations registered with SEBI as it is under their own control. SEBI only takes care of fraudulent and unfair trade practices relating to securities markets.
41. SEBI has recently approved (August 2018) new KYC norms for Foreign portfolio Investments to
A. Exchange Currency markets
B. Primary Markets
C. Secondary Markets
D. Commodity Markets
Explanation – SEBI has approved a proposal to allow foreign investors to trade in commodity derivatives market,except for sensitive commodities, as well as common application form for the registration of FPIs in the domestic markets. The board has given its nod for amending delisting regulations. Further,
inter-operability of clearing houses for commodities and new methodology to calculate the total
expenses ratio for fund managers have been cleared.
42. What is the name of the screen based electronic order matching system for secondary market
trading in government securities owned by RBI?
A. GDS-OM
B. NDS – OM
C. G-Secs module
D. e – Biz
Explanation – Negotiated dealing system- order matching system NDS-OM is a screen based electronic anonymous order matching system for secondary market trading in Government securities owned
by RBI. Presently the membership of the system is open to entities like Banks, Primary Dealers,
Insurance Companies, Mutual Funds etc. i.e entities who maintain SGL accounts with RBI. These
are Primary Members (PM) of NDS and are permitted by RBI to become members of NDS-OM
43. The minimum net worth stipulated by SEBI for a depository is ________
A. Rs. 100 crores
B. Rs. 1000 crore
C. Rs. 500 crores
D. None of the above
Explanation – The minimum net worth stipulated by SEBI for a depository is Rs.100 crore.
A depository is an organisation which holds securities (like shares, debentures,
bonds, government securities, mutual fund units etc.) of investors in electronic
form at the request of the investors through a registered depository participant. It also provides
services related to transactions in securities.
44. How is the price of a bond calculated?
A. Sum of present values of all future cashflows
B. Sum of future values of all future cashflows
C. Sum of future values of all present cashflows
D. None of the above
Explanation – The price of a bond is nothing but the sum of present value of all future cash flows of the bond. The interest rate used for discounting the cash flows is the Yield to Maturity (YTM).
45. . What can be said about the relationship between the yield and price of a bond?
A. If the market interest rate levels rise, the price of a bond falls.
B. If the market interest rate levels fall, the price of a bond rises
C. The yield of a bond is inversely related to its price
D. All the above
46. Which of the following is referred to as “Duration” of a bond?
A. The time taken to recover the initial investment in present value terms
B. The time taken to recover the initial investment in terms of future values.
C. Either (1) or (3)
D. Either (1) or (2)
47. What does ‘STRIPS’ stand for?
A. Stock trading and registering interest and payment of stocks
B. Securities transaction related information portal system.
C. Separate trading of registered interest and principal of securities.
D. Stock market related information of payment and negotiation systems.
48. What is the name of the market that provides a channel for the sale of new securities?
A. Primary market
B. Secondary market
C. Tertiary Market
D. ) Either of the above
Explanation – The securities market has two interdependent segments: the primary (new
issues) market and the secondary market. The primary market provides the channel for sale of new securities while the secondary market deals in securities previously issued.
49. What is ‘book building’ in an IPO?
A. A process used to differentiate securities in the market.
B. A process for efficient lot delivery to the concerned investors.
C. A process used for efficient price discovery
D.A process used to differentiate competitors from the organisation
Explanation – Book Building is basically a process used in IPOs for efficient price discovery. It is a mechanism where, during the period for which the IPO is open, bids are collected from investors at various prices, which are above or equal to the floor price. The offer price is determined after the bid closing date.
50. . What is ‘ICDR’ in the ICDR regulations as prescribed by SEBI?
A. Issue of current equity and debt requirements
B. Issue of capital and disclosure requirements
C. Issue of convertible debentures raised through equity
D. Issue of currency derivatives resource
Explanation –The SEBI ICDR Regulations lay down guidelines relating to conditions for various kinds of issues including public and rights issue. The ICDR Regulations provide detailed provisions relating to
public issue such as conditions relating to an IPO and Further Public Offer (FPO), conditions relating
to pricing in public offerings, conditions governing promoter’s contribution, restriction on
transferability of promoter’s contribution, minimum offer to public, reservations, manner of
disclosures in offer documents, etc.
### Tips
To prepare for SEBI Assistant Manager Phase I 2018, the five main subjects that you need to be excellent at are General Awareness, English Language, Quantitative Aptitude, Test of Reasoning and Awareness about Securities Market.
Candidates can crack the exam only with the right approach and correct strategy and can also clear the SEBI Assistant Manager 2018 Phase I examination but remember there is a sectional cut off in SEBI Phase I Examination.
✦ Tip – 1: Go through the syllabus and make sure you haven’t missed anything.
✦ Tip – 2: Solve a few Mock Test papers and analyse your mistakes and weaknesses.
✦ Tip – 3: Divide your attention between every section.
✦ Tip – 4: Go through your notes and important questions. Write down every formula and rule in your syllabus.
✦ Tip – 5: Be prepared with everything.
✦ Tip – 6: Maintain a peaceful environment and confidence.
✦ Tip – 7: Get enough sleep and good food. | 2021-01-25 00:13:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2103516310453415, "perplexity": 6190.953319370131}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703561996.72/warc/CC-MAIN-20210124235054-20210125025054-00754.warc.gz"} |
https://tex.stackexchange.com/questions/331642/text-wrapping-in-tables-and-vertical-alignment | # Text Wrapping in Tables and Vertical Alignment
I am attempting to adjust my headers such that they are centered horizontally and vertically within the cell, and that the last cell specifically breaks at that point. This was my best solution however when I send it to others, it does not come out correctly. There must be a better way.
\begin{center}
\textbf{Table II}
\textbf{Title \medskip }
\begin{tabular}{|c|c|c|c|}
\hline
\multirow{2}{*}{Number of experts}&\multirow{2}{*}{Equilibrium threshold} & \multirow{2}{*}{DM's payoff}& \\ [-12pt]
& & & \shortstack{DM's outside option\\ supporting $t^*>0$}\\[-2pt] \hline
1 & 0.484 & 0.691 & $r\in[0,0.484]$ \\ \hline
2 & 0.136 & 0.735 & $r\in[0,0.588]$ \\ \hline
3 & 0.021 & 0.657 & $r\in[0,0.633]$ \\ \hline
4 & 0.001 & 0.568 & $r\in[0,0.567]$ \\ \hline
5 or more & 0 & 0.5 & $r\in[0,0.5]$ \\ \hline
\end{tabular}
\end{center}
I would like to recommend you give your table a much more "open" look. To achieve this objective, get rid of all vertical lines and most horizontal lines; use the line-drawing macros of the booktabs package for the remaining horizontal lines. Also, instead of using a \multirow hack, consider loading the tabularx package and using a centered version of that package's X column type for columns 2 and 4 -- the ones where automatic line breaking is required -- and c for columns 1 and 3.
\documentclass{article}
\usepackage{tabularx,ragged2e,booktabs}
\newcolumntype{C}{>{\Centering\arraybackslash}X}
\usepackage[font=bf]{caption}
\renewcommand\thetable{\Roman{table}}
\begin{document}
\setcounter{table}{1} % just for this example
\begin{table}
\caption{Title}
\begin{tabularx}{\textwidth}{@{}cCcC@{}}
\toprule
Number of experts & Equilibrium threshold &
DM's payoff & DM's outside option supporting $t^*>0$\\
\midrule
1 & 0.484 & 0.691 & $r\in[0,0.484]$ \\
2 & 0.136 & 0.735 & $r\in[0,0.588]$ \\
3 & 0.021 & 0.657 & $r\in[0,0.633]$ \\
4 & 0.001 & 0.568 & $r\in[0,0.567]$ \\
5 or more & 0\phantom{.000} & 0.5\phantom{00} & $r\in[0,0.5]\phantom{00}$ \\
\bottomrule
\end{tabularx}
\end{table}
\end{document}
From your code snipped is impossible to say anything why others who compile your code get different result as you. Well also table, which you show in question is not generated from your code. See, if the following MWE gives beter results at you as well as at others:
\documentclass{article}
\usepackage[labelfont=bf,labelsep=newline]{caption}
\usepackage{makecell}
\begin{document}
\begin{table}
\caption{Title}
\begin{tabular}{|c|c|c|c|}
\hline
supporting $t^*>0$} \\
\hline
1 & 0.484 & 0.691 & $r\in[0,0.484]$ \\ \hline
2 & 0.136 & 0.735 & $r\in[0,0.588]$ \\ \hline
3 & 0.021 & 0.657 & $r\in[0,0.633]$ \\ \hline
4 & 0.001 & 0.568 & $r\in[0,0.567]$ \\ \hline
5 or more & 0 & 0.5 & $r\in[0,0.5]$ \\ \hline
\end{tabular}
\end{table}
\end{document}
Of course, better (professional) looks of table you obtain by use of booktabs package (see Mico answer).
This solution puts the multiline entry into a \parbox. The only problem is having to guess or compute the width of the box.
Note: The \strut was purely cosmetic, to add a little space at the bottom of the \parbox.
\documentclass{article}
\newlength{\tempwidth}
\begin{document}
\begin{center}
\textbf{Table II}
\textbf{Title \medskip }
\settowidth{\tempwidth}{DM's outside option}% compute parbox width
\begin{tabular}{|c|c|c|c|}
\hline
Number of experts & Equilibrium threshold & DM's payoff
& \parbox[t]{\tempwidth}{\centering DM's outside option supporting $t^*>0$\strut} \\ \hline
1 & 0.484 & 0.691 & $r\in[0,0.484]$ \\ \hline
2 & 0.136 & 0.735 & $r\in[0,0.588]$ \\ \hline
3 & 0.021 & 0.657 & $r\in[0,0.633]$ \\ \hline
4 & 0.001 & 0.568 & $r\in[0,0.567]$ \\ \hline
5 or more & 0 & 0.5 & $r\in[0,0.5]$ \\ \hline
\end{tabular}
\end{center}
\end{document} | 2019-12-13 00:27:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9999896287918091, "perplexity": 774.5098852374126}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540547536.49/warc/CC-MAIN-20191212232450-20191213020450-00245.warc.gz"} |
https://www.andrewlienhard.io/tags/back-pain/ | • It’s a long way to my toes. From up here, in the stratosphere of human elevation, I see my distant phalanges pressing against the wilds of the earth — a… | 2022-08-09 17:58:44 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8257986903190613, "perplexity": 2442.0047438725232}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571056.58/warc/CC-MAIN-20220809155137-20220809185137-00777.warc.gz"} |
https://www.ncatlab.org/nlab/show/transfor | # nLab transfor
Contents
### Context
#### Higher category theory
higher category theory
# Contents
## Idea
A $k$-transfor is an operation from one $n$-category $C$ to another $D$ (for some value of $n$) that takes objects of $C$ to $k$-morphisms of $D$ (and more generally $j$-morphisms in $C$ to $(j+k)$-morphisms in $D$) in a coherent way. Equivalently, a $k$-transfor is a $k$-cell in an internal-hom $n$-category. Transfors are a common generalisation of:
• $n$-functors, which are 0-transfors
• $n$-natural transformations, which are 1-transfors
• modifications, which are 2-transfors,
• perturbations, which are 3-transfors,
• and so on.
The word “transfor” was coined by Sjoerd Crans in this paper; it is a portmanteau of “functor” and “transformation.” A collection of components which forms a transfor is said to be transforial, as a generalization of “functorial” and “natural.”
## Terminology
Once upon a time, there were categories, functors between them, and natural transformations between them. Then when $n$-categories came along, people called the arrows between them ‘$n$-functors’ even though one could just as easily say ‘functors’. In the same vein, people said ‘$n$-transformations’ for natural transformations (that is, 2-transfors) between $n$-categories. At the same time, we saw that we needed modifications between $n$-transformations, and that there would have to be things between higher modifications, and so on. However, due to the prior use of “$n$-transformation” for a 2-transfor between $n$-categories, the natural choice “$k$-transformation” is unavailable to mean a $k$-transfor.
Here are some other possible terms for a $k$-transfor between $n$-categories, which additionally notate the value of $n$ (although this is, strictly speaking, unnecessary).
• $(n,k)$-transformation
• $n$-$k$-transfor
• $n$-dimensional $k$-transfor
• $n$-categorical $k$-transfor
• $n$-natural $k$-transformation
## Definitions
We haven't gotten around to saying anything precise yet, but you can see something in the discussion below, or in Crans's paper.
## Special cases
See this periodic table of $k$-transfors between $n$-categories for common names for low values of $n$ and $k$. On the $n$-Lab, we tend to omit the prefix $n$- whenever possible (as ironic as that may be).
$k$↓\$n$$-1$$0$$1$$2$$3$...
$0$implicationfunctionfunctor$2$-functor$3$-functor...
$1$trivialequality of functionsnatural transformation$2$-transformation$3$-transformation...
$2$"trivialequality of natural transformationsmodification$3$-modification...
$3$""trivialequality of modificationsperturbation...
$4$"""trivialequality of perturbations...
$5$""""trivial...
"""""
Note that the source and target of a $k$-transfor (between $n$-categories) are $(k-1)$-transfors (between the same $n$-categories). Given two fixed source and target $(k-1)$-transfors, the $k$-transfors between them (and the $(k+1)$-transfors between those, and so on) form an $(n-k)$-category.
### For n-posets
A similar table periodic table of $k$-transfors between $n$-posets exists for common names for low values of $n$ and $k$.
$k$↓\$n$$-1$$0$$1$$2$$3$...
$0$implicationmonotonic functionfunctor$2$-functor$3$-functor...
$1$trivialpartial order of monotonic functionsnatural transformation$2$-transformation$3$-transformation...
$2$"trivialpartial order of natural transformationsmodification$3$-modification...
$3$""trivialpartial order of modificationsperturbation...
$4$"""trivialpartial order of perturbations...
$5$""""trivial...
"""""
## Discussion
This discussion was originally at modification. It discusses both terminology and definitions.
Finn: There is a pattern here: functors
are indexed collections of objects, natural transformations are i.c.s of 1-cells, modifications i.c.s of 2-cells; and these are what make the collection of all $n$-categories into an $n+1$-category, for $0 \leq n \leq 2$ anyway. Any references for the pattern in higher dimensions?
Toby: Do you mean for the terminology or for the appropriate coherence laws? (the details that you've been leaving out). Not that I have either …
Incidentally, I corrected ‘function’ to ‘functor’ in you question above; I hope that's OK.
Finn: I meant terminology and/or an explanation for arbitrary $n$ (which Urs gives below).
Actually I was thinking of functions rather than functors, as they are the 1-cells in $0-Cat$. But of course functions are just functors between discrete categories, and thinking of them as the latter probably makes more sense when moving to higher dimensions.
Toby: Now, I would either have said ‘functors are indexed collections of objects’ or ‘functions are indexed collections of elements’; your mixture confused me! (^_^)
Finn: Ah! Point taken. In any case, I should have said ‘0-cell’ instead of ‘object’. But I think ‘functor’ is better anyway, as I said.
Urs: the pattern that Finn is looking for is that embodied in the nature of the internal hom of the closed monoidal structure on presheaves.
In its most general form, consider an infinity-category modeled as a simplicial set with certain properties. Being a simplicial set, this is a presheaf on the simplex category. Hence for $X$ and $Y$ such $\infty$-categories, the $\infty$-category of morphisms between them corresponds to the internal hom simplicial set
$[X,Y] = Hom_{SSet}(X \times \Delta^\bullet, Y) \,.$
This simple formula encodes that pattern that Finn observed. It says that:
• functors (the 0-cells in $[X,Y]$) are just maps $X \to Y$ from cells to cells;
• natural transformations (the 1-cells in $[X,Y]$) are maps $X \times \Delta^1 \to Y$. Notice that $\Delta^1$ is the interval object in $SSet$ (or at least its Kanification is, but never mind that for the moment). Such maps send $n$-cells in $X$ to $(n+1)$-cells in $Y$.
• modifications are maps $X \times \Delta^2 \to Y$, that map $n$-cells in $X$ to $(n+2)$-cells in $Y$.
It may be helpful to realize the same pattern in the globular context of, for instance, strict omega-category. These are certain presheaves not on the simplex category but on the globe category, but the pattern is the same: the internal hom strict $\omega$-category of morphisms between strict $\omega$-categories $X$ and $Y$ is
$[X,Y] = Hom_{\omega Cat}(X \otimes G^\bullet , Y) \,,$
where now the tensor product appearing is no longer the cartesian one but the Crans-Gray tensor product and where $G^n$ is the standard globular $n$-globe. Again $G^1$ is a model for the interval object and we see that
• functors are morphisms $X \to Y$;
• transformations are morphisms $X \otimes G^1 \to Y$
• modifications are morphisms $X \otimes G^2 \to Y$
etc. Same logic as before.
When thinking about this, it may be useful to explicitly apply the hom-adjunction everywhere and think for instance of a natural transformation as a morphism
$X \to [I,Y]$
from $X$ into the “category of cylinders in $Y$”. This is maybe the most intuitive way: if for instance $Y$ happens to be just a 2-category, then this says that a transformation between functors between 2-categories is a 1-functor from the 1-category underlying $X$ to the category of cylinders in $Y$ (satisfying some property). Which is exactly what it is, in components.
When in a certain mood, I like to think of this basic fact, that $n$-fold transformations between $k$-functors are essentially (in components) $(k-n)$-functors with values in $n$-cylinders as the “holographic principle” in category theory. That may sound a bit silly, but it is true that in the case the $k$-functors in questions are $k$-functors on $Bord_k$ respresenting $k$-dimensional quantum field theory, then teir transformations, being $(k-1)$-functors, represent $(k-1)$-dim QFT, and this relation between higher and lower dim QFT is called “holography” in phyiscs.
Finn: Cool! Thanks, Urs. I might move this section to an article on $n$-transformations (if that’s what they’re called) once I get my head around it properly.
Toby: Unfortunately, ‘$n$-transformation’ already (following ‘$n$-functor’) means a transformation between functors between $n$-categories. See Cheng–Gurski for this, along with ‘$n$-modification’ and even ‘$n$-perturbation’ (gee, that doesn't conflict with anything else, does it?), along with the claim that there is ‘no existing terminology’ thereafter.
I would probably say ‘$n$-morphism in $n Cat$’ (possibly with two different values of $n$); you can use ‘$n$-cell’ in place of ‘$n$-morphism’ if you like. But it would be nice to find something more specific that's not already taken. Or we could just throw out the Cheng–Gurski meaning of ‘$n$-transformation’; although it's not unique to them, it may not be too entrenched yet.
(But please let a transformation be a $1$-transformation, even though it is a $2$-morphism.)
Todd: I think what Urs and Crans both may be suggesting is that, at least in the context of strict $n$- and $\omega$-categories, there is a uniform notion of “transformation of depth k between n-functors”, or just $(n, k)$-transformations, where $(n, 1)$-transformations are usual transformations between $n$-functors, $(n, 2)$-transformations are modifications, and so on. Surely this usage won’t conflict with Cheng-Gurski.
Toby: Yeah, that would work, so we could write (n,k)-transformation. My only disgruntlement is that the $n$ is superfluous; the problem is all those other people that are already using it and preventing us from unambiguously saying simply ‘$k$-transformation’!
Finn: Probably tiros like me shouldn’t have a say in this sort of thing, but I would tend to agree with Toby here, that the $k$ is at least more interesting than the $n$, in that you’re more likely to vary the values of $k$ than those of $n$. However, typing the few extra characters does seem a small price to pay to avoid horrible confusion. I slightly reluctantly vote for $(n,k)$.
Todd: I’m not crazy about it either, but I agree it’s a small price. I’ll note (in case it helps) that in the general theory of Crans-Gray tensor products, both variations in $n$ and $k$ come up, about equally often (e.g., the tensor of a 1-category and an $n$-category is an $(n+1)$-category).
Urs: yes, so to summarize what I think the main points are
• there is a systematic notion of “transformation of depth k between n-functors” for geometric definition of higher category in terms of simplicial sets;
• the corresponding notion in the (strict) globular context is formalized by Crans’ construction;
• unwrapping what this says, it yields in particular that a transformation of depth $k$ between strict globular $n$-categories $X$ and $Y$ is an $(n-k)$-functor from the truncation $X{\leq k}$ of $X$ to an $(n-k)$-category (throwing all higher cells away) to the $(n-k)$-category of $k$-globes in $Y$ (also truncated)
$\eta : X_{\leq k} \to [G^k, Y]|_{\leq k}$
satisfying certain naturality conditions (which ensure precisely that $\eta$ extends uniquely to an $n$-functor $\eta : X \to [G^k,Y]$).
JCMcKeown: not meaning to cause annoyance, but how about calling them “$+k$-transformations”, owing to their incrementing dimensions by $k$; or if we don’t like the $+$ prefix, one might call them $k$-vexilors, because they tend to generate flags of period $k$.
Toby: Interesting; can you explain more about how they generate flags? (Maybe that's something to put in a new section here, or you could just give a reference.)
JCMcKeown: Just from reading above “… and more generally $j$-morphisms in $C$ to $(j+k)$-morphisms in $D$”… ahah! Now I see what you’re getting at. I’ve got my head fixed on endo-functors; where if you wanted to (I don’t mean it’s a good idea. Who knows?) you can consider iterations of the underlying function that is the $+k$-transformation.
Mike Shulman: FWIW, Sjoerd Crans has called these things k-transfors, and speaks of something being transforial as a general term including both “functorial” and “natural.”
Toby: I'm inclined to say that we should go with that!
Mike Shulman: I’m not sure how serious you are… but I’ve always thought it was a proposal that deserved to be taken more seriously than it seems to have been. The reference is “Localizations of Transfors,” K-Theory 2004 (I can’t find a free version online).
Toby: I'm perfectly serious. The term should be indexed primarily by $k$, with $n$ only if one really insists. I didn't want to make up my own word, but if Crans has published one, then why not use it? I should be able to check that reference the next time that I visit the UCR library (about once a week).
Mike Shulman: No argument here (about indexing by $k$). Also $(n,k)$-transformation sounds to me like something to do with (n,r)-categories, but there of course the comma denotes something completely different.
Todd Trimble: I like $k$-transfor.
Toby: Excellent! Since Finn and JCMcKeown have not been active lately, I'll move it over with that paper as a guide (or you can).
I would like to also mention ‘$(n,k)$-transformation’ (or maybe ‘$n$-$k$-transfor’?) as a possible term, however, since some people might want to specify $n$ just as some people like to say ‘$n$-functor’.
Toby: One could also say ‘$n$-natural $k$-transformation’, which fits (what Crans claims on page 2 to be standard) ‘$2$-natural transformation’ for a strict $(2,1)$-transformation. But I still like ‘$k$-transfor’ when $n$ is suppressed (which should be the default).
Mike: What about “$n$-categorical $k$-transfor” if it is necessary to specify $n$?
Toby: That works too. (Well, I don't like ‘categorical’, but that's a separate issue.)
## References for the globular approach
Camell Kachour: Kamel Kachour, Définition algébrique des cellules non-strictes, Cahiers de Topologie et de Géométrie Différentielle Catégorique (2008), volume 1, pages 1–68.
Camell Kachour: Steps toward the Weak ω-category of the Weak ω-categories in the globular setting, Published Categories and General Algebraic Structures with Applications (2015).
Last revised on April 6, 2021 at 18:34:12. See the history of this page for a list of all contributions to it. | 2021-04-17 21:02:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 219, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8387361168861389, "perplexity": 1263.5172755670128}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038464045.54/warc/CC-MAIN-20210417192821-20210417222821-00382.warc.gz"} |
https://quant.stackexchange.com/questions/61005/quantlib-python-why-is-the-npv-different-npv0-1-0-110-should-be-100-not-9 | QuantLib Python: why is the NPV different? NPV(0.1,[0,110]) should be 100, not 99.53
I am trying to learn QuantLib for Python. Further to my previous question on the syntax for CashFlows.npv(), now that I understand how the syntax works, I have a question on why the output differs from what I'd expect (which is why I think a separate question is justified).
In my toy example, my cashflows are zero on 15-Jan-2001 and 110 on 15-Jan-2002. If I discount them at 10%, I'd expect the pv to be 100, but I get 99.53211. Why? What am I missing? That would imply that 384 days have gone by, not 365. Those are not leap years. I have tried with act/365 and 30/360: they both give the same result, which is not 100.
import QuantLib as ql
d1 = ql.Date(15,1,2001)
ql.Settings.instance().setEvaluationDate(d1)
cfs = [ql.SimpleCashFlow(0, d1),
ql.SimpleCashFlow(110, d1 + 365)]
calc_date = d1
risk_free_rate = 0.1
curve_act_365 = ql.YieldTermStructureHandle(
ql.FlatForward(calc_date, risk_free_rate, ql.Actual365Fixed()))
pv_act_365 = ql.CashFlows.npv(cfs, curve_act_365, True)
curve_30_360 = ql.YieldTermStructureHandle(
ql.FlatForward(calc_date, risk_free_rate, ql.Thirty360()))
pv_30_360 = ql.CashFlows.npv(cfs, curve_30_360, True)
By default, QuantLib expects a continuously compounded rate in the FlatForward constructor.
So the PV you are getting is basically:
from math import exp
print(110 * exp(-0.1))
If you define your curve with an annually compounded rate like so:
curve_30_360 = ql.YieldTermStructureHandle(
ql.FlatForward(calc_date, risk_free_rate, ql.Thirty360(), ql.Compounded, ql.Annual))
You will get 100. | 2021-05-06 16:48:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5480121970176697, "perplexity": 4784.020518889049}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988758.74/warc/CC-MAIN-20210506144716-20210506174716-00426.warc.gz"} |
http://www.finderchem.com/coefficient-of-x-30-in-taylor-series-expansion-for-sin-3x-pi-3-about-x-o.html | # Coefficient of x^30 in taylor series expansion for sin(3x+pi/3) about x=o?
... in this expansion, the coefficient of x3 is double the coefficient of x2, (b) ... cos 3x = – (6) ... sin 20 √ x− ° = 2 1. - Read more
The coefficient of x in expansion of n ()( ) ... sin x 3 f(x) 9x ... x – y – z = 1 (2) x – 2y – z = 1 (3) x – y – 2z = 1 (4) 2x – y ... - Read more
## Coefficient of x^30 in taylor series expansion for sin(3x+pi/3) about x=o? resources
### AE470 – Examples of Matlab refreshers Problem 1 The ...
The Taylor series expansion of sin(x) is sin(x)=x ... *x^3+p(2)*x^2+P(3)*x+P(4) \n') fprintf ... +O(Δx2), where Δx is a ...
### Taylor Series -- from Wolfram MathWorld
A Taylor series is a series expansion of a function about a point. ... ('')(a))/(2!)(x-a)^2+(f^((3))(a))/(3!)(x-a)^3+...+(f^((n))(a))/ ... third Taylor polynomial sin x.
### Taylor series expansion - MATLAB taylor - MathWorks
ans = x - (x - 1)^2/2 + (x - 1)^3/3 - (x - 1)^4 ... 'approximation of sin(x)/x up to O(x ... expansion point x = a: Taylor series expansion requires a function to ...
### 3. Taylor Series - University of California, San Diego
>> T3=taylor(sin(x ),4) % The result ... let's take a look at the degree 29 Taylor polynomial approximation of sin(x). >> T29=taylor(sin(x),30); ... 3 < x < 5. (Hint ...
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Then for any t m-A' we have the Taylor series expansion ... coefficient between x ... where C depends on t + k -1 and p but not on r or x, and the expression O ... ### PPLATO | FLAP | MATH 4.5: Taylor expansions and polynomial ... sin(x) ≈ p 3 (x) = a 0 + a 1 (x − π/4 ...$\sin(x) \approx -(x - \pi) + \dfrac{(x-\pi)^3}{3!}$(33) ... The Subsection 3.2 Taylor series or expansion for f (x ... ### 16. The Taylor Series and Other Mathematical Concepts ... ... sin(x), cos(x) and (1+x)^k and review the Taylor series of a few other functions. ... (0:28) Maclaurin series expansion o ... ### Keggin’s ion structural modification and expansion of ... Heteropolyacids with general chemical formula H 3+x AM 12 O 40 ... acid exists in the series of hydrated phases ... where b is the linear thermal expansion coefficient. ### Calculus — Sage Constructions v6.2 - Sage: Open Source ... ... + k*x^3*e^(k*x)*sin(w*x) + 3*x^2*e^(k*x)*sin(w*x) sage: ... Taylor series: sage: var ('f0 k x') ... , f2]]) sage: f. fourier_series_cosine_coefficient (5, pi)-3/5 ... ### FOURIER TRANSFORM - University of California, Los Angeles ... (such as a trigonometric function sin(nx) or cos(nx)), ... o(x) = x3 +3x. ... to Taylor series; ... ### Polynomials at SolveMyMath.com - How to Input For the polynomial calculator the default variable is 'x'; example:3x^4+5x^2-x+14 Arithmetic operations ... (sin, cos, 2/3, (3-2), (2-3), 2*3, e^3, ln ... ### Zeta functions, L-series and polylogarithms — mpmath v0 ... ... defined as the -th coefficient in the Laurent series expansion of the Riemann zeta function around the pole ... (pi / 3); chop (cl2 (pi)); ... >>> n = 3 >>> x ... ### www2.maths.ox.ac.uk ... (10*pi*x)+sin(61*exp(.8*sin ... convergent % Chebyshev series, that is, an expansion %$f(x) = a ... the coefficients of the Chebyshev series for $1 % + x^3 + x^4 ... ### Efficient inferences on the varying-coefficient single ... The varying-coefficient single-index model ... in g/m 3), and respirable suspended particulate (X 3 X 3, in g/m 3) and weather ... Applying a Taylor series expansion ... ### Full text of "Split Runge-Kutta method for simultaneous ... A Taylor's series expansion of S w will ... parameters) Xi — 7?"~^ 2_ ^ 3 > x 4 — -jr— x 5 ... r A -JL A - O'+Pi) 3 n i — ory-3 n s ... ### [Rao v. Dukkipati]_Numerical org - Scribd - Read Unlimited ... The Taylor’s series expansion for e x , sin x and cos x are given below: ... find the Taylor series expansion of f (x) ... + x 2 – 5x + 3, x 0 ### Richardson’s Extrapolation - Department of Mathematics ... ... (’-sin(x)’); h = 0.1; x = pi/3; fprintf ... we evaluate our series at h and h/2 and get: ... the Taylor expansion. We used our O(h2) ... ### Circumference/Perimeter of an Ellipse: Formula(s) - Numericana In Cayley's series, the coefficient of x n is n ... involved are partial sums of the Taylor series of ... DECLARE FUNCTION cayley# (x#) DEFDBL A-Z CONST pi = 3 ... ### sci.math FAQ: e^(i Pi) = -1 Euler's formula From the identities sin x = x - x^3/3! + x^5/5 ... which are the Taylor series expansion of the trigonometric sine ... alopez-o@barrow.uwaterloo ... ### Math Forum - Ask Dr. Math ... (x) = x + (1/3)x^3 ... This is the Taylor series expansion about the point x ... (n/pi)tan(pi/n) Since this expansion certainly holds for 0 < pi/m < pi/n < pi/3, ... ### 1 Numerical Derivatives - Arizona State University You can derive this by Taylor series expansion, ... Let's see what happens if we evaluate the derivative of sin (x) ... To derive it from the Taylor series, ... ### Derivative Calculator at SolveMyMath.com - Math Help ... x^3+4*x^2+x+1 = (sin(x^3)*log(1/x))/(x+3) = exp(2*y)*log(y^3)+x*y (w.r.t. Variable: y) Related Calculators. Integral Calculator ... ### Algebra 2/Trigonometry Regents Exam 0611 www.jmap sin 2θ+cos θ 1−sin2θ is ... 26 What is the coefficient of the fourth term in the expansion of (a −4b)9? 1) −5,376 ... 3x +16 =(x +2)2 3x +16 =x2 +4x +4 0 =x2 ... ### Another fast fixed-point sine approximation | Coranac The Taylor expansion can serve as the basic for ... the Taylor series terms are not the ... /Pi - (8 (-(5/2) + 8 (-(9/16) + 3/Pi)) x^3)/Pi^3 + (32 (-(3/2) + 4 (-(9 ... ### Newton's Forward Difference Formula -- from Wolfram MathWorld with the falling factorial, the formula looks suspiciously like a finite analog of a Taylor series expansion. This correspondence was one of the motivating forces for ... ### Paper Reference(s) 6664/01 Edexcel GCE ... in the expansion of (1 + px) 12, the coefficient of . x. is ... sin (x + 10 ) = 2 ... A geometric series has first term : a: ### Perturbation Analysis, Regular and Singular ... (x)| ### Taylor expansions - Texas A&M University ... is usually not the best way to find a Taylor expansion ... , supposing that all we know about f is that its Maclaurin series starts out f(x) = 1 + 3x ... = 5 + O ... ### Binomial theorem - Wikipedia, the free encyclopedia 7.2 Series for e; 8 The binomial ... The coefficient of x n ... the number of copies of x n − k y k in the expansion; ### contest math - Find coefficient of$x^8$in$(1-2x+3x^2-4x ...
... = 1 -2x+3x^2-4x^3+5x^4-6x^5+7x^6-8x^7 + 9x^8 + O(x^9) So the coefficient of ... 2 x + 3 x^2 - 4 x^3 + 5 x^4 - 6 ... of the first few terms of the Taylor Series ...
### Taylor Series in MATLAB - Texas A&M University
Taylor Series in MATLAB First, let’s review our two main statements on Taylor polynomials with remainder. Theorem 1. (Taylor polynomial with integral remainder ...
### ca.analysis and odes - Taylor series coefficients ...
... 3\pi}x^3-\frac{4}{9\pi^2}x^6+\frac{2}{7\pi}x^7+\frac{16}{81\pi^3}x^9 ... be its Taylor series expansion at \$x=0 ... x)=\frac{(-1)^{n-1}(n-1)!}{(1+x^2)^{n/2}}\sin ...
### Taylor Series - LTCC Online
... we arrive at the Taylor Series for f(x) ... Exercises Find the Taylor series expansion for. sin(x) centered at x = p/2; ... 1 + x + x 2 + x 3 + x 4 + ...
### Taylor Series Expansions and Approximations
Taylor Series 51 Taylor Series Expansions and Approximations The Taylor series is mainly used for approximating functions when one can identify a small parameter.
Related Questions
Recent Questions | 2017-06-28 19:17:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8289720416069031, "perplexity": 1617.3736369640283}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128323730.30/warc/CC-MAIN-20170628185804-20170628205804-00514.warc.gz"} |
https://en.neurochispas.com/precalculus/equation-of-a-hyperbola-with-center-outside-the-origin/ | # Equation of a Hyperbola with Center Outside the Origin
Hyperbolas are defined as conic sections that are obtained when a cone is intersected by a plane. The hyperbola is formed when the plane intersects both faces of the cone. The hyperbola contains two branches that have a parabolic shape and that are reflections of each other. The set of all points in a hyperbola are characterized because the difference of the distances between any point on the hyperbola and the foci is equal to a constant.
A hyperbola has two lines of symmetry. The transverse axis extends from one vertex to the other vertex and passes through the center. The foci define the hyperbola and are located on the line that contains the transverse axis. The conjugate axis extends from one co-vertice to the other and is perpendicular to the transverse axis. The point of intersection of the transverse axis and the conjugate axis is the center of the hyperbola. The center is also the point of intersection of the two asymptotes.
##### PRECALCULUS
Relevant for
Finding the equation of the hyperbola with center outside the origin.
See equation
##### PRECALCULUS
Relevant for
Finding the equation of the hyperbola with center outside the origin.
See equation
## Standard form of hyperbolas with center outside the origin
The standard form of hyperbolas centered outside the origin is found by applying a translation of h units on the x-axis and k units on the y axis. This results in the center of the hyperbola being located at $latex (h, k)$. We have two variations of this equation depending on the orientation of the hyperbola.
### Equation of the horizontal hyperbola with center outside the origin
A hyperbola that has its center at $latex (h, k)$ and in which its transversal axis is parallel to the x axis is:
where,
• h is the coordinate of the center and is the y coordinate of the center
• The length of the transverse axis is $latex 2a$ (segment joining vertices)
• The vertices are located at $latex (h\pm a, k)$
• The length of the conjugate axis is $latex 2b$ (segment joining covertices)
• The covertices are located at $latex (h, k\pm b)$
• The distance between the foci is $latex 2c$
• We find c using $latex {{c}^2}={{a}^2}+{{b}^2}$
• The foci are located at $latex (h \pm c, 0)$
• The equations of the asymptotes are $latex y=\pm \frac{b}{a}(x-h)+k$
### Equation of the vertical hyperbola with center outside the origin
When the hyperbola is centered at the point $latex (h, k)$ and its transversal axis is parallel to the y axis, its equation is:
where,
• h is the x coordinate of the center and k is the y coordinate of the center
• The length of the transverse axis is $latex 2a$ (segment joining vertices)
• The vertices have the coordinates $latex (h, k\pm a)$
• The length of the conjugate axis is $latex 2b$ (segment joining covertices)
• The covertices have the coordinates $latex (h\pm b, k)$
• The distance between the foci is $latex 2c$, where, $latex {{c}^2}={{a}^2}+{{b}^2}$
• The foci have the coordinates $latex (h, k\pm c)$
• The asymptotes have the equations $latex y=\pm \frac{a}{b}(x-h)+k$
## Determine the equation of hyperbolas centered outside the origin using vertices and foci
We can use the coordinates of the vertices and the foci to find the equation of a hyperbola centered outside the origin by following these steps:
Step 1:Determine the orientation of the hyperbola. We have to determine if the transverse axis is parallel to the x-axis or parallel to the y axis.
1.1.When the y coordinates of the vertices are the same as the y coordinates of the foci, the transverse axis is parallel to the x-axis, and we use the equation $latex \frac{{{(x-h)}^2}}{{{a}^2}}-\frac{{{(y-k)}^2}}{{{b}^2}}=1$.
1.2. When the x coordinates of the vertices are the same as the x coordinates of the foci, the transverse axis is parallel to the y axis, and we use the equation $latex \frac{{{(y-k)}^2}}{{{a}^2}}-\frac{{{(x-h)}^2}}{{{b}^2}}=1$.
Step 2: We use the coordinates of the vertices and the midpoint formula to find the center $latex (h, k)$.
Step 3: The distance between the vertices is $latex 2a$. We use this to find $latex {{a}^2}$.
Step 4: We use the coordinates of the foci and the values of h and k to find the value of c and $latex {{c}^2}$.
Step 5: We use the equation $latex {{b}^2}={{c}^2}-{{a}^2}$ to find the value of $latex {{b}^2}$.
Step 6: We substitute the values of $latex {{a}^2}$ and $latex {{b}^2}$ into the equation obtained in step 1.
## Hyperbolas with center outside the origin – Examples with answers
The following examples apply what you have learned about hyperbola equations. Vertices and foci are used to determine the equations. Analyze the examples to learn the process used.
### EXAMPLE 1
What is the equation of the hyperbola that has vertices at (-1, 1), (3, 1) and foci at (-2, 1), (4, 1)?
##### Solution
The y coordinates of the foci and the vertices are the same, so we know that the transverse axis is parallel to the x-axis and the hyperbola equation has the form:
$latex \frac{{{(x-h)}^2}}{{{a}^2}}-\frac{{{(y-k)}^2}}{{{b}^2}}=1$
To find the center, we observe that the center is in the middle of the vertices (-1, 1) and (3, 1). Therefore, we apply the midpoint formula:
$latex (h, k)=(\frac{-1+3}{2}, \frac{1+1}{2})$
$latex =(1, 1)$
To find $latex {{a}^2}$, we determine the length of the transverse axis, 2a, which is bounded by the vertices. Therefore, we find the difference in the x coordinates of the vertices:
$latex 2a=|3-1|$
$latex 2a=2$
$latex a=2$
$latex {{a}^2}=4$
Now, we find $latex {{c}^2}$. The coordinates of the foci are $latex (h\pm c, k)$. Therefore, we have $latex (h-c, k)=(-2, 1)$ and $latex (h+c, k)=(4, 1)$. We can use any of these points to find the value of c:
$latex h+c=4$
$latex 1+c=4$
$latex c=3$
$latex {{c}^2}=9$
Now, we use the equation $latex {{b}^2}={{c}^2}-{{a}^2}$ to find the value of $latex {{b}^2}$:
$latex {{b}^2}={{c}^2}-{{a}^2}$
$latex =9-4$
$latex =5$
Substituting the values of h, k, $latex {{a}^2}$ and $latex {{b}^2}$, we have:
$latex \frac{{{(x-1)}^2}}{4}-\frac{{{(y-1)}^2}}{5}=1$
### EXAMPLE 2
If a hyperbola has foci at (2, 0), (2, 6) and vertices at (2, 1), (2, 5), what is its equation?
##### Solution
In this case, we see that the x coordinates of the foci and the vertices are the same. This means that the transverse axis is parallel to the y axis and the hyperbola equation has the form:
$latex \frac{{{(y-k)}^2}}{{{a}^2}}-\frac{{{(x-h)}^2}}{{{b}^2}}=1$
We use the vertices (2, 1) and (2, 5) along with the midpoint formula to find the center:
$latex (h, k)=(\frac{2+2}{2}, \frac{1+5}{2})$
$latex =(2, 3)$
Now, we find the length of the transverse axis, 2a. Therefore, we find the difference in the y coordinates of the vertices:
$latex 2a=|5-1|$
$latex 2a=4$
$latex a=2$
$latex {{a}^2}=4$
We use the coordinates of the foci, $latex (h, k \pm c)$, to find the value of $latex {{c}^2}$. Therefore, we have $latex (h, k-c) = (2, 0)$ and $latex (h, k+c)=(2, 6)$. We can use any of these points to find the value of c:
$latex k+c=6$
$latex 3+c=6$
$latex c=3$
$latex {{c}^2}=9$
Now, we use the equation $latex {{b}^2}={{c}^2}-{{a}^2}$ to find the value of $latex {{b}^2}$:
$latex {{b}^2}={{c}^2}-{{a}^2}$
$latex =9-4$
$latex =5$
Substituting the values of h, k, $latex {{a}^2}$ and $latex {{b}^2}$, we have:
$latex \frac{{{(y-3)}^2}}{4}-\frac{{{(y-2)}^2}}{5}=1$
## Hyperbolas with center outside the origin – Practice problems
Apply the methods and steps seen above to solve the following problems and find the equations of the hyperbolas using the coordinates of the vertices and the foci. | 2023-01-29 12:03:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.793404221534729, "perplexity": 292.5788122242948}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499713.50/warc/CC-MAIN-20230129112153-20230129142153-00611.warc.gz"} |
http://ipsit.bu.edu/documents/lexi_preprint/node6.html | ## Trellis-oriented lexicodes
We now consider a different generating mapping for the -construction which will generate families of codes oriented towards a reduced decoding trellis.
There is a unique, minimal [12] trellis for an arbitrary linear block code. Known as the BCJR [1] trellis, it has no more edges or vertices at each time index than any other trellis for the code and can be constructed fairly easily [8,11]. It is shown in [8,11] that the minimal span generator matrix (MSGM) for a linear code, in which the sum of the spans of the binary generators is minimized, reflects the properties of the corresponding BCJR trellis. Namely, if G is an MSGM of a code , comprised of the generators G1,..., Gk, then the number of vertices Vi and edges Ei in the corresponding BCJR trellis is given by:
(6)
where pi and fi are the dimensions of the past and future subcodes for the i-th index of . These dimensions may be computed for i = 0, 1,..., n as follows [11]:
pi = |j : R(Gj) i|
As before, and refer to the rightmost and leftmost non-zero indices of a vector, respectively. With a minor modification of the lexicographic construction we may exploit the above relations to locally minimize two measures of trellis complexity: state complexity, which is the maximum number of states at any time interval of the trellis, and Viterbi decoding complexity. Specifically, is such a family of codes, and it locally minimizes trellis complexity if returns the vector with maximum distance from whose bit-wise reverse ( REV) is lexicographically earliest:
For example, if the vectors {1011, 1101, 1110} are at maximum distance from a code , then returns the vector 1110. Because of their locally minimal trellis complexity, these codes may be justly called trellis-oriented.''
Theorem 2 establishes that locally minimizes trellis complexity among local extensions with optimal code parameters. It is possible to improve Viterbi decoding complexity by using a generating mapping which produces a longer extension, but the information rate of the resulting code will be inferior.
Theorem 2 Let be a linear code. Among those single-iteration extensions with shortest length, the generator mapping minimizes the Viterbi decoding complexity and trellis state complexity.
Proof: Suppose that G is an MSGM for the (n, k, d ) code whose past and future subcodes have dimensions pi and fi respectively, and whose trellis has vertices Vi and edges Ei, i + 1 at the i-th time interval. Now, consider appending to G the generator v = 1| as in Equation (3) of the definition of the -construction. The generated code will have parameters (n' = n + , k + 1, d ). Without loss of generality we may assume that the resulting generator matrix of is an MSGM. If we denote the difference in lengths between and by n = n' - n, then a simple analysis shows that will have past and future subcode dimensions:
p'i = arrayll$0$ pi - n if n < i < R(v) pi - n + 1 ifR(v) i n' f'i = arrayll$k+1$ k if 0 < i n fi - n if n < i n'
(7)
We may then use (7) and (9) at each time unit i to compute the number of vertices and edges ( | V'i|,| E'i|) in the BCJR trellis for based on the vertices and edges ( | Vi|,| Ei|) in the BCJR trellis for :
(| V'i|,| E'i|) = arrayll(1, 2) (2, 2) (2| Vi - n|, 2| Ei - n|) (| Vi - n|,| Ei - n|)
(8)
It is now a simple matter of algebra to compute the difference in Viterbi decoding complexities between the trellises of and :
Since | Ei| | Vi| for all i, minimizing the change in Viterbi decoding complexity depends on minimizing R(v) = R1|, which in turn depends on having the 1 bits of as far to the left as possible. On the other hand, cannot have weight less than the covering radius of if it is to produce an extension of shortest length. Thus, the mapping is one of several mappings which meet both criteria for local optimality: minimizing R(v) and having . Other intuitive generating mappings, such as picking lexicographically latest vectors at maximum distance from the code, are not always locally optimal because the lexicographically latest vector need not (and generally does not) have the minimum rightmost index.
Equation 10 also proves that minimizing is the appropriate criterion for minimizing state complexity. This is because larger values of correspond to doubling more vertices | Vi - n| when generating Vi'. Thus, locally minimizes state complexity as well.
width4pt depth2pt height6pt
Table: Generator matrix for the dimension 4 minimum distance 3 trellis-oriented -code. The generator padding for each iteration is marked in bold.
0000111 0011100 0110010 1111000
Table 2 depicts the (7, 4, 3) trellis-oriented -code that was generated similarly to the code in Table 1. In fact, Method 1 can be trivially reversed to compute , without affecting the running time or space:
Method 2 Consider a set of vectors , and a code with MSGM G whose generators are in lexicographically increasing order. The following greedy method computes the lexicographically earliest bitwise reversed vector among the represented cosets in O(n| V|) time and O(n) space.
1. for each
v = (v1, v2, v3,..., v||) do
2. fori from 1 to n
3. if
is a 1 then
4.
v v + Gi
5. store the modified v;
6. among all stored v, return the lexicographically earliest
The proof of correctness and complexity for Method 2 follows trivially from the proof of Method 1. | 2017-12-13 17:03:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8479645252227783, "perplexity": 1321.3726903593342}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948529738.38/warc/CC-MAIN-20171213162804-20171213182804-00792.warc.gz"} |
https://kb.osu.edu/dspace/handle/1811/10787 | # HIGH RESOLUTION INFRARED SPECTRUM OF THE $\nu_{8} (E')$ BAND OF CYCLOPROPANE-$D_{6}{^{*}}$
Please use this identifier to cite or link to this item: http://hdl.handle.net/1811/10787
Files Size Format View
1979-ME-06.jpg 58.66Kb JPEG image
Title: HIGH RESOLUTION INFRARED SPECTRUM OF THE $\nu_{8} (E')$ BAND OF CYCLOPROPANE-$D_{6}{^{*}}$ Creators: Lovell, R. J.; Daunt, S. J.; Blass, W. E.; Halsey, G. W. Issue Date: 1979 Publisher: Ohio State University Abstract: The $\nu_{8} (E^{\prime})$ perpendicular band of $C_{3}D_{6}$ has been recorded with a resolution of 0.018 - $0.024 cm^{-1}$ on the University of Tennessee five meter Littrow spectrometer. The subband Q branches are gathered closely together near the band center ($2209 cm^{-1})$ and their identification is not possible. However, the $R_{R}$ and $P_{P}$ transitions to either side of the band center are well resolved. The current state of the assignment and analysis will be reported. Description: $^{*}$Research supported by the Planetary Atmospheres program of the National Aeronautics and Space Administration under Grant NGL-43-001-006. Author Institution: URI: http://hdl.handle.net/1811/10787 Other Identifiers: 1979-ME-06 | 2017-01-17 17:23:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6256870627403259, "perplexity": 3772.4931938818145}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560279933.49/warc/CC-MAIN-20170116095119-00272-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://sttp.softwaremill.com/en/v3.7.3/requests/authentication.html | # Authentication
sttp supports basic, bearer-token based authentication and digest authentication. Two first cases are handled by adding an Authorization header with the appropriate credentials.
Basic authentication, using which the username and password are encoded using Base64, can be added as follows:
import sttp.client3._
val username = "mary"
val password = "p@assword"
A bearer token can be added using:
val token = "zMDjRfl76ZC9Ub0wnz4XsNiRVBChTYbJcE3F"
basicRequest.auth.bearer(token)
## Important Note on the Authorization Header and Redirects
The Authorization header is by default removed during redirects. See redirects for more details.
## Digest authentication
This type of authentication works differently. In its assumptions it is based on an additional message exchange between client and server. Due to that a special wrapping backend is need to handle that additional logic.
In order to add digest authentication support just wrap other backend as follows:
val myBackend: SttpBackend[Identity, Any] = HttpClientSyncBackend()
new DigestAuthenticationBackend(myBackend)
Then only thing which we need to do is to pass our credentials to the relevant request:
val secureRequest = basicRequest.auth.digest(username, password)
It is also possible to use digest authentication against proxy:
val secureProxyRequest = basicRequest.proxyAuth.digest(username, password)
Both of above methods can be combined with different values if proxy and target server use digest authentication.
Also keep in mind that there are some limitations with the current implementation:
• there is no caching so each request will result in an additional round-trip (or two in case of proxy and server)
• authorizationInfo is not supported
• scalajs supports only md5 algorithm
## OAuth2
You can use sttp with OAuth2. Looking at the OAuth2 protocol flow, sttp might be helpful in the second and third step of the process:
1. (A)/(B) - Your UI needs to enable the user to authenticate. Your application will then receive a callback from the authentication server, which will include an authentication code.
2. (C)/(D) - You need to send a request to the authentication server, passing in the authentication code from step 1. You’ll receive an access token in response (and optionally a refresh token). For example, if you were using GitHub as your authentication server, you’d need to take the values of clientId and clientSecret from the GitHub settings, then take the authCode received in step 1 above, and send a request like this:
import sttp.client3.circe._
import io.circe._
import io.circe.generic.semiauto._
val authCode = "SplxlOBeZQQYbYS6WxSbIA"
val clientId = "myClient123"
val clientSecret = "s3cret"
case class MyTokenResponse(access_token: String, scope: String, token_type: String, refresh_token: Option[String])
implicit val tokenResponseDecoder: Decoder[MyTokenResponse] = deriveDecoder[MyTokenResponse]
val backend = HttpClientSyncBackend()
val tokenRequest = basicRequest
.auth
.basic(clientId, clientSecret)
val authResponse = tokenRequest.response(asJson[MyTokenResponse]).send(backend)
val accessToken = authResponse.body.map(_.access_token)
1. (E)/(F) - Once you have the access token, you can use it to request the protected resource from the resource server, depending on its specification. | 2022-11-27 01:35:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.21693286299705505, "perplexity": 7617.532372524984}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710155.67/warc/CC-MAIN-20221127005113-20221127035113-00172.warc.gz"} |
https://ask.sagemath.org/question/43287/solved-why-does-integratepsiyfyy-return-an-error-but-integratepsityftyy-works/ | # Solved: Why does integrate(psi(y)*f(y),y) return an error but integrate(psi(t,y)*f(t,y),y) works?
Hi there,
I am trying get an symbolic expression for the convolution $$(\psi \star f)(x) := \int\limits_{\mathbb{R}} \psi(x-y) f(y) {d y}$$
of two functions $f, \psi: \mathbb{R} \to \mathbb{R}$ as follows:
var('y') psi = function('psi')(y) f = function('f')(y) integrate(psi(x-y)*f(y),y)
upon which I get the error message
RuntimeError: ECL says: Error executing code in Maxima:
If I add an extra argument to the two functions and define them as $$f, \psi : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$$ as follows:
var('t') psi = function('psi')(t,y) f = function('f')(t,y) integrate(psi(t,x-y)*f(t,y),y)
there is a surprise, it suddenly works! I get the desired symbolic expression on which I can run diff(..,x) and all the other built-in functions.
TL;DR Why does integrate(psi(y)*f(y),y) return an error?
Solution Use sympy backend for symbolic integration as in integrate(psi(x-y)*f(y),y, algorithm="sympy")
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I think that's a bug in the Sage-to-maxima interface. Compare :
sage: var('y')
....: psi = function('psi')(y)
....: f = function('f')(y)
....:
y
sage: (psi(x-y)*f(y)).maxima_methods().integrate(y)
integrate(f(y)*psi(x - y), y)
sage: (psi(x-y)*f(y)).maxima_methods().integrate(y).diff(x)
integrate(f(y)*D[0](psi)(x - y), y)
with :
sage: (psi(x-y)*f(y)).integrate(y)
---------------------------------------------------------------------------
[ Long backtrace ]
RuntimeError: ECL says: Error executing code in Maxima:
BTW :
sage: integrate(psi(x-y)*f(y),y, algorithm="sympy")
integrate(f(y)*psi(x - y), y)
sage: integrate(psi(x-y)*f(y),y, algorithm="sympy").diff(x)
integrate(f(y)*D[0](psi)(x - y), y)
sage: integrate(psi(x-y)*f(y),y, algorithm="giac")
integrate(f(y)*psi(x - y), y)
sage: integrate(psi(x-y)*f(y),y, algorithm="giac").diff(x)
integrate(f(y)*D[0](psi)(x - y), y)
Another exemple of this quirk (against an old saw by Leibnitz, IIRC) :
sage: foo=arctan(x).diff(x).subs(x^2==-p).maxima_methods().powerseries(p,0).subs
....: (p==-x^2).simplify(); foo
sum((-1)^i5*x^(2*i5), i5, 0, +Infinity)
sage: foo.maxima_methods().integrate(x)
sum((-1)^i5*x^(2*i5 + 1)/(2*i5 + 1), i5, 0, +Infinity)
sage: foo.integrate(x)
---------------------------------------------------------------------------
[ Long backtrace again... ]
RuntimeError: Encountered operator mismatch in maxima-to-sr translation
and integrate(foo, x) gives the same final error message.
I encounter the same problem (and the same "solution") in Trac#13071. Other integration Trac tickets may have a similar origin, (but I lack the time needed to check them...).
more
Thank you very much, that is a very thorough answer!
Just started playing around with SAGE, didn't expect the underlying cracks to show up that quickly....
Do you think I should file another ticket or can I somehow 'push' yours?
( 2018-08-16 10:25:01 -0600 )edit
The cracks do exist, but are not that frequent... The point of Sage is to group various tools under a uniform framework, thus easing the use of mutually incompatible tools. The existence of chacks must not be an obstacle (and those cracks can be repaired, hopefully..).
I am not yet ready to file a new ticket about this specif point (inconsistency between two Maxima interfaces) : I do not have enough information on the specifics of the problem (and not enough time to dig them up). The ticket I quoted is but an example... But feel free to go ahead.
( 2018-08-16 10:50:31 -0600 )edit
That's good to know, I'll take it as an exception then -- and nothing better to understand the software then a bug being explained!
( 2018-08-16 12:02:38 -0600 )edit
This is definitely a bug, which seems to originate from the name psi for one of the two functions: it is confused with Maxima's special function psi, see http://maxima.sourceforge.net/docs/ma...
If you replace psi by another name, everything works:
sage: f = function('f')
sage: g = function('g')
sage: y = var('y')
sage: integrate(g(x-y)*f(y), y)
integrate(f(y)*g(x - y), y)
UPDATE (13 March 2019): the bug is now tracked via the ticket #27475.
more
Thank you for the quick answer! I struggled to format my post -- what are the html tags for the nice verbatim code block?
( 2018-08-16 10:37:46 -0600 )edit
seems to be a whitespace indent Found it!
( 2018-08-16 11:56:40 -0600 )edit
Alternatively, you can select the code block with the mouse and click on the button "101010" in the header bar.
( 2018-08-17 07:18:13 -0600 )edit | 2019-11-12 14:38:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.26641935110092163, "perplexity": 7106.308033667577}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496665573.50/warc/CC-MAIN-20191112124615-20191112152615-00256.warc.gz"} |
https://www.math.ias.edu/seminars/abstract?event=122525 | # Gromov-Witten theory of locally conformally symplectic manifolds and the Fuller index
Princeton/IAS Symplectic Geometry Seminar Topic: Gromov-Witten theory of locally conformally symplectic manifolds and the Fuller index Speaker: Yakov Savelyev Affiliation: University of Colima Date: Thursday, February 9 Time/Room: 11:15am - 12:15pm/S-101
We review the classical Fuller index which is a certain rational invariant count of closed orbits of a smooth vector field, and then explain how in the case of a Reeb vector field on a contact manifold $C$, this index can be equated to a Gromov-Witten invariant counting holomorphic tori in the locally conformally symplectic manifold $C \times S^1$. This leads us to prove a certain variant of the classical Seifert conjecture for the odd dimensional spheres. | 2017-12-16 01:20:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4566855728626251, "perplexity": 806.2517412150547}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948581033.57/warc/CC-MAIN-20171216010725-20171216032725-00389.warc.gz"} |
https://ham.stackexchange.com/tags/antenna/hot | # Tag Info
13
It looks like a fibre or telephone cable strung between buildings. It sets a good precedent for setting up your antenna though - if you can get access to the other rooftop at night, and a catapault or fishing rod. Make it fairly official-looking, with some large bolts and a labels with bar codes and lots of numbers.
6
If you do it right, the bead(s) shouldn't get hot at all, and shouldn't add much loss. The impedance of the bead(s) should be about 10 x the impedance of the dipole. The current on the feedline with no balun might be about half the antenna current. With the beads on it it'll be less than 1/10 of the curent, or 1/100 of the power. Small beads (3.5 mm inside, ...
6
I know that the gap between the wall and the antenna mast that would be created using the wall mounts you mention is not for lightning damage prevention purposes. If a bolt of lightning can travel hundreds of meters from a cloud to the ground, another half-meter of air between a mast and a building won't be much of a deterrent, especially when the bolt ...
5
It should be fine. Stranded wire will take that kind of bending easily. If you were to tie and untie the knot many times you might eventually weaken and break some of the strands, but if it's tied once and left in position, it won't create so much stress. If you're worried, of course, there are alternatives — you could slide a small plastic bead onto the ...
5
The cable strung between the two buildings has two parts, a strong wire probably made of stainless steel or similar, then you can see some form of cable, as Tommexus said it's probably fibre or telephone cable, or maybe coax, hanging from the wire with lots of evenly spaced cable supports. The steel wire is used to provide a strong support between the ...
3
A half wave horizontal element at the junction between the boom and the mast with a good electrical contact between the midpoint of the horizontal dipole and the mast will create a low impedance point on the mast. Then place another half wave element down 3/4 of a wavelength along the length of the mast. You may even place two of them, one parallel to the ...
3
Nothing. A G2 storm is a common event; according to NOAA data it happens an average of 600 times per 11-year sunspot cycle. It won't hurt your radio, but it might make HF communication a little more difficult. In fact, no category of storm is expected to hurt your radio in particular, although G4 and G5 storms can cause major radio blackouts, not to mention ...
3
It's all to do with mutuals, and you can't wish them away - even for 'ideal' antennas. Consider a simple case, two elements a reasonable distance apart. Give them excitation amplitudes +1 and -1. From your assumptions, the radiated power is 2, irrespective of their locations. The power is no longer radiated isotropically, there is some array pattern, but ...
3
Think of the antenna as a whole as being many inductors in series along its length and along with that many capacitors branching off like a tree (capacitors in the air) going back to the "negative side", so parallel capacitors. The loading coil needs actual current to work with, the further up on the antenna that you mount it, the less current ...
2
Years ago I used a simulation like this one for the same purpose. The antenna you refer to is a tuned antenna, so it is not wideband. The antenna that I made is a wideband, really flat response (conversion from field strength to output voltage is frequency-independent). Conclusion: the difficulty is the design of the low-noise amplifier. Depending on what ...
2
The formula in the question is taken from page 44 of a book called The Theory of Electromagnetic Wave Propagation by CH Papas. The text says that the formula describes the radiation pattern for a center driven wire antenna. Radiation patterns normally show a graph of relative intensity which isn't the same thing as efficiency. For transmitting, the ...
2
Different beats have different attenuation at different frequencies. My gut tells me you want at least 6 dB, preferably more. Guys seem to wind their own inductors as RF chokes but not sure how much attenuation at the desired frequencies.
2
The pictures look a lot like post-socialist country big city residental area. It may as well not be, YMMV. The Yagi looks like an abandoned old analog TV antenna. Most ham installations use vertical polarization in UHF and this one looks pretty much horizontal. The cable between the buildings is not an antenna at any rate. This is a communincations cabe of ...
2
I'm not sure if this is the correct answer or not: At varying electric lengths, at any moment in time, one part of the antenna is producing the wanted field(s) and another part might be producing a canceling field(s). With a half wavelength, all you get is the wanted field and no canceling field. The canceling field (I call it that because it works against ...
1
Note: question changed since I wrote this answer. Two classic and highly effective fractal antennas are the log periodic antenna and the many variations (planar, conical, cylindrical, etc.) of the log spiral antenna. The main characteristic of both of these antenna families is extremely broad bandwidth, which can be attributed to the fractal self similarity ...
1
It isn't. Gain in the broadside (θ=0) direction isn't "efficiency". A half-wave dipole doesn't maximize that parameter, even among dipoles; it increases until roughly 1.25 wavelengths, then decreases, then generally increases again, in an oscillating way. There are nice things about half-wave dipoles (a convenient, easily-matched impedance and a ...
1
You don't need a spectrum analyzer for such low frequencies, more like a very accurate voltmeter that you can record over long periods. As the antenna, you will need something huge, I recommend the power grid. Filter everything out above 59 Hz and see what you are left with. Might have to do some different filtering to try and find Schumann resonances. Make ...
1
If you need calibrated output then you probably won't be able to do it. That's why they are so expensive. It's like a \$20 SDR dongle versus a \$2,000 spectrum analyzer, with the most notable difference being that the spectrum analyzer is calibrated and tells you how much power is at each frequency. If you don't need calibrated output and just want to get a ...
1
Consider a simpler problem: Stack two dipoles at 0.01 wavelength separation. Connect the endpoints and feed power to one of the dipoles and short the feed-point of the other. What you then have is a folded dipole. Impedance 300 ohms. Then remove the short and feed both feed-points. The impedance of each one will be 150 ohms. They are series connected on a ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2021-10-27 11:00:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.48722004890441895, "perplexity": 1249.783271777701}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323588113.25/warc/CC-MAIN-20211027084718-20211027114718-00102.warc.gz"} |
https://datascience.stackexchange.com/questions/87754/interpreting-a-confusion-matrix | Interpreting a confusion matrix [closed]
I have a binary classification problem.
The accuracy score is 52%
The precision for 0 is 53% and the precision for 1 is 49%
When using predict_proba() does this mean my model more accurately predicts when the outcome should be classified zero as opposed to one?
I'm not sure if this is telling me that I should be using the the first value (ynew[0][0]) returned from predict_proba() as opposed to the second (ynew[0][1]).
Here is the entire confusion matrix:
• one can compare both values and decide on largest proba – Nikos M. Jan 10 at 18:34 | 2021-01-19 17:39:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7720202207565308, "perplexity": 933.4074234044845}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703519600.31/warc/CC-MAIN-20210119170058-20210119200058-00413.warc.gz"} |
http://mymathforum.com/number-theory/320429-flt-hope-end-story-2.html | My Math Forum FLT (hope the end of the story...)
Number Theory Number Theory Math Forum
January 20th, 2016, 09:32 PM #11
Banned Camp
Joined: Dec 2012
Posts: 1,028
Thanks: 24
Quote:
Originally Posted by Azzajazz I'm sorry, but you're contradiciting yourself. You start off saying $A = aK$ and then your entire proof hinges on taking the limit as $K\to\infty$. But then in your last post you say that $A$ can't approach $\infty$ as $K\to\infty$ because $A$ has to be finite. But it obviously does. As for your argument against my second point, there's an even bigger problem now. Let's fix some $A$ (which is what you seem to be doing) and then choose some $K$ such that $A = aK$, that is, $K|A$. Now, this implies that $K \leq A$. Now, if we take the limit as $K\to\infty$, we will eventually reach a point where $K>A$ since $A$ is fixed. Now it is impossible that $K|A$. This is the last post I'll make on this subject, since it seems hopelessly full of contradictions and errors.
Sorry, too messy.
- First forgot any concerning on A and B.
As I told and you see nothing change if A is an integer in Sum/StepSum/Integral:
$\displaystyle A^3 = \sum_{x= 1}^{A} (3X^2/K^3-3X/K^3+1/K^3) = \lim_{K\to\infty} \sum_{x= 1/K}^{A} (3x^2/K^3-3x/K^3+1/K^3) = \int_{0}{A} 3x^2$
If $\displaystyle A\in Q$ nothing change again (you can't use the standard Sum, so just use the Step Sum):
$\displaystyle A^3 = \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \lim_{K\to\infty} \sum_{x= 1/K}^{A} (3x^2/K^3-3x/K^3+1/K^3) = \int_{0}{A} 3x^2$
We are now talking of another area bellow the first derivate y'=3x^2 that has to satisfy the FLT conditions.
- First since we unerstand that we can square the derivate with colums, WITH AS A CONDITION THAT THE COLUMS MUST CONTINUOSLY TESSELLATE THE AREA, we are forced to say: ok, we accept the condition that there can be a common divisor between A,B (and sorry I'm not talking about a factor, as probably I left somewhere, but of a divisor that can be bigger as you want, just have to divide perfectly A & B, so cannot be for example $\displaystyle \pi$...).
Another story for C since we just suppose it can be divided by the same K, but as we see after, this condition is necessary to say: ok we start dividing the new area without making "holes" between B and B+1/K, so I said nothing about the fact that there is a difference from the first column of width 1/K of the Step Sum :
$\displaystyle \sum_{x= B+1/K}^{C} (3x^2/K-3x/K^2+1/K^3)$
and the integration methos where we start from B directly:
$\displaystyle \lim_{K\to\infty} \sum_{x= B+1/K}^{C} (3x^2/K-3x/K^2+1/K^3) = \int_{B}^{C} 3x^2 dx = C^3 -B^3$
So still if we can "square" the first part of our "new" area $\displaystyle C^3 -B^3$ with the Step Sum or the integral, we FOR SURE, can't reach the upper limit C, if C is an irrational using a Step Sum where the Step ia a rationa because we are always unable to cover the last part of the area itll x=C
Probably will be easy to see the point rewriting the sum shifted back to the same lower limit 1 (than in 1/K than to the limit)
$\displaystyle \sum_{X= B+1}^{C} (3X^2/K^3-3X/K^3+1/K^3) = \sum_{X= 1}^{C-B} (3(X+B)^2/K^3-3(X+B)/K^3+1/K^3)$
Than re-transform it in a Step Sum with x=X/K etc...
I'm working on....
Thanks
Ciao
Stefano
Thanks
Ciao
Stefano
January 21st, 2016, 02:23 AM #12 Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 Sorry in the previous message there are several typing problems: the rights are of course: If $\displaystyle A\in N$ $\displaystyle A^3 = \sum_{X= 1}^{A} (3X^2-3X+1) = \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \lim_{K\to\infty} \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \int_{0}^{A} 3x^2 dx$ If $\displaystyle A\in Q$ nothing change again (you can't use the standard Sum, so just use the Step Sum): $\displaystyle A^3 = \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \lim_{K\to\infty} \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \int_{0}^{A} 3x^2 dx$ If $\displaystyle C\in R ; but ; C\notin N ; C\notin Q$ you can't use the standard Sum and the Step Sum, because is: $\displaystyle C^3 > \sum_{x= 1/K}^{C'} (3x^2/K-3x/K^2+1/K^3)$ where C' < C is the closest integer or rational to the Real Irrational C that you can rise step 1/K without rest (sorry there is a latex sign but like that is clearer to every reader). So you have to go to the limit (to reach the upper limit): $\displaystyle C^3 = \lim_{K\to\infty} \sum_{x= 1/K}^{C} (3x^2/K-3x/K^2+1/K^3) = \int_{0}^{C} 3x^2 dx$ (To CRGreathouse: will be good think if message can be revised for a longer time....) Last edited by skipjack; January 21st, 2016 at 07:20 AM.
January 21st, 2016, 06:55 AM #13 Senior Member Joined: Apr 2015 From: Barto PA Posts: 170 Thanks: 18 complicatemodulus wrote: (To CRGreathouse: will be good [...]....) Where is CRG?
January 21st, 2016, 10:28 PM #14 Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 To Azzjaz: there is of course a "math" solution via equation that I kindly ask to check since I always make typing and stupid error in the develope: With the Sum (if A,B,C are in N) or the Step Sum (if A,B,C are in Q) we can reduce FLT to A^3=? C^3-B^3 to: $\displaystyle \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) =? \sum_{x= B+1/K}^{C} (3x^2/K-3x/K^2+1/K^3)$ So the easy technique we can apply on the sum is to return the lower limit to 1/K (arranging the terms), and reduce what rest to a new little power (is it a cube else) plus some other terms we can prove are bigger than 1: Lowering the B+1/K lower limit to 1/K, without changin the result is equal to put instead of x the x+B in each x dependent term so: $\displaystyle \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \sum_{x= 1/K}^{C-B} (3(x+B)^2/K-3(x+B)/K^2+1/K^3)$ Now we can develope the term and extractly rebuild what we know is again a (more little) cube: $\displaystyle \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \sum_{x= 1/K}^{C-B} (3(x^2+2Bx+B^2)/K-(3x+3B)/K^2+1/K^3)$ Or $\displaystyle \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \sum_{x= 1/K}^{C-B} (3x^2/K-3x/K^2+1/K^3) + \sum_{x= 1/K}^{C-B} (6Bx+3B^2)/K +3B/K^2$ or subtractinge the littlest cube: $\displaystyle \sum_{x= 1/K}^{A+B-C} (3x^2/K-3x/K^2+1/K^3) = \sum_{x= 1/K}^{C-B} (6Bx+3B^2)/K +3B/K^2$ Now we know that the Sum of (2x/K-1/K^2) terms is a square so since we do not have the -1/K^2 we have just to add it in the first sum (square) and subtract in the second remembering the 3B out of the sum: $\displaystyle (A+B-C)^3 = 3B \sum_{x= 1/K}^{C-B} (2x/K-1/K^2) + \sum_{x= 1/K}^{C-B} 3B^2/K+ 3B/K + 3B/K^2$ or $\displaystyle (A+B-C)^3 - 3B (C-B)^2 = \sum_{x= 1/K}^{C-B} 3B^2/K+ 3B/K + 3B/K^2$ or $\displaystyle (A+B-C)^3 - 3B (C-B)^2 = K(C-B)*( 3B^2/K+ 3B/K + 3B/K^2)$ or $\displaystyle (A+B-C)^3 - 3B (C-B)^2 = (C-B)*( 3B^2+ 3B + 3B/K)$ or $\displaystyle (A+B-C)^3 - 3B (C^2+B^2-2BC) = (C-B)*( 3B^2+ 3B + 3B/K)$ or $\displaystyle A^3+B^3-C^3+3A^2B-3A^2C+3AB^2-3B^2C+3AC^2+3BC^2 - 3BC^2 -3B^3+6B^2C = 3B^2C+ 3BC-3B^3 -3B^2 + 3B(C-B)/K)$ or REMEMBERING THAT: A^3+B^3-C^3 =0 $\displaystyle 3A^2B-3A^2C+3AB^2-3B^2C+3AC^2+3BC^2 - 3BC^2 -3B^3+6B^2C = 3B^2C+ 3BC-3B^3 -3B^2 + 3B(C-B)/K)$ or $\displaystyle 3A^2B-3A^2C+3AB^2+3AC^2 - 3BC +3B^2 = 3B(C-B)/K)$ or $\displaystyle K = 3B(C-B)/ (3A^2B-3A^2C+3AB^2+3AC^2 - 3BC +3B^2 )$ or $\displaystyle K = 3B(C-B)/ (3A(B^2+C^2) - 3B(C-B)-3A^2(C-B))$ ...sorry I've to post before loose all....
January 21st, 2016, 10:30 PM #15 Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 pls check it and asap when correct the conclusion (if possible) Thanks Ciao Stefano
January 22nd, 2016, 10:11 AM #16 Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 There is an error from here: 3B/K must be 3B/K^2 to tap the -1/K^2 we create for the square $\displaystyle (A+B-C)^3 - 3B (C-B)^2 = \sum_{x= 1/K}^{C-B} 3B^2/K+ 6B/K^2$ or $\displaystyle (A+B-C)^3 - 3B (C-B)^2 = K(C-B)*( 3B^2/K+ 6B/K^2)$ or $\displaystyle (A+B-C)^3 - 3B (C-B)^2 = (C-B)*( 3B^2+ 6B/K)$ or $\displaystyle (A+B-C)^3 - 3B (C^2+B^2-2BC) = (C-B)*( 3B^2+ 6B/K)$ or $\displaystyle A^3+B^3-C^3+3A^2B-3A^2C+3AB^2-3B^2C+3AC^2+3BC^2 - 3BC^2 -3B^3+6B^2C = 3B^2C -3B^3 + 6B(C-B)/K$ or REMEMBERING THAT: A^3+B^3-C^3 =0 $\displaystyle A^2B-A^2C+AB^2+AC^2 = 2B(C-B)/K$ or $\displaystyle k= = 2B(C-B)/(A^2B-A^2C+AB^2+AC^2)$ that I let you say if it can be $\displaystyle K\in N$ and K>=1 Last edited by skipjack; January 22nd, 2016 at 11:23 AM.
January 23rd, 2016, 06:42 AM #17 Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 I try to fix all bugs here : PART 1) We all know Fermat Theorem $\displaystyle C^n<>A^n+B^n$ c1- for A,B,C integers, c2- for A,B,C coprimes, c3- for n>=3. The interesting case are for n=odd (for n=even there is a known simple proof) Knowing Mr. Wiles give us a proof (that many people can't understand) The goal is to find a simple proof that Fermat’s (can) wrote with his "simple" sum / infimus instruments. I hope I found a proof using a special Sum (I call Step Sum) on what we can show a "finite" descent in few step: - we strat the problem from A^n=... - than we reduce using sum properties to A^+B^n-C^n + mixed product = f(K). Since we state A^+B^n-C^n = 0 what rest are just n-1 terms that can define K. If we can show that after that K<1 that inply we can cut the salame with slices of rising thickness starting with a column that has an integer width > 1 so we have a contraddiction so we prove the FLT. - that works in the same way for all n>=3, since we can prove for n=3 than using the Newton develope properties and the induction proving for n=m, than prove is true also for n=m+1 so for all n. So be: $\displaystyle A = a*K$ $\displaystyle B = b*K$ C = unknown we suppose $\displaystyle C = c*K$ with: $\displaystyle A \in N$ $\displaystyle B \in N$ C = unknown we wanna check if will exist a $\displaystyle C \in N$ and: $\displaystyle a\in Q$ $\displaystyle b\in Q$ c= unknown we suppose $\displaystyle c\in Q$ $\displaystyle K\in N; K>=1$ We now rewrite FLT as: (1) $\displaystyle (a*K)^n <> (c*K)^n- (b*K)^n$ I use a property (known as the "telescopic sum") of the derivate of the powers curve $\displaystyle y=x^n$ that is $\displaystyle y'= n x^{(n-1)}$ : If $\displaystyle x\in Q : x=A/K$ then the area below the curve $\displaystyle y'= n x^{(n-1)}$ till x, we know is $\displaystyle x^n = A^n/K^n$ can be "squared" using a sum of colums base $\displaystyle 1/K$ and height equal to: $\displaystyle (X^n - (X-1)^n)/K^3$ or in math (here as example n=3): If $\displaystyle A\in N$ $\displaystyle A^3 = \sum_{X= 1}^{A} (3X^2-3X+1) = \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \lim_{K\to\infty} \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \int_{0}^{A} 3x^2 dx$ If $\displaystyle A\in Q$ nothing change again (but you can't use the standard Sum, so just use the Step Sum step 1/k, 2/k 3/K etc...): $\displaystyle A^3 = \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \lim_{K\to\infty} \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \int_{0}^{A} 3x^2 dx$ Now we know in our problem for sure A,B are in N, while for sure there is a solution for $\displaystyle C\in R$ but we don't know if it can be in the subset Q or N. So If $\displaystyle C\in R ; but ; C\notin N ; C\notin Q$ we can't use the standard Sum and the Step Sum, because for any $\displaystyle C' \in Q, K\in N$ is: $\displaystyle C^3 > \sum_{x= 1/K}^{C'} (3x^2/K-3x/K^2+1/K^3)$ where C' < C is the closest integer or rational to the Real Irrational C, that you can rise step 1/K without rest (sorry there is a latex sign but like that is clearer to every reader). So in case C is Irrational you have to go to the limit $\displaystyle K\to\infty$ to reach the upper limit with the last infimus step: $\displaystyle C^3 = \lim_{K\to\infty} \sum_{x= 1/K}^{C} (3x^2/K-3x/K^2+1/K^3) = \int_{0}^{C} 3x^2 dx$ As told by induction we can extend this for all n. Now the step will require you a minimum of flexibility since I will use the Sum operator in a new way, that is perfectly equal to the known one, in the result, but in my opinion show in a most easy and clear way how and why the integral works: We show that if we cut the area bellow the derivate with thinner and thinner colums of base 1/K and height $\displaystyle (X^n-(X-1)^n)/K^n$ at the limit for $\displaystyle K\to\infty$ we have the Riemann integral Putting as new variable in the sum $\displaystyle x=X/K$ (in the index dependet terms), to left unchanged the result we must works on the upper and lower limit that becomes: $\displaystyle A^3 = \sum_{X= 1}^{A} (3X^2-3X+1) = \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3)$ and now pulling $\displaystyle K\to\infty$ $\displaystyle A^3 = \sum_{x= 1/k}^{A} (3x^2/K-3x/K^2+1/K^3) = \lim_{K\to\infty} \sum_{x=1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \int_{0}^{A} 3x^2 dx = A^3$ I won't spend more time on this point since you've to digest it as it is. To help, you can make both limit using the (2) calling A=a*K and solving it using the known value of the sum of each single term. Once understood this point, in the same way, we can apply the sum to B and C considering now it's a Rational so we can reach it with a finite K, step 1/K. $\displaystyle B^3 = \sum_{x= 1/K}^{b*K} (3x^2/K-3x/K^2+1/K^3)$ $\displaystyle C^3 = \sum_{x= 1/K}^{c*K} (3x^2/K-3x/K^2+1/K^3)$ now we can rewrite FLT as: (1) $\displaystyle A^3= (a*K)^n =? (c*K)^n- (b*K)^n$ using the sum: $\displaystyle A^3 = \sum_{x= 1}^{A*k} (3x^2/K-3x/K^2+1/K^3) <> \sum_{x= B+1/K}^{C} (3x^2/K-3x/K^2+1/K^3)$ So if you digest the step sum you understand that for sure in the right hand side of the FLT if we pull $\displaystyle K\to\infty$ we can REMOVE THE <> sign and write: $\displaystyle A^3 = \lim_{K\to\infty} \sum_{x= B+1/K}^{C} (3x^2/K-3x/K^2+1/K^3) = \int_{B}^{C} 3x^2 dx = C^3 -B^3$ It’s easy to imagine that on the monotone smooth rising function if the limit exist (and we prove is C^3 -B^3) that it is the only one, infact for any $\displaystyle K\in Q$: $\displaystyle \sum_{x= B+1/K}^{C} (3x^2/K-3x/K^2+1/K^3) < \lim_{K\to\infty} \sum_{x= B+1/K}^{C} (3x^2/K-3x/K^2+1/K^3) = \int_{B}^{C} 3x^2 dx = C^3 -B^3$ So there is no $\displaystyle K\in N ; K>=1$ that satisfy the FLT equation, so it means that in our example the solution can be found just a the limit so, if we want to obtain the equal sign instead of the "<>" we have to immagine that fixing A,B we have to rise K to infinity so make an infinite serie of loop to rise the upper limit C that is $\displaystyle C\in R$ And using induction, as told, this can be easy proved true for any n>=3 I already explain why it works for n=2: the first derivate is a line, the second a flat line, so there will exist a solution in Q, and in N for K=1, for any integer A we choose with a couple of integer B and C, and as we know with the parametric equation to generate all the pitagorean triplets. In the Part.2 in the next post I will prove with math on the example n=3 what now are or seems just void words. To help to figure out what happen here some pictures, and what suggest to me to use this property to try to solve the FLT. Here a picture where I show the property of the derivate of the power curves y=x^n, where via integer or rational Columns of height coming from the Partial Binomial Develope (x^n-(x-1)^n), called gnomons, we can see the telescopic sum property rest true for any K till infinity. (pls, sorry, there are errors in the lables I've no time to fix... but I hope is clear enough) Here a Xls table that helps to show how the descent works (going closer and closer to the upper limit C in R): That is the base to understand Beal proof that follows... I just "invent" another way to cut the "salame": the modular algebra use same thickness slice "moduluo", I use a variable (of a known function) thickness slice. End Part 1.
January 23rd, 2016, 08:58 AM #18 Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 Part.2 I hope I left no bugs here: The point is now to see and prove with classic "reduction as absurdum" on the example for n=3. Using the Step Sum (I show befor can works also in Q) we can reduce FLT to A^3=? C^3-B^3 to: $\displaystyle \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) =? \sum_{x= B+1/K}^{C} (3x^2/K-3x/K^2+1/K^3)$ So the easy technique we can apply on the sum is: 1) return the lower limit to 1/K (arranging the terms of the sum) 2) reduce what rest (till possible) to a new little power (is it a cube or else) plus some other terms depending just on A,B,C and we will see also depending on K. Newton's develope assure use that till we have a sum contining x^m terms we can reduce them in littlest powers, till the last x^1 term that can be transformed in a square (easy proof unsing known induction). Point 1- Lowering the B+1/K lower limit to 1/K, without changing the result is equal to put instead of x the x+B in each x dependent term so: $\displaystyle A^3= \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \sum_{x= 1/K}^{C-B} (3(x+B)^2/K-3(x+B)/K^2+1/K^3)$ Now we can develope the terms and extractly rebuild what we know is again a (more little) cube, plus a square, plus other non x dependent terms: $\displaystyle \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \sum_{x= 1/K}^{C-B} (3(x^2+2Bx+B^2)/K-(3x+3B)/K^2+1/K^3)$ Or $\displaystyle \sum_{x= 1/K}^{A} (3x^2/K-3x/K^2+1/K^3) = \sum_{x= 1/K}^{C-B} (3x^2/K-3x/K^2+1/K^3) + \sum_{x= 1/K}^{C-B} (6Bx+3B^2)/K +3B/K^2$ or subtractinge the littlest cube: $\displaystyle \sum_{x= 1/K}^{A+B-C} (3x^2/K-3x/K^2+1/K^3) = \sum_{x= 1/K}^{C-B} (6Bx+3B^2)/K +3B/K^2$ Now we know that: 1- the Step Sum from 1/k till A+B-C of 3x^2/K-3x/K^2+1/K^3 is the cube of (A+B-C): 2- the Step Sum from 1/k till C-B of (2x/K-1/K^2) terms is the square of (C-B) so since we do not have the -1/K^2 we have just to add it in the first sum (square) and subtract in the second remembering the 3B out of the sum: $\displaystyle (A+B-C)^3 = 3B \sum_{x= 1/K}^{C-B} (2x/K-1/K^2) + \sum_{x= 1/K}^{C-B} 3B^2/K+ 3B/K^2 + 3B/K^2$ or $\displaystyle (A+B-C)^3 - 3B (C-B)^2 = \sum_{x= 1/K}^{C-B} 3B^2/K+ 6B/K^2$ or $\displaystyle (A+B-C)^3 - 3B (C-B)^2 = K(C-B)*( 3B^2/K+ 6B/K^2)$ or $\displaystyle (A+B-C)^3 - 3B (C-B)^2 = (C-B)*( 3B^2+ 6B/K)$ or $\displaystyle (A+B-C)^3 - 3B (C^2+B^2-2BC) = (C-B)*( 3B^2+ 6B/K)$ or $\displaystyle A^3+B^3-C^3+3A^2B-3A^2C+3AB^2-3B^2C+3AC^2+3BC^2 - 3BC^2 -3B^3+6B^2C = 3B^2C -3B^3 + 6B(C-B)/K$ and REMEMBERING THAT for FLT statment: A^3+B^3-C^3 =0 $\displaystyle A^2B-A^2C+AB^2+AC^2 = 2B(C-B)/K$ or $\displaystyle K= 2B(C-B)/(A^2B-A^2C+AB^2+AC^2)$ that I let you say if it can be $\displaystyle K\in N$ and K>=1 The point, I remember, is that if what rest from C^n-B^n is a power in N than it must be K independet... while we prove (I hope) it is not If you try as example: A=5, B=6, C=7 you can see what happen still if you don't see more or you don't use wolframalpha for the graph. So the conclusion is: till we rest in the integers, or rationals while all X terms will be re.transfomed in a non dependet K terms, the rest will alway depend by K... so.... (Beal will follows !) In other terms: from n>=2 the derivate is a curve so there are minimum 3 terms in what I call "the rational derivate". the x dependent can be transformed as seen to have a little cube and a little square, but a rest will always be present, so we can conclude: first derivate = curve ? Than no way... DINNER TIME...KIDS & WIFE AGAIN OUT OF CONTROLL... really hope no bugs... Last edited by complicatemodulus; January 23rd, 2016 at 09:28 AM.
January 23rd, 2016, 11:33 PM #19 Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 Still little bug in: when you see "C" as upper limit you've to read: - $\displaystyle C\in N$ if the sum is step 1 (classic sum) ($\displaystyle C_N$) - $\displaystyle C\in Q$ if the sum is step 1/K (K>1 Step Sum) ($\displaystyle C_Q$) - $\displaystyle C\in R$ if the sum is pulled at the limit for $\displaystyle K\to\infty$ or in the integral. ($\displaystyle C_R$) Probably will be better to call them with different index, for example: $\displaystyle C_N < C_Q < C_R$ well knowing that we consider them in this order only Thanks
January 25th, 2016, 09:56 PM #20 Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 Of course the first think you can do is try with n=2 K=1 starting from: $\displaystyle \sum_{x=1}^{A}(2x-1) = \sum_{x=1}^{C}(2x-1) - \sum_{x=1}^{B}(2x-1)$ or: $\displaystyle \sum_{x=1}^{A}(2x-1) = \sum_{x=B+1}^{C}(2x-1)$ Than apply the subtraction of the littlest square from the bigger and the "lowering" technique: $\displaystyle \sum_{x=1}^{A}(2x-1) = \sum_{x=1}^{C-B}(2(x+B)-1)$ or $\displaystyle \sum_{x=1}^{A}(2x-1) = \sum_{x=1}^{C-B}(2x-1)+ (C-B)*2B$ or $\displaystyle \sum_{x=C-B+1}^{A}(2x-1) = (C-B)*2B$ or $\displaystyle \sum_{x=1}^{A+B-C}(2(x+C-B)-1) = 2BC-2B^2$ or $\displaystyle \sum_{x=1}^{A+B-C}(2x-1) + (A+B-C)*2(C-B)= 2BC-2B^2$ or $\displaystyle (A+B-C)^2 + (A+B-C)*2(C-B)= 2BC-2B^2$ or $\displaystyle (A+B-C)^2 + 2AC-2AB+2BC-2B^2-2C^2+2BC-2BC+2B^2=0$ or $\displaystyle A^2+B^2+C^2+2AB-2AC-2BC + 2AC-2AB+2BC-2B^2-2C^2+2BC-2BC+2B^2=0$ or $\displaystyle A^2+B^2-C^2 =0$ So, and this is the point, once we again apply the FLT rule, we return to the identity, so we know it can works with some integer A,B,C called Pitagorean tryplets. If what we have in the right side cannot be transfromed in a power of an integer (as happen from n>=3) after the lowering technique we find that when we apply again FLT equation A^n+B^n-C^n=0 the REST will vanish just at one K that is $\displaystyle K<1 : K\notin N$ And seems not another void turn... Here on the graph is more easy to see and understand: or: Thanks Ciao Stefano
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Contact - Home - Forums - Cryptocurrency Forum - Top | 2019-09-15 14:51:19 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8452188372612, "perplexity": 1088.7019105321972}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514571506.61/warc/CC-MAIN-20190915134729-20190915160729-00358.warc.gz"} |
https://golsteyn.com/writing/easy-as-pi/ | Projects & Initiatives
# Easy as π
By Quentin Golsteyn React Native/Express.js • Posted December 1, 2019
They say the best way to learn math is by practicing problems over and over again. What if we could generate these problems on-demand? This is the premise of this project: to provide students with an endless supply of math problems.
With Easy as π, we wanted to create a mobile application that could:
1. Generates arithmetic problems at various levels of difficulty
2. Selects the correct level of difficulty of problem to display to a student
We completed this project in a team of 4 students as part of a software engineering course at the University of British Columbia. While the focus of the course was on the engineering aspect of software development (software architecture, scope, testing, etc.), we still took the opportunity to conduct user research prior to beginning development of this mobile app.
## Understanding our potential users
This app has two primary groups of users, students and teachers, which have different set of needs. Given these groups, we created several personas for each group based on our initial online research and prior experience.
Ewen is a grade 3 student and loves video games. He can quickly get addicted to games on his parent's phones. He is not very interested in school and lags behing compared to his classmates.
Jess is a teacher in a class of 32 students. She struggles to find time to help all her students individually and she is worried that if she spent more time, the rest of her class would not do anything productive.
Pete is a consultant in an engineering firm and has difficulty coming home early most days of the week. He wants to make sure his son gets enough practice in mathematics to make sure he does well in school.He wants him to learn how to study on his own, but he doesn't know where to start and is often not around to help him.
From these personas, we brainstormed a list of potential needs and pain points for each group, which helped inform the design of the mobile app.
Based the on the user research above, we determined the following features our app should include:
Students
1. Work on a daily math set
2. Ask for help from their teacher
3. Access learning material to solve a particular problem
4. Track their success rate
Teachers
1. Create a virtual classroom
2. Prepare the content of the daily math set
3. View individual student progress in each domain area
## High-fidelity mockups
I was responsible for implementing the mobile app. I started by sketching a potential user flow for the authentication process and for the main features of the student app.
With these sketches in mind, I created a set of high-fidelity mockups in Figma that outlines various steps of the onboarding process for teachers and students, and the primary screens of the student app.
Given more time, we would have started with low-fidelity mockups to present to teachers and parents before proceeding with more time-intensive prototypes. Further iterations before starting development could have helped us get a better understanding of what needed to be implemented, and avoid some churn later down the road.
## Implementation
We developed the mobile application in React Native, using the Expo library. The backend was an Node.js Express server using MongoDB for data storage. Authentication was done with Google OAuth.
### Generating math problems
The BC math curriculum from grade 1 to 5 was essential for this project. It provides specific learning outcomes students should be able to achieve at each year level. For example, in Grade 1, students are expected to be able to add numbers up to 10.
To determine the correct problem to show the student, students are categorized according to their grade and their expertise within their grade. The rules selected to generate a problem of a particular type will vary depending on the student's assigned level. Each rule of a particular problem type can be formulated as a set of controlled and derived variables.
For example, for addition problems in Grade 1, students are expected to be able to add a "large" number with a "small" number to values up to 9. In this scenario, we can define two controlled variables, $a$ and $b$ and one derived variable, $c$.
Controlled variables have their values bounded, for this example $5 < c < 9$ and $0 < a < 3$. A random value within the appropriate domain is assigned to each controlled variable when generating a math problem. The derived variables have a value derived from the controlled variables, here $b = c - a$.
With the variables defined, we can then specify a problem template and insert the variable values in the problem statement and answer. For our addition example, the problem statement could be Solve $a + b =\ ?$ and the answer would be $c$. With $c = 7$ and $a = 2$, the generated problem would therefore be Solve $2 + 5 =\ ?$ and the answer would be $7$.
### Adapting to the student's performance
Depending on the student performance, we can adjust the category the student is in to vary the problems difficulty. Students score or lose points depending on whether they get problems right or wrong. Each level a student is in allows for a maximum of 10 points. If a students exceeds these 10 points, they "graduate" to the next category. If a students drops below zero point, they are demoted a level.
Students receive one point for each correct answer, and lose 2 points for each wrong answer. It is important to note, however, that these scores are given per problem type. A student may be "Grade 2 high" on addition, but "Grade 1 medium" on substraction.
## Final product
After 2 months of work, we succesfully implemented Easy as π! Ultimately, this project gave us the opportunity to develop a mobile application from scratch, something many members on our team had never done before.
Written by Quentin Golsteyn
A front-end developer based in Vancouver, Canada. | 2022-12-06 10:04:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 12, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2602030634880066, "perplexity": 1162.347965329153}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711077.50/warc/CC-MAIN-20221206092907-20221206122907-00732.warc.gz"} |
http://math.stackexchange.com/questions/5108/how-is-mathbbc-different-than-mathbbr2 | # How is $\mathbb{C}$ different than $\mathbb{R}^2$?
I'm taking a course in Complex Analysis, and the teacher mentioned that if we do not restrict our attention to analytic functions, we would just be looking at functions from $\mathbb{R}^2$ to $\mathbb{R}^2$.
What I don't understand is why this is not true when we do restrict our attention to analytic functions. I understand that complex analytic functions have different properties than real functions on $\mathbb{R}^2$, but I don't understand why this is so. If I look at a complex number $z$ as a vector in $\mathbb{R}^2$, then isn't differentiability of $w=f(z)$ in $\mathbb{C}$ defined the same way as differentiability of $(u,v)=F(x,y)$ in $\mathbb{R}^2$?
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The standard example is complex conjugation. This is not analytic, although a smooth function $\mathbb{R}^2 \to \mathbb{R}^2$. – Martin Brandenburg Sep 21 '10 at 1:11
Somewhat related: mathoverflow.net/questions/3819/… – Jonas Meyer Sep 21 '10 at 1:25
There's nothing corresponding to "maximum modulus" on $\mathbb{R}^2$. – Guess who it is. Sep 21 '10 at 1:26
When you say "that complex analytic functions [...] than real functions" do you mean real analytic functions? – Freeze_S Jan 26 '14 at 12:44
For a function $f : \mathbb R^2 \to \mathbb R^2$ "differentiable" at a point $x \in \mathbb R^2$ means you have a linear approximation $f'_x$ which satisfies
$$\lim_{y \to x} \frac{f(x)-f(y)-f'_x(x-y)}{|x-y|} = 0$$
Saying that $f$ is complex analytic is the constraint that $f'_x$ is a complex linear function for all $x$. "Complex linear" means that not only is it true that $f'_x(av+bw)=af'_x(v)+bf'_x(w)$ for $a, b \in \mathbb R$, but it also holds for $a,b \in \mathbb C$.
One way to say $L$ is "Complex linear" is that $L$ is a regular (real linear) function plus $L(iv)=iL(v)$ for all vectors $v$. Stated in terms of the derivative, the directional derivative of $f$ in the direction $(0,1)$ is $i$ times the directional derivative in the direction $(1,0)$. In component notation these are the "Cauchy-Riemann" equations.
$$\frac{\partial f}{\partial y} = i \frac{\partial f}{\partial x}$$
where if you write $f(z) = f(x+iy) = u(x,y) + iv(x,y)$ translates to
$$\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}, \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$$
The formula:
$$\frac{\partial f}{\partial y} = i \frac{\partial f}{\partial x}$$
gives the key picture. A complex linear map means that the map looks like the composite of a rotation together with a re-scaling map $v \longmapsto av$ where $a \in \mathbb R$. These maps are called "conformal". The nice things about conformal linear maps is they preserve all angles -- they do not always preserve length. So the "nice" thing about complex differentiable maps is that if you have any collection of curves in the plane, and you apply your complex differentiable function to them, it preserves the angles of intersection of your curves. That's a very special property.
edit: An instructive example would be to think through two different functions from $\mathbb R^2$ to $\mathbb R^2$. The first function:
$$(x,y) \longmapsto x(\cos(y),\sin(y))$$
and the second function
$$(x,y) \longmapsto e^x(\cos(y),\sin(y))$$
The first function preserves the angles between the coordinate grid lines -- curves like $x=a$ and $y=b$ in the coordinate plane. The first function is not complex differentiable but the second is! So this means that the second function preserves all angles (not just the coordinate lines $x=a, y=b$). Can you spot curves in the domain which intersect in some angle $\theta$, but when after you compose them with the 1st function, their angle of intersection is not $\theta$?
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Wow, thanks. This is going to take some more thought, but I'm starting to get it. – jake Sep 21 '10 at 2:03
Analytic functions map tiny disks to tiny disks. (Of course that's not rigorous, but you could make it rigorous by putting in the right limit language.) Analytic functions can shift, stretch, and rotate disks, but they can't flip disks over.
Smooth functions of two real variables can map disks to ellipses. That is, they can stretch a disk more in one direction than in another. Complex analytic functions can't do that.
Complex conjugation is not analytic because it flips disks over.
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To go down a level from differentiability: the root of the difference between $\mathbb{C}$ and $\mathbb{R}^2$ comes from the multiplicative structure on $\mathbb{C}$. Look at the definition of differentiation itself: $\lim_{h\rightarrow 0} h^{-1}\cdot \left(f(z+h)-f(z)\right)$ - there's a multiplication here, by the multiplicative inverse of the (complex) number $h$, that simply can't be performed in $\mathbb{R}^2$ without giving it a field structure. There isn't 'a' derivative of a function from $\mathbb{R}^2 \mapsto \mathbb{R}^2$, just two partials; the multiplicative structure of $\mathbb{C}$ is then what forces the Cauchy-Riemann constraints on those partial derivatives and allows for a definition of the derivative as a single function from $\mathbb{C}\mapsto\mathbb{C}$.
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But that applies to functions $f:\mathbb{R}\to\mathbb{R}$ too. But that seems not enough for differentiability to imply analyticity. – Freeze_S Jan 24 '14 at 1:32 | 2015-08-29 19:30:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9056114554405212, "perplexity": 248.84832092219972}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644064538.25/warc/CC-MAIN-20150827025424-00065-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://ftp.aimsciences.org/article/doi/10.3934/dcds.2009.24.523 | Article Contents
Article Contents
# Population dynamical behavior of non-autonomous Lotka-Volterra competitive system with random perturbation
• In this paper, we consider a non-autonomous stochastic Lotka-Volterra competitive system $dx_i (t) = x_i(t)$[($b_i(t)$-$\sum_{j=1}^{n} a_{ij}(t)x_j(t))$$dt$$+ \sigma_i(t) d B_i(t)]$, where $B_i(t)$($i=1 ,\ 2,\cdots,\ n$) are independent standard Brownian motions. Some dynamical properties are discussed and the sufficient conditions for the existence of global positive solutions, stochastic permanence, extinction as well as global attractivity are obtained. In addition, the limit of the average in time of the sample paths of solutions is estimated.
Mathematics Subject Classification: Primary: 62F10, 34F05; Secondary: 92B05.
Citation: | 2022-12-06 17:11:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.45430174469947815, "perplexity": 648.7387616472816}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711111.35/warc/CC-MAIN-20221206161009-20221206191009-00629.warc.gz"} |
https://brilliant.org/problems/rps-002-eulers-totient/ | # RPS - Rimba's Problem Series #001
How many positive integers $$\leq 2015$$ which is not relatively prime with $$2015$$?
(Find other problem series in my profile!)
× | 2018-01-23 04:12:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5699269771575928, "perplexity": 5992.208784778598}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084891706.88/warc/CC-MAIN-20180123032443-20180123052443-00251.warc.gz"} |
http://skramm.blogspot.com/2014/04/state-machine-diagrams-with-graphviz.html | ## vendredi 11 avril 2014
### State machine diagrams with Graphviz
Once in a while, I need to draw a simple state machine diagram. These are a quick way to show in a visual way how a system works.
While these can be drawn with general drawing tools, or even with more dedicated tools, I usually prefer the textual way. Describing a drawing through some design language with an acceptable learning curve and letting some application do the drawing is IMO a better approach: editing the graph afterwards is just a matter of editing the source file.
Okay, so what tool ? Some are... funny (and I mean it!), but are not really usable for anything more than a small graph, or anything that needs to be edited many times.
No, this post is about Graphviz and it's associate set of tools. It does truely have some oddities buts its the best around.
Among its oddities, the default size units are in distance units. For an image, yes, no pixels here. Wait, that's not all ! The default units are inches. Inches !
I suppose this is for historical reasons and it seems that there is no option to change this. After thinking about it, once you go with distance, then, as metrication of image density is still not used, you might as well stay with inches.
Another thing, don't expect to be able to define precisely the position of nodes and edges: these are done by a placement algorithm, and adjusting this is not easy although some commands should help.
Nevertheless, say you want to describe some simple state machine. Just a light bulb connected to a switch.
#### Elementary graph
The associate state machine can be described by the following text file:
digraph g{
rankdir="LR";
edge[splines="curved"]
ON -> OFF;
OFF -> ON;
}
Assuming you have Graphviz correctly installed, the following shell command will generate the image:
dot -Tpng:cairo myfile.dot >myfile.png
#### Transitions
Ok, now lets add the transitions (the switch action). Lets call it "sw": if sw=1, the light bulb will be on, if 0, it will be off.
Ah. Here some problem appear. In the field of electrical engineering, the transitions between the states are frequently based on some boolean variable. Notation for the complement operation seems to be country-dependent. In France, this is usually expressed by a bar over the expression ("sw barre"), in Latex math-syntax, it will be $\bar{sw}$.
So, how can we manage this issue ?
First (and easiest), forget about the "bar" thing, and just go for plain text:
digraph g{
rankdir="LR";
edge[splines="curved"]
ON -> OFF [label="sw=0"];
OFF -> ON [label="sw=1"];
}
This is not very satisfying, it clutters the diagram.
Second solution: use Unicode. Graphviz natively supports it, and Unicode provides some special character that is supposed to handle this situation. So just enter:
digraph g{
rankdir="LR";
edge[splines="curved"]
OFF -> ON [label="sw"];
ON -> OFF [label="s̅w̅"];
}
(sorry, seems that the current hosting of this blog does not correctly display this, this is why the bar isn't exactly over the two letters).
In GTK+ based apps (Gedit, for instance), Unicode can be entered by hitting CTRL+SHIFT+U, then entering the desired character code (here '+0305') after each letter. Here, you need to do this manually after each letter of the label .
Unfortunately, the final rendering depends on the font used by the layout engine. It seems that the default png output of Graphviz does not use the Cairo library. Or if it does, it does not provide any control on the used font, so the final result looks quite ugly:
#### Direct insertion into Latex source file
If the graph image is intended to end up in a Latex source file, then check out the Graphviz package. It allows you to insert directly the graph command into the main Latex document. Unfortunately, this does not mean you suddenly have all the associated formatting power: this package only calls the 'dot' command himself, the only benefit is that you don't have to do it yourself and then import the image file into the Latex document. So for the issue detailed up here, it is of no help.
Another tool, dot2tex, has been specifically designed to have it all: direct editing of dot file inside Latex file and Latex formatting for labels and edges. Basically, it converts the dot file into PSTricks and/or PGF/TikZ format using some Python magic, then process it as regular Latex code.
Unfortunately, installation on my machine seems to suffer from some obscure Python bug, so I can't tell more at present! I hope to be able to try this soon.
Edit 2015/05: for more precise positioning of your nodes and vertices and better rendering, you'd better go off with a Latex-based solution. Tikz seems to be the easiest, see for example this sample.
For my own record, here are some relevant links: | 2019-04-21 17:12:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7847816944122314, "perplexity": 1674.8059196464126}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578531994.14/warc/CC-MAIN-20190421160020-20190421182020-00389.warc.gz"} |
https://www.physicsforums.com/threads/moment-of-inertia-of-a-square-problem-with-certain-method.798989/ | # Moment of Inertia of a Square: Problem with Certain Method
1. Feb 20, 2015
### DocZaius
For fun, I thought I would try to derive the moment of inertia of a square using different approaches (in each case, changing the differential area being integrated). Everything went well until I tried the approach of first considering the disk in the center of the square, then adding the bits at the corners.
Consider a square with sides of length L. The moment of inertia of the disk of radius L/2 in the center is:
$$I_{disk}=\frac{1}{2}MR^2=\frac{1}{2}M({\frac{L}{2}})^2=\frac{1}{8}ML^2$$
Now consider a corner of the square (attached image). If we figure out the contribution to the moment of inertia of that corner, we can account for the others by multiplying it by 4 because of radial symmetry. We will take arcs with differential thickness from the edge of the circle to the very corner of the square and add up their contributions to the moment of inertia. Here is the formula for the differential area:
$$\cos(\alpha)=\frac{\frac{L}{2}}{r}\\ \theta=\frac{\pi}{2}-2\alpha\\ \theta=\frac{\pi}{2}-2\arccos(\frac{L}{2r})\\ \\ da=r\,\theta\,dr\\ da=r(\frac{\pi}{2}-2\arccos(\frac{L}{2r}))dr$$
Now the integral for the contribution of the corner to the moment of inertia is:
$$\rho=\frac{M}{A}=\frac{M}{L^2}=\frac{dm}{da}\\ dm=\rho\,da=\rho \, r(\frac{\pi}{2}-2\arccos(\frac{L}{2r}))dr\\ I_{corner}=\int_M r^2 \, dm=\int_{\frac{L}{2}}^{\frac{\sqrt{2}}{2}L} r^2 \, \rho \, r(\frac{\pi}{2}-2\arccos(\frac{L}{2r}))dr\\$$
Which leads to the following for the total moment of inertia:
$$I_{square}=I_{disk}+4\,I_{corner}=\frac{1}{8}ML^2+4 \int_{\frac{L}{2}}^{\frac{\sqrt{2}}{2}L} r^2 \, \rho \, r(\frac{\pi}{2}-2\arccos(\frac{L}{2r}))dr$$
My problem is that the integral for the moment of inertia of a corner does not give me what it should ($\frac{1}{96}ML^2$) but rather gives me something with pi in it. I cannot figure out at what point I went wrong. Can someone see the error?
Thanks.
edit: Despite this not being homework, I realize it might belong in the homework forum. I wasn't sure and went with this forum. Mods, please move it if it needs to be moved.
#### Attached Files:
• ###### Untitled.png
File size:
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Last edited: Feb 20, 2015
2. Feb 20, 2015
### OldEngr63
I have not worked through the whole problem, but one place that looks suspect is the use of M. I see M as the mass of the disk, and the same symbol appearing again in the definition of rho for the corner MMOI calc. I think this may be part of your problem.
3. Feb 20, 2015
### DocZaius
Thank you! That might just be it.
4. Feb 20, 2015
### mathman
What do you have for the integral of the term involving arccos?
5. Feb 20, 2015
### DocZaius
That was it! Sorry mathman - I can't quite answer your question properly at the moment, but I plugged everything into wolfram alpha with M=L=1 and I got the right answer! (1/6 ML^2)
How funny that the bit I got wrong was in the easiest part of the derivation.
6. Feb 20, 2015
### DocZaius
To follow up on mathman's question (in case he still cares), the integral of the term involving arcos is:
$$\int_{\frac{L}{2}}^{\frac{\sqrt{2}}{2}L} r^2 \, \rho \, r(\frac{\pi}{2}-2\arccos(\frac{L}{2r}))dr=\frac{1}{384}ML^2(16-3\pi)$$
where I plugged in $\rho=\frac{M}{L^2}$
7. Feb 21, 2015
### mathman
I am confused. Your last 2 posts seem to contradict each other. One says you got the right answer, the other says the integral has a term involving π.
8. Feb 21, 2015
### Staff: Mentor
The integral has to have pi as part of its answer. The inner part has a pi-contribution as well (hidden in its area<->mass), and the final result for the square is rational.
9. Feb 21, 2015
### DocZaius
That's exactly right. Initially (post #1), I thought that I should not have pi in the integral involving arccos since my moment of inertia term for the circle did not have one. The reason I thought that was because my final answer needed to be rational: (1/6)ML^2. Therefore I thought my error had to do with the arccos integral yielding a pi. However, OldEngr63 pointed out that my error did not lie with my integral but instead came from the fact that I was using the wrong mass for the circle at the very beginning of my derivation. With the correct circle moment of inertia (which must have a pi in it when taken as a portion of the square's mass!) I now needed a pi in my arccos integral. But I always and correctly had it there! Thanks all.
Last edited: Feb 21, 2015
10. Feb 21, 2015
### DocZaius
Here is the corrected derivation:
To find the moment of inertia of a square: Consider a square with sides of length L. The moment of inertia of the disk of radius L/2 in the center is:
$$I_{disk}=\frac{1}{2}M_{disk}R^2=\frac{1}{2}(\frac{\pi R^2}{L^2}M_{square})R^2=\frac{1}{2}\frac{\pi (\frac{L}{2})^2}{L^2}M_{square}(\frac{L}{2})^2=\frac{1}{2}\frac{\pi}{4}M_{square}(\frac{L}{2})^2=\frac{\pi}{32}M_{square}L^2$$
Where $M_{square}=M$ from now on. Now consider a corner of the square (attached image). If we figure out the contribution to the moment of inertia of that corner, we can account for the others by multiplying it by 4 because of radial symmetry. We will take arcs with differential thickness from the edge of the circle to the very corner of the square and add up their contributions to the moment of inertia. Here is the formula for the differential area:
$$\cos(\alpha)=\frac{\frac{L}{2}}{r}\\ \theta=\frac{\pi}{2}-2\alpha\\ \theta=\frac{\pi}{2}-2\arccos(\frac{L}{2r})\\ \\ da=r\,\theta\,dr\\ da=r(\frac{\pi}{2}-2\arccos(\frac{L}{2r}))dr$$
Now the integral for the contribution of the corner to the moment of inertia is:
$$\rho=\frac{M}{A}=\frac{M}{L^2}=\frac{dm}{da}\\ dm=\rho\,da=\rho \, r(\frac{\pi}{2}-2\arccos(\frac{L}{2r}))dr\\ I_{corner}=\int_{M_{corner}} r^2 \, dm=\int_{\frac{L}{2}}^{\frac{\sqrt{2}}{2}L} r^2 \, \rho \, r(\frac{\pi}{2}-2\arccos(\frac{L}{2r}))dr\\$$
Which leads to the following for the total moment of inertia:
$$I_{square}=I_{disk}+4\,I_{corner}=\frac{\pi}{32}ML^2+4 \int_{\frac{L}{2}}^{\frac{\sqrt{2}}{2}L} r^2 \, \rho \, r(\frac{\pi}{2}-2\arccos(\frac{L}{2r}))dr=\frac{\pi}{32}ML^2+\frac{4}{384}ML^2(16-3\pi)=\frac{1}{6}ML^2$$
where $\rho=\frac{M}{L^2}$
#### Attached Files:
• ###### Untitled.png
File size:
3 KB
Views:
118
Last edited: Feb 21, 2015
11. Feb 22, 2015
### mathman
I(square)=ML2/6 . I don't understand your result.
12. Feb 22, 2015
### DocZaius
I was trying to find the moment of inertia of a (filled) square about its center. The answer to that is $\frac{1}{6}ML^2$. There are a few different ways to work it out, but the method I chose was to first find the moment of inertia for the disk contained in the square, and then do the corners. My most recent post above this one is that derivation. I arrive at the correct answer. Please let me know what you don't understand.
13. Feb 23, 2015
### mathman
Problem was with my web page display. Your typography is very large, so the right end of the expression was chopped off. | 2018-07-21 18:35:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8237466812133789, "perplexity": 481.1281807700773}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676592650.53/warc/CC-MAIN-20180721164755-20180721184755-00200.warc.gz"} |
https://www.gamedev.net/forums/topic/163370-lesson-7/ | • ### Popular Now
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#### Archived
This topic is now archived and is closed to further replies.
# Lesson 7
This topic is 5393 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
i''ve added a diffuselight source to my project with the help of lesson 7 but it seems that the lightsource rotates along with my cube. Yet i can find out why. May anyone has an idea? Grtz iTec
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You have to be carefull about the order you do your conversions. If I recall correctly first insert geometry, then do all the translations/rotations of geometry, then insert the light via glLightfv(light.name,GL_POSITION,light.position);.
If that doesn''t help, try doing the following:
glPushMatrix(); { // Do rotations + translations // Insert geometry } glPopMatrix(); // Insert Lights
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Don''t know anything about glPushMatrix() and glPopMatrix() yet but I insert the light in the initGL just like in the tutorial (only a bit modified)
glEnable(GL_TEXTURE_2D);
glClearColor(0.0f, 0.0f, 0.0f, 0.5f);
glClearDepth(1.0f);
glEnable(GL_DEPTH_TEST);
glDepthFunc(GL_LEQUAL);
glHint(GL_PERSPECTIVE_CORRECTION_HINT, GL_NICEST);
glLightfv(GL_LIGHT1, GL_AMBIENT, World.Light.Ambient.Color);
bool chk = false;
cDiffuseLight Current = (cDiffuseLight) *(World.Light.firstDiffuse());
do{
glLightfv(Current.Name, GL_DIFFUSE, Current.Color);
glLightfv(Current.Name, GL_POSITION,Current.Position);
glEnable(Current.Name);
if(Current.Volgende!=NULL){
cDiffuseLight *tc = Current.Volgende;
Current = (cDiffuseLight) *tc;
chk = true;
}else{
chk = false;
}
}while (chk);
glEnable(GL_LIGHTING);
return TRUE;
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Sorry, but it''s a bit difficult to see what part of your code you''re posting exactly.
quote:
glEnable(GL_TEXTURE_2D);
glClearColor(0.0f, 0.0f, 0.0f, 0.5f);
glClearDepth(1.0f);
glEnable(GL_DEPTH_TEST);
glDepthFunc(GL_LEQUAL);
glHint(GL_PERSPECTIVE_CORRECTION_HINT, GL_NICEST);
All this is initialisation stuff, and should only be done once (=beginning of your programm).
quote:
glLightfv(Current.Name, GL_POSITION,Current.Position);
While this is a part of code you''ll need to put into the main loop. You''ll have to insert the light every time you changed the orientation of the geometry, otherwise the changes will affect the light as well (which you don''t want obviously).
As for glPush/PopMatrix: These function store/restore the transformation matrix. That is: if you do glPushMatrix, then do some transformations (glRotate, glTranslate), then do glPopMatrix the original Matrix is restored. Resulting in geometry outside the Push/Pop block not being affected by any of the translations inside. | 2018-03-22 22:12:55 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2251991629600525, "perplexity": 5531.191592722442}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257648003.58/warc/CC-MAIN-20180322205902-20180322225902-00503.warc.gz"} |
https://space.stackexchange.com/questions/27116/does-the-size-of-a-satellite-have-an-effect-on-orbit | # Does the size of a satellite have an effect on orbit?
I understand that lighter satellites orbits decay faster. There is 2 satellites that weigh the same but 1 is huge and the other tiny in comparison. Would the size of the satellite make a difference on the orbit?
• A larger satellite experiences a larger atmospheric drag and has more area from which to absorb solar radiation. For this reason, a sheet of paper would probably decay faster than the same paper crumpled in a ball. May 6 '18 at 16:23
• For LEO satellites, drag would be different. But for GEO satellites, drag would be very, very small.
– Uwe
May 6 '18 at 19:21
Absolutely! The parameter that applies is the ballistic coefficient, the object's mass divided by its projected area modified by the drag coefficient. The drag force is given by$$F_{\text{drag}}=-\frac{1}{2} C_{\text{drag}} A \rho V^2 \,,$$where:
• $C_{\text{drag}}$ is the drag coefficient (at orbital speeds and atmospheric densities this is usually very close to $2$);
• $A$ is the projected area;
• $\rho$ is the atmospheric mass density; and
• $V$ is the velocity relative to the surrounding atmosphere.
The minus sign out front says the drag force is in the direction opposite the velocity direction. Acceleration, in this case deceleration, is $a_{\text{drag}}=\frac{F_{\text{drag}}}{m}$, where $m$ is the object's mass, so divide the expression for $F_{\text{drag}}$ by $m$ and you get acceleration. Within this new expression is $\frac{C_{\text{drag}}A}{m}$, the inverse of the ballistic coefficient,$$\text{BC} = \frac{m}{C_{\text{drag}}A} \,.$$The higher that number, i.e. the more massive the object per unit area, the lower the deceleration will be, and the longer it will take its orbit to decay.
Interestingly, the very small vertical gradient in atmospheric density shows up in decaying objects, if they aren't gravity-gradient stabilized (paper).
The slightly increased density on the lower side of the spacecraft produces more drag force per unit area there, and the spacecraft starts to rotate! | 2021-11-30 09:28:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6539115309715271, "perplexity": 606.6024970625958}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358966.62/warc/CC-MAIN-20211130080511-20211130110511-00558.warc.gz"} |
http://mathhelpforum.com/advanced-statistics/128840-proof-geometric-probability-distribution-print.html | # Proof of geometric probability distribution
• February 14th 2010, 06:40 PM
Matharch
Proof of geometric probability distribution
I would need help with this proof. It seems any way I try to prove I get stuck or mess up somewhere, I know it's easy but basically I suck at proving stuff. So a bit help would be appreciated.
$\sum_{ y=1}^{ \infty } {q}^{ y-1 }p=1$
• February 14th 2010, 07:51 PM
Focus
Well I assume that q=1-p, then you have that
$\sum_{n\geq 0} q^n p=p\sum_{n\geq 0} q^n=\frac{p}{1-q}=\frac{p}{p}$.
If you want to prove what the sum of a geometric series is then you can look it up on wikipedia. | 2015-01-25 03:59:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9371047616004944, "perplexity": 464.57089160658023}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422118059355.87/warc/CC-MAIN-20150124164739-00248-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://storage.guidotti.dev/course/text-mining-unimi-2019-2020/lect02-nlp.html | 2019/2020
## Natural Language Processing (NLP)
Natural Language Processing is the technology used to aid computers to understand the human’s natural language. Usually shortened as NLP, is a branch of artificial intelligence that deals with the interaction between computers and humans using the natural language.
The ultimate objective of NLP is to read, decipher, understand, and make sense of the human languages in a manner that is valuable.
A typical interaction between humans and machines using Natural Language Processing could go as follows:
1. A human talks to the machine
2. The machine captures the audio
3. Audio to text conversion takes place
4. Processing of the text’s data
5. Data to audio conversion takes place
6. The machine responds to the human by playing the audio file
Think of Google Translate, spellcheckers, or personal assistant applications such as OK Google, Siri, Cortana, and Alexa.
## Why is NLP difficult?
It’s the nature of the human language that makes NLP difficult. The rules that dictate the passing of information using natural languages are not easy for computers to understand.
Some of these rules can be high-leveled and abstract; for example, when someone uses a sarcastic remark to pass information.
On the other hand, some of these rules can be low-levelled; for example, using the character “s” to signify the plurality of items.
The ambiguity and imprecise characteristics of the natural languages are what make NLP difficult for machines to implement.
## What are the techniques used in NLP?
Syntactic analysis and semantic analysis are the main techniques used to convert the unstructured language data into a form that computers can understand.
### Syntactic analysis
• Apply grammatical rules
### Semantic analysis
• Understand the meaning (much harder)
## Regex
A regular expression (regex or regexp for short) is a special text string for describing a search pattern.
Typically, these patterns are used for four main tasks:
• Find text within a larger body of text
• Validate that a string conforms to a desired format
• Replace or insert text
• Split strings
## Example
Extract hashtags from the following tweet:
“It’s our job to #GoThere & tell the most difficult stories. Join us! For more breaking news updates follow @CNNBRK & Download our app. Email [email protected] to get involved in the new year.”
Hashtags are identified by the “#” symbol followed by one or more alphanumeric characters.
import re
re.findall(r'#\w+', text)
## ['#GoThere']
Extract callouts: strings identified by the “@” symbol followed by one or more alphanumeric characters.
re.findall(r'@\w+', text)
## ['@CNNBRK', '@cnn']
Oops… also part of the email address was extracted. We need to check for word boundaries.
re.findall(r'\[email protected]\w+', text)
## ['@CNNBRK']
Online tool: https://regex101.com
## Meta-characters: Character matches
Metacharacter Description
. wildcard, matches a single character
^ start of a string
$end of a string [] matches one of the set of characters within [] [a-z] matches one of the range of characters a, b, …, z [^abc] matches a character that is not a, b, or, c a|b matches either a or b, where a and b are strings () scoping for operators \ escape character for special characters (\t, \n, \b) ## Meta-characters: Character symbols Metacharacter Description \b matches word boundary \B matches where \b does not match \d any digit, equivalent to [0-9] \D any non-digit, equivalent to [^0-9] \s any whitespace, equivalent to [ \t\n\r\f\v] \S any non-whitespace, equivalent to [^ \t\n\r\f\v] \w alphanumeric character, equivalent to [a-zA-Z0-9_] \W non-alphanumeric, equivalent to [^a-zA-Z0-9_] ## Meta-characters: Quantifiers Metacharacter Description * matches zero or more occurrences + matches one or more occurrences ? matches zero or one occurrences {n} exactly n repetitions {n,} at least n repetitions {,n} at most n repetitions {m,n} at least m and at most n repetitions ## NLP Tasks ### Tokenization ## Tokenization Tokenization is the act of breaking up a sequence of strings into units such as words, keywords, phrases, symbols and other elements called tokens. One can think of token as parts like a word is a token in a sentence, and a sentence is a token in a paragraph. Tokenization is non trivial. ## How would you split this sentence into words? “Children shouldn’t drink a sugary drink before bed.” text.split(' ') # Naive ## ['Children', "shouldn't", 'drink', 'a', 'sugary', 'drink', 'before', 'bed.'] import nltk # NLTK has an in-built tokenizer! nltk.word_tokenize(text) ## ['Children', 'should', "n't", 'drink', 'a', 'sugary', 'drink', 'before', 'bed', '.'] ## How would you split sentences from a long text string? “This is the first sentence. A gallon of milk in the U.S. costs$2.99. Is this the third sentence? Yes, it is!”
# NLTK has an in-built sentence splitter too!
nltk.sent_tokenize(text)
## ['This is the first sentence.', 'A gallon of milk in the U.S.
## PRP$: pronoun, possessive ## her his mine my our ours their thy your ## Tagging Methods • default tagger: assigns the same tag to each token • regular expression tagger: assigns tags to tokens on the basis of matching patterns • supervised learning • unigram tagger: for each token, assign the tag that is most likely for that particular token • n-gram taggers: generalization of a unigram tagger whose context is the current word together with the part-of-speech tags of the n-1 preceding tokens • machine learning These methods can be combined using a technique known as backoff. Backoff is a method for combining models: when a more specialized model (such as a bigram tagger) cannot assign a tag in a given context, we backoff to a more general model (such as a unigram tagger). ## NLP Tasks ### Chunking ## Chunking Chunking is a process of extracting phrases from unstructured text. • Chunking works on top of POS tagging, it uses pos-tags as input and provides chunks as output. • Similar to POS tags, there are a standard set of Chunk tags like Noun Phrase (NP), Verb Phrase (VP), etc. • There are libraries which gives phrases out-of-box such as Spacy. NLTK provides a mechanism using regular expressions to generate chunks. ## Chunking with NLTK In order to create NP chunk, we define the chunk grammar using POS tags. We will define this using a regular expression. grammar = (''' NP: {<DT>?<JJ>*<NN>} V: {<VB[\w]?>} ''') The rule states that whenever the chunk finds: • an optional determiner (DT) followed by any number of adjectives (JJ) and then a noun (NN) then the chunk NP should be formed • a verb (VB, VBD, VBG, VBN, VBP, VBZ) then the chunk V should be formed text = "This is a simple example of chuncking a sentence" tagged = nltk.pos_tag(nltk.word_tokenize(text)) tree = nltk.RegexpParser(grammar).parse(tagged) for subtree in tree.subtrees(): print(subtree) ## (S ## This/DT ## (V is/VBZ) ## (NP a/DT simple/JJ example/NN) ## of/IN ## (V chuncking/VBG) ## (NP a/DT sentence/NN)) ## (V is/VBZ) ## (NP a/DT simple/JJ example/NN) ## (V chuncking/VBG) ## (NP a/DT sentence/NN) ## NLP Tasks ### Named Entity Recognition (NER) ## Named Entity Named Entities are definite chuncks that refer to specific types of real-world objects, such as organizations, persons, dates, and so on. ## Named Entity Recognition (NER) The goal of Named Entity Recognition (NER) is to identify all textual mentions of the Named Entities. This can be broken down into two sub-tasks: • identifying the boundaries of the Named Entity • identifying the type of the Named Entity The task is well-suited to the type of classifier-based approach that we saw for POS tagging and noun phrase chunking. In particular, we can: • extract chuncks corresponding to noun phrases • build a tagger that labels each chunck using the appropriate type based on the trainig data (unigram/n-gram tagger, …). ## Named Entity Recognition with NLTK NLTK provides a classifier that has already been trained to recognize Named Entities. Example: “European authorities fined Google a record$5.1 billion on Wednesday for abusing its power…”
nltk.ne_chunk(nltk.pos_tag(nltk.word_tokenize(text)))
## Tree('S', [Tree('GPE', [('European', 'JJ')]), ('authorities',
'NNS'), ('fined', 'VBD'), Tree('PERSON', [('Google', 'NNP')]), ('a',
'DT'), ('record', 'NN'), ('$', '$'), ('5.1', 'CD'), ('billion',
'CD'), ('on', 'IN'), ('Wednesday', 'NNP'), ('for', 'IN'), ('abusing',
'VBG'), ('its', 'PRP$'), ('power', 'NN'), ('...', ':')]) ## Named Entity Recognition with SpaCy SpaCy features an extremely fast statistical entity recognition system, that assigns labels to contiguous spans of tokens. import spacy nlp = spacy.load("en_core_web_sm") doc = nlp(text) for ent in doc.ents: print(ent.text, ent.label_) ## European NORP ## Google ORG ##$5.1 billion MONEY
## Wednesday DATE
## Take Home Concepts
• Regex are useful text strings to match patterns
• Case-folding, stop words and punctuation removal, are usually but not always a good idea
• Tokenization, sentence splitting, stemming and lemmatization are non-trivial tasks
• Part of Speech Tagging helps the machine understand how a word is used in a sentence
• Chunking is the process of extracting phrases from unstructured text
• Named Entity Recognition allows us to identify real-world objects in a text | 2020-07-14 18:26:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.18383316695690155, "perplexity": 7586.3165876932535}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593657151197.83/warc/CC-MAIN-20200714181325-20200714211325-00204.warc.gz"} |
https://stats.stackexchange.com/questions/468768/interpretation-of-ljung-box-tests-for-garch-models-from-the-rugarch-package-in | # Interpretation of Ljung-Box tests for GARCH models from the 'rugarch' package in R
I have used the 'rugarch' R package to fit a GARCH model, as:
model.garch = ugarchspec(mean.model=list(armaOrder=c(1,1)),variance.model=list(model = "sGARCH"),distribution.model = "norm")
ugarchfit(model.garch, data=my_data)
However, I am confused about the right interpretation of the Ljung-Box tests associated with my results. Specifically, this is what I have:
Weighted Ljung-Box Test on Standardized Residuals
------------------------------------
statistic p-value
Lag[1] 1.304 2.535e-01
Lag[2*(p+q)+(p+q)-1][14] 10.501 3.392e-06
Lag[4*(p+q)+(p+q)-1][24] 17.820 3.235e-02
d.o.f=5
H0 : No serial correlation
Weighted Ljung-Box Test on Standardized Squared Residuals
------------------------------------
statistic p-value
Lag[1] 0.1355 0.7128
Lag[2*(p+q)+(p+q)-1][5] 0.3466 0.9786
Lag[4*(p+q)+(p+q)-1][9] 0.4837 0.9986
d.o.f=2
Weighted ARCH LM Tests
------------------------------------
Statistic Shape Scale P-Value
ARCH Lag[3] 0.00900 0.500 2.000 0.9244
ARCH Lag[5] 0.03188 1.440 1.667 0.9974
ARCH Lag[7] 0.14606 2.315 1.543 0.9985
Given that some of the p-values from the "Weighted Ljung-Box Test on Standardized Residuals" are significant (with the exemption of Lag[1]), should I conclude that my GARCH model failed to correct for the temporal auto-correlation in my data?
Perhaps more importantly, how those results influence the overall assessment of the model given that the p-values from the "Weighted Ljung-Box Test on Standardized Squared Residuals" and the "Weighted ARCH LM Tests" are NOT significant? Thank you in advance!
A problem with applying any of these tests to standardized (squared) residuals from a GARCH model is that the test statistics have nonstandard distributions under the null. (They have their standard null distributions when applied to raw data, but not when applied to residuals of a GARCH model.)* As far as I know, this is not accounted for in the rugarch package. Hence, you should take the test results with a grain of salt. | 2021-06-23 00:19:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7226380109786987, "perplexity": 4701.701101410221}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488525399.79/warc/CC-MAIN-20210622220817-20210623010817-00421.warc.gz"} |
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## how to find the area of a right triangle
Now, consider a triangle that’s graphed in the coordinate plane. There are three primary methods used to find the perimeter of a right triangle. All rights reserved. So, with a hypotenuse of 6 (as the radius is given as 3), the sides of the triangle (and square) are ${3\sqrt{2}}$. ... Area Formula for Non-Right Triangles. Area = (1/2) * width * height The diagonal line also creates two right triangles, one at Point A. How do you find the perimeter of a right triangle? © 2006 -2021CalculatorSoup® For example, if one of the angles in a right triangle is #25^o#, the other acute angle is … Area of a triangle. From MathWorld--A Wolfram Web Resource. c = √(a2 + b2). This formula is known as the Pythagorean Theorem. 2. ab =. P = perimeter, See Diagram Below: Step 2: User will enter the three sides of the triangle a, b, c. Step 3: Calculating the Perimeter of the Triangle by calling the function we declared at the beginning of the main().This function will call the main function and execution of that main function will start here If we know side-angle-side information, solve for the missing side using the Law of Cosines. C Area of a Right Angled Triangle. K = area To find the area of a right triangle, multiply the lengths of the two sides that are perpendicular to each other (in other words, the two that form the right angle). Set D3 Measurement: Area of Polygons Blackline NAME DATE Run a display copy 2. The area of any other triangle can be found with the formula below. It is a right triangle because it has a right angle, not because it is facing to the right. When side lengths are given, add them together. For more information on right triangles see: Weisstein, Eric W. "Right Triangle." Moreover it allows specifying angles either in grades or radians for a more flexibility. Start by measuring the length of the base of the triangle. Let us know if you have any other suggestions! You can calculate angle, side (adjacent, opposite, hypotenuse) and area of any right-angled triangle and use it in real world to find height and distances. In our calculations for a right triangle we only consider 2 known sides to calculate the other 7 unknowns. Once we know sides a, b, and c we can calculate the perimeter = P, the semiperimeter = s, the area = K, and the altitudes: 30 15 in. That is side S C, 30 y a r d s long. The basic equation is a transformed version of a standard triangle height formula (a * h / 2). Method 1: https://www.calculatorsoup.com - Online Calculators. In this article, let us discuss what the area of a triangle is and different methods used to find the area of a triangle in coordinate geometry. Cite this content, page or calculator as: Furey, Edward "Right Triangles Calculator"; CalculatorSoup, A quick, informative and entertaining look at how to find the area of a right angled triangle. This formula is known as the Pythagorean Theorem. Finding the area of the triangle below: (Of course, this is a right triangle, so you could just use the two perpendicular sides as base and height.) area = (a * b) / 2 A right triangle is a triangle that has one angle equal to 90 degrees. Here you can enter two known sides or angles and calculate unknown side ,angle or area. hc = altitude of c. *Length units are for your reference only since the value of the resulting lengths will always be the same no matter what the units are. Click Create Assignment to assign this modality to your LMS. By finding the base and height of the triangle. This formula works for a right triangle as well, since the since of 90 is one. Enter any two values and press calculate to get the other values. Recall that the area formula for a triangle is given as $$Area=\dfrac{1}{2}bh$$, where $$b$$ is base and $$h$$ is height. Formulas used for calculations on this page: Trigonometric functions: Clearly: Height of the triangle = Width of the rectangle. Finding the Area of an Oblique Triangle Using the Sine Function. For example, if we know a and b we can calculate c using the Pythagorean Theorem. Because the right triangle legs are perpendicular to each other, one leg is taken as a base and the other is a right triangle height: area = a * b / 2. Area = ½ × (c) × (b × sin A) Which is (more simply): Area = 12 bc sin A. Explanation: An equilateral triangle has three congruent sides. If we know the width and height then, we can calculate the area of a right angled triangle using the below formula. On this page, you can solve math problems involving right triangles. Now that we can solve a triangle for missing values, we can use some of those values and the sine function to find the area of an oblique triangle. In the case of a right triangle a2 + b2 = c2. As we remember from basic triangle area formula, we can calculate the area by multiplying triangle height and base and dividing the result by two. Area = a*b/2, where a is height and b is base of the right triangle. Let us take the base and height of the triangle be x cm. Using the Pythagorean Theorem we get or and the area is. Such an equation, which gives a rule for … Likely the most commonly known equation for calculating the area of a triangle involves its base, b, and height, h. Solution for Find the area of each right triangle. From Let’s look at an example. 1. Area of right triangle formulas. % Progress . A = angle A b = side b Read the first problem with the class, and take a minute to review the definition of right triangle (a triangle that has a 90º angle). Pythagorean Theorem for Right Triangle: a, Perimeter of Right Triangle: P = a + b + c, Semiperimeter of Right Triangle: s = (a + b + c) / 2. Area of the triangle is a measure of the space covered by the triangle in the two-dimensional plane. Altitude. It measures 90 ° and has the hypotenuse, or longest side, opposite it. Before we step into the C Program to find Area of a Right Angled Triangle, Let see the definition and formula behind Area of a Right Angled Triangle. In the case of a right triangle a 2 + b 2 = c 2. Examples include: 3, 4, 5; 5, 12, 13; 8, 15, 17, etc. The height is the line perpendicular to the base, through the opposite vertex. C = angle C A right triangle is a special case of a There are multiple different equations for calculating the area of a triangle, dependent on what information is known. MEMORY METER. Find the area of the triangle in square units. This works because the area of a right triangle is exactly ___________ of the area of the rectangle. In our calculations for a right triangle we only consider 2 known sides to calculate the other 7 unknowns. 3. Area and perimeter of a right triangle are calculated in the same way as any other triangle. Our online tools will provide quick answers to your calculation and conversion needs. Free calculators and unit converters for general and everyday use. In short, to find the area of a triangle, all you need to do is take the area of a rectangle formula (A = b h) and divide it by 2. hb = altitude of b The perimeter is the sum of the three sides of the triangle and the area can be determined using the following equation: A =. Step-by-step explanations are provided for each calculation. MathWorld--A Wolfram Web Resource. B = angle B Two sides of isosceles right triangle are equal and we assume the equal sides to be the base and height of the triangle. The algorithm of this right triangle calculator uses the Pythagorean theorem to calculate the hypotenuse or one of the other two sides, as well as the Heron formula to find the area, and the standard triangle perimeter formula as described below. This implies an area of the square of 18 sq units and for the triangle, as one possible answer for the question, 9 sq units. An alternate formula for the area of a triangle. We are asked to find the perimeter of the triangle. Download: Use this right-triangle solver offline with our all-in-one calculator app for Android and iOS. a = side a Area = a*b/2, where a is height and b is base of the right triangle. Right Triangle. Finding the Area of a Right Triangle 1 Build this right triangle on your geoboard. The equilateral triangle can be broken into two right triangles, where the legs are and and the hypotenuses is . "Altitude." Khan Academy has a nifty drag tool that lets you see how the area of a triangle is found using the rectangle/parallelogram it's inscribed in. To calculate the area of a triangle, multiply the height by the width (this is also known as the 'base') then divide by 2. ha, hb, and hc. 2. Let's create a right triangle, C A S, with ∠ A as the right angle. The area formulas for all the different types of triangles like an area of an equilateral triangle, right-angled triangle, an isosceles triangle are given below. Then take half of that. The area of a triangle is given by where is the base and is the height. Weisstein, Eric W. Area equals half the product of two sides and the sine of the included angle. triangle where 1 angle is equal to 90 degrees. Solve for a missing side using the Pythagorean theorem. Right Triangle. Area of a isosceles right triangle, say A having base x cm and . They’re really not significantly different, though the derivation of … The hypotenuse is the largest side in a right triangle and is always opposite the right angle. ha = altitude of a In Geometry, a triangle is a 3 – sided polygon which has 3 edges and 3 vertices. 1. 4. sin(B) = b/c, cos(B) = a/c, tan(B) = b/a. The area of a right triangle can be found using the formula A = ½ bh. You can always use the distance formula, find the lengths of … By changing the labels on the triangle we can also get: Area = ½ ab sin C ; Area = ½ ca sin B; One more example: Easy to use calculator to solve right triangle problems. c = side c If we know one of these angles, we can easily substitute that value and find the missing one. To find the area of an equilateral triangle, we can use the Pythagorean Theorem to get the height of the triangle and then use formula A = 1 2 b h or we can use the following formula: The formula for the area of an equilateral triangle (with all sides congruent) is equal to For example, if we know a and b we can calculate c using the Pythagorean Theorem. Using the pronumerals A for area, b for base and h for height, we can write the formula for the area of a triangle as: Note: The rule (or equation) represents the relationship between the base and height of a triangle and its area. In a triangle, the base is one … Also, how to find the area of a triangle with 3 sides using Heron’s formula with examples. Using base and height. A right triangle is a special case of a scalene triangle, in which one leg is the height when the second leg is the base, so the equation gets simplified to: area = a * b / 2 That is, the sum of the two acute angles in a right triangle is equal to #90^o#. (Only right triangles have a hypotenuse).The other two sides of the triangle, AC and CB are referred to as the 'legs'. c = √ (a 2 + b 2). sin(A) = a/c, cos(A) = b/c, tan(A) = a/b Step 1: We declared the function with three arguments right after the header files. Area =_____ How did you figure it out? Sometimes it's not so obvious - you have other values given, not two legs. The distance from the top of the triangle to the base is called the height of the triangle. S, with ∠ a as the right angle are asked to find the area of the of! Two known sides to be the base of the space covered by the to! H / 2 ), with ∠ a as the right triangle. not two legs a and b can! We can easily substitute that value and find the lengths of … Start by measuring the length of the is... Be the base and height of the included angle consider a triangle, say a having x. Coordinate plane right after the header files a standard triangle height formula ( a 2 + 2. Easily substitute that value and find the area of the triangle. the length the! Missing one we assume the equal sides to be the base and is the line perpendicular to the base called. By measuring the length of the triangle., add them together other... Create Assignment to assign this modality to your how to find the area of a right triangle and conversion needs, opposite it solve for the side. Be the base and is the height for Android and iOS know one of these,... The opposite vertex one at Point a broken into two right triangles quick, and! Finding the area of a right triangle on your geoboard area and perimeter of triangle... Calculate the area of each right triangle is a right triangle is equal to 90 degrees, find the of! C using the Pythagorean Theorem we get or and the hypotenuses is three arguments right the. Angles and calculate unknown side, opposite it it measures 90 ° and the. Where the legs are and and the hypotenuses is how to find the area of a right triangle: area a. Values given, not because it is facing to the base, through the opposite vertex and has the,! With our all-in-one calculator app for Android and iOS ° and has the hypotenuse, or how to find the area of a right triangle side, it... Right triangle. triangle is exactly ___________ of the base and is line. Triangle can be found using the formula a = ½ bh height and b we can calculate using! Blackline NAME DATE Run a display copy 2 of Polygons Blackline NAME DATE Run a display copy 2 of! A having base x cm Assignment to assign this modality to your calculation and needs... Allows specifying angles either in grades or radians for a right triangle., you can two. Triangle is exactly ___________ of the base, through the opposite vertex through the opposite vertex how to find the area of a right triangle 2 substitute value... We know the width and height of the triangle = width of the.! 1/2 ) * width * height how do you find the area the! Equal and we assume the equal sides to calculate the area of any other suggestions: An triangle... Facing to the base of the rectangle of An Oblique triangle using the Pythagorean Theorem in. Formula works for a right triangle a2 + b2 = c2 triangle to the right formula... Base is called the height of the triangle. and and the area is as any other triangle be... Primary methods used to find the area of a right triangle a 2 b! Easily substitute that value and find the area of a isosceles right triangle we only consider known! Solve math problems involving right triangles see: Weisstein, Eric W. right triangle 1 Build this triangle. 15, 17, etc the function with three arguments right after the header.. Angles and calculate unknown side, angle or area to find the area of a triangle... With our all-in-one calculator app for Android and iOS and conversion needs s, with ∠ a the! With ∠ a as the right angle, not because it has a right angled using... Standard triangle height formula ( a * h / 2 ), opposite.! 90 degrees you can enter two known sides or angles and calculate unknown,! Measuring the length of the included angle obvious - you have any other suggestions Heron ’ s formula examples. Not two legs it 's not so obvious - you have other values given add! Can enter two known sides to calculate the other values explanation: An equilateral has. Arguments right after the header files and press calculate to get the other 7 unknowns how. Has a right triangle. well, since the since of 90 one! Of a right triangle. missing one here you can enter two known or. See: Weisstein, Eric W. right triangle we only consider 2 known to., or longest side, angle or area congruent sides a = ½ bh quick, informative and look. These angles, we can calculate c using the Sine of the.! Other values given, not two legs transformed version of a triangle that ’ s graphed in case!, the base, through the opposite vertex D3 Measurement: area of Polygons Blackline NAME DATE Run display. Two known sides or angles and calculate unknown side, opposite it the header.... Acute angles in a triangle that ’ s formula with examples these angles, we can calculate c using Pythagorean... That ’ s graphed in the case of a right triangle coordinate plane on triangles! D3 Measurement: area of a right angle can be found using the below formula covered. Not two legs: 3, 4, 5 ; 5, 12, 13 ; 8,,..., find the area of a right angled triangle. have any other triangle can be into... The line perpendicular to the right triangle, the sum of the space by. Offline with our all-in-one calculator app for Android and iOS because it has a right angled triangle.,... And calculate unknown side, angle or area solver offline with our all-in-one calculator app for and... Longest side, angle or area methods used to find the missing one get or and area... Angle or area and conversion needs grades or radians for a right triangle problems: of. Found using the Pythagorean Theorem … Start by measuring the length of the right triangle say. The since of 90 is one or longest side, opposite it know the and. The diagonal line also creates two right triangles, one at Point a b/2, where a is height b! Is the height is the base and height of the triangle be x and! Base of the included angle the included angle two right triangles, one at a... Are given, not two legs math problems involving right triangles more information on right triangles of... We know a and b is base of the triangle = width of base. Finding the base and height of the triangle is given by where is the line perpendicular the. Three arguments right after the header files line perpendicular to the base and is line. Unit converters for general and everyday use ( 1/2 ) * width * height how you. Header files it has a right triangle we how to find the area of a right triangle consider 2 known sides to calculate the values... The space covered by the triangle in square units in the case of a right triangle problems there are primary... Run a display copy 2 values given, not because it is facing to base. Two sides and the hypotenuses is in the coordinate plane and is the base height... Only consider 2 known sides to calculate the other 7 unknowns because it has a right.... Since of 90 is one step 1: we declared the function with three arguments right after the files... Given by where is the line perpendicular to the base of the triangle., etc you can solve problems. A right triangle as well, since the since of 90 is.. 2 known sides to calculate the area is on what information is known be broken into two triangles. … right triangle because it has a right triangle is equal to # 90^o.! Two known sides to be the base and height of the included angle create! An equilateral how to find the area of a right triangle has three congruent sides: use this right-triangle solver offline with our all-in-one app. The equilateral triangle has three congruent sides either in grades or radians for a right on..., angle or area 90 ° and has the hypotenuse, or side... We know side-angle-side information, solve for a right triangle because it a. Height of the triangle. or radians for a right triangle 1 Build this right.. For example, if we know a and b is base of the triangle ''... Three arguments right after the header files us take the base is one of Polygons Blackline NAME DATE a! Coordinate plane right triangles, where the legs are and and the hypotenuses is works for a flexibility. Also creates two right triangles see: Weisstein, Eric W. right triangle is a special case a... Are equal and we assume the equal sides to be the base is called the is. Each right triangle is given by where is the base and height of the triangle ''., angle or area different equations for calculating the area of the triangle. / 2 ) unknown,! Are and and the area of a triangle with 3 sides using ’. Entertaining look at how to find the perimeter of a right triangle we consider... Side lengths are given, not two legs here you can enter two known or! Pythagorean Theorem: we declared the function with three arguments right after the header files formula with.. B is base of the triangle. the top of the rectangle enter two known sides to be base!
Recent Posts | 2021-07-27 09:59:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6862828731536865, "perplexity": 452.4840446621881}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153223.30/warc/CC-MAIN-20210727072531-20210727102531-00428.warc.gz"} |
http://ncatlab.org/nlab/show/tangent+Lie+algebra | # nLab tangent Lie algebra
Given an algebraic group $G$ in characteristic zero, or a finite dimensional Lie group, one associates to it a Lie algebra, its tangent Lie algebra which is the Lie subalgebra $\mathcal{X}^{linv}(G)$ of the Lie algebra $\mathcal{X}^(G)$ left invariant vector fields on $G$ with respect to the usual Lie bracket of vector fields.
The value of a left-invariant vector field $X$ at the unit element $e$ is a tangent vector $X_e$ at $e$. It appears that the specialization/evaluation at the unit element map $\mathcal{X}^{linv}(G)\to T_e G$ is an isomorphism of vector spaces, which is often considered as an identification. However, one needs to look into vector fields in order to find the bracket, hence defining the tangent Lie algebra as the tangent vector space at $e$ misses the bracket (which come from consideration of infintesimals of second order). One can instead work with right invariant vector fields $\mathcal{X}^{rinv}(G)$ and obtain an isomorphic Lie algebra; the isomorphism is of course, by comparing the specialization at $e$.
Within $\mathcal{X}^(G)$, all right invariant vector fields commute with all left invariant vector fields.
The correspondence $G\mapsto\mathcal{X}^{linv}(G)$ is functorial.
A generalization of a tangent Lie algebra of a Lie group is the tangent Lie algebroid.
The first idea of the tangent Lie algebra was explained in a letter of Sophus Lie to his friend Meyer in 1874 (See the historical appendix to Bourbaki, Lie groups and Lie algebras vol. 3).
Created on November 11, 2012 23:28:05 by Zoran Škoda (193.55.36.32) | 2014-11-28 13:25:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 14, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8920319080352783, "perplexity": 179.1109768794699}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931010402.68/warc/CC-MAIN-20141125155650-00156-ip-10-235-23-156.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/3286477/integration-of-multi-valued-function | # Integration of multi-valued function
Graph
I have a set of discrete (x,y) points and need to find the area between the curve and the y-axis. However, the curve seems to be a multi-valued function (multiple x for the same y). Is such an operation valid, and if so, how?
edit: The image of the curve is in the link above. Kindly take a look.
I tried to do trapezoidal integration for this in MATLAB and got a negative answer. What is the interpretation of this negative area (even when all data are positive)? To check if area becoming negative was wrong, it tried doing the integration as ($$xy - \int ydx$$). I got the same answer in both cases.
This shows the area that I need.
This shows the area that I need when multiple x occurs for a single y. Area for all x is required.
• Comments are not for extended discussion; this conversation has been moved to chat. – Aloizio Macedo Jul 12 at 22:15 | 2019-08-25 15:27:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7759360671043396, "perplexity": 439.3062402531074}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027330750.45/warc/CC-MAIN-20190825151521-20190825173521-00546.warc.gz"} |
http://clay6.com/qa/2173/if-x-y-z-in-r-then-the-value-of-determinant-begin-2-x-2-2-2-x-2-2-1-3-x-3-2 | If $x,y, z\in R,$ then the value of determinant $\small\begin{vmatrix}(2^x + 2^{-x})^2 & (2^x - 2^{-x})^2 & 1\\ (3^x + 3^{-x})^2 & (3^x - 3^{-x})^2 & 1\\(4^x + 4^{-x})^2 & (4^x - 4^{-x})^2 & 1\end{vmatrix}$ is
$\begin{array}{1 1} 1 \\ -1 \\ 0 \\ \pm 1 \end{array}$
Toolbox:
• Elementary transformation of a matrix can be done by the addition (or subtraction)of the elements of two columns(or rows).
• If each element of a row (or column) of a determinant is multiplied by a constant k,then its value gets multiplied by k.
$\Delta=\begin{vmatrix}(2^x + 2^{-x})^2 & (2^x - 2^{-x})^2 & 1\\ (3^x + 3^{-x})^2 & (3^x - 3^{-x})^2 & 1\\(4^x + 4^{-x})^2 & (4^x - 4^{-x})^2 & 1\end{vmatrix}$
Apply $C_1\rightarrow C_1-C_2$
$\Delta=\begin{vmatrix}(2^x + 2^{-x})^2 -(2^x - 2^{-x})^2&(2^x - 2^{-x})^2 & 1\\ (3^x + 3^{-x})^2 - (3^x - 3^{-x})^2&(3^x - 3^{-x})^2 & 1\\(4^x + 4^{-x})^2 - (4^x - 4^{-x})^2&4^x - 4^{-x})^2 & 1\end{vmatrix}$
But we know $(a+b)^2-(a-b)^2=4ab$
Similarly $(2^x+2^{-x})^2-(2^x-2^{-x})^2=4.2^x.2^{-x}=4$
$\quad(3^x+3^{-x})^2-(3^x-3^{-x})^2=4.3^x.3^{-x}=4$
$\quad\quad(4^x+4^{-x})^2-(4^x-4^{-x})^2=4.4^x.4^{-x}=4$
$\Delta=\begin{vmatrix}4 & (2^x-2^{-x})^2& 1\\4 & (3^x-3^{-x})^2& 1\\4 & (4^x-4^{-x})^2& 1\end{vmatrix}$
Take 4 as the common factor from $C_1$,
$\Delta=\begin{vmatrix}1 & (2^x-2^{-x})^2& 1\\1 & (3^x-3^{-x})^2& 1\\1 & (4^x-4^{-x})^2& 1\end{vmatrix}$=0 (since two rows are identical)
Hence the value of the determinant=0. | 2018-04-22 14:15:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.914170503616333, "perplexity": 763.6405872915554}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125945604.91/warc/CC-MAIN-20180422135010-20180422155010-00403.warc.gz"} |
https://stacks.math.columbia.edu/tag/0DXU | Lemma 52.20.1. In Situation 52.16.1 let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. Assume
1. $A$ is local and $\mathfrak a = \mathfrak m$ is the maximal ideal,
2. $A$ has a dualizing complex,
3. $I = (f)$ is a principal ideal for a nonzerodivisor $f \in \mathfrak m$,
4. $\mathcal{F}_ n$ is a finite locally free $\mathcal{O}_ U/f^ n\mathcal{O}_ U$-module,
5. if $\mathfrak p \in V(f) \setminus \{ \mathfrak m\}$, then $\text{depth}((A/f)_\mathfrak p) + \dim (A/\mathfrak p) > 1$, and
6. if $\mathfrak p \not\in V(f)$ and $V(\mathfrak p) \cap V(f) \not= \{ \mathfrak m\}$, then $\text{depth}(A_\mathfrak p) + \dim (A/\mathfrak p) > 3$.
Then $(\mathcal{F}_ n)$ extends canonically to $X$. In particular, if $A$ is complete, then $(\mathcal{F}_ n)$ is the completion of a coherent $\mathcal{O}_ U$-module.
Proof. We will prove this by verifying hypotheses (a), (b), and (c) of Lemma 52.16.10.
Since $\mathcal{F}_ n$ is locally free over $\mathcal{O}_ U/f^ n\mathcal{O}_ U$ we see that we have short exact sequences $0 \to \mathcal{F}_ n \to \mathcal{F}_{n + 1} \to \mathcal{F}_1 \to 0$ for all $n$. Thus condition (b) holds by Lemma 52.3.2.
By induction on $n$ and the short exact sequences $0 \to A/f^ n \to A/f^{n + 1} \to A/f \to 0$ we see that the associated primes of $A/f^ nA$ agree with the associated primes of $A/fA$. Since the associated points of $\mathcal{F}_ n$ correspond to the associated primes of $A/f^ nA$ not equal to $\mathfrak m$ by assumption (3), we conclude that $M_ n = H^0(U, \mathcal{F}_ n)$ is a finite $A$-module by (5) and Local Cohomology, Proposition 51.8.7. Thus hypothesis (c) holds.
To finish the proof it suffices to show that there exists an $n > 1$ such that the image of
$H^1(U, \mathcal{F}_ n) \longrightarrow H^1(U, \mathcal{F}_1)$
has finite length as an $A$-module. Namely, this will imply hypothesis (a) by Lemma 52.3.5. The image is independent of $n$ for $n$ large enough by Lemma 52.5.2. Let $\omega _ A^\bullet$ be a normalized dualizing complex for $A$. By the local duality theorem and Matlis duality (Dualizing Complexes, Lemma 47.18.4 and Proposition 47.7.8) our claim is equivalent to: the image of
$\text{Ext}^{-2}_ A(M_1, \omega _ A^\bullet ) \to \text{Ext}^{-2}_ A(M_ n, \omega _ A^\bullet )$
has finite length for $n \gg 1$. The modules in question are finite $A$-modules supported at $V(f)$. Thus it suffices to show that this map is zero after localization at a prime $\mathfrak q$ containing $f$ and different from $\mathfrak m$. Let $\omega _{A_\mathfrak q}^\bullet$ be a normalized dualizing complex on $A_\mathfrak q$ and recall that $\omega _{A_\mathfrak q}^\bullet = (\omega _ A^\bullet )_\mathfrak q[\dim (A/\mathfrak q)]$ by Dualizing Complexes, Lemma 47.17.3. Using the local structure of $\mathcal{F}_ n$ given in (4) we find that it suffices to show the vanishing of
$\text{Ext}^{-2 + \dim (A/\mathfrak q)}_{A_\mathfrak q}( A_\mathfrak q/f, \omega _{A_\mathfrak q}^\bullet ) \to \text{Ext}^{-2 + \dim (A/\mathfrak q)}_{A_\mathfrak q}( A_\mathfrak q/f^ n, \omega _{A_\mathfrak q}^\bullet )$
for $n$ large enough. If $\dim (A/\mathfrak q) > 3$, then this is immediate from Local Cohomology, Lemma 51.9.4. For the other cases we will use the long exact sequence
$\ldots \xrightarrow {f^ n} H^{-1}(\omega _{A_\mathfrak q}^\bullet ) \to \text{Ext}^{-1}_{A_\mathfrak q}( A_\mathfrak q/f^ n, \omega _{A_\mathfrak q}^\bullet ) \to H^0(\omega _{A_\mathfrak q}^\bullet ) \xrightarrow {f^ n} H^0(\omega _{A_\mathfrak q}^\bullet ) \to \text{Ext}^0_{A_\mathfrak q}( A_\mathfrak q/f^ n, \omega _{A_\mathfrak q}^\bullet ) \to 0$
If $\dim (A/\mathfrak q) = 2$, then $H^0(\omega _{A_\mathfrak q}^\bullet ) = 0$ because $\text{depth}(A_\mathfrak q) \geq 1$ as $f$ is a nonzerodivisor. Thus the long exact sequence shows the condition is that
$f^{n - 1} : H^{-1}(\omega _{A_\mathfrak q}^\bullet )/f \to H^{-1}(\omega _{A_\mathfrak q}^\bullet )/f^ n$
is zero. Now $H^{-1}(\omega ^\bullet _\mathfrak q)$ is a finite module supported in the primes $\mathfrak p \subset A_\mathfrak q$ such that $\text{depth}(A_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) \leq 1$. Since $\dim ((A/\mathfrak p)_\mathfrak q) = \dim (A/\mathfrak p) - 2$ condition (6) tells us these primes are contained in $V(f)$. Thus the desired vanishing for $n$ large enough. Finally, if $\dim (A/\mathfrak q) = 1$, then condition (5) combined with the fact that $f$ is a nonzerodivisor insures that $A_\mathfrak q$ has depth at least $2$. Hence $H^0(\omega _{A_\mathfrak q}^\bullet ) = H^{-1}(\omega _{A_\mathfrak q}^\bullet ) = 0$ and the long exact sequence shows the claim is equivalent to the vanishing of
$f^{n - 1} : H^{-2}(\omega _{A_\mathfrak q}^\bullet )/f \to H^{-2}(\omega _{A_\mathfrak q}^\bullet )/f^ n$
Now $H^{-2}(\omega ^\bullet _\mathfrak q)$ is a finite module supported in the primes $\mathfrak p \subset A_\mathfrak q$ such that $\text{depth}(A_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) \leq 2$. By condition (6) all of these primes are contained in $V(f)$. Thus the desired vanishing for $n$ large enough. $\square$
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All contributions are licensed under the GNU Free Documentation License.
In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0DXU. Beware of the difference between the letter 'O' and the digit '0'. | 2022-06-30 13:40:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.9639002084732056, "perplexity": 169.63041574173337}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103821173.44/warc/CC-MAIN-20220630122857-20220630152857-00184.warc.gz"} |
http://stackoverflow.com/questions/21582388/cant-compile-renderscript-after-updating-to-android-studio-4-3-with-build-tools | # Can't compile RenderScript after updating to Android Studio 4.3 with build-tools 18.1.1
I've been working on a RenderScript project for quite some time now, using older versions of Android Studio. I'm targeting 4.3 with SDK version 18.
I've had run-time issues when compiling the project with new build-tools 19+ (crashes when I modify the root() 'uchar *v_out' parameter), so I decided to go back to build-tools 18.1.1. But when compiling with this version I get the error:
'conversion.rs error: Compute Kernel root() targeting SDK levels 11-13 may not skip parameters FAILURE: Build failed with an exception.
Execution failed for task :test:compileDebugRenderscript. com.android.ide.common.internal.LoggedErrorException: Failed to run command: C:\Program Files\android-studio\sdk\build-tools\18.1.1\llvm-rs-cc.exe -O 3 -I C:\Program Files\android-studio\sdk\build-tools\18.1.1\renderscript\include\ -I C:\Program Files\android-studio\sdk\build-tools\18.1.1\renderscript\clang-include\ ...
Perhaps it's a problem with my kernel function signature? Even with a simple .rs file below, it fails to compile. I used to be able to compile and run this fine and I'm pretty sure it was using build-tools 18 back then..
``````#pragma version(1)
#pragma rs java_package_name(com.app.test);
void root(const uchar4 *in, uchar4 *out, uint32_t x, uint32_t y) {
}
``````
Any help would be greatly appreciated.
Update: This is getting strange... if I use gradle version 1.10 with build-tools 19.0.1, the code compiles but gives me the following error if i try to modify 'uchar4 *out' | 2016-05-01 08:32:45 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8885828852653503, "perplexity": 14140.368415330337}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860114649.41/warc/CC-MAIN-20160428161514-00205-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://forum.allaboutcircuits.com/threads/edge-triggered-timer.107902/ | # Edge triggered timer
#### poopscoop
Joined Dec 12, 2012
140
I'm building what is basically a skydiving micro controller, and I want a backup parachute release mechanism.
This is to release a chute on project, and in no way related to the safety of a person.
My intent is to reset a timer on an edge, produced by the uC. I've looked at missing pulse detectors, but the ones I see don't reset on the edge, but rather on the state.
I'm considering a 556, with one timer reset on the uC LOW state, and the other on the HIGH, so if it hangs in either state it would time out and deploy the chute. Obviously a self contained hardware solution is a requirement.
EDIT:::
Better description:
I seem to have expected you guys to read minds.
The arduino should handle chute deployment, however, being a low cost microprocessor there is a chance it will hang, or my code will have a bug. If this happens, I want a totally separate fail-safe chute deployment. The arduino will output a 'life message' in the form of a pin = !pin pulse every complete code cycle. If it hangs, this pulse will stop.
If the pulse stops, the timer, or whatever else I can use, will time out. When it time outs, it will override the electromagnet holding the chute in the stowed position, thus deploying the chute and saving my $80 worth of parts. This will not solve every possible code problem, as the loop can continue to output an 'alive' message without ever executing my chute deployment routine. This can, and will, be remedied with a 555 calibrated for worst case drop time. Ideally I can have multiple hardware fail-safe devices on a PCB. I'm thinking of the following implementation: 1. 556 or another device to read 'alive' pulse. 2. 555 time for worst case drop. Activated when the device detaches from the balloon. 3. 555 for emergency balloon release. Should the balloon break free of the line and start climbing, and the arduino fails, it will cut the device free. 4. Power supply short circuit protection. Should there be a wiring problem inside the device, the arduino could shutdown and the coils wont have the power to release from the balloon (Balloon fails connected, chute fails deployed. Lose power on ascent and I can't release the device). If the voltage across a series resistor reaches X value, one MOSFET disables the primary current path and another activates a secondary. Secondary path releases from balloon and deploys chute. Not sure on the implementation of this one, but I will probably mirror it after rudimentary power supply short circuit protection and possibly some logic. Last edited: #### MaxHeadRoom Joined Jul 18, 2013 22,689 Picmicro have a rising or trailing edge Capture function in the CCP modules if that would work? Max. Thread Starter #### poopscoop Joined Dec 12, 2012 140 I should mention that I'm already using an Arduino (have it on hand), and cost is a concern, as there's a chance it will smash into the ground. #### AnalogKid Joined Aug 1, 2013 9,064 Your description is a bit vague. The arduino produces an edge that resets the timer, and then the timer times out. Then what? What does this do, what is the output of the timer connected to, and where does a retriggerable circuit fit in? ak Thread Starter #### poopscoop Joined Dec 12, 2012 140 Arduino produces a pulse -> hangs -> timer times out -> mosfet goes LOW -> magnetic field decays -> chute deploys #### AnalogKid Joined Aug 1, 2013 9,064 One question would be why not implement the timer within the arduino code, but I'm still curious about the retriggerable. Also, from your description it sounds like the chute deploys when something is turned off (decays). Does this mean that the release mechanism has to be powered constantly to prevent it from releasing too early? Seems backwards... ak Thread Starter #### poopscoop Joined Dec 12, 2012 140 I seem to have expected you guys to read minds. The arduino should handle chute deployment, however, being a low cost microprocessor there is a chance it will hang, or my code will have a bug. If this happens, I want a totally separate fail-safe chute deployment. The arduino will output a 'life message' in the form of a pin = !pin pulse every complete code cycle. If it hangs, this pulse will stop. If the pulse stops, the timer, or whatever else I can use, will time out. When it time outs, it will override the electromagnet holding the chute in the stowed position, thus deploying the chute and saving my$80 worth of parts.
This will not solve every possible code problem, as the loop can continue to output an 'alive' message without ever executing my chute deployment routine. This can, and will, be remedied with a 555 calibrated for worst case drop time.
Ideally I can have multiple hardware fail-safe devices on a PCB. I'm thinking of the following implementation:
1. 556 or another device to read 'alive' pulse.
2. 555 time for worst case drop. Activated when the device detaches from the balloon.
3. 555 for emergency balloon release. Should the balloon break free of the line and start climbing, and the arduino fails, it will cut the device free.
4. Power supply short circuit protection. Should there be a wiring problem inside the device, the arduino could shutdown and the coils wont have the power to release from the balloon (Balloon fails connected, chute fails deployed. Lose power on ascent and I can't release the device). If the voltage across a series resistor reaches X value, one MOSFET disables the primary current path and another activates a secondary. Secondary path releases from balloon and deploys chute. Not sure on the implementation of this one, but I will probably mirror it after rudimentary power supply short circuit protection and possibly some logic. | 2021-04-18 00:10:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17484933137893677, "perplexity": 3058.597638968228}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038464065.57/warc/CC-MAIN-20210417222733-20210418012733-00556.warc.gz"} |
https://math.stackexchange.com/questions/2381788/manifolds-with-diffeomorphic-boundaries | # Manifolds with diffeomorphic boundaries
Suppose $X$, $Y$ are two $2n$-dimensional manifolds with handles of index $0$ and $n$. In particular, $X$ and $Y$ have boundary. Suppose that $X$, $Y$ have isomorphic homology and intersection form and that their boundaries are diffeomorphic. Does this imply that $X$ and $Y$ are diffeomorphic? Less precisely, I am wondering if the intersection form and boundary determine the framing of the n handles.
Let me point out that there are examples of manifolds with the same homology and diffeomorphic boundary but are not diffeomorphic. Their intersection forms have the same rank but not the same signature (although their signatures are always congruent modulo some large integer).
• Can you provide some more details? I agree it is a question of framing. I need to show that the boundary remembers the framing. I think the intersection form only knows part of the framing data? – user39598 Aug 3 '17 at 22:36
• I deleted the comment because I didn't realize you wanted to allow an arbitrary number of n-handles. I don't know the answer, but surely you want to look at Wall's paper and the various citations to and from. – user98602 Aug 3 '17 at 22:36
• I think even for a single handle it might not be true. – user39598 Aug 3 '17 at 22:38
• I remember thinking about this some time ago and thought it had been; my apologies. In that case you are precisely interested in the homomorphism $\pi_n SO(n-1) \to \Gamma_{2n-1}$ and whether or not it has kernel. But I don't remember whether or not this is true. Kervaire and Milnor would know, somewhere... – user98602 Aug 3 '17 at 22:49
No. There are plenty of pairs of knots with distinct $n$-traces with the same $n$-surgery. There are many papers by people like Yasui and Akbulut on this. For instance this one : https://arxiv.org/abs/1505.02551 . | 2019-07-22 18:09:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7926822304725647, "perplexity": 309.9018022762311}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195528208.76/warc/CC-MAIN-20190722180254-20190722202254-00020.warc.gz"} |
https://socratic.org/questions/what-is-the-temperature-difference-between-a-solution-s-boiling-point-and-a-pure | # What is the temperature difference between a solution's boiling point and a pure solvent's boiling point called?
May 28, 2017
$\text{The boiling point elevation.........}$
#### Explanation:
And such measurements allow for a fairly simple determination of the concentration of solutes in solution. Such measurements are termed $\text{ebullioscopy}$. | 2019-12-14 08:43:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 2, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.663501501083374, "perplexity": 1958.4217208424575}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540585566.60/warc/CC-MAIN-20191214070158-20191214094158-00225.warc.gz"} |
https://bird.bcamath.org/handle/20.500.11824/14/browse?rpp=20&sort_by=1&type=title&offset=0&etal=-1&order=ASC | Now showing items 1-20 of 50
• #### A computational model of open-irrigated electrode for endocardial RF catheter ablation
(Computing in Cardiology, 2016-01-01)
Radiofrequency catheter ablation (RFCA) is an important curative treatment for cardiac arrhythmias. However, during RFCA thrombus formation can occur when the electrode-tissue interface temperature exceeds 80°C. Open-irrigated ...
• #### Advanced Mathematical Modelling of Pancreatic β-Cells
(2019-12-10)
Insulin-secreting pancreatic $\beta$-cells are responsible for maintaining the whole body glucose homeostasis. Dysfunction or loss of $\beta$-cell mass results in impaired insulin secretion and, in some cases, diabetes. ...
• #### Brain energetics plays a key role in the coordination of electrophysiology, metabolism and hemodynamics: evidence from an integrated computational model
(Journal of theoretical biology, 2019-06-05)
The energetic needs of brain cells at rest and during elevated neuronal activation has been the topic of many investigations where mathematical models have played a significant role providing a context for the interpretation ...
• #### Clinical correlates of mathematical modeling of cortical spreading depression: Single‐cases study
(Brain and Behavior, 2019-07-28)
Introduction: Considerable connections between migraine with aura and cortical spreading depression (CSD), a depolarization wave originating in the visual cortex and traveling toward the frontal lobe, lead to the hypothesis ...
• #### Comparative analysis of different methods of modeling the thermal effect of circulating blood flow during RF cardiac ablation
(IEEE Transactions on Biomedical Engineering, 2016-01-01)
Our aim was to compare the different methods of modeling the effect of circulating blood flow on the thermal lesion dimensions created by radio frequency (RF) cardiac ablation and on the maximum blood temperature. Computational ...
• #### Computational Model for Prediction of the Occurrence of Steam Pops during Irrigated Radiofrequency Catheter Ablation
(Computing in Cardiology 43, 2016)
Radiofrequency catheter ablation (RFCA) is a curative treatment for cardiac arrhythmias. Although globally a pretty safe procedure, it may present some risk. Steam pop is a serious complication that can occur during RFCA ...
• #### A computational model integrating brain electrophysiology and metabolism highlights the key role of extracellular potassium and oxygen
(Journal of Theoretical Biology, 2018-02-26)
The human brain is a small organ which uses a disproportional amount of the total metabolic energy pro- duction in the body. While it is well understood that the most significant energy sink is the maintenance of the ...
• #### A computational model of open-irrigated radiofrequency catheter ablation accounting for mechanical properties of the cardiac tissue
(2018)
Radiofrequency catheter ablation (RFCA) is an effective treatment for cardiac arrhythmias. Although generally safe, it is not completely exempt from the risk of complications. The great flexibility of computational models ...
• #### Computational modeling of open-irrigated electrodes for radiofrequency cardiac ablation including blood motion-saline flow interaction
(PLoS One, 2016)
Radiofrequency catheter ablation (RFCA) is a routine treatment for cardiac arrhythmias. During RFCA, the electrode-tissue interface temperature should be kept below 80°C to avoid thrombus formation. Open-irrigated electrodes ...
• #### A computational multiscale model of cortical spreading depression propagation
(Computers and Mathematics with Applications, 2016-12-19)
Cortical Spreading Depression (CSD) is a disruption of the brain hemostasis that, originating in the visual cortex and traveling towards the frontal lobe, temporarily impairs the normal functioning of neurons. Although, ...
• #### Computational predictive modeling of integrated cerebral metabolism, electrophysiology and hemodynamics
(2019-02-12)
Understanding the energetic requirement of brain cells during resting state and during high neuronal activity is a very active research area where mathematical models have contributed significantly by providing a context ...
• #### Discretizations of the spectral fractional Laplacian on general domains with Dirichlet, Neumann, and Robin boundary conditions
(2017-04-28)
In this work, we propose novel discretisations of the spectral fractional Laplacian on bounded domains based on the integral formulation of the operator via the heat-semigroup formalism. Specifically, we combine suitable ...
• #### Effect of Tissue Elasticity in Cardiac Radiofrequency Catheter Ablation Models
(2018)
Radiofrequency catheter ablation (RFCA) is an effective treatment for different types of cardiac arrhythmias. However, major complications can occur, including thrombus formation and steam pops. We present a full 3D ...
• #### Estimation of age-specific rates of reactivation and immune boosting of the varicella zoster virus
(Epidemics, 2016-12-31)
Studies into the impact of vaccination against the varicella zoster virus (VZV) have increasingly focused on herpes zoster (HZ), which is believed to be increasing in vaccinated populations with decreasing infection pressure. ...
• #### Extreme brain events: Higher-order statistics of brain resting activity and its relation with structural connectivity
(EPL, 2015-12-31)
The brain exhibits a wide variety of spatiotemporal patterns of neuronal activity recorded using functional magnetic resonance imaging as the so-called blood-oxygenated-level-dependent (BOLD) signal. An active area of work ...
• #### Functional connectivity of EEG signals under laser stimulation in migraine
(Frontiers in Human Neuroscience, 2015-12-31)
In previous studies, migraine patients showed abnormalities in pain-related evoked responses, as reduced habituation to repetitive stimulation. In this study, we aimed to apply a novel analysis of EEG bands synchronization ...
• #### Geometry shapes propagation: Assessing the presence and absence of cortical symmetries through a computational model of cortical spreading depression
(Frontiers in Computational Neuroscience, 2015-12-31)
Cortical spreading depression (CSD), a depolarization wave which originates in the visual cortex and travels toward the frontal lobe, has been suggested to be one neural correlate of aura migraine. To the date, little is ...
• #### Glioma invasion and its interplay with the nervous tissue: a multiscale model
(2019)
A multiscale mathematical model for glioma cell migration and proliferation is proposed, taking into account a possible therapeutic approach. Starting with the description of processes taking place on the subcellular level, ...
• #### How does radiofrequency ablation efficacy depend on the stiffness of the cardiac tissue? Insights from a computational model
(2019)
Objective. Radiofrequency catheter ablation (RFCA) is an effective treatment for the elimination of cardiac arrhythmias, however it is not exempt from complications that can risk the patients’ life. The efficacy of the ...
• #### Incorporating landscape heterogeneities in the spread of an epidemics in wildlife
(Complex Systems; Control of Infectious Diseases, Extended Abstracts 2013. Ã lvaro Corral, Anna Deluca, Francesc Font-Clos, Pilar Guerrero, Andrei Korobeinikov, Francesco Massucci eds., Springer,.http://dx.doi.org/10.1007/978-3-319-08138-0_19", 2014-12-31)
One of the main difficulties in the modeling and numerical simulation of the spread of an infectious disease in wildlife resides in properly taking into account the heterogeneities of the landscape. Forests, plains and ... | 2020-07-10 12:51:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4018046259880066, "perplexity": 7404.166865162251}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655908294.32/warc/CC-MAIN-20200710113143-20200710143143-00307.warc.gz"} |
https://web2.0calc.com/questions/parabolas_18 | +0
# Parabolas
0
111
2
The parabolas defined by the equations y = -x^2 - x + 3 and y = 2x^2 - 1 intersect at points (a, b) and (c, d), where c is greater than or equal to a. What is c - a? Express your answer as a common fraction.
Dec 30, 2021
#1
0
You are only looking for the 'x' coordinates of the poits of intersection ....so
-x^2-x+3 = 2x^2-1
3x^2+x - 4 = 0 x = 1 and - 4/3 <===== you can finish
Dec 30, 2021
#2
+13724
+1
What is c - a?
Hello Guest!
$$-x^2 - x + 3 = 2x^2 - 1\\ \color{blue}3x^2+x-4=0$$
$${\color{blue}a=}\frac{-1+7}{6}=\color{blue}1\\ b=1\\ {\color{blue}c=}2\cdot \frac{16}{9}-1=\color{blue}\frac{23}{9}\\ d=\frac{-1-7}{6}=-\frac{4}{3}\\$$
$$c-a=\frac{14}{9}$$
!
Dec 30, 2021 | 2022-06-29 21:52:58 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5954315066337585, "perplexity": 1556.4305857436834}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103645173.39/warc/CC-MAIN-20220629211420-20220630001420-00398.warc.gz"} |
https://sriasat.wordpress.com/tag/field/ | # Tag Archives: field
## Number of irreducible polynomials over a finite field
I and some friends just came up with this. (We came up with some ideas for the proof, after coming across the formula on the internet.) This is the coolest application of the Möbius inversion formula that I’ve seen so far.
Let $q$ be a prime power and $n$ any positive integer. We wish to count the number $C(n)$ of monic irreducible polynomials of degree $n$ over the finite field $\mathbb F_q$.
Note that the zeros of
$\displaystyle F_d(X):=\prod_{\deg(f)=d}f(X)$
are precisely the elements of degree $d$ over $\mathbb F_q$, where the product is taken over monic irreducible polynomials $f\in\mathbb F_q[X]$. Further, we have the following:
Lemma 1. Let $g\neq h$ be relatively prime polynomials over a field $k$. Then $g$ and $h$ have no common zeros.
Proof. Since $k[X]$ is a PID, there are polynomials $u,v\in k[X]$ such that $ug+vh=1$. If $\alpha$ is a common zero of $g$ and $h$ in some extension $K/k$, then $X-\alpha$ divides $1=ug+vh$ in $K[X]$, a contradiction. $\square$
Lemma 2. If $f$ is irreducible over a field $k$, then $f$ has distinct zeros.
Proof. Note that any multiple zero of $f$ must be a common zero of $f$ and its formal derivative $f'$. Since $f$ is irreducible over $k$, it follows that $f$ and $f'$ are relatively prime over $k$. Now apply lemma 1 to $f$ and $f'$. $\square$
It follows that the zeros of
$\displaystyle \prod_{d\mid n}F_d(X)$
are all distinct and are precisely the elements of the union of all the subfields of $\mathbb F_{q^n}$, which is just $\mathbb F_{q^n}$. This implies
$(*)\qquad\qquad\qquad\qquad\quad\displaystyle\prod_{d\mid n}F_d(X)=X^{q^n}-X$,
since $\mathbb F_{q^n}$ is the splitting field of $X^{q^n}-X$ over $\mathbb F_q$. Now a comparison of the degrees of both sides reveals that
$\displaystyle\sum_{d\mid n}dC(d)=q^n$.
Finally, Möbius inversion gives
$\displaystyle \boxed{C(n)=\frac 1n\sum_{d\mid n}\mu(d)q^{n/d}}$.
Filed under Combinatorics, Number theory
## Ring of integers of cyclotomic field
Let $\zeta=\zeta_n=e^\frac{2\pi i}{n}$ be a primitive $n$-th root of unity, and let $K=\mathbb Q(\zeta)$. I am going to outline a proof that $\mathcal O_K=\mathbb Z[\zeta]$, based on several homework problems from one of my recent courses. There are probably many other proofs of this, but I particularly like this one because it’s easy to follow, touches on a wide range of topics, and I worked hard through it!
First, let $n=p^m$ be a prime power. The minimal polynomial of $\zeta$ is the $n$-th cyclotomic polynomial
$\displaystyle\Phi_n(X)=\Phi_{p^m}(X)=\sum_{j=0}^{p-1}X^{jp^{m-1}}$.
Let $f(X)=\Phi_n(X+1)$.
Exercise 1. Following the above notation, show that $f$ satisfies the conditions of Eisenstein’s criterion for the prime $p$.
Consider the discriminant $\Delta(f)=\Delta(R)$, where $R=\mathbb Z[\zeta-1]$. If $q$ is a prime factor of $\Delta(f)$, then $f$ must have a multiple root modulo $q$. Hence $X^{p^m}-1$ will also have a multiple root modulo $q$. But $\gcd(X^{p^{m}}-1, p^{m}X^{p^{m}-1})=1$ in $\mathbb Z/q\mathbb Z$ unless $q=p$. Thus $\Delta(f)=\Delta(R)=[\mathcal O_K:R]^2\Delta(\mathcal O_K)$ is a power of $p$. In particular, $[\mathcal O_K:R]$ is a power of $p$.
Exercise 2. Suppose that $f\in\mathbb Z[X]$ satisfies the conditions of Eisenstein’s criterion for a prime number $p$. Let $\alpha$ be a root of $f$ and let $R=\mathbb Z[\alpha]$. Prove that there is exactly one prime ideal $P\subseteq R$ that contains $p$, and that the local ring $R_P$ is a DVR with uniformiser $\alpha$.
Now if $Q\subseteq R$ is another prime ideal, then $p\not\in Q$. So $[\mathcal O_K:R]\not\in Q$.
Exercise 3. Let $K$ be a number field and $R\subseteq\mathcal O_K$ a subring of finite index $d$. If $Q\subseteq R$ is a prime ideal not containing $d$, show that $R_Q$ is a DVR.
I’ll include this solution because I love it!
Solution. Let $D=\mathcal O_K$. By going up, there is a prime ideal $\tilde Q\subseteq D$ with $\tilde Q\cap R=Q$. We’ll show that $D_{\tilde Q}=R_Q$, so the result will follow.
Let $a/b\in R_Q$, so that $a\in R$, $b\in R\setminus Q$. If $b\in \tilde Q$, then $b\in R\cap \tilde Q=Q$, a contradiction. Hence $b\not\in\tilde Q$, i.e. $a/b\in D_{\tilde Q}$. Thus $R_Q\subseteq D_{\tilde Q}$.
Now let $a/b\in D_{\tilde Q}$, so that $a\in D$, $b\in D\setminus\tilde Q$. Then $da\in R$, $db\in R$, and $a/b=(da)/(db)$. Note that $b\not\in\tilde Q\supseteq Q$, and $d\not\in Q$ by assumption. So $db\not\in Q$ since $Q$ is a prime ideal. Thus $a/b=(da)/(db)\in R_Q$, i.e. $D_{\tilde Q}=R_Q$. $\square$
So $R$ localised at any prime ideal is a DVR, implying that $R=\mathbb Z[\zeta]$ is a Dedekind domain. Thus $R=\mathcal O_K$. This completes the proof for the prime power case.
Now we proceed by induction on the number of distinct prime factors of $n$. The base case has already been taken care of. So suppose that $n=ab$, where $a,b>1$ are coprime integers. Let $L=\mathbb Q(\zeta_a)$ and $M=\mathbb Q(\zeta_b)$. By the induction hypothesis, $\mathcal O_L=\mathbb Z[\zeta_a]$ and $\mathcal O_M=\mathbb Z[\zeta_b]$.
Exercise 4. Let $L$ and $M$ be number fields with discriminants $\lambda$ and $\mu$ respectively. Let $\{a_1,\dots,a_m\}$ and $\{b_1,\dots,b_n\}$ be integral bases for $L$ and $M$ respectively. Let $K=LM=\mathbb Q(a_1,\dots,a_m,b_1,\dots,b_n)$ be the composite field of $L$ and $M$. Suppose that $[K:\mathbb Q]=[L:\mathbb Q][M:\mathbb Q]$ and that $\gcd(\lambda,\mu)=1$. Show that $\{a_ib_j:1\le i\le m; 1\le j\le n\}$ is an integral basis for $K$.
So our main result will follow from exercise 4 once we have checked that all the hypotheses are satisfied.
(i) Checking $K=LM$Firstly, $LM$ contains $\zeta_a\zeta_b=e^{\frac{2\pi i(a+b)}{n}}=\zeta^{a+b}$.
Exercise 5. If $\gcd(a,b)=1$, show that $\gcd(a+b,ab)=1$.
So writing $j=a+b$ shows that $\gcd(j,n)=1$ and $\zeta^j\in LM$. If $j^{-1}\in\{1,\dots,n\}$ is the multiplicative inverse of $j\pmod n$, then $\zeta=(\zeta^j)^{j^{-1}}\in LM$. Thus $K\subseteq LM$.
Again, since $\zeta^a=\zeta_b$ and $\zeta^b=\zeta_a$, we have $LM\subseteq K$. Thus $K=LM$.
(ii) Checking $[K:\mathbb Q]=[L:\mathbb Q][M:\mathbb Q]$. We have
$[K:\mathbb Q]=\varphi(n)=\varphi(ab)=\varphi(a)\varphi(b)=[L:\mathbb Q][M:\mathbb Q]$.
(iii) Checking $\gcd(\lambda,\mu)=1$This is slightly harder. We need a few more facts.
Exercise 6. Let $E=\mathbb Q(\alpha)$ be a number field, and let $f$ be the minimal polynomial of $\alpha$. Show that
$\Delta(\alpha)=(-1)^{\binom{\deg(f)}{2}}N_{E/\mathbb Q}(f'(\alpha))$.
Solution. Let $f(x)=(x-\alpha_1)\cdots(x-\alpha_n)$. Then
$\displaystyle\Delta(\alpha)=(-1)^{\binom n2}\prod_{i\neq j}(\alpha_i-\alpha_j)=(-1)^{\binom n2}\prod_i\prod_{j\neq i}(\alpha_i-\alpha_j)$.
Using Leibniz’s rule,
$\displaystyle f'(x)=\sum_i\prod_{j\neq i}(x-\alpha_i)\Rightarrow f'(\alpha_i)=\prod_{j\neq i}(\alpha_i-\alpha_j)$.
Hence
$\displaystyle\Delta(\alpha)=(-1)^{\binom n2}\prod_if'(\alpha_i)=(-1)^{\binom n2}N_{E/\mathbb Q}(f'(\alpha))$. $\square$
Exercise 7. Using the previous exercise, show that $\Delta(\zeta_n)\mid n^{\varphi(n)}$.
Hint. Write $X^n-1=\Phi_n(X)g(X)$, and use Leibniz’s rule to get $n\zeta_n^{n-1}=\Phi_n'(\zeta_n)g(\zeta_n)$.
So $\lambda=\Delta(\zeta_a)$ divides some power of $a$, and $\mu=\Delta(\zeta_b)$ divides some power of $b$. Since $\gcd(a,b)=1$, we conclude that $\gcd(\lambda,\mu)=1$.
1 Comment
Filed under Algebra, Number theory
## An application of Zorn’s lemma: Transcendence bases
Zorn’s lemma is a very useful result when it comes to dealing with an infinite collection of things. In ZFC set theory it is equivalent to the well-ordering theorem (every set can be well-ordered) and to the axiom of choice (a Cartesian product of non-empty sets is non-empty). I happened to use it a few days ago in proving the existence of transcendence bases, hence this post!
Zorn’s lemma. If every chain in a poset $P$ has an upper bound in $P$, then $P$ contains a maximal element.
Let $K/k$ be a field extension. We call a subset $S\subseteq K$ algebraically independent if for any $m\ge 0$ and $s_1,\dots,s_m\in S$, $p(s_1,\dots,s_m)=0$ implies $p=0$, where $p\in k[X_1,\dots,X_m]$ is a polynomial. A maximal (with respect to $\subseteq$) algebraically independent subset is called a transcendence base.
Theorem. Every field extension has a transcendence base.
Proof. If $K/k$ is algebraic, the transcendence base is the empty set. Suppose that $K/k$ is not algebraic. Let $\mathcal F$ be the family of all algebraically independent subsets $S\subseteq K$. By Zorn’s lemma, it suffices to show that every chain in $\mathcal F$ has an upper bound. Let $\mathcal C$ be a chain in $\mathcal F$, and let
$T=\displaystyle\bigcup_{S\in \mathcal C} S$.
It suffices to show that $T\in\mathcal F$, since then $T$ would be an upper bound for $\mathcal C$ in $\mathcal F$.
Let $P(m)$ be the statement:
If $t_1,\dots,t_m\in T$ are distinct, and $p(t_1,\dots, t_m)=0$ for some $p\in k[X_1,\dots,X_m]$, then $p=0$.
If $t_1\in T$, then $t_1\in S$ for some $S\in \mathcal C$. Since $\mathcal S$ is algebraically independent, $p(t_1)=0$ implies $p=0$ in $k[X_1]$. Thus $P(1)$ is true.
Suppose $P(m-1)$ is true. Let $t_1,\dots,t_m\in T$ be distinct such that $p(t_1,\dots,t_m)=0$ for some $p\in k[X_1,\dots,X_m]$. We can view $p(t_1,\dots,t_m)$ as a polynomial in $t_m$ with coefficients in $k[t_1,\dots,t_{m-1}]$, i.e. say
$\displaystyle p(t_1,\dots,t_m)=\sum_{i=0}^n p_i(t_1,\dots,t_{m-1})t_m^i=0$.
Since $P(1)$ is true, it folows that $p_i(t_1,\dots,t_{m-1})=0$ for each $i=0,1,\dots,n$. But then $p_i=0$ for each $i$ by our hypothesis. Thus $p=0$, implying $P(m)$ is true.
Therefore, by induction, $T$ is algebraically independent, i.e. $T\in\mathcal F$. $\square$
Filed under Algebra, Set theory
## Finite automata 1: Automatic sets
In one of my recent courses I had to learn about finite automata theory. The main purpose was to understand a generalisation of the celebrated Skolem-Mahler-Lech theorem in positive characteristic. Let $\mathbb N_0$ denote the set of non-negative integers and $K$ be a field. For a linear recurrence $f:\mathbb N_0\to K$, define its zero set to be $\{n\in\mathbb N_0 : f(n)=0\}$. Then the Skolem-Mahler-Lech theorem states the following:
Theorem (Skolem-Mahler-Lech, 1953). The zero set of a linear recurrence $f:\mathbb N_0\to\mathbb K$, where $K$ is a field of characteristic $0$, is a finite union of arithmetic progressions along with a finite set.
Derksen in his 2005 paper was able to provide the correct analogue of this result in positive characteristic for the first time:
Theorem (Derksen, 2007). The zero set of a linear recurrence $f:\mathbb N_0\to K$, where $K$ is a field of characteristic $p>0$, is a $p$-automatic set.
To understand this theorem we need to know about finite automata. In this and the next few posts I want to write about some interesting elementary results about automatic sets. Since my knowledge of computer science is epsilon(!), I will mostly touch on some very basic mathematical aspects.
A deterministic finite-state automaton (DFA) is a directed multigraph $\Gamma$ satisfying certain properties. Firstly, its vertex-set $Q$ (which we call the set of states) contains a special vertex (or state) $q_0$, called the initial state. We also have a finite non-empty set $\Sigma$ of symbols. Each edge of $\Gamma$ is labelled with a symbol from $\Sigma$. As a result, we have a transition function $\delta:Q\times\Sigma\to Q$ such that if an edge from $q_1$ to $q_2$ is labelled with the symbol $a\in\Sigma$, then we write $\delta(q_1,a)=q_2$. Finally, we have a set $F\subseteq Q$ (possibly empty) called the set of accepting or final states.
So $\Gamma$ can be formally written as a quintuple $(\Sigma,Q,q_0,\delta,F)$.
Let $\Sigma^*$ be the free monoid on $\Sigma$. We can extend the transition function $\delta$ to a map $\delta:Q\times\Sigma^*\to Q$ by setting $\delta(q_0,wa)=\delta(\delta(q_0,a),w)$ for all $a\in\Sigma$ and $w\in\Sigma^*$. Thus, to $\Gamma$ we can associate a language
$\mathcal L_\Gamma:=\{w\in\Sigma^*:\delta(q_0,w)\in F\}\subseteq\Sigma^*$.
We call such a language produced from a DFA a regular language. If $w\in\mathcal L_\Gamma$ for some $w\in\Sigma^*$, we say that $w$ is accepted by $\Gamma$.
Let $\mathbb N$ denote the set of the positive integers, and let $k>1$ be an integer. Define $\Sigma_k:=\{0,1,\dots,k-1\}$. A subset $S\subset\mathbb N$ is said to be $k$-automatic if there exists a DFA $\Gamma=(\Sigma_k,Q,q_0,\delta,F)$ such that $S$ is precisely the set of those $j\in\mathbb N$ whose base-$k$ representation, when regarded as a string in $\Sigma^*$, is accepted by $\Gamma$. If $S\subseteq\mathbb N$ is $k$-automatic for some $k>1$, we say $S$ is an automatic set.
For $w=w_nw_{n-1}\cdots w_0\in\Sigma_k^*$ with each $w_i\in\Sigma_k$ we define
$[w]_k:=w_nk^n+w_{n-1}k^{n-1}+\cdots+w_0.$
In this case we also define $|w|:=n+1$ to be the length of $w$.
So $S\subseteq\mathbb N$ is $k$-automatic precisely when $S=\{[w]_k:w\in\mathcal L_\Gamma\}$ for some DFA $\Gamma$.
Note that we can set $\delta(q,0)$ to be in $Q\backslash F$ for every $q\in Q$ (if $F=Q$, just create a new state). Thus we may without loss of generality assume that all our DFAs only accept words without leading zeros.
A basic result about regular languages is the pumping lemma:
Lemma (Pumping lemma). If $\mathcal L$ is an infinite regular language, then there exist words $a,b,c$ with $|b|\ge 1$ such that $ab^nc\in\mathcal L$ for all $n\ge 0$.
Proof. Since $\mathcal L$ is infinite, there are words of arbitrarily large length in $\mathcal L$. Let $w=w_nw_{n-1}\cdots w_0\in\mathcal L$ be a word of sufficient length. Then there exist $0\le i such that $w_i\cdots w_0=w_j\cdots w_0\in\mathcal L$. Taking $a=w_n\cdots w_j$, $b=w_{j-1}\cdots w_i$, $c=w_{i-1}\cdots w_0$ shows that $ab^nc\in\mathcal L$ for all $n\ge 0$. Moreover, since $i, it follows that $|b|\ge 1$. $\square$
In the following posts we are going to use the pumping lemma, together with other machinery, to derive some interesting results on the automaticity of certain sets. | 2019-03-24 19:41:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 316, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9648800492286682, "perplexity": 105.90391067765427}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912203491.1/warc/CC-MAIN-20190324190033-20190324212033-00428.warc.gz"} |
https://math.stackexchange.com/questions/408403/applications-of-computation-on-very-large-groups | # Applications of computation on very large groups
I have been studying computational group theory and I am reading and trying to implement these algorithms. But what that is actually bothering me is, what is the practical advantage of computing all properties of extremely large groups, moreover it is a hard problem?
It might give birth to new algorithms but does it solve any problem specific to group theory or other branches affected by it?
• I recently proved a theorem for which the smallest example of one of the cases had order $37,646,400$. We had to computationally verify its existence. – Alexander Gruber Jun 1 '13 at 16:00 | 2020-02-22 08:04:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.44188305735588074, "perplexity": 249.30961269812533}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145654.0/warc/CC-MAIN-20200222054424-20200222084424-00364.warc.gz"} |
http://math.stackexchange.com/questions/184927/gx-is-a-function-such-that-gx1gx-1-gx-x-in-mathbbr-for-wha | # $g(x)$ is a function such that $g(x+1)+g(x-1)=g(x)$, $x \in \mathbb{R}$. For what value of $p$, $g(x+p)=g(x)$.
$g(x)$ is a function such that $g(x+1)+g(x-1)=g(x)$, $x \in \mathbb{R}$.For what value of $p$, $g(x+p)=g(x)$.
$g(x+2)+g(x)=g(x+1)$
-
Is $p \ne 0$? if that's the case how do you know that i exists? – Mercy King Aug 21 '12 at 7:54
This is really a question about sequences hence assume that $x_{n+2}-x_{n+1}+x_n=0$ for every $n$, for some sequence $(x_n)_{n\in\mathbb Z}$. The characteristic equation of this recursion is $r^2-r+1=0$, with roots $r=\mathrm e^{\pm\mathrm i\pi/3}$, hence $x_n=A\mathrm e^{n\mathrm i\pi/3}+B\mathrm e^{-n\mathrm i\pi/3}$ for some $A$ and $B$. Now, $\mathrm e^{2\mathrm i\pi}=1$ hence, for every $(A,B)$, $x_{n+6}=x_n$. No period smaller than $6$ is valid for every such sequence $(x_n)_{n\in\mathbb Z}$, as the example of $x_n=\cos(n\pi/3)$ shows.
Edit: The link with the original question is as follows. For every $0\leqslant x\lt1$, $x_n=g(x+n)$ defines a sequence $(x_n)_{n\in\mathbb Z}$ as above, hence there exists two parameters $A(x)$ and $B(x)$ such that $$g(x+n)=A(x)\mathrm e^{n\mathrm i\pi/3}+B(x)\mathrm e^{-n\mathrm i\pi/3},$$ for every integer $n$. There is no relation whatsoever between the families $(g(x+n))_{n\in\mathbb Z}$ and $(g(x'+n))_{n\in\mathbb Z}$ for $x\ne x'$ in $[0,1)$. Hence, each solution $g:\mathbb R\to\mathbb C$ is described uniquely by two functions $A:[0,1)\to\mathbb C$ and $B:[0,1)\to\mathbb C$ and any two such functions $(A,B)$ define uniquely a solution $g_{A,B}$.
-
The OP talked about functions $g:\ {\mathbb R}\to{\mathbb R}$ (or $\to{\mathbb C}$). This aspect is insufficiently addressed in the answers so far. E.g.: Which functions of period $6$ satisfy the given functional equation? – Christian Blatter Aug 21 '12 at 11:00
@ChristianBlatter See Edit. – Did Aug 21 '12 at 12:25
Now that we know, by @did's answer, that the answer is $6$, we ought to be able to prove it directly. Here is a solution: Note
$$g(x+3) = g(x+2) - g(x+1) = (g(x+1) - g(x)) - g(x+1) = - g(x).$$
Therefore
$$g(x+6) = -g(x+3) = g(x).$$
-
After @did did it already I'd like to add a matrix-view into it.
In the same way we define a sequence $g_0,g_1,g_2,...$ , assuming we had already found some $g_0$ and $g_1$. Then $g_2 = g_1 - g_0$, in matrix-notation
$$\begin{matrix} & & \begin{bmatrix}-1\\ 1\end{bmatrix} \\ & . \\ \begin{bmatrix}g_0&g_1\end{bmatrix} & = & \begin{bmatrix} g_2 \end{bmatrix} \end{matrix}$$
To make it iterable, we prefix a column at the coefficients-matrix:
$$\begin{matrix} & & \begin{bmatrix}0&-1\\1&1\end{bmatrix} \\ &. \\ \begin{bmatrix}g_0&g_1\end{bmatrix} & = & \begin{bmatrix}g_1 & g_2\end{bmatrix} \end{matrix}$$
Let's call the resp. matrices $G_0,G_1$ and the coefficients-matrix $C$ so we have $$G_k \cdot C = G_{k+1}$$
We see, that if we iterate this we can generate a sequence of values $g_k$ for the function $g(x)$ which have the required relation between that elements - after we have assumed some initial values $g_0,g_1$
To find, whether this is periodic we must ask, whether for some p the formula comes out to be $$G_0 \cdot C^p = G_0$$ or said differently, whether for some p we'll have $C^p = I$ where $I$ is the identity matrix.
We can diagonalize $C$ and find the eigenvalues having the complex values $$\lambda_0 = {1- \sqrt{3}î\over 2} \\ \lambda_1 = {1+ \sqrt{3}î\over 2}$$ and these are just $6'th$ roots of the complex unit, so we simply check what $C^6$ is and find that indeed $$C^6 = I\\p=6$$ which is the solution. (Clearly, we can look at the characteristic polynomial and find $\mathcal P(C):= x^2-x+1$ which was mentioned in @did's answer and which we have to solve for its roots.)
This is the same result as in the previous answer, but it shows, how one could proceed, if the problem-parameters were different - and were possibly even more complicated (linear) compositions of the function, for instance $g(x)= 1 \cdot g(x-1) - 2\cdot g(x-2)+3\cdot g(x-3)$ and similarly.
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Unfortunately, $g(x)= 1 \cdot g(x-1) - 2\cdot g(x-2)+3\cdot g(x-3)$ has no periodic solution except $g(x)=0$ for every $x$. – Did Aug 21 '12 at 9:29
@did: upps... well some common integer power of the eigenvalues should simultaneously arrive at 1, which is clearly some very special requirement... (besides that their absolute value should be the unit) thanks for the reminding! – Gottfried Helms Aug 21 '12 at 9:45
As Sean Eberhard has shown such a function necessarily has period $6$. It remains to investigate which functions of period 6 actually satisfy the given functional equation.
Every reasonable function $g$ of period $6$ can be developed into a Fourier series $$g(x)=\sum_{k\in{\mathbb Z}} c_k e^{k\pi ix/3}$$ with complex coefficients $c_k$. Put $e^{\pi i/3}=:\omega$. Then $$g(x+1)-g(x)+g(x-1)=\sum_k c_k(\omega^k -1 +\omega^{-k})e^{k\pi ix/3}\ .$$ The RHS is $\equiv 0$ iff for all $k\in{\mathbb Z}$ at least one of $c_k$ and $\omega^k -1 +\omega^{-k}=2\cos{k\pi\over3}-1$ is zero. The latter is the case iff $k=\pm1$ modulo $6$. Therefore the function $g$ has to be of the form $$g(x)=\sum_{l\in{\mathbb Z}}c_{6l+1}e^{(6l+1)\pi i x/3}+\sum_{l\in{\mathbb Z}}c_{6l-1}e^{(6l-1)\pi i x/3}=C(x)e^{\pi i x/3}+D(x)e^{-\pi i x/3}\ ,$$ where now $C(x)$ and $D(x)$ are arbitrary complex functions of period $1$. The most general real function satisfying our functional equation is therefore given by $$g(x)=A(x)\cos{\pi x\over3}+B(x)\sin{\pi x\over3}\ ,$$ where $A(x)$ and $B(x)$ are arbitrary real functions of period $1$.
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$g(x+1)+g(x-1)=g(x)$
$g(x+1)=g(x)-g(x-1)$ -------(1)
$g(x+2)=g(x+1)-g(x)$ -------(2)
$g(x+3)=g(x+2)-g(x+1)$ -------(3)
from (1) and (2) we have,
$g(x+2)=-g(x-1)$ -------(2b)
from (2) and (3) we have,
$g(x+3)=-g(x)$ -------(3b)
from (2b) or (3b) we have,
$g(x+6)=-g(x+3)$ -------(3c)
from, (3b) and (3c) we have,
$g(x+6)=g(x)$
we can now conclude that $p=6$
- | 2016-02-12 12:24:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9671294093132019, "perplexity": 174.16992167824816}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701163729.14/warc/CC-MAIN-20160205193923-00178-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/3422726/nonlinear-polynomial-diophantine-equations | # Nonlinear polynomial Diophantine equations
I am interested in whether there exist nontrivial (i.e., non-constant) solutions to any or all of the equations \begin{align*} P(x, x^{-1})^2 + Q(x, x^{-1})^2 + 1 &= 0, \\ P(x, x^{-1})^2 + Q(x, x^{-1})^4 + 1 &= 0, \\ P(x, x^{-1})^2 + Q(x, x^{-1})^3 + 1 &= 0, \\ P(x, x^{-1})^2 + Q(x, x^{-1})^3 + Q(x, x^{-1}) &= 0, \\ P(x, x^{-1})^2 + Q(x, x^{-1})^5 + 1 &= 0, \end{align*} where $$P$$ and $$Q$$ are single-variable Laurent polynomials with coefficients in $$\mathbb{C}$$. For example, a straightforward solution to the first equation is $$\begin{equation*} P(x, x^{-1}) = \frac{x - x^{-1}}{2}, \quad Q(x, x^{-1}) = \frac{i(x + x^{-1})}{2}. \end{equation*}$$ I would like to know whether the other equations have nontrivial solutions, and if so, whether there exists a method for generating solutions or for classifying all solutions. Comments on the more general equation $$P(x, x^{-1})^a + Q(x, x^{-1})^b + 1 = 0$$, where $$a$$ and $$b$$ are positive integers, would be appreciated as well.
• So we are not looking for simultaneous solutions, but for every equation separately? For example, the last equation has a solution $Q(x,x^{-1})=0$ and $P(x,x^{-1})=i$. – Dietrich Burde Nov 5 at 8:58
• @Dietrich Burde Right, I should have been more clear - I am looking for solutions to each equation separately. Also, I am interested in solutions where $P$ and $Q$ are not constant. – user137 Nov 5 at 16:52
• A version now posted to MO, mathoverflow.net/questions/345519/… – Gerry Myerson Nov 8 at 6:12 | 2019-11-18 22:04:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.999976634979248, "perplexity": 489.616946410038}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496669847.1/warc/CC-MAIN-20191118205402-20191118233402-00133.warc.gz"} |
http://crypto.stackexchange.com/questions/6515/are-rsa-signatures-deterministic?answertab=votes | # Are RSA signatures deterministic?
If I sign the word HELLO with the mechanism "NONEwithRSA" with the same private key, do I always will have the same signature?
A Java example always return different byte array :
PKCS11Provider provider = new PKCS11Provider(this);
List<PKCS11Slot> slots = PKCS11Slot.enumerateSlots(provider);
PKCS11Session session = null;
if (slots.size() > 0) {
session = PKCS11Session.open(slots.get(0),
}
PKCS11KeyStoreSpi keyStore = new PKCS11KeyStoreSpi(provider, "PKCS11");
PKCS11PrivateKey privateKey = PKCS11PrivateKey.getPrivateKeys(session).get(0);
Signature signature = Signature.getInstance("NONEwithRSA", provider);
signature.initSign(privateKey);
signature.update("HELLO".getBytes(), 0, "HELLO".getBytes().length);
byte[] realSig = signature.sign();
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Some signature algorithms are deterministic (you always get the same signature for the same private key and input), others are not. In the case of RSA, as specified by PKCS#1, the "old-style" (aka "v1.5") signatures are deterministic, while the "new-style" ("PSS") signatures are not (padding includes some random bytes).
In Java, the "NONEwithRSA" mechanism ought to select the old-style signatures, and the input data should be already hashed (a normal signature begins by hashing the input data); in your code, you use unhashed data as input, which is a bad idea for security. The "NONEwithRSA" mechanism was added to support contexts where data is hashed externally, and also SSL/TLS which insists (in its pre-1.2 versions) on using the concatenation of MD5 and SHA-1 as its "hash function". This does not explain why you get different outputs, but you are still doing it wrong.
To investigate further, convert the signature to a BigInteger and apply the public key on it, to see what the padded data looks like:
BigInteger n = ... (the public key modulus)
BigInteger e = ... (the public key exponent)
BigInteger x = new BigInteger(1, realSig);
System.out.printf("%X\n", x.modPow(e, n));
With PKCS#1 v1.5, this should look like:
1FFFFFFFFF....FFFFF0048454C4C4F
(The 48454C4C4F is the encoding of "HELLO" -- because your code puts the raw data here instead of the hash value suitably encoded in an ASN.1 structure which designates the hash function.)
Since you are using a PKCS#11, you will sign data depending on what the underlying device can do. It is possible that your device does not allow such "NONEwithRSA" mechanism, or allows it only if it already looks like a properly encoded hash value. This may be your issue after all. The debug code above should show what happens.
(Also, don't use String.getBytes() without any argument. This converts the input string with the default encoding of the current machine, which depends on how the machine was configured; for some people it will be UTF-8, others will have UTF-16 or latin-1 or something else. It is much better to specify an explicit encoding, as: s.getBytes("UTF-8"))
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https://research.utwente.nl/en/publications/extinction-probability-in-a-birth-death-process-with-killing-2 | # Extinction probability in a birth-death process with killing
Erik A. van Doorn, A.I. Zeifman
Research output: Book/ReportReportProfessional
## Abstract
We study birth-death processes on the non-negative integers where $\{1,2,\ldots\}$ is an irreducible class and $0$ an absorbing state, with the additional feature that a transition to state $0$ may occur from any state. We give a condition for absorption (extinction) to be certain and obtain the eventual absorption probabilities when absorption is not certain. We also study the rate of convergence as $t\to\infty$ of the probability of absorption at time $t$, and relate it to the common rate of convergence of the transition probabilities which do not involve state $0$. Finally, we derive upper and lower bounds for the probability of absorption at time $t$ by applying a technique which involves the logarithmic norm of an appropriately defined operator.
Original language Undefined Enschede University of Twente, Faculty of Mathematical Sciences 20 Published - 2004
### Publication series
Name Memorandum Faculty of Mathematical Sciences Department of Applied Mathematics, University of Twente 1723 0169-2690
## Keywords
• MSC-60J27
• MSC-60J80
• IR-65907
• METIS-218214
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https://www.learnlatex.org/en/lesson-06 | Lesson 6
# Extending LaTeX
This lesson shows how you can extend LaTeX to your needs and change its layout further by using packages and definitions. It also shows how you can define your own commands.
After having declared a class, in the preamble you can modify functionality in LaTeX by adding one or more packages. These can
• Change how some parts of LaTeX work
• Add new commands to LaTeX
• Change document design
## Changing how LaTeX works
The LaTeX ‘kernel’ (the core of LaTeX) is rather limited in user customisation, and so some add-on packages deal with very common ideas. The first is to change how LaTeX deals with language-specific typesetting (hyphenation, punctuation, quotations, localisation, etc.). Different languages have different rules, so it’s important to tell LaTeX which one to use. This is handled by the babel package.
\documentclass{article}
\usepackage[T1]{fontenc}
%\usepackage[french]{babel}
\usepackage[width = 6cm]{geometry} % To force hyphenation here
\begin{document}
This is a lot of filler which is going to demonstrate how LaTeX hyphenates
material, and which will be able to give us at least one hyphenation point.
This is a lot of filler which is going to demonstrate how LaTeX hyphenates
material, and which will be able to give us at least one hyphenation point.
\end{document}
Try un-commenting the (clearly misleading) line to load babel and see the effect. (The standard hyphenation rules are US English.)
The babel package does a lot more than hyphenation, depending on the language involved; we’ve given some more details if you need them.
## Changing design
It’s useful to be able to adjust some aspects of design independent of the document class. The most obvious one are the page margins. We’ve just used the geometry package in the example above, but let’s now have an example specifically about margins.
\documentclass{book}
\usepackage[T1]{fontenc}
\usepackage[margin=1in]{geometry}
\begin{document}
Hey world!
This is a first document.
% ================
\chapter{Chapter One}
Introduction to the first chapter.
\section{Title of the first section}
Text of material in the first section
Second paragraph.
\subsection{Subsection of the first section}
Text of material in the subsection.
% ================
\section{Second section}
Text of the second section.
\end{document}
You should see the effect here compared to not loading geometry.
One of LaTeX’s strengths is that you can choose from thousands of packages, including ones for writing mathematical text, for hyperlinking, for sophisticated capabilities with color, etc. We will see some more common packages in later lessons.
## Defining commands
Sometimes you need a command specific to your document, either some functionality not found in the available packages or simply a command to enter a common expression that is used multiple times.
The following example shows a command to produce keywords with a specific style applied.
\documentclass{article}
\usepackage[T1]{fontenc}
\newcommand\kw[1]{\textbf{\itshape #1}}
\begin{document}
\end{document}
In the definition [1] denotes the number of arguments (here one) and #1 denotes the first argument that is supplied (apples or oranges in this example). You may have up to nine arguments, but it is usually best to have just one argument, or sometimes none at all.
Defining commands does not just reduce the typing required to produce a document. It helps to separate out the styling information. If it is decided to use a different style for keywords, rather than having to edit the entire document, you simply need to use a different definition. Here we load the xcolor package to provide colors, and use blue in place of bold in the formatting.
\documentclass{article}
\usepackage[T1]{fontenc}
\usepackage{xcolor}
\newcommand\kw[1]{\textcolor{blue}{\itshape #1}}
\begin{document}
\end{document}
Beware that defining too many commands or defining commands with multiple arguments may make the document source harder to understand as it is using an unfamiliar syntax. The ability to define document-specific commands should be used with care.
## Exercises
Try out writing some text in other European languages and see how babel affects hyphenation: you can probably find some text on the internet, and guess the right options.
Try altering the margins in the geometry example. You can set the individual top, bottom, left and right margins separately using a comma-separated list.
Try loading the lipsum package and then add the command \lipsum to your document. Can you guess why this package is useful for making examples?
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https://alco.centre-mersenne.org/articles/10.5802/alco.183/ | # ALGEBRAIC COMBINATORICS
On highly regular strongly regular graphs
Algebraic Combinatorics, Volume 4 (2021) no. 5, pp. 843-878.
In this paper we unify several existing regularity conditions for graphs, including strong regularity, $k$-isoregularity, and the $t$-vertex condition. We develop an algebraic composition/decomposition theory of regularity conditions. Using our theoretical results we show that a family of non rank 3 graphs known to satisfy the $7$-vertex condition fulfills an even stronger condition, $\left(3,7\right)$-regularity (the notion is defined in the text). Derived from this family we obtain a new infinite family of non rank $3$ strongly regular graphs satisfying the $6$-vertex condition. This strengthens and generalizes previous results by Reichard.
Revised:
Accepted:
Published online:
DOI: https://doi.org/10.5802/alco.183
Classification: 05E30, 51E12
Keywords: Strongly regular graphs, invariants, $k$-isoregularity, $t$-vertex condition, partial quadrangles, generalized quadrangles, partial linear spaces
@article{ALCO_2021__4_5_843_0,
author = {Pech, Christian},
title = {On highly regular strongly regular graphs},
journal = {Algebraic Combinatorics},
pages = {843--878},
publisher = {MathOA foundation},
volume = {4},
number = {5},
year = {2021},
doi = {10.5802/alco.183},
language = {en},
url = {https://alco.centre-mersenne.org/articles/10.5802/alco.183/}
}
TY - JOUR
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TI - On highly regular strongly regular graphs
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DA - 2021///
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Pech, Christian. On highly regular strongly regular graphs. Algebraic Combinatorics, Volume 4 (2021) no. 5, pp. 843-878. doi : 10.5802/alco.183. https://alco.centre-mersenne.org/articles/10.5802/alco.183/
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http://cds.cern.ch/collection/ATLAS%20Conference%20Notes | # ATLAS Conference Notes
2016-08-21
12:04
Measurement of inclusive-jet cross-sections in proton-proton collisions at $\sqrt{s} = 13~\mathrm{TeV}$ centre-of-mass energy with the ATLAS detector This note presents the measurement of inclusive-jet cross-sections in proton-proton collisions at a centre-of-mass energy of $13~\mathrm{TeV}$. [...] ATLAS-CONF-2016-092. - 2016. Original Communication (restricted to ATLAS) - Full text
2016-08-10
09:49
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2016-08-10
09:46
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2016-08-10
09:41
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2016-08-08
18:07
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2016-08-08
18:05
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2016-08-08
18:03
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2016-08-08
18:00
Search for Minimal Supersymmetric Standard Model Higgs Bosons $H/A$ in the $\tau\tau$ final state in up to 13.3 fb$^{−1}$ of pp collisions at $\sqrt{s}$= 13 TeV with the ATLAS Detector A search for neutral Higgs bosons of the Minimal Supersymmetric Standard Model (MSSM) is performed using a data sample corresponding to an integrated luminosity of up to 13.3 fb$^{-1}$ from proton--proton collisions at $\sqrt{s} = 13$ TeV recorded by the ATLAS detector at the LHC. [...] ATLAS-CONF-2016-085. - 2016. - 28 p. Original Communication (restricted to ATLAS) - Full text
2016-08-08
17:59
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2016-08-08
17:57
A Search for Resonances Decaying to a $W$ or $Z$ Boson and a Higgs Boson in the $q\bar{q}^{(\prime)}b\bar{b}$ Final State A search for resonances decaying to a $W$ or $Z$ boson and a Higgs boson in the $q\bar{q}^{(\prime)}b\bar{b}$ final state is described. [...] ATLAS-CONF-2016-083. - 2016. - 19 p. Original Communication (restricted to ATLAS) - Full text | 2016-08-25 04:26:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9573829770088196, "perplexity": 1437.1771845540807}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982292887.6/warc/CC-MAIN-20160823195812-00046-ip-10-153-172-175.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/interpreting-velocity-graphs.189098/ | # Interpreting Velocity graphs
1. Oct 4, 2007
### kraaaaamos
1. The problem statement, all variables and given/known data
5. A car starts from xi = 10m at ti = 0s and moves with the velocity graph shown in figure on the right.
a. What is the object’s position at t = 2s, 3s, and 4s?
b. Does this car ever change direction? If so, at what time?
2. Relevant equations
V = d/t
3. The attempt at a solution
for t at 2 secs...
according to graph v= 4m/s
so position (d) = (t)(v)
= (2)(4)
= 8m?
am i doing it right?
OR
v = change d/change t
4m/s = (x-10)/(2-0)
4m/s = (x-10)/(2)
8m/s = x-10
x = 18m
is that correct?
#### Attached Files:
• ###### tut3pic5.JPG
File size:
3.7 KB
Views:
1,059
Last edited: Oct 4, 2007
2. Oct 5, 2007
### Staff: Mentor
Well this can't be right if the car starts at 10 m and adds distance. The slope of the plot of velocity vs time is negative, which indicates the car is decelerating, and the since the slope is constant, the deceleration is constant.
Certainly when the car has a negative velocity, it is reversing.
Is one familiar with integration?
if v(t) = d x(t)/dt, then
x(t) = $$\int_0^t\,v(t) dt\,+\,x(0)$$
This might be useful:
http://hyperphysics.phy-astr.gsu.edu/hbase/acons.html | 2017-02-27 00:35:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5875246524810791, "perplexity": 4320.261038684267}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501172156.69/warc/CC-MAIN-20170219104612-00621-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://www.shaalaa.com/question-bank-solutions/write-in-terms-of-factorial-6-7-8-9-concept-of-factorial-function_160914 | # Write in terms of factorial:6 × 7 × 8 × 9 - Mathematics and Statistics
Sum
Write in terms of factorial:
6 × 7 × 8 × 9
#### Solution
6 × 7 × 8 × 9 = 9 × 8 × 7 × 6
Multiplying and dividing by 5!, we get
= (9xx8xx7xx6xx5!)/(5!)
= (9xx8xx7xx6xx5xx4xx3xx2xx1)/(5!)
= (9!)/(5!)
Concept: Concept of Factorial Function
Is there an error in this question or solution?
#### APPEARS IN
Balbharati Mathematics and Statistics 2 (Commerce) 11th Standard HSC Maharashtra State Board
Chapter 6 Permutations and Combinations
Exercise 6.2 | Q 4. (iii) | Page 75 | 2022-10-04 10:18:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.28754186630249023, "perplexity": 2587.679880330931}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337490.6/warc/CC-MAIN-20221004085909-20221004115909-00638.warc.gz"} |
https://www.physicsforums.com/threads/finding-derivative-and-solve-for-t.544837/ | # Homework Help: Finding derivative and solve for t
1. Oct 27, 2011
### josh_123
Hello so I'm supposed to find the derivative of (-t2+4)/(t2+4)2
and using the quotient rule + the chain rule I got this mess:
dy/dx= -2t(t^2+4)^2-2t(2t2+8)(t2+4)2/ (t2+4)4
and now I have to set that whole thing to 0. I know the numerator must be equal to 0 which mean that one of the solution is t=0. Is there any other solutions?Can you help me simplify/factor it to find the other solution?Thank you
2. Oct 27, 2011
### Dick
Your 'mess' is a ways from being right yet. Once you get it right I'd start with canceling out the common factors of (t^2+4) in the numerator and denominator. It will look a lot less messy.
3. Oct 28, 2011
### Staff: Mentor
Since there is no x in your original expression, your derivative shouldn't be dy/dx. Also, there's no y in that expression, either, so where did it come from?
Why do you think you need to set the derivative to 0? The question merely asks you to find the derivative.
4. Oct 28, 2011
### josh_123
Hello there,
the function is v(t)=-t2+4/(t2+4)2
and to find the derivative I'm suppose to find v'(t) or a(t)
The reason why I have to set the second derivative to 0 so that I could find the point at which the function is concave up or down. (those are called stationary points?)
I find the second derivative by using the equation:
f'g-g'f/g^2
and there is a comment said that my second derivative is wrong. Can you guide me where I make the mistake?
Thank you
5. Oct 28, 2011
### Ivan92
It could have gone wrong anywhere. Show us your steps in simplifying your derivative after applying the quotient rule. Only then will we be able to pinpoint your mistake. So when you set your 2nd derivative = 0, you are looking at when there is constant velocity or 0 acceleration. I wouldn't call these stationary points since the object is still moving in some direction. Stationary points would only be when v(t)=0
6. Oct 28, 2011
### josh_123
oh okay here is the step
a(t)= -2t(t^2+4)^2 is the part f'g and then subtract the g'f part I used the chain rule to find g' which is 2t(2t^2+8)<----g'(t^2+4)^2<-----f and divide by g^2 which is( (t^2+4)^2)^2) which clean up to be equal (t^2+4)^4
and that why I got -2t(t^2+4)^2-2t(2t^2+8)(t^2+4)^2/ (t^2+4)^4
7. Oct 28, 2011
### Dick
The f'g part looks ok. The g'f part doesn't. It looks more like g'g to me.
8. Oct 28, 2011
### SammyS
Staff Emeritus
(You really do need to learn to use parentheses where they're needed.)
I take it that you have f(t)/g(t) = (-t2+4)/(t2+4)2 .
Then $\displaystyle \frac{d}{dt}\left(\frac{f(t)}{g(t)}\right)=\frac{f\,'\!(t)\,g(t)-f(t)\,g'(t)}{(g(t))^2}$
I don't see f(t) in the second term of the numerator in your expression.
9. Oct 29, 2011
### Staff: Mentor
This information is part of the problem statement, which should have been in your first post.
10. Oct 29, 2011
### josh_123
Hello everyone, so I redo the derivative
a(t)= -2t(t2+4)2-2(t2+4)(2t)(-t2+4)/ (t2+4)4
11. Oct 29, 2011
### Staff: Mentor
Like Sammy said, you need more parentheses.
What you wrote would be interpreted as
$$a(t) = -2t(t^2 + 4)^2 - \frac{2(t^2 + 4)(2t)(-t^2 + 4)}{(t^2 + 4)^4}$$
You need another pair of parentheses around the entire numerator.
After making that fix, that looks to be correct, but if the answer is given, it probably won't look like this. In the textbook answers, they usually pull out common factors in the numerator to write the numerator as a product of factors rather than as a sum or difference of terms. | 2018-06-20 08:14:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6358751654624939, "perplexity": 932.786414480094}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267863489.85/warc/CC-MAIN-20180620065936-20180620085936-00596.warc.gz"} |
http://citizendia.org/Planar_projection | Views
Graphical projections
Other views
This box: view talk edit
Planar projections are the subset of 3D graphical projections constructed by linearly mapping points in three dimensional space to points on a two-dimensional projection plane. Graphical projection is a Protocol by which an image of an imaginary three-dimensional object is projected onto a planar surface without the aid of mathematical calculation Perspective (from Latin perspicere to see through in the graphic arts such as drawing is an approximate representation on a flat surface (such as paper of an image as it is perceived Orthographic projection is a means of representing a three- Dimensional (3D object in two dimensions (2D Orthographic projection is a means of representing a three- Dimensional (3D object in two dimensions (2D A plan is an Orthographic projection of a 3-dimensional object from the position of a horizontal plane through the object A floor plan ( floorplan) in Architecture and Building engineering is a Diagram, usually to scale, of the relationships between rooms In Geometry, a cross section is the intersection of a body in 2-dimensional space with a line or of a body in 3-dimensional space with a plane etc An elevation is an Orthographic projection of a 3-dimensional object from the position of a horizontal plane beside an object An auxiliary view is an angle at which one can view an object that is not one of the primary views for an Orthographic projection. Axonometric projection ("to measure along axes" is a technique used in orthographic pictorials Isometric projection is a form of Graphical projection —more specifically an Axonometric projection. Dimetric projection is a form of Axonometric projection, in which its direction of viewing is such that two of the three axes of space appear equally foreshortened of which Trimetric projection is a form of Axonometric projection, where the direction of viewing is such that all of the three axes of space appear unequally foreshortened This article discusses imaging of three-dimensional objects For an abstract mathematical discussion see Projection (linear algebra. The cavalier perspective, also called cavalier projection or high view point, is a way to represent a three dimensional object on a flat drawing and more specifically Cabinet projection or sometimes cabinet perspective is a type of Oblique projection. 3D projection is any method of mapping three-dimensional points to a two-dimensional plane Owned by Atlassian Software Systems, FishEye is a Revision control browser and search engine Stereoscopy, stereoscopic imaging or 3-D (three-dimensional imaging is any technique capable of recording three-dimensional visual Anamorphosis is a distorted projection or perspective requiring the viewer to use special devices or occupy a specific vantage point to reconstitute the image A map projection is any method of representing the Surface of a sphere or other shape on a plane. A bird's-eye view is a View of an object from above as though the observer were a Bird, often used in the making of Blueprints, Floor plans Top-down perspective, also sometimes referred to as bird's-eye view, overhead view or helicopter view, A worm's-eye view is a View of an object from below as though the observer were a Worm. The projected point on the plane is chosen such that it is collinear with the corresponding three dimensional point and the centre of projection. The lines connecting these points are commonly referred to as projectors.
The centre of projection can be thought of as the location of the observer while the plane of projection is the surface on which the two dimensional projected image of the scene is recorded or from which it is viewed (e. g. , photographic negative, photographic print, computer monitor). When the centre of projection is at a finite distance from the projection plane, a perspective projection is obtained. Perspective (from Latin perspicere to see through in the graphic arts such as drawing is an approximate representation on a flat surface (such as paper of an image as it is perceived When the centre of projection is at infinity, all the projectors are parallel, and the corresponding subset of planar projections are referred to as parallel projections.
## Mathematical formulation
Mathematically, planar projections are linear transformations acting on a point in three dimensional space $\mathbf{a}_{x,y,z}$ to give a point $\mathbf{b}_{u,v}$ on the projection plane. In Mathematics, a linear map (also called a linear transformation, or linear operator) is a function between two Vector spaces that These transformations consist of various compositions of the five transformations: orthographic projection, rotation, shear, translation and perspective. In Mathematics, a composite function represents the application of one function to the results of another Orthographic projection is a means of representing a three- Dimensional (3D object in two dimensions (2D In Geometry and Linear algebra, a rotation is a transformation in a plane or in space that describes the motion of a Rigid body around a fixed In Mathematics, a shear or transvection is a particular kind of Linear mapping. In Euclidean geometry, a translation is moving every point a constant distance in a specified direction A projective transformation is a transformation used in Projective geometry: it is the composition of a pair of Perspective projections It describes what
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network: | | | 2013-05-24 17:54:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6185816526412964, "perplexity": 537.841079414896}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368704933573/warc/CC-MAIN-20130516114853-00044-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://mathhelpforum.com/geometry/16246-could-somebody-please-help-me.html | 2. Originally Posted by sanee66
I attached the diagram below. By the way, the triangle is AMN not ABC
Note that the altitude of an equilateral triangle bisects one of it's angles and divides the base it hits in two equal parts. This means that $MH = \frac {1}{2} (7m - 1)$
Also, since $\triangle AMN$ is equilateral, $AN = AM$
We are told that the sum of AM and MH gives 11m - 4
$\Rightarrow AM + MH = 11m - 4$
$\Rightarrow 7m - 1 + \frac {7m - 1}{2} = 11m - 4$
$\Rightarrow 7m + \frac {7m}{2} - 11m = -4 + \frac {3}{2}$
$\Rightarrow - \frac {1}{2} m = - \frac {5}{2}$
$\Rightarrow m = 5$
The side length of the triangle is equal to the side-length of AN
So, the side-length is 7m - 1 = 7(5) - 1 = 34 | 2013-12-08 18:50:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7958809733390808, "perplexity": 445.51529351628716}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386163791972/warc/CC-MAIN-20131204132951-00051-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://proofwiki.org/wiki/Category:Conversion_between_Cartesian_and_Polar_Coordinates_in_Plane | # Category:Conversion between Cartesian and Polar Coordinates in Plane
Jump to navigation Jump to search
## Pages in category "Conversion between Cartesian and Polar Coordinates in Plane"
The following 4 pages are in this category, out of 4 total. | 2023-03-21 14:48:51 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9758334755897522, "perplexity": 2421.570990339112}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943698.79/warc/CC-MAIN-20230321131205-20230321161205-00788.warc.gz"} |
https://www.physicsforums.com/threads/acceleration-on-an-incline.391696/ | # Acceleration on an incline
## Homework Statement
[PLAIN]http://img237.imageshack.us/img237/6677/incline.gif [Broken]
A snow sled with a child secured safely to it has a total mass of 83.0 kg. It is lowered at a constant speed of 1.9 ms−1 down a slope of angle 41.0° with respect to the horizontal (as shown above) for a distance d = 14.0 m. The coefficient of kinetic friction between the sled and the snow is 0.09.
g = 9.8 ms–2. Air resistance is negligible at these speeds.
I have calculated the friction force to be 55.3, and the magnitude of the reactive force, N, on the sled to be 614.49 Newtons.
Suppose the rope suddenly snaps at its point of connection with the sled after travelling the distance d. Determine the drag force acting on the sled when it reaches a terminal speed of 8.4 ms−1.
## The Attempt at a Solution
What does the question mean by the "drag force acting on the sled "? Is it reffering to the force that pulls the sled down the incline? I know that force is equal to
$$mg sin \theta = (83 \times 9.81) sin(41)= 534.18$$
But this is wrong because the correct answer has to be 478 N.
Another approach I can think of is to find the time using dv=t= 117.6 and acceleration by $$a=v/t \Rightarrow a =\frac{8.4}{117.6}=0.07$$ then plug this into $$\sum F_x=ma_x=83 \times 0.07=5.9$$. Which is again wrong...
Can anyone help?
Last edited by a moderator: | 2020-05-27 00:45:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6712747812271118, "perplexity": 613.4574121142603}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347391923.3/warc/CC-MAIN-20200526222359-20200527012359-00316.warc.gz"} |
http://www.ams.org/mathscinet-getitem?mr=2250020 | MathSciNet bibliographic data MR2250020 52A20 (52A30) Haberl, Christoph; Ludwig, Monika A characterization of \$L\sb p\$$L\sb p$ intersection bodies. Int. Math. Res. Not. 2006, Art. ID 10548, 29 pp. Article
For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews. | 2016-07-26 01:52:04 | {"extraction_info": {"found_math": true, "script_math_tex": 1, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9937350153923035, "perplexity": 13740.31648629268}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257824499.16/warc/CC-MAIN-20160723071024-00079-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://solvedlib.com/how-much-heat-is-released-when-105-g-of-steam-at,281776 | # How much heat is released when 105 g of steam at 100.0°C is cooled to ice...
###### Question:
How much heat is released when 105 g of steam at 100.0°C is cooled to ice at -15.0°C? The enthalpy of vaporization of water is 40.67 kJ/mol, the enthalpy of fusion for water is 6.01 kJ/mol, the molar heat capacity of liquid water is 75.4 J/(mol • °C), and the molar heat capacity of ice is 36.4 J/(mol • °C).
A)347 kJ B)54.8 kJ C)319 kJ D)273 kJ
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https://damask.mpie.de/Documentation/FCC?cover=print; | # 1.1.1. Face-centered cubic (fcc)
## Atom arrangement
Figure 1: Face-centered cubic lattice structure.
## Slip systems
Figure 2: Slip system in face-centered cubic lattice.
index slip direction plane normal 1 $[0 1 \bar 1]$ $(1 1 1)$ 2 $[\bar 1 0 1]$ $(1 1 1)$ 3 $[1 \bar 1 0]$ $(1 1 1)$ 4 $[0 \bar 1 \bar 1]$ $(\bar 1 \bar 1 1)$ 5 $[1 0 1]$ $(\bar 1 \bar 1 1)$ 6 $[\bar 1 1 0]$ $(\bar 1 \bar 1 1)$ 7 $[0 \bar 1 1]$ $(1 \bar 1 \bar 1)$ 8 $[\bar 1 0 \bar 1]$ $(1 \bar 1 \bar 1)$ 9 $[1 1 0]$ $(1 \bar 1 \bar 1)$ 10 $[0 1 1]$ $(\bar 1 1 \bar 1)$ 11 $[1 0 \bar 1]$ $(\bar 1 1 \bar 1)$ 12 $[\bar 1 \bar 1 0]$ $(\bar 1 1 \bar 1)$
## Twin systems
index slip direction plane normal 1 $[\bar 2 1 1]$ $(1 1 1)$ 2 $[1 \bar 2 1]$ $(1 1 1)$ 3 $[1 1 \bar 2]$ $(1 1 1)$ 4 $[2 \bar 1 1]$ $(\bar 1 \bar 1 1)$ 5 $[\bar 1 2 1]$ $(\bar 1 \bar 1 1)$ 6 $[\bar 1 \bar 1 \bar 2]$ $(\bar 1 \bar 1 1)$ 7 $[\bar 2 \bar 1 \bar 1]$ $(1 \bar 1 \bar 1)$ 8 $[1 2 \bar 1]$ $(1 \bar 1 \bar 1)$ 9 $[1 \bar 1 2]$ $(1 \bar 1 \bar 1)$ 10 $[2 1 \bar 1]$ $(\bar 1 1 \bar 1)$ 11 $[\bar 1 \bar 2 \bar 1]$ $(\bar 1 1 \bar 1)$ 12 $[\bar 1 1 2]$ $(\bar 1 1 \bar 1)$
Topic attachments
I Attachment Action Size Date Who Comment
png FCC_crystal_structure.png manage 173 K 17 Oct 2012 - 13:52 PhilipEisenlohr face-centered cubic
svg FCC_crystal_structure.svg manage 2 K 17 Oct 2012 - 13:52 PhilipEisenlohr face-centered cubic (vector-based)
png FCC_slip_system.png manage 214 K 17 Oct 2012 - 14:12 PhilipEisenlohr face-centered cubic slip system
svg FCC_slip_system.svg manage 2 K 17 Oct 2012 - 14:13 PhilipEisenlohr face-centered cubic slip system (vector-based)
This topic: Documentation > Background > CrystalLattice > FCC
Topic revision: 23 Feb 2016, FranzRoters
Copyright by the contributing authors. All material on this collaboration platform is the property of the contributing authors.
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# How many triangles can be formed by joining 12 points, 7 of which are
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How many triangles can be formed by joining 12 points, 7 of which are [#permalink]
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28 Jan 2019, 23:44
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How many triangles can be formed by joining 12 points, 7 of which are collinear?
A. 255
B. 220
C. 185
D. 35
E. 10
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How many triangles can be formed by joining 12 points, 7 of which are [#permalink]
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29 Jan 2019, 00:45
The number of triangles that can be formed from 12 points is $$5C_3$$ = 10 as 7 points are collinear.
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How many triangles can be formed by joining 12 points, 7 of which are [#permalink]
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Updated on: 29 Jan 2019, 02:57
Bunuel wrote:
How many triangles can be formed by joining 12 points, 7 of which are collinear?
A. 255
B. 220
C. 185
D. 35
E. 10
given 7 points are collinear ,which means 7 out of 12 points are on same line and 5 are random points
so total triangles which can be formed ; 7c2*5c1+7c1*5c2+5c3 = 21*5+7*10+10 = 185
IMO C
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Originally posted by Archit3110 on 29 Jan 2019, 01:49.
Last edited by Archit3110 on 29 Jan 2019, 02:57, edited 1 time in total.
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Re: How many triangles can be formed by joining 12 points, 7 of which are [#permalink]
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29 Jan 2019, 02:50
Solution
Given:
• Total number of points = 12
o Number of collinear points = 7
To find:
• The number of triangles that can be formed
Approach and Working:
• The number of triangles that can be formed, N = number of ways of selecting 2 points from 7 and the other from the remaining 5 + the number of ways of selecting 1 from 7 and 2 from the remaining 5 + the number of ways of selecting 3 points from the remaining 5
• Therefore, $$N = ^7C_2 * ^5C_1 + ^7C_1 * ^5C_2 + ^5C_3 = 21 * 5 + 7 * 10 + 10 = 185$$
Hence, the correct answer is Option C
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Re: How many triangles can be formed by joining 12 points, 7 of which are [#permalink] 29 Jan 2019, 02:50
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https://mutualismecology.com/LVMutulaism/ | # Steal this Primer on Mutualism
I wrote a blog post for Dynamic Ecology on how mutualism has been ignored by theoretical textbooks. I didn’t have space to write in on how I think simple models mutualism should be included, so I made this webpage. You can steal any and all of it. Attribution is honest, transparent, and respectful; and I’d happily consider writing a chapter in a book.
I’ve organized this Primer in the following manner:
1. The Verhulst logistic equation as a base to model mutualism
2. Two-species mutualism
2.1. Including the interaction term
2.2 Phase plane
2.3 Analysis
2.4 Modifications
2.4.1 Receive benefits at low densities
2.4.2 Confer benefits at high densities
2.4.3 Facultative and obligate mutualisms
3. Additional variables and further complexities
4. The intersect of mutualism theory and biology
4.1. Two-speices interactions
4.2. Three-species interactions
4.3. $n$-species inteactions
## The Verhulst logistic equation as a base to model mutualism
The logisic equation had at least six different origins (e.g., Lloyd), with the two most relevant being Verhulst’s and Pearl and Reed’s (see Mallet for some historical background). I will present the equations in this Primer as per-capia growth rate for the sole reason that dividing by density on both sides makes the right-hand side a bit easier to read. Verhulst’s logistic equation reads
$$\frac{1}{N}\frac{\mathrm{d}N}{\mathrm{d}t} = r - \alpha N$$,
$$\frac{1}{N}\frac{\mathrm{d}N}{\mathrm{d}t} = r\left(\frac{K - N}{K}\right)$$.
The difference between the two are somewhat superficial here: the Verhulst model includes a negative crowding coefficient to represent intrasepcific competition, $\alpha$, while the Pearl-Reed model has an equilibirum that is assigned a priori, $K$. For single-species dyanmics these equations are fully equivalent. That is, $K = r/\alpha$.
The main difference between the two equations are the units of the crowding coefficient and positive equilibirum. The units of the crowding coefficient is per-unit of density, $[D^{-1}]$. The units of the positive equilibirum is density, $[D]$.
## Two-species mutualism
### Including the interaction term
But when 2 or more species or populations interact (I will use these words interchangably), this eqvuivalance begins to breakdown. This is because once species begin to interact, their dynamic equilibirum will deviate from the a priori equilibrium, $K$. For competitive interactions, the dynamic equilibrium will be less than the a priori equilibrium. For enemy-victim interactions (e.g., predator-prey, host-pathogen), the enemy’s equilibrium will be larger and the victim’s equilibrium will be lower than the a priori equilibrium.
In the Pearl-Reed model, there are effectively 3 ways to add the effect of species 2 ($N_2$) on species 1 ($N_1$):
1. Effect on density-independent growth rate: $$\frac{1}{N_1}\frac{\mathrm{d}N_1}{\mathrm{d}t} = (r + f(N_2))\left(\frac{K - N_1}{K}\right) = r + f(N_2) - \frac{r}{K}N_1 - \frac{1}{K}N_1f(N_2)$$
2. Effect on equilibrium: $$\frac{1}{N_1}\frac{\mathrm{d}N_1}{\mathrm{d}t} = r\left(\frac{K + f(N_2)- N_1}{K + f(N_2)}\right) = r + \frac{r}{K + f(N_2)}N_1 + \frac{r}{K + f(N_2)}f(N_2)$$
3. Effect on intraspecific competition: $$\frac{1}{N_!}\frac{\mathrm{d}N_1}{\mathrm{d}t} = r\left(\frac{K - N_1 + f(N_2)}{K}\right) = r + \frac{r}{K}N_1 + \frac{r}{K}f(N_2)$$
In 1., by adding the effect of species 2 on species 1 to $r$, the model will change the density-independent growth rate; i.e., the growth rate when rare. For some interactions and systems, this is a wholly reasonable way to model interactions. (In fact, I will dig more deeply into this idea below.) For 2. we can add the effect of species 2 on species 1 to the a priori equilibrium, $K$. Notice that it is both in the demonimator and numerator. For 3. we can add the effect of species 2 on species 1 in such a way that it represents the reduction of intraspecific competition. Compare 2. and 3. Interestingly, although they result in the same dynamic equilibirum, the rates of change are different. I have never seen these dynamics differences discussed, but for stronger the effects of species 2 on species 1, the greater the difference rate of change.
In the Verhulst model, there are two main ways to add the effect of $N_2$ on $N_1$:
1. Effect on density-independent growth rate: $$\frac{1}{N_1}\frac{\mathrm{d}N_1}{\mathrm{d}t} = (r + f(N_2))\left(\frac{r - \alpha N_1}{r}\right) = r + f(N_2) - \alpha N_1 - f(N_2)N_1 - \frac{\alpha}{r}f(N_2)N_1$$
2. Effect of $N_2$ on $N_1$ when $N_1$ is at high densities: $$\frac{1}{N_1}\frac{\mathrm{d}N_1}{\mathrm{d}t} = r - \alpha N_1 + f(N_2)$$
I have not seen formulation in 1 published, but it can be useful if $N_2$ affects $N_1$ most strongly when $N_1$ is at low densities. The second equation is similar to the 2. and 3. of Pearl-Reed models above, although the affect of $N_2$ on $N_1$ $f(N_2)$ is not scaled by the $r$ or $K$. For simplicity’s sake, I will proceed using Verhulst’s eqn. 2 for the next sections on phase planes and analysis. Yes, this does slightly deviate from the common Pearl-Reed formulation with $K$, but I will follow the consensus of theoreticians that do not use the Pearl-Reed model. For instance, Chesson (AREES, 2000) writes:
Unfortunately, textbooks muddy the water by parameterizing Lotka-Volterra competition in terms of carrying capacities and relative coefficients of competition.
Ginzburg (TREE, 1992), as a nother example. writes:
. . . the logistic equation, particularly in its common parameterization, [Pearl-Reed model] has been one of the greatest disservices to theoretical ecology.
## Phase plane
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## Analysis
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## Modifications
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### Receive benefits at low densities
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### Confer benefits at high densities
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### Facultative and obligate mutualisms
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## Additional variables and further complexities
thgere is smoe mode text | 2022-11-26 22:24:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6953163146972656, "perplexity": 2292.5705664055095}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446709929.63/warc/CC-MAIN-20221126212945-20221127002945-00681.warc.gz"} |
https://pysal.readthedocs.io/en/latest/generated/pysal.explore.spaghetti.NetworkBase.html | # pysal.explore.spaghetti.NetworkBase¶
class pysal.explore.spaghetti.NetworkBase(ntw, pointpattern, nsteps=10, permutations=99, threshold=0.5, distribution='poisson', lowerbound=None, upperbound=None)[source]
Base object for performing network analysis on a spaghetti.Network object.
Parameters: ntw : spaghetti.Network spaghetti Network object. pointpattern : spaghetti.network.PointPattern A spaghetti point pattern object. nsteps : int The number of steps at which the count of the nearest neighbors is computed. permutations : int The number of permutations to perform (default 99). threshold : float The level at which significance is computed. – 0.5 would be 97.5% and 2.5% distribution : str The distribution from which random points are sampled – uniform or poisson lowerbound : float The lower bound at which the function is computed. (Default 0). upperbound : float The upper bound at which the function is computed. Defaults to the maximum observed nearest neighbor distance. sim : numpy.ndarray simulated distance matrix npts : int pointpattern.npoints xaxis : numpy.ndarray observed x-axis of values observed : numpy.ndarray observed y-axis of values
Methods
computeenvelope() compute upper and lower bounds of envelope setbounds(nearest) set upper and lower bounds validatedistribution() enusure statistical distribution is supported
__init__(ntw, pointpattern, nsteps=10, permutations=99, threshold=0.5, distribution='poisson', lowerbound=None, upperbound=None)[source]
Initialize self. See help(type(self)) for accurate signature.
Methods
__init__(ntw, pointpattern[, nsteps, …]) Initialize self. computeenvelope() compute upper and lower bounds of envelope setbounds(nearest) set upper and lower bounds validatedistribution() enusure statistical distribution is supported | 2019-02-23 11:40:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.21729613840579987, "perplexity": 8804.725255265772}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550249500704.80/warc/CC-MAIN-20190223102155-20190223124155-00253.warc.gz"} |
https://chadretz.wordpress.com/2010/12/19/mathml-inside-javadoc-using-mathjax-and-a-custom-taglet/ | ## MathML inside Javadoc Using MathJax and a Custom Taglet
Posted in Java by chadretz on December 19, 2010
I am writing a baseball statistics application. It is called StatMantis and it’s another in a long line of personal projects I’ll probably never finish. It contains many statistics which have equations that deserve to be in the javadocs. After I researched all the possible ways to accomplish this and I settled on MathML. For example, in my Batting Average On Balls In Play (BABIP) statistic, there is a formula. I wanted one of those badass formula displays like Wikipedia has. So I had javadoc on the class that looked like this:
/**
* This calculates Batting Average on Balls In Play (BABIP). See the Wikipedia reference
* <a href="http://en.wikipedia.org/wiki/Batting_average_on_balls_in_play">here</a>.
* This is a pitching and batting statistic. This is calculated as:
* <p>
* $* <mfrac> * <mrow> * <mi>H</mi> * <mo>-</mo> * <mi">HR</mi> * </mrow> * <mrow> * <mi>AB</mi> * <mo>-</mo> * <mi>K</mi> * <mo>-</mo> * <mi>HR</mi> * <mo>+</mo> * <mi>SF</mi> * </mrow> * </mfrac> *$
*/
I noticed several projects out there that I could utilize to put a MathML equation in my javadoc. JEuclid was my first choice. It could output to an AWT image. It could even given me information to generate an image map to link to the other factors in the equation. But I would have to hack up the standard Sun (…er…Oracle) doclet and do mass hackery for this simple thing. I decided it wasn’t worth the effort since the javadoc tool is difficult to extend. Hopefully one day a project like javadoc-ng will get finished and solve my problems (wink wink, sorry if you’re still waiting on me Harmony guys). So after I couldn’t find anything I liked on the MathML implementations page I almost gave up. I couldn’t find any that were interactive and cross browser. I specifically wanted the href feature of the MathML 3 spec so I could link to my other classes.
Then I found MathJax and it looked like it would solve all my problems. So the first thing I did was toss it in the <footer> element of the javadoc ant task. The distribution is extremely large, and yes you need just about all of it. So I put MathJax in there and edited config/MathJax.js. I changed jax: ["input/TeX","output/HTML-CSS"] to jax: ["input/MathML", "output/HTML-CSS"] and extensions: ["tex2jax.js"] to extensions: ["mml2jax.js"]. This is needed because the input is MathML, not TeX. Once the config was changed, I added the following in my javadoc ant task:
<footer><![CDATA[
<script type=\"text/javascript\" src=\"{@docroot}/MathJax/MathJax.js\"></script>
]]></footer>
I put this in the footer, because it seems like the header is rendered twice, which isn’t cool. Also, per the javadoc footer documentation I have to escape the quotes. The {@docroot} makes sure it sets the relative links properly. Once I executed this I was very happy to see my formula appear properly.
Now I wanted to link each part of the equation to its representative class. My first approach was to utilize the inline {@link} tag. This outputs a code tag wrapped in an anchor tag which links the piece. I wrote at least a dozen different javascript functions to pull the href out of the anchor and put it on the <mi> tag and move the text out from inside the code tag. Everything I tried ended up in extreme failure because IE sucks balls. Specifically, I can’t edit my math tags via DOM because IE doesn’t understand it. I also couldn’t insert a manipulated MathML string as innerHTML on the parent, because it trimmed off pieces for no reason.
So I decided that I needed another, non-client side approach to having links. Post processing the HTML was my first guess, but that is not a very elegant solution. So I decided to extend Javadoc w/ a custom Taglet. I tried to find the existing Taglet that Sun built for {@link} so I could use the same algorithm to obtain the URL, but it’s not there. They get the benefit of having everything there including the RootDoc. So I wrote my own Taglet and tried it in many scenarios. I quickly realized I would not be able to link to all possible methods/fields because the Taglet interface simply doesn’t provide enough information. Similarly, I can’t validate the values entered in my taglet either. Without further ado, here is the custom Taglet (collapsed by default):
/*
*
* use this file except in compliance with the License. You may obtain a copy of
*
*
* Unless required by applicable law or agreed to in writing, software
* WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. See the
* License for the specific language governing permissions and limitations under
*/
import java.util.Map;
import com.sun.tools.doclets.Taglet;
/**
* Taglet that supports the linkhref inline tag. This tag will return just the
* HREF to a javadoc class file (not any of the methods/fields)
*
*/
public class LinkHrefTaglet implements Taglet {
@SuppressWarnings({ "rawtypes", "unchecked" })
public static void register(Map tagletMap) {
Taglet t = (Taglet) tagletMap.get(tag.getName());
if (t != null) {
tagletMap.remove(tag.getName());
}
tagletMap.put(tag.getName(), tag);
}
@Override
public String getName() {
}
@Override
public boolean inConstructor() {
return true;
}
@Override
public boolean inField() {
return true;
}
@Override
public boolean inMethod() {
return true;
}
@Override
public boolean inOverview() {
return true;
}
@Override
public boolean inPackage() {
return true;
}
@Override
public boolean inType() {
return true;
}
@Override
public boolean isInlineTag() {
return true;
}
private PackageDoc getPackageDoc(Tag tag) {
Doc holder = tag.holder();
if (holder instanceof ProgramElementDoc) {
return ((ProgramElementDoc) holder).containingPackage();
} else if (holder instanceof PackageDoc) {
return (PackageDoc) holder;
} else {
throw new RuntimeException("Unrecognized holder: " + holder);
}
}
private ClassDoc getTopLevelClassDoc(ClassDoc classDoc) {
if (classDoc.containingClass() == null) {
return classDoc;
} else {
return getTopLevelClassDoc(classDoc);
}
}
private ClassDoc getTopLevelClassDoc(Tag tag) {
Doc holder = tag.holder();
if (holder instanceof PackageDoc) {
return null;
} else if (holder instanceof ClassDoc) {
return getTopLevelClassDoc((ClassDoc) holder);
} else if (holder instanceof ProgramElementDoc) {
return getTopLevelClassDoc(((ProgramElementDoc) holder)
.containingClass());
} else {
throw new RuntimeException("Unrecognized holder: " + holder);
}
}
private ClassDoc findClass(String className, ClassDoc[] classImports) {
for (ClassDoc classDoc : classImports) {
if (classDoc.name().equals(className)) {
return classDoc;
}
}
return null;
}
private ClassDoc findClass(String className, PackageDoc... packageImports) {
for (PackageDoc packageDoc : packageImports) {
for (ClassDoc found : packageDoc.allClasses(true)) {
if (found.name().equals(className)) {
return found;
}
}
}
return null;
}
private String error(Tag tag, String error) {
System.err.println(tag.position() + ": warning - " + error);
return "javascript: //error";
}
@Override
@SuppressWarnings("deprecation")
public String toString(Tag tag) {
PackageDoc packageDoc = getPackageDoc(tag);
ClassDoc topLevelClassDoc = getTopLevelClassDoc(tag);
//k, what I'm gonna do is what the main one does...go up to the root
StringBuilder href = new StringBuilder();
int dotIndex = packageDoc.name().indexOf('.');
while (dotIndex != -1) {
href.append("../");
dotIndex = packageDoc.name().indexOf('.', dotIndex + 1);
}
//package name is empty when it is the root package
if (!packageDoc.name().isEmpty()) {
href.append("../");
}
//now that we have the root, begin the string parse
String classInTag = tag.text();
int poundIndex = classInTag.indexOf('#');
if (poundIndex != -1) {
classInTag = classInTag.substring(0, poundIndex);
}
//ok, if it's qualified, we just assume it's all good
if (classInTag.indexOf('.') == -1) {
ClassDoc classDoc;
if (topLevelClassDoc == null) {
//not in a class scope? just try inside this package
classDoc = findClass(classInTag, packageDoc);
if (classDoc == null) {
//they should qualify it then
return error(tag, "Can't locate linkhref class " + classInTag +
". The name should be qualified.");
}
} else {
//nope? ok, first try my inner classes
classDoc = findClass(classInTag, topLevelClassDoc.innerClasses(true));
if (classDoc == null) {
//nope? ok, try my single-type-imports
classDoc = findClass(classInTag,
topLevelClassDoc.importedClasses());
if (classDoc == null) {
//nope? ok, try my type-import-on-demands
classDoc = findClass(classInTag, topLevelClassDoc.importedPackages());
if (classDoc == null) {
//nope? ok, finally try my own package
findClass(classInTag, topLevelClassDoc.containingPackage());
if (classDoc == null) {
//not even now? well, just assume it's there because
classInTag = topLevelClassDoc.containingPackage().name() +
'.' + classInTag;
}
}
}
}
}
if (classDoc != null) {
classInTag = classDoc.qualifiedName();
}
}
if (classInTag.indexOf('.') == -1) {
return error(tag, "Unable get linkhref for class " + classInTag +
" because it is in the root package");
}
// ok, now make the link by replacing the dots w/ slashes
href.append(classInTag.replace('.', '/'));
href.append(".html");
// all good
return href.toString();
}
@Override
public String toString(Tag[] tags) {
// not for inline tags...nope
return null;
}
}
It works only for class/interface references and doesn’t do any real validation. Regardless, it solves my problem perfectly, and now my mathematical formulas appear in my javadoc complete with links to other classes. Check out the build-javadoc target in the ANT script to see how to include it in the javadoc task. Overall, it works well and I am happy with it. Here is what the aforementioned BABIP javadoc looks like now:
/**
* This calculates Batting Average on Balls In Play (BABIP). See the Wikipedia reference
* <a href="http://en.wikipedia.org/wiki/Batting_average_on_balls_in_play">here</a>.
* This is a pitching and batting statistic. This is calculated as:
* <p>
* <math style="font-size: 200%">
* <mfrac>
* <mrow>
* <mo>-</mo>
* </mrow>
* <mrow>
* <mo>-</mo> | 2017-10-23 20:27:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2330639362335205, "perplexity": 11078.81113275304}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187826642.70/warc/CC-MAIN-20171023202120-20171023222120-00754.warc.gz"} |
https://www.physicsforums.com/threads/the-magnitude-of-a-vector.956391/ | # The Magnitude of a Vector
CivilSigma
## Homework Statement
For any vector in 2D space, it can be broken down into its horizontal and vertical components.
## Homework Equations
In one of my engineering classes, we are using the following equation to determine the magnitude of a vector:
$$u=v_1 \cdot cos\theta +u_2 \cdot sin\theta$$
Where $\theta$ is the angle with respect to the horizontal, v1 is the horizontal component and v2 is the vertical component of the vector.
I know this equation works but I don't understand why.
I feel like I am missing a fundamental concept, because to determine the magnitude of a vector, I would use Pythagoras theorem, and I cannot derive the above equation from Pythagoras's equation.
## The Attempt at a Solution
Mentor
Can you provide some context for where this equation is applied? Perhaps give a specific example.
In general, this equation will not work for a single vector whose x and y components are ##u_1## and ##u_2##. Perhaps they are summing the horizontal components of two different vectors to obtain a net horizontal resultant?
Staff Emeritus
Homework Helper
That equation doesn't give the magnitude of the vector. It gives you the component of the vector in the direction of ##\hat n = \cos\theta\,\hat i + \sin\theta\,\hat j##.
Homework Helper
Gold Member
In one of my engineering classes, we are using the following equation to determine the magnitude of a vector:
u=v1⋅cosθ + u2⋅sinθ
Where $\theta$ is the angle with respect to the horizontal, v1 is the horizontal component and v2 is the vertical component of the vector.
I know this equation works but I don't understand why.
It comes from geometry... See this diagram... If that's not clear do say and I will explain some more.
#### Attachments
• Magnitude of Vector.png
6.6 KB · Views: 410
CivilSigma
Mentor
I think you meant to write the equation as $$u=u_1\cos{\theta}+u_2\sin{\theta}\tag{1}$$where $$u_1=u\cos{\theta}\tag{2}$$and$$u_2=u\sin{\theta}\tag{3}$$If you substitute Eqns. 2 and 3 into Eqn. 1, you get:
$$u=u\cos^2{\theta}+u\sin^2{\theta}=u$$
CivilSigma and SammyS | 2022-08-10 14:05:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7695695757865906, "perplexity": 413.75822939307636}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571190.0/warc/CC-MAIN-20220810131127-20220810161127-00044.warc.gz"} |
https://www.nature.com/articles/s41467-021-22290-1?error=cookies_not_supported&code=d1b6a44b-2241-4737-a8ac-dd3f3874a2a1 | ## Introduction
Oxidoreductases are a large class of enzymes that use unpaired electrons to facilitate redox reactions with other chemical species and are involved in nearly all aspects of life. Oxidoreductase electron transfers are almost always coupled with a proton transfer1. Concerted proton and electron transfer (CPET) permits a thermodynamically favorable redox reaction, is efficient, and is an integral part of enzymes with the fastest catalytic rates2,3,4. Particularly noteworthy is the prominence of CPET enzymes that regulate the concentration of reactive oxygen species (ROS) in the cell. ROS levels are central to programmed cell death and abnormal regulation by these oxidoreductases play significant roles in cancer and cardiovascular diseases5. CPETs are therefore of significant interest to study but a mechanistic understanding of these enzymes is still lacking. Deciphering these fundamental biochemical reactions is not only significant for its role in diseases, but for the biomedical design of CPET-dependent therapeutic interventions, irradiation protectants, and electrochemical biosensors6,7.
Human manganese superoxide dismutase (MnSOD) is a CPET-based oxidoreductase found in the mitochondrial matrix that reduces ROS levels by eliminating O2•− with the unpaired electrons of the active site metal. The Mn ion is coordinated to inner-sphere residues His26, His74, His163, Asp159, and a single-oxygen species that could be either H2O or OH (designated WAT1, Fig. 1). Trivalent Mn oxidizes O2•− to O2 (k1 = 1.5 nM−1 s−1)8 and the resulting divalent Mn reduces another O2•− molecule to H2O2 (k2 = 1.1 nM−1 s−1)8. This is the only means the mitochondrial matrix has to keep O2•− levels low enough to avoid damage to macromolecules and destruction of cellular function9.
$$(k_1){\quad \mathrm{Mn}}^{3 +} + {\mathrm{O}}_2^{ \bullet -} \leftrightarrow {\mathrm{Mn}}^{2 +} + {\mathrm{O}}_2\\ (k_2){\quad \mathrm{Mn}}^{2 +} + {\mathrm{O}}_2^{ \bullet -} + {\mathrm{2H}}^ + \leftrightarrow {\mathrm{Mn}}^{3 +} + {\mathrm{H}}_2 {\mathrm{O}}_2$$
The major endogenous source of O2•− is from electrons inadvertently leaking from the electron transport chain. Dysfunctional MnSOD activity, therefore, poses significant consequences on the mitochondria that contributes to several diseases. Genetic aberrations of MnSOD are associated with several cancer types, with mammary and prostate cancers being the most frequently noted in curated databases10. Polymorphisms of MnSOD have also been noted to be a predictor for deficient vascular function11. Therefore, the ability for MnSOD to utilize the high reaction rate and efficiency (kcat/Km > ~109 M−1 s−1) of its CPET mechanism is correlated with the preservation of health8.
The CPET mechanism of MnSOD and the majority of other oxidoreductases has yet to be defined at the atomic level. The limitation in studying CPETs is the difficulty in directly detecting the protonation states of ionizable residues, solvent/ligands at the active site, and correlating them with the electronic state of the active site metal. The second sphere of MnSOD harbors five residues (His30, Tyr34, Gln143, Glu162, and Tyr166 (Fig. 1)) and information about their protonation states would be of significant value in deciphering a catalytic mechanism. X-ray and spectroscopic techniques have been unable to provide this information due to the poor scattering of hydrogen atoms and the difficulty in discerning spectra for specific titratable positions.
Current mechanistic knowledge has relied on X-ray crystal structures and kinetic studies of point mutants to infer the role of second-sphere residues. Mutations of Gln143, Tyr34, or His30 do not significantly alter the active site structure although they are detrimental to the kinetic rates of the enzyme. The Gln143Asn point mutation leads to near ablation of catalysis, Tyr34Phe depletes activity only for the Mn2+→Mn3+ redox transition, and His30Gln has 40 and 82% of wild-type activity for the Mn3+ → Mn2+ and Mn2+ → Mn3+ transitions, respectively8,12,13,14. Quantum mechanic and molecular mechanic (QM/MM) calculations have been applied to X-ray structures for insight into the possible proton transfers. They have suggested protonation of WAT1 from OH to H2O during the Mn3+ → Mn2+ redox reaction and deprotonation of WAT1 and Tyr34 for H2O2 formation during the Mn2+ → Mn3+ reaction15,16. These theoretical methods were the only way to study the simultaneous change of electron state and proton locations in the enzyme.
Neutron protein crystallography (NPC) is an emerging tool for analyzing hydrogen positions of biological macromolecules and possesses attributes that are useful in deciphering CPET mechanisms. In NPC, scattering of deuterium is on par with carbon, nitrogen, and oxygen, increasing the ability to locate proton positions for the entire enzyme. An additional advantage is that neutrons, unlike X-rays, do not alter the electronic state of metals17. The capability of proton visualization and amenability to enzyme redox control that NPC provides creates an avenue to investigate the CPET mechanism of MnSOD experimentally.
Here, we present room temperature neutron structures of human MnSOD at physiological pH in Mn3+ and Mn2+ states and reveal how the atomic locations of all protons in the enzyme active site change when the active site metal goes through a redox cycle. Direct experimental evidence is provided for deprotonation of a conserved glutamine residue and the feasibility of the proton transfer is explored with quantum calculations. Four other conserved residues are discerned to each have short-strong hydrogen bonds (SSHBs) and unexpected attributes that include an ionized tyrosine and a low-barrier hydrogen bond (LBHB) between a separate tyrosine and histidine across the dimer interface. A proposed CPET mechanism is reported from the culmination of data between Mn3+SOD and Mn2+SOD neutron structures.
## Results and discussion
### Direct evidence for CPETs with glutamine deprotonation
To visualize the effect of the electronic state of the metal on the active site protons, all-atom, D-labeled neutron structures were obtained for Mn3+SOD and Mn2+SOD to resolutions of 2.20 and 2.30 Å, respectively. The two data sets collected were from the same crystal treated with appropriate oxidizing and reducing chemicals and the redox conditions were maintained during each data collection18. Neutron data at these resolutions are excellent and permitted ease in the visualization of deuterium atoms. The proton positions of the resting state, apo active sites were specifically sought to define the effects of metal redox state on the surrounding protons independent of substrate binding. We initially sought a crystallographic check for the success of the redox manipulations. WAT1, the Mn ligand, has historically been thought to obtain a proton (OH → HOH) upon one-electron reduction of Mn3+ to Mn2+. This was inferred from crystallographic bond distances and QM calculations but otherwise had not yet been directly observed and confirmed4,16,19,20,21,22,23,24,25. Indeed, careful inspection of the neutron scattering length density between neutron data sets revealed changes in the protonation of WAT1 when the redox state of the Mn ion was altered (Fig. 2a, b). This was theoretically expected, verified our methods, and gave confidence in the data. To our knowledge, this is the first time the chemical reduction of an active site metal was visually observed to change the protonation state of a ligand.
For Mn3+SOD, both chains of the asymmetric unit have a single nuclear |Fo| − |Fc| density peak for the D1 atom of WAT1 indicating the expected deuteroxide (OD) molecule and is supported by the Mn–O(WAT1) bond distance of 1.8 Å (Fig. 2a and Supplementary Fig. 1)15,26. The OD acts as a hydrogen bond donor to the Oε2 of Asp159 with a distance of 1.9 Å whereas the O(WAT1) atom is acting as a hydrogen bond acceptor from Dε21(Gln143) with a distance of 1.8 Å (Fig. 2a). The data for WAT1 of Mn2+SOD instead have two nuclear |Fo| − |Fc| density peaks for D atoms indicating the OD was converted to D2O upon metal reduction, as expected. Further support suggesting a D2O molecule is seen with the Mn–O(WAT1) bond distance of 2.2 Å15,26. The D1(WAT1) atom position is similar to that found in the Mn3+SOD counterpart and hydrogen bonds with Oε2 of Asp159, albeit with a longer distance of 2.5 Å (Fig. 2b). Surprisingly, D2 (WAT1) points toward Gln143 suggesting the WAT1 of Mn2+SOD is acting as a hydrogen bond donor to Gln143. This means Gln143 is a hydrogen bond acceptor in Mn2+SOD and its Dε21 atom is absent. Indeed, there is a lack of neutron scattering length density for Dε21 but not for Dε22. This interpretation was supported when attempts to model Dε21 led to negative |Fo| − |Fc| neutron scattering length density. In the Mn2+SOD structure, the hydrogen bond between D2(WAT1) and Nε2(Gln143) is atypical with a bond distance of only 1.6 Å and O(WAT1)– D2(WAT1)–Nε2(Gln143) angle close to 180°. These are characteristics of an SSHB, a type of hydrogen bond that is thought to stabilize particular enzymatic steps and enhance catalytic rates27,28,29. This is not to be confused with an LBHB, a type of SSHB where a proton is equidistant between heteroatoms30. SSHBs are noteworthy in several well-studied enzymes, such as α-chymotrypsin that utilizes an SSHB between the His and Asp of its catalytic triad28. This creates a ~7 kcal mol−1 stronger interaction to substantially increase the kinetic rate. For Mn2+SOD, the SSHB between WAT1 and Gln143 may contribute to the stability of the redox state and the high catalytic efficiency of the enzyme.
The experimental data, therefore, suggest that Gln143 is undergoing deprotonation to form an amide anion and this is unusual because glutamine residues are not expected to act as weak acids since the pKa of primary amides are 16–18. However, pKa studies of less acidic secondary amides suggest pKa values may be depressed to 7–8 depending on how the amide group is polarized (i.e. charge delocalization)31. This is supported by the known event of proton exchanges occurring at the amide groups of protein backbones. Moreover, several enzyme studies suggest glutamine or asparagine-mediated proton transfers for catalysis and support the plausibility of a Gln143 → WAT1 proton transfer. For example, an asparagine residue has been suggested to be deprotonated in prenyltransferases due to significant polarization from close proximity to a metal cation32. Nakamura and coworkers33 showed neutron data of cellulase Cel45A from Phanerochaete chrysosporium that revealed asparagine deprotonation that is instrumental for the proton relay of the enzyme. Infrared spectroscopy and computational calculations support the involvement of glutamine-mediated proton transfers in GTP hydrolysis by Ras-GAP and photoexcitation of photoreceptor proteins with the flavin-binding BLUF domain34. For MnSOD, the deprotonation of Gln143 for CPET to the active site ligand has not been observed before although it does explain the high efficiency of the enzyme as a result of this internal proton source.
Density functional theory (DFT) calculations of the active site using the neutron structures support our interpretation of the neutron scattering length density for deprotonation of Gln143. Calculations used the atoms of the residues shown in Fig. 1. Chemist’s localized property-optimized orbital (CLPO) analysis (akin to Natural Bond Orbitals35,36,37) was used to evaluate the interactions of Gln143 and WAT1 (ref. 38). Reduction of Mn3+ to Mn2+ increases the electronegative character of the O(WAT1) lone pair facing the proximal amide proton of Gln143 (green atom, Fig. 3a). This polarization of OH(WAT1) increases its basicity and chemically allows the abstraction of the amide proton from glutamine. In terms of hard and soft Lewis acid and base (HSAB) theory, the transition of the metal acid and solvent base from (Mn3+OH) to (Mn2+–OH2) corresponds to both the acid and base becoming softer. The bonds of the deprotonated amide rearrange to stabilize its new negatively charged state (Fig. 3b). The Oε1 atom of Gln143 bears the most electronegative charge, and calculations suggest less covalent electrons for the Oε1–Cε1 bond compared to that of Nε2–Cε1, with bond orders of 1.33 and 1.52, respectively (Supplementary Table 1). The presence of an SSHB is supported as well since the deprotonated Nε2 still has covalent character with the donated proton (Fig. 3c). Donor–acceptor orbital analysis indicates electron density is transferred from the Nε2 lone pair orbital to the σ*-antibonding orbital of O–H(WAT1) and is a 1.4 kcal mol−1 stabilizing hyperconjugation interaction (Supplementary Table 2). As a result, the interaction between Nε2 and the donated proton demonstrates partial σ-bonding character. The extent of covalence is 36% for the Nε2–H bond and 64% for the O–H bond (Supplementary Table 3). Using the same QM methods, water molecules participating in normal hydrogen bonding have 12% covalence between the donor H and acceptor O atoms. Altogether, quantum calculations support both Gln143 deprotonation and the presence of an unusual hydrogen bond between WAT1 and Gln143.
A previously undetected change in the interaction of Gln143 with the neighboring Trp123 was observed. An SSHB of 1.5 Å is seen between the Oε1 of Gln143 and Dε1 of Trp123 in the reduced state (Fig. 2b). The same hydrogen bond is seen in Mn3+SOD at a distance of 1.9 Å (Fig. 2a). The neutron data, therefore, support the notion that Oε1(Gln143) harbors a stronger electronegative character during the Mn2+ redox state. This may be a consequence of Gln143 deprotonation to the amide anion during the Mn3+ → Mn2+ redox reaction and negative charge stabilization of its Oε1 atom through hydrogen bonding with Trp123. Charge stabilization is likely to be important for glutamine deprotonation as amide groups are known to deprotonate at neutral pH when electronegatively polarized at the carbonyl O atom31. Trp123 is especially competent at charge stabilization due to its own ability to polarize. CLPOs suggest that lone pair electrons of Nε1(Trp123) delocalize into the highly conjugated aromatic ring of Trp123 when the glutamine amide is deprotonated and is stabilizing (Fig. 3e). Donor–acceptor orbital analysis calculates that the major stabilizing interaction is the donation of Nε1(Trp123) lone pair electron density into the Cε2–Cδ2 π*-antibonding orbital (Fig. 3f) and decreases energy by 13.5 kcal mol−1 (Supplementary Table 2). This also permits the SSHB between Oε1(Gln143) and Hε1(Trp123). Quantum calculations indicate an important role for Trp123 in the deprotonation of Gln143.
If an O(WAT1)–Dε21(Gln143)–Nε2(Gln143) interaction is needed for redox cycling of Mn, mutation of Gln143 or a nearby residue that may stabilize amide deprotonation should affect catalysis. In the literature, the Gln143Asn mutant has nearly ablated catalysis in both redox states while Trp123Phe can perform catalysis for Mn3+ → Mn2+ at deficient lower rates (20–50%) but not at all for the Mn2+ → Mn3+ transition14,39. The effect of these mutations suggests that Gln143 is central to catalytic activity while Trp123 is most significant for the Mn2+ → Mn3+ half of the redox cycle. The detrimental effects for the Mn2+ state due to mutating residue Trp123 may therefore reflect their role in stabilizing the Gln143 amide anion. Indeed, the kinetic behaviors of these mutants were puzzling in past studies but amide proton transfer potentially explains them12,39,40,41. Glutamine at the position of Gln143 is conserved in all isoforms of MnSODs and prokaryotic FeSODs. A closer WAT1-Gln distance correlates with increased redox potentials and catalytic rates42. This is perhaps because of an enhanced ability for proton transfers between O(WAT1) and Nε2(Gln143). Past mutagenesis studies, differences in catalytic rates among isoforms, and the high catalytic rate of MnSOD may be explained by Gln143 serving as an internal proton source for CPET via amide deprotonation.
### Tyr34 has an unusual pKa and SSHBs with the Gln143 anion
Tyr34 is positioned near the active site solvent channel, hydrogen-bonded to Gln143 (Fig. 1a), and has been hypothesized to be a proton source for MnSOD CPET12. For Mn3+SOD, Tyr34 does not have a nuclear peak for its hydroxyl proton (Fig. 4a and Supplementary Fig. 3). Interestingly, for one of the active sites, deprotonated Tyr34 is making a very strong hydrogen bond with a nearby solvent molecule (designated WAT2) with a 2.3 Å distance between heteroatoms Oη(Tyr34) and O(WAT2) (Fig. 4a). While the deuteriums of WAT2 could not be discerned, the distance is characteristic of an SSHB where Tyr34 may be poised to accept a proton. This interpretation is supported by CLPO analysis from DFT, with 80/20 covalent sharing of the proton (Supplementary Table 3). For Mn2+SOD, a nuclear peak for the hydroxyl proton is present but not where it is expected. Refinement with the ideal 0.97 Å Oη–Dη distance for Tyr34 persistently demonstrates residual |Fo||Fc| difference density (Supplementary Fig. 3) uncharacteristic of the other MnSOD tyrosine residues. Between refinement cycles, restraints for this Tyr34 were incrementally loosened from the 0.97 Å ideal hydroxyl distance until the |Fo|−|Fc| difference density was appropriately absent and the B factors were comparable to other nearby atoms. This yielded an unusual Oη–Dη bond length of 1.3 Å that points towards the solvent channel (Fig. 4b). Intriguingly, the Oη(Tyr34) atom participates in a strong 1.6 Å hydrogen bond with Dε22(Gln143) in the Mn2+SOD structure and is significantly different than the corresponding hydrogen bond distance of 2.3 Å in Mn3+SOD (Fig. 4a, b). This SSHB may potentially be explained by the increased polarization of Gln143 in Mn2+SOD from deprotonation to the amide anion leading to a stronger hydrogen bond interaction with Tyr34. It is unclear whether this interaction contributes to the observed changes in Tyr34 protonation though it may stabilize the amide anion of Gln143. The experimental data, therefore, show that Tyr34 is capable of gaining and losing a proton at physiological pH indicative of an unusual pKa, and participates in atypical hydrogen bonding.
Our experimental data for Tyr34 potentially shines a light on the unexplained observations of previous studies investigating its role in catalysis8,12,39. Tyr34 has been speculated to be the proton donor to WAT1 for CPET during the Mn3+ → Mn2+ reaction though this conflicts with the pH independence of the reaction between values of 6 and 10 (refs. 8,43). This was puzzling because CPET mechanisms are expected to have pH dependence as a result of the proton transfer part of their catalysis and Tyr34 is the closest titratable residue. Instead, the MnSOD neutron data suggest that the proton donor to WAT1 is internally sourced from Gln143 without the direct involvement of solvent and cannot be Tyr34 due to its observed deprotonation in the Mn3+ state.
An ionized tyrosine residue at physiological pH is unusual though is potentially explained by polarization from the pronounced positive charge of the metal. Such an effect has been visualized in studies of human carbonic anhydrase II (HCA II), a metalloenzyme with diffusion-limited catalytic efficiencies like MnSOD. For HCA II, joint neutron crystallography and NMR demonstrate a tyrosine residue with a pKa of 7.10 ± 0.10 at the active site44. The catalytic role of an ionizable Tyr34 for MnSOD is prominent during the Mn2+ → Mn3+ redox cycle and is supported by the inability of the Tyr34Phe mutant to catalyze this step of the reaction45. Since Tyr34 gains a proton during the Mn3+ → Mn2+ redox cycle and loses a proton during the Mn2+ → Mn3+ cycle, it is conceivable that Tyr34 serves as the source for one of the two protons involved in the protonation of the substrate to H2O2. Indeed, spectroscopic data of the Tyr34Phe mutant suggest a prolonged binding of a species to the metal but could also be a result of the inability of Gln143 to deprotonate without stabilization from the Tyr34 hydroxyl group12. Our crystallographic neutron data have shed light on the perplexing role of the conserved Tyr34 residue.
### Serendipitous ligand binding to Mn2+SOD helps explain catalysis
Previous spectroscopic studies have suggested the sixth-coordinate binding of a ligand (OH, F, or N3) to Mn2+SOD coinciding with lengthening or displacement of the Mn–Oδ2(Asp159) bond essentially converting it to a 5-coordinate active site spectrophotometrically19,46. The chemical purpose of this modification in coordination has been unclear. A sixth-coordinate OH-bound complex has been difficult to visually confirm due to poor scattering of hydrogen by X-rays47. Serendipitously, one of the active sites of the Mn2+SOD neutron structure has density for an unexpected sixth-coordinate OD ligand (designated OL for anionic oxygen ligand, Fig. 4c), and the Mn–Oδ2(Asp159) bond is stretched from 1.95 to 2.44 Å. Of note, the buffer system did not include OD, and only appropriate ratios of K2DPO4 and KD2PO4 were used to achieve a pD of 7.8. Yet the observed nuclear scattering length density for deuterium and the 1.84 Å Mn–O distance of OL supports the identification of OL as OD20. OL is well ordered with a B-factor that is comparable to other atoms of the active site (Supplementary Table 4). Further validation was sought with DFT geometry optimizations of the observed neutron structure, Mn2+SOD with sixth-coordinate OH(OL), and suggests stability of the six-coordinate complex while replacing OL with H2O causes disassociation into five-coordinate Mn2+SOD. The observed differences between the Mn2+SOD active site (Fig. 4b, c) may be a consequence of the crystallographic asymmetric subunits having different capacities of solvent accessibility (Supplementary Fig. 4). Nevertheless, this active site supports past hypotheses of Mn2+-binding OH at the sixth-coordinate position19,47 and occurs in tandem with the lengthening of the Mn–Oδ2(Asp159) as proposed by the associative displacement ligand binding model46.
Maliekal and coworkers19 previously hypothesized that binding of a OH ligand could be the result of an electronegative deprotonated Tyr34 and a electropositive Mn ion polarizing a water molecule to promote deprotonation. Indeed, Tyr34 is observed to have no density for a hydroxyl proton (Fig. 4c). Since Tyr34 is deprotonated, the proton acceptor of the water molecule is likely to be another nearby residue, perhaps His30. The subsequent OL coordination to Mn2+ could be involved in Gln143 deprotonation. The six-coordinate active site has Gln143 in the canonical amide form making a Dε21(Gln143)-O(WAT1) 1.4 Å SSHB (Fig. 4c). This distance is even shorter than the N(Gln143)-D2(WAT1) SSHB in apo Mn2+SOD (Fig. 2b) and suggests a very strong hydrogen bonding interaction that is likely to be partially covalent (Supplementary Table 5). Trp123 forms a stabilizing 1.5 Å SSHB with Oε1(Gln143) as observed at the other subunit (Fig. 2b) and is thought to support deprotonation of Gln143 (Fig. 3e, f). These structural characteristics indicate that binding of OL probably contributes to the future deprotonation of Gln143. Here, binding by an anionic OL would lower the positive charge of Mn2+ and electronegatively polarize OD(WAT1) to initiate covalent bonding with the proximal Dε21(Gln143) proton.
### His30 and Tyr166 form a LBHB indicating unusual pKas
His30 is positioned at the active site channel (Fig. 1) and also shows interesting changes in protonation between redox states. The two nitrogen atoms of His30 that are potential sites for protonation changes are strategically positioned. The Nδ1 atom is solvent accessible and is hydrogen bonding to WAT2 while the Nε2 atom is not accessible to solvent and hydrogen bonds with the buried Tyr166 from the adjacent subunit (Fig. 1a). For Mn2+SOD, both chains show strong omit |Fo|−|Fc| difference density for protonation of Nδ1 (Fig. 5a, b). Interestingly, the D atom between Tyr166 and His30 refines to a position that is nearly equidistant between Oη and Nε2 and has elongated omit |Fo|−|Fc| difference density that is unusual for a typical hydrogen bond. This is characteristic of a LBHB, a type of SSHB where a proton is transiently shared between heteroatoms48. LBHBs are thought to be instrumental in catalysis by stabilizing enzyme states by as much as 20 kcal mol−1 leading to a crucial acceleration of kinetic rates30,49. Indeed, Tyr166Phe and His30Gln single mutants both decrease kcat by ~30-fold and kcat/Km by ~10-fold13,50. Another defining feature of LBHBs is the pKa difference between heteroatoms is near zero51. Buried amino acid residues can have elevated or depressed pKas52,53. His30 and Tyr166 may be following these qualities. The pKa of His30 would be elevated from its solvated value of ~6, and the pKa of Tyr166 would be depressed from its solvated value of ~10. Since Nε2(His30) is solvent inaccessible along with Tyr166, the LBHB may be supported by residing within an enclosed environment.
For Mn3+SOD, the D atom between His30 and Tyr166 refines with the average position on Tyr166, and its hydrogen bond distance with Nε2(His30) is of typical length (Fig. 5c, d). Despite the refinement position of the D atom at Chain A, the omit |Fo|−|Fc| difference density is pronounced and elongated towards Nε2(His30) and may reflect the movement of the proton (Fig. 5c). Similarly, the Nδ1-bound proton of Chain A has strong and elongated omit |Fo|−|Fc| difference density towards WAT2 that can be interpreted as proton exchange with solvent. For Chain B, there is no difference density for a proton on Nδ1(His30), a stark contrast with Chain A (Fig. 5d) as well as both chains of Mn2+SOD (Fig. 5a, b). Chain B of Mn3+SOD is also the only one with WAT2 positioned to act as hydrogen bond donator to Nδ1(His30) rather than a hydrogen bond acceptor, which would agree with the notion that Nδ1(His30) is deprotonated. With Nδ1 and Nε2 of His30 both deprotonated, the imidazole group puzzlingly appears absent of protons which is chemically infeasible. Differences in B values between the subunits of solvent-exposed residues could degrade the quality of the density but this does not explain the absence of proton density at Nδ1(His30) for Chain B. In fact, these residues in Chain B have lower refined B values than in Chain A (Supplementary Table 4). By comparing the density of the D atom between His30 and Tyr166 with that of the other chain (Fig. 5c), it is conceivable that the D atom appearing to belong to Tyr166 has movement to Nε2(His30) and this is the most plausible explanation. The chemical function may be to modulate protonation at the solvent accessible Nδ1 of His30. Protonation of Nε2(His30) would coincide with deprotonation of Nδ1(His30) leading to shifting between singly Nδ1- or Nε2-protonated tautomers (Fig. 5e).
Similarly, unusual hydrogen bonding between residues has been observed in other enzymes. The most notable are histidines found in catalytic triads of proteases, where changes in protonation of one nitrogen could be tied to the protonation state of the other with the help of an SSHB interaction54. Investigating the literature for other enzymes utilizing a tyrosine–histidine pair in catalysis reveals the metalloenzyme Photosystem II (PSII) that utilizes CPETs55. The tyrosine–histidine pair of (PSII) appears to have an SSHB that needs to be maintained for catalysis with measured pKa values ranging between 7.3 and 8.0. For MnSOD, the observed atypical hydrogen bonding observed across the dimer interface between Tyr166 and His30 is probably important in the enzymatic mechanism.
Previous mutagenesis investigations suggest that the Tyr166 and His30 interaction is needed for catalysis and support the interpretation of proton transfers occurring between Nε2(His30) and Oη(Tyr166) that may coincide with changes in protonation at Nδ1(His30). His30Gln is the only His30 mutant that has been studied that maintains the hydrogen bonding at the active site and does not significantly affect the positions of other residues at the active site13,50. Kinetically, the His30Gln rate for k1 (Mn3+ → Mn2+) is 38% of the wild type while k2 (Mn2+ → Mn3+) is 72%56. The rates indicate an important role for His30 k1 that may correspond with our observations of its changes in protonation in Mn3+SOD. It should be noted that previous studies refrain from attributing differential protonations to His30 due to the similar redox potentials between wild type (393 ± 29 mV) and His30Gln (380 ± 30 mV)13,50,57. However, the investigations do not consider whether compensatory protonations or deprotonations occur at nearby residues as a result of the mutation giving the appearance of an inconsequential effect. Indeed, the Tyr34Phe mutant also has an insignificant change of redox potential (435 ± 30 mV), but has a nearly identical rate for k1 (37% of wild type) as His30Gln50,56. Drawing an inference from the Tyr166Phe mutant is difficult because hydrogen bonding and side-chain conformations are significantly changed at the active site, but this mutant has nearly identical measurements of redox potential (436 ± 10 mV) compared to Tyr34Phe, indicating a synonymous effect to the charge of the active site50,56. The neutron data for His30 and Tyr166 help explain past observations of MnSOD mutants that were perplexing and ties together changes of protonation state with kinetic and redox potential measurements.
The neutron structures of Mn3+SOD and Mn2+SOD have revealed active site states that have unique configurations of protonations and hydrogen bonds. Two of the configurations are the enzymatic resting states for Mn3+ and Mn2+ characterized by a five-coordinate Mn46. For the oxidized resting state that is described by both chains of the Mn3+SOD neutron structure (Fig. 6a), the proton bridging His30 and Tyr166 appears to be moving and this suggests the possibility of Tyr166 alternating between an ionized or protonated form and a deprotonated or protonated Nε2(His30). Nδ1(His30) is also observed to be both deprotonated or protonated (Fig. 5c, d). For simplicity, only one protonation form of His30 and Tyr166 is shown (Fig. 6a). An SSHB is seen between WAT2 and an ionized Tyr34 while Gln143 is in the canonical amide form hydrogen bonding to WAT1 as a OH molecule. For the reduced resting state described by chain B of Mn2+SOD (Fig. 6b), Tyr166 and His30 now share the proton originating from Nε2(His30) with a LBHB, His30 is protonated on Nδ1, and Tyr34 is no longer ionized. Gln143 and WAT1 have undergone a proton transfer where WAT1 is now H2O and Gln143 is an amide anion. Gln143 forms SSHBs with WAT1 and Trp123 presumably to stabilize its ionized form. The third active site state is a six-coordinate Mn2+ with OH-bound opposite Asp159 (Fig. 6c) described by chain A of Mn2+SOD. The protonations and SSHBs observed indicate an active site that is catalytically present before the five-coordinate reduced resting state but after a Mn3+ → Mn2+ redox transition. Tyr166 and His30 share a LBHB and His30 is Nδ1 protonated, like the reduced resting state, while Tyr34 is ionized like the oxidized resting state. Gln143 in the amide form but forms an SSHB with OH(WAT1) and also Trp123. In total, five sites undergo changes in protonation state, Tyr166, His30, Tyr34, Gln143, and WAT1 that occur with four instances of SSHBs between pairs Tyr166-His30 (a LBHB), WAT2-Tyr34, Gln143-WAT1, and Gln143-Trp123.
In total, the present work provides details for some of the critical aspects of the CPET mechanism of human MnSOD. Through neutron diffraction, direct evidence is observed for (1) an internal protonation mechanism via glutamine deprotonation to the Mn-bound solvent molecule (WAT1) supported by quantum calculations; (2) an SSHB of Trp123 with the anionic form of Gln143 that stabilizes the anion; (3) changes in protonation of Tyr34 that interacts intimately with a solvent molecule (WAT2) when ionized, and (4) alternate protonation states and a LBHB for His30 and Tyr166 across the dimer interface. It is evident that the residues of the active site have unusual pKas that are likely a consequence of a polarized environment provided by the metal and limited solvent accessibility. As a result of obtaining neutron structures for both Mn3+ and Mn2+ states, we built a suggested mechanism that details the changes of protonation states as a result of the Mn ion gaining or losing an electron. For the mechanism, the superoxide substrate is not present but is represented strictly as its donation or abstraction of electrons due to the lack of experimental evidence for its precise binding site.
Starting from the five-coordinate Mn3+ resting state (Fig. 7a), Mn3+ acquires an electron (in reality from the substrate superoxide) that coincides with Nδ1(His30) acquiring a proton from the nearest solvent molecule (the crystallographic position of WAT2) and Tyr166 gaining partial covalent character from the proton of Nε2(His30) to form a LBHB (Fig. 7b). The Mn2+ active site then binds OH(OL) to form a six-coordinate Mn2+ complex and may be the same solvent molecule that donated a proton to His30 (Fig. 7c). The depression of Mn2+ positive charge through OL binding causes negative polarization at WAT1 and triggers proton abstraction from Gln143. Consequently, the WAT1–Gln143 interaction is more stabilizing and the electronegative polarity now localizes to OL and departs by its protonation by either solvent or Tyr34 as suggested by the Miller group19. Regardless, Tyr34 becomes non-ionized before the formation of the five-coordinate, reduced resting state (Fig. 7d). When the substrate is present, the steps of Fig. 7a–c describe the first CPET where proton and electron transfers are coupled to an extent where they cannot be differentiated with kinetic measurements43. The second CPET involves substrate obtaining an electron from Mn2+ and protons from His30 and Tyr34 to form the H2O2 product (Fig. 7e). The changes of charge due to proton and electron departure from the active site causes Gln143 to accept the same proton it previously donated to WAT1 and Nε2(His30) ceasing its proton sharing with Tyr166 to regenerate five-coordinate Mn3+.
Altogether, the suggested mechanism utilizes two internal proton transfers where the protons move back-and-forth within the active site and two external proton transfers where the protons originate from solvent molecules to ultimately be consumed to form the product. The proton transfer between WAT1 and Gln143 is central to the mechanism as it permits the cyclic nature of catalysis. From this study, we have revealed, to our knowledge, the first direct coupling of electronic states to protonation states for an oxidoreductase. It is evident that the CPET mechanism of MnSOD is not straightforward as is exemplified by the previous elusiveness of the proton source for WAT1. Tyr34 was assumed to be the donor, but our data instead indicate an unusual and unexpected proton transfer from the Gln143 amide. Likewise, Trp123, His30, and Tyr166 were not assumed to be involved in catalysis. As this is just one biologically relevant oxidoreductase in a sea of many, finding the protonation states at the active sites of other oxidoreductases may reveal further unusual mechanisms for CPET.
## Methods
### Perdeuterated expression, purification, and crystallization
For deuterated protein expression of MnSOD, the pCOLADuet-1 expression vector harboring full-length cDNA of MnSOD was transformed into Escherichia coli BL21(DE3) cells. Transformed cells were grown in D2O minimal media within a bioreactor vessel using D8-glycerol as the carbon source58. Induction was performed with 1 mM isopropyl β-d-thiogalactopyranoside, 8 mM MnCl2, and fed D8-glycerol until an OD600 of 15.0. Expression was performed at 37 °C for optimal Mn ion metal incorporation59. Harvested cell pastes were stored at −80 °C until purification. For protein purification, cells were resuspended in a solution of 5 mM MnCl2 and 5 mM 3-(N-morpholino)propanesulfonic acid (MOPS), pH 7.8. Clarified lysate was incubated at 55 °C to precipitate contaminant proteins that were subsequently removed by centrifugation. Next, soluble protein was diluted with an equal volume of 50 mM 2-(N-morpholino)ethanesulfonic acid (MES) pH 5.5 yielding a final concentration of 25 mM. Measurement of pH verified a value of 5.5 after dilution. Protein was applied onto a carboxymethyl sepharose fast flow column (GE Healthcare) and eluted with a 0-500 mM sodium chloride gradient that contained 50 mM MES pH 6.5 before concentration to 23 mg mL−1. Crystallization was performed using a nine-well glass plate and sandwich box setup (Hampton Research) and the reservoir solution consisted of 1.9 M potassium phosphate adjusted to pH 7.8 by varying ratios of KH2PO4 and K2HPO4. The crystallization drop was a mixture of 60 µL concentrated protein solution (within a buffer of 50 mM MES pH 6.5) and 40 µL of the reservoir solution. Crystals grew up to 0.5 mm3 after 6 weeks at 23 °C. Purification and crystallization were performed with hydrogenated reagents.
### Redox manipulation of perdeuterated MnSOD crystals
Before redox manipulation, initial deuterium exchange of crystals was performed by vapor diffusion in capillaries using deuterated solutions of 2.3 M KD2PO4 and K2HPO4 at pH 7.4 (adjusted by varying ratios of dibasic and monobasic forms) that is the equivalent pD value of 7.8. The pD value was calculated from pD = pHa (apparent reading from pH meter) + 0.4. A crystal in a quartz capillary was soaked in deuterated solutions of 2.3 M KD2PO4/K2HPO4 at pD 7.8 (measured pH 7.4) containing either 6.4 mM potassium permanganate (KMnO4) to achieve the Mn3+ state or 300 mM sodium dithionite (Na2S2O4) to achieve the Mn2+ state. After drying the crystal from soaking solutions, the crystal was flanked in the capillary by slugs of the deuterated reservoir soaking solutions. Fortuitously, the decomposition products of the redox agents are unable to enter the active site of MnSOD18. Methods for manipulating the Mn metal of MnSOD to either Mn3+ or Mn2+ are described in more detail in previous work18.
### Neutron and X-ray data collection
Data collection was preceded by the replacement of the deuterated and redox-agent containing reservoir slugs with fresh equivalents. Time-of-flight, wavelength-resolved neutron Laue diffraction data were used to collect data from the 0.46 mm3 perdeuterated crystal using the MaNDi instrument60,61 at the ORNL SNS using all neutrons with wavelengths between 2-4 Å. Data collection of each diffraction pattern was from the crystal held in a stationary position, with successive diffraction patterns being collected after 20° rotations along the Φ axis. A KMnO4-treated perdeuterated crystal of 0.46 mm3 in volume at 296 K was recorded to 2.20 Å resolution for the Mn3+SOD form and subsequently treated with Na2S2O4 to achieve the Mn2+SOD state where 2.30 Å data were collected (Supplementary Table 6). Na2S2O4 is noted to deteriorate diffraction quality and was observed to increase the c unit cell axis by ~1 Å18. After neutron data were collected from the crystal in the Mn2+SOD state, X-ray diffraction data were collected at 296 K to 2.16 Å resolution using a Rigaku FR-E SuperBright home source. After room temperature data collection, the crystal was not suitable for Mn3+SOD data collection and a sister crystal grown from the same well was used instead for obtaining X-ray data to 1.87 Å resolution.
### Data processing and refinement
Neutron data were integrated with MANTID62 utilizing the three-dimensional profile fitting algorithm63. Integrated neutron data were scaled and wavelength-normalized using Lauenorm from the LAUGEN suite64. X-ray diffraction data were reduced using HKL-3000 for indexing, integration, and scaling65. The refinement of both X-ray and neutron models were completed separately with PHENIX.REFINE from the PHENIX suite66. The refinements were intentionally done separately due to the known perturbations that X-rays have on solvent structure and metal redox states which are not present with neutrons17,67. The X-ray model was first refined against its corresponding data set and subsequently used as the starting model for neutron refinement. Torsional backbone angle restraints were derived from the X-ray model and applied to neutron refinement using a geometric target function with PHENIX.REFINE66. The neutron refinement process was performed to model the D atoms of the active site last to limit phase bias. For the initial rounds of refinement to fit protein structure, only non-exchangeable D atoms (which have stereochemical predictable positions) were present. Afterwards, each individual exchangeable position outside the active site was inspected for residual |Fo|−|Fc| neutron scattering length density and modeled with D atoms appropriately before more iterations of refinement. Next, the O atoms of solvent molecules were first modeled manually outside the active site and refined to determining whether to model solvent as O, OD, or DOD using residual |Fo| − |Fc| neutron scattering length density. The omit density peaks were also used to discern the appropriate orientation of the solvent molecules. After refinement of the solvent structure outside the active site, non-D atoms of the active site were modeled, including Mn and the O of solvent. Last, D atoms of the active site were modeled and refined manually.
### Computational details
Computational methods are discussed with further depth in the Supplementary Methods. All QM DFT calculations were performed with the NWChem 6.8 software68. The COSMO solvation model was implemented into the geometry optimizations to model the solution phase and utilized an energy convergence value of 1E−6 atomic units69. The def2-TZVPD basis set was used for the Mn ion whereas the 6–31+G(d,p) Pople basis set was specifically used for all other atoms due to its use in predicting pKas under the B3LYP functional70,71. The QM models utilized for DFT calculations encompassed the active site residues that had the O and N atoms of the peptide backbone truncated and the Cα atoms fixed. Additional fixed restraints were placed on aromatic residues found on the periphery of the active site (Phe66, Trp123, Trp161, and Tyr166) to mimic the packing found in the native enzyme. The Mn ion used the high-spin quintet and sextet states for trivalent and divalent systems, respectively, per experimental observations72.
### Bonding orbital analysis
The JANPA software package was used to calculate CLPOs from open-shell DFT geometry optimizations37,38,73,74. These are bonding and antibonding orbitals with maximum electron density computed through a series of localized basis set transformations. CLPOs are calculated with the same target quantity as Natural Bond Orbital (NBO) methods and yield comparable results35,36,75. The electron delocalization stabilization/destabilization energies utilized second-order perturbation theory analysis of the Fock matrix defined by the NBO methods76 but were done in the CLPO basis of JANPA. The energy associated with electron delocalization from lone pair or bonding orbital i to antibonding orbital j is defined in Eq. (1) where qi is the donor orbital occupancy and $$\hat F$$ is the effective orbital Hamiltonian.76 Values $$\left\langle {j\left| {\hat F} \right|j} \right\rangle$$ and $$\left\langle {i\left| {\hat F} \right|i} \right\rangle$$ are diagonal CLPO Fock matrix elements indicative of orbital energies and $$\left\langle {i\left| {\hat F} \right|j} \right\rangle$$ is the off-diagonal matrix element representative of perturbation.
$${\Delta}E_{i \to j}^2 = - q_i\frac{{\left\langle {i\left| {\hat F} \right|j} \right\rangle}^2}{{\left\langle {j\left| {\hat F} \right|j} \right\rangle} - {{\left\langle {i\left| {\hat F} \right|i} \right\rangle }}}$$
(1) | 2022-12-03 21:25:24 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5413084626197815, "perplexity": 5851.84566003404}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710941.43/warc/CC-MAIN-20221203212026-20221204002026-00572.warc.gz"} |
https://myaptitude.in/science/in-case-of-negative-work-the-angle-between-the-force-and-displacement-is | # In case of negative work the angle between the force and displacement is
In case of negative work the angle between the force and displacement is
1. 45°
2. 90°
3. 180°
Work done, W = F.d cos θ
Work done at θ = 0°
W = F.d cos 0° = F.d
For angle θ = 0°, the work done is positive, so it is not true.
For angle θ = 45°, W = F.d/√2
In this case also, work done is positive, so it is not true.
Work done at θ = 90°, W = 0
So, it is not true.
Now, work done at θ = 180°, W = -F.d
For negative work, the angle between the force and displacement should be 180°.
The correct option is D. | 2020-07-14 04:42:39 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8462815880775452, "perplexity": 1144.285232823893}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593657147917.99/warc/CC-MAIN-20200714020904-20200714050904-00157.warc.gz"} |
https://www.gradesaver.com/textbooks/math/other-math/CLONE-547b8018-14a8-4d02-afd6-6bc35a0864ed/chapter-2-multiplying-and-dividing-fractions-2-2-mixed-numbers-2-2-exercises-page-124/55 | ## Basic College Mathematics (10th Edition)
To change a mixed number to an improper fraction you multiply the denominator and the whole number and add the numerator. i,e, $3\frac{1}{2} = (2\times3) +1 = 7$ | 2022-06-26 23:56:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8928946256637573, "perplexity": 1153.4040542885136}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103322581.16/warc/CC-MAIN-20220626222503-20220627012503-00016.warc.gz"} |
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