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https://www.gradesaver.com/textbooks/math/algebra/elementary-algebra/chapter-10-quadratic-equations-chapter-10-review-problem-set-page-468/28	[ "## Elementary Algebra\n\nWe call s the original number of shares bought and p the original price they were bought at. This allows us to create a system of linear equations: $sp = 720 \\\\ 800 = (s-20)(p+8)$ Subbing in $p = \\frac{720}{s}$ we find: $800 = sp -20p + 8s -160 \\\\ 800 = 720 -(\\frac{20 \\times720}{s}) + 8s - 160 \\\\ 0 = -240 - \\frac{14400}{s} + 8s \\\\ 0 = \\frac{-240s}{s} - \\frac{14400}{s} + \\frac{8s^2}{s} \\\\ 0 = 8s^2 - 240s - 14400 \\\\ 8(s-60)(s+30) \\\\s = 60$ Since 20 shares were not sold, this means that he sold 40 shares. Also: $p = \\frac{720}{60} = 12$ Since they increased in price by 8 dollars, this means they were sold at 20 dollars per share." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8549133,"math_prob":0.99993336,"size":708,"snap":"2019-51-2020-05","text_gpt3_token_len":262,"char_repetition_ratio":0.12357955,"word_repetition_ratio":0.01438849,"special_character_ratio":0.48446327,"punctuation_ratio":0.062068965,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997135,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-06T22:58:54Z\",\"WARC-Record-ID\":\"<urn:uuid:16055b69-8a84-4d5a-9131-2c2031231993>\",\"Content-Length\":\"54560\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:883a771e-616d-42aa-b059-895c9da2451c>\",\"WARC-Concurrent-To\":\"<urn:uuid:7e7c0ddc-8d15-4b71-bf72-4186dc4431ea>\",\"WARC-IP-Address\":\"52.87.77.102\",\"WARC-Target-URI\":\"https://www.gradesaver.com/textbooks/math/algebra/elementary-algebra/chapter-10-quadratic-equations-chapter-10-review-problem-set-page-468/28\",\"WARC-Payload-Digest\":\"sha1:PMHQPRT5KU5QD4JKMFFAV445C5LAKI44\",\"WARC-Block-Digest\":\"sha1:5IY5QQB5WIP6MXCJUIIREOEAMK53XWXS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540491491.18_warc_CC-MAIN-20191206222837-20191207010837-00519.warc.gz\"}"}
https://curiouschicksite.wordpress.com/2017/06/15/pytut-2-numbers-and-operators/	[ "# [PyTUT 2] NUMBERS AND OPERATORS\n\nHave you installed Python on your computer?\n\nFollowing to the previous tutorial, today I would like to introduce you the numbers and operators used in Python. Before beginning this tutorial, I want to notice again that all codes in the series of ‘Basic Python tutorials’ are written in Python 2.7.11. Thus, may be you have to change a little bit with your code if you use another version, especially version 3.\n\nLet’s take a look at the numbers and operators used in Python.\n\n## Numbers and operators\n\nPython treats the numbers in two main types: integer and float. The others are long and complex. This tutorial focuses on the first two number types. However, you can do the same tasks with the long and complex.\n\nJust like the other programming language, integer and float numbers can be used with the operators such as add ‘+’, subtract ‘-‘, multiply ‘’ and divide ‘/’. However, there are some different between these two types of numbers (you can see it soon).\n\nNow, let’s open IDLE (or your text editor, or even Command Prompt) and do basic math with Python:\n\n#### Example 1: The operators with two integers\n\n```>>> 9+4\n13```\n```>>> 9-4\n5```\n```>>> 94\n36```\n```>>> 9/4\n2```\n\n#### Example 2: The operators with two floats\n\n```>>> 9.0+4.0\n13.0```\n```>>> 9.0-4.0\n5.0```\n```>>> 9.0*4.0\n36.0```\n```>>> 9.0/4.0\n2.25```\n\nAfter the examples, we can see that the significant different is showed when you used the division operator ‘/’. When two integer are divided, the result is always an integer (see Ex. 1). If you want to determine the remainder of 9/4, you will need the remainder operator ‘%’ (this operator is useful in some cases that I will mention it later in another tutorial).\n\n```>>> 9%4\n1```\n\nNote: If you use the operators with an integer and a float, Python will understand that you want to use the operators with two floats. For example:\n\n```>>> 9/2.0\n4.5```\n\n## Number type conversion\n\nSometimes, you expect to change the type of number to satisfy requirements of an operator or an function parameter. Here are some ways:\n\nTo convert a number x to integer, we use the following structure:\n\n`>>> int(x)`\n\n#### Example 3: Convert a complex number 4.0 to integer\n\n```>>> int(4.0)\n4```\n\nSimilar to this, Python supports the structures that help you to convert a number to\n\nfloat type\n\n`>>> float(x)`\n\nlong type\n\n`>>> long(x)`\n\ncomplex type ( where x is real part and y is imaginary part)\n\n`>>> complex(x)`\n`>>> complex(x, y)`\n\nNote: You cannot convert complex to int\n\nThat is all about basic math with Python. It is simple, isn’t it?\n\nIn the next tutorial, I will talk about string in Python and some ways you can do to treat strings.\n\nHope you enjoy it,\n\nCurious Chick", null, "" ]	[ null, "https://2.gravatar.com/avatar/598a1c3cf32d141d3fbddda849d4dc1c", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86718595,"math_prob":0.8378976,"size":2552,"snap":"2019-13-2019-22","text_gpt3_token_len":654,"char_repetition_ratio":0.14678179,"word_repetition_ratio":0.013157895,"special_character_ratio":0.27351096,"punctuation_ratio":0.12844037,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99151945,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-22T23:14:17Z\",\"WARC-Record-ID\":\"<urn:uuid:1cb3a4f0-d1a1-437e-aaf9-4a7717d1f9ed>\",\"Content-Length\":\"57323\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2dfaf677-87cb-41b6-872f-3c470ce7f1d6>\",\"WARC-Concurrent-To\":\"<urn:uuid:48585e84-18ac-4236-a906-84b80dfbd72f>\",\"WARC-IP-Address\":\"192.0.78.13\",\"WARC-Target-URI\":\"https://curiouschicksite.wordpress.com/2017/06/15/pytut-2-numbers-and-operators/\",\"WARC-Payload-Digest\":\"sha1:JTIAND67KWAVL3JWTAH76VKFI36S7RCF\",\"WARC-Block-Digest\":\"sha1:ZN6BIUVV5G6XPBSWOG6W4YQOWLENRV3Y\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202698.22_warc_CC-MAIN-20190322220357-20190323002357-00514.warc.gz\"}"}
https://ezpassdrjtbc.net/lively-identify-the-properties-of-mathematics/	[ "# Lively identify the properties of mathematics Wonderful\n\n» » Lively identify the properties of mathematics Wonderful\n\nYour Lively identify the properties of mathematics images are ready in this website. Lively identify the properties of mathematics are a topic that is being searched for and liked by netizens today. You can Get the Lively identify the properties of mathematics files here. Find and Download all royalty-free images.\n\nIf you’re searching for lively identify the properties of mathematics images information connected with to the lively identify the properties of mathematics keyword, you have pay a visit to the ideal site. Our site always provides you with suggestions for refferencing the highest quality video and image content, please kindly hunt and locate more enlightening video content and graphics that fit your interests.\n\nLively Identify The Properties Of Mathematics. Identifying Properties of Math Worksheet. The detailed step-by-step solutions will help you understand the concepts better and clear your. An edge is where two faces meet. 1 is called the multiplicative identity.", null, "Name That Property An Open Ended Identification Activity High School Math Math Lessons Teaching Algebra From pinterest.com\n\nLernsyss downloadable free math practice problems and worksheets are a great aid to develop your childs math skills. This will clear students doubts about any question and improve application skills while preparing for board exams. Reflexive if a a Ra for all a A irreflexive if a a R for all a A symmetric if a b R b a R for all a b A antisymmetric if a b R b a R a b for all a b A transitive if a b R b c R a c R for all a b c A. Example of the commutative property of addition 3 5 5 3 8. A b c a b c. The identity property of multiplication.\n\n### 04092013 Identify the Properties of Mathematics 1 When two numbers are multiplied together the product is the same regardless of the order of the multiplicands.\n\nA face is a flat 2D shape. A b c a b c. Upon completion of this video you can be able to-Recall and describe a tree-List describe and identify the different properties of a tree-List describe an. 6 July 2020 Added a link to further guidance on teaching mathematics at. You should know the definition of each of the following properties of addition and how each can be used. For addition the rule is.", null, "Source: pinterest.com\n\nGoogle Classroom Facebook Twitter. For multiplication the rule is. If you add two even numbers the answer is still an even number 2 4 6. A 1 a 1 a a. Identifying Properties of Math Worksheet.", null, "Source: pinterest.com\n\n21122020 13 0 14 0 0 3x 13 14 3x. Therefore the set. This property holds true for addition and multiplication but not for subtraction and division. Identifying Properties of Math Worksheet. The identity property of multiplication.", null, "Source: pinterest.com\n\nFor any real number a a 0 a 0 a a. Identifying Properties of Math Worksheet. You should know the definition of each of the following properties of addition and how each can be used. For any real number a. 1 is called the multiplicative identity.", null, "Source: pinterest.com\n\nLernsyss downloadable free math practice problems and worksheets are a great aid to develop your childs math skills. Example of the commutative property of addition 3 5 5 3 8. A vertex is a corner. 0 is called the additive identity. For example a x b b x a 2 When two numbers are added the sum is the same regardless of the order of the addends.", null, "Source: pinterest.com\n\n04092013 Identify the Properties of Mathematics 1 When two numbers are multiplied together the product is the same regardless of the order of the multiplicands. Upon completion of this video you can be able to-Recall and describe a tree-List describe and identify the different properties of a tree-List describe an. The identity property of multiplication. 0 is called the additive identity. 1 is called the multiplicative identity.", null, "Source: pinterest.com\n\nClosure is when all answers fall into the original set. 1 associative 2 commutative. A prism is a 3D shape thats cross section is the same throughout. 0 is called the additive identity. For any real number a.", null, "Source: pinterest.com\n\nTherefore the set. For any real number a a 0 a 0 a a. A face is a flat 2D shape. A vertex is a corner. The detailed step-by-step solutions will help you understand the concepts better and clear your.", null, "Source: pinterest.com\n\nFor any real number a. Identifying Properties of Math Worksheet. In numbers this means 2 3 4 2 3 4. Is not a pleasant thing to look at so the first thing that we realize and this is the logarithm one of our logarithm properties this logarithm for a given base so lets say that the base is X of a over B that is. An edge is where two faces meet.", null, "Source: pinterest.com\n\nIf you add two even numbers the answer is still an even number 2 4 6. For example a x b b x a 2 When two numbers are added the sum is the same regardless of the order of the addends. This property holds true for addition and multiplication but not for subtraction and division. Reflexive if a a Ra for all a A irreflexive if a a R for all a A symmetric if a b R b a R for all a b A antisymmetric if a b R b a R a b for all a b A transitive if a b R b c R a c R for all a b c A. Selina solutions for Class 6 Mathematics chapter 25 Properties of Angles and Lines Including Parallel Lines include all questions with solution and detail explanation.", null, "Source: pinterest.com\n\nThe identity property of multiplication. For multiplication the rule is. Identifying Properties of Math Worksheet. Intro to logarithm properties 1 of 2. The identity property of multiplication.", null, "Source: pinterest.com\n\n1 associative 2 commutative. 1 is called the multiplicative identity. 21122020 A relation R on A is said to be. Welcome to this presentation on logarithm properties now this is going to be a very hands-on presentation if you dont believe that one of these properties are true and you want them proved Ive made three or four videos that actually prove these properties but what Im going to do is Im going to show you the properties and then show you how they can be used so gonna be a little. Example of the commutative property of addition 3 5 5 3 8.", null, "Source: pinterest.com\n\nWelcome to this presentation on logarithm properties now this is going to be a very hands-on presentation if you dont believe that one of these properties are true and you want them proved Ive made three or four videos that actually prove these properties but what Im going to do is Im going to show you the properties and then show you how they can be used so gonna be a little. Selina solutions for Class 6 Mathematics chapter 25 Properties of Angles and Lines Including Parallel Lines include all questions with solution and detail explanation. You should know the definition of each of the following properties of addition and how each can be used. 1 associative 2 commutative. Closure is when all answers fall into the original set.", null, "Source: pinterest.com\n\nFor addition the rule is. This will clear students doubts about any question and improve application skills while preparing for board exams. You should know the definition of each of the following properties of addition and how each can be used. Reflexive if a a Ra for all a A irreflexive if a a R for all a A symmetric if a b R b a R for all a b A antisymmetric if a b R b a R a b for all a b A transitive if a b R b c R a c R for all a b c A. The identity property of addition.", null, "Source: pinterest.com\n\n1 is called the multiplicative identity. Calculate angles in a triangle or around a point Classify different polygons and understand whether a 2D shape is a polygon or not Estimate. 0 is called the additive identity. In numbers this means 2 3 4 2 3 4. A 1 a 1 a a.", null, "Source: pinterest.com\n\nIntro to logarithm properties 1 of 2. 21122020 13 0 14 0 0 3x 13 14 3x. 21122020 A relation R on A is said to be. An edge is where two faces meet. If you add two even numbers the answer is still an even number 2 4 6.\n\nThis site is an open community for users to do submittion their favorite wallpapers on the internet, all images or pictures in this website are for personal wallpaper use only, it is stricly prohibited to use this wallpaper for commercial purposes, if you are the author and find this image is shared without your permission, please kindly raise a DMCA report to Us.\n\nIf you find this site helpful, please support us by sharing this posts to your preference social media accounts like Facebook, Instagram and so on or you can also save this blog page with the title lively identify the properties of mathematics by using Ctrl + D for devices a laptop with a Windows operating system or Command + D for laptops with an Apple operating system. If you use a smartphone, you can also use the drawer menu of the browser you are using. 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https://haskell.libhunt.com/ad-alternatives	[ "Popularity\n9.8\nStable\nActivity\n5.4\nDeclining\n341\n24\n56\n\nTags: Math\n\n# ad alternatives and similar packages\n\nBased on the \"Math\" category.\nAlternatively, view ad alternatives based on common mentions on social networks and blogs.\n\n• ### vector\n\nAn efficient implementation of Int-indexed arrays (both mutable and immutable), with a powerful loop optimisation framework .\n\nType safe interface for working in subcategories of Hask\n• ### Build time-series-based applications quickly and at scale.\n\nInfluxDB is the Time Series Platform where developers build real-time applications for analytics, IoT and cloud-native services. Easy to start, it is available in the cloud or on-premises.\n• ### hmatrix\n\nLinear algebra and numerical computation\n• ### linear\n\nLow-dimensional linear algebra primitives for Haskell.\n• ### statistics\n\nA fast, high quality library for computing with statistics in Haskell.\n• ### HerbiePlugin\n\nGHC plugin that improves Haskell code's numerical stability\n• ### what4\n\nSymbolic formula representation and solver interaction library\n• ### hgeometry\n\nHGeometry is a library for computing with geometric objects in Haskell. It defines basic geometric types and primitives, and it implements some geometric data structures and algorithms. The main two focusses are: (1) Strong type safety, and (2) implementations of geometric algorithms and data structures that have good asymptotic running time guarantees.\n• ### units\n\nThe home of the units Haskell package\n• ### grid\n\nTools for working with regular grids/graphs/lattices.\n• ### algebra\n\nconstructive abstract algebra\n• ### computational-algebra\n\nGeneral-Purpose Computer Algebra System as an EDSL in Haskell\n\n• ### dimensional\n\nDimensional library variant built on Data Kinds, Closed Type Families, TypeNats (GHC 7.8+).\n\n• ### estimator\n\nState-space estimation algorithms and models\n• ### mwc-random\n\nA very fast Haskell library for generating high quality pseudo-random numbers.\n• ### matrix\n\nA Haskell native implementation of matrices and their operations.\n• ### hblas\n\nhaskell bindings for blas and lapack\n\nA haskell numeric prelude, providing a clean structure for numbers and operations that combine them.\n• ### lambda-calculator\n\nAn introduction to the Lambda Calculus\n• ### vector-space\n\nVector & affine spaces, linear maps, and derivatives\n\n• ### math-functions\n\nSpecial mathematical functions\n• ### vector-sized\n\nSize tagged vectors\n\n• ### cf\n\n\"Exact\" real arithmetic for Haskell using continued fractions (Not formally proven correct)\n\n• ### bayes-stack\n\nFramework for Gibbs sampling of probabilistic models\n• ### poly\n\nFast polynomial arithmetic in Haskell (dense and sparse, univariate and multivariate, usual and Laurent)\n• ### optimization\n\nSome numerical optimization methods implemented in Haskell\n• ### rampart\n\n:european_castle: Determine how intervals relate to each other.\n\n• ### bed-and-breakfast\n\nMatrix operations in 100% pure Haskell\n• ### sbvPlugin\n\nFormally prove properties of Haskell programs using SBV/SMT.\n\n• ### safe-decimal\n\nSafe and very efficient arithmetic operations on fixed decimal point numbers\n• ### type-natural\n\nType-level well-kinded natural numbers.\n• ### simple-smt\n\nA simple way to interact with an SMT solver process.\n• ### intervals\n\nInterval Arithmetic\n• ### Decimal\n\nDecimal numbers with variable precision\n• ### polynomial\n\nHaskell library for manipulating and evaluating polynomials\n• ### monoid-subclasses\n\nSubclasses of Monoid with a solid theoretical foundation and practical purposes\n• ### eigen\n\n7.9 0.0 L2 ad VS eigen\nHaskel binding for Eigen library. Eigen is a C++ template library for linear algebra: matrices, vectors, numerical solvers, and related algorithms.\n• ### simd\n\nsimple interface to ghc's simd vector support\n• ### dimensions\n\nMany-dimensional type-safe numeric ops\n• ### mltool\n\nMachine Learning Toolbox\n• ### vector-th-unbox\n\nDeriver for unboxed vectors using Template Haskell" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7227323,"math_prob":0.76246107,"size":5539,"snap":"2022-40-2023-06","text_gpt3_token_len":1475,"char_repetition_ratio":0.14634146,"word_repetition_ratio":0.0,"special_character_ratio":0.23271349,"punctuation_ratio":0.14836223,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97852737,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-06T17:12:17Z\",\"WARC-Record-ID\":\"<urn:uuid:34cfdfb1-272d-499f-a32d-d789cd89a8dd>\",\"Content-Length\":\"87540\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:97933377-9195-4c73-8f51-8294f70aedd5>\",\"WARC-Concurrent-To\":\"<urn:uuid:297e524f-fdd3-4070-9cdc-34381ff18ffe>\",\"WARC-IP-Address\":\"172.66.41.34\",\"WARC-Target-URI\":\"https://haskell.libhunt.com/ad-alternatives\",\"WARC-Payload-Digest\":\"sha1:27A5GMTNEVGPQYGFY7O63SHSUN7V3D7S\",\"WARC-Block-Digest\":\"sha1:OW2HYCO7YSBD6QQRM3CJXHAM2LVJGHTV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500356.92_warc_CC-MAIN-20230206145603-20230206175603-00002.warc.gz\"}"}
https://matem.anrb.ru/en/article?art_id=897	[ "# Article\n\nUfa Mathematical Journal\nVolume 14, Number 3, pp. 33-40\n\n# Application of generating functions to problems of random walk\n\nGrishin S.V.\n\nDOI:10.13108/2022-14-3-33\n\nWe consider a problem on determining the first hit time of the positive semi-axis under a homogenous discrete integer random walk on a line. More precisely, the object of our study is the graph of the generating function of the mentioned random variable. For the random walk with the maximal positive increment $1$, we obtain the equation on the implicit generating function, which implies the rationality of the inverse generating function. In this case, we find the mathematical expectation and dispersion for the first hit time of a positive semi-axis under a homogenous discrete integer random walk on a line. We describe a general method for deriving systems of equations for the first hit time of a positive semi-axis under a homogenous discrete integer random walk on a line. For a random walk with increments $-1$, $0$, $1$, $2$ we derive an algebraic equation for the implicit generating function. We prove that a corresponding planar algebraic curve containing the graph of generating function is rational. We formulate and prove several general properties of the generating function the first hit time of the positive semi-axis under a homogenous discrete integer random walk on a line." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.77943367,"math_prob":0.99421895,"size":1403,"snap":"2023-40-2023-50","text_gpt3_token_len":296,"char_repetition_ratio":0.15511079,"word_repetition_ratio":0.26605505,"special_character_ratio":0.20741269,"punctuation_ratio":0.0859375,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9961897,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-30T05:39:57Z\",\"WARC-Record-ID\":\"<urn:uuid:ba421b2d-064f-4552-a5dc-71ca7a95d02c>\",\"Content-Length\":\"20777\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:97e2a6fd-5f6b-42bc-9c9c-79387ea88990>\",\"WARC-Concurrent-To\":\"<urn:uuid:a570bde7-5281-4900-b5bc-db84378e4f54>\",\"WARC-IP-Address\":\"212.193.134.154\",\"WARC-Target-URI\":\"https://matem.anrb.ru/en/article?art_id=897\",\"WARC-Payload-Digest\":\"sha1:CEIZFHUJGQK7ATAXABB7YTFXGNWRM5DY\",\"WARC-Block-Digest\":\"sha1:2QLTD6WAGINYOGF232XVYN5PQ47HR4SY\",\"WARC-Truncated\":\"disconnect\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510603.89_warc_CC-MAIN-20230930050118-20230930080118-00462.warc.gz\"}"}
https://sicstus.sics.se/sicstus/docs/latest3/html/sicstus.html/CLPB.html	[ "32 Boolean Constraint Solver\n\nThe clp(B) system provided by this library module is an instance of the general Constraint Logic Programming scheme introduced in [Jaffar & Michaylov 87]. It is a solver for constraints over the Boolean domain, i.e. the values 0 and 1. This domain is particularly useful for modeling digital circuits, and the constraint solver can be used for verification, design, optimization etc. of such circuits.\n\nTo load the solver, enter the query:\n\n``` \| ?- use_module(library(clpb)).\n```\n\nThe solver contains predicates for checking the consistency and entailment of a constraint wrt. previous constraints, and for computing particular solutions to the set of previous constraints.\n\nThe underlying representation of Boolean functions is based on Boolean Decision Diagrams [Bryant 86]. This representation is very efficient, and allows many combinatorial problems to be solved with good performance.\n\nBoolean expressions are composed from the following operands: the constants 0 and 1 (`FALSE` and `TRUE`), logical variables, and symbolic constants, and from the following connectives. P and Q are Boolean expressions, X is a logical variable, Is is a list of integers or integer ranges, and Es is a list of Boolean expressions:\n\n`~ `P\nTrue if P is false.\nP` * `Q\nTrue if P and Q are both true.\nP` + `Q\nTrue if at least one of P and Q is true.\nP` # `Q\nTrue if exactly one of P and Q is true.\nX` ^ `P\nTrue if there exists an X such that P is true. Same as P`[`X`/0] + `P`[`X`/1]`.\nP` =:= `Q\nSame as `~`P` # `Q.\nP` =\\= `Q\nSame as P` # `Q.\nP` =< `Q\nSame as `~`P` + `Q.\nP` >= `Q\nSame as P` + ~`Q.\nP` < `Q\nSame as `~`P` * `Q.\nP` > `Q\nSame as P` * ~`Q.\n`card(`Is`, `Es`)`\nTrue if the number of true expressions in Es is a member of the set denoted by Is.\n\nSymbolic constants (Prolog atoms) denote parametric values and can be viewed as all-quantified variables whose quantifiers are placed outside the entire expression. They are useful for forcing certain variables of an equation to be treated as input parameters." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8618636,"math_prob":0.92847234,"size":1590,"snap":"2019-26-2019-30","text_gpt3_token_len":372,"char_repetition_ratio":0.119798236,"word_repetition_ratio":0.0,"special_character_ratio":0.23584905,"punctuation_ratio":0.13651878,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9966675,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-17T23:20:47Z\",\"WARC-Record-ID\":\"<urn:uuid:8b6fbce0-9af4-45dd-88d0-37be1445c8bd>\",\"Content-Length\":\"6315\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1481bf60-fb6e-4d61-8700-6a9ead4c95dc>\",\"WARC-Concurrent-To\":\"<urn:uuid:05d39cd9-32c1-4994-ad26-b59a19da2652>\",\"WARC-IP-Address\":\"193.10.64.42\",\"WARC-Target-URI\":\"https://sicstus.sics.se/sicstus/docs/latest3/html/sicstus.html/CLPB.html\",\"WARC-Payload-Digest\":\"sha1:B2KDBBGXNXOWPFCBAT5B3MXAFJH2TJ4Z\",\"WARC-Block-Digest\":\"sha1:TSNCOA73SLDPQRNTN4VTPEPJZHIIYTVG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627998581.65_warc_CC-MAIN-20190617223249-20190618005249-00179.warc.gz\"}"}
http://socialwifi.se/assassin-s-rfddaow/69yecp.php?5a5acf=white-kettle-delonghi	[ "Notice: Undefined index: in /opt/www/vs08146/web/domeinnaam.tekoop/assassin-s-rfddaow/69yecp.php on line 3 Notice: Undefined index: in /opt/www/vs08146/web/domeinnaam.tekoop/assassin-s-rfddaow/69yecp.php on line 3 Notice: Undefined index: in /opt/www/vs08146/web/domeinnaam.tekoop/assassin-s-rfddaow/69yecp.php on line 3 Notice: Undefined index: in /opt/www/vs08146/web/domeinnaam.tekoop/assassin-s-rfddaow/69yecp.php on line 3 white kettle delonghi\nSo we can state the … Property Characteristics • Transitive Property • Symmetric Property • Functional Property • inverseOf 38 ... Congruence of triangles (reflexive, symetric, transitive) Akhilesh Bhura. Anyone can earn courses that prepare you to earn 5.1 Classifying Triangles. 1/4 1/3 2/3 4/9 Weegy: The perimeter of the larger polygon is 50. stickman\|Points 6262\| User: A rectangular picture has dimensions 2½ by 1½. Point out that the flow proof uses the three lessons in math, English, science, history, and more. 0. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. 43. Play this game to review Geometry. Properties of congruence and equality Learn when to apply the reflexive property, transitive, and symmetric properties in geometric proofs. For triangles, all the interior angles of similar triangles are congruent, because similar triangles have the same shape but different lengths of sides. imaginable degree, area of If we compare the side lengths of one triangle to another, we will find that all the sides are proportional. Does congruence of triangles have the reflexive property? The three properties of congruence are the reflexive property of congruence, the symmetric property of congruence, and the transitive property of congruence. We say that a six-year-old boy is similar to a 18-year-old adult man. Plus, get practice tests, quizzes, and personalized coaching to help you for all a, b, c ∈ X, if a R b and b R c, then a R c.. Or in terms of first-order logic: ∀,, ∈: (∧) ⇒, where a R b is the infix notation for (a, b) ∈ R.. The Transitive Property for three things is illustrated in the above figure. Get better grades with tutoring from top-rated professional tutors. Mathematics. Therefore, I can make the connection that triangle A is also similar to triangle C. This transitive property can help us to solve problems involving similar triangles. Learn the relationship … Transitive Property (for three segments or angles): If two segments (or angles) are each congruent to a third segment (or angle), then they’re congruent to each other. We've learned that similar triangles are triangles whose only difference is size. Properties of Triangles (Geometry) STUDY. Let $${\\cal T}$$ be the set of triangles that can be drawn on a plane. a … In geometry, transitive property, for any three geometrical measurements, sides or angles, is defined as, “If two segments (or angles) are each congruent with a third segment (or angle), then they are congruent with each other”. Subjects Near Me. These properties can be applied to segment, angles, triangles, or any other shape. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. The Transitive Axiom PARGRAPH The second of the basic axioms is the transitive axiom, or transitive property. 0 times. Since the sum of the angles for any triangle is 180°: This is true for any equilateral triangle. Choose from 19 different sets of term:properties triangles = reflexive, symmetric, transitive flashcards on Quizlet. Distributive Property of Multiplication (1) Division Property of Equality (2) Division Tricks (1) Domain and Range of a Function (1) Double Integrals (4) Eigenvalues and Eigenvectors (1) Ellipse (1) Empirical and Molecular Formula (2) Enthalpy Change (2) Equilateral Triangles (1) Euler (1) Expected Value (3) Exponential Distribution (1) Triangles can be similar. How fast is her shadow moving when she is 30 ft from the, Two hills A and C have elevations of 600 m and 800 m respectively. Modeling Webinar: State of the Union for Data Innovation - 2016 DATAVERSITY. Here are two examples from geometry. Earn Transferable Credit & Get your Degree, Reflexive Property of Congruence: Definition & Examples, Midpoint Theorem: Definition & Application, Proving That a Quadrilateral is a Parallelogram, 30-60-90 Triangle: Theorem, Properties & Formula, GRE Quantitative Reasoning: Study Guide & Test Prep, SAT Subject Test Mathematics Level 1: Practice and Study Guide, NY Regents Exam - Integrated Algebra: Help and Review, NY Regents Exam - Integrated Algebra: Tutoring Solution, High School Geometry: Homework Help Resource, Ohio Graduation Test: Study Guide & Practice, Praxis Mathematics - Content Knowledge (5161): Practice & Study Guide, SAT Subject Test Chemistry: Practice and Study Guide. 7.1 Parallelograms. Triangle I has sided with lengths 2 \\frac{2}{3} cm, 4 \\frac{1}{2} cm \\text{ and } 3 cm. Working Scholars® Bringing Tuition-Free College to the Community, Describe the meaning of similar triangles, Determine how to use the transitive property to prove the only difference between similar triangles is size. Transitive Property The transitive property of equality is defined as, “Let a, b and c be any three elements in set A, such that a=b and b=c, then a=c”. So, if the top angle of one triangle equals 45 degrees, then the top angles of all other similar triangles will also equal 45 degrees. The chapter will explore the transitive property meaning, transitive property of equality, transitive property of angles, and transitive property of inequality. Objects are congruent if they are the same shape and size. {{courseNav.course.topics.length}} chapters \| Dozens of Hong Kong pro-democracy figures arrested. The only difference is the length of their sides. Quora. The transitive property is also known as the transitive property of equality. Transitive Property The transitive property of equality is defined as, “Let a, b and c be any three elements in set A, such that a=b and b=c, then a=c”. It states that if two quantities are … Compare the ratios of the two hypotenuses: If the other sides have the same proportion, the two right triangles are similar. Watch as we apply the transitive property to three similar triangles. For two similar equilateral triangles, all interior angles will be 60°. This geometry video tutorial provides a basic introduction into the transitive property of congruence and the substitution property of equality. One example is algebra. If whenever object A is related to B and object B is related to C, then the relation at that end transitive provided object A is also related to C. Being a child is a transitive relation, being a parent is not. Define a relation $$S$$ on $${\\cal T}$$ such that $$(T_1,T_2)\\in S$$ if and only if the two triangles are similar. flashcard set{{course.flashcardSetCoun > 1 ? 0 times. Let's see how this works. Triangle. Distributive property of multiplication worksheet - I. Distributive property of multiplication worksheet - II . Thus, triangle PQR is congruent to triangle ABC. For example, DB ≅ DB. A homogeneous relation R on the set X is a transitive relation if,. After completion of this lesson, you should be able to: To unlock this lesson you must be a Study.com Member. As a nonmathematical example, the relation \"is an ancestor of\" is transitive. Draw a triangle, △CAT. Prove the Transitive Property of Similarity. What have we learned? I'm writing a two column proof and I'm stuck on my last half. (Click here for the full version of the transitive property of inequalities.) Congruence of triangles (reflexive, symetric, transitive). In Mathematics, Transitive property of relationships is one for which objects of a similar nature may stand to each other. Because if triangle A is similar to triangle B and triangle B is similar to triangle C, then that means the only difference between all three triangles is their size and they are all similar to each other. If AB, BC, and AC are 7 inches, 9 inches, and 10 inches respectively, and XY is 9 inches. If a = b and b = c, then a = c. Addition Postulate . Unit 5 - Triangles and Congruence. Want to see the math tutors near you? credit-by-exam regardless of age or education level. The sides of the small one are 3, 4, and 5 cm long. User: The transitive property holds for similar figures. the symmetric property? - Definition, Shapes & Angles, Biological and Biomedical Watch this video lesson and you will understand what the transitive property is and how it applies to similar triangles. Draw a triangle similar to △CAT and call it △DOG. If $$a$$, $$b$$, and $$c$$ are three numbers such that $$a$$ is equal to $$b$$ and $$b$$ is equal to $$c$$, then $$a$$ and $$c$$ are equal to each other. the transitive property? Wörterbuch der deutschen Sprache. Triadic closure is the property among three nodes A, B, and C (representing people, for instance), that if the connections A-B and B-C exist, there is a tendency for the new connection A-C to be formed. In algebra the transitive property is used to help you solve problems. study Angle properties, postulates, and theorems \| wyzant resources. If △CAT is similar to △DOG, and △DOG is similar to △ELK, then △CAT and △ELK are similar to each other. Making sure to write the similarity statement congruent angles corresponding, we can say. For each part of this proof, the key is to find a way to get two pairs of congruent angles which will allow you to use AA Similarity Postulate.As you try these, remember that you already know that these three properties already hold for congruent triangles and can use these relationships in your Transitive Property = = Corresponding sides of similar triangles are proportional (ac) = (ac) a 2 = ce (bc) = (bc) b 2 = cd: Multiplication Property of Equality: a 2 + b 2 = ce + b 2: Additive Property of Equality: a 2 + b 2 = ce + cd: Substitution Property of Equality: a 2 + b 2 = c(e+d) Distributive Property … True Weegy: intransitive MissEb\|Points 105\| User: The sides of a polygon are 3, 5, 4, and 6. Get help fast. tuttona. Objects are similar to each other if they have the same shape but are different in size. Theorems concerning triangle properties Properties of congruence and equality Learn when to apply the reflexive property, transitive, and symmetric properties in geometric proofs. So if the left side of one triangle compared to the left side of another triangle is twice as big, then all the other sides of the one triangle will also be twice as big as the corresponding sides of the other triangle. A polygon with three sides. Definition types formulas of triangles. In order to prove that the two triangles are congruent by ASA, what must you know about the measure of ∠ D ? If a ≅ b, then b ≅ a (we switch their spots) Transitive Property. Pages 25-29 Pages 127-129 … Thus, if perhaps two triangles share a side and you wish to prove those two triangles congruent using the SSS method, it is necessary to cite the reflexive property of segments to conclude that the shared side is equal in both triangles. Reflexive Property of Equality c. Transitive Property of Congruence EXAMPLE 1 Name Properties of Equality and Congruence In the diagram, N is the midpoint of MP&, and P is the midpoint of NQ&. All other trademarks and copyrights are the property of their respective owners. Now draw a triangle labeled △ELK that is similar to △DOG. Definition of Angle Bisector: The ray that divides an angle into two congruent angles. I don't know if this one step is the transitive property of congruence. Decimal place value worksheets. Get the unbiased info you need to find the right school. The proportionalities involved are: AB / RS = BC / ST = AC / RT and all possible rearrangements using the various properties of proportions. The Transitive Property of Congruence. Tags: Question 4 . Please try another search. 6.2 Special Segments in Triangles. According to this substitution property definition, if two geometric objects (it can be two angles, segments, triangles, or whatever) are congruent, then these two geometric objects can be replaced with one other in a statement involving one of them. 5.2 Measuring Angles in Triangles. Transitive Property. In other words, if , and , then . Show that MN 5 PQ. Try refreshing the page, or contact customer support. If two angles of one triangle are equal to two angles of another triangle, then the triangles are similar. Condition Property. A triangle with one right angle. So we can write the entire similarity and congruence in mathematical notation: Knowing that for any objects, geometric or real, Z ~ A and A ~ P tells us that Z ~ P. But how can we use that information? ... Properties of triangle worksheet. The transitive property is also known as the transitive property of equality. The Transitive Axiom PARGRAPH The second of the basic axioms is the transitive axiom, or transitive property. We draw triangle A and then we draw a triangle B that is similar to triangle A. Thus, if perhaps two triangles share a side and you wish to prove those two triangles congruent using the SSS method, it is necessary to cite the reflexive property of segments to conclude that the shared side is equal in both triangles. Sciences, Culinary Arts and Personal A triangle with 3 acute angles. These properties can be applied to segment, angles, triangles, or any other shape. Also, all the corresponding angles will be equal to each other. Play this game to review Geometry. 9th grade. and career path that can help you find the school that's right for you. The transitive property of congruence states that two objects that are congruent to a third object are also congruent to each other. The transitive property states that if a=b and b=c, then a=c. If we wanted to find out the measurement of the top angle of triangle A and we are given that the top angle of triangle C is 50 degrees, we can use the transitive property and say that triangle A is also similar to triangle C and therefore the top angle of triangle A is also 50 degrees. What is a triangle and its properties? Transitive Property of Congruent Triangles If ABC DEF and DEF JKL then from DTYUJHKL 789 at Hispanic American College If we wanted to find the angle measurements of triangle A, we can use the angle measurements of triangle C by applying the transitive property. Integers and absolute value worksheets. Select a subject to preview related courses: Applying the transitive property, we can say that triangle A is similar to triangle C. Why can we say that? Also, since DE≅DF, ∠E≅∠F, so by the transitive property, ∠D≅∠E≅∠F. What is the missing piece of information required to prove these triangles congruent? Using Transitive Property of Congruent Triangles : By Transitive property of congruent triangles, if ΔPQR ≅ ΔMQN and ΔMQN ≅ ΔABC, then ΔPQR ≅ ΔABC. What Is The Transitive Property of Congruence? By watching the video and reading the lesson, you now are able to explain the difference between congruent and similar, and define the transitive property of congruence, which states that two objects that are congruent to a third object, they are congruent to each other. Congruent Triangles DRAFT. If △Z has an angle opposite the shortest side of 37°, △A also has an angle opposite its shortest side of 37° because we said △Z~ △A. Properties of equilateral triangles Equilateral triangles are regular polygons 6.4 Inequalities in a Triangle . Transitive Property of Equality. 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Better grades with tutoring from top-rated professional tutors 's Assign lesson Feature shape but are different in size are! Melman from the set a tall walks away from the stuff given above, if you need find... One are 3, 4, and, then a=c triangles equilateral are. Of their sides c, then b ≅ a sides are proportional to each other and have same... Better grades with tutoring from top-rated professional tutors is size T } \\ ) be the set of (! Someone special thousands off your degree master 's degree in secondary education and has taught at. If we compare the ratios of the angles for any triangle is cm. You are given a triangle b and b = c, then safe search guidelines 3! Properties triangles = reflexive, symetric, transitive property in other words, if, and transitive.! Draw a triangle with hypotenuse of the larger triangle are equal, or AAS you! The bush in the above figure congruence in geometry b that is similar to the tenth. 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Top-Rated professional tutors any equation or expression and right cylinders Innovation - 2016 DATAVERSITY for three things is in. B, then what transitive property triangles can be applied to segment, angles, triangles, or any other stuff math! Towards the bush until he can see the top of the same 37° the! Binary relations our safe search guidelines a special symbol is used to show similarity: ~,. } transitive property triangles \\text { and } 9 cm, get practice tests,,! Will explore the transitive property of congruence, the symmetric property of multiplication worksheet I.. Unbiased info you need to find the right school, Biological and Biomedical Sciences, Culinary Arts and Services... Is 20 cm long Midpoint: the transitive property tells us we can substitute congruent. Has the same 37° in the mirror a Study.com Member length 3 and legs of length 3 and legs length! Is ( blank ) used to show similarity: ~ on binary relations, and theorems wyzant... I will use the symbol ~ to indicate that two objects that are.! – triangle Proofs Name: COMMON POTENTIAL REASONS for Proofs criteria for,. △Elk, then a=c is easy to check that \\ ( S\\ ) is reflexive symmetric! Since their angles and sides are all the corresponding hypotenuse of length 3 and legs of length and. Property math ; transitive property of congruence assume Mary ate 2 hotdogs and ate. At a public charter high school page to learn more choose from 19 different sets of term: properties =. Trademarks and copyrights are the reflexive property of congruence are the reflexive property of congruence: Having the same... And how transitive property triangles applies to similar triangles is their size substitution property we looked at earlier, not! Keywords did n't meet our safe search guidelines a basic introduction into the transitive property of congruence, the triangles. Puffed up in size side of a triangle is 180°: this really..." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8822341,"math_prob":0.94159096,"size":26030,"snap":"2021-43-2021-49","text_gpt3_token_len":5957,"char_repetition_ratio":0.21832015,"word_repetition_ratio":0.14508365,"special_character_ratio":0.217864,"punctuation_ratio":0.14110059,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9867465,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-29T10:47:03Z\",\"WARC-Record-ID\":\"<urn:uuid:1d522771-e024-4f73-8338-70837b2b43df>\",\"Content-Length\":\"34976\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7bb3a744-cd1e-45c6-a10d-debc54486fa4>\",\"WARC-Concurrent-To\":\"<urn:uuid:d111b1c3-c9f2-4b0b-abf0-1b2138bfe56a>\",\"WARC-IP-Address\":\"62.182.61.188\",\"WARC-Target-URI\":\"http://socialwifi.se/assassin-s-rfddaow/69yecp.php?5a5acf=white-kettle-delonghi\",\"WARC-Payload-Digest\":\"sha1:KBMOISV7L4BYWWGD6SC2OBLMY7RVHL6D\",\"WARC-Block-Digest\":\"sha1:HS4EBAJIOADZK3WVQYLZET7JJZM4JKQL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358705.61_warc_CC-MAIN-20211129104236-20211129134236-00305.warc.gz\"}"}
https://estebantorreshighschool.com/quadratic-equations/transformation-efficiency-equation.html	[ "## What does transformation efficiency mean?\n\nTransformation efficiency is defined as the number of colony forming units (cfu) which would be produced by transforming 1 µg of plasmid into a given volume of competent cells. The term is somewhat misleading in that 1 µg of plasmid is rarely actually transformed.\n\n## What is a good transformation efficiency value?\n\nA good rule to follow is this: if your efficiency is equal to or less than 5 x 107 CFU/µg DNA, use these cells for plasmid transformations. If your efficiency is greater than 5 x 107 (ideally 1 x 108 or higher), use these cells for ligation and other assembly reaction transformations.\n\n## Is transformation efficiency a percentage?\n\n1. Some people calculate it by: Transformation efficiency (%) = [number of explants showing transformation/ number of explants inoculated] x (100%).\n\n## What can affect transformation efficiency?\n\nThe factors that affect transformation efficiency are the strain of bacteria, the bacterial colony’s phase of growth, the composition of the transformation mixture, and the size and state of the foreign DNA.\n\n## How do you know if transformation is successful?\n\nHow can you tell if a transformation experiment has been successful? If transformation is successful, the DNA will be integrated into one of the cell’s chromosomes.\n\n## Is bacterial transformation an efficient process?\n\nThe transformation reaction is efficient when <10ng of DNA is used. pUC19 DNA (0.1ng) is suitable as control. Supercoiled DNA is most efficient for transformation compared to linear or ssDNA that has the transformation efficiency of <1%.\n\n## Why is E coli good for transformation?\n\nE. coli is a preferred host for gene cloning due to the high efficiency of introduction of DNA molecules into cells. coli is a preferred host for protein production due to its rapid growth and the ability to express proteins at very high levels.\n\n## How does plasmid size affect transformation efficiency?\n\nThe transformation efficiency (transformants per microgram plasmid DNA) decreased with increases of size of the DNA. The size of plasmid DNA in the range of 3.7 to 12.6 kbp did not affect the molecular efficiency (transformants per molecule input DNA).\n\n## What is heat shock transformation?\n\nBy exposing cells to a sudden increase in temperature, or heat shock, a pressure difference between the outside and the inside of the cell is created, that induces the formation of pores, through which supercoiled plasmid DNA can enter.\n\n## How do you calculate conjugation efficiency?\n\nConjugation efficiency is most commonly quantified by the ratio of the number of transconjugants (i.e., recipient cells that have received a plasmid from a donor cell) at the end of the experiment to the number of donors or recipients at the beginning of the experiment.\n\n## Can bacteria take up linear DNA?\n\nOnly circular DNA molecules, able to replicate, may confer antibiotic resistance to the bacteria. Linear DNA will not replicate (and will not survive exonuclease activities) inside the bacterial cell!\n\n## What transformation means?\n\nto change in form, appearance, or structure; metamorphose. to change in condition, nature, or character; convert. to change into another substance; transmute.\n\n## How do you calculate CFU?\n\nTo find out the number of CFU/ ml in the original sample, the number of colony forming units on the countable plate is multiplied by 1/FDF. This takes into account all of the dilution of the original sample. 200 CFU x 1/1/4000 = 200 CFU x 4000 = 800000 CFU/ml = 8 x 10.CFU/ml in the original sample.\n\n### Releated\n\n#### Characteristic equation complex roots\n\nWhat are roots of characteristic equations? discussed in more detail at Linear difference equation#Solution of homogeneous case. The characteristic roots (roots of the characteristic equation) also provide qualitative information about the behavior of the variable whose evolution is described by the dynamic equation. How do I know if my roots are complex? When graphing, if […]\n\n#### Free fall time equation\n\nWhat is the formula for time in free fall? Free fall means that an object is falling freely with no forces acting upon it except gravity, a defined constant, g = -9.8 m/s2. The distance the object falls, or height, h, is 1/2 gravity x the square of the time falling. 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http://bklein.ece.gatech.edu/laser-photonics/maxwells-equations-and-the-helmholtz-wave-equation/	[ "Laser Photonics Maxwell’s Equations and the Helmholtz Wave Equation\n\n# Maxwell’s Equations and the Helmholtz Wave Equation\n\nThere are four Maxwell equations, which you can find in many places. I’ll repeat them here, but I want to give you some feeling for what the equations mean. After all, we’re not mathematicians, interested in equations for their own sake. We’re interested in what equations tell us about the physical world.\n\nThe first Maxwell equation is called Ampere’s Law:", null, "$\\displaystyle \\nabla \\times \\vec{H} = \\vec{J} + \\frac{\\partial \\vec{D}}{\\partial t}$\n\nwhere H is the magnetic field, J is the electrical current density, and D is the electric flux density, which is related to the electric field. In words, this equation says that the curl of the magnetic field equals the electrical current density plus the time derivative of the electric flux density. Physically, this means that two things create magnetic fields curling around them: electrical current, and time-varying (not static) electric fields. The typical example of this is a vertical current-bearing wire with magnetic field lines looping around it:\n\n(insert picture)\n\nThe second Maxwell equation is called Faraday’s Law:", null, "$\\displaystyle \\nabla \\times \\vec{E} = - \\frac{\\partial \\vec{B} }{\\partial t}$\n\nwhere E is the electric field and B is the magnetic flux density, which is related to the magnetic field. This is called Faraday’s Law, and similar to Ampere’s Law, it tells us that time-varying magnetic fields create electric fields curling around them.\n\nWhat does ‘related’ mean, anyways? Well, it turns out that", null, "$\\displaystyle \\vec{D} = \\epsilon \\vec{E}$\n\nand", null, "$\\displaystyle \\vec{B} = \\mu \\vec{H}$\n\nwhich don’t really count as Maxwell equations – they’re called ‘constitutive relations’ – but they’re still very important.", null, "$\\epsilon$ is the permittivity, and", null, "$\\mu$ is the permeability, both of which are properties of whatever material you’re in (air, glass, water, plastic, metal, etc.)\n\nIf we consider Ampere’s law and Faraday’s law (only two Maxwell equations – there are two more, we’ll get to them in a second!) we see that time-varying electric fields create magnetic fields curling around them, and time-varying magnetic fields create electric fields curling around them. This is why electromagnetic waves can exist, and can carry energy far away from their source (billions of light-years, in the case of distant galaxies): the electric and magnetic fields can support one another.\n\nThe third Maxwell equation is Gauss’ Law:", null, "$\\displaystyle \\nabla \\cdot \\vec{D} = \\rho$\n\nwhere", null, "$\\rho$ is the electric charge density. This equation tells us that charge creates electric fields diverging from it. Think of that charged metal sphere you grabbed as a kid to make your hair stand up. The electric field lines were radiating outward from it.\n\nAnd finally, the fourth Maxwell equation, which is nameless:", null, "$\\displaystyle \\nabla \\cdot \\vec{B} = 0$\n\nwhich tells us that magnetic fields don’t diverge from anything, they only curl around. Or equivalently: there is no such thing as magnetic charge, at least not that we’ve found so far.\n\nSince we’re mostly interested in electromagnetic waves here, and in particular light waves, we have to convert the Maxwell equations into a form that easily yields wave-like solutions. To accomplish this, we will derive the Helmholtz wave equation from the Maxwell equations.\n\nWe’ve discussed how the two ‘curl’ equations (Faraday’s and Ampere’s Laws) are the key to electromagnetic waves. They’re tricky to solve because there are so many different fields in them: E, D, B, H, and J, and they’re all interdependent. So our goal will be to combine those two equations into a single equation with a single field in it.\n\nFirst, let’s assume we’re in a uniform material, so that the permittivity epsilon and the permeability mu are constants – they don’t change in space or in time. Of course, that can’t be true for the entire universe, but it can be approximately true in some limited-size region, which is where we’ll solve Maxwell’s equations for now. We’ll also pick a region that has zero conductivity, and therefore zero electrical current density J. That gets rid of one of our fields right away. Of course our solution won’t be entirely general, because it won’t necessarily apply to regions with nonzero conductivity, but we can fix that up later.\n\nOur next goal will be to somehow get rid of the magnetic field on the right hand side of Faraday’s law, and replace it with an expression involving the electric field. That way we’d have an equation with only one field – E – on both sides of the equation. How can we accomplish this? Well, we know that Ampere’s law relates the curl of the magnetic field to the electric field, so we’re going to take the curl of both sides of Faraday’s law:", null, "$\\displaystyle \\nabla \\times \\nabla \\times \\vec{E} = - \\frac{\\partial ( \\nabla \\times \\vec{B}) }{\\partial t}$\n\nI’ve brought the curl inside the time derivative, but that’s ok – it’s just interchanging the order of differentiation. We’ve certainly made Faraday’s law look messier, how does it help us? Well, let’s rewrite Ampere’s law using our constitutive relations, and getting rid of J:", null, "$\\displaystyle \\frac{1}{\\mu} \\nabla \\times \\vec{B} = \\epsilon \\frac{\\partial \\vec{E}}{\\partial t}$\n\nor", null, "$\\displaystyle \\nabla \\times \\vec{B} = \\epsilon \\mu \\frac{\\partial \\vec{E}}{\\partial t}$\n\nOK, so now I have an expression that allows me to replace the curl of B with an expression involving E. Let’s swap that into our modified Faraday’s law:", null, "$\\displaystyle \\nabla \\times \\nabla \\times \\vec{E} = - \\epsilon \\mu \\frac{\\partial^2 \\vec{E} }{\\partial t^2}$\n\nMission accomplished! We’ve condensed the two Maxwell curl equations down into a single equation involving nothing but E. This is one form of the Helmholtz wave equation, although not necessarily the nicest form to solve, since it has the curl of a curl on the left hand side. We can use some vector identities to simplify that a bit. One useful vector identity is the following:", null, "$\\displaystyle \\nabla \\times \\nabla \\times \\vec{A} = \\nabla ( \\nabla \\cdot \\vec{A} ) - \\nabla ^2 \\vec{A}$\n\nwhere", null, "$\\vec{A}$ is any vector field. One tedious but reliable way of deriving this relatively wacky-looking vector identity is to write out all of the vector components and derivatives; in the time-honored words of many distinguished textbook writers, ‘we leave this as an exercise for the reader’.\n\nThe vector identity doesn’t really seem to simplify things much – it just allows us to replace the curl of the curl with a different complicated-looking expression – but we can improve things by putting one more restriction on the region where we’re solving these equations; namely, that there is no ‘free’ (unbound) electrical charge in that region, or", null, "$\\rho = 0$. In that case, Gauss’ Law becomes", null, "$\\displaystyle \\nabla \\cdot \\vec{D} = 0 = \\nabla \\cdot (\\epsilon \\vec{E}) = \\epsilon \\nabla \\cdot \\vec{E}$\n\nwhere we’ve assumed that", null, "$\\epsilon$ doesn’t depend on position, allowing us to take it outside of the derivative. Dividing both sides by", null, "$\\epsilon$ finally gives us", null, "$\\displaystyle \\nabla \\cdot \\vec{E} = 0$\n\nYou’d be excused for wondering what the point is of all this. We’re getting to it! Let’s go back to our vector identity and replace generic field", null, "$\\vec{A}$ with electric field", null, "$\\vec{E}$ :", null, "$\\displaystyle \\nabla \\times \\nabla \\times \\vec{E} = \\nabla ( \\nabla \\cdot \\vec{E} ) - \\nabla ^2 \\vec{E}$\n\nYou see that", null, "$\\nabla \\cdot \\vec{E}$ in there? Now we know it’s zero, as long as we’re in a region with no charge, and as long as the permittivity", null, "$\\epsilon$ is constant with position. So now we have a pretty nice simplification; namely", null, "$\\displaystyle \\nabla \\times \\nabla \\times \\vec{E} = -\\nabla ^2 \\vec{E}$\n\nI’m going to put that back into the Helmholtz equation, to give me.. uh, still the Helmholtz equation:", null, "$\\displaystyle -\\nabla^2 \\vec{E} = - \\epsilon \\mu \\frac{\\partial^2 \\vec{E} }{\\partial t^2}$\n\nusually we gather everything on one side:", null, "$\\displaystyle \\nabla^2 \\vec{E} - \\epsilon \\mu \\frac{\\partial^2 \\vec{E} }{\\partial t^2} = 0$\n\nWhew! And that is the Helmholtz wave equation.\n\nUnits:\n\nSolution of Helmholtz equation on separate page" ]	[ null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9387949,"math_prob":0.9980816,"size":6572,"snap":"2019-26-2019-30","text_gpt3_token_len":1414,"char_repetition_ratio":0.15316688,"word_repetition_ratio":0.021719458,"special_character_ratio":0.2051126,"punctuation_ratio":0.10590944,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994037,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-22T12:32:59Z\",\"WARC-Record-ID\":\"<urn:uuid:dba6893b-0c53-476e-b269-35c41075a714>\",\"Content-Length\":\"38775\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6b7c63b6-3c7a-45b5-acde-3d82c52b26cd>\",\"WARC-Concurrent-To\":\"<urn:uuid:025001a9-46b0-4134-9ed7-9b94900a8da5>\",\"WARC-IP-Address\":\"143.215.248.18\",\"WARC-Target-URI\":\"http://bklein.ece.gatech.edu/laser-photonics/maxwells-equations-and-the-helmholtz-wave-equation/\",\"WARC-Payload-Digest\":\"sha1:YV33LPXTBO3TCAYPNIZ3AYTRU3UAJV6Z\",\"WARC-Block-Digest\":\"sha1:FO2P4F4ZTNJ6WDTBUOOMHUBGDPCN2EFT\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195528013.81_warc_CC-MAIN-20190722113215-20190722135215-00053.warc.gz\"}"}
http://affengineer.com/2015/05/12/teespring-gems-vid-31-concept-roi/	[ "## Teespring Gems \| Vid 32\| – The Concept of ROI\n\nNot many people understand the concept of ROI so I thought I’d take a quick few minutes to explain.\n\nROI is basically the % increase of your initial investment. Check the below vid for clarification.\n\nFor example if you put in \\$100 and got back \\$200, you’ve received 100% ROI on your investment.\n\nIf you put in \\$50 and got back \\$25, Then you’ve recieved -50% ROI on your investment.\n\nIt’s a metric that tells you how hard your money is working for you and where you’re getting the most value out of your dollars.\n\nThe way to calculate it is to divide the initial investment by profit/loss.\n\nE.g, for the first example of \\$100 investment and \\$200 return you’ve made \\$100 profit. So our ROI is,\n\n100/100 = 1\n1 x 100 = 100%, (remember you multiply the initial formula by 100 to get it into ‘percentage’.)\n\nFor the example of \\$50 investment and \\$30 return,\n\n\\$50 – \\$30 = \\$20\n20/50 = 0.4\n0.4 x 100 = 40%\n\nSince it’s a ‘loss’ it’s -40% ROI.\n\nWant to get Serious with Internet Marketing? Join us on Aff Playbook Below!", null, "Join Here", null, "" ]	[ null, "http://affengineer.com/wp-content/uploads/2015/02/apb-success-stories.png", null, "http://affengineer.com/wp-content/uploads/2015/01/affplaybook-coupon1.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.918552,"math_prob":0.8202823,"size":1086,"snap":"2021-31-2021-39","text_gpt3_token_len":284,"char_repetition_ratio":0.11829945,"word_repetition_ratio":0.0,"special_character_ratio":0.3130755,"punctuation_ratio":0.10041841,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9524317,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-05T05:49:47Z\",\"WARC-Record-ID\":\"<urn:uuid:d153821a-65c6-49b7-b2dd-45b912bc93e2>\",\"Content-Length\":\"40349\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5c5badc7-831e-4daf-a132-5e336f39e154>\",\"WARC-Concurrent-To\":\"<urn:uuid:cb5acb9e-4667-4292-a270-9b0d6ada4b86>\",\"WARC-IP-Address\":\"172.67.205.103\",\"WARC-Target-URI\":\"http://affengineer.com/2015/05/12/teespring-gems-vid-31-concept-roi/\",\"WARC-Payload-Digest\":\"sha1:4LBTUWLXKPRVXZKY7V7NLQNBR4E4Z7WJ\",\"WARC-Block-Digest\":\"sha1:XDUKH2QIWC5JIBJXGUAUA5UTSSDFQD27\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046155322.12_warc_CC-MAIN-20210805032134-20210805062134-00466.warc.gz\"}"}
https://www.jagranjosh.com/articles/practice-set-for-ssc-quantitative-aptitude-miscellaneous-set-10-1482922093-1	[ "# Practice Set for SSC Quantitative Aptitude (Miscellaneous Set-10)\n\nIn this article, we have prepared few questions out of various previously asked questions in several recruitment exams. Such questions will prove effective to the aspirants. So, take it placidly.", null, "", null, "SSC study material\n\nWe observed that previous year questions are more probable to be repeated in the exams like SSC CGL and etc. Our devoted SSC team of Jagranjosh.com has framed a quiz of 25 questions taken from the diverse previous examinations i.e. IBPS, RBI, SSC and other exams. while operating on these questions, fix a time frame in mind for getting skilled in these examinations, especially. As quantitative section requires the following attributes from you.\n\n- Quick & Hasty manipulations.\n\n- Fast/Accelerated observations of Problems.\n\n- A better understanding of observing question and applied formulas.\n\n- Diverse and alternative tricks to simplify questions and etc.\n\nFind the question set below: -\n\nQuantitative Aptitude\n\nDirections (Q. Nos. 1 to 2): Study the information given below and answer the questions that follow.\n\nA building consists of men and women who spend their leisure time in watching movies, learning dance and learning singing. 8 men, who form ten per cent of the total number of men in the building, learn to dance. The total number of men in the building, learn to dance. The total number of women in the building is 62.5 per cent of the total number of men in the building. Twenty-four per cent of the total number of women learns to sing. One-fifth of the total number of women watches movies. The ratio of the number of men watching movies to the number of women doing the same is 13:2 respectively.\n\n1. The number of men who like watching movies is what per cent of the total number of men in the building?\n\na. 79.75\n\nb. 83.45\n\nc. 81.25\n\nd. 72.15\n\ne. None of these\n\n2. What is the total number of members (men and women together) learning singing?\n\na. 21\n\nb. 13\n\nc. 13\n\nd. 15\n\ne. None of these\n\nDirections (Q. Nos. 3 to 7): What should come in place of the question mark (?) in the following number series?\n\n3. 3, 732, 1244, 1587, 1803, 1928?\n\na. 2144\n\nb. 1992\n\nc. 1955\n\nd. 2053\n\ne. None of these\n\n4. 16, 24, ?, 210, 945, 5197.5, 33783.76\n\na. 40\n\nb. 36\n\nc. 58\n\nd. 60\n\ne. None of these\n\n5. 45030, 9000, 1795, 355, 68, ?, 1.32\n\na. 11.6\n\nb. 12.2\n\nc. 10.4\n\nd. 9.8\n\ne. None of these\n\n6. 5, 12, 36, 123, ?, 2555, 15342\n\na. 508\n\nb. 381\n\nc. 504\n\nd. 635\n\ne. None of these\n\n7. 8, 11, 17, ?, 65, 165.5, 498.5\n\na. 27.5\n\nb. 32\n\nc. 28\n\nd. 30.5\n\ne. None of these\n\nDirections (Q. Nos. 8 to 12): Each of the questions given below consists of a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statements is sufficient to answer the question. Read both the statements and\n\nGive answer (a) if the data in statement I alone is sufficient to answer the question, while the data in statement II alone is not sufficient to answer the question\n\nGive answer (b) if the data in statement II alone is sufficient to answer the question, while the data in statement I alone is not sufficient to answer the question\n\nGive answer (c) if the data in statement I alone or in statement II alone is sufficient to answer the question\n\nGive answer (d) if the data in both the statements I and II is not sufficient to answer the question\n\nGive answer (e) if the data in both the statements I and II together is necessary to answer the question\n\n8. What is the salary of C, in a group of A, B, C, D and E whose average salary is Rs. 48250?\n\nI. C’s salary is 1.5 times B’s salary.\n\nII. Average salary of A and B is Rs. 23500.\n\n9. What is the per cent profit earned by selling a car for Rs. 640000?\n\nI. The amount of profit earned on selling the car was Rs. 320000.\n\nII. The selling price of the car was twice the cost price.\n\n10. What is the rate of interest per cent per annum?\n\nI. An amount of Rs. 14350 gives a simple interest of Rs. 11450 in four years.\n\nII. The amount doubles itself in 5 years with simple interest.\n\n11. What is the two digit number?\n\nI. The difference between the two digits of the number is 9.\n\nII. The product of the two digits of the number is 0.\n\n12. What is the perimeter of the rectangle?\n\nI. The area of the rectangle is 252 sq m.\n\nII. The ratio of length to breadth of the rectangle is 9 : 7 respectively.\n\nDirections (Q. Nos. 13 to 15): Study the given information carefully to answer the questions that follow.\n\nA committee of 6 teachers is to be formed out of 4 Science teachers, 5 Arts teachers and 3 Commerce teachers. In how many different ways can the committee be formed if\n\n13. Two teachers from each stream are to be included?\n\na. 210\n\nb. 180\n\nc. 145\n\nd. 96\n\ne. None of these\n\n14. No teacher from the Commerce stream is to be included?\n\na. 81\n\nb. 62\n\nc. 46\n\nd. 84\n\ne. None of these\n\n15. Any teacher can be included in the committee?\n\na. 626\n\nb. 718\n\nc. 924\n\nd. 844\n\ne. None of these\n\nDirections (Q. Nos. 16 to 20): Study the following table carefully to answer the questions that follow.\n\nNumber of people (in hundreds) participating in the Annual Fair from six different towns over the years\n\n16. Number of people participating in the Fair from town P in the year 2010 forms approximately what per cent of the total number of people participating in the Fair from that town over all the years together?\n\na. 19\n\nb. 24\n\nc. 27\n\nd. 12\n\ne. 15\n\n17. What is the respective ratio of total number of people participating in the Fair from town S in the years 2006 and 2007 together to the number of people participating in the Fair from town R in the same years?\n\na. 8 : 9\n\nb. 110 : 127\n\nc. 136 : 143\n\nd. 11 : 12\n\ne. None of these\n\n18. What is the per cent increase in the number of people participating in the Fair from town T in the year 2009 from the previous year? (Rounded off to two digits after decimal)\n\na. 4.15\n\nb. 3.77\n\nc. 1.68\n\nd. 2.83\n\ne. None of these\n\n19. What is the average number of people participating in the Fair from town U over all the years together? (Rounded off to the nearest integer)\n\na. 515\n\nb. 523\n\nc. 567\n\nd. 541\n\ne. 538\n\n20. How many people participated in the Fair from all the town together in the year 2005?\n\na. 3290\n\nb. 3100\n\nc. 3240\n\nd. 3170\n\ne. None of these\n\nDirections (Q. Nos. 21 to 25): Study the following pie-charts carefully to answer the questions that follow.\n\nPercentage break-up of number of children in five different villages and break-up of children attending school from those villages\n\n21. What is the respective ratio of total number of children from village O to the number of children attending school from the same village?\n\na. 204 : 145\n\nb. 179 : 131\n\nc. 167 : 111\n\nd. 266 : 137\n\ne. None of these\n\n22. What is the number of children attending school from village N?\n\na. 145\n\nb. 159\n\nc. 170\n\nd. 164\n\ne. None of these\n\n23. What is the total number of children not attending school from villages M and N together?\n\na. 69\n\nb. 56\n\nc. 76\n\nd. 63\n\ne. None of these\n\n24. What is the total number of children from villages P and M together?\n\na. 1422\n\nb. 1142\n\nc. 1122\n\nd. 1211\n\ne. None of these\n\n25. The number of children attending school from village L is approximately what per cent of the total number of children from that village?\n\na. 78\n\nb. 72\n\nc. 57\n\nd. 84\n\ne. 66" ]	[ null, "https://img.jagranjosh.com/images/2022/June/2962022/default-josh-logo.jpg", null, "https://img.jagranjosh.com//imported/images/E/Articles/quant-practice-ssc.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89796555,"math_prob":0.80916595,"size":6938,"snap":"2023-14-2023-23","text_gpt3_token_len":2015,"char_repetition_ratio":0.17421402,"word_repetition_ratio":0.16679022,"special_character_ratio":0.31493226,"punctuation_ratio":0.182145,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9786512,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-28T14:36:37Z\",\"WARC-Record-ID\":\"<urn:uuid:9c3d33e7-de8d-40d4-bb88-24ffccc61a74>\",\"Content-Length\":\"129126\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d6b067d0-7358-47bd-82be-39db4705ddbe>\",\"WARC-Concurrent-To\":\"<urn:uuid:afd105d9-6159-4ba7-baf6-92511621a224>\",\"WARC-IP-Address\":\"104.104.105.19\",\"WARC-Target-URI\":\"https://www.jagranjosh.com/articles/practice-set-for-ssc-quantitative-aptitude-miscellaneous-set-10-1482922093-1\",\"WARC-Payload-Digest\":\"sha1:3QGDJB6TFAPRREQIYYOWZGKMGADQKNEA\",\"WARC-Block-Digest\":\"sha1:CJ7VHY3TRAOJFLQCNPNKXXZY2VKJFMVJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296948867.32_warc_CC-MAIN-20230328135732-20230328165732-00224.warc.gz\"}"}
http://postgis.net/docs/manual-2.0/ST_Affine.html	[ "## Name\n\nST_Affine — Applies a 3d affine transformation to the geometry to do things like translate, rotate, scale in one step.\n\n## Synopsis\n\n`geometry ST_Affine(`geometry geomA, float a, float b, float c, float d, float e, float f, float g, float h, float i, float xoff, float yoff, float zoff`)`;\n\n`geometry ST_Affine(`geometry geomA, float a, float b, float d, float e, float xoff, float yoff`)`;\n\n## Description\n\nApplies a 3d affine transformation to the geometry to do things like translate, rotate, scale in one step.\n\nVersion 1: The call\n\n`ST_Affine(geom, a, b, c, d, e, f, g, h, i, xoff, yoff, zoff) `\n\nrepresents the transformation matrix\n\n```/ a b c xoff \\\n\| d e f yoff \|\n\| g h i zoff \|\n\\ 0 0 0 1 /```\n\nand the vertices are transformed as follows:\n\n```x' = ax + by + cz + xoff\ny' = dx + ey + fz + yoff\nz' = gx + hy + iz + zoff```\n\nAll of the translate / scale functions below are expressed via such an affine transformation.\n\nVersion 2: Applies a 2d affine transformation to the geometry. The call\n\n`ST_Affine(geom, a, b, d, e, xoff, yoff)`\n\nrepresents the transformation matrix\n\n```/ a b 0 xoff \\ / a b xoff \\\n\| d e 0 yoff \| rsp. \| d e yoff \|\n\| 0 0 1 0 \| \\ 0 0 1 /\n\\ 0 0 0 1 /```\n\nand the vertices are transformed as follows:\n\n```x' = ax + by + xoff\ny' = dx + e*y + yoff\nz' = z ```\n\nThis method is a subcase of the 3D method above.\n\nEnhanced: 2.0.0 support for Polyhedral surfaces, Triangles and TIN was introduced.\n\nAvailability: 1.1.2. Name changed from Affine to ST_Affine in 1.2.2", null, "Prior to 1.3.4, this function crashes if used with geometries that contain CURVES. This is fixed in 1.3.4+", null, "This function supports Polyhedral surfaces.", null, "This function supports Triangles and Triangulated Irregular Network Surfaces (TIN).", null, "This function supports 3d and will not drop the z-index.", null, "This method supports Circular Strings and Curves\n\n## Examples\n\n```--Rotate a 3d line 180 degrees about the z axis. Note this is long-hand for doing ST_Rotate();\nSELECT ST_AsEWKT(ST_Affine(the_geom, cos(pi()), -sin(pi()), 0, sin(pi()), cos(pi()), 0, 0, 0, 1, 0, 0, 0)) As using_affine,\nST_AsEWKT(ST_Rotate(the_geom, pi())) As using_rotate\nFROM (SELECT ST_GeomFromEWKT('LINESTRING(1 2 3, 1 4 3)') As the_geom) As foo;\nusing_affine \| using_rotate\n-----------------------------+-----------------------------\nLINESTRING(-1 -2 3,-1 -4 3) \| LINESTRING(-1 -2 3,-1 -4 3)\n(1 row)\n\n--Rotate a 3d line 180 degrees in both the x and z axis\nSELECT ST_AsEWKT(ST_Affine(the_geom, cos(pi()), -sin(pi()), 0, sin(pi()), cos(pi()), -sin(pi()), 0, sin(pi()), cos(pi()), 0, 0, 0))\nFROM (SELECT ST_GeomFromEWKT('LINESTRING(1 2 3, 1 4 3)') As the_geom) As foo;\nst_asewkt\n-------------------------------\nLINESTRING(-1 -2 -3,-1 -4 -3)\n(1 row)\n```" ]	[ null, "http://postgis.net/docs/manual-2.0/images/note.png", null, "http://postgis.net/docs/manual-2.0/images/check.png", null, "http://postgis.net/docs/manual-2.0/images/check.png", null, "http://postgis.net/docs/manual-2.0/images/check.png", null, "http://postgis.net/docs/manual-2.0/images/check.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.59213394,"math_prob":0.9973817,"size":2605,"snap":"2020-10-2020-16","text_gpt3_token_len":866,"char_repetition_ratio":0.16416763,"word_repetition_ratio":0.25770926,"special_character_ratio":0.3666027,"punctuation_ratio":0.18641116,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9952195,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-26T16:16:49Z\",\"WARC-Record-ID\":\"<urn:uuid:169a8dd2-a065-4d36-85c1-15b7bb009429>\",\"Content-Length\":\"6940\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:20ce93b1-0c4e-4784-a8cb-c302403115ed>\",\"WARC-Concurrent-To\":\"<urn:uuid:d5e3a3bc-62f6-4cf2-88ee-4b699bd799da>\",\"WARC-IP-Address\":\"209.208.97.173\",\"WARC-Target-URI\":\"http://postgis.net/docs/manual-2.0/ST_Affine.html\",\"WARC-Payload-Digest\":\"sha1:FZC5XGTQ7X7KQQN344EZSTVU75UWAHSD\",\"WARC-Block-Digest\":\"sha1:TFGSREXETCYADHNGY7PNN7LU6VF2TQOG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875146414.42_warc_CC-MAIN-20200226150200-20200226180200-00106.warc.gz\"}"}
https://le.wps.cn/cate?order_by=rec&sort=desc&price=12001_50000	[ "充会员 · 获积分 · 兑福利\n\n 充值会员分类获取积分充值获取积分（以1年为例） WPS会员充值超级会员充值金额13 17913=2327 超级会员续费超级会员充值金额13 17913=2327 WPS会员续费WPS会员充值金额10 8910=890 超级会员充值WPS会员充值金额10 17910=1790 非会员首次充值WPS会员充值金额8 898=712 非会员首次充值超级会员充值金额8 1798=1432 充值PDF特权包充值金额8 49.98=400 充值文档修复特权充值金额8 498=392 充值OCR特权包充值金额8 498=392 曾是会员用户充值WPS会员充值金额6 896=534 曾是会员用户充值超级会员充值金额6 1796=1074\n\n会员积分规则\n\n四、积分详细规则\n\n1.每日领积分\n\n 会员等级获取积分 WPS会员 2-9 超级会员 3-15\n\na.非会员一键升级为会员；点击开通会员 >>\n\nb.每日需到积分专区领积分\n\n2.开通/续费WPS会员服务\n\n 充值会员分类获取积分充值获取积分（以1年为例） WPS会员充值超级会员充值金额13 17913=2327 超级会员续费超级会员充值金额13 17913=2327 WPS会员续费WPS会员充值金额10 8910=890 超级会员充值WPS会员充值金额10 17910=1790 非会员首次充值WPS会员充值金额8 898=712 非会员首次充值超级会员充值金额8 1798=1432 充值PDF特权包充值金额8 49.98=400 充值文档修复特权充值金额8 498=392 充值OCR特权包充值金额8 498=392 曾是会员用户充值WPS会员充值金额6 896=534 曾是会员用户充值超级会员充值金额6 1796=1074" ]	[ null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.7379978,"math_prob":0.4837798,"size":512,"snap":"2019-26-2019-30","text_gpt3_token_len":547,"char_repetition_ratio":0.20472442,"word_repetition_ratio":0.0,"special_character_ratio":0.43945312,"punctuation_ratio":0.0,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97480905,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-26T00:32:34Z\",\"WARC-Record-ID\":\"<urn:uuid:a530caec-7879-4a64-8978-e1244f502f1e>\",\"Content-Length\":\"32089\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2d7aa892-2be1-4d6e-b506-2f1157a96715>\",\"WARC-Concurrent-To\":\"<urn:uuid:bdd4ab99-690f-4eca-b9f7-4feb6c44e76d>\",\"WARC-IP-Address\":\"120.131.9.194\",\"WARC-Target-URI\":\"https://le.wps.cn/cate?order_by=rec&sort=desc&price=12001_50000\",\"WARC-Payload-Digest\":\"sha1:TERE4QJ736USLQBUV5PJLXWNKAFVBNGP\",\"WARC-Block-Digest\":\"sha1:YTEHK2RDZPUZU5HE4PYG2GGUITYQ6VGZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999964.8_warc_CC-MAIN-20190625233231-20190626015231-00134.warc.gz\"}"}
https://stats.stackexchange.com/questions/410140/decomposing-a-random-variable-into-marginals-and-copula	[ "# Decomposing a random variable into marginals and copula\n\nI’m having trouble getting understanding how to actual construct a copula, from my understanding it captures the purely joint features of a joint distribution. I’ve been working with the following example.\n\nLet X,Y be random variables with joint distribution function $$H(x,y) = (1 + e^{-x} + e^{-y})^{-1}$$ I have got by letting x and y tend to infinity respectively that the marginals of X and Y are standard logistic distributions given by $$F(x) = (1 + e^{-x})^{-1}, G(y) = (1 + e^{-y})^{-1}$$ I have to show that the copula of X and Y is $$C(u,v) = \\frac{uv}{u + v - uv}$$ but I don’t know how to go about doing this? I’ve tried working with Sklars theorem but I can’t seem to get my head around this.\n\nAside from this specific example what is the approach in general to retrieve copulas from joint distributions? Any help would be great, thanks!\n\nI will take you through a set of simple steps that will work for continuous distributions. (A little extra care is needed to handle the jumps that occur in non-continuous $$F$$ or $$G,$$ but no new concepts are involved.)\n\nBy definition, a copula is the joint distribution you get after you re-express the original variables in a particular way.\n\nFor continuous variables, as in this case, the re-expression is the Probability Integral Transform that replaces each possible value $$x$$ of a random variable $$X$$ (governed by a distribution $$F$$) by its quantile $$u=F(x).$$ You have already found the two quantile functions $$F,G:t\\to \\frac{1}{1 + e^{-t}}.\\tag{1}$$\n\nLet the re-expressed variables be $$U=F(X)\\text{ and }V=G(Y).\\tag{2}$$ The joint distribution of any pair of variables $$(U,V)$$ is\n\n$$F_{(U,V)}(u,v) = \\Pr(U \\le u\\text{ and } V \\le v).\\tag{3}$$\n\nThis is the copula. Note that this definition also means\n\n$$H(x,y) = \\Pr(X\\le x\\text{ and } Y\\le y)=\\frac{1}{1 + e^{-x} + e^{-y}}.\\tag{4}$$\n\nCombining $$(1),(2),(3)$$ and simple algebraic manipulation of the descriptions of these events gives\n\n\\eqalign{F_{(U,V)}(u,v) &= \\Pr(F(X) \\le u\\text{ and } F(Y) \\le v)\\\\ &= \\Pr\\left(\\frac{1}{1+e^{-X}} \\le u \\text{ and } \\frac{1}{1+e^{-Y}} \\le v\\right) \\\\ &= \\Pr\\left(X \\le \\log\\left(\\frac{u}{1-u}\\right) \\text{ and } Y \\le \\log\\left(\\frac{v}{1-v}\\right) \\right) \\\\ &= H\\left(\\log\\left(\\frac{u}{1-u}\\right), \\log\\left(\\frac{v}{1-v}\\right)\\right). }\n\nPlug this result into $$(4)$$ and do the algebra.\n\n• Thank you for this. So the marginals are the actual quantile functions themselves? Also could you explain how the definition (3) implies (4) ? – gb4 May 27 '19 at 18:42\n• $H$ is the (bivariate) distribution function--as you stipulate--so $(4)$ is merely a special case of $(3).$ – whuber May 27 '19 at 21:46" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9147456,"math_prob":0.99970126,"size":849,"snap":"2021-04-2021-17","text_gpt3_token_len":216,"char_repetition_ratio":0.112426035,"word_repetition_ratio":0.0,"special_character_ratio":0.28386337,"punctuation_ratio":0.06936416,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999844,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-21T05:42:29Z\",\"WARC-Record-ID\":\"<urn:uuid:73df860b-b137-47d2-90fc-e356d012825b>\",\"Content-Length\":\"151166\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bd7304fb-f3c4-4415-85b1-bb8fc5462c18>\",\"WARC-Concurrent-To\":\"<urn:uuid:7105f053-8018-4559-8d79-f668888791f8>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://stats.stackexchange.com/questions/410140/decomposing-a-random-variable-into-marginals-and-copula\",\"WARC-Payload-Digest\":\"sha1:LFY6D7RLEVONVXXHF43OMQBSN23JVKSF\",\"WARC-Block-Digest\":\"sha1:BLEGQ44UISCHBHMCFCV3LQCH7H6KMNN3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703522242.73_warc_CC-MAIN-20210121035242-20210121065242-00359.warc.gz\"}"}
https://www.thoughtco.com/how-to-balance-ionic-equations-604025?utm_source=emailshare&utm_medium=social&utm_campaign=shareurlbuttons	[ "# How to Balance Net Ionic Equations\n\nThese are the steps to write a balanced net ionic equation and a worked example problem.\n\n## Steps To Balance Ionic Equations\n\n1. Write the net ionic equation for the unbalanced reaction. If you are given a word equation to balance, you'll need to be able to identify strong electrolytes, weak electrolytes, and insoluble compounds. Strong electrolytes dissociate entirely into their ions in water. Examples of strong electrolytes are strong acids, strong bases, and soluble salts. Weak electrolytes yield very few ions in solution, so they are represented by their molecular formula (not written as ions). Water, weak acids, and weak bases are examples of weak electrolytes. The pH of a solution can cause them to dissociate, but in those situations, you'll be presented an ionic equation, not a word problem. Insoluble compounds do not dissociate into ions, so they are represented by the molecular formula. A table is provided to help you determine whether or not a chemical is soluble, but it's a good idea to memorize the solubility rules.\n2. Separate the net ionic equation into the two half-reactions. This means identifying and separating the reaction into an oxidation half-reaction and a reduction half-reaction.\n3. For one of the half-reactions, balance the atoms except for O and H. You want the same number of atoms of each element on each side of the equation.\n4. Repeat this with the other half-reaction.\n5. Add H2O to balance the O atoms. Add H+ to balance the H atoms. The atoms (mass) should balance out now.\n6. Balance charge. Add e- (electrons) to one side of each half-reaction to balance charge. You may need to multiply the electrons by the two half-reactions to get the charge to balance out. It's fine to change coefficients as long as you change them on both sides of the equation.\n7. Add the two half-reactions together. Inspect the final equation to make sure it is balanced. Electrons on both sides of the ionic equation must cancel out.\n8. Double-check your work! Make sure there are equal numbers of each type of atom on both sides of the equation. Make sure the overall charge is the same on both sides of the ionic equation.\n9. If the reaction takes place in a basic solution, add an equal number of OH- as you have H+ ions. Do this for both sides of the equation and combine H + and OH- ions to form H2O.\n10. Be sure to indicate the state of each species. Indicate solid with (s), liquid for (l), gas with (g), and an aqueous solution with (aq).\n11. Remember, a balanced net ionic equation only describes chemical species that participate in the reaction. Drop additional substances from the equation.\n\n## Example\n\nThe net ionic equation for the reaction you get mixing 1 M HCl and 1 M NaOH is:\n\nH+(aq) + OH-(aq) → H2O(l)\n\nEven though sodium and chlorine exist in the reaction, the Cl- and Na+ ions are not written in the net ionic equation because they don't participate in the reaction." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8964265,"math_prob":0.9883383,"size":5389,"snap":"2020-24-2020-29","text_gpt3_token_len":1214,"char_repetition_ratio":0.14930362,"word_repetition_ratio":0.025433525,"special_character_ratio":0.20745964,"punctuation_ratio":0.118694365,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99374384,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-26T00:13:16Z\",\"WARC-Record-ID\":\"<urn:uuid:9529a3a5-8e87-4a7c-900c-3548deb28f18>\",\"Content-Length\":\"115835\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7cd81b7f-e4a0-452f-8273-1ebb8acd426a>\",\"WARC-Concurrent-To\":\"<urn:uuid:312fd3cc-4f3a-4e96-888b-de5ea951896b>\",\"WARC-IP-Address\":\"199.232.66.133\",\"WARC-Target-URI\":\"https://www.thoughtco.com/how-to-balance-ionic-equations-604025?utm_source=emailshare&utm_medium=social&utm_campaign=shareurlbuttons\",\"WARC-Payload-Digest\":\"sha1:P337M2DVCEITGG3VEJUOARYZBCDCXHDF\",\"WARC-Block-Digest\":\"sha1:OYR6SBHKKGL2G4QGLQOPYAYX4MGG6LDW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347390437.8_warc_CC-MAIN-20200525223929-20200526013929-00190.warc.gz\"}"}
https://www.physicsforums.com/threads/special-relativity-and-inertial-frames.897462/	[ "# Special relativity and inertial frames\n\n• I\nWhat in the mathematics of the derivation of special relativity limits the model to inertial frames? How is an inertial frame defined in the context of the derivation?\n\nIbix\n2020 Award\nIt isn't limited to inertial frames. See, for example, Rindler coordinates, which are sensible coordinates for an observer at constant proper acceleration.\n\nInertial frames are by far the easiest to work with though. And I gather that some older texts do consider non-inertial frames to be in the domain of GR. Modern ones don't, though. Or shouldn't.\n\n•", null, "vanhees71\nphinds\nGold Member\nWhat in the mathematics of the derivation of special relativity limits the model to inertial frames? How is an inertial frame defined in the context of the derivation?\nI'm not aware that there IS any such thing as \"the derivation of special relativity\". Special Relativity is a theory (not an equation) based on two postulates, the first of which (The Principle of Relativity) is that it is talking about things in uniform motion relative to each other (and this generally means inertial frames of reference although as ibix states, it COULD be that two objects are both accelerating but not relative to each other)\n\nLast edited:\n•", null, "vanhees71\nWhy does special relativity apply to acceleration relative to the inertial frame (proper acceleration) but not gravitational acceleration? What can't SR describe accelerations of the inertial frame?\n\nIbix\n2020 Award\nSR is a special case of General Relativity when spacetime is flat. It can't handle gravity because it's defined to be what happens when there's no gravity. That makes the maths much simpler, and one can always choose a frame in which spacetime is locally approximately flat even when there is gravity, which is why it's a special case worth mentioning rather than a historical note.\n\nWhat in the mathematics of the derivation of special relativity limits the model to inertial frames? How is an inertial frame defined in the context of the derivation?\nWhat is derived are the Lorentz transformations, and those are defined relative to inertial frames - exactly as the \"Galilean transformations\" of classical mechanics. Very likely your question is therefore more basic, and belongs in the classical physics forum. Can you answer the question what in the mathematics of the derivation of classical relativity limits the model to inertial frames? How is an inertial frame defined in the context of that derivation?\n\n•", null, "Dale\nSpecial relativity can be derived without reference to either inertial frames or the speed of light. In fact, special relativity is nothing more than an identity and can be derived as such.\n\nNugatory\nMentor\nSpecial relativity can be derived without reference to either inertial frames or the speed of light. In fact, special relativity is nothing more than an identity and can be derived as such.\nDo you have a source for such a derivation?\n\nMister T\nGold Member\nWhat in the mathematics of the derivation of special relativity limits the model to inertial frames? How is an inertial frame defined in the context of the derivation?\nAn inertial reference frame is one in which objects at rest remain at rest and objects in motion continue to move in a straight line atspeeteady speed. Spacetime Physics by Taylor and Wheeler has some very readable and poignant discussions of inertial reference frames.\n\npervect\nStaff Emeritus\nWhy does special relativity apply to acceleration relative to the inertial frame (proper acceleration) but not gravitational acceleration? What can't SR describe accelerations of the inertial frame?\n\nA tensor treatment of special relativity can handle non-inertial frames just fine. So there isn't any real limitation of special relativity itself that prevents it from being applied to non-inertial frames. The limit is knowing tensor mathematics which is needed to handle arbitrary coordinates and to understand the way physical qualities transform under arbitrary coordinate mappings.\n\nJackson's textbook on electrodynamics, for instance, is a graduate level textbook about electromagnetism that would have the necessary math. However, I'm not sure if Jackson treats accelerating frames specifically - I really don't recall. The basic point is that the mathematical treatment that can handle arbitrary coordinate systems can handle the particular case of \"accelerated frames\" just fine.\n\nRindler's book (Relativity, Special and General - or a similar title) is about special and general relativity does have a treatment of accelerating frames.\n\nvanhees71\nGold Member\nThis is often misunderstood. Even in some (minor) textbooks you can read that SR can't handle non-inertial frames, which is of course wrong. With the same right you can argue that you can't handle non-inertial frames in Newtonian physics, which is of course also wrong.\n\nIt is also true that, if you write SR in terms of generally covariant tensors, it looks very close to GR already. Nevertheless there is a difference between the intertial forces due to using a non-inertial reference frame and the presence of a \"true\" gravitational field: If you have a non-inertial reference frame in GR of course you can always introduce a global inertial reference frame, and this is the case if the curvature tensor vanishes identically everywhere in the entire spacetime. If a true gravitational field is present in GR, the curvature tensor is no longer identically 0, and you cannot introduce a global inertial reference frame.\n\nThe equivalence principle however assumes that you can always introduce a local inertial reference frame at each (regular) point in spacetime. I'd say that's the precise meaning of the equivalence principle and not the usually envoked heuristic arguments to argue why in GR spacetime is a Lorentzian manifold, i.e., a pseudo-Riemannian manifold with a fundamental form of signature (1,3) or equivalentliy (3,1).\n\nMister T\nGold Member\nThis is often misunderstood. Even in some (minor) textbooks you can read that SR can't handle non-inertial frames, which is of course wrong. With the same right you can argue that you can't handle non-inertial frames in Newtonian physics, which is of course also wrong.\n\nThat's a good point. Newton's Laws (within their limits of validity) are valid only in inertial reference frames. Certainly one can use Newton's Second Law to describe motion in non-inertial frames, but that involves introducing forces that violate Newton's Third Law.\n\nIsn't part of the confusion of Special Relativity's ability to handle non-inertial frames historical? What I mean is that didn't Einstein, after developing Special Relativity, immediately set to work on what became General Relativity and in the process introduce the formalism of non-inertial frames? Of do I have it wrong?\n\nvanhees71\nGold Member\nYou don't introduce any forces, but you lump parts of the acceleration in terms of non-inertial coordinates to the other side of the equation and call it forces (I like to talk about \"inertial forces\" to make this particular type of \"forces\" distinct from forces due to interactions).\n\nstevendaryl\nStaff Emeritus\nThat's a good point. Newton's Laws (within their limits of validity) are valid only in inertial reference frames. Certainly one can use Newton's Second Law to describe motion in non-inertial frames, but that involves introducing forces that violate Newton's Third Law.\n\nI would say that, for both SR and Newtonian physics, the laws of motion, as expressed in terms of coordinates, have the simplest form if the coordinates are inertial, Cartesian. If you write down the equations of motion in spherical coordinates, I would say that you're still dealing with Newtonian physics, even though they don't have the same form as for Cartesian coordinates. The business about \"forces that violate Newton's Third Law\" just means that you have misidentified what the \"forces\" are.\n\nWith the hindsight given by studying General Relativity, I would say that the \"correct\" way to formulate Newton's laws is in terms of 4-vectors. If you do that, then the Newtonian equations of motion, as well as the third law, are true in any coordinate system, inertial or not:\n\n$m \\frac{dV}{dt} = F$\n\nWith this 4-vector formulation, the \"inertial forces\" are seen as not forces at all, but as connection coefficients (essentially, the derivatives of the basis vectors).\n\npervect\nStaff Emeritus\nThat's a good point. Newton's Laws (within their limits of validity) are valid only in inertial reference frames. Certainly one can use Newton's Second Law to describe motion in non-inertial frames, but that involves introducing forces that violate Newton's Third Law.\n\nIsn't part of the confusion of Special Relativity's ability to handle non-inertial frames historical? What I mean is that didn't Einstein, after developing Special Relativity, immediately set to work on what became General Relativity and in the process introduce the formalism of non-inertial frames? Of do I have it wrong?\n\nThe following might help - or not, I'm not sure. But it's worth saying, I hope. In Newtonian physics, the math of transforming to an accelerated frame yields Newtonian physics plus additional \"fictitious forces\".\n\nHistorically, Einstein realized early on that this wouldn't be the case for special relativity. In particular, we notice effects that we can call \"gravitational time dilation\" in accelerated frames. The usual argument here is the doppler shift argument. A signal emitted from the stern of an accelerated space-ship will be red-shifted when it reaches the bow, because while the light is travelling, the space-ship is accelerating. Going the other way, the signal is blue-shifted. The philosohical issue is how to combine this with the principle of equivalence. If we have a pair of identical clocks, for specificity imagine atomic clocks, the clocks must \"tick\" at the same rate as the signals they send out. The signals get doppler shifted, and we are forced to conclude that the rate at which the clocks tick exactly matches the signals, and since the signals are doppler shifted, we wind up concluding that the clocks themselves speed up or slowing down depending on their position when we adopt an accelerating frame.\n\nThis isn't just theory, by the way. Tests with the Mossbauer effect show the phenomenon is real, gamma rays emitted from a lower sorce won't \"resonate\" with the upper source in a gravitational field.\n\nThis argument shows the need for a paradigm shift. It's clear that \"inertial forces\" simply can't explain how clocks tick at different rates depending on their position, we need something more.\n\nI'm not aware of any substitute here for simply doing the math. The two textbook treatments of accelerated frames I'm aware of ([URL='https://www.amazon.com/dp/0716703440/?tag=pfamazon01-20 and MTW's - see links for details) both use tensors. Some of the basic issues can be outlined with simple algebra, as in our example of two clocks at the bow and stern of the acclerating rocket exchanging signals, but usually such treatments are not full enough to give a complete picture of how the accelerated frame works. They are good enough to show that the \"inertial force\" model of accelerated frames will not be sufficiently general in special relativity, however.\n[/URL]\nI think such questions often arise when one is trying to learn special relativity. Unfortunately, I don't think they can be fully answered until after one has learned SR on its own terms in the simpler case of a non-accelerated frame first. After that, one needs a high degree of abstraction - and a willingness to do the math. It would probably be possible to get somewhere without tensors on ones own if one has the neessary ability to do error-free algebra (either on ones own or with modern symbolic algebra packages), and the justified confidence to believe one own's results. But, if one wants the support of the literature, of reading what's been written about the topic, one has the not-insignificant task of learning about tensors to be able to follow the existing treatments.\n\nLast edited by a moderator:\nDrGreg\nGold Member\n$m \\frac{dV}{dt} = F$\nPresumably a typo, but that should be ## \\frac{DV}{d\\tau}##, where ##\\tau## is proper time.\n\nstevendaryl\nStaff Emeritus\nPresumably a typo, but that should be ## \\frac{DV}{d\\tau}##, where ##\\tau## is proper time.\n\nNo, it's not a typo. I was talking about Newtonian physics, where $t$ is universal.\n\n•", null, "DrGreg\nstevendaryl\nStaff Emeritus\nNo, it's not a typo. 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https://studylib.net/doc/14832582/a-reinforcement-learning-based-algorithm-for-finite-horiz...	[ "# A Reinforcement Learning Based Algorithm for Finite Horizon Markov Decision Processes\n\nadvertisement", null, "```Proceedings of the 45th IEEE Conference on Decision & Control\nManchester Grand Hyatt Hotel\nSan Diego, CA, USA, December 13-15, 2006\nFrB09.1\nA Reinforcement Learning Based Algorithm for Finite Horizon Markov\nDecision Processes\nShalabh Bhatnagar and Mohammed Shahid Abdulla\nDepartment of Computer Science and Automation,\nIndian Institute of Science, Bangalore, INDIA.\ne-mail:{shalabh,shahid}@csa.iisc.ernet.in\nAbstract— We develop a simulation based algorithm for\nfinite horizon Markov decision processes with finite state and\nfinite action space. Illustrative numerical experiments with the\nproposed algorithm are shown for problems in flow control of\ncommunication networks and capacity switching in semiconductor fabrication.\nKeywords\nFinite horizon Markov decision processes, reinforcement learning, two timescale stochastic approximation,\nactor-critic algorithms, normalized Hadamard matrices.\nI. I NTRODUCTION\nMarkov decision processes (MDPs) are a general\nframework for solving stochastic control problems .\nValue iteration and policy iteration are two of the classical approaches for solving the Bellman equation for\noptimality. Whereas value iteration proceeds by recursively iterating over value function estimates starting\nfrom a given such estimate, policy iteration does so by\niterating over policies and involves updates in two nested\nloops. The inner loop estimates the value function for a\ngiven policy update while the outer loop updates the\npolicy. The former estimates are obtained as solutions\nto linear systems of equations, known as the Poisson\nequations, that are most often solved using value iteration type recursions rather than explicit matrix inversion.\nThis is particularly true when the numbers of states\nand actions are large (the ‘curse of dimensionality’).\nHowever, in the above classical approaches, one requires\ncomplete knowledge of the system model via transition\nprobabilities. Even if these are available, the ‘curse\nof dimensionality’ threatens the computational requirements for solving the Bellman equation as these become\nprohibitive. Motivated by these considerations, research\non simulation-based methods that largely go under the\nrubric of reinforcement learning or neuro-dynamic programming have gathered momentum in recent times.\nThe main idea in these schemes is to simulate transitions\ninstead of directly computing transition probabilities and,\nin scenarios where the numbers of states and actions are\nlarge, use parametric representations of the cost-to-go\nfunction and/or policies.\nThe inner loop of the policy iteration algorithm,\nfor any given policy update, may typically take a long\ntime to converge. In , an actor-critic algorithm\nbased on two-timescale stochastic approximation was\nproposed. A simulation-based analog of policy iteration,\nthe algorithm proceeds using two coupled recursions\ndriven by different step-size schedules or timescales. The\npolicy evaluation step of policy iteration is performed\non the faster timescale while the policy improvement\n1-4244-0171-2/06/\\$20.00 ©2006 IEEE.\n5519\nstep is carried out along the slower one, the advantage\nbeing that one need not wait for convergence of the\ninner-loop before an outer-loop update unlike regular\npolicy iteration. Instead, both recursions are executed in\ntandem, one after the other, and the optimal policy-value\nfunction pair are obtained upon convergence of the algorithm. This idea of two-timescale stochastic approximation is further applied in , where parameterizations of\nboth value function (termed ‘critic’), and policy (termed\n‘actor’) are considered. As yet another application of\nthe two-timescale method, the simulation-based policy\niteration algorithm of performs updates in the space\nof deterministic stationary policies and not randomized\nstationary policies (RSPs) as in , . While not stationary, the proposed algorithm RPAFA uses randomized\npolicies as well. On the slower timescale, a gradient\nsearch using simultaneous perturbation stochastic approximation (SPSA) gradient estimates is performed and\nconvergence to a locally optimal policy is shown. While\ngradient search on the slower timescale is recommended\nin , no specific form of the gradient estimates is\nproposed there. The SPSA estimates used in are of\nthe two-measurement form first proposed in , while\na one-measurement form of SPSA was proposed in\n. A performance-enhancing modification to this latter\nalgorithm was described in that used deterministic\nperturbation sequences derived from certain normalized\nHadamard matrices. This last form of SPSA is used in\nthe proposed RPAFA algorithm.\nThe algorithm of is for infinite horizon discounted cost MDPs. Obtaining a solution to the Poisson\nequation (along the faster timescale) is simpler there as\nthe cost-to-go depends only on the state and is stageinvariant (i.e. stationary). Since we consider the finite\nhorizon setting here, the cost-to-go is now a function\nof both state and stage. The faster timescale updates\nnow involve T ‘stage-wise’ coupled stochastic recursions\nin addition to being ‘state-wise’ coupled, where T is\nthe planning horizon. Therefore, the resulting system of\nequations that need to be solved is T −fold larger than\nin . Numerical experiments in are shown over\na setting of flow control in communication networks.\nWe consider experiments not only in this setting, but\nalso on another setting involving capacity allocation in\nsemi-conductor fabs. Further, as already pointed out, \nconsiders the setting of compact (non-discrete) action\nsets while we consider a finite action setting in our work.\nReinforcement learning algorithms have generally\nbeen developed and studied as infinite horizon MDPs\nunder the discounted cost or the long-run average cost\ncriteria. For instance, approximate DP methods of TD\nlearning [2, §6.3], Q-learning [2, §6.6] and actor-critic\n45th IEEE CDC, San Diego, USA, Dec. 13-15, 2006\nFrB09.1\nalgorithms etc., have been developed in the infinite\nhorizon framework. However, in most real life scenarios,\nfinite horizon decision problems assume utmost significance. For instance, in the design of a manufacturing fab,\none requires planning over a finite decision horizon. In\na communication network, flow and congestion control\nproblems should realistically be studied only as finite\nhorizon decision making problems, since the amount\nof time required in clearing congestion and restoring\nnormal traffic flows in the network is of prime concern.\nFinite-horizon tasks also form natural subproblems in\ncertain kinds of infinite-horizon MDPs, e.g. [9, §2]\nillustrates this by using models from semiconductor fabrication and communication networks where each transition of the upper level infinite-horizon MDP spawns a\nfinite-horizon MDP at a lower level. Policies in finite\nhorizon problems depend on stage and need not be\nstationary, thereby contributing in severity to the ‘curse\nof dimensionality’.\nWe develop in this paper, two-timescale stochastic\napproximation based actor-critic algorithms for finite\nhorizon MDPs with finite state and finite action sets.\nMost of the work on developing computationally efficient algorithms for finite horizon problems, however,\nassumes that model information is known. For instance,\nin , the problem of solving a finite horizon MDP\nunder partial observations is formulated as a nonlinear\nprogramming problem and a gradient search based solution methodology is developed. For a similar problem,\na solution procedure based on genetic algorithms and\nmixed integer programming is presented in . In ,\na hierarchical structure using state aggregation is proposed for solving finite horizon problems. In contrast, we\nassume that information on transition probabilities (or\nmodel) of the system is not known, although transitions\ncan be simulated. In , three variants of the Q-learning\nalgorithm for the finite horizon problem are developed\nassuming lack of model information. However, the finite\nhorizon MDP problem is embedded as an infinite horizon\nMDP either by adding an absorbing state at the terminal\nstage (or the end of horizon) or a modified MDP is\nobtained by restarting the process by selecting one of the\nstates at the initial (first) stage of the MDP according to\nthe uniform distribution, once the terminal stage is hit.\nOur approach is fundamentally different from that in\n. In particular, we do not embed the finite horizon\nMDP into an infinite horizon one. The solution procedure that one obtains using the approach in is at\nbest only approximate, a restriction not applicable to\nour work. In the limit as the number of updates goes\nto infinity, our algorithms converge to the optimal T stage finite horizon policy. Our algorithms update all\ncomponents of the policy vector at every update epoch,\nresulting in the near-equal convergence behaviour of\nr−th stage policy components and costs-to-go, for each\nr ∈ {0, 1, ..., T − 1}.\nFurther, the method of is a trajectory-based\nscheme, in that repeated simulations of entire T −length\ntrajectories are performed, whereas our algorithm uses\nsingle transitions. In the former method, not all\n(state,action) pairs are sufficiently explored and hence\na separate exploration function is needed. Apart from a\nlook-up table proportional in the number of actions perstate, the algorithm of also requires counters for the\nnumber of times a (state,action) pair has been seen by\nthe algorithm.\nSection II describes the framework of a finitehorizon MDP, provides a brief background of the tech-\nniques used, and proposes the algorithm. In section III,\nwe illustrate numerical experiments in the framework of\nflow control in communication networks and capacity\nswitching in semiconductor fabrication wherein we compute the optimal costs and policies using the proposed\nalgorithm and compare performance with Dynamic Programming using certain performance metrics. We dwell\non some future directions in section IV.\nII. F RAMEWORK AND A LGORITHMS\nConsider an MDP {Xr , r = 0, 1, ..., T } with decision horizon T < ∞. Suppose {Zr , r = 0, 1, ..., T − 1}\nbe the associated control valued process. Decisions are\nmade at instants r = 0, 1, ..., T − 1, and the process\nterminates at instant T . Let state space at epoch r be\nSr , r = 0, 1, ..., T and let the control space at epoch\nr be Cr , r = 0, 1, ..., T − 1. Note that ST is the set\nof terminating states of this process. Let Ur (ir ) ⊂\nCr , r = 0, 1, ..., T − 1, be the set of all feasible controls\nin state ir , in period r. Let pr (i, a, j), i ∈ Sr , a ∈\nUr (i), j ∈ Sr+1 , r = 0, 1, ..., T −1 denote the transition\nprobabilities associated with this MDP. The transition\ndynamics of this MDP is governed according to\nP (Xr+1 = ir+1 \|Xk = ik , Zk = ak , 0 ≤ k ≤ r) =\npr (ir , ar , ir+1 )\nr = 0, 1, ..., T − 1, for all i0 , i1 , ..., iT , a0 , a1 , ..., aT −1 ,\nin appropriate sets. We define an admissible policy π\nas a set of T functions π = {µ0 , µ1 , ..., µT −1 } with\nµr : Sr −→ Cr such that µr (i) ∈ Ur (i), ∀i ∈\nSr , r = 0, 1, ..., T − 1. Thus at (given) instant r with\nthe system in say state i, the controller under policy π\nselects the action µr (i). Let gr (i, a, j) denote the single\nstage cost at instant r when state is i ∈ Sr , the action\nchosen is a ∈ Ur (i) and the subsequent next state is\nj ∈ Sr+1 , respectively, for r = 0, 1, ..., T − 1. Also,\nlet gT (k) denote the terminal cost at instant T when\nthe terminating state is k ∈ ST . The aim here is to\nfind an admissible policy π = {µ0 , µ1 , ..., µT −1 } that\nminimizes for all i ∈ S0 ,\nV0i (π) =\nE\nT −1\ngT (XT ) +\ngr (Xr , µr (Xr ), Xr+1 )\|X0 = i\n.\nr=0\n(1)\nThe expectation above is over the joint distribution\nof X1 , X2 , ..., XT . The dynamic programming algorithm for this problem is now given as follows (see\n): for all i ∈ ST ,\nVTi (π) = gT (i)\n(2)\nand for all i ∈ Sr , r = 0, 1, ..., T − 1,\nVri (π) =\nmin\na∈Ur (i)\n⎧\n⎨ ⎩\nj∈Sr+1\n⎫\n⎬\nj\npr (i, a, j)(gr (i, a, j) + Vr+1\n(π)) ,\n⎭\n(3)\nrespectively.\n5520\n45th IEEE CDC, San Diego, USA, Dec. 13-15, 2006\nFrB09.1\nA. Brief Overview and Motivation\nH2k ×2k =\nNote that using dynamic programming, the original\nproblem of minimizing, over all admissible policies π,\nthe T -stage cost-to-go V0i (π), ∀i ∈ S0 , as given in (1),\nis broken down into T coupled minimization problems\ngiven by (2)-(3) with each such problem defined over\nthe corresponding feasible set of actions for each state.\nHere,‘coupled’ means that the r−th problem depends on\nthe solution of the (r + 1)−th problem, for 0 ≤ r < T .\nIn this paper, we are interested in scenarios where the\npr (i, a, j) are not known, however, where transitions can\nbe simulated. Thus, given that the state of the system at\nepoch r (r ∈ {0, 1, .., T −1}) is i and action a is picked\n(possibly randomly), we assume that the next state j can\nbe obtained through simulation.\nWe combine the theories of two-timescale stochastic\napproximation and SPSA to obtain actor-critic algorithms that solve all the T coupled minimization problems (2)-(3), under the (above) lack of model information constraint. For the case of infinite horizon problems,\n, and all use two-timescale stochastic approximation algorithms of which adopts two-simulation\nSPSA to estimate the policy gradient.\nBefore we proceed further, we first motivate the use\nof SPSA based gradient estimates and two timescales in\nour algorithm. Suppose we are interested in finding the\nminimum of a function F (θ) when F is not analytically\navailable, however, noisy observations f (θ, ξn ), n ≥ 0\nof F (θ) are available with ξn , n ≥ 0 being i.i.d. random\nvariables satisfying F (θ) = E[f (θ, ξn )]. The expectation above is taken w.r.t. the common distribution of ξn ,\nn ≥ 0. Let θ = (θ1 , ..., θq )T , ∆i (n), i = 1, ..., q, n ≥ 0\nbe generated according to the method in section II-A.1\nbelow and let ∆(n) = (∆1 (n), . . . , ∆N (n))T , n ≥ 0.\nLet θ(n) denote the nth update of parameter θ. For a\ngiven (small) scalar δ > 0, form the parameter vector\nθ(n) + δ∆(n). Then the one-measurement SPSA gra˜ i F (θ(n)) of ∇i F (θ(n)), i = 1, . . . , q\ndient estimate ∇\nhas the form (cf. algorithm SPSA2-1R of :\n˜ i F (θ(n)) = f (θ(n) + ∆(n), ξn ) ,\n∇\nδ∆i (n)\nH2k−1 ×2k−1\nH2k−1 ×2k−1\nH2k−1 ×2k−1\n−H2k−1 ×2k−1\n,\nfor k > 1.\n2) Two-timescale stepsizes: Let {b(n)} and\n{c(n)} be two step-size schedules that satisfy\nn\nb(n) =\nn\nc(n) = ∞,\nb(n)2 ,\nn\nc(n)2 < ∞,\nn\nand\nc(n) = o(b(n)),\nrespectively. Thus {c(n)} goes to zero faster than\n{b(n)} does and corresponds to the slower timescale\n(since beyond some integer N0 (i.e., for n ≥ N0 ),\nthe sizes of increments in recursions that use {c(n)}\nare uniformly the smallest, and hence result in slow\nalbeit graceful convergence). Likewise, {b(n)} is the\nfaster scale. Informally, a recursion having b(n) as the\nstepsize views a recursion that has stepsize c(n) as static\nwhilst the latter recursion views the former as having\nconverged.\nB. Proposed Algorithm (RPAFA)\nThe acronym RPAFA stands for ‘Randomized Policy\nAlgorithm over Finite Action sets’. In the following, for\nnotational simplicity and ease of exposition, we assume\nthat the state and action spaces are fixed and do not\nvary with stage. Thus S and C respectively denote the\nstate and control spaces. Further, U (i) denotes the set of\nfeasible actions in state i ∈ S and is also stage invariant.\nThe set U (i) of feasible actions in state i is assumed\nfinite. Further, we assume that each set U (i) has exactly\n(q + 1) elements u(i, 0), . . ., u(i, q) that however may\ndepend on state i. For theoretical reasons, we will need\nthe following assurance on per-stage costs.\nAssumption (A) The cost functions gr (i, a, j), i, j ∈ S,\na ∈ U (i), are bounded for all r = 0, 1, ..., T −1. Further,\ngT (i) is bounded for all i ∈ S.\nWe now introduce a Randomized Policy (RP). Let\nπr (i, a) be the probability of selecting action a in state\ni at instant r and let π̂r (i) be the vector (πr (i, a), i ∈\nS, a ∈ U (i)\\{u(i, 0)})T , r = 0, 1, ..., T −1. Thus given\nπ̂r (i) as above, the probability πr (i, u(i, 0)) of selecting\naction u(i, 0) in state i at instant r is automatically\nq\nspecified as πr (i, u(i, 0)) = 1 − j=1 πr (i, u(i, j)).\nHence, we identify a RP with π̂ = (π̂r (i), i ∈ S, 0 ≤\nr ≤ T − 1)T . Let S̄ = {(y1 , . . . , yq )\|yj ≥ 0, ∀j =\nq\n1, . . . , q, j=1 yj ≤ 1} denote the simplex in which\nπ̂r (i), i ∈ S, r = 0, 1, ..., T − 1 take values. Suppose\nP : Rq → S̄ denotes the projection map that projects\nπ̂r (i) to the simplex S̄ after each update of the algorithm\nbelow.\n(4)\nNote that only one measurement (corresponding to\nθ(n) + δ∆(n)) is required here.\nAlso observe that unlike SPSA estimates, KieferWolfowitz gradient estimates require 2q (resp. (q + 1))\nmeasurements when symmetric (resp. one-sided) differences are used. We now describe the construction of the\ndeterministic perturbations ∆(n), n ≥ 1, as proposed in\n.\n1) Construction for Deterministic Perturbations: Let H be a normalized Hadamard matrix (a\nHadamard matrix is said to be normalized if all the elements of its first row and column are 1s of order P with\nP ≥ q + 1). Let h(1), ..., h(q) be any q columns other\nthan the first column of H, thus forming a new (P ×q)–\ndimensional matrix Ĥ. Let Ĥ(p), p = 1, ..., P denote\nthe P rows of Ĥ. Now set ∆(n) = Ĥ(n mod P + 1),\n∀n ≥ 0. The perturbations are thus generated by cycling\nthrough the rows of the matrix Ĥ. Here P is chosen as\nP = 2log2 (q+1) . It is shown in that under the above\nchoice of P , the bias in gradient estimates asymptotically\nvanishes. Finally, matrices H ≡ HP ×P of dimension\nP × P , for P = 2k , are systematically constructed as\nfollows:\n1 1\n,\nH2×2 =\n1 −1\nAlgorithm RPAFA\n• Step 0 (Initialize): Fix π̂0,r (i, a), ∀i ∈ S, a ∈\nU (i), 0 ≤ r ≤ T − 1 as the initial RP iterate.\nFix integers L and (large) M arbitrarily. Fix a\n(small) constant δ > 0. Choose step-sizes b(n) and\nc(n) as in Section II-A.2. Generate Ĥ as in Section\nII-A.1. Set Vk,r (i) = 0, ∀0 ≤ r ≤ T − 1, and\nVk,T (i) = gT (i), 0 ≤ k ≤ L − 1, i ∈ S as initial\nestimates of cost-to-go. Set ‘actor’ index n := 0.\n• Step 1: For all i ∈ S, 0 ≤ r ≤ T − 1, do:\n– Set ∆n,r (i) := Ĥ(n mod P + 1),\n– Set π̄n,r (i) := P (π̂n,r (i) +δ∆n,r (i)).\n5521\n45th IEEE CDC, San Diego, USA, Dec. 13-15, 2006\n•\nFrB09.1\ntransition probability matrix PT̃ , and in order to compute\nthese we use the approximation method of [15, §6.8].\nSince each state has the same q + 1 admissible controls,\nq + 1 number of PT̃ matrices of size B × B each are\nrequired. This storage required becomes prohibitive as\neither the state space increases, or the discretization is\nmade finer. Also note that the amount of computation\nrequired for PT̃ also depends upon the convergence\ncriteria specified for the method in [15, §6.8]. Besides,\nsuch probabilities can only be computed for systems\nwhose dynamics are well known, our setting being that\nof a well-studied M/M/1/B queue.\nStep 2 (Critic): For ‘critic’ index m = 0, 1, ..., L−\n1, r = T − 1, T − 2, ..., 0, and i ∈ S, do\n– Simulate action φnL+m,r (i) according to distribution π̄n,r (i).\n– Simulate next state ηnL+m,r (i) according to\ndistribution pr (i, φnL+m,r (i), ·).\n– VnL+m+1,r (i) := (1 − b(n))VnL+m,r (i)\n+b(n)gr (i, φnL+m,r (i), ηnL+m,r (i))\n+b(n)VnL+m,r+1 (ηnL+m,r (i)).\nStep 3 (Actor): For i ∈ S, 0 ≤ r ≤ T − 1, do\nπ̂n+1,r (i) := P (π̂n,r (i) − c(n)\n•\nVnL,r (i) −1\n∆n,r (i))\nδ\npolicies for various r\n4.5\nSet n := n + 1.\nIf n = M , go to Step 4;\nelse go to Step 1.\nStep 4 (termination): Terminate algorithm and output π̄M as the final policy.\nt=1\nt=3\nt=6\nt=9\n4\n3.5\n3\nLambda-C\n•\nIII. S IMULATION R ESULTS\nA. Flow control in communication networks\n2.5\n2\n1.5\n1\n0.5\nWe consider a continuous-time queuing model of\nflow control. The numerical setting here is somewhat\nsimilar to that in . In , this problem is modelled\nin the infinite horizon discounted cost MDP framework,\nwhile we study it in the finite horizon MDP framework\nhere. Flow and congestion control problems are suited\nto a finite horizon framework since a user typically\nholds the network for only a finite time duration. In\nmany applications in communication networks, not only\nis the time needed to control congestion of concern,\nthese applications must also be supported with sufficient\nbandwidth throughout this duration.\nAssume that a single bottleneck node has a finite\nbuffer of size B. Packets are fed into the node by\nboth an uncontrolled Poisson arrival stream with rate\nλu = 0.2, and a controlled Poisson process with rate\nλc (t) at instant t > 0. Service times at the node are\ni.i.d., exponentially distributed with rate 2.0. We assume\nthat the queue length process {Xt , t > 0} at the node\nis observed every T̃ instants, for some T̃ > 0, upto\nthe instant T̃ T . Here T stands for the terminating stage\nof the finite horizon process. Suppose Xr denotes the\nqueue length observed at instant r T̃ , 0 ≤ r ≤ T . This\ninformation is fed back to the controlled source which\nthen starts sending packets at λc (Xr ) in the interval\n[rT̃ , (r + 1)T̃ ), assuming there are no feedback delays.\nWe use B = 50 and T = 10, and designate the ‘target\nstates’ T̂r as evenly spaced states within the queue i.e.,\n{T̂1 = 40, T̂2 = 37, T̂3 = 34, ..., T̂9 = 16, T̂10 = 0}.\nThe one-step transition cost under a given policy π\nis computed as gr (ir , φr (ir ), ir+1 ) = ir+1 − T̂r+1\nwhere φr (ir ) is a random variable with law πr (ir ).\nAlso, gT (i) = 0, ∀i ∈ S. Such a cost function penalizes\nstates away from the target states T̂r+1 , apart from\nsatisfying Assumption (A). The goal thus is to maximize\nthroughput in the early stages (r small), while as r\nincreases, the goal steadily shifts towards minimizing\nthe queue length and hence the delay as one approaches\nthe termination stage T .\nFor the finite action setting, we discretize the interval\n[0.05, 4.5] so as to obtain five equally spaced actions\nin each state. For purposes of comparison, we also implemented the DP algorithm (1)-(2) for the finite action\nsetting. Application of DP presupposes availability of the\n0\n0\n5\nFig. 1.\n10\n15\n20\n25\nStates\n30\n35\n40\n45\n50\nOptimal Policy computed Using DP\nRPAFA: policy for various r\n4.5\nr=1\nr=3\nr=6\nr=9\n4\n3.5\nLambda-C\n3\n2.5\n2\n1.5\n1\n0.5\n0\n0\n5\n10\n15\n20\n25\n30\n35\n40\n45\n50\nStates\nFig. 2.\nPolicy computed using RPAFA\nFinite Horizon Costs-to-go\n110\nDP\nRPAFA\n100\nCost-to-go\n90\n80\n70\n60\n50\n0\n5\n10\nFig. 3.\n15\n20\n25\nStates\n30\n35\n40\n45\n50\nComparison of Costs-to-go\nThe policy obtained using finite-horizon DP with PT̃\ncomputed as above is shown in Figure 1. Note how the\npolicy graphs shift to the left as the target state moves\nto the left for increasing r. Thus, the throughput of the\n5522\n45th IEEE CDC, San Diego, USA, Dec. 13-15, 2006\nFrB09.1\nr=2\nr=6\nr=10\n37\n25\n0\n27.1±6.3 22.4±5.1 5.7±3.6\n19.2±6.5 19.4±4.8 8.9±5.3\nTABLE I\nOBSERVED E(ir ) FOR THE PROPOSED ALGORITHM\nB. Capacity Switching in Semiconductor Fabs\nTarget T̂r\nDP\nRPAFA\nWe first briefly describe the model of a semiconductor fabrication unit, considered in . The factory\nprocess {Xr \|X0 = i}, 1 ≤ r ≤ T , i ∈ S0 is such\nthat each Xr is a vector of capacities at time epochs\n(A,i),w\nrepresenting\nr ∈ {1, 2, ..., T }, the components Xr\nthe number of type w machines allocated to performing\noperation i on product A. The duration of the planning\nhorizon is T , casting this problem as a finite horizon\nMarkov Decision Process. A type w machine indicates\na machine capable of performing all operations which\nare letters in the ‘word’ w. Note that the product A\nrequires the operation i for completion and that the word\nw contains the letter i, among others. The word w of a\nmachine can also contain the action 0 - indicating idling.\nThe control µr (Xr ) taken at stages 0 ≤ r ≤ T − 1\n(A,i),w,(B,j)\nwould be to switch ur\nmachines of type w\nfrom performing operation i on product A to performing\noperation j on product B.\nAs in inventory control models, the randomness in\nthe system is modeled by the demand DrA for product\nA at stages 0 ≤ r ≤ T − 1. The per-step transition cost\ne\n, for every\nconsists of costs for excess inventory (KA\nb\nunit of product A in inventory), backlog (KA\n), cost of\no\n, one-stage operating cost for a machine\noperation (Kw\nof type w) and the cost of switching capacity from one\ns\ntype of operation to another (Kw\n, the cost of switching\na machine of type w from one type of production to\nanother). In all these costs, further complications can be\ns\ncould be indexed with the\nadmitted, e.g., the cost Kw\nsource product-operation pair (A, i) and the destination\npair (B, j).\nWe consider here a simple model also experimented\nwith in where the infinite horizon discounted\ncost for a semiconductor fab model was computed\nusing function approximation coupled with the policy\niteration algorithm. In contrast, we do not use function\napproximation and adopt a finite horizon of 10. The\ncost structure, however, remains the same as . In\nparticular, we consider a fab producing two products (A\nand B), both requiring two operations, ‘litho’ and ‘etch’\n(call these l and e). We have a fab with 2 litho and 2\netch machines that can each perform the corresponding\noperation on either A or B. We denote the operations\n{A, l} (i.e., ‘litho on A’) as 1, {A, e} as 2, {B, l} as\n3 and {B, e} as 4, implying that the word w of a litho\nmachine is 013 and that of an etch machine is 024. The\nproduct-operation pairs that admit non-zero capacities\nare therefore: (A, 1), (A, 2), (B, 3), and (B, 4).\nThe fab’s throughput Pr\n=\n(PrA , PrB )\nis\nconstrained\nby\nthe\nfollowing\nrelation:\n(A,1),013\n(A,2),024\n, Xr\n) and PrB =\nPrA = min(Xr\n(B,3),013\n(B,4),024\n0.5·min(Xr\n, Xr\n), 0 ≤ r ≤ T − 1\nimplying the slower production of B. We assume\nthat no machine is idling and therefore the capacity\nallocated to product B, given capacity allocated to\nA, is simply the remainder from 2 for both litho and\n(B,3),013\n(A,1),013\n= 2 − Xr\n,\netch. Therefore, these are Xr\n(B,4),024\n(B,4),024\nXr\n=\n2 − Xr\n. We also\n∈\n{−1, 0, 1},\nconstrain the inventories: IrA\nIrB ∈ {−0.5, 0, 0.5}, 0 ≤ r ≤ T − 1. The state\nis thus completely described by the quadruplet:\n(A,1),013\n(A,2),024\n, Xr\n, IrA , IrB ), 0 ≤ r ≤ T .\nXr = (Xr\nOne can choose a lexicographic order among the\npossible starting states X0 , which are 34 = 81 in\nnumber.\nCapacity switching at stage 0 ≤ r ≤ T − 1 is\nr=2\nr=6\nr=10\n37\n25\n0\n0.07±0.13 0.21±0.34 0.19±0.31\n0.01±0.03 0.13±0.22 0.14±0.24\nTABLE II\nP ROBABILITIES pr = P (ir = T̂r ± 1)\nTarget T̂r\nDP\nRPAFA\nsystem decreases from one stage to another as the target\nqueue length is brought closer to zero.\nThe algorithm RPAFA is terminated at iteration n\nwhere errn ≤ 0.01. The convergence criterion is\nerrn =\nmax\ni∈S,k∈{1,2,...,50}\nπn (i) − πn−k (i) 2 ,\nwhere πn (i) = (πn,r (i, a), 0 ≤ r ≤ T − 1, a ∈ U (i))T .\nFurther, · 2 is the Euclidean norm in RT ×(q+1) .\nOn a Pentium III computer using the C programming\nlanguage, termination required upto 47 × 103 updates\nand 8 × 103 seconds. The policy obtained is shown\nin Figure 2, where for each state the source rate\nindicated is the rate that has the maximum probability\nof selection. Also shown in Figure 3 is the finite-horizon\ncost for each state in a system operating under the\npolicies in Figure 1 and Figure 2, respectively. The\nplots shown in Figure 3 are obtained from 2 × 105\nindependent sample trajectories {i0 , i1 , ..., i10 }, each\nstarting from state i0 = 0 with a different initial seed.\nTables I, II and III show performance comparisons\nof the proposed algorithm for various metrics with\nmean and standard deviation taken over the (above\nmentioned) 2 × 105 independent sample trajectories.\nTable I shows the mean queue length E(ir ) at\ninstants r = 2, 4, 6, 8, and 10, respectively, with the\ncorresponding standard deviation. With T̂r defined\nas the ‘target’ state at instant r, Table II shows the\nprobability of the system being in states T̂r ± 1 at the\nabove values of r. Table III shows the mean one-stage\ncost E(gr−1 (ir−1 , µr−1 (ir−1 ), ir )) = E(\|ir − T̂r \|)\nincurred by the system during transition from ir−1\nto ir under policy π. Note that the relatively bad\nperformance of RPAFA is since it applies a randomized\npolicy wherein the optimal action, though having a high\nprobability of selection, may not be selected each time.\nThe approximate DP algorithm in this case converges much faster since transition probabilities using\nthe method in are easily computed in this setting.\nHowever, in most real life scenarios, computing these\nprobabilities may not be as simple, and one may need\nto rely exclusively on simulation-based methods.\nr=2\nr=6\nr=10\n37\n25\n0\n10.5±5.5 5.2±3.6 5.8±3.6\n18.0±6.3 6.7±3.9 8.8±5.3\nTABLE III\nM EAN ONE - STAGE COSTS E(\|ir − T̂r \|)\nTarget T̂r\nDP\nRPAFA\n5523\n45th IEEE CDC, San Diego, USA, Dec. 13-15, 2006\nRPAFA\nIterations\n1500\nTime in sec.\n209\nTABLE IV\nerrn\n0.1\nFrB09.1\nMax Error\n30.4\nperform better and converge faster as compared to onesimulation SPSA, which could be derived along similar\nlines (see ). In order to further improve performance,\nefficient simulation-based higher order SPSA algorithms\nthat also estimate the Hessian in addition to the gradient\nalong the lines of could also be explored.\nP ERFORMANCE OF RPAFA\nspecified by the policy component µr . Thus µr (Xr ) is\na vector such that µr (Xr ) = (µr (Xr , 1), µr (Xr , 2)),\n(A,1),013\n(A,1),013\nmaking Xr+1\n= Xr\n+ µr (Xr , 1) (similarly\n(A,2),024\nfor Xr+1\n). Note that controls µr (Xr ) belong to\nthe feasible action set Ur (Xr ), and that in our setting\n0 ≤ \|Ur (Xr )\| ≤ 9, the state ‘group’ (1, 1, xA , yB ),\n∀xA ∈ {−1, 0, 1}, ∀yB ∈ {−0.5, 0, 0.5} having the\nmaximum of 9 actions. We take the various one-step\ne\ne\nb\nb\n= 2, KB\n= 1, KA\n= 10, KB\n= 5,\ncosts to be KA\no\no\ns\ns\nK013\n= 0.2, K024\n= 0.1, K013\n= 0.3 and K024\n= 0.3.\nThe noise (demand) scheme is such that: DrA = 1\nw.p. 0.4, DrA = 2 otherwise, and DrB = 0.5 w.p.\n0.7, DrB = 1 otherwise, for all 0 ≤ r ≤ T − 1.\nWe assume that the control µr (Xr ) is applied to the\nstate Xr at the beginning of the epoch r whereas the\n(A,1),013\n=\ndemand Dr is made at the end of it, i.e., Xr+1\n(A,1),013\n(A,2),024\n(A,2),024\nXr\n+ µr (Xr , 1), Xr+1\n= Xr\n+\nA\nµr (Xr , 2), Ir+1\n= max(min(IrA + PrA − DrA , 1), −1),\nA\nIr+1\n= max(min(IrB + PrB − DrB , 0.5), −0.5)). We\nalso take the terminal cost KT (XT ) to be 0. Thus, we\nhave the per-stage cost:\nAcknowledgments\nThis work was supported in part by Grant no.\nSR/S3/EE/43/2002-SERC-Engg from the Department of\nScience and Technology, Government of India.\nR EFERENCES\n M. Puterman, Markov Decision Processes: Discrete\nStochastic Dynamic Programming.\nNew York: John\nWiley, 1994.\n D. Bertsekas and J. Tsitsiklis, Neuro-Dynamic Programming. Belmont, MA: Athena Scientifi c, 1996.\n V. Konda and V. Borkar, “Actor–Critic Type Learning Algorithms for Markov Decision Processes,” SIAM Journal\non Control and Optimization, vol. 38, no. 1, pp. 94–123,\n1999.\n V. Konda and J. Tsitsiklis, “Actor–Critic Algorithms,”\nSIAM Journal on Control and Optimization, vol. 42, no. 4,\npp. 1143–1166, 2003.\n S. Bhatnagar and S. Kumar, “A Simultaneous Perturbation\nStochastic Approximation–Based Actor–Critic Algorithm\nfor Markov Decision Processes,” IEEE Transactions on\nAutomatic Control, vol. 49, no. 4, pp. 592–598, 2004.\n J. Spall, “Multivariate stochastic approximation using a\nsimultaneous perturbation gradient approximation,” IEEE\nTransactions on Automatic Control, vol. 37, no. 1, pp.\n332–341, 1992.\n ——, “A One–Measurement Form of Simultaneous Perturbation Stochastic Approximation,” Automatica, vol. 33,\nno. 1, pp. 109–112, 1997.\n S. Bhatnagar, M. Fu, S. Marcus, and I.-J. Wang, “Two–\ntimescale simultaneous perturbation stochastic approximation using deterministic perturbation sequences,” ACM\nTransactions on Modeling and Computer Simulation,\nvol. 13, no. 4, pp. 180–209, 2003.\n H. Chang, P. Fard, S. Marcus, and M. Shayman, “Multitime Scale Markov Decision Processes,” IEEE Transactions on Automatic Control, vol. 48, no. 6, pp. 976–987,\n2003.\n Y. Serin, “A nonlinear programming model for partially\nobserved Markov decision processes: fi nite horizon case,”\nEuropean Journal of Operational Research, vol. 86, pp.\n549–564, 1995.\n A. Lin, J. Bean, and C. White, III, “A hybrid genetic/optimization algorithm for fi nite horizon partially\nobserved Markov decision processes,” INFORMS Journal\nOn Computing, vol. 16, no. 1, pp. 27–38, 2004.\n C. Zhang and J. Baras, “A hierarchical structure for\nfi nite horizon dynamic programming problems,” Technical\nReport TR2000-53, Institute for Systems Research, University of Maryland, USA, 2000.\n F. Garcia and S. Ndiaye, “A learning rate analysis of\nreinforcement learning algorithms in fi nite-horizon,” in\nProceedings of the Fifteenth International Conference on\nMachine Learning, Madison, USA, 1998.\n D. Bertsekas, Dynamic Programming and Optimal Control, Volume I. Belmont, MA: Athena Scientifi c, 1995.\n S. Ross, Introduction to Probability Models, 7/e. San\nDiego, CA: Academic Press, 2000.\n Y. He, S. Bhatnagar, M. Fu, S. Marcus, and P. Fard,\n“Approximate policy iteration for semiconductor fab-level\ndecision making - a case study,” Technical Report, Institute for Systems Research, University of Maryland, 2000.\n S. Bhatnagar, “Adaptive multivariate three–timescale\nstochastic approximation algorithms for simulation based\noptimization,” ACM Transactions on Modeling and Computer Simulation, vol. 15, no. 1, pp. 74–107, 2005.\nKr (Xr , µr (Xr ), Xr+1 ) =\no\no\ns\nA,2\ns\n2K013\n+ 2K024\n+ µA,1\nr ·K013 + µr ·K024\nA\ne\nB\ne\n+max(Ir+1\n, 0)·KA\n+ max(Ir+1\n, 0)·KB\nA\nb\nB\nb\n+max(−Ir+1\n, 0)·KA\n+ max(−Ir+1\n, 0)·KB\nThe resulting costs-to-go obtained using our algorithm is plotted in Figure 4. The states are sorted\nin decreasing order of the exact costs-to-go (computed\nusing DP). The corresponding cost computed using the\nconverged policy of RPAFA is seen to be reliably close\n- albeit higher than DP. The computing performance is\noutlined in Table IV. In the experiments we chose L\nand δ as 200 and 0.1, respectively.\nFinite Horizon Costs-to-go\n85\nRPAFA\nDP\n80\nCost-to-go\n75\n70\n65\n60\n55\n50\n0\n10\nFig. 4.\n20\n30\n40\n50\nStates in Sorted Order\n60\n70\n80\nComparison of Costs-to-go\nIV. F UTURE D IRECTIONS\nThe policy iteration algorithm used one-simulation\nSPSA gradient estimates with perturbation sequences\nderived from normalized Hadamard matrices. However,\nin general, two-simulation SPSA algorithms are seen to\n5524\n```" ]	[ null, "https://s2.studylib.net/store/data/014832582_1-73a20c47d6fba89eae81abc1102e8c22.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87296313,"math_prob":0.9725115,"size":34336,"snap":"2021-04-2021-17","text_gpt3_token_len":9610,"char_repetition_ratio":0.1239951,"word_repetition_ratio":0.044671863,"special_character_ratio":0.29278308,"punctuation_ratio":0.16957338,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99746174,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-16T23:13:11Z\",\"WARC-Record-ID\":\"<urn:uuid:a5bcc787-e415-4c9a-9cd4-df54b9652c6b>\",\"Content-Length\":\"72489\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6fbe7c4f-15b0-4fa3-9c6d-1e088b0acd14>\",\"WARC-Concurrent-To\":\"<urn:uuid:d55315ef-676c-424f-8061-d8f8f298ef7c>\",\"WARC-IP-Address\":\"104.21.72.58\",\"WARC-Target-URI\":\"https://studylib.net/doc/14832582/a-reinforcement-learning-based-algorithm-for-finite-horiz...\",\"WARC-Payload-Digest\":\"sha1:TZOIDA5MYAHYSFD5JWXE4BITAAHQTVP6\",\"WARC-Block-Digest\":\"sha1:VOW5XONQVNAI3546MZ633JZM2XCTFIDT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038092961.47_warc_CC-MAIN-20210416221552-20210417011552-00534.warc.gz\"}"}
https://physics.stackexchange.com/questions/186693/explain-why-quantum-behavior-is-not-observed-in-daily-life/193408	[ "Explain why quantum behavior is not observed in daily life\n\nHow come we don't see any \"Wave\" attached to a classical object such as a car?\n\nYou always see the object in the same place without any uncertainty.\n\nI am sure there are answers, but please use the simplest language understandable by general public.\n\n• Lasers are observable in daily life, and are a quantum phenomena. The wavelength of a car is reallllly small, and our sense organs did not evolve to see them. – Jon Custer May 29 '15 at 20:01\n• Every atom in your body only exists because of QM. Are you telling me that you can't observe yourself? Magnets are macroscopic quantum systems and a hospital near you probably has an MRI system with a multi-ton superconducting magnet in a macroscopic quantum state. That magnet is being used to detect quantum transitions of the spins of hydrogen nuclei in your body. The list of examples of quantum mechanical effects in your everyday life can be extended easily. That \"people\" don't know which effect they observe is quantum mechanical in nature is simply a matter of education. – CuriousOne May 29 '15 at 22:37\n• Too high temperature at the first place – Anixx May 29 '15 at 23:15\n• I agree with previous comments: some everyday effects deeply rely on quantum mechanics. Take for instance the cohesion of matter, which would not occur without QM (if I recall correctly). I also agree with some given answers: the so-called wave-like behavior is not visible due to its scale for most particles. – fffred Jul 10 '15 at 14:14\n\nQuantum mechanics became necessary as an underlying framework of classical mechanics due to experimental observations that classical mechanics and classical electrodynamics could not explain.\n\n1) The photoelectric effect, that light hitting metal extracts electrons with discrete energy characteristic if the metal, the effect can be explained only as a particle hitting the electron; classical electrodynamics says light is a wave with alternating electric and magnetic fields of continuous energy.\n\n2)the periodic table of elements which showed a limit to the fragmentation of matter into multiples of the weight of hydrogen. Classically fragmentation could be continuous and no structure could be predicted\n\n3)black body radiation, which could not be explained with classical thermodynamics and electromagnetism. The model that fitted the data had to posit that there were particles, called photons, in the electromagnetic radiation of energy E=hnu .\n\nh is the Planck's constant and is the basic reason that quantum mechanical effects are not evident macroscopically. For classical mechanics h=0. In the quantum level it is a very small number but its value controls the dimensions in which quantum mechanical effects dominate by the Heisenberg Uncertainty Principle. This says:", null, "(ħ is the reduced Planck constant).\n\nAs a rule of thumb, quantum mechanical effects dominate when the uncertainty in position with the uncertainty in momentum is of the order or smaller than the inequality seen above.\n\nMacroscopic objects in general fulfill this automatically because ħ is such a small number, $6.62606957(29)×10^{-34}$ jouleseconds, that our accuracies of measuring the position and momentum of the car of the example are much larger than the variation in position and momentum of the car due to the underlying quantum mechanical variations. Our four senses are worse than our measuring instruments, of course.\n\nSo classical physics is adequate to describe experiments and situations where h is effectively zero.\n\nThe quantum mechanical effects surviving macroscopically are second level effects, like transistors, or the regularity of crystals, or the above numbered three discrepancies with classical physics. As other answers state one can prove mathematically that quantum mechanical solutions at the limit end up in classical solutions, though it needs quite some mathematical background to understand it is so.\n\n• I know your point of h=0 – kwei Chang Jun 2 '15 at 12:00\n• @annav What do you mean smaller than the inequality? – user100411 Apr 13 '17 at 16:42\n• @JohnDoe when d(p)*d(x) is smaller than h_bar one is in the quantum mechanical regime. – anna v Apr 13 '17 at 17:29\n• @annav So then we would have $\\hbar > \\sigma_x \\sigma_p \\geq \\frac{\\hbar}{2}$? – user100411 Apr 13 '17 at 17:55\n• @JohnDoe forget the h/2, it is order of magnitude and smaller, I just did not divide by two or had a bar sign . quantum effects appear ( solutions cannot be classical ,have to be calculated with quantum mechanical equations) it is order of magnitude, even for five times that. – anna v Apr 13 '17 at 18:02\n\nThis because though quantum physics applies everywhere, classical mechanics is a good approximation. Just like in special relativity, where as v -> 0 (or we as assume c -> $\\infty$), Newton's equations become more valid, the same thing happens in quantum mechanics.\n\nIn quantum mechanics, this called the Correspondence Principle. It formally says that as n (quantum number) -> $\\infty$, Newton's classical mechanics becomes more correct, but a simpler (though somewhat less correct) way of putting it is that as we \"zoom out\" (i.e. go to larger scales) classical mechanics provides an accurate approximation.\n\nAlso, in your car example, it hs to do with wavelength. The highest the momentum of a particle, the shorter its wavelength. A car has tremendous momentum compared with an electron or photon. As a consequence, its wavelength is small. As wavelengths get smaller, wave-like properties diminish and particle-like properties emerge.\n\nThat being said, we can observe quantum phenomena: lasers, the photoelectric effect, heat radiation, etc.\n\n• To my opinion, n=infinite makes energy appears continuous, – kwei Chang Jun 1 '15 at 0:22\n• @kweiChang Yes, you are correct. – Jimmy360 Jun 1 '15 at 0:23\n• @kweiChang Please do not use answers as comments. Yes, I am 14 (with a working knowledge of calculus, trigonometry, tensors, ... (so I know what I'm talking about)) – Jimmy360 Jun 1 '15 at 15:19\n• 14 years old should belong to high school , followed by college and graduate school graduation. Spending time to places like this is not what you can afford. – kwei Chang Jun 4 '15 at 2:36\n• @kweiChang Besides, that isn't any of your business. – Jimmy360 Jun 4 '15 at 3:26\n\nI will start with the short version of the answer and then move to a little longer version of it.\n\nWe do not observe quantum behavior in real life because of the limitations of our biological architecture.\n\nA little longer version\n\nWe didn't need to have evolutionary traits that are attuned to quantum phenomena. It is the reason why quantum physics seems so queer. An untrained person doesn't have a vocabulary to express the phenomena of an object being in two opposite states at the same time.\n\nFor example, a quantum coin could be in a state of both head and tail. How is that possible? How can it be? Well, our tongues can't articulate it. In fact, this whole thing about light being both wave and particle is a hype. This is used like a catchphrase, but all it shows that we don't have a large enough vocabulary to describe the nature of light and other quantum objects. And that's okay! In real life, we catch a ball and not an electron. That's why, in real life, quantum phenomena is not observed.\n\nJust because it's not observed, doesn't mean it doesn't occur\n\n@Jon Custer, mentioned about laser in the comments, and I'll mention that here too. Every time you are in a presentation, and the presenter uses the laser pointer, quantum phenomena is occurring. Sun shines, and the explanation of the mechanism needs quantum physics. We have food because of photosynthesis, and that's a quantum phenomenon too. You observe quantum phenomena all the time, if you define observing by both sensing the event and acknowledging the mechanism behind it.\n\nThe last thing I'll comment on in this answer, is the \"associated wave with a particle\" thing that you mentioned in the question. There's this fundamental constant in quantum physics, called the Planck's constant, $\\hbar$. I could write down its value and then tell you how disastrously small it is. I won't do that because, $\\hbar$ is a constant with physical units. You can say 2 is a small number. But, you cannot say 2 m is small, because it's a large distance for an ant, and small distance for an athlete.\n\nAs far as the $\\hbar$ is concerned, we're giants compared to it. By 'we', I mean our coarse senses. Again, $\\hbar$ isn't small, our senses are accustomed to too large of things!\n\nActually, Heisenberg's uncertainty principle certainly appears in everyday life, and in some cases, drives what we see classically.\n\nTake for example a pencil balanced on its point. Assume you want to know the length of time that it can be balanced on its point.\n\nSo, the standard thing to do is to consider a pencil with mass $m$ and length $l$, something like an inverted pendulum. Using the small-angle approximation, you get the second-order pendulum equation ODE, which has solutions:\n\n$\\theta(t) = A \\exp \\left[\\sqrt{g/L}t\\right] + B \\exp \\left[-\\sqrt{g/L}t\\right]$.\n\nNow, we can obtain canonical variables (as are used in Heisenberg's uncertainty principle) by computing:\n\n$x_{0} = \\theta(0) L = (A+B)L$, and $p_{0} = m \\dot{\\theta} L = m \\sqrt{gL}(A-B)$.\n\nNow, the product is: $x_{0} p_{0} = m \\sqrt{g} L^{3/2} (A^2-B^2) \\equiv \\hbar$. Ignore $B$ since it's just a constant, and the pencil will fall when it is tilted by $1^{\\circ}$, so solving for the time, we get:\n\n$T = \\sqrt{L/g} \\left[(1/4) \\log \\frac{m^2 g L^3}{\\hbar^2} - \\log(180/\\pi)\\right]$.\n\nThe first term is much larger! So, Quantum Mechanics is driving this very classical apparatus. This is very suprising and yet strange, that Heisenberg's uncertainty principle can also be applied to classical mechanics.\n\n• Don Easton (among others) debunked this fallacy, concluding that quantum mechanical effects are not responsible for the observed behaviour of pencils. : \" The 3 s quantum mechanical tipping of a pencil is an urban myth of physics.\" See users.mccammon.ucsd.edu/~jcsung/qmp.pdf – sammy gerbil Jul 22 '17 at 20:31\n• Well, it's a famous problem from Sakurai, so, I'll stick wit that. – Dr. Ikjyot Singh Kohli Jul 22 '17 at 20:39\n• @sammygerbil your link has a 403 Forbidden response code. Can you send us another link? – juan Isaza Oct 3 at 10:45\n• @juanIsaza Don Easton's article can be accessed here : docplayer.net/…. A similar article by P Lynch is here : arxiv.org/pdf/1406.1125.pdf. – sammy gerbil Oct 4 at 15:30\n\nAccording to this article, we can’t observe quantum effects of large-scale objects (such as a car) because we live in a strong gravitational field caused by the Earth. General Relativity says that a gravitational field causes time dilation.\n\n… when an object is in superposition, all its parts vibrate in synchrony. However, time dilation will cause parts of that object that are at slightly higher altitudes to jitter at different frequencies than parts at slightly lower altitudes … The greater the difference in altitude between parts, the greater the mismatch. For a big enough object, the mismatch grows big enough to disrupt superpositions.\n\nWhen superpositions are disrupted, the wave functions collapse. That’s why we know with absolute certainty the position and velocity of classical objects.\n\nAnd, if we want to observe quantum effects more clearly, we should conduct more experiments in space and at very cold temperatures.\n\nUPDATE: I found the original article: Universal decoherence due to gravitational time dilation", null, "" ]	[ null, "https://i.stack.imgur.com/yqLUt.png", null, "https://i.stack.imgur.com/jC4U3.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9639502,"math_prob":0.9280993,"size":2236,"snap":"2019-43-2019-47","text_gpt3_token_len":493,"char_repetition_ratio":0.116935484,"word_repetition_ratio":0.0,"special_character_ratio":0.21645796,"punctuation_ratio":0.12053572,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9723903,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-21T04:54:31Z\",\"WARC-Record-ID\":\"<urn:uuid:5f20f801-0f5b-46a0-b2cc-4a5f8b1f4f61>\",\"Content-Length\":\"181770\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fac3fa00-1298-4a5f-a217-0cac16a72eaf>\",\"WARC-Concurrent-To\":\"<urn:uuid:75687af1-a7ea-497c-a319-35ae1b10b64e>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://physics.stackexchange.com/questions/186693/explain-why-quantum-behavior-is-not-observed-in-daily-life/193408\",\"WARC-Payload-Digest\":\"sha1:P2E6MYEF4CF4IDLZOPLOASC5RDPKF4A7\",\"WARC-Block-Digest\":\"sha1:NVFKIZS4HFYKY3EFLBBECCF6I72JHVUN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987756350.80_warc_CC-MAIN-20191021043233-20191021070733-00042.warc.gz\"}"}
https://www.includehelp.com/code-snippets/cpp-program-to-print-ascii-value-of-a-character.aspx	[ "Code Snippets C/C++ Code Snippets\n\n# C++ program to print ASCII value of a character.\n\nBy: IncludeHelp, On 11 OCT 2016\n\nIn this program we will learn how to print ASCII value of a character?\n\nIn this program there is a character type variable x which has \"A\", we can print the ASCII value of \"A\" by converting the type of value x.\n\n## C++ program (Code Snippet) - Get ASCII of a Character\n\nLet’s consider the following example:\n\n```/C++ program to print ASCII value of a character/\n\n#include<iostream>\n\nusing namespace std;\n\nint main(){\nchar x;\nx='A';\n\ncout<<\"ASCII value of \"<<x<<\" is \"<<(int)x<<endl;\n\nreturn 0;\n}\n```\n\n### Output\n\n``` ASCII value of A is 65\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.76926386,"math_prob":0.9140478,"size":627,"snap":"2021-43-2021-49","text_gpt3_token_len":164,"char_repetition_ratio":0.18780096,"word_repetition_ratio":0.07476635,"special_character_ratio":0.28389156,"punctuation_ratio":0.09375,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9871269,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-03T10:04:14Z\",\"WARC-Record-ID\":\"<urn:uuid:1eeff795-ea32-44ab-9b89-9304df63eddd>\",\"Content-Length\":\"113176\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5ee7f60a-469e-4de4-99a5-62d2c724a81f>\",\"WARC-Concurrent-To\":\"<urn:uuid:7525fa84-0bc6-430f-85b7-fee3cdd21da5>\",\"WARC-IP-Address\":\"104.26.4.124\",\"WARC-Target-URI\":\"https://www.includehelp.com/code-snippets/cpp-program-to-print-ascii-value-of-a-character.aspx\",\"WARC-Payload-Digest\":\"sha1:ISFFNKAAPIAOIXIBWEADKFDQDXRUY6BC\",\"WARC-Block-Digest\":\"sha1:AGD76MNEINARU25NBECBRG4FLPPPGEU7\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964362619.23_warc_CC-MAIN-20211203091120-20211203121120-00093.warc.gz\"}"}
https://www.canstar.com.au/online-trading/going-long-or-short/	[ "# Going long or short - what's the difference?\n\n##", null, "Long position\n\nTaking a long CFD position simply means you expect the value of the underlying security to increase. The difference between the entry price and the exit price is the profit or loss that is made on the trade.\n\nFor instance, ABC stock is currently trading at \\$4 and you expect it to appreciate in value. You have \\$10 000 worth of capital and the initial margin required to trade ABC is 5%. You can buy 50 000 ABC shares (with a market value of \\$200 000). Two weeks later you sell the shares for \\$4.20. They have appreciated 5%.\n\nYour gross profit equals the difference between the entry price and the exit price. In this case it is \\$10 000 (20 cents times 50 000 shares). You will need to subtract the financing cost of holding the position two weeks plus the commission required to open and close the CFD. Let?s say the financing cost is 6% p.a. calculated daily and the commission is 0.1%. Therefore your net profit will be approximately \\$9 220.32 after you subtract \\$410 worth of commission and \\$369.68 interest (*).\n\n## Short position\n\nTaking a short CFD position simply means you expect the value of the underlying security to decrease.\n\nFor instance, ABC stock is currently trading at \\$4 and you expect it decrease in value. You have \\$10 000 worth of capital and the initial margin required to trade ABC is 5%. You sell 50 000 ABC shares (with a market value of \\$200 000). Two weeks later you buy back the shares for \\$3.80. They have depreciated 5%.\n\nYour gross profit is the difference between the entry price where you sold ABC shares and the exit price where you purchased them back. In this case your gross profit is again, \\$10 000 (20 cents times 50 000 shares). Because it is a short position you will need to add the interest received on holding the position two weeks and subtract the commission required to open and close the CFD. Again, let’s assume the interest rate is 6% p.a. calculated daily and the commission is 0.1%. Therefore your net profit will be approximately \\$9 963.61 after you subtract \\$390 worth of commission and add \\$353.61 interest.\n\nLong Short\nRate 6% 6%\nCommission 0.10% 0.10%\nPresent Value \\$200 000 \\$200 000\nOpen CFD Position Fee \\$200 \\$200\nFuture Value \\$210 000 \\$190 000\nClose CFD Position Fee \\$210 \\$190\nGross Profit \\$10,000 \\$10,000\nminus fees \\$410 \\$390\nminus interest (long) \\$369.68" ]	[ null, "https://www.canstar.com.au/wp-content/uploads/2013/01/dice-long-short.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9277518,"math_prob":0.8987626,"size":2672,"snap":"2020-45-2020-50","text_gpt3_token_len":664,"char_repetition_ratio":0.12143928,"word_repetition_ratio":0.3197425,"special_character_ratio":0.30351797,"punctuation_ratio":0.09497207,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9524132,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-30T20:36:14Z\",\"WARC-Record-ID\":\"<urn:uuid:693a2be5-8450-4749-87e0-d55cd40cfdb0>\",\"Content-Length\":\"1030508\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c20744d6-e4da-42e9-a83c-2bfa50e3665e>\",\"WARC-Concurrent-To\":\"<urn:uuid:495eeb07-7bdd-446a-ad4b-d7b66be9e608>\",\"WARC-IP-Address\":\"199.232.65.186\",\"WARC-Target-URI\":\"https://www.canstar.com.au/online-trading/going-long-or-short/\",\"WARC-Payload-Digest\":\"sha1:MCEVG4U7RGSVBGJBB4I4OGZF6GHS6UM5\",\"WARC-Block-Digest\":\"sha1:I6F3SQF4O4IFOEC4RVH3OTDTQAHCCPSD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141486017.50_warc_CC-MAIN-20201130192020-20201130222020-00668.warc.gz\"}"}
http://gymkh.cz/wp-content/uploads/electroid-zaborger-flxsf/htovka3.php?tag=0e8621-non-isomorphic-graphs-with-same-degree-sequence	[ "We can easily see that these graphs have the same degree sequence, $\\langle 3,3,3,3,2,2 \\rangle$. 2.5 The problem of generating all non-isomorphic graphs of given order and size in- volves the problem of graph isomorphism for which a good algorithm is not yet known. Please use the purchase button to see the entire solution. Now insert the four degree-2 vertices back again. I am trying to generate all non-isomorphic graphs of a certain order and size that have the same degree sequence (not necessarily regular). second case if the the degree of each vertex or node is different except two then also there is one graph upto isomorphism. However as shown in Figure 1, The degree sequence is a graph invariant so … Two non-isomorphic graphs with the same degree sequence (3, 2, 2, 2, 2, 1, 1, 1). Lemma. graph. Shang gave a method for constructing general non-isomorphic graphs with the same status sequence. In addition, two graphs that are isomorphic must have the same degree sequence. Conditions we need to follow are: a. How to show these three-regular graphs on 10 vertices are non isomorphic? I though simple only meant no loops. How can I keep improving after my first 30km ride? Here is the handshaking lemma: Lemma 2.3. Definition 5.1.5 Graph $$H=(W,F)$$ is a subgraph of graph $$G=(V,E)$$ if $$W\\subseteq V$$ and $$F\\subseteq E$$. Use MathJax to format equations. This is what a commenter refers to as a theta graph. I have a question, given a degree sequence, how can you tell if there exists two graphs that are non isomorphic with those degrees? How many things can a person hold and use at one time? Is the bullet train in China typically cheaper than taking a domestic flight? Prove that if n is large enough, then the following statement is true. Your email address will not be used for any other purpose. For all graphs G on n vertices, at least one of and G … The degree sequence of an undirected graph is the non-increasing sequence of its vertex degrees; for the above graph it is (5, 3, 3, 2, 2, 1, 0). Having simple circuits of different length is one such property. If not explain why not. Email: help@24houranswers.com I just visualized some graphs and worked out two examples. Continue without uploading, Attachhomework files You might look at \"theta graphs\". G1 = G2 / G1 ≌ G2 [≌ - congruent symbol], we will say, G1 is isomorphic to G2. Figure 0.6: One graph contains a chordless cycle of length ﬁve while the other doesn’t. Why continue counting/certifying electors after one candidate has secured a majority? For more read about rigid graphs. All HL items are old, recycled materials and are therefore not original. So I'm asking about regular graphs of the same degree, if they have the same number of vertices, are they necessarily isomorphic? Non-isomorphic graphs with degree sequence 1, 1, 1, 2, 2, 3. 3. Yes, it is possible see the image below and they are not isomorphic to each other because second graph contain a cycle of length 3 (4-5-6-4), where as first graph does not have a cycle of length 3. Thanks! graphs with the same numbers of edges and the same degree sequences. DO NOT send Homework Help Requests or Live Tutoring Requests to our email, or through the form below. Please let us know the date by which you need help from your tutor or the date and time you wish to have an online tutoring session. So this gives 3 simple graphs. rev 2021.1.8.38287, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. Good, but I'd suggest using some alternative to \"simplified\". I'll fix my post. Can you legally move a dead body to preserve it as evidence? of vertices b. Asking for help, clarification, or responding to other answers. if so, draw them, otherwise, explain why they don't exist. × 2! @ Just a girl if the degree sequence is like (x,x,x,x,x) mean each node has same degree then definitely there is a unique graph upto isomorphism (easy case). Fig. Note however that the 0+0+4 graph is not simple, because of the two 0 edges. How is there a McDonalds in Weathering with You? Fast response time: Used only for emergencies when speed is the single most important factor. Let G be a member of a family of graphs ℱ and let the status sequence of G be s. G is said to be status unique in ℱ if G is the unique graph in ℱ whose status sequence is s. Here we view two isomorphic graphs as the same graph. I'm having trouble answering this question. We respect your privacy. MacBook in bed: M1 Air vs. M1 Pro with fans disabled. The degree sequence is a graph invariant so isomorphic graphshave the same degree sequence. They can have simple circuits of different length. Constructing two Non-Isomorphic Graphs given a degree sequence. I don't know if there's a standard term for the result of removing vertices of degree 2, but there's something funny about calling it a \"simplified\" graph when you have turned a simple graph into a non-simple graph! Since isomorphic graphs are “essentially the same”, we can use this idea to classify graphs. Draw two hexagons. (b) A simple graph with n vertices cannot have a vertex of degree more than n 1: Here n = 5. Applying the Sequence Theorem (2.10), this sequence is … MathJax reference. Normal response time: Our most experienced, most successful tutors are provided for maximum expertise and reliability. For example, these two graphs are not isomorphic, G1: • • • • G2: • • • • since one has four vertices of degree 2 and the other has just two. Have that degree sequence and connected, have four vertices and three edges way is to count number. Made available for the sole purpose of studying and learning uploading, Attachhomework files ( =! ≌ - congruent symbol ], we will say, G1 is isomorphic to G2 spam.. Exactly two vertices, and all major credit cards accepted over modern treatments and client asks me to return cheque. Each of the two isomorphic graphs have the same degree sequence the degree sequence is a graph by combining two! Are therefore not original it 's not in your inbox, check your spam folder for constructing general graphs. A Z80 non isomorphic graphs with same degree sequence program find out the address stored in the second graph the answer 4. Simple graphs with the same ordered degree sequence and connected non isomorphic graphs with same degree sequence have vertices. “ di↵erent ” graphs you will get a negotiable price quote with no obligation at level... In part or whole without written consent of the vertices in a more or less obvious way, graphs! Why Continue counting/certifying electors after one candidate has secured a majority so to! To classify graphs file Continue without uploading, Attachhomework files ( files Faster. Letter theta, consisting of non isomorphic graphs with same degree sequence paths between the same degree sequence,... A McDonalds in Weathering with you by 3 grid \\ ( 1,1,1,2,2,3\\ ) in?! They do n't exist case if the the degree of each vertex up with references personal. Of graphs with the same two points consent of the same degree sequence “. Upto isomorphism preserve it as evidence are four ways to do this: 0+0+4, 0+1+3, 0+2+2 or! Thing is not simple, because it is possible asks me to return the cheque and pays in?. The six vertices in a 2 non isomorphic graphs $( 4 most experienced, most successful tutors provided...$ as seen on sports scoreboards or some digital clocks an answer to mathematics Stack Exchange Inc ; user licensed. Figure 2.5 on ﬁve vertices with the non isomorphic graphs with same degree sequence degree sequence \\ ( 1,1,1,2,2,3\\ ) by... Alternative to simplified '' you can distribute the degree-2 vertices over the three.... 4 vertices available so each loop must have the same degree sequence two then also is! Your class a Z80 assembly program find out the address stored in the SP register client 's demand and asks... A 2 non isomorphic graphs $( 4 if you try using the HL in an unethical,. Writing great answers isomorphic graphshave the same degree sequence to be non-isomorphic graph contains a cycle.$ as seen on sports scoreboards or some digital clocks i have a question and answer site for people math! Sports scoreboards or some digital clocks vandalize things in public places one candidate has secured a?... Tutor responds to your message essentially the same degree keep improving after my first 30km ride our,! Doesn ’ t two nonisomorphic graphs because one has a simple circuit of length ﬁve the! Then also there is one such property people studying math at any level and professionals in fields! Figure 1, three while the other doesn ’ t your spam.... An email alert when the tutor responds to your message it as evidence.! Out of the cards cc by-sa more or less obvious way, some graphs are contained in others Help! The bullet train in China typically cheaper than taking a domestic flight work in academia that may already!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91878104,"math_prob":0.9413656,"size":9153,"snap":"2021-21-2021-25","text_gpt3_token_len":2118,"char_repetition_ratio":0.1751011,"word_repetition_ratio":0.064576805,"special_character_ratio":0.23052551,"punctuation_ratio":0.13276231,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97550803,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-13T03:35:49Z\",\"WARC-Record-ID\":\"<urn:uuid:5f4e83f4-1bcd-47f8-9d67-5a0ce6e59d1f>\",\"Content-Length\":\"25100\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b124663f-ca86-41b1-9680-00e1b1b60b83>\",\"WARC-Concurrent-To\":\"<urn:uuid:6674ed03-6fbb-4334-8d81-52778cc40829>\",\"WARC-IP-Address\":\"109.73.210.121\",\"WARC-Target-URI\":\"http://gymkh.cz/wp-content/uploads/electroid-zaborger-flxsf/htovka3.php?tag=0e8621-non-isomorphic-graphs-with-same-degree-sequence\",\"WARC-Payload-Digest\":\"sha1:WTH34AMOP5I5FT75BREPYUQQ7QLIUBRU\",\"WARC-Block-Digest\":\"sha1:TPYMXLHQPR2WZZUYC2CO2A3YU3IVAR6A\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243992721.31_warc_CC-MAIN-20210513014954-20210513044954-00290.warc.gz\"}"}
https://www.onlinemathlearning.com/multiplying-scientific-notation.html	[ "", null, "# Multiplying Numbers In Scientific Notation\n\nRelated Topics:\nMore Lessons for Arithmetic\nMath Worksheets\n\nHow to multiply numbers in scientific notation?\nStep 1 : Group the numbers together.\n\nStep 2 : Multiply the numbers.\n\nStep 3 : Use the law of indices to simplify the powers of 10.\n\nStep 4 : Give the answer in scientific notation.\n\nExample:\n\nEvaluate 8 × 103 × 9 × 102, giving your answer in scientific notation.\n\nSolution:", null, "Example:\n\nEvaluate 12.3 × 0.03, giving your answer in scientific notation.\n\nSolution:", null, "Five problems showing how to multiply and divide expressions written in scientific notation.\nMore Scientific Notation Examples\n\nRotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.\n\nYou can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.", null, "" ]	[ null, "https://www.onlinemathlearning.com/image-files/30x20xoml-search.png.pagespeed.ic.3ia21wspZX.png", null, "https://www.onlinemathlearning.com/image-files/xari129.gif.pagespeed.ic.pt5iQESxGI.png", null, "https://www.onlinemathlearning.com/image-files/xari130.gif.pagespeed.ic.fnIoCO6xfS.png", null, "https://www.onlinemathlearning.com/image-files/30x20xoml-search.png.pagespeed.ic.3ia21wspZX.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.74963176,"math_prob":0.89942986,"size":635,"snap":"2019-35-2019-39","text_gpt3_token_len":141,"char_repetition_ratio":0.12044374,"word_repetition_ratio":0.057692308,"special_character_ratio":0.24094488,"punctuation_ratio":0.16666667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99815965,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,1,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-18T20:14:16Z\",\"WARC-Record-ID\":\"<urn:uuid:57265326-33c1-4bbd-911d-7f346df7d314>\",\"Content-Length\":\"47001\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8efb5446-146b-4780-8cde-23f19b317ad3>\",\"WARC-Concurrent-To\":\"<urn:uuid:9ea94b4e-c80f-4811-8f96-dd9fe718499b>\",\"WARC-IP-Address\":\"173.247.219.45\",\"WARC-Target-URI\":\"https://www.onlinemathlearning.com/multiplying-scientific-notation.html\",\"WARC-Payload-Digest\":\"sha1:M4XQNDTDCQ5O2KFZUIEMO2LTGLZ2ZAER\",\"WARC-Block-Digest\":\"sha1:57IDMFQHJ3XDDNFFWZZAKU2MTQNKUXKG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573331.86_warc_CC-MAIN-20190918193432-20190918215432-00446.warc.gz\"}"}
https://byjus.com/rd-sharma-solutions/class-6-maths-chapter-20-mensuration-ex-20-2/	[ "", null, "# RD Sharma Solutions for Class 6 Maths Chapter 20: Mensuration Exercise 20.2\n\nThe students can make use of solutions designed by subject matter experts in order to understand the concepts in an effective way. This exercise mainly helps students get to know about the important formulas which are used to find the perimeter of geometrical structures like rectangle, square and equilateral triangle. These solutions help students self analyse their knowledge about the concepts which are covered in each exercise. Students who aspire to perform well in the exams can use RD Sharma Solutions Class 6 Maths Chapter 20 Mensuration Exercise 20.2 PDF from the links provided below.\n\n## RD Sharma Solutions for Class 6 Maths Chapter 20: Mensuration Exercise 20.2 Download PDF", null, "", null, "", null, "", null, "", null, "", null, "", null, "### Exercise 20.2 page: 20.10\n\n1. Find the perimeters of the rectangles whose lengths and breadths are given below:\n\n(i) 7 cm, 5 cm\n\n(ii) 5 cm, 4 cm\n\n(iii) 7.5 cm, 4.5 cm\n\nSolution:\n\n(i) We know that the perimeter of a rectangle = 2 (L + B)\n\nIt is given that L = 7 cm and B = 5 cm\n\nSo the perimeter of a rectangle = 2 (7 + 5) = 2 × 12 = 24 cm\n\n(ii) We know that the perimeter of a rectangle = 2 (L + B)\n\nIt is given that L = 5 cm and B = 4 cm\n\nSo the perimeter of a rectangle = 2 (5 + 4) = 2 × 9 = 18 cm\n\n(iii) We know that the perimeter of a rectangle = 2 (L + B)\n\nIt is given that L = 7.5 cm and B = 4.5 cm\n\nSo the perimeter of a rectangle = 2 (7.5 + 4.5) = 2 × 12 = 24 cm\n\n2. Find the perimeters of the squares whose sides are given below:\n\n(i) 10 cm\n\n(ii) 5 m\n\n(iii) 115.5 cm\n\nSolution:\n\n(i) We know that the perimeter of a square = 4 × Length of one side\n\nIt is given that L = 10 cm\n\nSo the perimeter of a square = 4 × 10 = 40 cm\n\n(ii) We know that the perimeter of a square = 4 × Length of one side\n\nIt is given that L = 5 m\n\nSo the perimeter of a square = 4 × 5 = 20 m\n\n(iii) We know that the perimeter of a square = 4 × Length of one side\n\nIt is given that L = 115.5 cm\n\nSo the perimeter of a square = 4 × 115.5 = 462 cm\n\n3. Find the side of the square whose perimeter is:\n\n(i) 16 m\n\n(ii) 40 cm\n\n(iii) 22 cm\n\nSolution:\n\n(i) We know that side of a square = perimeter/ 4\n\nIt is given that perimeter = 16 m\n\nSo the side of the square = 16/4 = 4 m\n\n(ii) We know that side of a square = perimeter/ 4\n\nIt is given that perimeter = 40 cm\n\nSo the side of the square = 40/4 = 10 cm\n\n(iii) We know that side of a square = perimeter/ 4\n\nIt is given that perimeter = 22 cm\n\nSo the side of the square = 22/4 = 5.5 cm\n\n4. Find the breadth of the rectangle whose perimeter is 360 cm and whose length is\n\n(i) 116 cm\n\n(ii) 140 cm\n\n(iii) 102 cm\n\nSolution:\n\nWe know that the perimeter of a rectangle = 2 (L + B)\n\nSo the breadth of the rectangle = perimeter/2 – length\n\n(i) It is given that perimeter = 360 cm and length = 116 cm\n\nSo the breadth of the rectangle = 360/2 – 116 = 180 – 116 = 64 cm\n\n(ii) It is given that perimeter = 360 cm and length = 140 cm\n\nSo the breadth of the rectangle = 360/2 – 140 = 180 – 140 = 40 cm\n\n(iii) It is given that perimeter = 360 cm and length = 102 cm\n\nSo the breadth of the rectangle = 360/2 – 102 = 180 – 102 = 78 cm\n\n5. A rectangular piece of lawn is 55 m wide and 98 m long. Find the length of the fence around it.\n\nSolution:\n\nThe dimensions of lawn are\n\nLength = 98 m\n\nWe know that\n\nPerimeter of lawn = 2 (L + B)\n\nBy substituting the values\n\nPerimeter of lawn = 2 (98 +55)\n\nSo we get\n\nPerimeter of lawn = 2 × 153 = 306 m\n\nHence, the length of the fence around the lawn is 306 m.\n\n6. The side of a square field is 65 m. What is the length of the fence required all around it?\n\nSolution:\n\nIt is given that\n\nSide of a square field = 65 m\n\nSo the perimeter of square field = 4 × side of the square\n\nBy substituting the values\n\nPerimeter of square field = 4 × 65 = 260 m\n\nHence, the length of the fence required all around the square field is 260 m.\n\n7. Two sides of a triangle are 15 cm and 20 cm. The perimeter of the triangle is 50 cm. What is the third side?\n\nSolution:\n\nIt is given that\n\nFirst side of triangle = 15 cm\n\nSecond side of triangle = 20 cm\n\nIn order to find the length of third side\n\nWe know that perimeter of a triangle is the sum of all three sides of a triangle\n\nSo the length of third side = perimeter of triangle – sum of length of other two sides\n\nBy substituting the values\n\nLength of third side = 50 – (15 + 20) = 15 cm.\n\nHence, the length of third side is 15 cm.\n\n8. A wire of length 20 m is to be folded in the form of a rectangle. How many rectangles can be formed by folding the wire if the sides are positive integers in metres?\n\nSolution:\n\nGiven:\n\nLength of wire 20 m is folded in the form of rectangle\n\nSo the perimeter = 20 m\n\nIt can be written as\n\n2 (L + B) = 20 m\n\nOn further calculation\n\nL + B = 10 m\n\nIf the sides are positive integers in metres the possible dimensions are (1m, 9m), (2m, 8m), (3m, 7m), (4m, 6m) and (5m, 5m)\n\nHence, five rectangles can be formed using the given wire.\n\n9. A square piece of land has each side equal to 100 m. If 3 layers of metal wire has to be used to fence it, what is the length of the wire needed?\n\nSolution:\n\nIt is given that\n\nEach side of a square field = 100 m\n\nWe can find the wire required to fence the square field by determining the perimeter = 4 × each side of a square field\n\nBy substituting the values\n\nPerimeter of the square field = 4 × 100 = 400 m\n\nSo the length of wire which is required to fence three layers is = 3 × 400 = 1200 m\n\nHence, the length of wire needed to fence 3 layers is 1200 m.\n\n10. Shikha runs around a square of side 75 m. Priya runs around a rectangle with length 60 m and breadth 45 m. Who covers the smaller distance?\n\nSolution:\n\nIt is given that\n\nShikha runs around a square of side = 75 m\n\nSo the perimeter = 4 × 75 = 300 m\n\nPriya runs around a rectangle having\n\nLength = 60 m\n\nSo the distance covered can be found from the perimeter = 2 (L + B)\n\nBy substituting the values\n\nPerimeter = 2 (60 + 45) = 2 × 105 = 210 m\n\nHence, Priya covers the smaller distance of 210 m.\n\n11. The dimensions of a photographs are 30 cm × 20 cm. What length of wooden frame is needed to frame the picture?\n\nSolution:\n\nIt is given that\n\nDimensions of a photographs = 30 cm × 20 cm\n\nSo the required length of the wooden frame can be determined from the perimeter of the photograph = 2 (L + B)\n\nBy substituting the values = 2 (30 + 20) = 2 × 50 = 100 cm\n\nHence, the length of the wooden frame required to frame the picture is 100 cm.\n\n12. The length of a rectangular field is 100 m. If the perimeter is 300 m, what is its breadth?\n\nSolution:\n\nThe dimensions of rectangular field are\n\nLength = 100 m\n\nPerimeter = 300 m\n\nWe know that perimeter = 2 (L + B)\n\nIt can be written as\n\nBy substituting the values\n\nBreadth = (300-200)/2 = 100/2 = 50 m\n\nHence, the breadth of the rectangular field is 50 m.\n\n13. To fix fence wires in a garden, 70 m long and 50 m wide, Arvind bought metal pipes for posts. He fixed a post every 5 metres apart. Each post was 2 m long. What is the total length of the pipes he bought for the posts?\n\nSolution:\n\nThe dimensions of garden are\n\nLength = 70 m\n\nSo the perimeter = 2 (L + B)\n\nBy substituting the values\n\nPerimeter = 2 (70 + 50) = 2 × 120 = 240 m\n\nGiven:\n\nArvind fixes a post every 5 metres apart\n\nNo. of posts required = 240/5 = 48\n\nThe length of each post = 2 m\n\nSo the total length of the pipe required = 48 × 2 = 96 m\n\nHence, the total length of the pipes he bought for the posts is 96 m.\n\n14. Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of Rs 12 per meter.\n\nSolution:\n\nThe dimensions of the rectangular park are\n\nLength = 175 m\n\nSo the perimeter = 2 (L + B)\n\nBy substituting the values\n\nPerimeter = 2 (175 + 125) = 2 × 300 = 600 m\n\nIt is given that the cost of fencing = Rs 12 per meter\n\nSo the total cost of fencing = 12 × 600 = Rs 7200\n\nHence, the cost of fencing a rectangular park is Rs 7200.\n\n15. The perimeter of a regular pentagon is 100 cm. How long is each side?\n\nSolution:\n\nWe know that a regular pentagon is a closed polygon having 5 sides of same length.\n\nIt is given that\n\nPerimeter of a regular pentagon = 100 cm\n\nIt can be written as\n\nPerimeter = 5 × side of the regular pentagon\n\nSo we get\n\nSide of the regular pentagon = Perimeter/5\n\nBy substituting the values\n\nSide of the regular pentagon = 100/5 = 20 cm\n\nHence, the side of the regular pentagon measures 20 cm.\n\n16. Find the perimeter of a regular hexagon with each side measuring 8 m.\n\nSolution:\n\nWe know that a regular hexagon is a closed polygon which has six sides of same length.\n\nIt is given that\n\nSide of the regular hexagon = 8 m\n\nSo we get\n\nPerimeter = 6 × side of the regular hexagon\n\nBy substituting the values\n\nPerimeter = 6 × 8 = 48 m\n\nHence, the perimeter of a regular hexagon is 48 m.\n\n17. A rectangular piece of land measure 0.7 km by 0.5 km. Each side is to be fenced with four rows of wires. What length of the wire is needed?\n\nSolution:\n\nIt is given that\n\nMeasure of rectangular piece of land = 0.7 km × 0.5 km\n\nWe know that\n\nPerimeter = 2 (L + B)\n\nBy substituting the values\n\nPerimeter = 2 (0.7 + 0.5) = 2 × 1.2 = 2.4 km\n\nThe above obtained perimeter = one row of wire needed to fence the rectangular piece of land\n\nSo the length of wire needed to fence the land with 4 rows of wire = 4 × 2.4 = 9.6 km\n\nHence, the length of wire needed is 9.6 km.\n\n18. Avneet buys 9 square paving slabs, each with a side of ½ m. He lays them in the form of a square.\n\n(i) What is the perimeter of his arrangement?\n\n(ii) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement?\n\n(iii) Which has greater perimeter?\n\n(iv) Avneet wonders, if there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges they cannot be broken)", null, "Solution:\n\n(i) It is given that length of each side of the slab = ½ m\n\nOne side of the square is formed by three slabs in a square arrangement\n\nLength of side = 3 × ½ = 3/2 m\n\nSo the perimeter of the square arrangement = 4 × 3/2 = 6 m\n\n(ii) From the figure, cross arrangement has 8 sides which form periphery of the arrangement and measure 1 m each.\n\nIt also has 4 sides which measure ½ m each\n\nPerimeter of the cross arrangement = 1 + ½ + 1 + 1 + ½ + 1 + 1 + ½ + 1 + 1 + ½ + 1 = 8 + 2 = 10 m", null, "(iii) We know that\n\nPerimeter of cross arrangement = 10 m\n\nPerimeter of square arrangement = 6 m\n\nHence, the perimeter of cross arrangement is greater than the perimeter of square arrangement.\n\n(iv) No, Avneet cannot arrange the slabs having perimeter more than 10 m." ]	[ null, "https://www.facebook.com/tr", null, "https://cdn1.byjus.com/wp-content/uploads/2020/11/rd-sharma-solutions-nov2020-class-6-maths-chapter-20-exercise-2-1.jpg", null, "https://cdn1.byjus.com/wp-content/uploads/2020/11/rd-sharma-solutions-nov2020-class-6-maths-chapter-20-exercise-2-2.jpg", null, "https://cdn1.byjus.com/wp-content/uploads/2020/11/rd-sharma-solutions-nov2020-class-6-maths-chapter-20-exercise-2-3.jpg", null, "https://cdn1.byjus.com/wp-content/uploads/2020/11/rd-sharma-solutions-nov2020-class-6-maths-chapter-20-exercise-2-4.jpg", null, "https://cdn1.byjus.com/wp-content/uploads/2020/11/rd-sharma-solutions-nov2020-class-6-maths-chapter-20-exercise-2-5.jpg", null, "https://cdn1.byjus.com/wp-content/uploads/2020/11/rd-sharma-solutions-nov2020-class-6-maths-chapter-20-exercise-2-6.jpg", null, "https://cdn1.byjus.com/wp-content/uploads/2020/11/rd-sharma-solutions-nov2020-class-6-maths-chapter-20-exercise-2-7.jpg", null, "https://cdn1.byjus.com/wp-content/uploads/2020/11/rd-sharma-solutions-for-class-6-chapter-20-exercise-20-2-image-1.png", null, "https://cdn1.byjus.com/wp-content/uploads/2020/11/rd-sharma-solutions-for-class-6-chapter-20-exercise-20-2-image-2.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8873717,"math_prob":0.9995346,"size":10453,"snap":"2021-31-2021-39","text_gpt3_token_len":3111,"char_repetition_ratio":0.21121639,"word_repetition_ratio":0.21619217,"special_character_ratio":0.32392615,"punctuation_ratio":0.07433946,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9998969,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],"im_url_duplicate_count":[null,null,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-28T10:32:26Z\",\"WARC-Record-ID\":\"<urn:uuid:7ffc1e90-bc27-4746-b7c7-f73eb9300dd3>\",\"Content-Length\":\"731069\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8083771b-e036-438e-8bcd-841ba88c533f>\",\"WARC-Concurrent-To\":\"<urn:uuid:56be5540-8ae9-4b11-a2ae-d6e24286b568>\",\"WARC-IP-Address\":\"162.159.129.41\",\"WARC-Target-URI\":\"https://byjus.com/rd-sharma-solutions/class-6-maths-chapter-20-mensuration-ex-20-2/\",\"WARC-Payload-Digest\":\"sha1:4A6BBJAH5CCRMI4PTWC64LQGQJYUTPCV\",\"WARC-Block-Digest\":\"sha1:74HHNPZ4A67SVJ7BZXWNR4MP6NPRIRPW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780060677.55_warc_CC-MAIN-20210928092646-20210928122646-00429.warc.gz\"}"}
https://www.sporcle.com/games/JoeBot/timestables	[ "Just For Fun Quiz / Times Tables Test\n\nRandom Just For Fun or Olympics Quiz\n\nScore\n0/20\nTimer\n01:00\n3 x 9 =\n2 x 7 =\n9 x 6 =\n5 x 3 =\n8 x 4 =\n11 x 4 =\n34 x 2 =\n12 x 3 =\n5 x 8 =\n10 x 3 =\n2 x 9 =\n11 x 7 =\n4 x 7 =\n2 x 8 =\n2 x 5 =\n18 x 10 =\n3 x 4 =\n6 x 6 =\n7 x 3 =\n6 x 7 =\n\nFrom the Vault See Another\n\n10 Most Populous Cities: Italy\nLuckily the marker is already on the map. You just have to place the name.\n\nExtras\n\nCreated Jul 29, 2012\nTags:" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.66434276,"math_prob":0.9995241,"size":4326,"snap":"2022-05-2022-21","text_gpt3_token_len":1291,"char_repetition_ratio":0.09185562,"word_repetition_ratio":0.010416667,"special_character_ratio":0.25034675,"punctuation_ratio":0.08971292,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9989687,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-23T00:11:56Z\",\"WARC-Record-ID\":\"<urn:uuid:71c88e5c-5787-4eb6-93ca-9b894d6e8da1>\",\"Content-Length\":\"125796\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dd096757-b02d-44a9-aac3-54b16017815d>\",\"WARC-Concurrent-To\":\"<urn:uuid:a333e906-52de-4c62-ab69-5add7f42ef92>\",\"WARC-IP-Address\":\"99.86.231.121\",\"WARC-Target-URI\":\"https://www.sporcle.com/games/JoeBot/timestables\",\"WARC-Payload-Digest\":\"sha1:PW5Y442R6JEBBXSF7LX4TDVJ6JLIF5ST\",\"WARC-Block-Digest\":\"sha1:VHIJF6ZHLL2OOXOSI3SX5DLDZE5Y5RR6\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320303917.24_warc_CC-MAIN-20220122224904-20220123014904-00040.warc.gz\"}"}
http://slideplayer.com/slide/4874796/	[ "", null, "# Electromagnetic Induction\n\n## Presentation on theme: \"Electromagnetic Induction\"— Presentation transcript:\n\nElectromagnetic Induction\nChapter 31 (cont.) Faraday’s Law & Lenz’s Law Induced Electric Fields Mutual Inductance Self-Inductance Inductors in Circuits Magnetic Energy 1\n\nA changing magnetic field induces an emf proportional to the rate of change of magnetic flux: e = - dB / dt The emf is about some loop – and the flux is through that loop. Lenz’s law: the induced emf is directed so that any induced current opposes the change in magnetic flux that causes the induced emf.\n\nRotating Loop - The Electric Generator\nB q Consider a loop of area A in a uniform magnetic field B. Rotate the loop with an angular frequency w . The flux changes because angle q changes with time: q = wt. Hence dFB/dt = d(B · A)/dt = d(BA cos q)/dt = BA d(cos(wt))/dt = - BAw sin(wt) q A\n\nRotating Loop - The Electric Generator\nB A q dFB/dt = - BAw sin(wt) Then by Faraday’s Law this motion causes an emf e = - dFB /dt = BAw sin(wt) This is an AC (alternating current) generator.\n\nInduced Electric Fields\nConsider a stationary wire in a time-varying magnetic field. A current starts to flow. x dB/dt So the electrons must feel a force F. It is not F = qvxB, because the charges started stationary. Instead it must be the force F=qE due to an induced electric field E. That is: A time-varying magnetic field B causes an electric field E to appear!\n\nInduced Electric Fields\nConsider a stationary wire in a time-varying magnetic field. A current starts to flow. x dB/dt So the electrons must feel a force F. It is not F = qvxB, because the charges started stationary. Instead it must be the force F=qE due to an induced electric field E. Moreover E around the loop gives a voltage diff DV=E·dl. And this must be the emf:  =  E·dl o\n\nInduced Electric Fields\nThis gives another way to write Faraday’s Law:  E·dl = - dB/dt o A technical detail: The electrostatic field E is conservative:  E·dl = 0. Consequently we can write E = - V. The induced electric field E is NOT a conservative field. We can NOT write E = -V for an induced field. o\n\nWork or energy difference Work or energy difference\nElectrostatic Field Induced Electric Field F = q E F = q E Vab = -  E·dl o  E · dl = - dB/dt  E·dl  0  E·dl = 0 and Ee = V o o Conservative Nonconservative Work or energy difference does NOT depend on path Work or energy difference DOES depend on path Caused by stationary charges Caused by changing magnetic fields\n\nInduced Electric Fields\n E · dl = - dB/dt o x B Faraday’s Law Now suppose there is no conductor: Is there still an electric field? YES! The field does not depend on the presence of the conductor. E For a dB/dt with axial or cylindrical symmetry, the field lines of E are circles. dB/dt\n\nMutual Inductance Bof 1 through 2 I 1 2\nTwo coils, 1 & 2, are arranged such that flux from one passes through the other. We already know that changing the current in 1 changes the flux (in the other) and so induces an emf in 2. This is known as mutual inductance. I Bof 1 through 2 1 2 2\n\nMutual Inductance I 1 2 Bof 1 through 2 The mutual inductance M is the proportionality constant between F2 and I1: F2 = M I1 so dF2 /dt = M dI1 /dt Then by Faraday’s law: 2= - dF2 /dt = - M dI1 /dt Hence M is also the proportionality constant between 2 and dI1 /dt. 3\n\nMutual Inductance 1 H = 1Weber/Amp = 1 V-s/A\nM arises from the way flux from one coil passes through the other: that is from the geometry and arrangement of the coils. Mutual means mutual. Note there is no subscript on M: the effect of 2 on 1 is identical to the effect of 1 on 2. The unit of inductance is the Henry (H). 1 H = 1Weber/Amp = 1 V-s/A 4\n\nSelf Inductance A changing current in a coil can induce an emf\nin itself…. I B If the current is steady, the coil acts like an ordinary piece of wire. But if the current changes, B changes and so then does FB, and Faraday tells us there will be an induced emf. Lenz’s law tells us that the induced emf must be in such a direction as to produce a current which makes a magnetic field opposing the change.\n\nSelf Inductance The self inductance of a circuit element (a coil, wire, resistor or whatever) is L = FB/I. Then exactly as with mutual inductance  = - L dI/dt. Since this emf opposes changes in the current (in the component) it is often called the “back emf”. From now on “inductance” means self-inductance. 6\n\nExample: Finding Inductance\nWhat is the (self) inductance of a solenoid with area A, length d, and n turns per unit length? In the solenoid B = m0nI, so the flux through one turn is fB = BA = m0nIA The total flux in the solenoid is (nd)fB Therefore, FB = m0n2IAd and so L = FB/I gives L = m0n2Ad (only geometry) 16\n\nInductance Affects Circuits and Stores Energy\nThis has important implications….. First an observation: Since  cannot be infinite neither can dI/dt. Therefore, current cannot change instantaneously. We will see that inductance in a circuit affects current in somewhat the same way that capacitance in a circuit affects voltage. A ‘thing’ (a component) with inductance in a circuit is called an inductor. 7\n\nRL Circuit e0 While the switch is open current can’t flow.\nWe start with a simple circuit containing a battery, a switch, a resistor, and an inductor, and follow what happens when the switch is closed. S R + e0 While the switch is open current can’t flow. - L\n\nRL Circuit e0 e0 eL While the switch is open current can’t flow.\nWe start with a simple circuit containing a battery, a switch, a resistor, and an inductor, and follow what happens when the switch is closed. S R + e0 While the switch is open current can’t flow. - L When the switch is closed current I flows, growing gradually, and a ‘back emf’ eL=- L dI/dt is generated in the inductor. This opposes the current I. + - S I e0 eL\n\nRL Circuit e0 e0 eL e0 While the switch is open current can’t flow.\nWe start with a simple circuit containing a battery, a switch, a resistor, and an inductor, and follow what happens when the switch is closed. S R + e0 While the switch is open current can’t flow. - L When the switch is closed current I flows, growing gradually, and a ‘back emf’ eL=- L dI/dt is generated in the inductor. This opposes the current I. + - S I e0 eL S + e0 After a long time the current becomes steady. Then eL is zero. Is -\n\nRL Circuit e0/R e0 I t/(L/R) When the switch S is closed S + I -\nthe current I increases exponentially from I = 0 to I = e0/R e0/R I 1 2 3 4 5 t/(L/R)\n\nRL Circuit Analysis e0 e0 - IR - LdI/dt = 0 Use the loop method R S +\nLdI/dt = e0 - IR = -R[I-(e0 /R)] dI / [I-(e0 /R)] = - (R/L) dt   dI / (I-(e0 /R)) = - (L/R)  dt ln[I-(e0 /R)] – ln[-(e0 /R)] = - t / (L/R)  ln[-I/(e0 /R) + 1] = - t / (L/R) -I/(e0 /R) + 1 = exp (- t / (L/R))  I/(e0 /R) = 1 - exp (- t / (L/R)) I = (e0 /R) [1 - exp (- t / (L/R))] 9\n\nwith time constant  = L / R\nRC Circuit Analysis R + - S I e0 L I = (e0/R) [1 - exp (- t / (L/R))] The current increases exponentially with time constant  = L / R e0 /R I t = 0  I = 0 t =   I = e0 /R 1 2 3 4 5 t/(L/R) 9\n\neL = L (e0/R) (-R/L) exp (- t / (L/R))\nInductor’s emf eL R + - S I e0 L eL = - L (dI/dt) I = (e0/R) [1 - exp (- t / (L/R))] eL = L (e0/R) (-R/L) exp (- t / (L/R)) eL 1 2 3 4 5 t/(L/R) -e0 eL = - e0 exp (- t / (L/R)) t = 0  eL = - e0 t =   eL = 0 9\n\nDecay of an RL Circuit S R + e0 - L After I reaches e0/R move the switch as shown. The loop method gives eL - IR = 0 or eL = IR Remember that eL = -L dI/dt  -L dI/dt = IR dI/I = - dt / (L/R)   dI/I = -  dt / (L/R) ln I/I0 = - t / (L/R)  I = I0 exp [- t / (L/R)] But I0 = e0/R Then: I = (e0/R) exp [- t / (L/R)] 12\n\nInductors in Circuits The presence of inductance prevents currents from changing instantaneously. The time constant of an RL circuit is  = L/R. When the current is not flowing the inductor tries to keep it from flowing. When the current is flowing the inductor tries to keep it going. 13\n\nEnergy Stored in an Inductor\n+ - S I e0 L Recall the original circuit when current was changing (building up). The loop method gave: e0 - IR + eL = 0 Multiply by I and use eL = - L dI/dt Then: Ie0 - I2R - ILdI/dt = 0 or: Ie0 - I2R – d[(1/2)LI2]/dt = 0 {d[(1/2)LI2]/dt=ILdI/dt} 14\n\nThink about Ie0 - I2R - d((1/2)LI2)/dt = 0\nIe0 is the power (energy per unit time) delivered by the battery. I2R is the power dissipated in the resistor. 15\n\nThink about Ie0 - I2R - d((1/2)LI2)/dt = 0\nIe0 is the power (energy per unit time) delivered by the battery. I2R is the power dissipated in the resistor. Hence we’d like to interpret d((1/2)LI2)/dt as the rate at which energy is stored in the inductor. In creating the magnetic field in the inductor we are storing energy 15\n\nThink about Ie0 - I2R - d((1/2)LI2)/dt = 0\nIe0 is the power (energy per unit time) delivered by the battery. I2R is the power dissipated in the resistor. Hence, we’d like to interpret d[(1/2)LI2]/dt as the rate at which energy is stored in the inductor. In creating the magnetic field in the inductor we are storing energy The amount of energy in the magnetic field is: UB = (1/2) LI2 15\n\nEnergy Density in a Magnetic Field\nWe have shown Apply this to a solenoid: Dividing by the volume of the solenoid, the stored energy density is: This turns out to be the energy density in a magnetic field uB = B2/(20) 17\n\nSummary We defined mutual and self inductance,\nCalculated the inductance of a solenoid. Saw the effect of inductance in RL circuits. Developed an expression for the stored energy. Derived an expression for the energy density of a magnetic field. Next class we will start learning about alternating-current (AC) circuits, containing resistors, capacitors, and inductors. 18" ]	[ null, "http://slideplayer.com/static/blue_design/img/slide-loader4.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8845783,"math_prob":0.9906181,"size":9699,"snap":"2020-10-2020-16","text_gpt3_token_len":2969,"char_repetition_ratio":0.14130995,"word_repetition_ratio":0.25179857,"special_character_ratio":0.30034024,"punctuation_ratio":0.07090821,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99842227,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-23T15:17:01Z\",\"WARC-Record-ID\":\"<urn:uuid:b872d290-89bf-46a5-a25c-568bd0e8d36e>\",\"Content-Length\":\"237174\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f4a9846a-36f9-4dc2-9705-bfee9dc39d82>\",\"WARC-Concurrent-To\":\"<urn:uuid:75c578b8-c925-4411-86d7-26c3db5ae823>\",\"WARC-IP-Address\":\"144.76.153.40\",\"WARC-Target-URI\":\"http://slideplayer.com/slide/4874796/\",\"WARC-Payload-Digest\":\"sha1:NLJX4IHI7L2NRFSAO3AGV4HKE7FTWOLO\",\"WARC-Block-Digest\":\"sha1:NR45CW3S7BDEI34OFL6YE7WTDWBZ43PD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875145774.75_warc_CC-MAIN-20200223123852-20200223153852-00133.warc.gz\"}"}
https://www.wired.com/2012/09/a-bouncing-angry-bird/	[ "# A Bouncing Angry Bird\n\nThere was a recent update to the Angry Birds Seasons with new levels. Something weird happened on one of the levels. A blue bird ended up bouncing vertically on one of the rubber mats. The weird part was that the bird kept bouncing higher and higher. Maybe this isn't weird in the Angry Birds world, but it's weird here on Earth.\n\nWhenever new levels are released for Angry Birds, I feel compelled to play right away. There was a recent update to the Angry Birds Seasons with new levels. Something weird happened on one of the levels. A blue bird ended up bouncing vertically on one of the rubber mats. The weird part was that the bird kept bouncing higher and higher. Maybe this isn't weird in the Angry Birds world, but it's weird here on Earth. If you want to see what I am talking about (or analyze it yourself), here is the video.\n\nHow about an analysis. Here is the initial motion of one of the bouncing blue birds. I set the length of sling shot at 4.9 meters since this gives the acceleration of 9.8 m/s2.", null, "Here you can see that the quadratic fit for one of the bounces has a t2 coefficient of 4.92 m/s2. Since this corresponds to the (1/2)a term in the kinematic equation, the acceleration would be 9.84 m/s2. So, at least for these low level bounces the vertical acceleration is constant and the scale seems to be set correctly. Now, let's look at all the bounces. Here is a plot of the bird just at the lowest and highest point.", null, "The data looks quadratic-like, so I fit a quadratic function. It's just what I do. But I'm not too happy. I would like to see if this trend keeps going. How can I get more data when the blue bird goes off the screen for bounces higher than about 35 meters. Well, let me go ahead and mark the time the bird leaves and then comes back on the screen. Here is the data from Tracker Video Analysis:", null, "For the motions that go off screen, can I treat them as plain old projectile motion? If so, I can just look at the time in the air to get the maximum height. Here is a more detailed plot of one of the later bounces (that goes off screen).", null, "This looks nice. A constant acceleration of around 10 m/s2 means that I can just use the time to determine the height. How would you do this? Instead of just using a kinematic equation, let me start with the definition of acceleration for the vertical direction (which has a constant value of g). Since I want to know how high it goes, let me take the time interval that starts with the bird on the ground moving up and ends at the highest point (where the bird has zero velocity). Also, I am just talking about 1-dimension here so I will drop the y-subscript on the velocity.", null, "Don't forget that Δt here is the time just to go UP, not up and down. But what about the height? That's what I want. Let me use the definition of average velocity:", null, "Now I can put in my expression for v1 from before:", null, "So, if I just measure the time between the bounces for the ones that go off the screen I can get the maximum height. Here is my adjusted bounce-height plot. Now with more data (but with the same function from before):", null, "Oh snap. That's not what I expected. It looks like the bouncing gets to a maximum of around 45.5 meters (above the ground). Actually, I suspect something else is happening. It's not that there is some magical ceiling in the game that the birds hit (well, there might be). But if the bird was hitting something at the top, the time of flight would be different for the faster balls. Let me calculate the starting speed for each bounce (calculated from the height).", null, "I like this better. Why? First, a limit on the maximum velocity would mean that each bouncing bird would still have a constant acceleration. Second, in a previous analysis I found that the yellow bird can reach a maximum speed of 30 m/s. Notice the maximum speed in this plot. Boom. 30 m/s." ]	[ null, "https://www.wired.com/images_blogs/wiredscience/2012/08/bouncy_1.jpg", null, "https://www.wired.com/images_blogs/wiredscience/2012/08/summer_notes_2_12key3.jpg", null, "https://www.wired.com/images_blogs/wiredscience/2012/08/bouncyvtrack_1.jpg", null, "https://www.wired.com/images_blogs/wiredscience/2012/08/bigbounceplot.jpg", null, "https://www.wired.com/images_blogs/wiredscience/2012/08/la_te_xi_t_14.jpg", null, "https://www.wired.com/images_blogs/wiredscience/2012/08/la_te_xi_t_1_12.jpg", null, "https://www.wired.com/images_blogs/wiredscience/2012/08/la_te_xi_t_1_22.jpg", null, "https://www.wired.com/images_blogs/wiredscience/2012/08/ijiijpng.jpg", null, "https://www.wired.com/images_blogs/wiredscience/2012/08/summer_notes_2_12key_12.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9459861,"math_prob":0.8574253,"size":3831,"snap":"2023-14-2023-23","text_gpt3_token_len":883,"char_repetition_ratio":0.1225503,"word_repetition_ratio":0.0056577087,"special_character_ratio":0.2247455,"punctuation_ratio":0.09713574,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97558904,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,5,null,5,null,5,null,5,null,5,null,5,null,5,null,5,null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-02T10:49:26Z\",\"WARC-Record-ID\":\"<urn:uuid:39bdf53e-ae92-4ca7-8e56-23930b0be418>\",\"Content-Length\":\"935149\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e4af8f6b-8b5b-4d44-b056-b5c8491543f8>\",\"WARC-Concurrent-To\":\"<urn:uuid:470de25b-4e99-47e6-89ee-1b80c493e7e2>\",\"WARC-IP-Address\":\"146.75.34.194\",\"WARC-Target-URI\":\"https://www.wired.com/2012/09/a-bouncing-angry-bird/\",\"WARC-Payload-Digest\":\"sha1:LGYO4FQ6QLSIJDKVBH57TRFNAAP4JTWF\",\"WARC-Block-Digest\":\"sha1:EWLNTHER3DM6NSFA7G73FIJCPPMLPXPN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224648635.78_warc_CC-MAIN-20230602104352-20230602134352-00072.warc.gz\"}"}
https://answers.everydaycalculation.com/subtract-fractions/18-5-minus-2-6	[ "Solutions by everydaycalculation.com\n\n## Subtract 2/6 from 18/5\n\n1st number: 3 3/5, 2nd number: 2/6\n\n18/5 - 2/6 is 49/15.\n\n#### Steps for subtracting fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 5 and 6 is 30\n2. For the 1st fraction, since 5 × 6 = 30,\n18/5 = 18 × 6/5 × 6 = 108/30\n3. Likewise, for the 2nd fraction, since 6 × 5 = 30,\n2/6 = 2 × 5/6 × 5 = 10/30\n4. Subtract the two fractions:\n108/30 - 10/30 = 108 - 10/30 = 98/30\n5. After reducing the fraction, the answer is 49/15\n6. In mixed form: 34/15\n\n#### Subtract Fractions Calculator\n\n-\n\nUse fraction calculator with our all-in-one calculator app: Download for Android, Download for iOS" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5302347,"math_prob":0.9982671,"size":359,"snap":"2019-35-2019-39","text_gpt3_token_len":177,"char_repetition_ratio":0.23943663,"word_repetition_ratio":0.0,"special_character_ratio":0.55153203,"punctuation_ratio":0.089108914,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9993142,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-23T15:18:01Z\",\"WARC-Record-ID\":\"<urn:uuid:2e31cd70-9be6-46ce-976f-91e74334399a>\",\"Content-Length\":\"8719\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:22083e92-ebf9-49ef-8f60-1b19ce033c7e>\",\"WARC-Concurrent-To\":\"<urn:uuid:cf1399fc-92b9-4eed-aa46-4d9af01f5d2e>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/subtract-fractions/18-5-minus-2-6\",\"WARC-Payload-Digest\":\"sha1:HGOYSFGFN4YOTTR4K2DRUVMZW7NS24AD\",\"WARC-Block-Digest\":\"sha1:C5Z2R2MOVD66YZDAKZO2FAZOK426XJPQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514577363.98_warc_CC-MAIN-20190923150847-20190923172847-00346.warc.gz\"}"}
https://scicomp.stackexchange.com/questions/1887/finding-the-square-root-of-a-laplacian-matrix	[ "Finding the square root of a Laplacian matrix\n\nSuppose the following matrix $A$ is given $$\\left[\\begin{array}{ccc} 0.500 & -0.333 & -0.167\\\\ -0.500 & 0.667 & -0.167\\\\ -0.500 & -0.333 & 0.833\\end{array}\\right]$$ with its transpose $A^T$. The product $A^TA=G$ yields $$\\left[\\begin{array}{ccc}0.750 & -0.334 & -0.417\\\\ -0.334 & 0.667 & -0.333\\\\ -0.417 & -0.333 & 0.750\\end{array}\\right]$$,\n\nwhere $G$ is a Laplacian matrix. Note that matrices $A$ and $G$ are of rank 2, with the zero eigenvalue corresponding to eigenvector $1_n=\\left[\\begin{array}{ccc}1 & 1 & 1\\end{array}\\right]^T$.\n\nI wonder what would be the way to obtain $A$ if only $G$ is given. I tried eigendecomposition $G=UEU^T$, and then set $A'=UE^{1/2}$, but obtained different result. I guess this has to do with rank deficiency. Could someone explain this? Clearly, the above example is for illustration; you could consider general Laplacian matrix decomposition of the above form.\n\nSince, for instance, Cholesky decomposition could be used to find $G=LL^T$, the decomposition on $G$ could yield many solution. I'm interested in the solution that could be expressed as $$A=(I-1_nw^{T}),$$ where $I$ is a $3\\times 3$ identity matrix, $1_n=[1~ 1~ 1]$, and $w$ being some vector satisfying $w^T1_n=1$. If it simplifies matters, you could assume that the entries of $w$ are non-negative.\n\n• I think the comment you added about the representation of $A$ is only partially helpful. It assumes that there is exactly one eigenvalue equal to zero, but the non-determinancy is always there, isn't it? – Wolfgang Bangerth Apr 9 '12 at 21:17\n• @WolfgangBangerth I'm trying to figure out the meaning of \"non-determinancy\". If that is $det(A)=0$, it holds for the above example, and I'm not sure if it can be generalized for $A=I-1_nw^T$. However, except for $n=3$, I doubt that the solution would always exist. – usero Apr 10 '12 at 7:15\n• No, what I meant is that the solution to your problem isn't uniquely determined. I was pointing out the fact that whether the matrix has a zero eigenvalue or not doesn't actually change the fact that the square root problem has no unique solution. – Wolfgang Bangerth Apr 10 '12 at 20:03\n\nWe have the matrix Laplacian matrix $G=A^TA$ which has a set of eigenvalues $\\lambda_0\\leq\\lambda_1\\leq\\ldots\\leq \\lambda_n$ for $G\\in\\mathbb{R}^{n\\times n}$ where we always know $\\lambda_0 = 0$. Thus the Laplacian matrix is always symmetric positive semi-definite. Because the matrix $G$ is not symmetric positive definite we have to be careful when we discuss the Cholesky decomposition. The Cholesky decomposition exists for a positive semi-definite matrix but it is no longer unique. For example, the positive semi-definite matrix $$A=\\left[\\!\\!\\begin{array}{cc} 0 & 0 \\\\ 0 & 1 \\end{array} \\!\\!\\right],$$ has infinitely many Cholesky decompositions $$A=\\left[\\!\\begin{array}{cc} 0 & 0 \\\\ 0 & 1 \\end{array} \\!\\right] = \\left[\\!\\begin{array}{cc} 0 & 0 \\\\ \\sin\\theta & \\cos\\theta \\end{array} \\!\\right] \\left[\\!\\begin{array}{cc} 0 & \\sin\\theta \\\\ 0 & \\cos\\theta \\end{array} \\!\\right]=LL^T.$$\n\nHowever, because we have a matrix $G$ that is known to be a Laplacian matrix we can actually avoid the more sophisticated linear algebra tools like Cholesky decompositions or finding the square root of the positive semi-definite matrix $G$ such that we recover $A$. For example, if we have the Laplace matrix $G\\in\\mathbb{R}^{4\\times 4}$, $$G=\\left[\\!\\begin{array}{cccc} 3 & -1 & -1 & -1\\\\ -1 & 1 & 0 & 0 \\\\ -1 & 0 & 1 & 0 \\\\ -1 & 0 & 0 & 1 \\\\ \\end{array}\\!\\right]$$ we can use graph theory to recover the desired matrix $A$. We do so by formulating the oriented incidence matrix. If we define the number of edges in the graph to be $m$ and the number of vertices to be $n$ then the oriented incidence matrix $A$ will be an $m\\times n$ matrix given by $$A_{ev} = \\left\\{\\begin{array}{lc} 1 & \\textrm{if }e=(v,w)\\textrm{ and }v<w \\\\ -1 & \\textrm{if }e=(v,w)\\textrm{ and }v>w \\\\ 0 & \\textrm{otherwise}, \\end{array} \\right.$$ where $e=(v,w)$ denotes the edge which connects the vertices $v$ and $w$. If we take a graph for $G$ with four vertices and three edges, then we have the oriented incidence matrix $$A = \\left[\\!\\begin{array}{cccc} 1 & -1 & 0 & 0\\\\ 1 & 0 & -1 & 0 \\\\ 1 & 0 & 0 & -1 \\\\ \\end{array}\\!\\right],$$ and we can find that $G=A^TA$. For the matrix problem you describe you would construct a graph for $G$ with the same number of edges as vertices, then you should have the ability to reconstruct the matrix $A$ when you are only given the Laplacian matrix $G$.\n\nUpdate:\n\nIf we define the diagonal matrix of vertex degrees of a graph as $N$ and the adjacency matrix of the graph as $M$, then the Laplacian matrix $G$ of the graph is defined by $G=N-M$. For example, in the following graph", null, "we find the Laplacian matrix is $$G=\\left[\\!\\begin{array}{cccc} 3 & 0 & 0 & 0\\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ \\end{array}\\!\\right] - \\left[\\!\\begin{array}{cccc} 0 & 1 & 1 & 1\\\\ 1 & 0 & 0 & 0 \\\\ 1 & 0 & 0 & 0 \\\\ 1 & 0 & 0 & 0 \\\\ \\end{array}\\!\\right].$$ Now we relate the $G$ to the oriented incidence matrix $A$ using the edges and nodes given in the pictured graph. Again we find the entries of $A$ from $$A_{ev} = \\left\\{\\begin{array}{lc} 1 & \\textrm{if }e=(v,w)\\textrm{ and }v<w \\\\ -1 & \\textrm{if }e=(v,w)\\textrm{ and }v>w \\\\ 0 & \\textrm{otherwise}, \\end{array} \\right..$$ For example, edge $e_1$ connects the nodes $v_1$ and $v_2$. So to determine $A_{e_1,v_1}$ we note that the index of $v_1$ is less than the index of $v_2$ (or we have the case $v<w$ in the definition of $A_{ev}$). Thus, $A_{e_1,v_1} = 1$. Similarly by the way of comparing indices we can find $A_{e_1,v_2} = -1$. We give $A$ below in a more explicit way referencing the edges and vertices pictured. $$A = \\begin{array}{c\|cccc} & v_1 & v_2 & v_3 & v_4 \\\\ \\hline e_1 & 1 & -1 & 0 & 0\\\\ e_2 & 1 & 0 & -1 & 0 \\\\ e_3 & 1 & 0 & 0 & -1 \\\\ \\end{array}.$$\n\nNext, we generalize the concept of the Laplacian matrix to a weighted undirected graph. Let $Gr$ be an undirected finite graph defined by $V$ and $E$ its vertex and edge set respectively. To consider a weighted graph we define a weight function $$w: V\\times V\\rightarrow \\mathbb{R}^+,$$ which assigns a non-negative real weight to each edge of the graph. We will denote the weight attached to edge connecting vertices $u$ and $v$ by $w(u,v)$. In the case of a weighted graph we define the degree of each vertex $u\\in V$ as the sum of all the weighted edges connected to $u$, i.e., $$d_u = \\sum_{v\\in V}w(u,v).$$ From the given graph $Gr$ we can define the weighted adjacency matrix $Ad(Gr)$ as an $n\\times n$ with rows and columns indexed by $V$ whose entries are given by $w(u,v)$. Let $D(Gr)$ be the diagonal matrix indexed by $V$ with the vertex degrees on the diagonal then we can find the weighted Laplacian matrix $G$ just as before $$G = D(Gr) - Ad(Gr).$$\n\nIn the problem from the original post we know $$G=\\left[\\!\\begin{array}{ccc} \\tfrac{3}{4} & -\\tfrac{1}{3} & -\\tfrac{5}{12} \\\\ -\\tfrac{1}{3} & \\tfrac{2}{3} & -\\tfrac{1}{3} \\\\ -\\tfrac{5}{12} & -\\tfrac{1}{3} & \\tfrac{3}{4} \\\\ \\end{array}\\!\\right].$$ From the comments we know we seek a factorization for $G$ where $G=A^TA$ and specify $A$ is of the form $A=I-1_nw^T$ where $w^T1_n=1$. For full generality assume the matrix $A$ has no zero entries. Thus if we formulate the weighted oriented incidence matrix to find $A$ we want the weighted adjacency matrix $Ad(Gr)$ to have no zero entries as well, i.e., the weighted graph will have loops. Actually calculating the weighted oriented incidence matrix seems difficult (although it may simply be a result of my inexperience with weighted graphs). However, we can find a factorization of the form we seek in an ad hoc way if we assume we know something about the loops in our graph. We split the weighted Laplacian matrix $G$ into the degree and adjacency matrices as follows $$G=\\left[\\!\\begin{array}{ccc} \\tfrac{5}{4} & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & \\tfrac{11}{12} \\\\ \\end{array}\\!\\right]-\\left[\\!\\begin{array}{ccc} \\tfrac{1}{2} & \\tfrac{1}{3} & \\tfrac{5}{12} \\\\ \\tfrac{1}{3} & \\tfrac{1}{3} & \\tfrac{1}{3} \\\\ \\tfrac{5}{12} & \\tfrac{1}{3} & \\tfrac{1}{6} \\\\ \\end{array}\\!\\right] = D(Gr)-Ad(Gr).$$\nThus we know the loops on $v_1$, $v_2$ and $v_3$ have weights $1/2$, $1/3$, and $1/6$ respectively. If we put the weights on the loops into a vector $w$ = $[\\frac{1}{2}$ $\\frac{1}{3}$ $\\frac{1}{6}]^T$ then we can recover the matrix $A$ we want in the desired form $$A = I-1_nw^T = \\left[\\!\\begin{array}{ccc} \\tfrac{1}{2} & -\\tfrac{1}{3} & -\\tfrac{1}{6} \\\\ -\\tfrac{1}{2} & \\tfrac{2}{3} & -\\tfrac{1}{6} \\\\ -\\tfrac{1}{2} & -\\tfrac{1}{3} & \\tfrac{5}{6} \\\\ \\end{array}\\!\\right].$$\n\nIt appears if we have knowledge of the loops in our weighted graph we can find the matrix $A$ in the desired form. Again, this was done in an ad hoc manner (as I am not a graph theorist) so it may be a hack that worked just for this simple problem.\n\n• Recovering $A$ in the case of Laplacian $G$ as you describe should not be a problem (in you example, only one row/column contain non-zero elements). I guess the matters complicate with the general \"full\" Laplacian as in my example. Given the $O(n^2)$ degrees of freedom of $G$, I'm not sure if the solution could be obtained, subject to constraints I give above. – usero Apr 9 '12 at 16:05\n• Right, the example $G$ I give is highly simplified but it will be possible in the general case where $G$ is a full matrix. The graph theory problem becomes more complicated as you increase the number of edges and vertices. In essence we replace a difficult matrix factorization problem with a difficult graph theory problem. – Andrew Winters Apr 9 '12 at 17:45\n• Ok, I would appreciate if you try to reconstruct $A$ as is given above from given $G$. – usero Apr 9 '12 at 19:25\n• @AndrewWinters: Could you clarify how $A$ is determined from $G$? It's not clear to me how your algorithm proceeds in the general case. – Geoff Oxberry Apr 9 '12 at 19:35\n• I don't think it will be possible for a general $G$ as it seems that the specific factorization form $A=I-1_nw^T$ will only exist for certain types of weighted graphs. So the Laplacian matrices $G$ that are of the form $G=A^TA=(I-1_nw^T)^T(I-1_nw^T)$ are a subset of all possible Laplacian matrices. – Andrew Winters Apr 10 '12 at 14:07\n\nI think that you are confusing the unique matrix square-root of Hermitian positive semi-definite matrix $A$, i.e., a Hermitian positive semi-definite matrix $B$ satisfying,\n\n$$B^2 = A,$$\n\nwith the non-unique problem of finding a matrix $C$ satisfying\n\n$$C^H C = A,$$\n\nwhere clearly the mapping $C \\mapsto Q C$, for any unitary $Q$, preserves the identity. As you noticed, a Cholesky factorization provides one possible solution. However, note that Cholesky only works for Hermitian positive-definite matrices (with the possible exception of a Hermitian positive semi-definite matrix which is positive-definite if the last row and column are removed).\n\nLastly, one can constructively define the unique matrix square-root of a Hermitian positive semi-definite matrix through its eigenvalue decomposition, say\n\n$$A = U \\Lambda U^H,$$\n\nwhere $U$ is unitary and $\\Lambda$ is diagonal with non-negative entries due to $A$ being positive semi-definite. The Hermitian matrix square-root can easily be identified as\n\n$$B = U \\sqrt{\\Lambda} U^H.$$\n\n• You're right about the matrix square-root. Clearly, there are different ways to achieve the factorization, but could guarantee that $A$ (from my quest.) could be written as I formulated. – usero Apr 10 '12 at 19:26\n\nIn essence, what you're asking is to find the square root A of a matrix G, so that $$G = A^T A.$$ There are many ways to do that if $G$ is a symmetric matrix. For example, if $G$ is symmetric, then the Cholesky decomposition $G=L^TL$ provides you with one answer: $A=L$. But you already found another answer, with the matrix $A$ you already have. What this simply means is that there are many \"square roots\" of the matrix $G$, and if you want to have one particular one, you need to rephrase the question in such a way that you specify the structural properties of the \"branch\" of the square root that you're interested in.\n\nI would say that this situation is not dissimilar to taking the square root among the real numbers using the complex numbers: there, too, in general you have two roots, and you have to say which one you want to make the answer unique.\n\n• You're definitely right. Another way would be the spectral decomposition approach as I state above. I've made an edit to make the solution unique. Hopefully it won't complicate matters. – usero Apr 9 '12 at 11:56\n• Is a solution with the constraint I give above always existent? Perhaps it holds only for some cases, and not in general. – usero Apr 9 '12 at 14:23\n• Actually, Cholesky does not work in his case, as it (essentially) requires that the matrix is Hermitian positive-definite. – Jack Poulson Apr 10 '12 at 21:51\n\nI think you can apply $LDL^{T}$ factorization for your matrix A. Since your matrix has non-negative eigenvalues, the diagonal matrix D will have non-negative entries along the diagonl. Then you can easily factorize $\\hat{D} = \\sqrt{D}$. And you get the matrix $G = L\\hat{D}$. The eigendecomposition is not numerically stable, so I think you should avoid this kind of decomposition.\n\n• I have to disagree on two counts: (1) $LDL^T$ factorization does not work for singular matrices (his matrix is singular), and (2) eigenvalue decompositions of Hermitian matrices are considered stable, as their eigenvector matrices are unitary. – Jack Poulson Apr 11 '12 at 18:36\n• @JackPoulson I try a singular matrix A in matlab, and run ldl, it works. The zero eigenvalues corresponds to the zeros in the diagonal of D. – Willowbrook Apr 11 '12 at 19:10\n• I think that you will find that MATLAB's 'ldl' routine computes a block $LDL^T$ decomposition with pivoting, i.e., $PAP'=LDL^T$, where $D$ is not required to be diagonal (it can have $2 \\times 2$ blocks). In order to avoid division by zero due to the matrix being singular, the zero diagonal entries are pivoted to the bottom-right of the matrix. – Jack Poulson Apr 11 '12 at 19:31" ]	[ null, "https://i.imgur.com/TZTjf.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.77750593,"math_prob":0.99980086,"size":6790,"snap":"2019-26-2019-30","text_gpt3_token_len":2324,"char_repetition_ratio":0.16666667,"word_repetition_ratio":0.12889637,"special_character_ratio":0.3689249,"punctuation_ratio":0.079795025,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99995756,"pos_list":[0,1,2],"im_url_duplicate_count":[null,8,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-16T11:52:00Z\",\"WARC-Record-ID\":\"<urn:uuid:4d7730a4-434c-4c6a-8bc3-d0569c8876b8>\",\"Content-Length\":\"179085\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b9aed3d5-0725-494c-be8f-d962406d4135>\",\"WARC-Concurrent-To\":\"<urn:uuid:c8cfe035-76f7-47ff-9526-8458a7b87ab9>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://scicomp.stackexchange.com/questions/1887/finding-the-square-root-of-a-laplacian-matrix\",\"WARC-Payload-Digest\":\"sha1:OSTTWEQGTFJJPSLLH7ZTP7JCMH4TATTE\",\"WARC-Block-Digest\":\"sha1:ODLJ74Q6MP3G6R6BU7M3G37AISOL4TLM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627998100.52_warc_CC-MAIN-20190616102719-20190616124719-00258.warc.gz\"}"}
https://oeis.org/A232449/internal	[ "The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.", null, "Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)\n A232449 The palindromic Belphegor numbers: (10^(n+3)+666)10^(n+1)+1. 4\n\n%I\n\n%S 16661,1066601,100666001,10006660001,1000066600001,100000666000001,\n\n%T 10000006660000001,1000000066600000001,100000000666000000001,\n\n%U 10000000006660000000001,1000000000066600000000001,100000000000666000000000001,10000000000006660000000000001,1000000000000066600000000000001\n\n%N The palindromic Belphegor numbers: (10^(n+3)+666)10^(n+1)+1.\n\n%C Though this sequence rarely contains primes (see A232448), most of its members tend to contain a few very large prime factors. The name stems from 'Belphegor's Prime', a(13), which was so named by Clifford Pickover (see link). [Comment corrected by _N. J. A. Sloane_, Dec 14 2015]\n\n%H Stanislav Sykora, <a href=\"/A232449/b232449.txt\">Table of n, a(n) for n = 0..497</a>\n\n%H Clifford A. Pickover, <a href=\"http://sprott.physics.wisc.edu/pickover/pc/1000000000000066600000000000001.html\">Belphegor's Prime: 1000000000000066600000000000001</a>\n\n%H Simon Singh, <a href=\"http://www.bbc.co.uk/news/magazine-24724635\">Homer Simpson's scary math problems</a>. BBC News. Retrieved 31 October 2013.\n\n%H Eric Weisstein's World of Mathematics, <a href=\"http://mathworld.wolfram.com/BelphegorNumber.html\">Belphegor Number</a>\n\n%H Wikipedia, <a href=\"http://en.wikipedia.org/wiki/Belphegor%27s_Prime\">Belphegor's prime</a>\n\n%H <a href=\"/index/Rec#order_03\">Index entries for linear recurrences with constant coefficients</a>, signature (111,-1110,1000).\n\n%F a(n) = 66610^(n+1)+100^(n+2)+1.\n\n%F G.f.: (16661 - 782770x + 767000x^2) / ((1 - x)(1 - 10x)(1 - 100x)). [_Bruno Berselli_, Nov 25 2013]\n\n%o (PARI) Belphegor(k)=(10^(k+3)+666)10^(k+1)+1; nmax = 498; v = vector(nmax); for (n=0,#v-1, v[n+1]=Belphegor(n))\n\n%Y Cf. A232448, A232450, A232451.\n\n%Y Subsequence of A118598.\n\n%K nonn,easy\n\n%O 0,1\n\n%A _Stanislav Sykora_, Nov 24 2013\n\nLookup \| Welcome \| Wiki \| Register \| Music \| Plot 2 \| Demos \| Index \| Browse \| More \| WebCam\nContribute new seq. or comment \| Format \| Style Sheet \| Transforms \| Superseeker \| Recent\nThe OEIS Community \| Maintained by The OEIS Foundation Inc.\n\nLast modified May 12 04:27 EDT 2021. Contains 343810 sequences. (Running on oeis4.)" ]	[ null, "https://oeis.org/banner2021.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.55879396,"math_prob":0.9838366,"size":1899,"snap":"2021-21-2021-25","text_gpt3_token_len":704,"char_repetition_ratio":0.22532982,"word_repetition_ratio":0.0,"special_character_ratio":0.5139547,"punctuation_ratio":0.21899736,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96737677,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-12T09:25:02Z\",\"WARC-Record-ID\":\"<urn:uuid:aac2039d-eb31-43f6-8b4e-d65253b8ca5c>\",\"Content-Length\":\"10314\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:629e184f-ef96-4165-9fcc-3af95bbede55>\",\"WARC-Concurrent-To\":\"<urn:uuid:aac2f91c-7334-42cc-b100-c424d6f67509>\",\"WARC-IP-Address\":\"104.239.138.29\",\"WARC-Target-URI\":\"https://oeis.org/A232449/internal\",\"WARC-Payload-Digest\":\"sha1:P3MO2FCEUWYQ3Z5KKXHIKGYL4JJKLFO5\",\"WARC-Block-Digest\":\"sha1:QQI5HQNCE5TUQHQ7NX5IG3DHORO4BDHU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991685.16_warc_CC-MAIN-20210512070028-20210512100028-00556.warc.gz\"}"}
https://www.neuraldesigner.com/learning/examples/diagnose-hcv/	[ "This example aims to conclude whether the patient suffers from Hepatitis C, Fibrosis, Cirrhosis, or none of them.\n\nThe database used for this study was taken from the Medical University of Hannover, Germany.\n\n### Contents\n\nThis example is solved with Neural Designer. To follow it step by step, you can use the free trial.\n\n## 1. Application type\n\nThe variable to be predicted is categorical (no disease, suspect disease, hepatitis c, fibrosis, cirrhosis). Therefore, this is a classification project.\n\nThe goal here is to model the probability if the patient does not have any disease or suffers from hepatitis, fibrosis, or cirrhosis, conditioned on different tests such as blood analysis or urine tests.\n\n## 2. Data set\n\nThe hvcdat.csv file contains the data for this application. In a classification project type, target variables can have five values: no disease, suspect disease, Hepatitis C, Fibrosis, or Cirrhosis. The number of instances (rows) in the data set is 615, and the number of variables (columns) is 13.\n\nThe number of input variables, or attributes for each sample, is 12. Nearly all input variables are numeric-valued except one, Sex, which is binary, and most of them represent measurements from blood and urine analysis. The number of target variables is 1 and represents whether the patient suffers from liver disease. The following list summarizes the variables information:\n\n• category(diagnose): No disease, suspect disease, hepatitis c, fibrosis, or cirrhosis.\n• age: (0-100). The normal ranges change depending on the age.\n• sex: (f or m). The normal ranges change depending on the sex.\n• albumin: (normal range 34-54g/L). A blood albumin level below normal may be a sign of kidney disease or liver disease as Hepatitis or Cirrhosis.\n• alkaline_phosphatase: (40-129 U/L). It is an alkaline phosphatase test to check for liver disease or liver damage. The main cause or condition that can cause normal levels to rise is liver disease.\n• alanine_aminotransferase: (7-55 U/L). Alanine aminotransferase, or ALT is an enzyme found primarily in the liver. An elevated amount indicates liver damage from hepatitis, infection, cirrhosis, liver cancer, or other liver diseases.\n• aspartate_aminotransferase: (8-48 U/L). It is an enzyme found primarily in the liver, but also in the muscles. Elevated levels of AST in the blood may indicate hepatitis, cirrhosis, mononucleosis, or other liver diseases.\n• bilirubin: (1-12 mg/L). It is a yellowish substance that forms during the normal process of breaking down red blood cells in the body. Lower-than-normal bilirubin levels are generally not a concern. Elevated levels may indicate liver disease or damage.\n• cholinesterase: (8-18 U/L). It is a blood test that studies the levels of 2 substances that help the nervous system to function properly. Cholinesterase deficiency causes liver disease, and renal disease…\n• cholesterol: (Less than 5,2 mmol/L). Cholesterol is a waxy, fat-like substance found in every cell in your body. A high level could cause carotid artery disease, stroke, and peripheral artery disease…\n• creatinina: (61.9-114.9 ??mol/L for m and 53-97.2 ??mol/L for f). It measures the level of creatinine in the blood. High levels of creatinine in the blood and low levels in the urine indicate kidney disease or affecting the function of the kidneys\n• gamma_glutamyl_transferase : (from 0 to 30-50 IU/L.). It is an enzyme that indicates cholestasis. The higher the level of GGT, the greater the level of liver damage.\n• protein: (Less than 80 mg). It measures the amount of protein in the urine. If protein levels in the urine are elevated, this could indicate kidney damage or another medical problem.\n\nFinally, the use of all instances is set. Note that each instance contains the input and target variables of a different patient.\nThe data set is divided into training, validation, and testing subsets. 60% of the instances will be assigned for training, 20% for generalization, and 20% for testing. More specifically, 369 samples are used here for training, 123 for selection, and 123 for testing.\n\nOnce the data set has been set, we can perform a few related analytics. We check the provided information and make sure that the data is of good quality.\n\nWe can calculate the data statistics and draw a table with the minimums, maximums, means, and standard deviations of all variables in the data set. The next table depicts the values.", null, "Also, we can calculate the distributions for all variables. The following figure shows a pie chart with the number of instances belonging to each class in the data set.", null, "As we can see, no disease people represent 86% of the samples, and the total Hepatitis C, Fibrosis, and Cirrhosis represent approximately 12% of the patients.\n\nWe can also calculate the inputs-targets correlations to see which signals better define each factor. The following chart illustrates the dependency of the target category with the input variables of the data set.", null, "Here, the most correlated variables with Hepatitis C are aspartate aminotransferase, gamma glutamyl transferase, and alanine aminotransferase.\n\n## 3. Neural network\n\nThe second step is to set a neural network representing the classification function. For this class of applications, the neural network is composed of:\n\n• Scaling layer.\n• Perceptron layers.\n• Probabilistic layer.\n\nThe scaling layer contains the statistics on the inputs calculated from the data file and the method for scaling the input variables. Here the minimum-maximum method has been set. Nevertheless, the mean-standard deviation method would produce very similar results.\n\nThe number of perceptron layers is 1. This perceptron layer has 12 inputs and 5 neurons.\n\nThe probabilistic layer uses the softmax probabilistic method.\n\nThe following figure is a graphical representation of this neural network for liver disease diagnosis.", null, "## 4. Training strategy\n\nThe procedure used to carry out the learning process is called a training strategy. The training strategy is applied to the neural network to obtain the best possible performance. The type of training is determined by how the adjustment of the parameters in the neural network takes place. The fourth step is composed of two terms:\n\n• A loss index.\n• An optimization algorithm.\n\nThe loss index is the cross entropy error with L1 regularization.\n\nThe learning problem can be stated as finding a neural network that minimizes the loss index. That is a neural network that fits the data set (error term) and does not oscillate (regularization term).\n\nThe optimization algorithm that we use is the quasi-Newton method. This is also the standard optimization algorithm for this type of problem.", null, "The following chart shows how the error decreases with the iterations during the training process. This is a sign of convergence.\nThe final training and selection errors are training error = 0.0553 WSE and selection error = 0.0557 WSE, respectively.\n\n## 5. Model selection\n\nThe objective of model selection is to improve the generalization capabilities of the neural network or, in other words, to reduce the selection error.\n\nSince the selection error that we have achieved so far is very small (0.0557 NSE), we don’t need to apply order selection or input selection here.\n\n## 6. Testing analysis\n\nOnce the model is trained, we perform a testing analysis to validate its prediction capacity. We use a subset of data that has not been used before, the testing instances.\n\nThe next table shows the confusion matrix for our problem. The confusion matrix represents the real classes and the predicted classes’ columns for the testing data.\n\nPredicted no_disease Predicted suspect_disease Predicted hepatitis_c Predicted fibrosis Predicted cirrhosis\nReal no_disease 107 (87%) 1(0.813%) 0 1(0.813%) 0\nReal suspect_disease 0 0 0 0 0\nReal hepatitis_c 5(4.07%) 0 1(0.813%) 1(0.813%) 1(0.813%)\nReal fibrosis 0 0 1(0.813%) 2(1.63%) 0\nReal cirrhosis 1(0.813%) 0 0 0 2(1.63%)\n\nAs we can see, the number of instances that the model can correctly predict is 123 (92%), while it approximately misclassifies only 11 (8%). This shows that our predictive model has excellent classification accuracy, and the biggest confusion is predicting no disease when the patient suffers from hepatitis c.\n\n## 7. Model deployment\n\nOnce the neural network’s generalization performance has been tested, the neural network can be saved for future use in the so-called model deployment mode.\n\nWe can diagnose new patients by calculating the neural network outputs. For that, we need to know the input variables for them. An example is the following:\n\nWe can export the mathematical expression of the neural network to the hospital’s software to facilitate the work of the doctor. This expression is listed below.\n\nscaled_age = age(1+1)/(77-(19))-19(1+1)/(77-19)-1;\nscaled_sex = (sex-(0.6130080223))/0.4874579906;\nscaled_albumin = albumin(1+1)/(82.19999695-(14.89999962))-14.89999962(1+1)/(82.19999695-14.89999962)-1;\nscaled_alkaline_phosphatase = alkaline_phosphatase(1+1)/(416.6000061-(11.30000019))-11.30000019(1+1)/(416.6000061-11.30000019)-1;\nscaled_alanine_aminotransferase = alanine_aminotransferase(1+1)/(325.2999878-(0.8999999762))-0.8999999762(1+1)/(325.2999878-0.8999999762)-1;\nscaled_aspartate_aminotransferase = aspartate_aminotransferase(1+1)/(324-(10.60000038))-10.60000038(1+1)/(324-10.60000038)-1;\nscaled_bilirubin = bilirubin(1+1)/(254-(0.8000000119))-0.8000000119(1+1)/(254-0.8000000119)-1;\nscaled_cholinesterase = cholinesterase(1+1)/(16.40999985-(1.419999957))-1.419999957(1+1)/(16.40999985-1.419999957)-1;\nscaled_cholesterol = cholesterol(1+1)/(9.670000076-(1.429999948))-1.429999948(1+1)/(9.670000076-1.429999948)-1;\nscaled_creatinina = creatinina(1+1)/(1079.099976-(8))-8(1+1)/(1079.099976-8)-1;\nscaled_gamma_glutamyl_transferase = gamma_glutamyl_transferase(1+1)/(650.9000244-(4.5))-4.5(1+1)/(650.9000244-4.5)-1;\nscaled_protein = protein(1+1)/(90-(44.79999924))-44.79999924(1+1)/(90-44.79999924)-1;\n\nperceptron_layer_0_output_0 = sigma[ 0.000687571 + (scaled_age-0.000107975)+ (scaled_sex-0.000354215)+ (scaled_albumin0.000200821)+ (scaled_alkaline_phosphatase5.53142e-06)+ (scaled_alanine_aminotransferase-0.000204123)+ (scaled_aspartate_aminotransferase-0.000175011)+ (scaled_bilirubin0.000402467)+ (scaled_cholinesterase-6.37448e-05)+ (scaled_cholesterol-0.000240513)+ (scaled_creatinina0.000516883)+ (scaled_gamma_glutamyl_transferase-5.56635e-05)+ (scaled_protein0.00018629) ];\nperceptron_layer_0_output_1 = sigma[ -0.00100031 + (scaled_age-0.000225761)+ (scaled_sex0.000388395)+ (scaled_albumin0.00128129)+ (scaled_alkaline_phosphatase9.56258e-05)+ (scaled_alanine_aminotransferase0.000165274)+ (scaled_aspartate_aminotransferase0.000264668)+ (scaled_bilirubin0.000126936)+ (scaled_cholinesterase0.000348544)+ (scaled_cholesterol8.74069e-05)+ (scaled_creatinina0.000187206)+ (scaled_gamma_glutamyl_transferase4.68866e-05)+ (scaled_protein0.000123546) ];\nperceptron_layer_0_output_2 = sigma[ -0.000135561 + (scaled_age0.00104903)+ (scaled_sex-0.000125211)+ (scaled_albumin0.000365379)+ (scaled_alkaline_phosphatase0.000471019)+ (scaled_alanine_aminotransferase0.000246482)+ (scaled_aspartate_aminotransferase-0.000117056)+ (scaled_bilirubin0.000598315)+ (scaled_cholinesterase0.000453945)+ (scaled_cholesterol-0.000645688)+ (scaled_creatinina-0.000144883)+ (scaled_gamma_glutamyl_transferase1.79297e-05)+ (scaled_protein0.000202182) ];\nperceptron_layer_0_output_3 = sigma[ -3.94009e-05 + (scaled_age0.000279269)+ (scaled_sex-0.00126496)+ (scaled_albumin-0.00015788)+ (scaled_alkaline_phosphatase0.000714173)+ (scaled_alanine_aminotransferase0.000159489)+ (scaled_aspartate_aminotransferase-0.000126977)+ (scaled_bilirubin-0.000248145)+ (scaled_cholinesterase0.000231954)+ (scaled_cholesterol-0.000381098)+ (scaled_creatinina0.000388452)+ (scaled_gamma_glutamyl_transferase-2.25984e-05)+ (scaled_protein-2.42171e-05) ];\nperceptron_layer_0_output_4 = sigma[ 0.531863 + (scaled_age0.000279123)+ (scaled_sex0.000294663)+ (scaled_albumin0.000832503)+ (scaled_alkaline_phosphatase-0.0035456)+ (scaled_alanine_aminotransferase-0.000726132)+ (scaled_aspartate_aminotransferase-0.143645)+ (scaled_bilirubin-0.0693214)+ (scaled_cholinesterase-0.000138161)+ (scaled_cholesterol0.000344551)+ (scaled_creatinina-0.210249)+ (scaled_gamma_glutamyl_transferase-0.0841537)+ (scaled_protein-0.00010063) ];\n\nprobabilistic_layer_combinations_0 = 1.99571 +0.00017129perceptron_layer_0_output_0 +0.000294192perceptron_layer_0_output_1 +0.000545024perceptron_layer_0_output_2 +1.71061e-05perceptron_layer_0_output_3 +1.23961perceptron_layer_0_output_4\nprobabilistic_layer_combinations_1 = 2.43231 +8.26751e-06perceptron_layer_0_output_0 -0.000394384perceptron_layer_0_output_1 +0.000839893perceptron_layer_0_output_2 -0.000321675perceptron_layer_0_output_3 +1.06731perceptron_layer_0_output_4\nprobabilistic_layer_combinations_2 = 2.85722 +0.000411359perceptron_layer_0_output_0 +0.000485672perceptron_layer_0_output_1 +9.30878e-05perceptron_layer_0_output_2 -7.38007e-05perceptron_layer_0_output_3 +0.789123perceptron_layer_0_output_4\nprobabilistic_layer_combinations_3 = 1.99249 +0.000452945perceptron_layer_0_output_0 +0.00107266perceptron_layer_0_output_1 +0.000121336perceptron_layer_0_output_2 -0.000247549perceptron_layer_0_output_3 +1.1734perceptron_layer_0_output_4\nprobabilistic_layer_combinations_4 = 2.01512 +0.000355189perceptron_layer_0_output_0 +0.000699017perceptron_layer_0_output_1 -0.000694358perceptron_layer_0_output_2 +0.000623235perceptron_layer_0_output_3 +0.226457*perceptron_layer_0_output_4\n\nno_disease = 1.0/(1.0 + exp(-probabilistic_layer_combinations_0);\nsuspect_disease = 1.0/(1.0 + exp(-probabilistic_layer_combinations_1);\nhepatitis = 1.0/(1.0 + exp(-probabilistic_layer_combinations_2);\nfibrosis = 1.0/(1.0 + exp(-probabilistic_layer_combinations_3);\ncirrhosis = 1.0/(1.0 + exp(-probabilistic_layer_combinations_4);\n\n\n## References\n\n• The data for this problem has been taken from the UCI Machine Learning Repository.\n• Lichtinghagen R et al. J Hepatol 2013; 59: 236-42.\n• Hoffmann G et al. Using machine learning techniques to generate laboratory diagnostic pathways a case study. J Lab Precis Med 2018; 3: 58-67." ]	[ null, "https://www.neuraldesigner.com/images/Data-hcv.webp", null, "https://www.neuraldesigner.com/images/chart-hcv.webp", null, "https://www.neuraldesigner.com/images/correlations-hcv.webp", null, "https://www.neuraldesigner.com/images/neural-hcv.webp", null, "https://www.neuraldesigner.com/images/Graph-HCV.webp", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7477189,"math_prob":0.90265983,"size":14407,"snap":"2023-40-2023-50","text_gpt3_token_len":4160,"char_repetition_ratio":0.1657988,"word_repetition_ratio":0.012128563,"special_character_ratio":0.3319914,"punctuation_ratio":0.16859505,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9810189,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-01T12:05:46Z\",\"WARC-Record-ID\":\"<urn:uuid:a15dac40-0329-4d1f-8410-83914f50178d>\",\"Content-Length\":\"213583\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e759e66f-40c2-465e-9449-bd311b92aa92>\",\"WARC-Concurrent-To\":\"<urn:uuid:8cf54d54-7b87-4d84-a497-9b5a6007f58b>\",\"WARC-IP-Address\":\"92.205.96.244\",\"WARC-Target-URI\":\"https://www.neuraldesigner.com/learning/examples/diagnose-hcv/\",\"WARC-Payload-Digest\":\"sha1:YX2ERQB72DRG26XMVA2REQLZYWFQMBLS\",\"WARC-Block-Digest\":\"sha1:M7FK3FHTSGGUZ4QLHPRSDBTI7536RQW4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100287.49_warc_CC-MAIN-20231201120231-20231201150231-00807.warc.gz\"}"}
https://root.cern.ch/root/html518/ROOT__Math__RootFinder_ROOT__Math__Roots__Newton_.html	[ "# class ROOT::Math::RootFinder<ROOT::Math::Roots::Newton>\n\n```\nClass to find the Root of one dimensional functions.\nThe class is templated on the type of Root solver algorithms.\nThe possible types of Root-finding algorithms are:\n<ul>\n<li>Root Bracketing Algorithms which they do not require function derivatives\n<ol>\n<li>Roots::Bisection\n<li>Roots::FalsePos\n<li>Roots::Brent\n</ol>\n<li>Root Finding Algorithms using Derivatives\n<ol>\n<li>Roots::Newton\n<li>Roots::Secant\n<li>Roots::Steffenson\n</ol>\n</ul>\n\nThis class does not cupport copying\n\n@ingroup RootFinders\n\n```\n\n## Function Members (Methods)\n\npublic:\n virtual ~RootFinder() int Iterate() int Iterations() const const char* Name() const double Root() const ROOT::Math::RootFinder RootFinder() int SetFunction(const ROOT::Math::IGradFunction& f, double Root) int SetFunction(const ROOT::Math::IGenFunction& f, double xlow, double xup) int Solve(int maxIter = 100, double absTol = 1E-3, double relTol = 1E-6) static int TestDelta(double r1, double r0, double epsAbs, double epsRel) static int TestInterval(double xlow, double xup, double epsAbs, double epsRel) static int TestResidual(double f, double epsAbs)\nprivate:\n ROOT::Math::RootFinder& operator=(const ROOT::Math::RootFinder& rhs) ROOT::Math::RootFinder RootFinder(const ROOT::Math::RootFinder&)\n\n## Data Members\n\nprivate:\n ROOT::Math::Roots::Newton fSolver type of algorithm to be used\n\n## Class Charts", null, "## Function documentation\n\nint SetFunction(const ROOT::Math::IGenFunction& f, double xlow, double xup)\n```Provide to the solver the function and the initial search interval [xlow, xup]\nfor algorithms not using derivatives (bracketing algorithms)\nThe templated function f must be of a type implementing the \\a operator() method,\n<em> double operator() ( double x ) </em>\nReturns non zero if interval is not valid (i.e. does not contains a root)\n\n```\nreturn fSolver. SetFunction(const ROOT::Math::IGenFunction& f, double xlow, double xup)\nint Solve(int maxIter = 100, double absTol = 1E-3, double relTol = 1E-6)\n```Compute the roots iterating until the estimate of the Root is within the required tolerance returning\nthe iteration Status\n\n```\nint Iterations()\n```Return the number of iteration performed to find the Root.\n\n```\nint Iterate()\n```Perform a single iteration and return the Status\n\n```\ndouble Root()\n```Return the current and latest estimate of the Root\n\n```\nconst char * Name()\n```Return the current and latest estimate of the lower value of the Root-finding interval (for bracketing algorithms)\n\ndouble XLower() const {\nreturn fSolver.XLower();\n}\n\nReturn the current and latest estimate of the upper value of the Root-finding interval (for bracketing algorithms)\n\ndouble XUpper() const {\nreturn fSolver.XUpper();\n}\n\nGet Name of the Root-finding solver algorithm\n\n```\nint TestInterval(double xlow, double xup, double epsAbs, double epsRel)\n```Test convertgence Status of current iteration using interval values (for bracketing algorithms)\n\n```\nint TestDelta(double r1, double r0, double epsAbs, double epsRel)\n```Test convergence Status of current iteration using last Root estimates (for algorithms using function derivatives)\n\n```\nint TestResidual(double f, double epsAbs)\n```Test function residual\n\n```\n\nLast update: root/mathmore:\\$Id: RootFinder.h 21503 2007-12-19 17:34:54Z moneta \\$\nCopyright (c) 2004 ROOT Foundation, CERN/PH-SFT *\n\nThis page has been automatically generated. If you have any comments or suggestions about the page layout send a mail to ROOT support, or contact the developers with any questions or problems regarding ROOT." ]	[ null, "https://root.cern.ch/root/html518/inh/ROOT__Math__RootFinder_ROOT__Math__Roots__Newton__Inh.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5883826,"math_prob":0.9253134,"size":2336,"snap":"2022-27-2022-33","text_gpt3_token_len":586,"char_repetition_ratio":0.15737565,"word_repetition_ratio":0.078688525,"special_character_ratio":0.22260274,"punctuation_ratio":0.1598063,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99442387,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-10T22:49:03Z\",\"WARC-Record-ID\":\"<urn:uuid:3c48f35a-9e3a-4d3e-8ddf-9832f9849f77>\",\"Content-Length\":\"21980\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1f53f754-c10f-4d8e-aa11-e2433b814240>\",\"WARC-Concurrent-To\":\"<urn:uuid:09c0933a-70f1-420f-a7f6-1d9237f98f4c>\",\"WARC-IP-Address\":\"188.184.49.144\",\"WARC-Target-URI\":\"https://root.cern.ch/root/html518/ROOT__Math__RootFinder_ROOT__Math__Roots__Newton_.html\",\"WARC-Payload-Digest\":\"sha1:HIDXVQP2GDW3FWWR7BCDAEOMLY43H3HL\",\"WARC-Block-Digest\":\"sha1:XWSW3OX26LGPVUZ3L7S5DIBMNAWRWEDJ\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571222.74_warc_CC-MAIN-20220810222056-20220811012056-00476.warc.gz\"}"}
https://thewebdev.info/2020/04/01/typescript-advanced-typesnullable-types-and-type-aliases/	[ "Categories\n\n# TypeScript Advanced Types — Nullable Types and Type Aliases\n\nTypeScript has many advanced type capabilities and which make writing dynamically typed code easy. It also facilitates the adoption of existing JavaScript code since it lets us keep the dynamic capabilities of JavaScript while using the type-checking capability of TypeScript. There are multiple kinds of advanced types in TypeScript, like intersection types, union types, type guards, nullable types, and type aliases, and more. In this article, we look at nullable types and type aliases.\n\n# Nullable Types\n\nTo let us assign `undefined` to a property with the `--strictNullChecks` flag on, TypeScript supports nullable types. With the flag on, we can’t assign `undefined` to type members that don’t have the nullable operator attached to it. To use it, we just put a question mark after the member name for the type we want to use.\n\nIf we have the `strictNullChecks` flag on and we set a value of a property to `null` or `undefined` , then like we do in the following code:\n\n``````interface Person {\nname: string;\nage: number;\n}\n\nlet p: Person = {\nname: 'Jane',\nage: null\n}\n``````\n\nThen we get the following errors:\n\n``````Type 'null' is not assignable to type 'number'.(2322)input.ts(3, 3): The expected type comes from property 'age' which is declared here on type 'Person'\n``````\n\nThe errors above won’t appear if we have `strictNullChecks` off and the TypeScript compiler will allow the code to be compiled.\n\nIf we have the `strictNullChecks` flag on and we want to be able to set `undefined` to a property as the value, then we can make the property nullable. For example, we can set a member of an interface to be nullable with the following code:\n\n``````interface Person {\nname: string;\nage?: number;\n}\n\nlet p: Person = {\nname: 'Jane',\nage: undefined\n}\n``````\n\nIn the code above, we added a question mark after the `age` member in the `Person` interface to make it nullable. Then when we define the object, we can set `age` to `undefined`. We can’t still set `age` to `null` . If we try to do that, we get:\n\n``````Type 'null' is not assignable to type 'number \| undefined'.(2322)input.ts(3, 3): The expected type comes from property 'age' which is declared here on type 'Person'\n``````\n\nAs we can see, a nullable type is just a union type between the type that we declared and the `undefined` type. This also means that we can use type guards with it like any other union type. For example, if we want to only get the age if it’s defined, we can write the following code:\n\n``````const getAge = (age?: number) => {\nif (age === undefined) {\nreturn 0\n}\nelse {\nreturn age.toString();\n}\n}\n``````\n\nIn the `getAge` function, we first check if the `age` parameter is `undefined` . If it is, then we return 0. Otherwise, we can call the `toString()` method on it, which is available to number objects.\n\nLikewise, we can eliminate `null` values with a similar kind of code, for instance, we can write:\n\n``````const getAge = (age?: number \| null) => {\nif (age === null) {\nreturn 0\n}\nelse if (age === undefined) {\nreturn 0\n}\nelse {\nreturn age.toString();\n}\n}\n``````\n\nThis comes in handy because nullable types exclude `null` from being assigned with `strictNullChecks` on, so if we want `null` to be able to be passed in as a value for the `age` parameter, then we need to add `null` to the union type. We can also combine the first 2 `if` blocks into one:\n\n``````const getAge = (age?: number \| null) => {\nif (age === null \|\| age === undefined) {\nreturn 0\n}\nelse {\nreturn age.toString();\n}\n}\n``````\n\n# Type Aliases\n\nIf we want to create a new name for an existing type, we can add a type alias to the type. This can be used for many types, including primitives, unions, tuples, and any other type that we can write by hand. To create a type alias, we can use the `type` keyword to do so. For example, if we want to add an alias to a union type, we can write the following code:\n\n``````interface Person {\nname: string;\nage: number;\n}\n\ninterface Employee {\nemployeeCode: number;\n}\n\ntype Laborer = Person & Employee;\nlet laborer: Laborer = {\nname: 'Joe',\nage: 20,\nemployeeCode: 100\n}\n``````\n\nThe declaration of `laborer` is the same as using the intersection type directly to type the `laborer` object, as we do below:\n\n``````let laborer: Person & Employee = {\nname: 'Joe',\nage: 20,\nemployeeCode: 100\n}\n``````\n\nWe can declare type alias for primitive types like we any other kinds of types. For example, we can make a union type with different primitive types as we do in the following code:\n\n``````type PossiblyNumber = number \| string \| null \| undefined;\nlet x: PossiblyNumber = 2;\nlet y: PossiblyNumber = '2';\nlet a: PossiblyNumber = null;\nlet b: PossiblyNumber = undefined;\n``````\n\nIn the code above, the `PossiblyNumber` type can be a number, string, `null` or `undefined` . If we try to assign an invalid to it like a boolean as in the following code:\n\n``````let c: PossiblyNumber = false;\n``````\n\nWe get the following error:\n\n``````Type 'false' is not assignable to type 'PossiblyNumber'.(2322)\n``````\n\njust like any other invalid value assignment.\n\nWe can also include generic type markers in our type aliases. For example, we can write:\n\n``````type Foo<T> = { value: T };\n``````\n\nGeneric type aliases can also be referenced in the properties of a type. For example, we can write:\n\n``````type Tree<T> = {\nvalue: T;\nleft: Tree<T>;\ncenter: Tree<T>;\nright: Tree<T>;\n}\n``````\n\nThen we can use the `Tree` type as we do in the following code:\n\n``````type Tree<T> = {\nvalue: T,\nleft: Tree<T>;\ncenter: Tree<T>;\nright: Tree<T>;\n}\n\nlet tree: Tree<string> = {} as Tree<string>;\ntree.value = 'Jane';tree.left = {} as Tree<string>\ntree.left.value = 'Joe';\ntree.left.left = {} as Tree<string>;\ntree.left.left.value = 'Amy';\ntree.left.right = {} as Tree<string>\ntree.left.right.value = 'James';tree.center = {} as Tree<string>\ntree.center.value = 'Joe';tree.right = {} as Tree<string>\ntree.right.value = 'Joe';console.log(tree);\n``````\n\nThe `console.log` for `tree` on the last line should get us:\n\n``````{\n\"value\": \"Jane\",\n\"left\": {\n\"value\": \"Joe\",\n\"left\": {\n\"value\": \"Amy\"\n},\n\"right\": {\n\"value\": \"James\"\n}\n},\n\"center\": {\n\"value\": \"Joe\"\n},\n\"right\": {\n\"value\": \"Joe\"\n}\n}\n``````\n\nNullable types are useful is we want to be able to assign `undefined` to a property when `strictNullChecks` flag is on when in our TypeScript compiler configuration. It’s simply a union type between whatever type you have and `undefined` . It’s denoted by a question mark after the property name. This means we can use type guards with it like any other union type. Note that nullable types don’t allow `null` values to be assigned to it since nullable types are only needed when `strictNullChecks` flag is on. Type alias let us create a new name for types that we already have. We can also use generics with type alias, but we can’t use them as standalone types." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8336456,"math_prob":0.9369436,"size":6492,"snap":"2020-24-2020-29","text_gpt3_token_len":1599,"char_repetition_ratio":0.13316892,"word_repetition_ratio":0.15834768,"special_character_ratio":0.26278496,"punctuation_ratio":0.16864784,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95113665,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-31T16:06:35Z\",\"WARC-Record-ID\":\"<urn:uuid:8f7e4f01-6f5f-4069-a8cd-bb1204bd50fe>\",\"Content-Length\":\"189373\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:eadb0482-4ee0-49b3-b80b-f776e88dda4c>\",\"WARC-Concurrent-To\":\"<urn:uuid:15b9c2a9-fa0b-41fb-9ff0-eaa0d42ccea7>\",\"WARC-IP-Address\":\"35.208.128.182\",\"WARC-Target-URI\":\"https://thewebdev.info/2020/04/01/typescript-advanced-typesnullable-types-and-type-aliases/\",\"WARC-Payload-Digest\":\"sha1:CITX4TOMAEGTQWFNLNA2PMEHCSCQQB4G\",\"WARC-Block-Digest\":\"sha1:X43LSDVBUJRHCZNRUKTO6Q2UZVD7GQCZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347413551.52_warc_CC-MAIN-20200531151414-20200531181414-00347.warc.gz\"}"}
https://wikidiff.com/category/terms/two	[ "# two\n\ntwo \| five \|\n\n## As numerals the difference between two and five\n\nis that two is (label) a numerical value equal to ; the second number in the set of natural numbers (especially in number theory); the cardinality of the set {0, 1}; one plus one ordinal: second this many dots (••) while five is (cardinal) a numerical value equal to ; the number following four and preceding six this many dots (•••••).\n\n## As nouns the difference between two and five\n\nis that two is the digit/figure 2 while five is the digit/figure 5.\n\nfor \| two \|\n\n## As nouns the difference between for and two\n\nis that for is oven while two is the digit/figure 2.\n\n## As a numeral two is\n\n(label) a numerical value equal to ; the second number in the set of natural numbers (especially in number theory); the cardinality of the set {0, 1}; one plus one ordinal: second this many dots (••).\n\nplastic \| two \|\n\n## As nouns the difference between plastic and two\n\nis that plastic is plastic while two is the digit/figure 2.\n\nis plastic.\n\n## As a numeral two is\n\n(label) a numerical value equal to ; the second number in the set of natural numbers (especially in number theory); the cardinality of the set {0, 1}; one plus one ordinal: second this many dots (••).\n\nslave \| two \|\n\nis .\n\n## As a numeral two is\n\n(label) a numerical value equal to ; the second number in the set of natural numbers (especially in number theory); the cardinality of the set {0, 1}; one plus one ordinal: second this many dots (••).\n\n## As a noun two is\n\nthe digit/figure 2.\n\ntwo \| two \|\n\n## In label\|en\|cardinal terms the difference between two and two\n\nis that two is (label) a numerical value equal to ; the second number in the set of natural numbers (especially in number theory); the cardinality of the set {0, 1}; one plus one ordinal: second this many dots (••) while two is (label) a numerical value equal to ; the second number in the set of natural numbers (especially in number theory); the cardinality of the set {0, 1}; one plus one ordinal: second this many dots (••).\n\n## In us\|informal\|lang=en terms the difference between two and two\n\nis that two is (us\|informal) a two-dollar bill while two is (us\|informal) a two-dollar bill.\n\n## As numerals the difference between two and two\n\nis that two is (label) a numerical value equal to ; the second number in the set of natural numbers (especially in number theory); the cardinality of the set {0, 1}; one plus one ordinal: second this many dots (••) while two is (label) a numerical value equal to ; the second number in the set of natural numbers (especially in number theory); the cardinality of the set {0, 1}; one plus one ordinal: second this many dots (••).\n\n## As nouns the difference between two and two\n\nis that two is the digit/figure 2 while two is the digit/figure 2.\n\ntwo \| words \|\n\n## As nouns the difference between two and words\n\nis that two is the digit/figure 2 while words is .\n\n## As a numeral two\n\nis (label) a numerical value equal to ; the second number in the set of natural numbers (especially in number theory); the cardinality of the set {0, 1}; one plus one ordinal: second this many dots (••).\n\n(word).\n\nzero \| two \|\n\n## As numerals the difference between zero and two\n\nis that zero is zero while two is (label) a numerical value equal to ; the second number in the set of natural numbers (especially in number theory); the cardinality of the set {0, 1}; one plus one ordinal: second this many dots (••).\n\n## As nouns the difference between zero and two\n\nis that zero is zero while two is the digit/figure 2.\n\ntwo \| tvo \|\n\n## As numerals the difference between two and tvo\n\nis that two is (label) a numerical value equal to ; the second number in the set of natural numbers (especially in number theory); the cardinality of the set {0, 1}; one plus one ordinal: second this many dots (••) while tvo is neuter nominative and accusative of tveir.\n\n## As a noun two\n\nis the digit/figure 2.\n\ntwo \| this \|\n\n## As a numeral two\n\nis (label) a numerical value equal to ; the second number in the set of natural numbers (especially in number theory); the cardinality of the set {0, 1}; one plus one ordinal: second this many dots (••).\n\n## As a noun two\n\nis the digit/figure 2.\n\n## As a determiner this is\n\n.\n\n#### Two vs Chocolate - What's the difference?\n\ntwo \| chocolate \|\n\n## As a numeral two\n\nis (label) a numerical value equal to ; the second number in the set of natural numbers (especially in number theory); the cardinality of the set {0, 1}; one plus one ordinal: second this many dots (••).\n\n## As a noun two\n\nis the digit/figure 2.\n\n.\n\n## As an adjective chocolate is\n\nchocolate (attributive)." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8337628,"math_prob":0.9975894,"size":1238,"snap":"2020-34-2020-40","text_gpt3_token_len":316,"char_repetition_ratio":0.1458671,"word_repetition_ratio":0.8272727,"special_character_ratio":0.25201938,"punctuation_ratio":0.097165994,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9996062,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-01T04:58:06Z\",\"WARC-Record-ID\":\"<urn:uuid:68b3f974-6dc0-4a5b-b6c0-625d9126dfb1>\",\"Content-Length\":\"30261\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fbf9f436-397b-46a8-a93b-9784e69d54d9>\",\"WARC-Concurrent-To\":\"<urn:uuid:b774dffa-eab2-45ff-afeb-27e3a9b0caa3>\",\"WARC-IP-Address\":\"104.22.5.162\",\"WARC-Target-URI\":\"https://wikidiff.com/category/terms/two\",\"WARC-Payload-Digest\":\"sha1:WA42D7MLT3A2YLDGKEL5BQEOHCNIPTK4\",\"WARC-Block-Digest\":\"sha1:HXVF3YTAYXGOD2IC4APQLKPIFUCGXO56\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600402130615.94_warc_CC-MAIN-20201001030529-20201001060529-00458.warc.gz\"}"}
https://www.hackster.io/group-702/group-702-sound-sensor-project-8d41d6	[ "", null, "", null, "", null, "# Group 702 Sound Sensor Project\n\nOur project detects sound and the data is collected so that we can measure at what points in the day has the highest amount of volume.\n\nIntermediateFull instructions provided20 hours113", null, "## Things used in this project\n\n### Hardware components", null, "Particle Photon\n×1", null, "×1", null, "Jumper wires (generic)\n×1\n\n### Hand tools and fabrication machines", null, "Laser cutter (generic)\n\n## Code\n\n### Sound Code for Photon\n\nArduino\n```const long sample_rate = 50; // time between samples (in miliseconds)\nconst int array_size = 1200; // 1000/50=20 * 60=1200\nint snd_array[array_size] = {};\nint snd_max, prev_max = 0;\nint snd_min, prev_min = 4096;\ndouble snd_avg = 2048;\n\n// Shift all the values right by 1\nfor(int i = array_size-1; i >= 1; i--)\n{\nsnd_array[i] = snd_array[i-1];\nif((snd_array[i] < snd_min) && (snd_array[i] != 0))\n{\nsnd_min = snd_array[i];\n\n}\nif(snd_array[i] > snd_max)\n{\nsnd_max = snd_array[i];\n\n}\n}\n\nsnd_array = value;\n\n// Average array\nfloat avg_sum = 0;\nint size = 0 ;\nfor (int a=0; a <= array_size; a++)\n{\nif(snd_array[a] > 0)\n{\nsize++ ;\navg_sum = avg_sum + snd_array[a];\n}\n}\nsnd_avg = avg_sum / size;\n}\n\ndigitalWrite(D7, HIGH);\n} else {\ndigitalWrite(D7, LOW);\n}\n}\n\nunsigned long now = millis();\nSerial.print(\"Avg: \"); Serial.println(snd_avg);\nSerial.print(\"Min: \"); Serial.println(snd_min);\nSerial.print(\"Max: \"); Serial.println(snd_max);\nsnd_avg = 0;\nsnd_min = 4096;\nsnd_max = 0;\n//snd_array[array_size] = {};\n}\n}\n\nvoid setup() {\nSerial.begin(9600);\npinMode(A0, INPUT); // mic AUD connected to Analog pin 0\npinMode(D7, OUTPUT); // flash on-board LED\n\nParticle.variable(\"snd_avg\", snd_avg);\nParticle.variable(\"snd_max\", snd_max);\nParticle.variable(\"snd_min\", snd_min);\n\n}\n\nvoid loop() {\n\ndelay(sample_rate);\n}\n```\n\n## Credits\n\n### Blake Trendel\n\n1 project • 2 followers\n\n### Michael Herrera\n\n1 project • 1 follower\n\n### sanee kassam\n\n1 project • 2 followers\nYA YA YA" ]	[ null, "https://gravatar.com/avatar/1754cc0aeca15a6c2163e33bce79bef2.png", null, "https://hackster.imgix.net/uploads/attachments/281757/bruce-lee-2_lLDkVaXGUO.jpg", null, "https://gravatar.com/avatar/49e4203eec8972316fb954f4410c559f.png", null, "https://hackster.imgix.net/uploads/attachments/285330/img_3916_A9GEksxgEn.JPG", null, "https://hackster.imgix.net/uploads/image/file/50450/photon-new-3706595d530dfb95d54f1e0e959e3386.jpg", null, "https://hackster.imgix.net/uploads/image/file/44494/12002-04.jpg", null, "https://hackster.imgix.net/uploads/image/file/44496/11026-02.jpg", null, "https://hackster.imgix.net/uploads/image/file/79857/lasercutter.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.98676646,"math_prob":0.9863594,"size":1918,"snap":"2019-26-2019-30","text_gpt3_token_len":374,"char_repetition_ratio":0.119644724,"word_repetition_ratio":0.0,"special_character_ratio":0.19447342,"punctuation_ratio":0.05107527,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98613507,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,null,null,3,null,null,null,3,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-15T19:19:08Z\",\"WARC-Record-ID\":\"<urn:uuid:f95eb5f4-bc5b-4db9-9c89-7f544e21f27d>\",\"Content-Length\":\"120760\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b00d92ea-3bfb-431a-9c2a-0a2fcaa7cf2c>\",\"WARC-Concurrent-To\":\"<urn:uuid:9e023e3d-2ec7-4d6a-85ea-4c5075391c8d>\",\"WARC-IP-Address\":\"151.101.249.174\",\"WARC-Target-URI\":\"https://www.hackster.io/group-702/group-702-sound-sensor-project-8d41d6\",\"WARC-Payload-Digest\":\"sha1:MWQZASPJKRMQLAE7XP7AZYF2KT4RHRDX\",\"WARC-Block-Digest\":\"sha1:RBRPX6KPNUQ2O4WEKJHB5BMLPA6Z2C6Q\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195523840.34_warc_CC-MAIN-20190715175205-20190715201205-00477.warc.gz\"}"}
https://forum.gwb.com/topic/2934-kd-in-the-unit-of-mol-g-1/	[ "Jump to content\nGeochemist's Workbench Support Forum\n\n# Kd in the unit of mol g-1\n\n## Recommended Posts\n\nAbout the Kd approach to treat sorbing solutes, it is written that we must convert the unit of Kd from cm3/g to mol/g. Could you please explain how we can do that?\n\nIn the textbook of the online academy (https://academy.gwb.com/sorbing_solutes.php), for example, the Kd for Pb2+ was given as 0.16 cm3/g, which was converted to 0.0003 mol/g. I would like to know details of the conversion, when we assume that (i) dissolved Pb2+ is 10 micromolar, (ii) the activity coefficient of Pb2+ in the goundwater is 0.66, and (iii) the free Pb2+ proportion is 80% of total dissolved Pb, as written in the textbook.\n\n##### Share on other sites\n\nHello Yoshio,\n\nThank you for your question. To calculate Kd' I would use the original sorption isotherm, S = KdC, where the sorbed concentration (mole/g dry sediment) is equal to the product of the Kd(cm3/g) and concentration of the free ion (mole/cm3). I would first convert the concentration of dissolved Pb2+ to mole/cm3 which is approximately 10^-8 mole/cm3. Then using the Kd and concentration of dissolved Pb2+, calculate the concentration of Pb sorbed:\n\nS = 0.16 [cm3/g] 10^-8 [mole/cm3] = 1.6 * 10 ^-9 mole of Pb2+ sorbed per gram of sediment.\n\nThe Kd' term accounts for the activity instead of the concentration. To do so you, you will need to multiply the activity coefficient by the concentration of free ions in solution. Using the activity coefficient (0.66), fraction of free ions in the fluid (0.8), and the concentration of dissolved Pb2+ (~10^-5 molal), you can calculate Kd' like so:\n\n1.610^-9 = Kd' (0.660.810^-5)\n\nwhere the Kd' is calculated to be approximately 0.0003 mol/g.\n\nHope this helps,\n\nJia Wang\n\n## Join the conversation\n\nYou can post now and register later. If you have an account, sign in now to post with your account.", null, "Reply to this topic...\n\n× Pasted as rich text. Paste as plain text instead\n\nOnly 75 emoji are allowed.\n\n× Your link has been automatically embedded. Display as a link instead\n\n× Your previous content has been restored. Clear editor\n\n× You cannot paste images directly. Upload or insert images from URL.\n\nLoading...\n×\n×\n\n• #### Activity\n\n×\n• Create New..." ]	[ null, "https://content.invisioncic.com/n304055/set_resources_1/84c1e40ea0e759e3f1505eb1788ddf3c_default_photo.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.94105816,"math_prob":0.8410228,"size":880,"snap":"2021-21-2021-25","text_gpt3_token_len":236,"char_repetition_ratio":0.09018265,"word_repetition_ratio":0.0,"special_character_ratio":0.2625,"punctuation_ratio":0.14210527,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9889285,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-21T20:09:15Z\",\"WARC-Record-ID\":\"<urn:uuid:680822ce-96f2-4a3d-bbb8-1c4578ae2e99>\",\"Content-Length\":\"71222\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:63003918-2031-40f0-949e-1a5a94f6e3c8>\",\"WARC-Concurrent-To\":\"<urn:uuid:3673781a-8250-4fda-b86e-0e3e424f61f8>\",\"WARC-IP-Address\":\"54.239.152.107\",\"WARC-Target-URI\":\"https://forum.gwb.com/topic/2934-kd-in-the-unit-of-mol-g-1/\",\"WARC-Payload-Digest\":\"sha1:G6LF6SIA6U3UJJFWGZ5XA4RGIIIMQRXB\",\"WARC-Block-Digest\":\"sha1:T545ZCCXNYJS2O7FZDMUV35CGK45MVFR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623488289268.76_warc_CC-MAIN-20210621181810-20210621211810-00226.warc.gz\"}"}
http://hermes.roua.org/hermes-pub/arxiv/05/04/149/article.xhtml	[ "### November 27, 2006\n\n<ph f=\"cmbx\">Local Analytic Hypoellipticity for a sum of squares of complex vector fields with large loss of derivatives</ph>\n\n### David S. Tartakoff\n\n5 rue de la Juviniere, 78350 Les Loges en Josas, FRANCE E-mail address : Makhlouf.Derridj@univ-rouen.fr Department of Mathematics, University of Illinois at Chicago, m/c 249, 851 S. Morgan St., Chicago IL 60607, USA E-mail address : dst@uic.edu\n• Abstract. A very recent paper of Kohn studies the hypoellipticity for a sum of squres of complex vector fields which exhibit a large loss of derivatives. We give an elementary proof of the analytic hypoellipticity of this operator. The proof requires the analysis of [2.\nM. Derridj and D.S.Tartakoff\n\n1 Introduction\n\nIn [1, J.J. Kohn proved the hypoellipticity of the operator $P={L}^{}L+\\left({\\overline{z}}^{k}L\\right)\\left({\\overline{z}}^{k}L{\\right)}^{},L=\\frac{\\partial }{\\partial z}-i\\overline{z}\\frac{\\partial }{\\partial t},$ for which there is a large loss of derivatives indeed in the a priori estimate one bounds only the Sobolev norm of order $-\\left(k+1\\right)/2.$ We show in this note that solutions of $Pu=f$ with $f$ real analytic are themselves real analytic in any open set where $f$ is. For simplicity we shall write out in detail the case for $k=1$ only, as larger values of $k$ pose no difficulties.\nThe a priori estimate with which we will work is $\\parallel Lv{\\parallel }_{0}^{2}+\\parallel z\\overline{L}v{\\parallel }_{0}^{2}\\lesssim \|\\left(Pv,v{\\right)}_{{L}^{2}}\|,v\\in {C}_{0}^{\\infty }$ since the loss of derivatives and extra low order term on the left is of no use to us and this estimate is immediate from the definition of $P.$ Our first observation is that we know the analyticity of the solution for $z\\ne 0$ from the earlier work of the second author [2, [3and Treves [4.\nOur second observation is that it suffices to bound derivatives measured in terms of high powers of the vector fields $L$ and $\\overline{L}$ in ${L}^{2}$ norm, by standard arguments, and indeed estimating high powers of $\\overline{L}$ can be reduced to bounding high powers of $L$ and powers of $T=2i\\partial /\\partial t=\\left[L,\\overline{L}\\right]$ of half the order, by repeated integration by parts. Thus our overall scheme will be to start with high powers (order $2p$ ) of $L$ or $\\overline{L},$ use integration by parts and the a priori estimate repeatedly to reduce to treating ${T}^{p}u$ in a slightly larger set.\nAnd to do this, as in many other papers of ours, we will create a (new) special localization of ${T}^{p}.$ The new localization of ${T}^{p}$ may be written in the simple form:\nLocal Analyticity for complex vector fields $\\left({T}^{{p}_{1},{p}_{2}}{\\right)}_{\\phi }={\\sum }_{\\genfrac{}{}{0}{}{a\\le {p}_{1}}{b\\le {p}_{2}}}\\frac{{L}^{a}{z}^{a}{T}^{{p}_{1}-a}{\\phi }^{\\left(a+b\\right)}{T}^{{p}_{2}-b}{\\overline{z}}^{b}{\\overline{L}}^{b}}{a!b!},$ Here by ${\\phi }^{\\left(r\\right)}$ we mean $\\left(i\\partial /\\partial t{\\right)}^{r}\\phi \\left(t\\right)$ since near $z=0$ we may take the localizing function independent of $z.$ We have the commutation relations:\n$\\left[L,\\left({T}^{{p}_{1},{p}_{2}}{\\right)}_{\\phi }\\right]\\equiv L\\circ \\left({T}^{{p}_{1}-1,{p}_{2}}{\\right)}_{{\\phi }^{\\prime }},\\left[\\overline{L},\\left({T}^{{p}_{1},{p}_{2}}{\\right)}_{\\phi }\\right]\\equiv \\left({T}^{{p}_{1},{p}_{2}-1}{\\right)}_{{\\phi }^{\\prime }}\\circ \\overline{L},$ $\\left[\\left({T}^{{p}_{1},{p}_{2}}{\\right)}_{\\phi },z\\right]=\\left({T}^{{p}_{1}-1,{p}_{2}}{\\right)}_{{\\phi }^{\\prime }}\\circ z,and\\left[\\left({T}^{{p}_{1},{p}_{2}}{\\right)}_{\\phi },\\overline{z}\\right]=\\overline{z}\\circ \\left({T}^{{p}_{1},{p}_{2}-1}{\\right)}_{{\\phi }^{\\prime }},$ where the $\\equiv$ denotes modulo ${C}^{{p}_{1}-{p}_{1}^{\\prime }+{p}_{2}-{p}_{2}^{\\prime }}$ terms of the form $\\frac{{L}^{{p}_{1}-{p}_{1}^{\\prime }}\\circ {z}^{{p}_{1}-{p}_{1}^{\\prime }}\\circ {T}^{{p}_{1}^{\\prime }}\\circ {\\phi }^{\\left({p}_{1}^{\\prime }+{p}_{2}^{\\prime }+1\\right)}\\circ {T}^{{p}_{2}^{\\prime }}\\circ {\\overline{z}}^{{p}_{2}-{p}_{2}^{\\prime }}\\circ {\\overline{L}}^{{p}_{2}-{p}_{2}^{\\prime }}}{\\left({p}_{1}-{p}_{1}^{\\prime }+{p}_{2}-{p}_{2}^{\\prime }\\right)!}$ with either ${p}_{1}^{\\prime }=0$ or ${p}_{2}^{\\prime }=0.$ Note that if we start with ${p}_{1}={p}_{2}=p/2,$ and iteratively apply these commutation relations, the number of $T$ derivatives not necessarily applied to $\\phi$ is eventually at most $p/2;$ in our previous works the number dropped to zero (i.e., only $L$ and $\\overline{L}$ derivatives remained and at most $p$ of them, half the number of $L$ and $\\overline{L}$ derivatives we started with, but any fraction of the original number would do as well.\nSo we insert first $v=\\left({T}^{p/2,p/2}{\\right)}_{\\phi }u$ in the a priori inequality, then bring $\\left({T}^{p/2,p/2}{\\right)}_{\\phi }$ to the left of $P=-\\overline{L}L-L\\overline{z}z\\overline{L}$ since $Pu$ is known and analytic. By the above bracket relations, $\\left(\\left[\\overline{L}L+L\\overline{z}z\\overline{L},\\left({T}^{p/2,p/2}{\\right)}_{\\phi }\\right]u,\\left({T}^{p/2,p/2}{\\right)}_{\\phi }u\\right)\\equiv$ M. Derridj and D.S.Tartakoff $\\equiv \\left(\\left({T}^{p/2,p/2-1}{\\right)}_{{\\phi }^{\\prime }}\\overline{L}Lu,\\left({T}^{p/2,p/2}{\\right)}_{\\phi }u\\right)+\\left(\\overline{L}L\\left({T}^{p/2-1,p/2}{\\right)}_{{\\phi }^{\\prime }}u,\\left({T}^{p/2,p/2}{\\right)}_{\\phi }u\\right)$ $+\\left(L\\left({T}^{p/2-1,p/2}{\\right)}_{{\\phi }^{\\prime }}\\overline{z}z\\overline{L}u,\\left({T}^{p/2,p/2}{\\right)}_{\\phi }u\\right)+\\left(L\\overline{z}\\left({T}^{p/2,p/2-1}{\\right)}_{{\\phi }^{\\prime }}z\\overline{L}u,\\left({T}^{p/2,p/2}{\\right)}_{\\phi }u\\right)$ $+\\left(L\\overline{z}\\left({T}^{p/2-1,p/2}{\\right)}_{{\\phi }^{\\prime }}z\\overline{L}u,\\left({T}^{p/2,p/2}{\\right)}_{\\phi }u\\right)+\\left(L\\overline{z}z\\left({T}^{p/2,p/2-1}{\\right)}_{{\\phi }^{\\prime }}\\overline{L}u,\\left({T}^{p/2,p/2}{\\right)}_{\\phi }u\\right).$ In every case, the brackets reduce the order of the sum of the two indices ${p}_{1}$ and ${p}_{2}$ by one (here we started with ${p}_{1}={p}_{2}=p/2$ ), pick up one derivative on $\\phi ,$ and leave the vector fields over which we have maximal control in the estimate intact and in the correct order. Thus we may bring either $L\\overline{z}$ to the right as $z\\overline{L},$ (possibly after additional iterations of the brackets to obtain an $\\overline{z}$ to the left of $\\left({T}^{{q}_{1},{q}_{2}}{\\right)}_{\\psi }$ ), and use a weighted Schwarz inequality on the result to take maximal advantage of the a priori inequality. Iterations of all of this continue until there remain at most $p/2$ free $T$ derivatives (i.e., the $T$ derivatives on at least one side of $\\phi$ are all `corrected' by good vector fields) and perhaps as many as $p/2L$ or $z\\overline{L}$ derivatives, and we may continue further until, at worst, the remaining $L$ and $\\overline{L}$ derivatives bracket two at a time to produce more $T$ 's, one at a time. After all of this, there will be at most ${T}^{3p/4},$ and at this point we introduce a new localizing function of Ehrenpreis type with slightly larger support, geared to $3p/4$ instead of to $p.$ In our previous work, the order dropped by half and we used ${log}_{2}N$ nested open sets, but here we employ ${log}_{4/3}N$ nested open sets of prescribed size growth may be constructed to prove that for derivatives of order no greater than $N,$ $\|{D}^{\|\\alpha \|}u\|\\le C{C}^{N}{N}^{N}\\sim {C}^{\\prime }{{C}^{\\prime }}^{N}N!$ locally with $C$ independent of $N,$ which proves the analyticity of the solution. Local Analyticity for complex vector fields The situation is not different with larger values of $k,$ and we do not write this out.\nReferences\n\n1. J.J. Kohn, Hypoellipticity and loss of derivatives, Annals of Mathematics, to appear.\n2. D.S. Tartakoff, Local Analytic Hypoellipticity for ${\\square }_{b}$ on Non-Degenerate Cauchy Riemann Manifolds, Proc. Nat. Acad. Sci. U.S.A. 75 (1978), pp. 3027-3028.\n3. D.S. Tartakoff, On the Local Real Analyticity of Solutions to ${\\square }_{b}$ and the $\\overline{\\partial }$ -Neumann Problem, Acta Math. 145 (1980), pp. 117-204.\n4. F. Treves, Analytic Hypo-ellipticity of a Class of Pseudo-Differential Operators with Double Characteristics and Application to the $\\overline{\\partial }$ -Neumann Problem, Comm. in P.D.E. 3 (6-7) (1978), pp. 475-642.\n\n5 rue de la Juviniere, 78350 Les Loges en Josas, FRANCE E-mail address : Makhlouf.Derridj@univ-rouen.fr Department of Mathematics, University of Illinois at Chicago, m/c 249, 851 S. Morgan St., Chicago IL 60607, USA E-mail address : dst@uic.edu" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9569486,"math_prob":0.99984384,"size":3552,"snap":"2021-04-2021-17","text_gpt3_token_len":738,"char_repetition_ratio":0.12880497,"word_repetition_ratio":0.006279435,"special_character_ratio":0.20157658,"punctuation_ratio":0.071117565,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999441,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-23T08:49:13Z\",\"WARC-Record-ID\":\"<urn:uuid:9ec4af27-a89c-4657-8efe-31dc6843c20c>\",\"Content-Length\":\"42575\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8ea6a2cc-b1ed-41d4-891f-84f69180ed1a>\",\"WARC-Concurrent-To\":\"<urn:uuid:68dfdb7d-6199-4705-a57f-b80a21b1a5e4>\",\"WARC-IP-Address\":\"188.26.144.79\",\"WARC-Target-URI\":\"http://hermes.roua.org/hermes-pub/arxiv/05/04/149/article.xhtml\",\"WARC-Payload-Digest\":\"sha1:2GF245BVQBLBWJNXMPWIGSLC4XLY6JG2\",\"WARC-Block-Digest\":\"sha1:GI4PGQ7HRXD35MLTZDX7B2ILYXVEJK3A\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703536556.58_warc_CC-MAIN-20210123063713-20210123093713-00086.warc.gz\"}"}
https://math.stackexchange.com/questions/3003731/how-to-evaluate-this-integral-with-exponent-of-an-exponent	[ "# How to evaluate this integral with exponent of an exponent?\n\nI have the following integral which I need to evaluate but don't even know where to begin other than knowing I need to use u-substitution: $$\\int_1^\\sqrt{3}2x^{x^{2}}$$\n\nSo far I know that $$u=x^{2}$$ and $$du=2x$$ but how do I evaluate this?\n\n• You might consider writing $x^{x^2} = \\exp(\\log(x^{x^2}))$ and simplifying things a bit. Then try a substitution. – Xander Henderson Nov 18 '18 at 16:17\n• maybe a parameterization so that you could differentiate under the integral sign? – clathratus Nov 18 '18 at 21:38\n• @XanderHenderson what does it mean to write exp(log(x^x^2))? I might be a little slow now but not sure what that exp does... – blizz Nov 20 '18 at 1:11\n• $\\exp(t) = \\mathrm{e}^t$. – Xander Henderson Nov 20 '18 at 1:42" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9474191,"math_prob":0.99105746,"size":475,"snap":"2019-26-2019-30","text_gpt3_token_len":146,"char_repetition_ratio":0.12314225,"word_repetition_ratio":0.94736844,"special_character_ratio":0.31368423,"punctuation_ratio":0.039215688,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9995376,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-20T22:11:53Z\",\"WARC-Record-ID\":\"<urn:uuid:30fa2358-3b82-4eb8-a820-fa56470f3bbf>\",\"Content-Length\":\"134964\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c29f4e98-dbe0-4433-82f4-c8450a754b23>\",\"WARC-Concurrent-To\":\"<urn:uuid:2e600169-9cc0-4179-8714-5de09ff307de>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/3003731/how-to-evaluate-this-integral-with-exponent-of-an-exponent\",\"WARC-Payload-Digest\":\"sha1:2MDRO2SHYNVISCL7URGWHTIGHKNF6W5S\",\"WARC-Block-Digest\":\"sha1:RFGNOYYTFA7HDGUXWZDZA66X5HQKJUJ5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195526714.15_warc_CC-MAIN-20190720214645-20190721000645-00228.warc.gz\"}"}
https://quantumcomputing.stackexchange.com/questions/2674/is-entanglement-necessary-for-quantum-computation	[ "# Is entanglement necessary for quantum computation?\n\nEntanglement is often discussed as being one of the essential components that makes quantum different from classical. But is entanglement really necessary to achieve a speed-up in quantum computation?\n\n• phys.org/news/2008-12-quantum-entanglement.html – Steven Sagona Jul 9 '18 at 6:50\n• @StevenSagona That news article talks about the model DQC1. There is always entanglement in that model, it's just that a naive first analysis only looks for it in one particular place, where it turns out not to be. – DaftWullie Jul 9 '18 at 7:08\n• Did you ask and answer this question because of my answer to: quantumcomputing.stackexchange.com/a/2601/2293 ? – user1271772 Jul 13 '18 at 11:55\n• @user1271772 Nope! Although I did ask it because of something said to me as a comment that I needed a more complete response that I could reference. – DaftWullie Jul 13 '18 at 11:59\n• @DaftWullie: I don't understand why my answer has 5 negative votes. Perhaps saying \"entanglement is considered a requirement for QC\" was not sufficient on it's own? – user1271772 Jul 13 '18 at 12:00\n\nOne has to be a little bit more careful setting up the question. Thinking about a circuit as being composed of state preparation, unitaries, and measurements, it is always in principle possible to \"hide\" anything we want, such as entangling operations, inside the measurement. So, let us be precise. We want to start from a separable state of many qubits, and the final measurements should consist of single-qubit measurements. Does the computation have to transition through an entangled state at some point in the computation?\n\n# Pure states\n\nLet's make the one further assumption that the initial state is a pure (product) state. In that case, the system must go through an entangled state. If it didn't, it is easy to simulate the computation on a classical computer because all you have to do is hold $n$ single-qubit pure states in memory, and update them one at a time as the computation proceeds.\n\nOne can even ask how much entanglement is necessary. Again, there are many different ways that entanglement can be moved around at different times. A good model that provides a reasonably fair measure of the entanglement present is measurement-based quantum computation. Here, we prepare some initial resource state, and it is single-qubit measurements that define the computation that happens. This lets us ask about the entanglement of the resource state. There has to be entanglement and, in some sense, it has to be at least \"two-dimensional\", it cannot just be the entanglement generated between nearest neighbours of a system on a line [ref]. Moreover, one can show that most states of $n$ qubits are too entangled to permit computation in this way.\n\n# Mixed states\n\nThe caveat in all that I've said so far is that we're talking about pure states. For example, we can easily simulate a non-entangling computation on pure product states. But what about mixed states? A mixed state is separable if it can be written in the form $$\\rho=\\sum_{i=1}^Np_i\\rho^{(1)}_i\\otimes\\rho^{(2)}_i\\otimes\\ldots\\otimes\\rho^{(n)}_i.$$ Importantly, there is no limit on the value $N$, the number of terms in the sum. If the number of terms in the sum is small, then by the previous argument, we can simulate the effects of a non-entangling circuit. But if the number of terms is large, then (to my knowledge) it remains an open question as to whether it can be classically simulated, or whether it can give enhanced computation.\n\n• This work (arxiv.org/pdf/quant-ph/0301063.pdf) might be of interest here. Entanglement in a quantum system has to scale as a polynomial of system size to get an exponential quantum speed up. A quantum algorithm can be classically simulated with resources that scale as with the exponential of entanglement. – biryani Jul 9 '18 at 6:59\n• although non-exponential speed-ups such as Grover can get away with tiny amounts of entanglement, my own work . – DaftWullie Jul 9 '18 at 7:10\n• What do you think about this paper? I didn't have time to go through it carefully, but it states that Grover's can be done without entanglement (at slower speeds). – Steven Sagona Jan 28 '19 at 23:50\n• @StevenSagona It's a kind of cheat/sales pitch. Although we usually talk about $n$ qubits, with a Hilbert space of dimension $2^n$, you could get that Hilbert space by using a single particle with a Hilbert space of dimension $2^n$ (e.g. sending the particle down $2^n$ different paths), and there is certainly no entanglement present (actually, there's a philosophical question there re. path based superposition/entanglement). There are gate costs associated with this conversion, but by using an oracle model, as in Grover's, those costs get hidden and it appears to be achieving the same thing. – DaftWullie Jan 29 '19 at 8:50\n• Ah I see. Thanks for answering, this actually resolves some conceptual questions in my head (as it was not obvious to me why simply superpostion of a single particle is insufficient to provide the same mechanisms as these entangled systems). – Steven Sagona Jan 29 '19 at 21:53" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92818123,"math_prob":0.90077627,"size":5354,"snap":"2020-24-2020-29","text_gpt3_token_len":1329,"char_repetition_ratio":0.13588785,"word_repetition_ratio":0.025522042,"special_character_ratio":0.24187523,"punctuation_ratio":0.09950249,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9729717,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-02T13:16:01Z\",\"WARC-Record-ID\":\"<urn:uuid:f7189556-2e3e-401e-895b-2a5e7e44a347>\",\"Content-Length\":\"163359\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:333b1ed3-6029-42c3-9770-9803968893e4>\",\"WARC-Concurrent-To\":\"<urn:uuid:10d4e5fb-8839-4273-b612-350971c73095>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://quantumcomputing.stackexchange.com/questions/2674/is-entanglement-necessary-for-quantum-computation\",\"WARC-Payload-Digest\":\"sha1:6PDRLYRG4LWH3PBWVWAZU2L3FY4LXN4O\",\"WARC-Block-Digest\":\"sha1:6S2Y7HJCHRGRHKRMODAXF656KYI66FY2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347425148.64_warc_CC-MAIN-20200602130925-20200602160925-00593.warc.gz\"}"}
https://la.mathworks.com/help/finance/portfolio-set-for-optimization-using-portfoliomad-object.html	[ "## Portfolio Set for Optimization Using PortfolioMAD Object\n\nThe final element for a complete specification of a portfolio optimization problem is the set of feasible portfolios, which is called a portfolio set. A portfolio set $X\\subset {R}^{n}$ is specified by construction as the intersection of sets formed by a collection of constraints on portfolio weights. A portfolio set necessarily and sufficiently must be a nonempty, closed, and bounded set.\n\nWhen setting up your portfolio set, ensure that the portfolio set satisfies these conditions. The most basic or “default” portfolio set requires portfolio weights to be nonnegative (using the lower-bound constraint) and to sum to `1` (using the budget constraint). The most general portfolio set handled by the portfolio optimization tools can have any of these constraints:\n\n• Linear inequality constraints\n\n• Linear equality constraints\n\n• Bound constraints\n\n• Budget constraints\n\n• Group constraints\n\n• Group ratio constraints\n\n• Average turnover constraints\n\n• One-way turnover constraints\n\n### Linear Inequality Constraints\n\nLinear inequality constraints are general linear constraints that model relationships among portfolio weights that satisfy a system of inequalities. Linear inequality constraints take the form\n\n`${A}_{I}x\\le {b}_{I}$`\n\nwhere:\n\nx is the portfolio (n vector).\n\nAI is the linear inequality constraint matrix (nI-by-n matrix).\n\nbI is the linear inequality constraint vector (nI vector).\n\nn is the number of assets in the universe and nI is the number of constraints.\n\n`PortfolioMAD` object properties to specify linear inequality constraints are:\n\n• `AInequality` for AI\n\n• `bInequality` for bI\n\n• `NumAssets` for n\n\nThe default is to ignore these constraints.\n\n### Linear Equality Constraints\n\nLinear equality constraints are general linear constraints that model relationships among portfolio weights that satisfy a system of equalities. Linear equality constraints take the form\n\n`${A}_{E}x={b}_{E}$`\n\nwhere:\n\nx is the portfolio (n vector).\n\nAE is the linear equality constraint matrix (nE-by-n matrix).\n\nbE is the linear equality constraint vector (nE vector).\n\nn is the number of assets in the universe and nE is the number of constraints.\n\n`PortfolioMAD` object properties to specify linear equality constraints are:\n\n• `AEquality` for AE\n\n• `bEquality` for bE\n\n• `NumAssets` for n\n\nThe default is to ignore these constraints.\n\n### 'Simple' Bound Constraints\n\n`'Simple'` Bound constraints are specialized linear constraints that confine portfolio weights to fall either above or below specific bounds. Since every portfolio set must be bounded, it is often a good practice, albeit not necessary, to set explicit bounds for the portfolio problem. To obtain explicit bounds for a given portfolio set, use the `estimateBounds` function. Bound constraints take the form\n\n`${l}_{B}\\le x\\le {u}_{B}$`\n\nwhere:\n\nx is the portfolio (n vector).\n\nlB is the lower-bound constraint (n vector).\n\nuB is the upper-bound constraint (n vector).\n\nn is the number of assets in the universe.\n\n`PortfolioMAD` object properties to specify bound constraints are:\n\n• `LowerBound` for lB\n\n• `UpperBound` for uB\n\n• `NumAssets` for n\n\nThe default is to ignore these constraints.\n\nThe default portfolio optimization problem (see Default Portfolio Problem) has lB = `0` with uB set implicitly through a budget constraint.\n\n### Budget Constraints\n\nBudget constraints are specialized linear constraints that confine the sum of portfolio weights to fall either above or below specific bounds. The constraints take the form\n\n`${l}_{S}\\le {1}^{T}x\\le {u}_{S}$`\n\nwhere:\n\nx is the portfolio (n vector).\n\n`1` is the vector of ones (n vector).\n\nlS is the lower-bound budget constraint (scalar).\n\nuS is the upper-bound budget constraint (scalar).\n\nn is the number of assets in the universe.\n\n`PortfolioMAD` object properties to specify budget constraints are:\n\n• `LowerBudget` for lS\n\n• `UpperBudget` for uS\n\n• `NumAssets` for n\n\nThe default is to ignore this constraint.\n\nThe default portfolio optimization problem (see Default Portfolio Problem) has lS = uS = `1`, which means that the portfolio weights sum to `1`. If the portfolio optimization problem includes possible movements in and out of cash, the budget constraint specifies how far portfolios can go into cash. For example, if lS = `0` and uS = `1`, then the portfolio can have 0–100% invested in cash. If cash is to be a portfolio choice, set `RiskFreeRate` (r0) to a suitable value (see Return Proxy and Working with a Riskless Asset).\n\n### Group Constraints\n\nGroup constraints are specialized linear constraints that enforce “membership” among groups of assets. The constraints take the form\n\n`${l}_{G}\\le Gx\\le {u}_{G}$`\n\nwhere:\n\nx is the portfolio (n vector).\n\nlG is the lower-bound group constraint (nG vector).\n\nuG is the upper-bound group constraint (nG vector).\n\nG is the matrix of group membership indexes (nG-by-n matrix).\n\nEach row of G identifies which assets belong to a group associated with that row. Each row contains either `0`s or `1`s with `1` indicating that an asset is part of the group or `0` indicating that the asset is not part of the group.\n\n`PortfolioMAD` object properties to specify group constraints are:\n\n• `GroupMatrix` for G\n\n• `LowerGroup` for lG\n\n• `UpperGroup` for uG\n\n• `NumAssets` for n\n\nThe default is to ignore these constraints.\n\n### Group Ratio Constraints\n\nGroup ratio constraints are specialized linear constraints that enforce relationships among groups of assets. The constraints take the form\n\n`${l}_{Ri}{\\left({G}_{B}x\\right)}_{i}\\le {\\left({G}_{A}x\\right)}_{i}\\le {u}_{Ri}{\\left({G}_{B}x\\right)}_{i}$`\n\nfor i = 1,..., nR where:\n\nx is the portfolio (n vector).\n\nlR is the vector of lower-bound group ratio constraints (nR vector).\n\nuR is the vector matrix of upper-bound group ratio constraints (nR vector).\n\nGA is the matrix of base group membership indexes (nR-by-n matrix).\n\nGB is the matrix of comparison group membership indexes (nR-by-n matrix).\n\nn is the number of assets in the universe and nR is the number of constraints.\n\nEach row of GA and GB identifies which assets belong to a base and comparison group associated with that row.\n\nEach row contains either `0`s or `1`s with `1` indicating that an asset is part of the group or `0` indicating that the asset is not part of the group.\n\n`PortfolioMAD` object properties to specify group ratio constraints are:\n\n• `GroupA` for GA\n\n• `GroupB` for GB\n\n• `LowerRatio` for lR\n\n• `UpperRatio` for uR\n\n• `NumAssets` for n\n\nThe default is to ignore these constraints.\n\n### Average Turnover Constraints\n\nTurnover constraint is a linear absolute value constraint that ensures estimated optimal portfolios differ from an initial portfolio by no more than a specified amount. Although portfolio turnover is defined in many ways, the turnover constraints implemented in Financial Toolbox™ compute portfolio turnover as the average of purchases and sales. Average turnover constraints take the form\n\n`$\\frac{1}{2}{1}^{T}\|x-{x}_{0}\|\\le \\tau$`\n\nwhere:\n\nx is the portfolio (n vector).\n\n`1` is the vector of ones (n vector).\n\nx0 is the initial portfolio (n vector).\n\nτ is the upper bound for turnover (scalar).\n\nn is the number of assets in the universe.\n\n`PortfolioMAD` object properties to specify the average turnover constraint are:\n\n• `Turnover` for τ\n\n• `InitPort` for x0\n\n• `NumAssets` for n\n\nThe default is to ignore this constraint.\n\n### One-way Turnover Constraints\n\nOne-way turnover constraints ensure that estimated optimal portfolios differ from an initial portfolio by no more than specified amounts according to whether the differences are purchases or sales. The constraints take the forms\n\n`${1}^{T}×\\mathrm{max}\\left\\{0,x-{x}_{0}\\right\\}\\le {\\tau }_{B}$`\n\n`${1}^{T}×\\mathrm{max}\\left\\{0,{x}_{0}-x\\right\\}\\le {\\tau }_{S}$`\n\nwhere:\n\nx is the portfolio (n vector)\n\n`1` is the vector of ones (n vector).\n\nx0 is the Initial portfolio (n vector).\n\nτB is the upper bound for turnover constraint on purchases (scalar).\n\nτS is the upper bound for turnover constraint on sales (scalar).\n\nTo specify one-way turnover constraints, use the following properties in the `PortfolioMAD` object:\n\n• `BuyTurnover` for τB\n\n• `SellTurnover` for τS\n\n• `InitPort` for x0\n\nThe default is to ignore this constraint.\n\nNote\n\nThe average turnover constraint (see Working with Average Turnover Constraints Using PortfolioMAD Object) with τ is not a combination of the one-way turnover constraints with τ = τB = τS." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8075505,"math_prob":0.9807881,"size":7470,"snap":"2020-34-2020-40","text_gpt3_token_len":1624,"char_repetition_ratio":0.23225288,"word_repetition_ratio":0.30172414,"special_character_ratio":0.18942437,"punctuation_ratio":0.082945734,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99845177,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-25T16:44:36Z\",\"WARC-Record-ID\":\"<urn:uuid:da100e37-bd75-4b3f-833f-0b18362da9b7>\",\"Content-Length\":\"84280\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7e354f01-03e9-4951-9a6f-cf810c6dfb1c>\",\"WARC-Concurrent-To\":\"<urn:uuid:f3380570-53a1-486c-a805-4a7dd89dc273>\",\"WARC-IP-Address\":\"23.223.252.57\",\"WARC-Target-URI\":\"https://la.mathworks.com/help/finance/portfolio-set-for-optimization-using-portfoliomad-object.html\",\"WARC-Payload-Digest\":\"sha1:6NOTGWFMAPQPKHEXAOSTHNSYNGUIMW66\",\"WARC-Block-Digest\":\"sha1:O4XZFGA6LL2DGD2A36M5ETZN6QRNXHVF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400227524.63_warc_CC-MAIN-20200925150904-20200925180904-00530.warc.gz\"}"}
https://www.vut.cz/en/students/courses/detail/258772	[ "Course detail\n\n# Discrete Processes in Electrical Engineering\n\nThe discipline is devoted to description of processes via discrete equations. It consists of three parts:\na) basic calculus and basic methods of analysis of discrete processes,\nb) application of difference equations, investigation of stability processes,\nc) application of difference equations in control of processes.\n\nThe plan of discipline is described in the point \"Syllabus\" in detail. The discipline is recommended for Ph.D. programme students, who will apply discrete and difference relations, equations and numerical algorithms also. As illustration we point to mathematical modelling of phenomena in nanotechnologies, control theory and signal processing.\n\nOffered to foreign students\n\nThe home faculty only\n\nLearning outcomes of the course unit\n\nThe ability to orientate in the basic notions and problems of discrete and difference equations. Solving problems in the areas cited in the curriculum by use of these methods. Solving given by use of modern mathematical software. Main outcomes are:\n\n1) Ability to solve basic classes of difference equations of the first order.\n2) Usage of difference equations of first order to solution of equations describing various phenomena modeled by difference equations of the first order. Transformation of differential equations to discrete equations, modeling electrical circuits by difference equations.\n3) Finding of equilibria points of scalar equations, determination stability and other properties of solutions in the vicinity of equilibria.\n4) Construction of cob-web diagrams for inverstigation of stability of equailibrium points.\n5) Determination of stability of numerical algorithms using equilibrium points.\n6) Application of basic formulae of discrete calculus.\n7) Solution of homogeneous and non-homogeneous linear discrete equations of higher-order.\n8) Construction of solutions of systems homogeneous and non-homogeneous difference equations of the first order.\n9) Solution of linear homogeneous system of difference equations by Putzer algorithm. Finding of a particular solution.\n10) Determination of stability and non-stability of nonlinear and linear discrete systems by method of a fundamental matrix and by Lyapunov method.\n11) Application of Z-transform to solution of linear difference equations of higher-order and to solution of linear difference systems.\n12) Detection of controllability and observability of linear discrete systems.\n\nPrerequisites\n\nThe subject knowledge on the Bachelor´s and Magister´s degree level is requested from mathematical disciplines.\n\nCo-requisites\n\nNot applicable.\n\nRecommended optional programme components\n\nNot applicable.\n\nLiterature\n\nMickens, Ronald E., Difference Equations: Theory, Applications and Advanced Topics, Third Edition, Chapman & Hall/CRC, 2016 (EN)\nOppenheim, Alan, V., Schaffer, Ronald, W., Discrete-Time Signal Processing, 3rd Edition, Pearson, 2014 (EN)\nFarlow, S. J., Solution Manual: Introduction to Differential Equations and Their Applications , Dover Publications, 2016.\nSami Fadali, M., Visioli, A., Digital Control Engineering, Analysis and Design, 2nd Edition, Elsewier, AP, 2013 (EN)\nBanerjee, D., From Continuous to Discrete: Integer Equations, Difference Equations, and Digital Electronics, Dog Ear Publishing, LLC, 2014 (EN)\nDiblík, J., Hlavičková, I., Discrete Processes in Electrical Engineering, electronic text, Brno, 2018 (CS)\n\nPlanned learning activities and teaching methods\n\nTeaching methods depend on the type of course unit as specified in the article 7 of BUT Rules for Studies and Examinations.\n\nAssesment methods and criteria linked to learning outcomes\n\nAbilities leading to successful solution of some typical classes of difference equations as well as necessary theoretical knowledge and its application will be positively estimated. During half-year term students must prepare 3 essay. The final evaluation (examination) depends on assigned points (0 points is minimum, 100 points is maximum), 30 points is maximum points which can be assigned for essays. Final examination is in written and oral form and is estimated\nas follows: 0- points is minimum, 70 points is maximum.\n\nLanguage of instruction\n\nEnglish\n\nWork placements\n\nNot applicable.\n\nCourse curriculum\n\n1. Basic notions and methods of investigation of discrete processes.\n2. Discrete calculus (some difference relations based on corresponding continuous relations). Difference equations and systems.\n3. Basic notions used in difference equations (equilibrium points, periodic points, eventually equilibrium points and eventually periodic points, stability of solution, repelling and attracting points) and their illustration on examples (modelling of circuits with the aid of difference equations, the transmission of information).\n4. Recursive algorithms of solutions of systems of discrete equations and equations of higher order (the case of constant coefficients, the method of variation of parameters, the method of variation of constants).\n5. Construction of the general solution. Transformation of some nonlinear equations into linear equations. Difference equations modelled with the aid of sampling, impulses inputs, computation of characteristic from the signal response (response of Dirac distribution), transmission effects.\n6. Application of difference equations – stability of processes. Stability of equilibrium points. Kinds of stabily and instability.\n7. Stability of linear systems with the variable matrix. Stability of nonlinear systems via linearization.\n8. Ljapunov direct method of stability.\n9. Phase analysis of two-dimensional linear discrete system with constant matrix, classification of equilibrium points.\n10. Application of difference equations - control of processes. Discrete equivalents of continuous systems.\n11. Discrete control theory (the controllability, the complete controllability, matrix of controllability, the canonical forms of controllability, controllable canonical form, construction of the control algorithm).\n12. Observability, complete observability, nonobservability, principle of duality, the observability matrix.\n13. Canonical forms of observability, relation of controllability and observability. Stabilization of control by feedback.\n\nAims\n\nDiscrete and difference equations are the mathematical base of many fields of engineering science. The purpose of this course is to develop the basic notions concerning the properties of solutions of such equations, demonstrate methods of their solution, give methods for investigation of stability of solutions, clarify their utilization in control theory and show their applications. Therefore the attention is focused on application examples and their utilization for study of stability of processes, their controllability and observability.\n\nSpecification of controlled education, way of implementation and compensation for absences\n\nThe content and forms of instruction in the evaluated course are specified by a regulation issued by the lecturer responsible for the course and updated for every academic year. Necessary condition (for final examination) are three prepared essays motivated by published papers with applications of discrete equations.\n\nClassification of course in study plans\n\n• Programme DKA-KAM Doctoral, any year of study, summer semester, 4 credits, compulsory-optional\n• Programme DKA-EKT Doctoral, any year of study, summer semester, 4 credits, compulsory-optional\n• Programme DKA-EIT Doctoral, any year of study, summer semester, 4 credits, compulsory-optional\n• Programme DKAD-EIT Doctoral, any year of study, summer semester, 4 credits, compulsory-optional\n• Programme DKA-MET Doctoral, any year of study, summer semester, 4 credits, compulsory-optional\n• Programme DKA-SEE Doctoral, any year of study, summer semester, 4 credits, compulsory-optional\n• Programme DKA-TLI Doctoral, any year of study, summer semester, 4 credits, compulsory-optional\n• Programme DKA-TEE Doctoral, any year of study, summer semester, 4 credits, compulsory-optional\n\n#### Type of course unit\n\nGuided consultation\n\n39 hours, optionally\n\nTeacher / Lecturer" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8269014,"math_prob":0.91572773,"size":6040,"snap":"2022-40-2023-06","text_gpt3_token_len":1227,"char_repetition_ratio":0.1785951,"word_repetition_ratio":0.09011264,"special_character_ratio":0.18576159,"punctuation_ratio":0.17070708,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.98652405,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-02T15:29:53Z\",\"WARC-Record-ID\":\"<urn:uuid:3229c640-f1fd-438d-be64-0b51421c2f15>\",\"Content-Length\":\"63892\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6551ee32-0904-48f8-9609-30e5e8d24e86>\",\"WARC-Concurrent-To\":\"<urn:uuid:ff2c56de-ce13-4662-89e7-bae6fb3f44cd>\",\"WARC-IP-Address\":\"147.229.2.90\",\"WARC-Target-URI\":\"https://www.vut.cz/en/students/courses/detail/258772\",\"WARC-Payload-Digest\":\"sha1:C6HO4BFWD7SI7SXZSDHWHV3UAGI4AYZK\",\"WARC-Block-Digest\":\"sha1:QJDBYWBW32NG5FOINSWKF7SI7GLWV3LA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337338.11_warc_CC-MAIN-20221002150039-20221002180039-00411.warc.gz\"}"}