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https://mathoverflow.net/questions/35127/why-does-the-algebraic-condition-of-flatness-on-the-structure-sheaves-give-a-go	[ "# Why does the algebraic condition of flatness on the structure sheaves give a good definition of family?\n\nHartshorne remarks that is is something of a mystery as to why the algebraic condition of flatness on the structure sheaves gives a good definition of a family (see below). Are there any known enlightening explanations that help serve to unravel this mystery? Below is Hartshorne's introductory motivation to flat families containing said remark:\n\nFor many reasons it is important to have a good notion of an algebraic family of varieties or schemes. The most naive definition would be just to take the fibres of a morphism. To get a good notion, however, we should require that certain numerical invariants remain constant in a family, such as the dimension of the fibres. It turns out that if we are dealing with non- singular (or even normal) varieties over a field, then the naive definition is already a good one. Evidence for this is the theorem (9.13) that in such a family, the arithmetic genus is constant.\n\nOn the other hand, if we deal with nonnormal varieties, or more general schemes, the naive definition will not do. So we consider a flat family of schemes, which means the fibres of a flat morphism, and this is a very good notion. Why the algebraic condition of flatness on the structure sheaves should give a good definition of a family is something of a mystery. But at least we will justify this choice by showing that flat families have many good properties, and by giving necessary and sufficient conditions for flatness in some special cases. In particular, we will show that a family of closed subschemes of projective space (over an integral scheme) is flat if and only if the Hilbert polynomials of the fibres are the same. -- Hartshorne, Algebraic Geometry, 1977, III.9.5, p. 256\n\n• Here's something \"technically useful\" (& underlies Hilbert schemes, ur-example of flatness for moduli problems): if $X \\rightarrow S$ is map of schemes and $j:Z \\hookrightarrow X$ is closed immersion defined by q-coh ideal $I$ such that $Z$ is flat over $S$, then for any $f:S' \\rightarrow S$ the exact sequence $0 \\rightarrow I \\rightarrow O_X \\rightarrow j_{\\ast}O_Z \\rightarrow 0$ has $S$-flat right term, so $f^{\\ast}I \\rightarrow O_{X'}$ is injective. Thus, ideal of $Z' \\hookrightarrow X'$ is the \"abstract\" pullback $f^{\\ast}I$ treating $I$ as \"abstract\" $O_X$-module. This is important. – BCnrd Aug 10 '10 at 16:11\n• To \"explain\" ubiquity of flatness, use Theorem 23.1 in Matsumura's Commutative Ring Theory book: if $f:X \\rightarrow Y$ is map from regular scheme to Cohen-Macaulay scheme (e.g., smooth varieties over field) then \"dimension formula\" ${\\rm{dim}} O_ {X,x} = {\\rm{dim}} O_ {Y,f(x)} + {\\rm{dim}} O_ {X_ {f(x)}, x}$ is equivalent to flatness of $f$ at $x$. A great non-trivial example is q-finite map between smooth varieties of same dimension. Serre told me that the geometric significance of flatness was entirely due to Grothendieck (Serre invented it as a purely algebraic device for GAGA). – BCnrd Aug 10 '10 at 16:18\n• Correction: in 2nd comment above, $Y$ should be regular, and $X$ Cohen-Macalay (or just assume both regular so can save memory for other things...but the CM case is useful for surfaces...) – BCnrd Aug 10 '10 at 16:19\n• Section II.3.4 in Eisenbud/Harris contains an elaborate introduction to flat families of schemes. Especially Proposition II.29 might help you. It says roughly that for a flat family + other conditions the fiber of a closed point 0 is the limit of the fibers of points b where b goes to 0. – C. Jost Aug 11 '10 at 8:25\n• Thanks to all for the many interesting remarks, here and below. – Bill Dubuque Aug 17 '10 at 18:54\n\nHere is an elementary and intuitive explanation. The fiber of a map is locally a tensor product: if $X=\\text{Spec} S$ and $Y=\\text{Spec} R$ and the ring map is $R \\to S$, then the fiber at a point $p \\in Y$ is the Spec of $R_p/pR_p \\otimes_R S$.\n\nFlatness is exactly the condition that makes tensor products behave like a dream (almost by definition), it preserves a lot of useful structures. Many algebraic results with geometric consequences go like this: let $(P)$ be a reasonable property and $f: R\\to S$ a flat local homomorphism. Then $S$ satisfies $(P)$ if and only if $R$ and the fiber at the closed point satisfy $(P)$ (these are called Grothendieck localization problem).\n\nI am not a historian, but I suspect that was how flatness arised: people wanted certain nice things to be true, and were naturally lead to flatness (see BCnrd's comment below for the precise history).\n\n• Hailong, the history is that for his GAGA theorems, Serre needed an algebraic mechanism to explain how passing from algebraic to analytic settings (such as for vector bundles, coherent sheaves, etc.) preserved various exactness properties. In this case the algebraic and analytic local rings at a common point were known to be noetherian with the same completion, and completion was known to be preserve exactness under tensoring (Artin-Rees). This led Serre to isolate flatness as an interesting algebraic concept (incorporating exactness of localization), and then it took on a life of its own. – BCnrd Aug 10 '10 at 19:12\n• Dear Brian, thanks for the precise history. – Hailong Dao Aug 10 '10 at 19:27\n• $Quot(R/p) \\otimes_R S$, right? – Martin Brandenburg Aug 25 '10 at 9:39\n• Oh, I forgot to localize! – Hailong Dao Aug 26 '10 at 19:36\n• I think the following precise statement holds: Let $\\phi : A \\to B$ be a morphism of noetherian rings then $B$ is flat iff the preimage $Spec A/ \\mathfrak{p}A \\otimes_A B$ coincides with $Spec B/\\phi(\\mathfrak{p})B$ for every $\\mathfrak{p}\\in SpecA$ . Which means that the two possible ways of taking preimages of closed subsets coincide (first is the usual pullback as schemes and the second is pulling back the ideal and looking at the closed subscheme defined by it). Strangely enough, I have never contemplated this subtle point about preimages. – Saal Hardali Jun 6 '16 at 23:55\n\nThere is also the following (probably unhistorical) point of view (it is a version of Hailong Dao's answer). Namely, you don't have to work with flat families at all, so if you want, you can just declare all morphisms to be families'. The problem with this approach is that this is a family ofderived' objects. Here's an example:\n\nLet $S$ be a scheme, and let $F$ be a coherent sheaf on S. When is it a `family' of its fibers? If it is flat, it definitely deserves to be called a family of vector spaces (a vector bundle). But even if it is not flat, you can still view it as a family, but the family of what? The (derived) fibers of $F$ are no longer vector spaces, they are complexes of vector spaces (precisely because $F$ fails to be flat), so we can view $F$ as a nice family of complexes of vector spaces, even though $F$ itself is a sheaf, not a complex.\n\nTo summarize: by all means, let's forget about flatness and declare any morphism to be a family... of some kind of derived objects. If we now want members of the family to be actual objects (schemes, vector spaces, sheaves, or whatever it is we are trying to include in a family), flatness is forced on you more or less by definition.\n\nThere is a nice computational perspective in Bayer and Mumford's What Can Be Computed in Algebraic Geometry? pages 4,5." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9488219,"math_prob":0.98149824,"size":1704,"snap":"2019-51-2020-05","text_gpt3_token_len":370,"char_repetition_ratio":0.12411765,"word_repetition_ratio":0.048442908,"special_character_ratio":0.20481221,"punctuation_ratio":0.10060976,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99524146,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-15T11:31:46Z\",\"WARC-Record-ID\":\"<urn:uuid:916f6d4d-be1a-434d-9677-af8c03c635b3>\",\"Content-Length\":\"139067\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c8bdcfb8-6436-4de7-9cd9-6f04369fdcc3>\",\"WARC-Concurrent-To\":\"<urn:uuid:2e4c074a-562d-4597-bd76-b6a1676650ff>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://mathoverflow.net/questions/35127/why-does-the-algebraic-condition-of-flatness-on-the-structure-sheaves-give-a-go\",\"WARC-Payload-Digest\":\"sha1:QNB7LD2MH5F7CYU3Z3AZGPZRDEDRHGBN\",\"WARC-Block-Digest\":\"sha1:6C3B573A4YUSJA7QJLEOVWAGGG5OG6ID\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575541307813.73_warc_CC-MAIN-20191215094447-20191215122447-00142.warc.gz\"}"}
https://solvedlib.com/n/activity-6-3-4-consider-a-thimbar-of-length-20-cm-whose,19293825	[ "# Activity 6.3.4. Consider a thimbar of length 20 cm whose density is distributed according to the function (r) = 4+0.lz,\n\n###### Question:\n\nActivity 6.3.4. Consider a thimbar of length 20 cm whose density is distributed according to the function (r) = 4+0.lz, where z = 0 represents the left end of the bar, Assume that p is measured in g/cm and z is measured in cm_ Find the total mass, M, of the bar; Without doing any calculations, do you expect the center of mass of the bar to be equal to 10, less than 10,or greater than 10? Why? Compute the exact center of mass of the bar. What is the average density of the bar? Now consider a different density function, given by plz) = 4e0.0207322,also for a bar of Tlength 20 cm whose left end is at z = 0. Plot both e(z) and p(z) on the same axes. Without doing any calculations, which bar do you expect to have the greater center of mass? Why? Compute the exact center of mass of the bar described in (e) whose density function iS p(c) 0207321 Check the result against the prediction you made in (e): prt SC delete home nunt Toc 8 Dacaadeca", null, "", null, "#### Similar Solved Questions\n\n##### 72sin (0) lecnaov ;WnmibenceebNoial IYou want\n72 sin (0) le cnaov ; Wnmibenceeb Noial IYou want...\n##### Write a program that reads the integer numbers in a text file (numbers.txt) and stores the...\nWrite a program that reads the integer numbers in a text file (numbers.txt) and stores the numbers to a binary file (binaryNumber.dat). (5 pts) The number.txt file contains the following numbers: 1 2 3 4 5 6 7 8 9 0 7. Write a program that reads the integer numbers in a text file (numbers.txt) and ...\n##### A 400.0-N sign hangs from the end of uniform strut The strut is 4.0 long and weighs 600.0 N. The strut is supported by hinge at the wall and by cable whose other end is tied t0 the wall at point 3.0 m above the left end of the stut: (A) Draw all the forces acting on the strut: Find the tension in the supporting cable: Find the force (magnitude and direction) that the hinge puts on the strul50m#om40in\nA 400.0-N sign hangs from the end of uniform strut The strut is 4.0 long and weighs 600.0 N. The strut is supported by hinge at the wall and by cable whose other end is tied t0 the wall at point 3.0 m above the left end of the stut: (A) Draw all the forces acting on the strut: Find the tension in th...\n##### Uantdaam _ mcticn?NcolercrdinStem EgWSua5 47^iloneEegte caI10 ajonznarticim \|ude058-17\nuantda am _ mcticn?Ncoler crdin Stem Eg WSua 5 47^ ilone Eegte caI 10 ajonz narticim \| ude 058-17..." ]	[ null, "https://cdn.numerade.com/ask_images/f3111260624b4fa78404ac70e0021483.jpg ", null, "https://cdn.numerade.com/previews/704057c1-ed1a-4d55-a8b5-9c271a103a84_large.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.94795567,"math_prob":0.9565998,"size":8055,"snap":"2023-14-2023-23","text_gpt3_token_len":1968,"char_repetition_ratio":0.17749348,"word_repetition_ratio":0.015253203,"special_character_ratio":0.23985103,"punctuation_ratio":0.13114753,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9833604,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-08T12:26:52Z\",\"WARC-Record-ID\":\"<urn:uuid:d890634e-9f8b-46b4-b3a2-5e74c9edc9e8>\",\"Content-Length\":\"124601\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d7faf97e-9661-43c5-9ac9-f32a4c00fd58>\",\"WARC-Concurrent-To\":\"<urn:uuid:2a037790-bc9e-497b-b910-eac0b54804be>\",\"WARC-IP-Address\":\"172.67.132.66\",\"WARC-Target-URI\":\"https://solvedlib.com/n/activity-6-3-4-consider-a-thimbar-of-length-20-cm-whose,19293825\",\"WARC-Payload-Digest\":\"sha1:YKJ7KQVUNY3IQHOU3YOZZLAQZT2XXCQ4\",\"WARC-Block-Digest\":\"sha1:37NLYC2DTF7B7FEZMMAOAZ7AT5ZDMVQF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224654871.97_warc_CC-MAIN-20230608103815-20230608133815-00079.warc.gz\"}"}
http://math.rwinters.com/18012A/calendar.htm	[ "Math 18.01A-18.02A Concourse Calendar of topics and HW assignments -- Fall 2018 and IAP 2019\nlast updated Tuesday, February 5, 2019 9:35 AM\nThe topics and assignments will change as the course proceeds!! Check back frequently.\n\nN Date Topics Assignments & References [Solutions]\n1 Wed, Sept 5 Review of differentiation [Simmons Chapters 1-3]\nLinear and quadratic approximations.\n\nRead: SN-A (Approximation);\nRead text: 2.6 (through pg. 77);\nRead SN-MVT (Mean Value Theorem).\nRead text: 12.2, 12.3 (examples 1-3, remark 1).\n\nProblem Set #1 (due Thurs, Sept 13)\n\n2 Thurs, Sept 6 Higher order (Taylor and Maclaurin) polynomial approximations.\n3 Mon, Sept 10 Extended Mean Value Theorem and error estimation.\n4 Tues, Sept 11 Indeterminate forms, L'Hospital's rule, growth rate of functions.\n5 Wed, Sept 12 Definite integral; summation notation. First Fundamental Theorem of Calculus, Second Fundamental Theorem, ln x as an integral, Leibnitz' Rule.\n\nRead: SN-PI (Properties of Integrals)\nRead: SN-FT (Second Fundamental Theorem).\nRead text: 6.3-6.7; 7.1-7.6\nRead SN-AV (Average Value).\n\nProblem Set #2 (due Fri, Sept 21)\n\n6 Thurs, Sept 13 Problem set questions; functions defined by integrals. Geometric applications: area, signed area, average value of a function; area between curves.\n7 Mon, Sept 17 Integral applications: volumes, volumes of solids of revolution, arclength, surface area.\n8 Tues, Sept 18 Examples of volumes, surface area, arclength using integration.\n9 Wed, Sept 19 Integral applications: work, mass, pumping. Integration techniques: basic integrals, substitution, trig. integrals,\n\nReferences: Text sections 7.7 (work and energy); 10.1-10.6 (Integration: substitution, trig. integrals, completing the square, partial fractions and rational functions.)\n\nProblem Set #3 (due Fri, Sept 28)\n\n10 Thurs, Sept 20 Problem set questions; Integration techniques: basic integrals, substitution, trig. integrals, powers and products of trig. functions.\n11 Mon, Sept 24 Integration techniques: powers and products of trig. functions; dealing with quadratic expressions in integrals (trigonometric substitutions), completing the square; integration of rational functions via partial fractions.\n12 Tues, Sept 25 Integration techniques: partial fractions.\n13 Wed, Sept 26 Integration techniques: partial fractions, integration by parts, reduction formulas.\n\nRead text: 10.7-10.9 (integration by parts, miscellaneous other methods, numerical integration); SN: F; 12.4 (improper integrals); Read SN-INT (Improper Integrals)\n\nProblem Set #4 (due Fri, Oct 5)\n\n14 Thurs, Sept 27 Numerical integration; improper integrals (infinite domain, discontinuous integrands), comparison test, limit comparison test, p-test for integrals. Problem set questions.\n15 Mon, Oct 1 Numerical integration; improper integrals (infinite domain, discontinuous integrands), comparison test, limit comparison test, p-test for integrals.\n\nExam 1 Practice Questions Solutions\n(use same username/password as solutions)\n\n16 Tues, Oct 2\n\nExam 1 - Topics covered may include: Solutions\n\n1) Linear, quadratic, Taylor (nth order) approximations and series: a) by formula; b) by manipulating known series;\n\n2) L'Hôpital's Rule and indeterminate forms (limits);\n\n3) Definite integrals in the calculation of area, volume, arclength, average value, work;\n\n4) First and Second Fundamental Theorems of Calculus, and Leibnitz' Rule\n\n5) Integration techniques: a) substitution; b) trig. substitutions; c) powers and products of trig. functions; partial fractions; integration by parts; reduction formulas.\n\n17 Wed, Oct 3 Sequences of real numbers, convergence of a sequence. Infinite series, geometric series\n\nRead relevant portions of text: 13.1 – 13.8 (sequences and series).\n\nSupplement on Integral Test and Comparison Tests\n\nProblem Set #5 (due Mon, Oct 15)\n\n18 Thurs, Oct 4 Problem Set questions\n19 Wed, Oct 10 Harmonic series, convergence tests (divergence test, p-test, comparison test, integral test, limit comparison test, ratio test); Alternating series, absolute vs. conditional convergence.\n20 Thurs, Oct 11\n21 Mon, Oct 15 Ratio Test for absolute convergence. Power series, radius and interval of convergence.\n\nRead relevant portions of text: 14.1-14.5\n\nProblem Set #6 (due Monday, Oct 22)\n\n22 Tues, Oct 16 Taylor series revisited.\n23 Wed, Oct 17\n\nError estimation using Taylor's Theorem with Remainder.\n\nPractice Questions for 18.01A Final Exam (Exam #2)\n\nSolutions to 18.01A Exam #2 Practice Questions\n\n24 Thurs, Oct 18\n\nTaylor's Theorem with Remainder, continued.\n\n25 Mon, Oct 22\n\nCoordinates, vectors (start of 18.02A); coordinate-free vector proofs, parameterized lines.\n\nLecture Notes #1\n\nRead text: 17.3, 18.1, 18.2, 18.3, 18.4.\nRW-Vectors (Supplement on Vectors, dot product, projections, cross product)\n\nProblem Set #7 (due Wednesday, Oct 31)\n\n26 Tues, Oct 23 18.01A Final Exam, a 2-hour exam\n27 Wed, Oct 24\n\nCoordinates, vectors; coordinate-free vector proofs, parameterized lines; dot product.\n[Lecture Notes on Coordinates & Vectors]\n\n28 Thurs, Oct 25 Dot product, scalar and vector projection, equation of a plane; introduction to cross-product.\n[Lecture Notes on Dot Product, Cross Product, Planes, Area, and Volumes]\n\nLecture Notes #2\n\nRead text: 17.1, 17.2, 17.4, 17.5; 18.02 Supplementary Notes: SN-D (Determinants), SN-M (Matrices and Linear Algebra)\n\nRW-Vectors (Supplement on Vectors, dot product, projections, cross product)\n\nSupplement on Solving Systems of Linear Equations via Row Reduction\n\nSupplement: Vector and matrix forms of a system of linear equations, matrix algebra, inverse matrices, and related facts\n\nProblem Set #8 (due Thurs, Nov 8)\n\n29 Mon, Oct 29\n\nCross product; lines and planes; determinants, distance, area, volume.\n\n30 Tues, Oct 30\n31 Wed, Oct 31\n\nMatrix methods, row reduction, inverse matrices.\n[Lecture Notes on vector and matrix forms of a system of linear equations, matrix algebra, inverse matrices, and related facts]\n\n32 Thurs, Nov 1\n33 Mon, Nov 5\n\nParametric equations and vector derivatives; velocity vector, acceleration vector, speed, unit tangent vector, curvature.\n[Lecture Notes on Parameterized Curves and Vector-Valued Functions]\n\nLecture Notes #3\n\nRead text: 17.1-17.2, 17.4-17.6; RW-ParamCurves (Supplement on parametrized curves and vector-valued functions); SN-K (Kepler's 2nd Law)\n\nProblem Set #9 (due Thurs, Nov 15)\n\nExam 3 Practice Questions Solutions\n\n34 Tues, Nov 6\n\nParametric equations, continued.\nKepler's second law (read the supplement, not covered in class)\n\n35 Wed, Nov 7\n\nGraphs of surfaces, cross-sections. Functions of several variables, graph vs. level sets (contours). [Lecture Notes on Paramaterized Surfaces, Tangent Vectors, Partial Derivatives]\n\nLecture Notes #4\n\n36 Thurs, Nov 8\n\nFunctions of several variables, level curves and surfaces. Limits and continuity, differentiability, and partial derivatives. Tangent plane, linear approximation.\n\n37 Mon, Nov 12\n(to be rescheduled)\n\nTangent plane, linear approximation, increments and differentials, rate of change of a function along a parameterized curve: basic Chain Rule. Gradient and directional derivative. [Notes on limits, continuity, differentiability and linear approximation]\n\nLecture Notes #5\n\nRead 19.1, 19.2 (Partial Derivatives)\nSN-TA (Tangent Approximation)\n19.6 (Chain Rule)\n\nNotes on Gradients, Chain Rule, Implicit Differentiation, and Higher Order Derivatives\n\nProblem Set #10 (due Thurs, Nov 29)\n\n38 Tues, Nov 13 Chain Rule, gradient vector, directional derivatives. Find normal vectors using the gradient vector.\n39 Wed, Nov 14\n\nImplicit differentiation.\n[Lecture Notes on Differentials, the Chain Rule, Gradients, Directional Derivative, and Normal Vectors]\n\nLecture Notes #6\n\n40 Thurs, Nov 15 Exam 3\nTopics include: Coordinates, vectors, vector operations and tools (adding, scaling, dot product, cross product), lines, planes, intersections, distance, area, volume; matrix methods for solving systems of linear equations, row reduction, parameterization, matrix inverse; parameterized curves, velocity, acceleration, unit tangent vector, unit normal vector, curvature, differentiation of vector-valued functions.\n41 Mon, Nov 19\n\nGeneral Chain Rule, implicit differentiation, higher order derivatives, equality of mixed partials (Clairaut's Theorem).\n\n42 Tues, Nov 20\n\nQuadratic approximation; critical points, second derivative test; unconstrained optimization.\n\nLecture Notes #7\n\n43 Mon, Nov 26\n\nLecture Notes on Extrema of Functions of Several Variables\n\nRead:\n19.7 (Maximum and Minimum Problems)\n19.8 (Constrained Maxima and Minima, Lagrange Multipliers)\n19.10 (Implicit Functions)\nRW-Chain Rule and Implicit Differentiation\nSN-LS (Least Squares Interpolation)\nSN-N (Non-independent Variables, sections 1-3)\n\nProblem Set #11 (due Thurs, Dec 6)\n\n44 Tues, Nov 27\n\nQuestions\n\n45 Wed, Nov 28\n\nMax-min problems, Method of Least Squares. Constrained optimization, Method of Lagrange Multipliers.\n\nLecture Notes #8\n\n46 Thurs, Nov 29\n\nMethod of Lagrange Multipliers, continued - economics applications.\n\n47 Mon, Dec 3\n\nConstrained optimization w/multiple constraints; Non-independent variables.\n\nPractice Exam #4 Questions Solutions\n48 Tues, Dec 4\n\nDouble integrals and their applications - volume, mass, area, averaging, centroid. Evaluation of double integrals via iterated single integrals, the Fubini Theorem.\n\nLecture Notes #9\n\nRead 20.1 (Volumes)\n20.2 (Double Integrals and Iterated Integrals)\nSN-I (Limits in Iterated Integrals)\n20.3 (Applications of Double Integrals)\n16.1 (Polar Coordinates)\n16.2 (Graphs of Polar Equations)\n20.4 (Double Integrals in Polar Coordinates)\nSN-CV (Change of Variables in Double Integrals)\n\nProblem Set #12 (due 1st week IAP)\n\n49 Wed, Dec 5\n\nPolar coordinates, double integrals in polar coordinates; additional applications. Weighted averages, center of mass; change of variables in double integrals.\n\n50 Thurs, Dec 6 Exam 4: Topics may include partial derivatives, linear approximation, differentials, gradient vector, normal vectors to curves and surfaces, directional derivative, the Chain Rule, implicit differentiation, stationary points, second derivative test, optimization, Method of Lagrange Multipliers, constrained optimization with one or more constraints, partial derivatives in the case of non-independent (constrained) variables.\n51 Mon, Dec 10 Polar coordinates, double integrals in polar coordinates; additional applications. Weighted averages, center of mass; change of variables in double integrals.\n\nLecture #10 Notes\n\n52 Tues, Dec 11 Examples and techniques for change of variables; problem set questions\n53 Wed, Dec 12\n\nWeighted averages, center of mass; change of variables in double integrals.\n\nLecture #11 Notes (first part re: change of variables in double integrals)\n\nHoliday Break\n*\n\nIAP Schedule (starting Monday, January 7):\nLecture/Recitation: Mon, Wed 12:00pm to 1:00pm; Tues, Thurs, 1:00pm to 2:30pm in 16-160.\nAdditional recitation time informally afternoons in Concourse Lounge or the classroom.\n\nIAP Schedule (Mon, Jan 14 onward):\nLecture: Mon, Wed 12:00pm to 1:30pm; Tues 1:00pm to 3:00pm; Recitation: Thurs 1:00pm to 2:30pm.\nAdditional recitation time informally afternoons in Concourse Lounge or the classroom.\n\nPset 13: Vector fields, integration along curves, line integrals, work, conservative vector fields, and work.\n\nNotes V1 (Plane Vector Fields)\nNotes V8 (Vector Fields in Space)\nNotes V11 (Line Integrals in Space)\nNotes V2 (Gradient Fields and Exact Differentials)\nNotes V12 (Gradient Fields in Space)\n\nSimmons text, sections 21.1 and 21.2\nand posted Lecture Notes.\n\nProblem Set #13 (due Fri, Jan 11)\n\n54 Mon, Jan 7\n\nVector fields; integration along curves in R2 and R3. Line integrals and work.\n\nLecture #11 Notes\n\nHere's a website that has a good java-based tool for showing vector fields and flows: http://math.rice.edu/~dfield/dfpp.html. The current version requires you to download a Java executable file to your own computer and to run it locally on your own machine. You can customize various options. You can also print the graphs. [How Java is called varies on computer and operating system, so we may have to provide some additional documentation.]\n55 Tues, Jan 8 Gradient and conservative fields.\n56 Wed, Jan 9\n\nPotential functions; test for exactness; algebraic definitions of curl and divergence.\n\nPset 14: Green's Theorem (ordinary and normal form), its corollaries and applications; algebraic definitions of divergence and curl of a vector field.\n\nSimmons text, section 21.3\nNotes V3 (Two-dimensional Flux)\nNotes V4 (Green's Theorem in Normal Form)\n\nProblem Set #14 (due Wed, Jan 16)\n\n57 Thurs, Jan 10 Green's Theorem; corollaries and applications of Green's Theorem.\n\n2D-Flux: normal form of Green's Theorem.\n\nLecture #12 Notes\n\n58 Mon, Jan 14 Review of double integrals; introduction to triple integrals and applications.\nTriple integrals in rectangular, cylindrical, and spherical coordinates.\n\nLecture #10 Notes\n\nPset 15: Integration of functions over regions in space using Cartesian, cylindrical, and spherical coordinates; applications in physics.\n\nSimmons text, sections 20.5, 20.6, and 20.7\nNotes CV (Changing Variables in Multiple Integrals)\nNotes G (Gravitational Attraction)\n\nProblem Set #15 (due Tues, Jan 22)\n\n59 Tues, Jan 15 Triple integrals (again) in rectangular, cylindrical, and spherical coordinates.\n\nExam #5 Topics & Practice Questions (18.02A - IAP) Solutions\n\n60 Wed, Jan 16\n\nTriple integrals applications - moment of intertia; gravitational attraction.\nApplications, continued. Change of variable in triple integrals and Jacobian determinants.\n\nLecture #11 Notes\n\n61 Thurs, Jan 17 Exam 5 covering vector fields, curl and divergence of a vector field; integration along curves, line integrals, Green's Theorem; triple integrals in Cartesian, cylindrical, and spherical coordinates; applications of integration (area, volume, mass, average value, weighted averages, centroids, center of mass, moment of inertia, etc.).\n\nPset 16: Parameterized surfaces, surface area, surface integrals and their applications, flux of a vector field through a surface, Divergence Theorem.\n\nNotes V7 (Laplace's Equation and Harmonic Functions)\n15.5 (Surface Integrals)\nNotes V9 (Surface Integrals)\nRW-Supplement on surface integrals\nNotes V10 (The Divergence Theorem)\n\nProblem Set #16 (due Tues, Jan 29)\n\n62 Tues, Jan 22 Parameterization of surfaces; surface integrals; flux of a vector field through a surface.\nSupplement on surface integrals\n\nLecture #13 Notes\n\n63 Wed, Jan 23 Surface integrals, continued; Divergence Theorem.\n64 Thurs, Jan 24 Applications and interpretations of the Divergence Theorem. Stokes' Theorem.\n\nLecture #14 Notes\n\nPset 17: Divergence Theorem, Stokes' Theorem, applications to physics.\n\nNotes V10 (The Divergence Theorem)\nNotes V13 (Stokes' Theorem)\nNotes V14 (Some Topological Questions)\nNotes V15 (Relations to Physics)\nRW-Orthogonal Curvilinear Coordinates - Div, grad, curl, and Laplacian\n\nProblem Set #17 (due Thurs, Jan 31)\n\n65 Mon, Jan 28 More Stokes' Theorem.\n66 Tues, Jan 29 Applications and interpretations of Stokes' Theorem.\n67 Wed, Jan 30 Div, grad, curl, and Laplacian in curvilinear coordinates\n68 Thurs, Jan 31 Generalized Stokes' Theorem in terms of forms; Final Exam review.\n\n Practice exams from mainstream 18.02: Practice Final Exam A Solutions to Practice Final Exam A Practice Final Exam B Solutions to Practice Final Exam B\n\n(Some topics on these practice exams might not appear on Friday's exam, but they are a reasonable approximation.)\n\nFri, Feb 1 Final Exam - Friday, February 1 in 16-160 from 9:00am to 12:00pm\n\nReturn to main Math 18.01A-02A Concourse page" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.785847,"math_prob":0.8485129,"size":12973,"snap":"2019-13-2019-22","text_gpt3_token_len":3313,"char_repetition_ratio":0.14604056,"word_repetition_ratio":0.04703367,"special_character_ratio":0.2606182,"punctuation_ratio":0.22181222,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.982983,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-21T23:46:34Z\",\"WARC-Record-ID\":\"<urn:uuid:af315ab9-f332-4dab-a233-3395c4bff411>\",\"Content-Length\":\"36601\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5f598958-daf8-4b73-ba44-011c1ebfc5b2>\",\"WARC-Concurrent-To\":\"<urn:uuid:4923d596-4486-4dcc-b7a2-360716331c68>\",\"WARC-IP-Address\":\"143.95.79.223\",\"WARC-Target-URI\":\"http://math.rwinters.com/18012A/calendar.htm\",\"WARC-Payload-Digest\":\"sha1:3JTCA3PBOAPAHYTZLA7QMVRRLFUWSQYC\",\"WARC-Block-Digest\":\"sha1:T4F6LY5XTZYZEQ2NC6BYSN2Y3H3XKMVN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232256586.62_warc_CC-MAIN-20190521222812-20190522004812-00462.warc.gz\"}"}
https://artofproblemsolving.com/wiki/index.php/1989_USAMO_Problems/Problem_5	[ "# 1989 USAMO Problems/Problem 5\n\n## Problem\n\nLet", null, "$u$ and", null, "$v$ be real numbers such that", null, "$$(u + u^2 + u^3 + \\cdots + u^8) + 10u^9 = (v + v^2 + v^3 + \\cdots + v^{10}) + 10v^{11} = 8.$$ Determine, with proof, which of the two numbers,", null, "$u$ or", null, "$v$, is larger.\n\n## Solution\n\nThe answer is", null, "$v$.\n\nWe define real functions", null, "$U$ and", null, "$V$ as follows:", null, "\\begin{align} U(x) &= (x+x^2 + \\dotsb + x^8) + 10x^9 = \\frac{x^{10}-x}{x-1} + 9x^9 \\\\ V(x) &= (x+x^2 + \\dotsb + x^{10}) + 10x^{11} = \\frac{x^{12}-x}{x-1} + 9x^{11} . \\end{align} We wish to show that if", null, "$U(u)=V(v)=8$, then", null, "$u .\n\nWe first note that when", null, "$x \\le 0$,", null, "$x^{12}-x \\ge 0$,", null, "$x-1 < 0$, and", null, "$9x^9 \\le 0$, so", null, "$$U(x) = \\frac{x^{10}-x}{x-1} + 9x^9 \\le 0 < 8 .$$ Similarly,", null, "$V(x) \\le 0 < 8$.\n\nWe also note that if", null, "$x \\ge 9/10$, then", null, "\\begin{align} U(x) &= \\frac{x-x^{10}}{1-x} + 9x^9 \\ge \\frac{9/10 - 9^9/10^9}{1/10} + 9 \\cdot \\frac{9^{9}}{10^9} \\\\ &= 9 - 10 \\cdot \\frac{9^9}{10^9} + 9 \\cdot \\frac{9^9}{10^9} = 9 - \\frac{9^9}{10^9} > 8. \\end{align} Similarly", null, "$V(x) > 8$. It then follows that", null, "$u, v \\in (0,9/10)$.\n\nNow, for all", null, "$x \\in (0,9/10)$,", null, "\\begin{align} V(x) &= U(x) + V(x)-U(x) = U(x) + 10x^{11}+x^{10} -9x^9 \\\\ &= U(x) + x^9 (10x -9) (x+1) < U(x) . \\end{align} Since", null, "$V$ and", null, "$U$ are both strictly increasing functions over the nonnegative reals, it then follows that", null, "$$V(u) < U(u) = 8 = V(v),$$ so", null, "$u, as desired.", null, "$\\blacksquare$\n\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.", null, "" ]	[ null, "https://latex.artofproblemsolving.com/9/a/d/9ad99798ec4c38e165cf517cb9e02b1c9e824103.png ", null, "https://latex.artofproblemsolving.com/a/9/f/a9f23bf124b6b2b2a993eb313c72e678664ac74a.png ", null, "https://latex.artofproblemsolving.com/7/4/7/747d227c5c93ab38b45ffe98794d68ac188986dd.png ", null, "https://latex.artofproblemsolving.com/9/a/d/9ad99798ec4c38e165cf517cb9e02b1c9e824103.png ", null, "https://latex.artofproblemsolving.com/a/9/f/a9f23bf124b6b2b2a993eb313c72e678664ac74a.png ", null, "https://latex.artofproblemsolving.com/a/9/f/a9f23bf124b6b2b2a993eb313c72e678664ac74a.png ", null, "https://latex.artofproblemsolving.com/e/2/b/e2bbebb3bd73f1ae5c64098ab0244f739abf7ca4.png ", null, "https://latex.artofproblemsolving.com/1/2/d/12d58aa29201da09d8e620f8698e3a37547f6b4a.png ", null, "https://latex.artofproblemsolving.com/1/f/5/1f5ec8e2b4de4d798c17a6493147af0a3b272f47.png ", null, "https://latex.artofproblemsolving.com/7/0/7/7074ae0a95013a4f6045ac1b928149f0110af38e.png ", null, "https://latex.artofproblemsolving.com/e/d/7/ed7ffe0563c6df8b378ba45bf11c11e0c3f79c42.png ", null, "https://latex.artofproblemsolving.com/1/b/6/1b667f99b88cc4dd636e45b0cdf1bf965992738a.png ", null, "https://latex.artofproblemsolving.com/c/c/e/cce24baa4db0ce59b69d97c7576c0120f35ad1af.png ", null, "https://latex.artofproblemsolving.com/1/4/d/14dd5bcd5d074df8011693f56cd0faa1362fe9bd.png ", null, "https://latex.artofproblemsolving.com/4/3/0/43018098f9a03499b21d8adcc101ba68e7dbf3bc.png ", null, "https://latex.artofproblemsolving.com/a/1/b/a1b8d3aedf346b5aa699546e25156e20d990103d.png ", null, "https://latex.artofproblemsolving.com/6/4/4/6445c800f83d07dc795bce8f22b5a0970f2b8829.png ", null, "https://latex.artofproblemsolving.com/c/d/e/cdef99f026ca46b44a997cf5f8d85076428f2118.png ", null, "https://latex.artofproblemsolving.com/7/b/8/7b801995147880225d6c060cc681c033d06a980b.png ", null, "https://latex.artofproblemsolving.com/a/f/1/af1fdd1c9c151d28f320d685ba8e804116434177.png ", null, "https://latex.artofproblemsolving.com/e/3/7/e375fcb7506e388abd2d43bbe2e5720165362463.png ", null, "https://latex.artofproblemsolving.com/f/2/8/f2814c6f6ef5bcdca44e299fbe3286f96eb48a6a.png ", null, "https://latex.artofproblemsolving.com/7/b/0/7b0965a08065dd3d4b7c4e1a8589d1da595cd56c.png ", null, "https://latex.artofproblemsolving.com/1/2/d/12d58aa29201da09d8e620f8698e3a37547f6b4a.png ", null, "https://latex.artofproblemsolving.com/e/2/b/e2bbebb3bd73f1ae5c64098ab0244f739abf7ca4.png ", null, "https://latex.artofproblemsolving.com/d/d/e/ddeb36defeb65d2a05aa127bdc978a2ec1a4ef25.png ", null, "https://latex.artofproblemsolving.com/9/a/c/9ac520c83016d5022c055f0fe3089af59adba0cd.png ", null, "https://latex.artofproblemsolving.com/3/b/b/3bbb9e25e5be095e317771a9c5d00877ee9c8012.png ", null, "https://wiki-images.artofproblemsolving.com//8/8b/AMC_logo.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8400274,"math_prob":1.0000091,"size":749,"snap":"2023-40-2023-50","text_gpt3_token_len":191,"char_repetition_ratio":0.12885906,"word_repetition_ratio":0.0,"special_character_ratio":0.2563418,"punctuation_ratio":0.16438356,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000093,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58],"im_url_duplicate_count":[null,null,null,null,null,10,null,null,null,null,null,null,null,null,null,null,null,10,null,10,null,10,null,null,null,10,null,null,null,10,null,10,null,10,null,10,null,10,null,10,null,10,null,10,null,10,null,null,null,null,null,10,null,10,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-22T22:01:48Z\",\"WARC-Record-ID\":\"<urn:uuid:143c57ee-d112-43db-8f5e-4d324f843da0>\",\"Content-Length\":\"39676\",\"Content-Type\":\"application/http; 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https://www.neetprep.com/question/69463-following-data-ofH-following-reactionsCsOCOgHkJCsHOCOgHgHkJWhat-mole-composition-mixture-steam-oxygen-passedover-coke--K-keeping-temperature-constant------------/54-Chemistry--Thermodynamics/650-Thermodynamics	[ "From the following data of $∆H$, of the following reactions,\n\nWhat is the mole composition of the mixture of steam and oxygen on being passed over coke at 1273 K, keeping temperature constant.\n\n1. 0.5 : 1\n\n2. 0.6 : 1\n\n3. 0.8 : 1\n\n4. 1 : 1\n\nConcept Questions :-\n\nEnthalpy and It's Type\nHigh Yielding Test Series + Question Bank - NEET 2020\n\nDifficulty Level:\n\nState which of the following statements is true ?\n\n1. First law of thermodynamics is not adequate in predicting the direction of the process.\n\n2. In an exothermic reaction, the total enthalpy of the products is greater than that of reactants.\n\n3. The standard enthalpy of diamond is zero at 298 K and 1 atm pressure.\n\n4. It is possible to calculate the value of $∆H$ for the reaction ${H}_{2}\\left(g\\right)+B{r}_{2}\\left(l\\right)\\to 2HBr\\left(g\\right)$ from the bond enthalpy data.\n\nConcept Questions :-\n\nFirst Law of Thermodynamics\nHigh Yielding Test Series + Question Bank - NEET 2020\n\nDifficulty Level:\n\nIf $∆H$ of a reaction is 100 kJ mol-1, then the activation energy for forward reaction must be-\n\n1. Greater than 100 kJ mol-1\n\n2. Less than 100 kJ mol-1\n\n3. Equal to 100 kJ mol-1\n\n4. None is correct\n\nConcept Questions :-\n\nEnthalpy and It's Type\nHigh Yielding Test Series + Question Bank - NEET 2020\n\nDifficulty Level:\n\nThe intermediate SiH2 is formed in the thermal decomposition of silicon hydrides. Calculate $∆{H}_{f}^{\\circ }$ of SiH2 given the following reactions\n\n1. 353 kJ/mol\n\n2. 321 kJ/mol\n\n3. 198 kJ/mol\n\n4. 274 kJ/mol\n\nConcept Questions :-\n\nThermochemistry\nHigh Yielding Test Series + Question Bank - NEET 2020\n\nDifficulty Level:\n\nA certain vessel X has water and nitrogen gas at a total pressure of 2 atm. and 300 K. All the contents of the vessel are transferred to another vessel Y having half the capacity of the vessel X. The pressiure of N2 in this vessel was 3.8 atm. at 300 K. The vessel Y is heated to 320 K and the total pressure observed was 4.32 atm. Calculate the enthalpy of vapourisation of water assuming it to be independent of temperature. Also assume the volume occupied by the gases in a vessel is equal to the volume of the vessel.\n\n1. 39.637 kJ mol-1\n\n2. 19.531 kJ mol-1\n\n3. 396.37 kJ mol-1\n\n4. 3.9127 kJ mol-1\n\nConcept Questions :-\n\nEnthalpy and It's Type\nHigh Yielding Test Series + Question Bank - NEET 2020\n\nDifficulty Level:\n\nFor a reaction, is given by the equation in the temperature range 300-600 K. The enthalpy of the reaction at 300 K -s -25.0 KJ. Calculate the enthalpy of the reaction at 450 K.\n\nAlso\n\n1. -10.12 kJ\n\n2. -28.32 kJ\n\n3. -18.72 kJ\n\n4. -8.21 kJ\n\nConcept Questions :-\n\nEnthalpy and It's Type\nHigh Yielding Test Series + Question Bank - NEET 2020\n\nDifficulty Level:\n\nThe heat of combustion of ethylene at ${17}^{\\circ }C$ and at constant volume is -332.19 kcals. What is the value at constant pressure, given that water is in liquid state ?\n\n1. -131.25 k cals\n\n2. -412.23 k cals\n\n3. -534.12 k cals\n\n4. -333.35 k cals\n\nConcept Questions :-\n\nEnthalpy and It's Type\nHigh Yielding Test Series + Question Bank - NEET 2020\n\nDifficulty Level:\n\nThe enthalpies of the following reactions are shown alongwith.\n\nCalculate the O-H bond energies for the hydroxyl radical.\n\n1. 223.18 kJ mol-1\n\n2. 423.38 kJ mol-1\n\n3. 513.28 kJ mol-1\n\n4. 113.38 kJ mol-1\n\nConcept Questions :-\n\nThermochemistry\nHigh Yielding Test Series + Question Bank - NEET 2020\n\nDifficulty Level:\n\nThe bond dissociation enthalpy of gaseous H2, Cl2 and HCl are 435, 243 and 431 kJ mol-1, respectively. Calculate the enthalpy of formation of HCl gas.\n\n1. -92 kJ mol-1\n\n2. -82 kJ mol-1\n\n3. -21 kJ mol-1\n\n4. -55 kJ mol-1\n\nConcept Questions :-\n\nEnthalpy and It's Type\nHigh Yielding Test Series + Question Bank - NEET 2020\n\nDifficulty Level:\n\nThe standard enthalpy of combustion at ${25}^{\\circ }C$ of hydrogen, cyclohexene (C6H10) and cyclohexane (C6H12) are -241, -3800 and -3920 kJ mol-1, respectively. Calculate the standard enthalpy of hydrogenation of cyclohexene.\n\n1. -131 kJ mol-1\n\n2. -155 kJ mol-1\n\n3. -167 kJ mol-1\n\n4. -121 kJ mol-1\n\nConcept Questions :-\n\nThermochemistry" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8003565,"math_prob":0.96518546,"size":2961,"snap":"2020-24-2020-29","text_gpt3_token_len":936,"char_repetition_ratio":0.1795739,"word_repetition_ratio":0.011090573,"special_character_ratio":0.33198243,"punctuation_ratio":0.16595745,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99273187,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-25T07:01:05Z\",\"WARC-Record-ID\":\"<urn:uuid:c600bdf2-a75a-475e-9069-863bafb7790c>\",\"Content-Length\":\"457127\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b5491998-e57f-44a6-af53-dd61232a9654>\",\"WARC-Concurrent-To\":\"<urn:uuid:9ed038b4-a817-486f-b161-1639973562d0>\",\"WARC-IP-Address\":\"13.235.211.187\",\"WARC-Target-URI\":\"https://www.neetprep.com/question/69463-following-data-ofH-following-reactionsCsOCOgHkJCsHOCOgHgHkJWhat-mole-composition-mixture-steam-oxygen-passedover-coke--K-keeping-temperature-constant------------/54-Chemistry--Thermodynamics/650-Thermodynamics\",\"WARC-Payload-Digest\":\"sha1:APSOT6OZYAHITD3WJEOB2LARM5Q3HBOC\",\"WARC-Block-Digest\":\"sha1:IXC3USTEGRIOBUEQ6FQQRZ6H63432DOB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347388012.14_warc_CC-MAIN-20200525063708-20200525093708-00475.warc.gz\"}"}
http://dannevang.net/wp-content/uploads/2020/boom-boom-mxpkejf/page.php?110780=6th-grade-math-word-problems	[ "", null, "", null, "## 6th grade math word problems\n\nR.15 Ratios and rates: word problems. 0 Time elapsed Time. share to google . Some of the worksheets for this concept are Word problem practice workbook, Decimals work, Multiplying decimals word problems, Fraction and decimal word problems no problem, Grade 5 decimals word problems, 8 fractions, 501 math word problems, Exercise work. Our team Don't Pass on to third parties any identifiable information about mathskills4kids.com users. 6th Grade Math Common Core Math Task Cards include SKILLS PRACTICE and real-world WORD PROBLEMSWhat is Included:This resource includes 9 sets (360 task cards) of 6th Grade Common Core skills practice and real-world word problems, student answer sheets, and answer keys. These sixth grade math worksheets cover most of the core math topics previous grades, including conversion worksheets, measurement worksheets, mean, median and range worksheets, number patterns, exponents and a variety of topics expressed as word problems. We feature a series of word problems from beginner to more advanced. Full curriculum of exercises and videos. 6th grade English Learner ... Thoughtfully paced outlines help when teaching children how to solve math word problems from simple equations to multi-step word problems requiring conversions. The word problems are solved with the help of tape diagrams, block diagrams or bar model (Singapore Math) Related Topics: More Math Word Problems Algebra Word Problems More Singapore Math Word Problems Example: These sixth grade math worksheets cover most of the core math topics previous grades, including conversion worksheets, measurement worksheets, mean, median and range worksheets, number patterns, exponents and a variety of topics expressed as word problems. Sixth Grade Math Word Problems Worksheets - Multi-Step Math Word Problems Worksheet #6 Teacher Resources made by Other Teachers: Differentiated Multi-Step Math Word Problems 5th Grade … PRINTABLE MATH WORKSHEETS FOR GRADE 6. Before working through the worksheets, discuss with your children any phrases or vocabulary that they may be unsure of. For example, the fraction word problems 6th grade students often have to tackle will challenge even the most competent adult. GRADE 6 MATH WORD PROBLEMS WITH ANSWERS. If your kids can vividly estimate reasonably, then there’ll be no doubt that their accuracy in math will increase, thus math experts. This totals over 900 pages and includes over 250 resource That is the reason that these worksheets are generally published out and distributed in … Free worksheets for ratio word problems Find here an unlimited supply of worksheets with simple word problems involving ratios, meant for 6th-8th grade math. Enjoy these math word problem packets. Math Word Problem Worksheets Read, explore, and solve over 1000 math word problems based on addition, subtraction, multiplication, division, fraction, decimal, ratio and more. Estimate recipes when cooking, baking, etc. If your child has entered the world of fraction word problems and is looking for a challenge, try to stump him with these problems… share to google . Solving math problems can intimidate sixth-graders but it shouldn't. ... Word problems on sum of the angles of a triangle is 180 degree. 6th Grade Math Skill Practice. and \"You have 15 yards of ribbon for your gift boxes. For the second answer, the solution is simpler: First, convert 20% to a decimal to get 0.20. So each box gets 0.75 yards of ribbon. (b) What is the ratio of the number of blue pencils to the total number of pencils? Some of the worksheets for this concept are Word problem practice workbook, Decimals work, Multiplying decimals word problems, Fraction and decimal word problems no problem, Grade 5 decimals word problems, 8 fractions, 501 math word problems, Exercise work. Worksheets > Math > Grade 5 > Word problems. Grade 6 Percents Word Problems Name: _____ Class: _____ Question 1 Father gave me some money. Printable Math Worksheets for Grade 6. Estimate the number of people you’ll invite for your coming event, depending on the budget available. Word Problem Worksheets. These word problems help children hone their reading and analytical skills; understand the real-life application of math operations and other math topics. Worksheet - 2. grade 6 math word problem worksheets with answers, Solving multi step word problems exercises with answers, Word problems with extra or missing information, Distance word problems with different directions. Learn sixth grade math for free—ratios, exponents, long division, negative numbers, geometry, statistics, and more. Free worksheets for ratio word problems Find here an unlimited supply of worksheets with simple word problems involving ratios, meant for 6th-8th grade math. The worksheets on this page combine the skills necessary to solve all four types of problems covered previously (addition word problems, subtraction word problems, multiplication word problems and division word problems) and they require students to determine which operation is appropriate for solving the each problem. Grade 6 Math Word Problems with Answers. Whether at home, in the market, on the street or among friends, our activities will always be surrounded around estimation. Great stuff and not boring! Estimate and know how to manage or spend your precious time. Share skill. Let's pick 6 as a possible greatest common multiple because it's the largest factor of 12 other than 12. Word problems are an important part of math practice at nearly every grade level. How much ribbon will each of your 20 gift boxes get? Worksheet - 4. Math problems for sixth-grade students should be read to them. I could choose between 15% of 1,500\\$ or 25% of 1,000\\$. 1 Solutions. Improve your math knowledge with free questions in \"Multi-step word problems\" and thousands of other math skills. On worksheet No. Some of the worksheets for this concept are Word problem practice workbook, Sample work from, Decimals practice booklet table of contents, Grade 6 math word problems with percents, Grade 3 mixed math problems and word problems work, Fractionwordproblems, Percent word problems, 501 math word problems. This zip file includes every resource I have created for sixth grade math in five complete year bundles: 70 assessments, 58 stick-n-solve foldables + 68 vocab diagrams, 67 math stations, 13 choice boards, and 6 bingo games for the whole year. In a significant way, our super amazing problem solving and estimation worksheets will help your young math learners to quickly understand the relevance of estimation skills in math concepts and real life. Similar: Proportions word problems - using decimals Ratio word problems You would set up the problem as follows: 0.20 x 50 marbles = 10 blue marbles. Six can go into 36 six times (6 x 6), 12 can go into 36 three times (12 x 3), and 18 can go into 36 twice (18 x 2), but 24 cannot. Middle school brings challenging math concepts. 6th Grade Math Skill Practice. Worksheet - 1. ... Word problems on sum of the angles of a triangle is 180 degree. ... Sixth grade . You're suppose to protect downloaded contents and take it for personal or classroom use.Special rule : Teachers can use our contents to teach in class. Students apply their knowledge of GCF and LCM to solve word problems. But before we begin solving these word problems, we need to; Finally, check the reasonableness of your answer by comparing it with the one you estimated above. Your email address and other information will NEVER be given or sold to a third party. In this lesson, we will learn how to solve percent word problems • using block diagrams or bar models or tape diagrams \", Other problems require only a basic knowledge of percentages, as well as how to convert percentages to decimals, such as: \"Jasmine has 50 marbles in a bag. Some of the worksheets displayed are 501 math word problems, Addition word problems, Proportions word problems, Math mammoth south african version grade 6 sample, Word problems involving discount, All decimal operations with word problems, Word problem practice workbook, Cc 1st grade word problems. What then are those peculiar strategies to consider when faced with situations of problem solving and estimation? Peter had 12 pencils. Using a few simple formulas and a bit of logic can help students quickly calculate answers to seemingly intractable problems. The worksheets provide a great and simple way to do quick formative assessments for an entire math class. https://www.thoughtco.com/6th-grade-math-word-problems-2312642 ZB9. It should be noted that, not only are these problem solving and estimation skills a key part in math concepts, but are equally an important approach to boast your kid’s mental math skills, logical and creative thinking abilities. What’s the other number I’m thinking of? The multiples of 6 are 6, 12, 18, 24, 30, and 36. I could choose between 15% of 1,500\\$ or 25% of 1,000\\$. Please click the following links to get math printable math worksheets for grade 6. The following are some examples of 6th Grade Math Word Problems that deals with ratio and proportions. Carefully read the entire problem, twice, in order to better understand its key words. How many meters will it travel in 12 minutes. Learn sixth grade math for free—ratios, exponents, long division, negative numbers, geometry, statistics, and more. On this PDF, your students will solve problems such as: \"Your brother traveled 117 miles in 2.25 hours to come home for school break. Integer Worksheets Unit Rate Word Problem Worksheet 1 (Integers) - This 13 problem worksheet features word problems where you will calculate the unit rate for everyday situations like … Full curriculum of exercises and videos. ThoughtCo uses cookies to provide you with a great user experience. ... Word problems : GCF and LCM 6.NS.B.4 - Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. These worksheets provide students with real world word problems that students can solve with grade 5 math concepts. 7 of them were blue and the rest were red. (You stop at 36 because the problem says that this number is the least common multiple.) Printable 6th Grade Math Word Problem Worksheets – The 6th grade Math worksheet is among the most demanded worksheets among secondary school trainees since it helps them learn to be issue solvers. 6th Grade > CCSS 6th Grade Math > Word problems : ratios. You will now find grade leveled problems in sets and skill based word problems as well. ... Word problems : ratios. \", Print PDF Solutions: Worksheet No. 6th Grade Math Word Problems - Displaying top 8 worksheets found for this concept.. Enjoy. This is true as we keep on using the phrase “Let’s say…….”. Along with the word problems are mixed review materials. These worksheets are pdf files.. Moreover, with estimation skills, they can quickly determine whether their answer is within a reasonable range or not. The worksheets on this page combine the skills necessary to solve all four types of problems covered previously (addition word problems, subtraction word problems, multiplication word problems and division word problems) and they require students to determine which operation is appropriate for solving the each problem. He had 45 blue pencils. Given that estimation skill enhances kid’s mental math competency, your 6th grader will be able to arrive at reasonable or concrete answers within a twinkle of an eye. Math Word Problem Worksheets Read, explore, and solve over 1000 math word problems based on addition, subtraction, multiplication, division, fraction, decimal, ratio and more. Count on our printable 6th grade math worksheets with answer keys for a thorough practice. She teaches fifth-grade math and is the Kindergarten-6th Grade Math Department Chair at St. Anne’s-Belfield School in Charlottesville, Virginia. These word problems help children hone their reading and analytical skills; understand the real-life application of math operations and other math topics. Improve your math knowledge with free questions in \"Multiply and divide decimals: word problems\" and thousands of other math skills. So, problem solving and estimation skills will help your kids to easily; Our grade 6 math word problem worksheets with answers are a perfect example for kids to grab vital strategies, best for solving estimation word problems for 6th grade. Worksheet - 3. Word Problems 6th Grade Worksheet 6th Grade Math Worksheets with ... #196026 That is the reason that these worksheets are generally published out and distributed in … Deb Russell is a school principal and teacher with over 25 years of experience teaching mathematics at all levels. These word problems worksheets are appropriate for 4th Grade, 5th Grade, 6th Grade, and 7th Grade. Download these sixth grade word problems books. In level 1 , the problems ask for a specific ratio (such as, \" Noah drew 9 hearts, 6 stars, and 12 circles. He had 45 blue pencils. Isolate the variable by dividing \"r\" from each side of the equation to yield the revised formula, r = t ÷ d. Plug in the numbers to get: r = 117 ÷ 2.25, yielding r = 52 mph. Which amount did I choose? The answers for each worksheet are provided in the next slide right after each worksheet. For the first problem on this worksheet, you need to know that the factors of 12 are 1, 2, 3, 4, 6, and 12; and the multiples of 12 are 12, 24, 36. Most importantly, these problem solving and estimation skills will not only strengthen kid’s skills on basic math operations, but will prepare them for areas of advanced math, such as probability, statistics, geometry and algebra. Having understood the problem well, endeavor to estimate the answer before solving. A good book on problem solving with very varied word problems and strategies on how to solve problems. PRINTABLE MATH WORKSHEETS FOR GRADE 6. If the dentist fills any cavities, an additional charge of \\$100 per cavity gets added to the bill If the dentist finds v cavities, what will the cost of the visit be? The Word Problem Worksheets listed below will provide help for students who need to practice solving math word problems. 6th Grade Word Problems With Fractions And Decimal - Displaying top 8 worksheets found for this concept.. Grade 6 maths word problems with answers are presented. 20% of the marbles are blue. Math Worksheets with Word Problems for Every Grade and Skill Level On this page, you'll find worksheets for 1st to 6th Grade, Middle School, and High School. Each grade has weekly word problems for kids to solve. Worksheet - 4. (a ) What is the ratio of the number of blue pencils to the number of reds? See more ideas about word problem worksheets, math word problems, math words. Most at times, math word problems require a step-by-step solving procedure. To the Teacher These worksheets are the same ones found in the Chapter Resource Masters for Glencoe Math Connects, Course 1 .The answers to these worksheets are available at the end Showing top 8 worksheets in the category - 6th Grade Math Word Problems. the completed Word Problem Practice Workbookcan help you in reviewing for quizzes and tests. You simply use the basic formula: rate times the time equals distance, or r * t = d (where \"\" is the symbol for multiplication.). Then, multiply the number of marbles (50) by 0.20. All these exercises have been created to generate interest and eagerness for more math word problems operations. Math worksheets: Proportions word problems. Sixth Grade Math Word Problems Worksheets - Multi-Step Math Word Problems Worksheet #6 Teacher Resources made by Other Teachers: Differentiated Multi-Step Math Word Problems 5th Grade … These books combine mixed work with a few word problems so kids are engaged. Problem-Solving: The Heart of Singapore Math When I was in school, my math classes were typical of what one would expect a “traditional” math … Improve your math knowledge with free questions in \"Ratios and rates: word problems\" and thousands of other math skills. Grade 6 Ratios Word Problems Name: _____ Class: _____ 1. Word Problems 6th Grade Worksheet 6th Grade Math Worksheets with ... #196026 If you're seeing this message, it means we're having trouble loading external resources on our website. Example 1: The price of a visit to the dentist is \\$50. Showing top 8 worksheets in the category - Grade 6 Math Word Problems. Therefore the answer is 18, as 18 is the largest common multiple that can go into 36. Grade 6 Algebra Word Problems How to write algebraic expressions from word problems? How many marbles are blue? Worksheets > Math > Grade 6 > Proportions > Proportions word problems. Count on our printable 6th grade math worksheets with answer keys for a thorough practice. Printable 6th Grade Math Word Problem Worksheets – The 6th grade Math worksheet is among the most demanded worksheets among secondary school trainees since it helps them learn to be issue solvers. Plan your 60-minute lesson in Math or Number Sense and Operations with helpful tips from Andrea Palmer. Primary Grade Challenge Math by Edward Zaccaro. Sixth Grade Math Worksheets for January : Math for Week of January 6: Math for Week of January 13: Math for Week of January 20: Math for Week of January 27: Free Math Worksheets; Addition : Division: Hundreds Charts : Multiplication : Word Problems : Subtraction : Place Value : Feedback or Contact Us; Have students work the problems, fill in their answers in the provided blank spaces, then explain how they would arrive at the solutions for questions where they are having difficulty. For the second problem, you don't even need to use a formula—just basic math and some common sense. 12 and my other number have a greatest common factor of 6 and their least common multiple is 36. Conversely, if you know the speed (rate) that a person is traveling as well as the distance, you can calculate the time he traveled. Sixth Grade Math Word Problems. Worksheet - 1. If the dentist fills any cavities, an additional charge of \\$100 per cavity gets added to the bill If the dentist finds v cavities, what will the cost of the visit be? Math Word Problems Made Easy: Grade 6 is designed to help you help students sharpen their problem-solving abilities (and share a chuckle or two along the way). Example 1: The price of a visit to the dentist is \\$50. This is mathskills4kids.coma premium math quality website with original Math activities and other contents for math practice.We provide 100% free Math ressources for kids from Preschool to Grade 6 to improve children skills. share to facebook share to twitter Questions. The problem involves simple division: 15 yards of ribbon divided by 20 boxes, can be shortened as 15 ÷ 20 = 0.75. Problem 1 : A train travels 18 km/hr. This is relevant to our multi steps word problems exercise. Grade 6 Math Word Problems with Answers. Rates & Ratios Word problems : ratios Write a ratio Word problems : unit rates Question 2 Mike had 180 blue and red pencils. To solve the first equation on the worksheet, use the basic formula: rate times the time = distance, or r t = d. In this case, r = the unknown variable, t = 2.25 hours, and d = 117 miles. Worksheet - 2. This book is divided into three main sections to help you: The Fantastic Five-Step Process The first section describes a simple five-step problem … (A) 3600 (B) 4200 (C) 5100. Showing top 8 worksheets in the category - Grade 6 Math Word Problems. 2 Solution. Grade 6 math word problem worksheets with answers - Estimation word problems for 6th Grade are made of the following Math skills for kids: estimate to solve word problems, multi steps word problems, identifying word problems with extra or missing information, distance direction to starting point word problems, using logical reasoning to find the order, guest and check word problems. Worksheet - 3. A good book on problem solving with very varied word problems and strategies on how to solve problems. GRADE 6 MATH WORD PROBLEMS WITH ANSWERS. In sixth grade, completing word problems can help students apply the new math concepts they're learning, like proportions and statistics, to real-life situations. When solving, show a step-by-step calculation, making visible diverse operation signs where necessary. Rates & Ratios Word problems : ratios Write a ratio Word problems : unit rates What percentage of his pencils was red? Some of these problems are challenging and need more time to solve. Also detailed solutions and full explanations are included. By using ThoughtCo, you accept our, Quiz 8th-Graders With These Math Word Problems, Subtraction of Fractions With Common Denominators, Free Math Word Problem Worksheets for Fifth-Graders, Halloween Math Worksheets & Printable Activities, Thanksgiving Math Worksheets and Activities for Kids, Multiplication Word Problems With Printable Worksheets, Math Glossary: Mathematics Terms and Definitions, Find the Equivalent Fractions - Worksheets. The following math word problems are specific for children in the sixth grade and are divided into the main math categories: Number Concepts, Patterns and Algebra, Geometry and … Jul 31, 2017 - Explore BodyGlam HeadtoToe's board \"6th grade worksheets\" on Pinterest. This will prevent careless distractions and as well encourage you to accomplish your task. Problem 1 : A train travels 18 km/hr. Includes chapters on: Sequences, Problem-solving, Money, Percents, Algebraic Thinking, Negative Numbers, Logic, Ratios, Probability, Measurements, Fractions, Division. Includes chapters on: Sequences, Problem-solving, Money, Percents, Algebraic Thinking, Negative Numbers, Logic, Ratios, Probability, Measurements, Fractions, Division. The free, printable worksheets below involve problems such as these, as well as other important problems, such as determining the largest common factor, calculating percentages, and more. What percentage of his pencils was red? mathskills4kids.com uses cookies to ensure we give you the best experience on our website. Two numbers N and 16 have LCM = 48 and GCF = 8. ... First Math Word Problems 1st Grade Word Problems 2nd Grade Word Problems 3rd Grade Word Problems 4th Grade Word Problems 5th Grade Word Problems 6th Grade Word Problems. Not boring! \", Print Solutions PDF: Worksheet No. 4th grade share to facebook share to twitter Questions. ... Sixth grade . Each box gets the same amount of ribbon. At this point, they will be required to apply logical reasoning and estimation skills. Math Worksheets with Word Problems for Every Grade and Skill Level On this page, you'll find worksheets for 1st to 6th Grade, Middle School, and High School. Printable Math Worksheets for Grade 6. Sixth Grade Word Problems 6th Grade Math Worksheets With Answers September 20, 2020 by admin 21 Posts Related to Sixth Grade Word Problems 6th Grade Math … What’s the average speed that he was traveling?\" Below are three versions of our grade 6 math worksheet on solving proportions word problems. Which amount did I choose? Some of the worksheets displayed are Word problem practice workbook, Sample work from, Decimals practice booklet table of contents, Grade 6 math word problems with percents, Grade 3 mixed math problems and word problems work, Fractionwordproblems, Percent word problems, 501 math word problems. EdHelper's huge library of math worksheets, math puzzles, and word problems is a resource for teachers and parents. I would recommend these exercise for 5th grade, 6th grade, and 7th grade math students. if you want to stay within a budget. Each lesson plan includes an outline for introducing the lesson, teacher modeling, guided practice, and materials and preparation lists to support learning (A) 3600 (B) 4200 (C) 5100. Grade 6 Algebra Word Problems How to write algebraic expressions from word problems? On this page you will find a selection of Christmas math worksheets and regular math worksheets decorated with Christmas ornaments.. Some of the Christmas math worksheets may be fairly large due to the number of images included. The following are some examples of 6th Grade Math Word Problems that deals with percentages. 2, students solve problems that involve a little bit of logic and a knowledge of factors, such as: \"I’m thinking of two numbers, 12 and another number. Find out more about cookies. 6th Grade Word Problems With Fractions And Decimal - Displaying top 8 worksheets found for this concept.. 0 … © Copyright 2018mathskills4kids.com - All rights reserved. Percent Word Problems can be solved using different methods. EdHelper's huge library of math worksheets, math puzzles, and word problems is a resource for teachers and parents. Primary Grade Challenge Math by Edward Zaccaro. Some of the worksheets displayed are Word problem practice workbook, Sample work from, Decimals practice booklet table of contents, Grade 6 math word problems with percents, Grade 3 mixed math problems and word problems work, Fractionwordproblems, Percent word problems, 501 math word problems. Word Problem Assessment: Flying Through Fourth Grade In this assessment, your students will flex their math muscles as they use all four operations to solve a series of word problems. 6th Grade > CCSS 6th Grade Math > Word problems : GCF and LCM. Please click the following links to get math printable math worksheets for grade 6. Merry Christmas from Math-Drills.Com where our gift to you is a page of Christmas math worksheets! Explain to students that you can find the rate (or speed) that someone is traveling if you know the distance and time that she traveled. Grade 6 math word problem worksheets with answers - Estimation word problems for 6th Grade are made of the following Math skills for kids: estimate to solve word problems, multi steps word problems, identifying word problems with extra or missing information, distance direction to starting point word problems, using logical reasoning to find the order, guest and check word problems." ]	[ null, "http://www.dannevang.com/wp-content/uploads/2016/03/bord.jpg", null, "http://www.dannevang.com/wp-content/themes/Lamoon/images/placeholders/940x220.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91570765,"math_prob":0.8821669,"size":26449,"snap":"2021-04-2021-17","text_gpt3_token_len":5605,"char_repetition_ratio":0.2161089,"word_repetition_ratio":0.30139536,"special_character_ratio":0.21426141,"punctuation_ratio":0.13051434,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9946307,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,10,null,10,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-16T08:52:22Z\",\"WARC-Record-ID\":\"<urn:uuid:54402584-30eb-4cde-85d5-fb9a174211e1>\",\"Content-Length\":\"54212\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:86d1ba90-c11d-45ff-b971-f5d943f72095>\",\"WARC-Concurrent-To\":\"<urn:uuid:9b8ed49b-c938-4e10-93d6-1cd60f0a3ff6>\",\"WARC-IP-Address\":\"94.231.97.39\",\"WARC-Target-URI\":\"http://dannevang.net/wp-content/uploads/2020/boom-boom-mxpkejf/page.php?110780=6th-grade-math-word-problems\",\"WARC-Payload-Digest\":\"sha1:SABJNLB4CUQ2EUYG6OE5HCB6Q6NJEKT2\",\"WARC-Block-Digest\":\"sha1:HTZH6SSJ5LBGUYTHT3U36U2LUCFW3OIY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038088731.42_warc_CC-MAIN-20210416065116-20210416095116-00488.warc.gz\"}"}
https://beautyfll.com/is-1-7-oz-equal-50ml/	[ "# Is 1.7 oz equal 50ml?", null, "50 ml equals 1.69 ounces, or there are 1.69 ounces in 50 milliliters.\n\nAlso , How many cups does 4 oz equal? U.S. Standard to Metric\n\nSpoon & Cups Fluid Oz. Metric\n1/2 cup 4 ounces 120 mL\n3/4 cup 6 ounces 180 mL\n1 cup 8 ounces or 1/2 pint 240 mL\n1 1/2 cups 12 ounces 350 mL\n\nIs 1 oz the same as 30ml?\n\noz) use 1 fl. oz. = 30 mL to convert.\n\nAlso to know is, Is 50ml 1.5 oz? Liquid Volume Unit Conversion Table\n\nMilliliter or CC Ounce (Approx.) mL or CC (Approx.)\n15 mL 0.51 oz (1/2 oz) 15 mL\n30 mL 1 oz 30 mL\n50 mL 1.7 oz 48 mL\n60 mL 2 oz 60 mL\n\nIs 50 mg the same as 50ml?\n\nThere are no ml in 50 mg. ml is a measure of volume, 1/1000 of a liter. This is a finite amount of SPACE (Volume).\n\n## How much is 4 oz of meat?\n\n4 oz of raw, lean meat is about 3 ounces after cooking. 3 oz of grilled fish is the size of a checkbook. A medium apple, peach, or orange is about the size of a tennis ball.\n\n## How many cups does 10 ounces equal?\n\n10 oz = 1.25 cups\n\nThus, you can take 12.5 percent of 10 oz to get the same answer.\n\n## How many cups is 10 and 3/4 oz?\n\noz: [US customary cups] = [10 3/4] / 8 = 1.344.\n\n## How do you measure 1 oz?\n\nDetermining How Much Is an Ounce\n\n1. 1 oz = 2 tablespoons.\n2. 8 fl oz = 1 cup.\n3. 32 ounces = 1 quart.\n4. 16 tablespoons = 1 cup.\n5. 1 gram = . 035 oz.\n6. 100 grams = 3.5 oz.\n\n## Does 2 tbsp equal 1 oz?\n\nHow Many Tablespoons are in a Fluid Ounce? There are 2 tablespoons in a fluid ounce, which is why we use this value in the formula above.\n\n## Which unit of measure is equivalent to 1 oz?\n\nDifferences between the U.S. and Imperial Systems\n\nUnit of Measurement Imperial System Metric Equivalent\n1 ounce 1 (fluid) oz. 28.41 mL\n1 gill 5 (fluid) oz. 142.07 mL\n1 cup Not commonly used\n1 pint 20 (fluid) oz. 568.26 mL\n\n## What is 500ml in ounces?\n\nConvert 500 Milliliters to Ounces\n\n500 Milliliters (ml) 17.637 Ounces (oz)\n1 ml = 0.035274 oz 1 oz = 28.350 ml\n\n## Is oz or mL bigger?\n\n1 Fluid ounce (oz) is equal to 29.5735296 milliliters (ml). … For example, to find out how many ml there are in 16 fluid ounces, multiply 29.5735296 by 16, that makes 473.176473 ml in 16 fluid ounces.\n\n## Is 40ml 2 oz?\n\nMilliliters to US Fluid Ounces table\n\nMilliliters US Fluid Ounces\n40 mL 1.35 us fl oz\n41 mL 1.39 us fl oz\n42 mL 1.42 us fl oz\n43 mL 1.45 us fl oz\n\n• 22 juil. 2018\n\n## Is 10mg the same as 10 ml?\n\nA Milliliter (mL) is a unit of measure for a very small amount of fluid. It means that for every 1 milliliter of a substance , there are 10 milligrams of something else in that milliliter.\n\nWhat percent is 10 mg ml?\n\n10 mg = 0.01 ml 210 mg = 0.21 ml 700 mg = 0.7 ml\n80 mg = 0.08 ml 280 mg = 0.28 ml 1400 mg = 1.4 ml\n\n• 27 mars 2020\n\n## Is 5mL the same as 5mg?\n\nIn respect to this, is 5ml the same as 5mg? mL is a measurement of volume, while mg is a measurement of mass. For instance, if you have a solution that contains 10mg of medication in 5mL and you want to give 5mg of medication, then “give over have” or 5/10 or one half, so half of 5mL is 2.5 mL.\n\n## How much is a milligram of liquid?\n\nOne milliliter (British spelling: millilitre) (ml) is 1/1000 of a liter and is a unit of volume. 1 milligram (mg) is 1/1000 of a gram and is a unit of mass/weight.\n\nWhat is 10 mg converted to ML?\n\n10 mg = 0.01 ml 210 mg = 0.21 ml 700 mg = 0.7 ml\n20 mg = 0.02 ml 220 mg = 0.22 ml 800 mg = 0.8 ml\n\n• 16 mars 2020\n\n## How can I measure 3 ounces of meat without a scale?\n\nYou can also use your hand to measure food portions of meat and produce. For example, a single 3-ounce serving of chicken, beef, or fish is roughly the size of your palm. A 1-cup serving of fruit or vegetables is roughly the size of your closed fist. A single serving of cheese is about the size of your thumb.\n\n## How much is 3 oz of meat in cups?\n\n3 oz to cups = 0.375 cup in 3 oz. 4 oz to cups = 0.5 cup in 4 oz.\n\n## How much is 3 oz of meat?\n\nTwo servings, or 6 oz., of lean meat (poultry, fish, shellfish, beef) should be a part of a daily diet. Measure the right amount with your palm. One palm size portion equals 3 oz., or one serving.\n\n## How do you measure fl oz?\n\nDetermining How Much Is an Ounce\n\n1. 1 oz = 2 tablespoons.\n2. 8 fl oz = 1 cup.\n3. 32 ounces = 1 quart.\n4. 16 tablespoons = 1 cup.\n5. 1 gram = . 035 oz.\n6. 100 grams = 3.5 oz.\n\n## How many cups is 8 fl ounces?\n\n8 oz equals 1.00 cups. 1 ounce is equivalent to 0.125 cups, and there are 1.00 cups in 8 ounces.\n\n## Is 10 ounces bigger than 2 cups?\n\nFor basic culinary purposes, a cup is 8 fluid ounces, so 10 ounces would be one and one quarter cups.\n\nWe will be happy to hear your thoughts", null, "Enable registration in settings - general" ]	[ null, "https://beautyfll.com/wp-content/themes/rehub-theme/images/default/noimage_450_350.png", null, "https://beautyfll.com/wp-content/uploads/2021/02/Logo-beautyfll.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8377559,"math_prob":0.99367535,"size":4467,"snap":"2022-27-2022-33","text_gpt3_token_len":1575,"char_repetition_ratio":0.124355815,"word_repetition_ratio":0.12058212,"special_character_ratio":0.38459817,"punctuation_ratio":0.14120781,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99161637,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-02T14:09:06Z\",\"WARC-Record-ID\":\"<urn:uuid:96df4942-3eb7-4dd6-ba52-3bd11f6a049e>\",\"Content-Length\":\"105100\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e8c6e66f-78c3-4dd8-8d98-dd5728b0def1>\",\"WARC-Concurrent-To\":\"<urn:uuid:3eade23f-7406-4e16-9f6d-b44de93d2d7b>\",\"WARC-IP-Address\":\"79.137.36.139\",\"WARC-Target-URI\":\"https://beautyfll.com/is-1-7-oz-equal-50ml/\",\"WARC-Payload-Digest\":\"sha1:EPKYD3BTNQ3WOT7EMBA6INEKU6I55NXW\",\"WARC-Block-Digest\":\"sha1:QQTH4XYOHYOIT6YT2NVLT2OWQTCCHJRQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104141372.60_warc_CC-MAIN-20220702131941-20220702161941-00269.warc.gz\"}"}
https://dubbsong.github.io/dev/2019/10/02/codewars-6kyu-10/	[ "# likes.js\n\n### Problem\n\n• You probably know the ‘like’ system from Facebook.\n• Facebook의 ‘like’ 시스템을 알고 있을 것이다.\n• We want to create the text that should be displayed next to such an item.\n• 옆에 표시할 텍스트를 작성하려고 한다.\n• Implement a function, which must take in input array, containing the names of people who like an item.\n• like를 누른 사람들의 이름을 포함하는 배열을 가져오는 함수를 작성한다.\n• For 4 more names, the number in ‘and 2 others’ simply increases.\n• 이름이 4개 이상인 경우, ‘and 2 others’와 같이 단순히 증가시킨다.\n\n#### For example\n\nlikes([]) => `no one likes this`\n\nlikes([‘Sam’]) => `Sam likes this`\n\nlikes([‘Ralph’, ‘Lauren’]) => `Ralph and Lauren like this`\n\nlikes([‘Ralph’, ‘Lauren’, ‘Leo’]) => `Ralph, Lauren and Leo like this`\n\nlikes([‘Sam’, ‘Ralph’, ‘Lauren’, ‘Leo’]) => `Sam, Ralph and 2 others like this`\n\n### Solution 01\n\n``````function likes(arr) {\nif (arr.length === 0) return 'no one likes this';\nif (arr.length === 1) return arr + ' likes this';\nif (arr.length === 2) return arr + ' and ' + arr + ' like this';\nif (arr.length === 3) return arr + ', ' + arr + ' and ' + arr + ' like this';\nreturn arr + ', ' + arr + ' and ' + (arr.length - 2) + ' others like this';\n}\n\nlikes([]); // no one likes this\nlikes(['Sam']); // Sam likes this\nlikes(['Ralph', 'Lauren']); // Ralph and Lauren like this\nlikes(['Ralph', 'Lauren', 'Leo']); // Ralph, Lauren and Leo like this\nlikes(['Sam', 'Ralph', 'Lauren', 'Leo']); // Sam, Ralph and 2 others like this\n``````\n\n### Solution 02\n\n``````function likes(arr) {\nswitch(arr.length) {\ncase 0:\nreturn 'no one likes this';\ncase 1:\nreturn `\\${arr} likes this`;\ncase 2:\nreturn `\\${arr} and \\${arr} like this`;\ncase 3:\nreturn `\\${arr}, \\${arr} and \\${arr} like this`;\ndefault:\nreturn `\\${arr}, \\${arr} and \\${arr.length - 2} others like this`;\n}\n}\n\nlikes([]); // no one likes this\nlikes(['Sam']); // Sam likes this\nlikes(['Ralph', 'Lauren']); // Ralph and Lauren like this\nlikes(['Ralph', 'Lauren', 'Leo']); // Ralph, Lauren and Leo like this\nlikes(['Sam', 'Ralph', 'Lauren', 'Leo']); // Sam, Ralph and 2 others like this\n``````\n\n### Solution 03\n\n``````function likes(arr) {\nreturn {\n0: 'no one likes this',\n1: `\\${arr} likes this`,\n2: `\\${arr} and \\${arr} like this`,\n3: `\\${arr}, \\${arr} and \\${arr} like this`,\n4: `\\${arr}, \\${arr} and \\${arr.length - 2} others like this`,\n}[Math.min(4, arr.length)];\n}\n\nlikes([]); // no one likes this\nlikes(['Sam']); // Sam likes this\nlikes(['Ralph', 'Lauren']); // Ralph and Lauren like this\nlikes(['Ralph', 'Lauren', 'Leo']); // Ralph, Lauren and Leo like this\nlikes(['Sam', 'Ralph', 'Lauren', 'Leo']); // Sam, Ralph and 2 others like this\n``````\n\n` 오 많 배`" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5680997,"math_prob":0.955218,"size":2567,"snap":"2020-45-2020-50","text_gpt3_token_len":876,"char_repetition_ratio":0.22863832,"word_repetition_ratio":0.36214954,"special_character_ratio":0.39267626,"punctuation_ratio":0.19144982,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9833013,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-22T03:44:21Z\",\"WARC-Record-ID\":\"<urn:uuid:58c754cd-03b6-42ab-847a-e52382e82c95>\",\"Content-Length\":\"35769\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f71ae993-b46e-4b90-bb75-776b4ee396bb>\",\"WARC-Concurrent-To\":\"<urn:uuid:40394aea-dcdd-48bc-b0cf-fab98afdc865>\",\"WARC-IP-Address\":\"185.199.109.153\",\"WARC-Target-URI\":\"https://dubbsong.github.io/dev/2019/10/02/codewars-6kyu-10/\",\"WARC-Payload-Digest\":\"sha1:PRSB5FFVWSI6L5VJ2S3L44ODGVUYGP4L\",\"WARC-Block-Digest\":\"sha1:V2HWNUTRZHXWUCFIPTFSZM62NOKGFAKN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107878879.33_warc_CC-MAIN-20201022024236-20201022054236-00227.warc.gz\"}"}
https://iccl.inf.tu-dresden.de/web/LATPub366	[ "# Pinpointing in the Description Logic EL\n\nFranz Baader, Rafael Peñaloza, Boontawee Suntisrivaraporn\nPinpointing in the Description Logic EL\nProceedings of the 30th German Conference on Artificial Intelligence (KI2007), volume 4667 of Lecture Notes in Artificial Intelligence, 52-67, 2007. Springer\n• KurzfassungAbstract\nAxiom pinpointing has been introduced in description logics (DLs) to help the user understand the reasons why consequences hold by computing minimal subsets of the knowledge base that have the consequence in question. Until now, the pinpointing approach has only been applied to the DL ALC and some of its extensions. This paper considers axiom pinpointing in the less expressive DL EL+, for which subsumption can be decided in polynomial time. More precisely, we consider an extension of the pinpointing problem where the knowledge base is divided into a static part, which is always present, and a refutable part, of which subsets are taken. We describe an extension of the subsumption algorithm for EL+ that can be used to compute all minimal subsets of (the refutable part of) a given TBox that imply a certain subsumption relationship. The worst-case complexity of this algorithm turns out to be exponential. This is not surprising since we can show that a given TBox may have exponentially many such minimal subsets. However, we can also show that the problem is not even output polynomial, i.e., unless P=NP, there cannot be an algorithm computing all such minimal sets that is polynomial in the size of its input and output. In addition, we show that finding out whether there is such a minimal subset within a given cardinality bound is an NP-complete problem. In contrast to these negative results, we also show that one such minimal subset can be computed in polynomial time. Finally, we provide some encouraging experimental results regarding the performance of a practical algorithm that computes one (small, but not necessarily minimal) subset that has a given subsumption relation as consequence.\n• Forschungsgruppe:Research Group: Automatentheorie\nThe final publication is available at Springer.\n@inproceedings{ BaaPenSun-KI-07,\ntitle = {Pinpointing in the Description Logic $\\mathcal{EL}$}," ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88159955,"math_prob":0.9203517,"size":2926,"snap":"2022-05-2022-21","text_gpt3_token_len":687,"char_repetition_ratio":0.11156742,"word_repetition_ratio":0.12765957,"special_character_ratio":0.20813397,"punctuation_ratio":0.10788382,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98170936,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-20T04:27:26Z\",\"WARC-Record-ID\":\"<urn:uuid:9ddc93d4-6523-44c6-bdda-d24d296ffe59>\",\"Content-Length\":\"38085\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:06a1b3f9-2e4b-4436-89d1-c8e4d4a19d25>\",\"WARC-Concurrent-To\":\"<urn:uuid:b34260bf-0d76-43d1-9869-10c3df10fe66>\",\"WARC-IP-Address\":\"141.76.16.67\",\"WARC-Target-URI\":\"https://iccl.inf.tu-dresden.de/web/LATPub366\",\"WARC-Payload-Digest\":\"sha1:3ZU4KEJ5LPKH7LLOTU4F62YG5G5DVH52\",\"WARC-Block-Digest\":\"sha1:VAWAF3L7NMJRR3YQAORQVYWP5VZYCKHL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662531352.50_warc_CC-MAIN-20220520030533-20220520060533-00589.warc.gz\"}"}
https://goprep.co/q23-the-differential-equation-which-represents-the-family-of-i-1nlo3s	[ "Q. 23\n\n# Mark the correct alternative in each of the following:The differential equation which represents the family of curves y = eCx isA. y1 = C2 yB. xy1 – ln y = 0C. x ln y = yy1D. y ln y = xy1\n\ny = eCx\n\nTaking log both sides we get,\n\nlog y = log eCx\n\nlog y = Cx log e log ax = x log a\n\nlog y = Cx --(1) log e = 1\n\nDifferentiate w.r.t x we get,", null, "Put value of C in (1) we get,", null, "", null, "Which is y ln y = xy1 = (D)\n\nRate this question :\n\nHow useful is this solution?\nWe strive to provide quality solutions. Please rate us to serve you better.\nRelated Videos", null, "", null, "Interactive Quiz on DIfferential Calculus50 mins", null, "", null, "Functional Equations - JEE with ease48 mins\nTry our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts\nDedicated counsellor for each student\n24X7 Doubt Resolution\nDaily Report Card\nDetailed Performance Evaluation", null, "view all courses", null, "" ]	[ null, "https://gradeup-question-images.grdp.co/liveData/PROJ32365/1559797512725274.png", null, "https://gradeup-question-images.grdp.co/liveData/PROJ32365/1559797513477560.png", null, "https://gradeup-question-images.grdp.co/liveData/PROJ32365/1559797514183544.png", null, "https://grdp.co/cdn-cgi/image/quality=90,fit=contain,width=350,f=auto/https://gs-post-images.grdp.co/2020/10/022186a09c6921f3f0da0b5e239e205d17cd077ef35f87398aef5bde9a634516poster-high-webp.png", null, "https://gs-post-images.grdp.co/2020/8/group-2x-img1598864374422-78.png-rs-high-webp.png", null, "https://grdp.co/cdn-cgi/image/quality=90,fit=contain,width=350,f=auto/https://gs-post-images.grdp.co/2020/10/4a9234fcadab3795f046a4663a99560b295e60494b8cfa70d2f2f9dde60e787dposter-high-webp.png", null, "https://gs-post-images.grdp.co/2020/8/group-2x-img1598864374422-78.png-rs-high-webp.png", null, "https://grdp.co/cdn-cgi/image/height=128,quality=80,f=auto/https://gs-post-images.grdp.co/2020/8/group-7-3x-img1597928525711-15.png-rs-high-webp.png", null, "https://gs-post-images.grdp.co/2020/8/group-img1597139979159-33.png-rs-high-webp.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.79035324,"math_prob":0.9824484,"size":1523,"snap":"2020-45-2020-50","text_gpt3_token_len":412,"char_repetition_ratio":0.16129032,"word_repetition_ratio":0.021505376,"special_character_ratio":0.24097176,"punctuation_ratio":0.08791209,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.952257,"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,1,null,1,null,1,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-28T09:09:18Z\",\"WARC-Record-ID\":\"<urn:uuid:5fdf11df-ba8a-43b5-b296-837d1d60bb84>\",\"Content-Length\":\"632703\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5912b777-a1a9-4a00-a6c8-56c7ce403856>\",\"WARC-Concurrent-To\":\"<urn:uuid:6600d7fa-317d-4301-a103-d3b44d5512a9>\",\"WARC-IP-Address\":\"104.18.25.35\",\"WARC-Target-URI\":\"https://goprep.co/q23-the-differential-equation-which-represents-the-family-of-i-1nlo3s\",\"WARC-Payload-Digest\":\"sha1:L5WKU7XFCANYO5BBXE5F5RTSPE6KMMO3\",\"WARC-Block-Digest\":\"sha1:RPJORJWBQSFRZM6Z6JCVM5BP4CCTQVKE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107897022.61_warc_CC-MAIN-20201028073614-20201028103614-00239.warc.gz\"}"}
https://nanoscalereslett.springeropen.com/articles/10.1186/1556-276X-8-30	[ "# Displacing, squeezing, and time evolution of quantum states for nanoelectronic circuits\n\n## Abstract\n\nThe time behavior of DSN (displaced squeezed number state) for a two-dimensional electronic circuit composed of nanoscale elements is investigated using unitary transformation approach. The original Hamiltonian of the system is somewhat complicated. However, through unitary transformation, the Hamiltonian became very simple enough that we can easily treat it. By executing inverse transformation for the wave function obtained in the transformed system, we derived the exact wave function associated to the DSN in the original system. The time evolution of the DSN is described in detail, and its corresponding probability density is illustrated. We confirmed that the probability density oscillates with time like that of a classical state. There are two factors that drive the probability density to oscillate: One is the initial amplitude of complementary functions, and the other is the external power source. The oscillation associated with the initial amplitude gradually disappears with time due to the dissipation raised by resistances of the system. These analyses exactly coincide with those obtained from classical state. The characteristics of quantum fluctuations and uncertainty relations for charges and currents are also addressed.\n\n## Background\n\nThe technical range of nanoscale is 1 to 999 nm, but people often refer to nanosize when an element is smaller than about 100 nm, where quantum effects are dominant instead of classical ones. Nanophysics and nanoelectronics have been rapidly developed thanks to the advancement of relevant technologies such as crystal growth and lithography, which facilitate sophisticated experiments for nanosystems [1, 2]. A recent conspicuous trend in the community of electronic device is that the integrated circuits and components are miniaturized towards atomic-scale dimensions . We can confirm from many experiments and theories associated with nanoscale elements that the quantum effects become prominent when the transport dimension reaches a critical value which is the Fermi wavelength, while at the same situation, the classical theory for the motion of charges and currents is invalid. Not only quantum dot and quantum wire but also the quantum characteristics of electronic circuits involving nanoscale elements are important as a supporting theory for nanometer electronic technology and quantum information technology. For this reason, quantum effects in electronic circuits with nanoscale elements have been widely studied in recent years.\n\nThe simple quantum model of a lossless inductor-capacitor (LC) circuit have been suggested firstly by Louisell . Zhang et al. investigated the quantum properties of two-dimensional electronic circuits which have no power source . The quantum behavior of charges and currents for an LC circuit and a resistor-inductor-capacitor (RLC) linear circuit driven by a power source have been studied by several researchers. If a circuit contains resistance, the electronic energy of the system dissipates with time. In this case, the system is described by a time-dependent Hamiltonian. Another example of the systems described by time-dependent Hamiltonian is electronic circuits driven by time-varying power sources. The quantum problem of time-dependent Hamiltonian systems attracted great concern in the community of theoretical physics and chemistry for several decades [4, 6, 7].\n\nThe study of electronic characteristics of charge carriers in nanoelectronic circuits is basically pertained to a physical problem. There are plentiful reports associated with the physical properties of miniaturized two-loop (or two-dimensional) circuits and more high multi-loop circuits including their diverse variants. Various applications which use two-loop circuits include a switch-level resistor-capacitor (RC) model of an n-transistor (see Figure 3 of ), a design of a prototype of current-mode leapfrog ladder filters (Sect. 3 of ), and a port-Hamiltonian system , whereas higher loop circuits can be used as a transmission line model for multiwall carbon nanotube and a filter circuit for electronic signals (Sect. 5 of ).\n\nIn this paper, we derive quantum solutions of a two-dimensional circuit coupled via RL and investigate its displaced squeezed number state (DSN) . We suppose that the system is composed of nanoscale elements and driven by a time-varying power source. The unitary transformation method which is very useful when treating time-dependent Hamiltonian systems in cases like this will be used. We can obtain the wave functions of DSN by first applying the squeezing operator in those of the number state and then applying the unitary displacement operator. Under displaced quantum states of circuit electrodynamics, conducting charges (or currents) exhibit collective classical-like oscillation. The fluctuations and uncertainty relations for charges and currents will be evaluated in the DSN without approximation.\n\nDisplaced squeezed number states, which are the main topic in this work, belong to nonclassical states that have been objects of many investigations. The statistical properties of these states exhibit several pure quantum effects which have no classical analogues, including the interference in the phase space , the revival/collapse phenomenon , and sub-Poissonian statistics . The position representation of these states with overall phases is derived by Moller et al. for the simple harmonic oscillator by employing geometric operations in phase space . The effects of quantum interference between two distinct DSNs prepared to be out of phase with respect to each other are investigated by Faisal et. al., discussing various nonclassical properties in connection with quantum number distribution, purity, quadrature squeezing, W-function, etc. .\n\n## Methods and results\n\n### Simplification via unitary transformation\n\nLet us consider two loops of RLC circuit, whose elements are nanosized, that are coupled with each other via inductance and resistance as shown in Figure 1. Using Kirchhoff’s law, we obtain the classical equations of motion for charges of the system :\n\n$L 1 d 2 q 1 d t 2 + R 1 d q 1 dt + q 1 C 1 + L 0 d 2 q 1 d t 2 − d 2 q 2 d t 2 + R 0 d q 1 dt − d q 2 dt = E ( t ) ,$\n(1)\n$L 2 d 2 q 2 d t 2 + R 2 d q 2 dt + q 2 C 2 − L 0 d 2 q 1 d t 2 − d 2 q 2 d t 2 − R 0 d q 1 dt − d q 2 dt = 0 ,$\n(2)\n\nwhere q j (j=1,2; hereafter, this convention will be used for all j) are charges stored in the capacitances C j , respectively, and $E(t)$ is an arbitrary time-varying voltage source connected in loop 1. If we consider not only the existence of $E(t)$ but also the mixed appearance of q1 and q2 in these two equations, it may be not an easy task to treat the system directly. If the scale of resistances are sufficiently large, the system is described by an overdamped harmonic oscillator, whereas the system becomes an underdamped harmonic oscillator in the case of small resistances. In this paper, we consider only the underdamped case.\n\nFor convenience, we suppose that R0/L0=R1/L1=R2/L2β. Then, the classical Hamiltonian of the system can be written as\n\n$H = e − βt p 1 2 2 L 1 + p 2 2 2 L 2 − 1 2 ( k 1 p 1 − k 2 p 2 ) 2 + e βt q 1 2 2 C 1 + q 2 2 2 C 2 − E ( t ) q 1 ,$\n(3)\n\nwhere p j are canonical currents of the system, and k j =(1/L j )(1/L0+1/L1+1/L2)−1/2. From Hamilton’s equations, we can easily see that p j are given by\n\n$p 1 = e βt [( L 0 + L 1 ) q ̇ 1 − L 0 q ̇ 2 ],$\n(4)\n$p 2 = e βt [( L 0 + L 2 ) q ̇ 2 − L 0 q ̇ 1 ].$\n(5)\n\nIf we replace classical variables q j and p j in Equation 3 with their corresponding operators, $q ̂ j$ and $p ̂ j$, the classical Hamiltonian becomes quantum Hamiltonian:\n\n$Ĥ = e − βt p ̂ 1 2 2 L 1 + p ̂ 2 2 2 L 2 − 1 2 ( k 1 p ̂ 1 − k 2 p ̂ 2 ) 2 + e βt q ̂ 1 2 2 C 1 + q ̂ 2 2 2 C 2 − E ( t ) q ̂ 1 ,$\n(6)\n\nwhere $p ̂ j =−iℏ∂/∂ q j$. Now, we are going to transform $H ̂$ into a simple form using the unitary transformation method, developed in for a two-loop LC circuit, in order to simplify the problem. Let us first introduce a unitary operator\n\n$Û A = Û A 1 Û A 2 ,$\n(7)\n\nwhere\n\n$Û A 1 = exp i ℏ ln C 2 C 1 1 / 8 + i β 4 ℏ t ( p ̂ 1 q ̂ 1 + q ̂ 1 p ̂ 1 ) × exp i ℏ ln C 1 C 2 1 / 8 + i β 4 ℏ t ( p ̂ 2 q ̂ 2 + q ̂ 2 p ̂ 2 ) ,$\n(8)\n$Û A 2 = exp − iφ ℏ ( p ̂ 1 q ̂ 2 − p ̂ 2 q ̂ 1 ) ,$\n(9)\n\nwith\n\n$φ = 1 2 tan − 1 2 k 1 k 2 C 2 C 1 k 1 2 − 1 L 1 − C 1 C 2 k 2 2 − 1 L 2 − 1 .$\n(10)\n\nUsing Equation 7, we can transform the Hamiltonian such that\n\n$Ĥ A = Û A − 1 Ĥ Û A −iℏ Û A − 1 ∂ Û A ∂t .$\n(11)\n\nA straightforward algebra after inserting Equation 6 into the above equation gives\n\n$Ĥ A ( q ̂ 1 , p ̂ 1 , q ̂ 2 , p ̂ 2 ,t)= Ĥ A 1 ( q ̂ 1 , p ̂ 1 ,t)+ Ĥ A 2 ( q ̂ 2 , p ̂ 2 ,t),$\n(12)\n\nwhere\n\n$Ĥ Aj ( q ̂ j , p ̂ j , t ) = p ̂ j 2 2 μ j + β 4 ( q ̂ j p ̂ j + p ̂ j q ̂ j ) + 1 2 C 1 C 2 q ̂ j 2 − q ̂ j E ( t ) C 1 C 2 1 / 4 e βt / 2 cos φ − Π 2 δ j , 2 ,$\n(13)\n\nwith\n\n(14)\n(15)\n\nOne can see from Equation 13 that the coupled term involving $q ̂ 1 q ̂ 2$ in the original Hamiltonian is decoupled through this transformation. However, the Hamiltonian still contains linear terms that are expressed in terms of $q ̂ j E(t)$, which are hard to handle when developing a quantum theory of the system. To remove these terms, we introduce another unitary operator of the form\n\n$Û B = Û B 1 Û B 2 ,$\n(16)\n$Û B 1 = exp i ℏ [ p 1 p ( t ) q ̂ 1 + p 2 p ( t ) q ̂ 2 ] × exp − i ℏ [ q 1 p ( t ) p ̂ 1 + q 2 p ( t ) p ̂ 2 ] ,$\n(17)\n$Û B 2 =exp − iβ 4 ℏ ( μ 1 q ̂ 1 2 + μ 2 q ̂ 2 2 ) ,$\n(18)\n\nwhere q j p (t) and p j p (t) are classical particular solutions of the firstly transformed system described by $Ĥ A$ in the charge and the current spaces, respectively. From basic Hamiltonian dynamics with the use of Equation 12, we see that q j p (t) and p j p (t) satisfy the time-dependent classical equations that are given by\n\n$q ̈ jp (t)+ ω j 2 q jp (t)− E ( t ) μ j C 1 C 2 4 e βt / 2 cos φ − Π 2 δ j , 2 =0,$\n(19)\n$p ̈ jp (t)+ ω j 2 p jp (t)− E ̇ (t) C 1 C 2 4 e βt / 2 cos φ − Π 2 δ j , 2 =0,$\n(20)\n\nwhere\n\n$ω j = 1 μ j C 1 C 2 − β 2 4 1 / 2 .$\n(21)\n\nThen, the second transformation yields\n\n$Ĥ B ( q ̂ 1 , p ̂ 1 , q ̂ 2 , p ̂ 2 , t ) = Û B − 1 Ĥ A Û B − iℏ Û B − 1 ∂ Û B ∂t = Ĥ B 1 ( q ̂ 1 , p ̂ 1 , t ) + Ĥ B 2 ( q ̂ 2 , p ̂ 2 , t ) ,$\n(22)\n\nwhere\n\n$Ĥ Bj ( q ̂ j , p ̂ j ,t)= p ̂ j 2 2 μ j + 1 2 μ j ω j 2 q ̂ j 2 + L jp (t),$\n(23)\n\nwith\n\n$L jp (t)= 1 2 μ j q ̇ jp 2 (t)− 1 2 C 1 C 2 q jp 2 (t).$\n(24)\n\nThe finally transformed Hamiltonian, Equation 22, is very simple and no longer involves linear terms that contain $E(t)$. If we neglect $L jp (t)$, this is exactly the same as that of the two-dimensional simple harmonic oscillator of frequencies ω j . We will use this formula in order to develop DSN, which is a typical nonclassical quantum state.\n\nIf we regard that the transformed Hamiltonian is very simple, the quantum dynamics in the transformed system may be easily developed. Let us write the Schrödinger equations for elements of the transformed Hamiltonian as\n\n$iℏ ∂ ψ n j B ( q j , t ) ∂t = Ĥ Bj ψ n j B ( q j ,t),$\n(25)\n\nwhere $ψ n j B ( q j ,t)$ represent number state wave functions for each component of the decoupled systems described by $Ĥ Bj$.\n\nBy means of the usual annihilation operator,\n\n$â j = μ j ω j 2 ℏ q ̂ j + i p ̂ j 2 ℏ μ j ω j ,$\n(26)\n\nand the creation operator $â j ‡$ defined as the Hermitian adjoint of $â j$, one can identify the initial wave functions of the transformed system in number state such that\n\n$ψ n 1 , n 2 B ( q 1 , q 2 ,0)= ψ n 1 B ( q 1 ,0) ψ n 2 B ( q 2 ,0),$\n(27)\n\nwhere\n\n$ψ n j B ( q j , 0 ) = μ j ω j ℏΠ 1 / 4 1 2 n j n j ! H n j μ j ω j ℏ 1 / 2 q j × exp − μ j ω j 2 ℏ q j 2 .$\n(28)\n\nThis formula of wave functions will be used in the next section in order to derive the DSN of the system.\n\n### Displaced squeezed number state\n\nThe DSNs are defined by first squeezing the number states and then displacing them. Like squeezed states, DSNs exhibit nonclassical properties of the quantum field in which the fluctuation of a certain observable can be less than that in the vacuum state. This state is a generalized quantum state for dynamical systems and, in fact, equivalent to excited two-photon coherent states in quantum optics. If we consider that DSNs generalize and combine the features of well-known important states such as displaced number states (DNs) , squeezed number states , and two-photon coherent states (non-excited) , the study of DSNs may be very interesting. Different aspects of these states, including quantal statistics, entropy, entanglement, and position space representation with the correct overall phase, have been investigated in [17, 23, 25].\n\nTo obtain the DSN in the original system, we first derive the DSN in the transformed system according to its exact definition. Then, we will transform it inversely into that of the original system. The squeeze operator in the transformed system is given by\n\n$Ŝ j ( z j )=exp − 1 2 ( z j ∗ â j 2 − z j â j ‡ 2 ) ,$\n(29)\n\nwhere\n\n$z j = r j e i ϕ j .$\n(30)\n\nUsing the Baker-Campbell-Hausdorff relation that is given by \n\n$exp 1 2 ℏ [ a q ̂ 2 + ic ( q ̂ p ̂ + p ̂ q ̂ ) − b p ̂ 2 ] = 1 cosh θ − c θ sinh θ × exp a 2 θℏ sinh θ cosh θ − c θ sinh θ − 1 q ̂ 2 × exp − i ℏ ln cosh θ − c θ sinh θ q ̂ p ̂ × exp − b 2 θℏ sinh θ cosh θ − c θ sinh θ − 1 p ̂ 2 ,$\n(31)\n\nwhere $θ= c 2 − ab$, the squeeze operator can be rewritten as\n\n$Ŝ j ( z j ) = 1 cosh r j + cos ϕ j sinh r j × exp i μ j ω j 2 ℏ sin ϕ j sinh r j cosh r j + cos ϕ j sinh r j q ̂ j 2 × exp − i ℏ ln cosh r j + cos ϕ j sinh r j q ̂ j p ̂ j × exp − i 2 μ j ω j ℏ sin ϕ j sinh r j cosh r j + cos ϕ j sinh r j p ̂ j 2 .$\n(32)\n\nLet us express the DSN in the transformed system in the form\n\n$ψ s , n 1 , n 2 B ( q 1 , q 2 ,t)= ψ s , n 1 B ( q 1 ,t) ψ s , n 2 B ( q 2 ,t),$\n(33)\n\nwhere $ψ s , n j B ( q j ,t)$ represent two decoupled states which are drivable from\n\n$ψ s , n j B ( q j ,t)= T ̂ Bj ( q ̂ j , p ̂ j ,t) D ̂ j ( α j ) Ŝ j ( z j ) ψ n j B ( q j ,0).$\n(34)\n\nHere, $D ̂ j ( α j )$ are displacement operators in the transformed system, which are given by\n\n$D ̂ j ( α j )=exp( α j â j ‡ − α j ∗ â j ),$\n(35)\n\nwhere α j is an eigenvalue of $â j$ at initial time. By considering Equation 26, we can confirm that\n\n$α j = μ j ω j 2 ℏ q jc (0)+ i p jc ( 0 ) 2 ℏ μ j ω j ,$\n(36)\n\nwhere q j c (t) and p j c (t) are classical solutions of the equation of motion in charge and current spaces, respectively, for the finally transformed system. If we regard that the complementary functions of the equation of motion in the firstly transformed system are the same as the classical solutions of the finally transformed system, q j c (t) and p j c (t) can also be complementary functions of the firstly transformed system. The other operators $T ̂ Bj ( q ̂ j , p ̂ j ,t)$ are time-displacement operators:\n\n$T ̂ Bj ( q ̂ j , p ̂ j ,t)=exp − i ℏ ∫ 0 t Ĥ Bj ( q ̂ j , p ̂ j , t ′ ) d t ′ .$\n(37)\n\nAt first, the action of squeezing operator in wave functions of the initial number state gives\n\n$Ŝ j ( z j ) ψ n j B ( q j , 0 ) = μ j ω j ℏΠ 1 / 4 1 2 n j n j ! G bj n j G aj H n j μ j ω j ℏ G cj 1 / 2 q j × exp − μ j ω j 2 ℏ G dj q j 2 ,$\n(38)\n\nwhere\n\n$G aj =cosh r j + e i ϕ j sinh r j ,$\n(39)\n$G bj = cosh r j + e − i ϕ j sinh r j cosh r j + e i ϕ j sinh r j ,$\n(40)\n$G cj = cosh 2 r j + sinh 2 r j +2cos ϕ j cosh r j sinh r j ,$\n(41)\n$G dj = 1 − i sin ϕ j sinh r j ( cosh r j + e i ϕ j sinh r j ) ( cosh r j + cos ϕ j sinh r j ) ( cosh r j + e i ϕ j sinh r j ) .$\n(42)\n\nThe evaluation of the other actions of the operators in Equation 34 may be easily performed using Equation 31 and the relation \n\n$exp c ℏ 2 ∂ 2 ∂ q 2 h ( q ) = 1 4 Π ℏ 2 c ∫ − ∞ ∞ exp − ( y − q ) 2 4 c ℏ 2 h ( y ) dy ,$\n(43)\n\ntogether with the eighth formula of 7.374 in (see Appendix Appendix 1), yielding\n\n(44)\n\nwhere\n\n$h aj =cos( ω j t)+i G dj sin( ω j t),$\n(45)\n$h bj =1− 2 i sin ( ω j t ) h aj G cj .$\n(46)\n\nHere, the time evolution of complementary functions are\n\n$q jc (t)= q jc (0)cos( ω j t)+ p jc ( 0 ) μ j ω j sin( ω j t),$\n(47)\n$p jc (t)= p jc (0)cos( ω j t)− μ j ω j q jc (0)sin( ω j t).$\n(48)\n\nThe transformed system reduces to a two-dimensional undriven simple harmonic oscillator in the limit $E(t)=0$. Our result in Equation 44 is exact, and in this limit, we can easily confirm that some errors in Equation 45 in are corrected (see Appendix Appendix 2).\n\nThe wave function associated to the DSN in the transformed system will be transformed inversely to that of the original system in order to facilitate full study in the original system. This is our basic strategy. Thus, we evaluate the DSN in the original system from\n\n$ψ s , n , m ( q 1 , q 2 ,t)= Û A Û B ψ s , n , m B ( q 1 , q 2 ,t).$\n(49)\n\nUsing the unitary operators given in Equations 7 and 16, we derive\n\n$ψ s , n 1 , n 2 ( q 1 , q 2 , t ) = μ 1 μ 2 ω 1 ω 2 ℏ 2 Π 2 4 1 2 n 1 + n 2 n 1 ! n 2 ! ( h b 1 G b 1 ) n 1 ( h b 2 G b 2 ) n 2 h a 1 G a 1 h a 2 G a 2 e βt / 2 × exp i ℏ e βt / 2 [ p 1 p ( t ) Q 1 + p 2 p ( t ) Q 2 ] H n 1 μ 1 ω 1 ℏ h a 1 2 h b 1 G c 1 [ e βt / 2 Q 1 − q 1 p ( t ) − q 1 c ( t ) ] × H n 2 μ 2 ω 2 ℏ h a 2 2 h b 2 G c 2 [ e βt / 2 Q 2 − q 2 p ( t ) − q 2 c ( t ) ] × exp − ∑ j = 1 2 μ j 2 ℏ ω j h aj [ G dj cos ( ω j t ) + i sin ( ω j t ) ] + iβ 2 [ e βt / 2 Q j − q jp ( t ) ] 2 × exp ∑ j = 1 2 μ j ω j 2 ℏ h aj 2 [ e βt / 2 Q j − q jp ( t ) ] G dj q jc ( 0 ) + i p jc ( 0 ) ω j μ j − q jc 2 ( 0 ) G dj cos ( ω j t ) × exp − ∑ j = 1 2 i p jc 2 ( 0 ) sin ( ω j t ) 2 μ j ω j h aj ℏ + i q jc ( 0 ) p jc ( 0 ) ℏ 1 2 − i G dj sin ( ω j t ) h aj × exp − i ℏ ∫ 0 t [ L 1 p ( t ′ ) + L 2 p ( t ′ ) ] d t ′ .$\n(50)\n\nThis is the full expression of the time evolution of wave functions for the DSN. If we let r→0, the squeezing effects disappear, and consequently, the system becomes DN. Of course the above equation reduces, in this limit, to that of the DN.\n\nTo see the time behavior of this state, we take a sinusoidal signal as a power source, which is represented as\n\n$E(t)= E 0 cos(Ωt+δ).$\n(51)\n\nThen, the solution of Equations 19 and 20 is given by\n\n(52)\n(53)\n(54)\n(55)\n\nwhere\n\n$M j (t)= C 1 C 2 4 4 E 0 e βt / 2 β 4 + 16 ( Ω 2 − ω j 2 ) 2 + 8 β 2 ( Ω 2 + ω j 2 ) .$\n(56)\n\nThe probability densities $\| ψ s , n 1 , n 2 ( q 1 , q 2 ,t) \| 2$ are plotted in Figures 2 and 3 as a function of q1 and t under this circumstance. As time goes by, the overall probability densities gradually converge to the origin where q1=0 due to the dissipation of energy caused by the existence of resistances in the circuit. If there are no resistances in the circuit, the probability densities no longer converge with time. An electronic system in general loses energy by the resistances, and the lost energy changes to thermal energy. Actually, Figure 2 belongs to DN due to the condition r1=r2=0 supposed in it. The wave function used in Figure 2a is not displaced and is consequently the same as that of the number state. Figure 2b is distorted by the effect of displacement. From Figure 2c,d, you can see that the exertion of a sinusoidal power source gives additional distortion. The frequency of $E(t)$ is relatively large for Figure 2c whereas it is small for Figure 2d.\n\nYou can see the effects of squeezing from Figure 3. The probability densities in the DSN are more significantly distorted than those of the DN. We can see from Figure 3b,c that the time behavior of probability densities is highly affected by external power source. If there is no power source in the circuit, the displacement of charge, specified with an initial condition, may gradually disappear according to its dissipation induced by resistances in the circuit. This is the same as that interpreted from the DN and exactly coincides with classical analysis of the system.\n\nWhile various means and technologies to generate squeezed and/or displaced light are developed in the context of quantum optics after the seminal work of Slusher et al. for observing squeezed light in the mid 1980s, (displaced) squeezed number state with sufficient degree of squeezing for charges and currents in a circuit quantum electrodynamics is first realized not long ago by Marthaler et al. as far as we know. The circuit they designed not only undergoes sufficiently low dissipation but its potential energy also contains a positive quartic term that leads to achieving strong squeezing. Another method to squeeze quantum states of mechanical oscillation of charge carriers in a circuit is to use the technique of back-action evasion [33, 34] that is originally devised in order to measure one of two arbitrary conjugate quadratures with high precision beyond the standard quantum limit.\n\nThough it is out of the scope of this work, the superpositions of any two DSNs may also be interesting topics to study, thanks to their nonclassical features that have no classical analogues. The quantum properties such as quadrature squeezing, quantum number distribution, purity, and the Mandel Q parameter for the superposition of two DSNs out of phase with respect to each other are studied in the literatures (see, for example, ).\n\n### Quantum fluctuations\n\nNow let us see the quantum fluctuations and uncertainty relations for charges and currents in the DSN for the original system. It is well known that quantum energy and any physical observables are temporarily changed due to their quantum fluctuations. The theoretical study for the origin and background physics of quantum fluctuations have been performed in by introducing stochastic and microcanonical quantizations.\n\nIf we consider the method of consecutive unitary transformation, the expectation value for an arbitrary operator $O ̂ j$ in the original system can be evaluated from\n\n$〈 ψ s , n 1 , n 2 ( t ) \| O ̂ j \| ψ s , n 1 , n 2 ( t ) 〉 = 〈 ψ n 1 , n 2 B ( 0 ) \| Ŝ j ‡ D ̂ j ‡ T ̂ Bj ‡ Û B ‡ Û A ‡ O ̂ j Û A Û B T ̂ Bj D ̂ j Ŝ j \| ψ n 1 , n 2 B ( 0 ) 〉 .$\n(57)\n\nUsing this relation, the expectation value of charges $q ̂ j$ and currents $p ̂ j$ is derived to be\n\n$〈 ψ s , n 1 , n 2 ( t ) \| q ̂ 1 \| ψ s , n 1 , n 2 ( t ) 〉 = C 1 C 2 4 e − βt / 2 [ Y 1 ( t ) cos φ + Y 2 ( t ) sin φ ] ,$\n(58)\n$〈 ψ s , n 1 , n 2 ( t ) \| q ̂ 2 \| ψ s , n 1 , n 2 ( t ) 〉 = C 2 C 1 4 e − βt / 2 [ − Y 1 ( t ) sin φ + Y 2 ( t ) cos φ ] ,$\n(59)\n$〈 ψ s , n 1 , n 2 ( t ) \| p ̂ 1 \| ψ s , n 1 , n 2 ( t ) 〉 = − C 2 C 1 4 e βt / 2 [ Y 1 ( t ) cos φ + Y 2 ( t ) sin φ ] ,$\n(60)\n$〈 ψ s , n 1 , n 2 ( t ) \| p ̂ 2 \| ψ s , n 1 , n 2 ( t ) 〉 = C 1 C 2 4 e βt / 2 [ Y 1 ( t ) sin φ − Y 2 ( t ) cos φ ] ,$\n(61)\n\nwhere\n\n$Y j (t)= ℏ 2 μ j ω j ( α j e − i ω j t + α j ∗ e i ω j t )+ q jp (t),$\n(62)\n$Y j ( t ) = μ j ℏ 2 ω j α j e − i ω j t ( β / 2 + i ω j ) + α j ∗ e i ω j t ( β / 2 − i ω j ) − p jp ( t ) .$\n(63)\n\nThe expectation value of square of $q ̂ j$ and $p ̂ j$ can also be obtained form the same method, and we listed them in Appendix Appendix 3. In fact, Equations 58 and 59 are the same as the classically predicted amount of charges qcl,1 and qcl,2 in C1 and C2 in the original system, respectively. If we consider that α j are given by Equation 36, qcl,1 and qcl,2 can be rewritten, after a little evaluation, in the form\n\n$q cl , 1 = C 1 C 2 4 e − βt / 2 { [ q 1 c ( t ) + q 1 p ( t ) ] cos φ + [ q 2 c ( t ) + q 2 p ( t ) ] sin φ } ,$\n(64)\n$q cl , 2 = C 2 C 1 4 e − βt / 2 { [ q 2 c ( t ) + q 2 p ( t ) ] cos φ − [ q 1 c ( t ) + q 1 p ( t ) ] sin φ } .$\n(65)\n\nWe illustrated qcl,1 and qcl,2 in Figure 4 as a function of time. To understand the time behavior of these quantities, it may be worth to recall that complementary functions, q j c (t), and particular solutions, q j p (t), are not associated to the original system but to the firstly transformed system. We can also easily confirm from similar evaluation that the time behavior of canonical conjugate currents pcl,j are represented in terms of q j c (t), p j c (t), and p j p (t) (see Appendix Appendix 4).\n\nThe definition of quantum fluctuations for any quantum operator $O ̂ j$ in the DSN is given by\n\n(66)\n\nUsing this, we obtain the fluctuations of charges and currents as\n\n(67)\n(68)\n(69)\n(70)\n\nAs we have seen before, the expectation values associated to charges and currents are represented in terms of complementary functions, q j c (t) and p j c (t), and particular solutions q j p (t) and p j p (t). The amplitude of complementary functions is determined from the strength of displacements, whereas the particular solutions are determined by the power source $E(t)$ (see Equations 19 and 20). However, all of the fluctuations do not involve such solutions. This means that the displacement and the electric power source do not affect to the fluctuations of charges and currents.\n\nThe uncertainty products $( Δ q ̂ j ) s ( Δ p ̂ j ) s$ between charges and their conjugate currents can be easily identified by means of Equations 67 to 70. For the case of the DN that are given from the limit r1=r2→0, we have F1=F2=0 and $F j = ω j 2 + β 2 /4$. Then, the uncertainty products become\n\n(71)\n(72)\n\nThese are the same as the uncertainty products in the number states and are always larger than $ℏ/2$, preserving the uncertainty principle. Thus, we can conclude that the uncertainty products in the DN are the same as those of the ordinary number states. Evidently, the uncertainty principle is inherent in quantum mechanical context described by canonical variables. The results, Equations 71 and 72 with n1=n2=0, are exactly the same as Equations 29 and 30 of , respectively. Moreover, for R1=R2=R3→0 (i.e., β→0), the above two equations reduce to Equations 52 and 53 in , which are evaluated in ordinary number state. Hence, this work includes all the results of both (no power source) and (no resistances) as special cases. The fluctuations and uncertainty product in the DN and in the DSN are plotted in Figure 5. We can adjust the uncertainty (or fluctuation) of a quadrature to be small at the expense of broadening that of another quadrature, or vice versa. The uncertainty $( Δ q ̂ 1 ) s$ in the case of this figure is larger than $( Δ q ̂ 1 ) c$, while $( Δ p ̂ 1 ) s$ is smaller than $( Δ p ̂ 1 ) c$ due to the squeezing effect. Therefore, it is relatively difficult for us to know the precise value of charge q1, while we can find out the conjugate current p1 more precisely. However, the relevant uncertainty product in the DSN is nearly unaltered from that in the DN.\n\n## Conclusions\n\nIn summary, the time evolution of the DSN for the two-dimensional electronic circuit composed of nanoscale elements and driven by a power source is investigated using unitary transformation method. Two steps of the unitary transformation are executed: We removed the cross term involving $p ̂ 1 p ̂ 2$ in the original Hamiltonian from the first step, and the linear terms represented in terms of $q ̂ j E(t)$ in the firstly transformed Hamiltonian are eliminated by second unitary transformation.\n\nWe can see from Equation 6 that the original Hamiltonian is time-dependent. When treating a time-dependent Hamiltonian system dynamically, one usually employs classical solutions of the equation of motion for a given system (or for a system similar to a given system) [6, 7]. We also introduced such classical solutions in Equations 19 to 20 and in Equations 47 to 48. Among them, particular solutions q j p and p j p are important in developing quantum theory of the system involving external power source since they are crucial factors that lead the transformed Hamiltonian to be simple so that we can easily treat it.\n\nSince the transformed system is just the same as the one that consists of two independent simple harmonic oscillators, provided that we can neglect the trivial terms $L jp (t)$ in the transformed Hamiltonian, we easily identified the complete quantum solutions in the DSN in the transformed system. We also obtained the wave functions of the DSN in the original system via the technique of inverse transformation, as shown in Equation 50. If we regard the fact that the probability does not reflect the phase of a wave function, the overall phase of these states is relatively unimportant for many cases. However, in some applications such as the computation of expectation values using generating or characteristic functions given in , the exact knowledge of overall phase is crucial. For r1=r2=0, the wave function in the DSN exactly reduces to that of the DN.\n\nWe analyzed the probability densities in the DN and in the DSN from Figures 2 and 3, respectively, with the choice of sinusoidal signal source. The probability densities in the DN given in Figure 2b,c,d oscillate with time. Moreover, their time behaviors are more or less distorted. The probability density, however, does not oscillate when there are no displacement and no signal of power source (see Figure 2a). The probability densities in the DSN are distorted much more significantly than those of the DN.\n\nThe time behavior of probability densities of quantum states, both the DN and the DSN, is highly affected by external driving power source. When there is no external power source($E(t)$=0), the displacement of charges, specified with a certain initial condition, gradually disappears as time goes by like a classical state.\n\nThe fluctuations and uncertainty products of charges and currents are derived in the DSN, and it is shown that their value is independent of the size of the particular solutions q j p (t) and p j p (t). From this, together with the fact that q j p (t) and p j p (t) are determined by the characteristics of $E(t)$, it is clear that the electric power source does not affect on the fluctuation of canonical variables. If we ignore the time dependence of F j (t) and $F j (t)$, $( Δ q ̂ j ) s$ decrease exponentially with time, whereas $( Δ p ̂ j ) s$ increase exponentially.\n\nFrom Equations 64 and 65, we can see that the time behavior of q j is determined by two factors: One is displacement and the other is the signal of power source. For better understanding of this, recall that the amplitude of complementary functions gives displacement of the system, and the particular solutions are closely related to external driving force (i.e., in this case, the power source).\n\nIn this paper, we did not consider thermal effects for the system. The thermal effects, as well as dissipation, may be worth to be considered in the studies of quantum fluctuations of electronic circuits with nanosize elements because the practical circuits are always working in thermal states with the presence of damping. It may therefore be a good theme to investigate DSNs with thermalization as a next task, and we plan to investigate it in the near future.\n\n## Appendix 1\n\n### The eighth formula of 7.374 in \n\n$∫ − ∞ ∞ e − ( x − y ) 2 H n ( ax ) dx = Π 1 / 2 ( 1 − a 2 ) n / 2 H n ay ( 1 − a 2 ) 1 / 2 .$\n(73)\n\n## Appendix 2\n\n### Correction of Equation 45 of \n\nThe second line of Equation 45 of needs to be corrected as\n\n$exp ⋯ + x ( x 0 F 2 + i p 0 ) cos t B − ⋯ → exp ⋯ + x x 0 F 2 + i p 0 B − ⋯ .$\n(74)\n\nBesides, among various functions that appeared in Equation 45 of , $F 3$ (Equation 23) and A (Equation 46) should be altered as\n\n$F 3 = cosh r + e − iϕ sin ϕ sinh r cosh r + e iϕ sin ϕ sinh r → F 3 = cosh r + e − iϕ sinh r cosh r + e iϕ sinh r ,$\n(75)\n$A = 1 − 2 i sin t F 4 2 B B − 2 i sin t / F 4 2 B → A = 1 − 2 i sin t F 4 2 B .$\n(76)\n\nFor the convenience of comparison, we provide a list of correspondences between our notations and the notations used in :\n\n$G aj ⇔ F 1 , G bj ⇔ F 3 , G cj ⇔ ( F 4 ) 2 , G dj ⇔ F 2 , h aj ⇔ B , h bj ⇔ A.$\n(77)\n\n## Appendix 3\n\n### Expectation value of $q ̂ j 2$ and $p ̂ j 2$\n\nAccording to the rule, Equation 57, for evaluating expectation values, we also have the expectation value of square of charges and currents as\n\n(78)\n(79)\n(80)\n(81)\n\nwhere\n\n$F j (t)=cosh(2 r j )+sinh(2 r j )cos( ϕ j −2 ω j t),$\n(82)\n$F j ( t ) = ( β 2 / 4 + ω j 2 ) cosh ( 2 r j ) + [ ( β 2 / 4 − ω j 2 ) cos ( ϕ j − 2 ω j t ) − β ω j sin ( ϕ j − 2 ω j t ) ] sinh ( 2 r j ) .$\n(83)\n\n## Appendix 4\n\n### Classical currents\n\nThrough the same vein as that of the calculation of qcl,1 and qcl,2 given in Equations 64 and 65, we can evaluate classical currents pcl,1 and pcl,2 from their quantum expectation value given in Equations 60 and 61. Thus, we have\n\n$p cl , 1 = C 2 C 1 4 e βt / 2 { [ p 1 c ( t ) − μ 1 β q 1 c ( t ) / 2 + p 1 p ( t ) ] cos φ + [ p 2 c ( t ) − μ 2 β q 2 c ( t ) / 2 + p 2 p ( t ) ] sin φ } ,$\n(84)\n$p cl , 2 = C 1 C 2 4 e βt / 2 { [ p 2 c ( t ) − μ 2 β q 2 c ( t ) / 2 + p 2 p ( t ) ] cos φ − [ p 1 c ( t ) − μ 1 β q 1 c ( t ) / 2 + p 1 p ( t ) ] sin φ } .$\n(85)\n\n## Abbreviations\n\nDN:\n\ndisplaced number state.\n\n## References\n\n1. 1.\n\nSohn LL, Kouwenhoven LP, Schön G: Mesoscopic Electron Transport. Kluwer: Dordrecht; 1997.\n\n2. 2.\n\nAndo T, Arakawa Y, Furuya K, Komiyama S, Nakashima H: Mesoscopic Physics and Electronics. Springer: Berlin; 1998.\n\n3. 3.\n\nLouisell WH: Quantum Statistical Properties of Radiation. New York: Wiley; 1973.\n\n4. 4.\n\nZhang S, Choi JR, Um CI, Yeon KH: Quantum uncertainties of mesoscopic inductance-resistance coupled circuit. 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Phys Lett A 1997, 229: 135–143. [Equation 45] [Equation 45] 10.1016/S0375-9601(97)00183-7\n\n31. 31.\n\nSlusher RE, Hollberg LW, Yurke B, Mertz JC, Valley JF: Observation of squeezed states generated by four-wave mixing in an optical cavity. Phys Rev Lett 1985, 55: 2409–2412. 10.1103/PhysRevLett.55.2409\n\n32. 32.\n\nMarthaler M, Schön G, Shnirman A: Photon-number squeezing in circuit quantum electrodynamics. Phys Rev Lett 2008, 101: 147001.\n\n33. 33.\n\nSzorkovszky A, Doherty AC, Harris GI, Bowen WP: Mechanical squeezing via parametric amplification and weak measurement. Phys Rev Lett 2011, 107: 213603.\n\n34. 34.\n\nClerk AA, Marquardt F, Jacobs K: Back-action evasion and squeezing of a mechanical resonator using a cavity detector. New J Phys 2008, 10: 095010. 10.1088/1367-2630/10/9/095010\n\n35. 35.\n\nObada A-SF, Abd Al-Kader GM: Superpositions of squeezed displaced Fock states: properties and generation. 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Phys Lett A 2004, 324: 145–151. 10.1016/j.physleta.2004.02.074\n\n## Author information\n\nAuthors\n\n### Corresponding author\n\nCorrespondence to Jeong Ryeol Choi." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8972423,"math_prob":0.99133795,"size":30574,"snap":"2021-43-2021-49","text_gpt3_token_len":7123,"char_repetition_ratio":0.15155381,"word_repetition_ratio":0.040987033,"special_character_ratio":0.23448028,"punctuation_ratio":0.1266217,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99837285,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-05T18:22:41Z\",\"WARC-Record-ID\":\"<urn:uuid:570bb720-42d0-40e4-a4ae-fd9981dcd9ac>\",\"Content-Length\":\"1049576\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:331d3d20-7532-4b4d-afe3-8f537986c2e6>\",\"WARC-Concurrent-To\":\"<urn:uuid:7c237949-e9a8-4086-b428-1c039f02911f>\",\"WARC-IP-Address\":\"146.75.32.95\",\"WARC-Target-URI\":\"https://nanoscalereslett.springeropen.com/articles/10.1186/1556-276X-8-30\",\"WARC-Payload-Digest\":\"sha1:6BKKLKGBFWH3FGE62HP7MDO5YRC76ZSJ\",\"WARC-Block-Digest\":\"sha1:AVSW7ZTL7XFLVC544MIACX3V6SDEEORP\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363215.8_warc_CC-MAIN-20211205160950-20211205190950-00007.warc.gz\"}"}
https://springerplus.springeropen.com/articles/10.1186/s40064-016-2068-x	[ "# Dynamics of almost periodic Schoener’s competition model with time delays and impulses\n\n## Abstract\n\nIn real world, the ecological systems are usually perturbed by human exploitation activities such as planting and harvesting and so on. In order to obtain a more accurate description for such phenomenon, the impulsive differential equations play an important role. This paper is concerned with a kind of almost periodic Schoener’s competition model with pure-delays and impulsive effects. By using the comparison theorem and the Lyapunov functional method of the impulsive differential equations, some sufficient conditions are obtained for the persistence and existence of a unique uniformly asymptotically stable positive almost periodic solution in a class of impulsive Schoener’s competition model with pure-delays. The main results in this paper improve some results in recent years. And the method used in this paper provides a possible and effective method to study the permanence and existence of a unique uniformly asymptotically stable positive almost periodic solution of the models with impulsive perturbations in biological populations. Finally, an example and numerical simulations are given to illustrate the feasibility and effectiveness of our main results.\n\n## Background\n\nOne of the most interest topics in mathematics ecology concerns the uniform persistence, almost periodic oscillations and global stability of ecology system. It is well known that a lot of Lotka–Volterra competitive systems have been discussed. Chen (1988) proposed and discussed a more practical competitive model as follows:\n\n\\begin{aligned} \\left\\{ \\begin{array}{lll} \\dot{x}_{1}(t)=x_{1}(t)\\bigg [\\frac{a_{10}(t)}{x_{1}(t)+m_{1}(t)} -a_{11}(t)x_{1}(t)-a_{12}(t)x_{2}(t)-c_{1}(t)\\bigg ],\\\\ \\dot{x}_{2}(t)=x_{2}(t)\\bigg [\\frac{a_{20}(t)}{x_{2}(t)+m_{2}(t)}-a_{21}(t)x_{1}(t) -a_{22}(t)x_{2}(t)-c_{2}(t)\\bigg ]. \\end{array}\\qquad \\right. \\end{aligned}\n(1)\n\nIn biological terms, $$x_1$$ and $$x_2$$ stand for the spatial densities of each species and $$c_1$$ and $$c_2$$ are their respective net death rate. The coefficients $$a_{11}$$ and $$a_{22}$$ are intra-specific competition rates whereas $$a_{12}$$ and $$a_{21}$$ are inter-specific competition rates. The asymptotic behaviors of the solution to the system (1) have been studied in Liu et al. (2006).\n\nIt is well known that the assumption of almost periodicity of the coefficients in systems is a way of incorporating the time-dependent variability of the environment, especially when the various components of the environment are periodic with not necessary commensurate periods (e.g., climate change, food supplies, mating habits and harvesting). For this reason, the assumption of almost periodicity is more realistic, more important and more general when we consider the effects of the environmental factors. In recent years, there are many scholars concerning with the Schoener’s competition system. Topics such as existence, uniqueness and global attractivity of positive periodic solutions or almost periodic solutions of the system were extensively investigated, and many excellent results have been derived (see Xue et al. 2015; Tian et al. 2015; Zhang et al. 2015; Liu and Xu 2004; Liu et al. 2006; Li and Yang 2009; Gan and Lin 2012; Wu et al. 2009; Zhang et al. 2012 and the references cited therein).\n\nOn the other hand, many evolution processes are characterized by the fact that at certain moments of time they experience a change of state abruptly. These processes are subject to short-term perturbations whose duration is negligible in comparison with the duration of the process. Consequently, it is natural to assume that these perturbations act instantaneously, that is, in the form of impulses. Thus impulsive differential equations, that is, differential equations involving impulse effects, appear as a natural description of observed evolution phenomena of real world problems. The ecological systems are often deeply perturbed by human exploitation activities such as planting and harvesting and so on, which makes them unsuitable to be considered continually. In recent years, the impulsive problems in ecological systems have been intensively investigated (see Lakshmikantham et al. 1989; Stamov 2012; Samoilenko and Perestyuk 1995; Bainov and Simeonov 1993; Jin et al. 2005; Stamov 2009; Liu and Chen 2007; He et al. 2010; Zhang et al. 2014; Zhang and Li 2011 for more detail). For instance, Zhang et al. (2012) studied the following almost periodic Schoener’s competition model with pure-delays and impulsive effects:\n\n\\begin{aligned} \\left\\{ \\begin{array}{lll} \\dot{x}_{1}(t)=x_{1}(t)\\bigg [\\frac{a_{10}(t)}{x_{1}(t-\\tau _{10})+m_{1}(t)} -a_{11}(t)x_{1}(t-\\tau _{11})-a_{12}(t)x_{2}(t-\\tau _{12})-c_{1}(t)\\bigg ],\\\\ \\dot{x}_{2}(t)=x_{2}(t)\\bigg [\\frac{a_{20}(t)}{x_{2}(t-\\tau _{20})+m_{2}(t)}-a_{21}(t)x_{1}(t-\\tau _{21}) -a_{22}(t)x_{2}(t-\\tau _{22})-c_{2}(t)\\bigg ],\\quad t\\ne \\tau _k,\\\\ \\Delta x_1(\\tau _k)=h_{1k}x_1(\\tau _k),\\\\ \\Delta x_2(\\tau _k)=h_{2k}x_2(\\tau _k),\\quad k\\in {\\mathbb {Z}}^+:=\\{0,1,\\ldots \\}, \\end{array} \\right. \\end{aligned}\n(2)\n\nwhere $$x_{1}(t)$$, $$x_{2}(t)$$ are population densities of species $$x_{1}$$, $$x_{2}$$ at time t, respectively, $$\\tau _{ij}$$ are all nonnegative integers, $$a_{ij}$$, $$m_{i}$$ and $$c_{i}$$, are all positive almost periodic functions, $$i=1,2$$, $$j=0,1,2$$, $$h_{1k}, h_{2k}\\ge 0$$ are almost periodic sequences, $$0=\\tau _0<\\tau _1< \\tau _2< \\cdots< \\tau _k< \\tau _{k+1} < \\cdots$$, are impulse points with $$\\lim _{k\\rightarrow +\\infty }\\tau _k=+\\infty$$, and the set of sequences $$\\{\\tau _{k}^{j}\\}, \\tau _{k}^{j} =\\tau _{k+j}-\\tau _{k}, k\\in {\\mathbb {Z}}^+, j\\in {\\mathbb {Z}}$$ is uniformly almost periodic (see Definition 1 in “Preliminaries” section).\n\nThe permanence and almost periodicity of all species in multispecies community are very important in biological populations. In recent years, the permanence and almost periodic solution of the models in biological populations has been studied by many authors (see Zhang 2013, 2014; Du and Lv 2013; Meng and Chen 2006; Lin and Chen 2009; Zhou et al. 2013; Zhang et al. 2014; Xia et al. 2008; Xia 2013; Liao and Zhang 2012; Liao and Xu 2014 and the references cited therein). In these cases, the comparison theorems and the Liapunov functional method of differential equations have been applied to obtain the permanence and almost periodic solutions. However, we find few papers referring to the discontinuous systems (i.e., impulsive systems).\n\nRecently, there are some scholars concerning with the permanence and almost periodic solution of the impulsive models in biological populations, see Zhang et al. (2012, 2014), He et al. (2010). For example, He et al. (2010) considered the following impulsive differential equation model of plankton allelopathy\n\n\\begin{aligned} \\left\\{ \\begin{array}{lll} \\dot{x}_{1}(t)=x_{1}(t)\\bigg [r_1(t) -a_{11}(t)x_{1}(t)-a_{12}(t)x_{1}(t)-b_{1}(t)x_{1}(t)x_2(t)\\bigg ],\\\\ \\dot{x}_{2}(t)=x_{2}(t)\\bigg [r_1(t) -a_{21}(t)x_{1}(t)-a_{22}(t)x_{2}(t)-b_{2}(t)x_{1}(t)x_2(t)\\bigg ],\\quad t\\ne t_k,\\\\ \\Delta x_1(t_k)=h_{1k}x_1(t_k),\\\\ \\Delta x_2(t_k)=h_{2k}x_2(t_k),\\quad k\\in {\\mathbb {Z}}^+. \\end{array} \\right. \\end{aligned}\n(3)\n\nBy using the relation between the solutions of impulsive system and the corresponding non-impulsive system, the authors transformed impulsive system (3) into a continuous system. Further, by means of the comparison theorems and the Liapunov functional method of differential equations without impulses, the authors obtained some sufficient conditions ensuring the existence of a unique uniformly asymptotically stable positive almost periodic solution of system (3).\n\nNext, by using a similar method as that in He et al. (2010), the authors in Zhang et al. (2012, 2014) studied the permanence and almost periodic solution of system (2) and the following impulsive multispecies mutualism system:\n\n\\begin{aligned} \\left\\{ \\begin{array}{ll} \\dot{x}_{i}(t)=x_{i}(t)\\bigg [r_i(t) -a_{i}(t)x_{i}(t-\\tau _i(t))+\\sum \\nolimits _{j=1,j\\ne i}^nb_{ij}(t)\\frac{x_j(t)}{1+x_j(t)}\\bigg ],\\quad t\\ne t_{k},\\\\ \\Delta x_i(t_k)=h_{ik}x_i(t_k),\\quad k\\in {\\mathbb {Z}}^+,\\quad i=1,2,\\ldots ,n. \\end{array} \\right. \\end{aligned}\n(4)\n\nIt is worthwhile to note that the main results of paper (Zhang et al. 2012, 2014; He et al. 2010) indicates the impulsive coefficients $$h_{ik}$$ in system (2)–(4) satisfying the following condition:\n\n(F) :\n\n$$H_{i}(t):=\\prod _{0<t_{k}<t}(1+h_{ik})$$, $$i=1,2$$ or $$i=1,2,\\ldots ,n\\,$$ (for system (4)), $$k\\in {\\mathbb {Z}}^+$$ is almost periodic function and there exist positive constants $$H_{i}^{u}$$ and $$H_{i}^{l}$$ such that $$H_{i}^{l}\\le H_{i}(t)\\le H_{i}^{u}$$.\n\n### Remark 1\n\nObviously, condition (F) is too harsh. For example, if the impulse coefficient $$h_{ik}\\equiv 0.3\\,(i=1,2)$$ in system (2)–(4), then (F) is invalid. Therefore, the main results in papers (Zhang et al. 2012, 2014; He et al. 2010) are difficult to apply to more extensive model with impulsive effects.\n\nIn Zhang et al. (2012, 2014), He et al. (2010), although the authors considered the impulsive system, but still used the research method of continuous systems. Stimulated by this, by means of the comparison theorem and the Lyapunov functional method of the impulsive differential equations (Lakshmikantham et al. 1989; Stamov 2012), the main purpose of this paper is to establish some sufficient conditions which guarantee the permanence and existence of a unique uniformly asymptotically stable positive almost periodic solution of system (2). The main results obtained in this paper remove the harsh condition (F) and provide a possible and effective method to study the permanence and existence of a unique uniformly asymptotically stable positive almost periodic solution of the models with impulsive perturbations in biological populations.\n\nLet $${\\mathbb {R}}$$ and $${\\mathbb {Z}}$$ denote the sets of real numbers and integers integers, respectively. Related to a continuous function f, we use the following notations:\n\n$$f^l=\\inf _{s\\in {\\mathbb {R}}}f(s),\\quad f^u=\\sup _{s\\in {\\mathbb {R}}}f(s).$$\n\nThe organization of this paper is as follows. In “Preliminaries” section, we give some basic definitions and necessary lemmas which will be used in later sections. In “Permanence” section, by using the comparison theorem of the impulsive differential equations (Lakshmikantham et al. 1989), we give the permanence of system (2). In “Almost periodic solution” section, we study the existence of a unique uniformly asymptotically stable positive almost periodic solution of system (2) by applying the Lyapunov method of the impulsive differential equations (Stamov 2012). Finally, an example and numerical simulations are given to illustrate that our results are feasible.\n\n## Preliminaries\n\nNow, let us state the following definitions and lemmas, which will be useful in proving our main result.\n\nBy $${\\mathbb {I}}$$, $${\\mathbb {I}}=\\{\\{\\tau _{k}\\}\\in {\\mathbb {R}}: \\tau _{k}<\\tau _{k+1}$$, $$k\\in {\\mathbb {Z}}$$, $$\\lim _{k\\rightarrow \\pm \\infty }\\tau _{k}=\\pm \\infty \\}$$, we denote the set of all sequences that are unbounded and strictly increasing. Let $$\\Omega \\subset {\\mathbb {R}}$$, $$\\Omega \\ne \\emptyset$$, $$\\tau =\\max \\{2\\tau _{ij}, i=1,2, j=0,1,2\\}$$, $$\\xi _{0}\\in {\\mathbb {R}}$$, introduce the following notations:\n\n$$PC(\\xi _{0})$$ is the space of all functions $$\\phi :[\\xi _{0}-\\tau ,\\xi _{0}]\\rightarrow \\Omega$$ having points of discontinuity at $$\\mu _{1},\\mu _{2},\\ldots \\in [\\xi _{0}-\\tau ,\\xi _{0}]$$ of the first kind and left continuous at these points.\n\nFor $$J\\subset {\\mathbb {R}}$$, $$PC(J, {\\mathbb {R}})$$ is the space of all piecewise continuous functions from J to $${\\mathbb {R}}$$ with points of discontinuity of the first kind $$\\tau _{k}$$, at which it is left continuous.\n\nLet $$\\phi _{1}, \\phi _{2}\\in PC(0)$$. Denote by $$x_{i}(t)=x_{i}(t;0,\\phi _{i})$$, $$x_{i}\\in \\Omega$$, $$i=1,2$$ the solution of system (2) satisfying the initial conditions\n\n$$0\\le x_{i}(s;0,\\phi _{i})=\\phi _{i}(s)<+\\infty ,\\quad s\\in [-\\tau ,0],\\quad x_{i}(0+0;0,\\phi _{i})=\\phi _{i}(0)>0.$$\n(5)\n\nBy the basic theories of impulsive differential equations with delay in Stamov (2012), system (2) and (5) has a unique solution. Since the solution of system (2) and (5) is a piecewise continuous function with points of discontinuity of the first kind $$\\tau _{k}$$, $$k\\in {\\mathbb {Z}}$$ we adopt the following definitions for almost periodicity.\n\n### Definition 1\n\n(Stamov 2012) The integer number p is said to be an $$\\epsilon$$-almost period of $$\\{\\tau _k\\}$$, if for $$k\\in {\\mathbb {Z}}$$, $$\|\\tau _{k+p}-\\tau _k\|<\\epsilon .$$\n\n### Definition 2\n\n(Stamov 2012) The set of sequences $$\\{\\tau _{k}^{j}\\}, \\tau _{k}^{j} =\\tau _{k+j}-\\tau _{k}, k\\in {\\mathbb {Z}}, j\\in {\\mathbb {Z}}, \\{\\tau _{k}\\}\\in {\\mathbb {I}}$$ is said to be uniformly almost periodic if for arbitrary $$\\epsilon >0$$ there exists a relatively dense set of $$\\epsilon$$-almost periods common for any sequences.\n\n### Definition 3\n\n(Stamov 2012) The function $$\\varphi \\in PC({\\mathbb {R}},{\\mathbb {R}})$$ is said to be almost periodic, if the following hold:\n\n1. (1)\n\nThe set of sequences $$\\{\\tau _{k}^{j}\\}, \\tau _{k}^{j}=\\tau _{k+j}- \\tau _{k}, k\\in {\\mathbb {Z}}, j\\in {\\mathbb {Z}}, \\{\\tau _{k}\\}\\in {\\mathbb {I}}$$ is uniformly almost periodic.\n\n2. (2)\n\nFor any $$\\epsilon >0$$ there exists a real number $$\\delta >0$$ such that if the points $$t'$$ and $$t''$$ belong to one and the same interval of continuity of $$\\varphi (t)$$ and satisfy the inequality $$\|t'-t''\|<\\delta$$, then $$\|\\varphi (t')-\\varphi (t'')\|<\\epsilon$$.\n\n3. (3)\n\nFor any $$\\epsilon >0$$ there exists a relatively dense set T such that if $$\\eta \\in T$$ , then $$\|\\varphi (t+\\eta )-\\varphi (t)\|<\\epsilon$$ for all $$t\\in {\\mathbb {R}}$$ satisfying the condition $$\|t-\\tau _{k}\|>\\epsilon$$, $$k\\in {\\mathbb {Z}}$$. The elements of T are called $$\\epsilon$$-almost periods.\n\n### Lemma 1\n\n(Stamov 2012) Let $$\\{\\tau _{k}\\}\\in {\\mathbb {I}}$$ Then there exists a positive integer. A such that on each interval of length 1, we have no more than A elements of the sequence $$\\{\\tau _{k}\\}$$ , i.e.,\n\n$$i(s,t)\\le A(t-s)+A,$$\n\nwhere i(st) is the number of the points $$\\tau _{k}$$ in the interval (st).\n\nTheoretically, one can investigate the existence, uniqueness and stability of almost periodic solution for functional differential equations by using Lyapunov functional as follows (Stamov 2012, $$\\mathrm {P}_{109}$$):\n\nLet $${\\mathbb {R}}^{n}$$ be the n-dimensional Euclidean space with elements $$x=(x_{1},\\ldots ,x_{n})^{T}$$ and norm $$\|x\|_{0}=\\sum _{i=1}^{n}\|x_{i}\|$$, $$C=C([-\\tau ,0],{\\mathbb {R}}^{n})$$, $${\\mathbb {B}}\\in {\\mathbb {R}}^+$$. Denote $$C_{\\mathbb {B}}=\\{\\varphi \\in C:\\Vert \\varphi \\Vert <{\\mathbb {B}}\\}$$, with $$\\Vert \\varphi \\Vert =\\sup _{s\\in [-\\tau ,0]}\|\\varphi (s)\|_{0}$$.\n\nConsider the system of impulsive differential equations with delay:\n\n\\begin{aligned} \\left\\{ \\begin{array}{lll} \\dot{x}(t)=f(t,x_t),\\quad t\\ne \\tau _k,\\\\ \\Delta x(\\tau _k)=I_{k}(x(\\tau _k)), \\end{array} \\right. \\end{aligned}\n(6)\n\nwhere $$t\\in {\\mathbb {R}}$$, $$\\{\\tau _k\\}\\in {\\mathbb {I}}$$, $$f(t,\\varphi )$$ is continuous in $$(t,\\varphi )\\in {\\mathbb {R}}\\times C_{\\mathbb {B}}$$ and almost periodic in t uniformly for $$\\varphi \\in C_{\\mathbb {B}}$$, $$\\forall \\rho >0$$, $$\\exists M(\\rho )>0$$ such that $$\|f(t,\\varphi )\|\\le M(\\rho )$$ as $$t\\in {\\mathbb {R}}$$, $$\\varphi \\in C_{\\mathbb {\\rho }}$$, while $$x_{t}\\in C_{\\mathbb {B}}$$ is defined as $$x_{t}(s)=x(t+s)$$ for $$s\\in [-\\tau ,0]$$, $$I_k: D\\rightarrow {\\mathbb {R}}^n$$, $$k\\in {\\mathbb {Z}}$$, D is an open set in $${\\mathbb {R}}^n$$.\n\nIntroduce the following conditions:\n\n$$(C_1)$$ :\n\nThe sequence $$\\{I_k(x)\\}$$, $$k\\in {\\mathbb {Z}}$$ is almost periodic uniformly with respect to $$x\\in D$$.\n\n### Lemma 2\n\n(Stamov 2012, $${\\mathrm {P}}_{109}$$) Suppose that there exists a Lyapunov functional $$V(t,\\phi ,\\psi )$$ defined on $${\\mathbb {R}}\\times C_{{\\mathbb {B}}}\\times C_{\\mathbb {B}}$$ satisfying the following conditions:\n\n1. (1)\n\n$$u(\\Vert \\phi -\\psi \\Vert )\\le V(t,\\phi ,\\psi )\\le v(\\Vert \\phi -\\psi \\Vert )$$, where $$u, v\\in {\\mathcal {P}}$$ with $${\\mathcal {P}}=\\{u:{\\mathbb {R}}^{+}\\rightarrow {\\mathbb {R}}^{+}\|u$$ is continuous increasing function and $$u(s)\\rightarrow 0$$ as $$s\\rightarrow 0\\}$$.\n\n2. (2)\n\n$$\|V(t,\\bar{\\phi },\\bar{\\psi })-V(t,\\hat{\\phi },\\hat{\\psi })\|\\le {\\mathrm {L}}(\\Vert \\bar{\\phi }-\\hat{\\phi }\\Vert +\\Vert \\bar{\\psi }-\\hat{\\psi }\\Vert )$$, where $${\\mathrm {L}}>0$$ is a constant.\n\n3. (3)\n\nFor $$t=\\tau _k$$, $$V(t^+,\\phi +I_k(\\phi ),\\psi +I_k(\\psi ))\\le V(t,\\phi ,\\psi )$$; For $$t\\ne \\tau _k$$, $$\\dot{V}_{(2.2)}(t,\\phi ,\\psi )\\le -\\gamma V(t,\\phi ,\\psi )$$, $$\\forall k\\in {\\mathbb {Z}}$$, where $$\\gamma >0$$ is a constant.\n\nMoreover, one assumes that system (6) has a solution that remains in a compact set $$S\\subset D$$. Then system (6) has a unique almost periodic solution which is uniformly asymptotically stable.\n\n### Remark 2\n\nFrom the proof of Lemma 2, it is not difficult to prove that condition (1) of Lemma 2 can be replaced by the following condition:\n\n1. (1)′\n\n$$u(\|\\phi (0)-\\psi (0)\|_{0})\\le V(t,\\phi ,\\psi )\\le v(\\Vert \\phi -\\psi \\Vert )$$, where $$u, v\\in {\\mathcal {P}}$$ with $${\\mathcal {P}}=\\{u:{\\mathbb {R}}^{+}\\rightarrow {\\mathbb {R}}^{+}\|u$$ is continuous increasing function and $$u(s)\\rightarrow 0$$ as $$s\\rightarrow 0\\}$$.\n\n## Permanence\n\nIn this section, we establish a permanence result for system (2).\n\n### Lemma 3\n\n(Lakshmikantham et al. 1989) Assume that $$x\\in PC({\\mathbb {R}})$$ with points of discontinuity at $$t=\\tau _k$$ and is left continuous at $$t=\\tau _k$$ for $$k\\in {\\mathbb {Z}}^+$$, and\n\n$$\\left\\{ \\begin{array}{lll} \\dot{x}(t)\\le f(t,x(t)), &\\quad t\\ne \\tau _k,\\\\ x(\\tau _k^+)\\le I_{k}(x(\\tau _k)), &\\quad k\\in {\\mathbb {Z}}^+, \\end{array} \\right.$$\n(7)\n\nwhere $$f\\in C({\\mathbb {R}}\\times {\\mathbb {R}},{\\mathbb {R}})$$, $$I_k\\in C({\\mathbb {R}},{\\mathbb {R}})$$ and $$I_k(x)$$ is nondecreasing in x for $$k\\in {\\mathbb {Z}}^+$$. Let $$u^(t)$$ be the maximal solution of the scalar impulsive differential equation\n\n$$\\left\\{ \\begin{array}{lll} \\dot{u}(t)=f(t,u(t)),&\\quad t\\ne \\tau _k,\\\\ u(\\tau _k^+)=I_{k}(u(\\tau _k))\\ge 0, &\\quad k\\in {\\mathbb {Z}}^+,\\\\ u(t_0^+)=u_0 \\end{array} \\right.$$\n(8)\n\nexisting on $$[t_0,\\infty )$$. Then $$x(t_0^+)\\le u_0$$ implies $$x(t)\\le u^(t)$$ for $$t\\ge t_0$$.\n\n### Remark 3\n\nIf the inequalities (7) in Lemma 3 is reversed and $$u_(t)$$ is the minimal solution of system (8) existing on $$[t_0,\\infty )$$, then $$x(t_0^+)\\ge u_0$$ implies $$x(t)\\ge u_(t)$$ for $$t\\ge t_0$$.\n\nFor arbitrary $$a, b>0$$, $$h_k\\ge 0$$, we give the following notations:\n\n\\begin{aligned} \\xi: &= \\ln \\sup _{k\\in {\\mathbb {Z}}}\\frac{1}{1+h_k},\\quad \\alpha :=a-\\xi A,\\quad \\theta :=\\inf _{k\\in {\\mathbb {Z}}}\\tau _k^1,\\quad \\eta :=\\inf _{k\\in {\\mathbb {Z}}}\\bigg \\{\\prod _{j=0}^1\\frac{1}{1+h_{j+k}},1\\bigg \\},\\\\ W(t,s) &=\\left\\{ \\begin{array}{ll} e^{-a(t-s)}, &{}\\tau _{k-1}<s<t<\\tau _{k}; \\\\ \\prod \\nolimits _{j=m}^{k+1}\\frac{1}{1+h_j} e^{-a(t-s)}, \\qquad &{}\\tau _{m-1}<s\\le \\tau _{m}<\\tau _{k}<t\\le \\tau _{k+1}, \\end{array} \\right. \\end{aligned}\n\nwhere A is defined as that in Lemma 1, $$\\tau _k^1=\\tau _{k+1}-\\tau _k$$ is defined as that in Definition 2.\n\n### Lemma 4\n\nAssume that $$a, b>0$$, $$h_k\\ge 0$$, then the following impulsive logistic equation\n\n\\begin{aligned} \\left\\{ \\begin{array}{lll} \\dot{x}(t)=x(t)\\big [ a-bx(t)\\big ],\\quad t\\ne \\tau _k,\\\\ \\Delta x(\\tau _k)=h_{k}x(\\tau _k),\\quad k\\in {\\mathbb {Z}}^+ \\end{array} \\right. \\end{aligned}\n(9)\n\nhas a unique globally asymptotically stable positive almost periodic solution $$x^$$ , which can be expressed as follows:\n\n$$\\frac{\\alpha }{e^{\\xi A}b}\\le x^(t)=\\bigg [b\\int _{-\\infty }^{t}W(t,s)\\,{\\mathrm {d}}s\\bigg ]^{-1}\\le \\frac{a}{\\eta b(1-e^{-a\\theta })}.$$\n(10)\n\n### Proof\n\nLet $$u=\\frac{1}{x}$$, then system (9) changes to\n\n$$\\left\\{ \\begin{array}{lll} \\frac{{\\mathrm {d}}u(t)}{{\\mathrm {d}}t}= -au(t)+b,&\\quad t\\ne \\tau _k,\\\\ \\Delta u(\\tau _k)=-\\frac{h_k}{1+h_{k}}u(\\tau _k),&\\quad k\\in {\\mathbb {Z}}^+. \\end{array} \\right.$$\n(11)\n\nTogether with the system (11) we consider the linear system\n\n$$\\left\\{ \\begin{array}{lll} \\frac{{\\mathrm {d}}u(t)}{{\\mathrm {d}}t}= -au(t),&\\quad t\\ne \\tau _k,\\\\ \\Delta u(\\tau _k)=-\\frac{h_k}{1+h_{k}}u(\\tau _k),&\\quad k\\in {\\mathbb {Z}}^+. \\end{array} \\right.$$\n(12)\n\nNow let us consider the equation\n\n$$\\frac{{\\mathrm {d}}u(t)}{{\\mathrm {d}}t}= -au(t),\\quad \\tau _{k-1}< t\\le \\tau _{k}$$\n\nand its solution\n\n$$u(t)=u(s)e^{-a(t-s)},\\quad \\tau _{k-1}<s<t\\le \\tau _{k}.$$\n\nThen from Stamov (2012), the Cauchy matrix of the linear system (12) is\n\n$$W(t,s)=\\left\\{ \\begin{array}{ll} e^{-a(t-s)}, &{}\\tau _{k-1}<s<t<\\tau _{k}; \\\\ \\prod \\nolimits _{j=m}^{k+1}\\frac{1}{1+h_j} e^{-a(t-s)}, \\quad &{}\\tau _{m-1}<s\\le \\tau _{m}<\\tau _{k}<t\\le \\tau _{k+1} \\end{array} \\right.$$\n\nand the solution of system (12) is in the form\n\n$$u(t;t_{0};u(t_{0}))=W(t,t_{0}) u(t_{0}),\\quad t_{0}\\in {\\mathbb {R}}.$$\n\nTherefore, system (11) has a solution\n\n$$u(t;t_{0};u(t_{0}))=W(t,t_0)u(t_0)+b\\int _{t_0}^{t}W(t,s)\\,{\\mathrm {d}}s.$$\n\nLetting $$t_0\\rightarrow -\\infty$$ in the above equation ($$W(t,t_0)u(t_0)\\rightarrow 0$$), then by Stamov (2012) we have\n\n$$u(t)=b\\int _{-\\infty }^{t}W(t,s)\\,{\\mathrm {d}}s$$\n\nis a solution of system (11) and is almost periodic. Then system (9) has a almost periodic solution $$x^(t)$$ which can be expressed by (10). By Lemma 1, we have from (10) that\n\n\\begin{aligned} x^(t) \\ge \\bigg [b\\int _{-\\infty }^{t}e^{i(s,t)\\xi }e^{-a(t-s)}\\,{\\mathrm {d}}s\\bigg ]^{-1} \\ge \\bigg [b\\int _{-\\infty }^{t}e^{\\xi A}e^{-\\alpha (t-s)}\\,{\\mathrm {d}}s\\bigg ]^{-1} =\\frac{\\alpha }{e^{\\xi A}b}. \\end{aligned}\n\nOn the other hand,\n\n\\begin{aligned} x^(t) \\le \\bigg [b\\int _{t-\\theta }^{t}W(t,s)\\,{\\mathrm {d}}s\\bigg ]^{-1} \\le \\bigg [b\\int _{t-\\theta }^{t}\\eta e^{-a(t-s)}\\,{\\mathrm {d}}s\\bigg ]^{-1} =\\frac{a}{\\eta b(1-e^{-a\\theta })}. \\end{aligned}\n\nNext, we shall prove that the uniqueness and stability of $$x^(t)$$ of system (9). Suppose that x(t) is another positive solution of system (9). Define a function\n\n$$V(t)=\|\\ln x^(t)-\\ln x(t)\|, \\quad \\forall t\\in {\\mathbb {R}}.$$\n\nFor $$t\\ne \\tau _k$$, $$k\\in {\\mathbb {Z}}^+$$, calculating the upper right derivative of V(t) along the solution of system (9), we have\n\n$$D^+V(t)={\\mathrm {sgn}}[x^(t)-x(t)]\\bigg [\\frac{\\dot{x}^(t)}{x^(t)}-\\frac{\\dot{x}(t)}{x(t)}\\bigg ] =-b\|x^(t)-x(t)\|.$$\n(13)\n\nFor $$t=\\tau _k$$, $$k\\in {\\mathbb {Z}}^+$$, we have\n\n\\begin{aligned} V(\\tau _k^+) &= \|\\ln x^(\\tau _k^+)-\\ln x(\\tau _k^+)\|\\\\&= \\left\| \\ln \\frac{(1+h_{k})x^(\\tau _k)}{(1+h_{k})x(\\tau _k)}\\right\| =\|\\ln x^(\\tau _k)-\\ln x(\\tau _k)\|=V(\\tau _k). \\end{aligned}\n\nTherefore, V is non-increasing. Integrating (13) from 0 to t leads to\n\n$$V(t)+b\\int _{0}^t\|x(s)-x^(s)\|\\,{\\mathrm {d}}s \\le V(0)<+\\infty , \\quad \\forall t\\ge 0,$$\n\nthat is,\n\n$$\\int _{0}^{+\\infty }\|x(s)-x^(s)\|\\,{\\mathrm {d}}s<+\\infty ,$$\n\nwhich implies that\n\n$$\\lim _{s\\rightarrow +\\infty }\|x(s)-x^(s)\|=0.$$\n\nThus, the almost periodic solution of system (9) is globally asymptotically stable. This completes the proof.\n\n### Lemma 5\n\nAssume that $$a, b>0$$, $$h_k\\ge 0$$, then every solution x of the following system with delay\n\n$$\\left\\{ \\begin{array}{lll} \\dot{x}(t)\\le x(t)\\big [ a-bx(t-\\tau )\\big ],\\quad t\\ne \\tau _k,\\\\ \\Delta x(\\tau _k)= h_{k}x(\\tau _k),\\quad k\\in {\\mathbb {Z}}^+ \\end{array} \\right.$$\n(14)\n\nsatisfies\n\n$$\\lim \\sup _{t\\rightarrow \\infty }x(t)\\le M:=\\frac{a}{\\eta B(1-e^{-a\\theta })},$$\n\nwhere $$B=\\inf _{t\\in {\\mathbb {R}}}b\\prod _{\\tau _k\\in [t-\\tau ,t)}(1+h_k)^{-1}e^{-a\\tau }$$, $$\\theta :=\\inf \\nolimits_{k\\in {\\mathbb {Z}}}\\tau_k^1$$ and $$\\eta :=\\inf \\nolimits _{k\\in {\\mathbb {Z}}}\\prod _{j=1}^2\\frac{1}{1+h_{j+k}}$$.\n\n### Proof\n\nFrom system (14), we have\n\n\\begin{aligned} \\left\\{ \\begin{array}{lll} \\dot{x}(t)\\le ax(t),\\quad t\\ne \\tau _k,\\\\ \\Delta x(\\tau _k)=h_{k}x(\\tau _k),\\quad k\\in {\\mathbb {Z}}^+ \\end{array} \\right. \\end{aligned}\n\nis equivalent to\n\n\\begin{aligned} \\left\\{ \\begin{array}{lll} \\frac{{\\mathrm {d}}}{{\\mathrm {dt}}}[x(t)e^{-at}]\\le 0,\\quad t\\ne \\tau _k,\\\\ \\Delta x(\\tau _k)=h_{k}x(\\tau _k),\\quad k\\in {\\mathbb {Z}}^+. \\end{array} \\right. \\end{aligned}\n(15)\n\nFor some $$t\\in [0,+\\infty )$$ and $$t\\ne \\tau _k$$, $$k\\in {\\mathbb {Z}}^+$$, consider interval $$[t-\\tau ,t)$$. Assume that $$\\tau _1<\\tau _2<\\cdots <\\tau _j$$ are the impulse points in $$[t-\\tau ,t)$$. Integrating the first inequality of system (15) from $$t-\\tau$$ to $$\\tau _1$$ leads to\n\n$$x(\\tau _1)e^{-a\\tau _1}\\le x(t-\\tau )e^{-a(t-\\tau )}.$$\n\nIntegrating the first inequality of system (15) from $$\\tau _{1}$$ to $$\\tau _2$$ leads to\n\n$$x(\\tau _{2})e^{-a\\tau _{2}}\\le x(\\tau _{1}^+)e^{-a\\tau _{1}}=(1+h_{1})x(\\tau _{1})e^{-a\\tau _{1}} \\le (1+h_{1})x(t-\\tau )e^{-a(t-\\tau )}.$$\n\nIntegrating the first inequality of system (15) from $$\\tau _{2}$$ to $$\\tau _3$$ leads to\n\n$$x(\\tau _{3})e^{-a\\tau _{3}}\\le x(\\tau _{2}^+)e^{-a\\tau _{2}}=(1+h_{2})x(\\tau _{2})e^{-a\\tau _{2}} \\le (1+h_{1})(1+h_{2})x(t-\\tau )e^{-a(t-\\tau )}.$$\n\nRepeating the above process, integrating the first inequality of system (15) from $$\\tau _j$$ to t leads to\n\n$$x(t)e^{-at}\\le x(\\tau _j^+)e^{-a\\tau _j}=(1+h_j)x(\\tau _j)e^{-a\\tau _j}\\le \\prod _{\\tau _k\\in [t-\\tau ,t)}(1+h_k)x(t-\\tau )e^{-a(t-\\tau )}.$$\n\nThen\n\n$$x(t-\\tau )\\ge \\prod _{\\tau _k\\in [t-\\tau ,t)}(1+h_k)^{-1}e^{-a\\tau }x(t).$$\n(16)\n\nSubstituting (16) into system (14) leads to\n\n\\begin{aligned} \\left\\{ \\begin{array}{lll} \\dot{x}(t)\\le x(t)\\big [ a-Bx(t)\\big ],\\quad t\\ne \\tau _k,\\\\ \\Delta x(\\tau _k)=h_{k}x(\\tau _k),\\quad k\\in {\\mathbb {Z}}^+. \\end{array} \\right. \\end{aligned}\n\nConsider the auxiliary system\n\n\\begin{aligned} \\left\\{ \\begin{array}{lll} \\dot{z}(t)=z(t)\\big [ a-Bz(t) \\big ],\\quad t\\ne \\tau _k,\\\\ z(\\tau _k^+)=(1+h_{k})z(\\tau _k),\\quad k\\in {\\mathbb {Z}}^+. \\end{array} \\right. \\end{aligned}\n(17)\n\nBy Lemma 3, $$x(t)\\le z(t)$$, where z(t) is the solution of system (17) with $$z(0^+)=x(0^+)$$. By Lemma 4, system (17) has a unique globally asymptotically stable positive almost periodic solution $$z^$$ which can be expressed as follows:\n\n\\begin{aligned} z^(t)=\\bigg [B\\int _{-\\infty }^{t}W(t,s)\\,{\\mathrm {d}}s\\bigg ]^{-1} \\le \\bigg [B\\int _{t-\\theta }^{t}W(t,s)\\,{\\mathrm {d}}s\\bigg ]^{-1} \\le \\frac{a}{\\eta B(1-e^{-a\\theta })}:=M. \\end{aligned}\n\nThen for any constant $$\\epsilon >0$$, there exists $$T_1>0$$ such that $$x(t)\\le z(t)<z^(t)+\\epsilon \\le M+\\epsilon$$ for $$t>T_1$$. So\n\n$$\\lim \\sup _{t\\rightarrow \\infty }x(t)\\le M.$$\n\nThis completes the proof. $$\\square$$\n\n### Lemma 6\n\nAssume that $$a, b>0$$, $$h_k\\ge 0$$, then every solution x of the following system with delay\n\n\\begin{aligned} \\left\\{ \\begin{array}{lll} \\dot{x}(t)\\ge x(t)\\big [ a-bx(t-\\tau )\\big ],\\quad t\\ne \\tau _k,\\\\ \\Delta x(\\tau _k)=h_{k}x(\\tau _k),\\quad k\\in {\\mathbb {Z}}^+ \\end{array} \\right. \\end{aligned}\n(18)\n\nsatisfies\n\n$$\\lim \\inf _{t\\rightarrow \\infty }x(t)\\ge N:=\\frac{a-\\xi A}{e^{\\xi A}D},$$\n\nwhere $$D=\\sup _{t\\in {\\mathbb {R}}}b\\prod _{\\tau _k\\in [t-\\tau ,t)}(1+h_k)^{-1}e^{-(a-bM)\\tau }$$, $$\\xi :=\\ln \\sup \\nolimits _{k\\in {\\mathbb {Z}}}\\frac{1}{1+h_k}$$ and A is defined as that in Lemma 1.\n\n### Proof\n\nAccording to Lemma 5, there exist $$\\epsilon >0$$ and $$T_2>0$$ such that\n\n$$x(t)\\le M+\\epsilon \\quad \\text { for } t\\ge T_{2}.$$\n\nFrom system (18), we have\n\n\\begin{aligned} \\left\\{ \\begin{array}{lll} \\dot{x}(t)\\ge [a-b(M+\\epsilon )]x(t),\\quad t\\ne \\tau _k,\\quad t\\ge T_{2},\\\\ \\Delta x(\\tau _k)=h_{k}x(\\tau _k),\\quad k\\in {\\mathbb {Z}}^+ \\end{array} \\right. \\end{aligned}\n\nis equivalent to\n\n\\begin{aligned} \\left\\{ \\begin{array}{lll} \\frac{{\\mathrm {d}}}{\\mathrm {dt}}[x(t)e^{-[a-b(M+\\epsilon )]t}]\\ge 0,\\quad t\\ne \\tau _k\\quad t\\ge T_{2},\\\\ \\Delta x(\\tau _k)=h_{k}x(\\tau _k),\\quad k\\in {\\mathbb {Z}}^+. \\end{array} \\right. \\end{aligned}\n\nSimilar to the argument as that in (16), we have\n\n$$bx(t-\\tau )\\le b\\prod _{\\tau _k\\in [t-\\tau ,t)}(1+h_k)^{-1}e^{-[a-b(M+\\epsilon )]\\tau }x(t):=D_\\epsilon x(t),\\quad t\\ge T_{2}.$$\n(19)\n\nSubstituting (19) into system (18) leads to\n\n\\begin{aligned} \\left\\{ \\begin{array}{lll} \\dot{x}(t)\\ge x(t)\\big [ a-D_\\epsilon x(t)\\big ],\\quad t\\ne \\tau _k,\\quad t\\ge T_{2},\\\\ \\Delta x(\\tau _k)=h_{k}x(\\tau _k),\\quad k\\in {\\mathbb {Z}}^+. \\end{array} \\right. \\end{aligned}\n\nConsider the auxiliary system\n\n\\begin{aligned} \\left\\{ \\begin{array}{lll} \\dot{z}(t)=z(t)\\big [ a-D_\\epsilon z(t) \\big ],\\quad t\\ne \\tau _k,\\quad t\\ge T_{2},\\\\ z(\\tau _k^+)=(1+h_{k})z(\\tau _k),\\quad k\\in {\\mathbb {Z}}^+. \\end{array} \\right. \\end{aligned}\n(20)\n\nBy Remark 3, $$x(t)\\ge z(t)$$, where z(t) is the solution of system (20) with $$z(0^+)=x(0^+)$$. By Lemma 4, system (20) has a unique globally asymptotically stable positive almost periodic solution $$z^$$ which can be expressed as follows:\n\n\\begin{aligned} z^(t)=\\bigg [D_\\epsilon \\int _{-\\infty }^{t}W(t,s)\\,{\\mathrm {d}}s\\bigg ]^{-1} \\ge \\bigg [D_\\epsilon \\int _{t-\\theta }^{t}W(t,s)\\,{\\mathrm {d}}s\\bigg ]^{-1} \\ge \\frac{a-\\xi A}{e^{\\xi A}D_\\epsilon }. \\end{aligned}\n\nLetting $$\\epsilon \\rightarrow 0$$ in the above inequality leads to\n\n$$z^(t) \\ge \\frac{a-\\xi A}{e^{\\xi A}D}:=N.$$\n\nSimilar to the argument as that in Lemma 5, it follows that\n\n$$\\lim \\inf _{t\\rightarrow \\infty }x(t)\\ge N.$$\n\nThis completes the proof. $$\\square$$\n\n### Remark 4\n\nWhen $$h_{ik}(i=1,2)\\equiv 0$$ in systems (14) and (18), then Lemmas 5, 6 change to the corresponding results in Nakata and Muroya (2010). So Lemmas 5, 6 extend the corresponding result in Nakata and Muroya (2010).\n\nLet\n\n$$\\eta _i:=\\inf _{k\\in {\\mathbb {Z}}}\\prod _{j=0}^1\\frac{1}{1+h_{i(j+k)}},\\quad \\xi _i:=\\ln \\sup _{k\\in {\\mathbb {Z}}}\\frac{1}{1+h_{ik}},\\quad i=1,2.$$\n\n### Proposition 1\n\nEvery solution $$(x_1,x_2)^T$$ of system (2) satisfies\n\n$$\\lim \\sup \\limits _{t\\rightarrow \\infty }x_i(t)\\le M_i,\\quad i=1,2,$$\n\nwhere $$M_1$$ and $$M_2$$ are defined as that in (21) and (22), respectively.\n\n### Proof\n\nFrom system (1), we have\n\n\\begin{aligned} \\left\\{ \\begin{array}{lll} \\dot{x}_{1}(t)\\le x_{1}(t)\\big [r_1^u -a_{11}^lx_{1}(t-\\tau _{11})\\big ],\\quad t\\ne \\tau _k,\\\\ x_1(\\tau _k^+)=(1+h_{1k})x_1(\\tau _k),\\quad k\\in {\\mathbb {Z}}^+, \\end{array} \\right. \\end{aligned}\n\nwhere $$r_1^u:=\\sup _{t\\in {\\mathbb {R}}}\\big \|\\frac{a_{10}(t)}{m_{1}(t)}-c_{1}(t)\\big \|$$. By Lemma 5, we have\n\n$$\\lim \\sup _{t\\rightarrow \\infty }x_1(t)\\le M_1:=\\frac{r_1^u}{\\eta _1 B_1(1-e^{-r_1^u\\theta })},$$\n(21)\n\nwhere $$B_1=\\inf _{t\\in {\\mathbb {R}}}a_{11}^l\\prod _{\\tau _k\\in [t-\\tau _{11},t)}(1+h_{1k})^{-1}e^{-r_1^u\\tau _{11}}$$. Similarly, ones obtain that\n\n$$\\lim \\sup _{t\\rightarrow \\infty }x_2(t)\\le M_2:=\\frac{r_2^u}{\\eta _2 B_2(1-e^{-r_2^u\\theta })},$$\n(22)\n\nwhere $$r_2^u:=\\sup _{t\\in {\\mathbb {R}}}\\big \|\\frac{a_{20}(t)}{m_{2}(t)}-c_{2}(t)\\big \|$$, $$B_2=\\inf _{t\\in {\\mathbb {R}}}a_{22}^l\\prod _{\\tau _k\\in [t-\\tau _{22},t)}(1+h_{2k})^{-1}e^{-r_2^u\\tau _{22}}$$. This completes the proof. $$\\square$$\n\n### Proposition 2\n\nLet $$N_1$$ and $$N_2$$ are defined as that in (23) and (24), respectively. Then every solution $$(x_1,x_2)^T$$ of system (2) satisfies\n\n$$\\lim \\inf \\limits _{t\\rightarrow \\infty }x_i(t)\\ge N_i,\\quad i=1,2,$$\n\nif the following condition holds:\n\n$$(H_{1})$$ :\n\n$$r_1^l:=\\inf _{t\\in {\\mathbb {R}}}\\big [\\frac{a_{10}(t)}{M_1+m_{1}(t)} -a_{12}(t)M_2-c_{1}(t)\\big ]\\ge \\xi _1A$$, $$r_2^l:=\\inf _{t\\in {\\mathbb {R}}}\\big [\\frac{a_{20}(t)}{M_2+m_{2}(t)} -a_{21}(t)M_1-c_{2}(t)\\big ]\\ge \\xi _2A.$$\n\n### Proof\n\nAccording to Proposition 1 and $$(H_1)$$, there exist $$\\epsilon >0$$ and $$T_3>0$$ such that\n\n\\begin{aligned}&x_i(t)\\le M_i+\\epsilon \\quad \\text { for } t\\ge T_{3},\\quad i=1,2,\\\\&r_1^l(\\epsilon ):=\\inf _{t\\in {\\mathbb {R}}}\\bigg [\\frac{a_{10}(t)}{(M_1+\\epsilon )+m_{1}(t)} -a_{12}(t)(M_2+\\epsilon )-c_{1}(t)\\bigg ]\\ge \\xi _1A,\\\\&r_2^l(\\epsilon ):=\\inf _{t\\in {\\mathbb {R}}}\\bigg [\\frac{a_{20}(t)}{(M_2+\\epsilon )+m_{2}(t)} -a_{21}(t)(M_1+\\epsilon )-c_{2}(t)\\bigg ]\\ge \\xi _2A. \\end{aligned}\n\nFrom system (1), we have\n\n\\begin{aligned} \\left\\{ \\begin{array}{lll} \\dot{x}_{1}(t)\\ge x_{1}(t)\\big [r_{1}^l(\\epsilon ) -a_{11}^ux_{1}(t-\\tau _{11})\\big ],\\quad t\\ne \\tau _k, \\; t>T_3,\\\\ x_1(\\tau _k^+)=(1+h_{1k})x_1(\\tau _k),\\quad k\\in {\\mathbb {Z}}^+. \\end{array} \\right. \\end{aligned}\n\nBy Lemma 6 and the arbitrariness of $$\\epsilon$$, one has\n\n$$\\lim \\inf _{t\\rightarrow \\infty }x_1(t)\\ge N_1:=\\frac{r_{1}^l-\\xi _1 A}{e^{\\xi _1 A}D_1},$$\n(23)\n\nwhere $$D_1:=\\sup _{t\\in {\\mathbb {R}}}a_{11}^u\\prod _{\\tau _k\\in [t-\\tau _{11},t)}(1+h_{1k})^{-1}e^{-(r_1^l-a_{11}^uM_1)\\tau _{11}}$$. Similarly, we have\n\n$$\\lim \\inf _{t\\rightarrow \\infty }x_2(t)\\ge N_2:=\\frac{r_{2}^l-\\xi _2 A}{e^{\\xi _2 A}D_2},$$\n(24)\n\nwhere $$D_2:=\\sup _{t\\in {\\mathbb {R}}}a_{22}^u\\prod _{\\tau _k\\in [t-\\tau _{22},t)}(1+h_{2k})^{-1}e^{-(r_2^l-a_{22}^uM_2)\\tau _{22}}$$. This completes the proof. $$\\square$$\n\n### Remark 5\n\nIn view of $$(H_1)$$ in Proposition 2, the values of impulse coefficients $$h_{ik}\\,(i=1,2)$$ and the number of the impulse points $$\\tau _k$$ in each interval of length 1 have negative effect on the permanence of system (2).\n\nBy Propositions 1, 2, we have\n\n### Theorem 1\n\nAssume that $$(H_1)$$ holds, then system (2) is permanent.\n\n### Remark 6\n\nTheorem 1 is a permanence result of system (2) without (F). So Theorem 1 improves the corresponding result in Zhang et al. (2012). Further, Theorem 1 provides a possible and effective method to study the permanence of the models with impulsive perturbations and pure-delays in biological populations.\n\n### Remark 7\n\nFrom the proof of Propositions 1, 2, we know that under the conditions of Theorem 1, the set $$S=\\{(x_1,x_2)^T\\in {\\mathbb {R}}^2:N_i\\le x_i\\le M_i, i=1,2\\}$$ is an invariant set of system (2).\n\n## Almost periodic solution\n\nThe main result of this paper concerns the existence of a unique uniformly asymptotically stable positive almost periodic solution for system (2).\n\nFor convenience, we introduce some notations as follows:\n\n\\begin{aligned} \\alpha _{1} &= {} \\frac{a_{10}^{l}N_{1}}{\\big (M_{1}+m_{1}^{u}\\big )^{2}}+a_{11}^{l}N_{1} -\\bigg (\\frac{\\sqrt{\\tau _{10}}a_{10}^{u}M_{1}}{\\big (N_{1}+m_{1}^{l}\\big )^{2}}\\bigg )^{2} -\\frac{(\\tau _{10}+\\tau _{11})a_{10}^{u}a_{11}^{u}M_{1}^{2}}{\\big (N_{1}+m_{1}^{l}\\big )^{2}} -\\tau _{11}a_{11}^{u2}M_{1}^{2},\\\\ \\alpha _{2} &= \\frac{a_{20}^{l}N_{2}}{\\big (M_{2}+m_{2}^{u}\\big )^{2}}+a_{22}^{l}N_{2} -\\bigg [\\frac{\\sqrt{\\tau _{20}}a_{20}^{u}M_{2}}{\\big (N_{2}+m_{2}^{l}\\big )^{2}}\\bigg ]^{2} -\\frac{(\\tau _{20}+\\tau _{22})a_{20}^{u}a_{22}^{u}M_{2}^{2}}{\\big (N_{2}+m_{2}^{l}\\big )^{2}} -\\tau _{22}a_{22}^{u2}M_{2}^{2},\\\\ \\beta _{1} &= \\frac{\\tau _{20}a_{20}^{u}a_{21}^{u}M_{1}M_{2}}{\\big (N_{2}+m_{2}^{l}\\big )^{2}}+\\tau _{22}a_{21}^{u}a_{22}^{u}M_{1}M_{2} +a_{21}^{u}M_{1},\\\\ \\beta _{2} &= \\frac{\\tau _{10}a_{10}^{u}a_{12}^{u}M_{1}M_{2}}{\\big (N_{1}+m_{1}^{l}\\big )^{2}} +\\tau _{11}a_{11}^{u}a_{12}^{u}M_{1}M_{2}+a_{12}^{u}M_{2}, \\end{aligned}\n\nwhere $$M_1$$, $$M_2$$, $$N_1$$ and $$N_2$$ are defined as that in “Permanence” section.\n\n### Theorem 2\n\nAssume that $$(H_1)$$ holds. Suppose further that\n\n$$(H_2)$$ :\n\nThere exist two positive constants $$\\lambda _{1}$$ and $$\\lambda _{2}$$ such that $$\\lambda _{1}\\alpha _{1}>\\lambda _{2}\\beta _{1}$$ and $$\\lambda _{2}\\alpha _{2}>\\lambda _{1}\\beta _{2}$$.\n\nThen system (2) admits a unique uniformly asymptotically stable almost periodic solution.\n\n### Proof\n\nLet $$x_{i}(t)=e^{z_{i}(t)}$$, $$i=1, 2$$, then system (2) is transformed into\n\n\\begin{aligned} \\left\\{ \\begin{array}{lll} \\dot{z}_{1}(t)=\\frac{a_{10}(t)}{e^{z_{1}(t-\\tau _{10})}+m_{1}(t)} -a_{11}(t)e^{z_{1}(t-\\tau _{11})}-a_{12}(t)e^{z_{2}(t-\\tau _{12})}-c_{1}(t),\\\\ \\dot{z}_{2}(t)=\\frac{a_{20}(t)}{e^{z_{2}(t-\\tau _{20})}+m_{2}(t)}-a_{21}(t)e^{z_{1}(t-\\tau _{21})} -a_{22}(t)e^{z_{2}(t-\\tau _{22})}-c_{2}(t),\\quad t\\ne \\tau _k,\\\\ e^{z_1(\\tau _k^+)}=(1+h_{1k})e^{z_1(\\tau _k)},\\\\ e^{z_2(\\tau _k^+)}=(1+h_{2k})e^{z_2(\\tau _k)},\\quad k\\in {\\mathbb {Z}}^+. \\end{array} \\right. \\end{aligned}\n(25)\n\nSuppose that $$Z(t)=(z_{1}(t),z_{2}(t))^{T}$$ and $$Z^{}(t)=(z_{1}^{}(t),z_{2}^{}(t))^{T}$$ are any two solutions of system (25). Consider the product system of system (25)\n\n\\begin{aligned} \\left\\{ \\begin{array}{lll} \\dot{z}_{1}(t)=\\frac{a_{10}(t)}{e^{z_{1}(t-\\tau _{10})}+m_{1}(t)} -a_{11}(t)e^{z_{1}(t-\\tau _{11})}-a_{12}(t)e^{z_{2}(t-\\tau _{12})}-c_{1}(t),\\\\ \\dot{z}_{2}(t)=\\frac{a_{20}(t)}{e^{z_{2}(t-\\tau _{20})}+m_{2}(t)}-a_{21}(t)e^{z_{1}(t-\\tau _{21})} -a_{22}(t)e^{z_{2}(t-\\tau _{22})}-c_{2}(t),\\\\ \\dot{z}_{1}^(t)=\\frac{a_{10}(t)}{e^{z_{1}^(t-\\tau _{10})}+m_{1}(t)} -a_{11}(t)e^{z_{1}^(t-\\tau _{11})}-a_{12}(t)e^{z_{2}^(t-\\tau _{12})}-c_{1}(t),\\\\ \\dot{z}_{2}^(t)=\\frac{a_{20}(t)}{e^{z_{2}^(t-\\tau _{20})}+m_{2}(t)}-a_{21}(t)e^{z_{1}^(t-\\tau _{21})} -a_{22}(t)e^{z_{2}^(t-\\tau _{22})}-c_{2}(t),\\quad t\\ne \\tau _k,\\\\ e^{z_1(\\tau _k^+)}=(1+h_{1k})e^{z_1(\\tau _k)},\\\\ e^{z_2(\\tau _k^+)}=(1+h_{2k})e^{z_2(\\tau _k)},\\\\ e^{z_1^(\\tau _k^+)}=(1+h_{1k})e^{z_1^(\\tau _k)},\\\\ e^{z_2^(\\tau _k^+)}=(1+h_{2k})e^{z_2^(\\tau _k)},\\quad k\\in {\\mathbb {Z}}^+. \\end{array} \\right. \\end{aligned}\n(26)\n\nSet $$S_1=\\{\\phi =(z_{1t},z_{2t})^{T}\\in C([-\\tau ,0],{\\mathbb {R}}^{2}):\\ln N_i\\le z_{it}\\le \\ln M_i, t\\in {\\mathbb {R}}^{+}, i=1,2\\}$$, which is an invariant set of system (26) directly from Remark 7.\n\nConstruct a Lyapunov functional $$V(t)=V(t,\\phi ,\\psi )=V\\big (t,(z_{1t},z_{2t})^{T},(z_{1t}^{},z_{2t}^{})^{T}\\big )$$ defined on $${\\mathbb {R}}^{+}\\times S_1\\times S_1$$ as follows:\n\n$$V(t,\\phi ,\\psi )= V_{1}(t,\\phi ,\\psi )+V_{2}(t,\\phi ,\\psi )+V_{3}(t,\\phi ,\\psi )+V_{4}(t,\\phi ,\\psi ),$$\n\nwhere\n\n\\begin{aligned} V_{1}(t,\\phi ,\\psi ) &= \\lambda _{1}\|z_{1}(t)-z_{1}^{}(t)\|+\\lambda _{2}\|z_{2}(t)-z_{2}^{}(t)\|,\\\\ V_{2}(t,\\phi ,\\psi )&= \\lambda _{1}\\bigg (\\frac{a_{10}^{u}M_{1}}{\\big (N_{1}+m_{1}^{l}\\big )^{2}}\\bigg )^{2} \\int _{-2\\tau _{10}}^{-\\tau _{10}}\\int _{t+s}^{t}\\big \|z_{1}(r)-z_{1}^{}(r)\\big \|\\,{\\mathrm {d}}r\\,{\\mathrm {d}}s\\\\&\\quad+\\,\\lambda _{1}\\frac{a_{10}^{u}a_{11}^{u}M_{1}^{2}}{\\big (N_{1}+m_{1}^{l}\\big )^{2}} \\int _{-\\tau _{11}-\\tau _{10}}^{-\\tau _{11}}\\int _{t+s}^{t}\\big \|z_{1}(r)-z_{1}^{}(r)\\big \|\\,{\\mathrm {d}}r\\,{\\mathrm {d}}s\\\\&\\quad+\\,\\lambda _{1}\\frac{a_{10}^{u}a_{11}^{u}M_{1}^{2}}{\\big (N_{1}+m_{1}^{l}\\big )^{2}}\\int _{-\\tau _{10}-\\tau _{11}}^{-\\tau _{10}}\\int _{t+s}^{t}\\big \|z_{1}(r)-z_{1}^{}(r)\\big \|\\,{\\mathrm {d}}r\\,{\\mathrm {d}}s\\\\&\\quad+\\,\\lambda _{1}a_{11}^{u2}M_{1}^{2}\\int _{-2\\tau _{11}}^{-\\tau _{11}}\\int _{t+s}^{t}\\big \|z_{1}(r)-z_{1}^{}(r)\\big \|\\,{\\mathrm {d}}r\\,{\\mathrm {d}}s\\\\&\\quad+\\,\\lambda _{2}\\frac{a_{20}^{u}a_{21}^{u}M_{1}M_{2}}{\\big (N_{2}+m_{2}^{l}\\big )^{2}}\\int _{-\\tau _{21}-\\tau _{20}}^{-\\tau _{21}}\\int _{t+s}^{t}\\big \|z_{1}(r)-z_{1}^{}(r)\\big \|\\,{\\mathrm {d}}r\\,{\\mathrm {d}}s\\\\&\\quad+\\,\\lambda _{2}a_{21}^{u}a_{22}^{u}M_{1}M_{2}\\int _{-\\tau _{21}-\\tau _{22}}^{-\\tau _{21}}\\int _{t+s}^{t}\\big \|z_{1}(r)-z_{1}^{}(r)\\big \|\\,{\\mathrm {d}}r\\,{\\mathrm {d}}s,\\\\ V_{3}(t,\\phi ,\\psi )&= \\lambda _{1}\\frac{a_{10}^{u}a_{12}^{u}M_{1}M_{2}}{\\big (N_{1}+m_{1}^{l}\\big )^{2}} \\int _{-\\tau _{12}-\\tau _{10}}^{-\\tau _{12}}\\int _{t+s}^{t}\\big \|z_{2}(r)-z_{2}^{}(r)\\big \|\\,{\\mathrm {d}}r\\,{\\mathrm {d}}s\\\\&\\quad+\\,\\lambda _{1}a_{11}^{u}a_{12}^{u}M_{1}M_{2}\\int _{-\\tau _{12}-\\tau _{11}}^{-\\tau _{12}}\\int _{t+s}^{t}\\big \|z_{2}(r)-z_{2}^{}(r)\\big \|\\,{\\mathrm {d}}r\\,{\\mathrm {d}}s\\\\&\\quad+\\,\\lambda _{2}\\bigg (\\frac{a_{20}^{u}M_{2}}{\\big (N_{2}+m_{2}^{l}\\big )^{2}}\\bigg )^{2}\\int _{-2\\tau _{20}}^{-\\tau _{20}}\\int _{t+s}^{t}\\big \|z_{2}(r)-z_{2}^{}(r)\\big \|\\,{\\mathrm {d}}r\\,{\\mathrm {d}}s\\\\&\\quad+\\,\\lambda _{2}\\frac{a_{20}^{u}a_{22}^{u}M_{2}^{2}}{\\big (N_{2}+m_{2}^{l}\\big )^{2}}\\int _{-\\tau _{22}-\\tau _{20}}^{-\\tau _{22}}\\int _{t+s}^{t}\\big \|z_{2}(r)-z_{2}^{}(r)\\big \|\\,{\\mathrm {d}}r\\,{\\mathrm {d}}s\\\\&\\quad+\\,\\lambda _{2}\\frac{a_{20}^{u}a_{22}^{u}M_{2}^{2}}{\\big (N_{2}+m_{2}^{l}\\big )^{2}}\\int _{-\\tau _{20}-\\tau _{22}}^{-\\tau _{20}}\\int _{t+s}^{t}\\big \|z_{2}(r)-z_{2}^{}(r)\\big \|\\,{\\mathrm {d}}r\\,{\\mathrm {d}}s\\\\&\\quad+\\,\\lambda _{2}a_{22}^{u2}M_{2}^{2}\\int _{-2\\tau _{22}}^{-\\tau _{22}}\\int _{t+s}^{t}\\big \|z_{2}(r)-z_{2}^{}(r)\\big \|\\,{\\mathrm {d}}r\\,{\\mathrm {d}}s,\\\\ V_{4}(t,\\phi ,\\psi )&= \\lambda _{2}a_{21}^{u}M_{1}\\int _{t-\\tau _{21}}^{t}\\big \|z_{1}(s)-z_{1}^{}(s)\\big \|\\,{\\mathrm {d}}s +\\lambda _{1}a_{12}^{u}M_{2}\\int _{t-\\tau _{12}}^{t}\\big \|z_{2}(s)-z_{2}^{}(s)\\big \|\\,{\\mathrm {d}}s. \\end{aligned}\n\nBy the definitions of $$S_1$$ and V, there is some large enough positive constant K such that\n\n$$V(t,\\phi ,\\psi )\\le K.$$\n\nSimilar to the argument as that in Zhang et al. (2012), we have\n\n$$V(t,\\phi ,\\psi )\\ge \\underline{\\lambda }\|\\phi (0)-\\psi (0)\|_{0},$$\n(27)\n\nwhere $$\\underline{\\lambda }:=\\min \\{\\lambda _{1},\\lambda _{2}\\}$$,\n\n$$V(t,\\phi ,\\psi )\\le \\overline{\\lambda }\\Vert \\phi -\\psi \\Vert ,$$\n(28)\n\nwhere\n\n\\begin{aligned} \\overline{\\lambda }:&= \\lambda _{1}+\\lambda _{2}+2\\lambda _{1}\\tau ^{2}\\bigg (\\frac{a_{10}^{u}M_{1}}{\\big (N_{1}+m_{1}^{l}\\big )^{2}}\\bigg )^{2} +2\\lambda _{1}\\tau ^{2}\\frac{a_{10}^{u}a_{11}^{u}M_{1}^{2}}{\\big (N_{1}+m_{1}^{l}\\big )^{2}}\\\\&\\quad+\\,2\\lambda _{1}\\tau ^{2}\\frac{a_{10}^{u}a_{11}^{u}M_{1}^{2}}{\\big (N_{1}+m_{1}^{l}\\big )^{2}}+2\\lambda _{1}\\tau ^{2}a_{11}^{u2}M_{1}^{2} +2\\lambda _{2}\\tau ^{2}\\frac{a_{20}^{u}a_{21}^{u}M_{1}M_{2}}{\\big (N_{2}+m_{2}^{l}\\big )^{2}}\\\\&\\quad+\\,2\\lambda _{2}\\tau ^{2}a_{21}^{u}a_{22}^{u}M_{1}M_{2} +2\\lambda _{1}\\tau ^{2}\\frac{a_{10}^{u}a_{12}^{u}M_{1}M_{2}}{\\big (N_{1}+m_{1}^{l}\\big )^{2}} +2\\lambda _{1}\\tau ^{2}a_{11}^{u}a_{12}^{u}M_{1}M_{2}\\\\&\\quad+\\,2\\lambda _{2}\\tau ^{2}\\bigg (\\frac{a_{20}^{u}M_{2}}{\\big (N_{2}+m_{2}^{l}\\big )^{2}}\\bigg )^{2} +2\\lambda _{2}\\tau ^{2}\\frac{a_{20}^{u}a_{22}^{u}M_{2}}{\\big (N_{2}+m_{2}^{l}\\big )^{2}} +2\\lambda _{2}\\tau ^{2}\\frac{a_{20}^{u}a_{22}^{u}M_{2}^{2}}{\\big (N_{2}+m_{2}^{l}\\big )^{2}}\\\\&\\quad+\\,2\\lambda _{2}\\tau ^{2}a_{22}^{u2}M_{2}^{2} +\\lambda _{2}\\tau a_{21}^{u}M_{1} +\\lambda _{1}\\tau a_{12}^{u}M_{2}, \\end{aligned}\n\nand for $$\\forall \\,\\bar{\\phi }=(\\bar{z}_{1t},\\bar{z}_{2t})^{T}, \\bar{\\psi }=(\\bar{z}_{1t}^{},\\bar{z}_{2t}^{})^{T}, \\hat{\\phi }=(\\hat{z}_{1t},\\hat{z}_{2t})^{T}, \\hat{\\psi }=(\\hat{z}_{1t}^{},\\hat{z}_{2t}^{})^{T}\\in S_1$$, it follows that\n\n$$\\big \|V(t,\\bar{\\phi },\\bar{\\psi })-V(t,\\hat{\\phi },\\hat{\\psi })\\big \| \\le \\overline{\\lambda }(\\Vert \\bar{\\phi }-\\hat{\\phi }\\Vert +\\Vert \\bar{\\psi }-\\hat{\\psi }\\Vert ).$$\n\nSo condition (2) in Lemma 2 is satisfied. In view of (27)–(28), let $$u, v\\in C({\\mathbb {R}}^{+},\\,{\\mathbb {R}}^{+})$$, $$u(s)=\\underline{\\lambda }s$$, $$v(s)=\\overline{\\lambda }s$$, thus condition (1) $$'$$ in Remark 2 is satisfied.\n\nFrom article Zhang et al. (2012), for $$t\\ne \\tau _k$$, $$k\\in {\\mathbb {Z}}^+$$, calculating the upper right derivative of V along the solution of system (26), we have\n\n$$D^{+}V(t,\\phi ,\\psi )\\le -\\gamma V(t,\\phi ,\\psi ),$$\n(29)\n\nwhere $$\\gamma :=\\frac{\\chi \\underline{\\lambda }\|\\phi (0)-\\psi (0)\|_0}{K}$$, $$\\chi :=\\min \\{\\frac{\\Theta }{\\lambda _1},\\frac{\\Theta }{\\lambda _2}\\}$$ and $$\\Theta :=\\min \\{\\lambda _{1}\\alpha _{1}-\\lambda _{2}\\beta _{1},\\lambda _{2}\\alpha _{2}-\\lambda _{1}\\beta _{2}\\}$$.\n\nFor $$t=\\tau _k$$, $$k\\in {\\mathbb {Z}}^+$$, we have\n\n\\begin{aligned} V(\\tau _k^+,\\phi ,\\psi )&= V_{1}(\\tau _k^+,\\phi ,\\psi )+V_{2}(\\tau _k^+,\\phi ,\\psi )+V_{3}(\\tau _k^+,\\phi ,\\psi )+V_{4}(\\tau _k^+,\\phi ,\\psi ) \\\\&= V_{1}(\\tau _k^+,\\phi ,\\psi )+V_{2}(\\tau _k,\\phi ,\\psi )+V_{3}(\\tau _k,\\phi ,\\psi )+V_{4}(\\tau _k,\\phi ,\\psi ) \\\\&= \\sum _{i=1}^2\\lambda _{i}\|z_{i}(\\tau _k^+)-z_{i}^{}(\\tau _k^+)\|+V_{2}(\\tau _k,\\phi ,\\psi )+V_{3}(\\tau _k,\\phi ,\\psi )+V_{4}(\\tau _k,\\phi ,\\psi ) \\\\&= \\sum _{i=1}^2\\lambda _{i}\|z_{i}(\\tau _k)-z_{i}^{}(\\tau _k)\|+V_{2}(\\tau _k,\\phi ,\\psi )+V_{3}(\\tau _k,\\phi ,\\psi )+V_{4}(\\tau _k,\\phi ,\\psi ) \\\\&= V_{1}(\\tau _k,\\phi ,\\psi )+V_{2}(\\tau _k,\\phi ,\\psi )+V_{3}(\\tau _k,\\phi ,\\psi )+V_{4}(\\tau _k,\\phi ,\\psi ) \\\\&= V(\\tau _k,\\phi ,\\psi ). \\end{aligned}\n(30)\n\nIn view of (29)–(30), condition (3) in Lemma 2 is satisfied.\n\nBy Lemma 2, system (2) admits a unique uniformly asymptotically stable positive almost periodic solution $$(z_{1}(t),z_{2}(t))^{T}$$. This completes the proof. $$\\square$$\n\n### Remark 8\n\nWithout (F), system (2) also admits a unique uniformly asymptotically stable positive almost periodic solution. So Theorem 2 extends the corresponding result in Zhang et al. (2012). Further, Theorem 2 gives the sufficient conditions for the uniform asymptotical stability of a unique positive almost periodic solution of system (2). Therefore, Theorem 2 provides a possible method to study the existence, uniqueness and stability of positive almost periodic solution of the models with impulsive perturbations and pure-delays in biological populations.\n\n### Remark 9\n\nIn the last two decades, the method of constructing a Lyapunov functional has been extensively used in the study of stability of the deterministic models (Xue et al. 2015; Tian et al. 2015; Zhang et al. 2015; Liu and Xu 2004; Liu et al. 2006; Li and Yang 2009; Gan and Lin 2012; Wu et al. 2009; Zhang et al. 2012). However, there have been numerous relevant works using the Lyapunov functional method in stochastic systems, see Shang (2014, 2015, 2016). The methods used in this paper can be extended to study the permanence and existence of a unique uniformly asymptotically stable positive almost periodic solution of the stochastic models with impulsive perturbations in biological populations.\n\n## An example and numerical simulations\n\n### Example 1\n\nConsider the following Schoener’s competition model with pure-delays and impulsive effects:\n\n\\begin{aligned} \\left\\{ \\begin{array}{ll} \\dot{x}_{1}(t)=x_{1}(t)\\bigg \\{\\frac{1}{x_{1}(t-0.0001)+2}-a_{11}(t)x_{1}(t-0.0001)-a_{12}(t)x_{2}(t-0.0002)-c_{1}(t)\\bigg \\},\\\\ \\dot{x}_{2}(t)=x_{2}(t)\\bigg \\{\\frac{1}{x_{2}(t-0.0001)+2}-a_{21}(t)x_{1}(t-0.0002)-a_{22}(t)x_{2}(t-0.0001)-c_{2}(t)\\bigg \\},\\quad t\\ne \\tau _{k},\\\\ \\Delta x_{1}(\\tau _{k})=0.5x_{1}(\\tau _{k}),\\\\ \\Delta x_{2}(\\tau _{k})=0.4x_{2}(\\tau _{k}),\\quad \\{\\tau _{k}:k\\in {\\mathbb {Z}}\\}\\subset \\{5k:k\\in {\\mathbb {Z}}\\}, \\end{array} \\right. \\end{aligned}\n\nwhere $$a_{11}(t)=a_{22}(t)=0.35+0.05\\cos (\\sqrt{3}t)$$, $$a_{12}(t)=a_{21}(t)\\equiv 0.0001$$, $$c_{1}(t)=0.30005+0.00005\\sin (\\sqrt{2}t)$$, $$c_{2}(t)=0.30005+0.00005\\cos (\\sqrt{2}t)$$, $$t\\in {\\mathbb {R}}$$. Then the above system is permanent and has a unique uniformly asymptotically stable almost periodic solution.\n\n### Proof\n\nObviously, $$a_{10}^{u}=a_{20}^{u}=1$$, $$a_{10}^{l}=a_{20}^{l}=1$$, $$a_{11}^{u}=a_{22}^{u}=0.4$$, $$a_{11}^{l}=a_{22}^{l}=0.3$$, $$a_{12}^{u}=a_{21}^{u}=0.0001$$, $$a_{12}^{l}=a_{21}^{l}=0.0001$$, $$c_{1}^{u}=c_{2}^{u}=0.3001$$, $$c_{1}^{l}=c_{2}^{l}=0.3$$, $$\\theta =2$$, $$\\eta _1\\approx 0.4$$, $$\\eta _2=4$$, $$\\xi _1\\approx 0.3$$, $$\\xi _2\\approx 0.7$$, $$A=1$$. By calculation, we obtain that $$y_{1}^{}\\approx 0.7256$$, $$y_{2}^{}\\approx 0.5421$$, $$\\min \\{r_{1}^{l},r_{2}^{l}\\}\\ge 0.03>0$$, $$y_{1}=y_{2}\\ge 0.01$$. So $$(H_1)$$ holds. Further, we also get that\n\n$$\\alpha _{1}\\ge 0.03,\\quad \\alpha _{2}\\ge 0.02,\\quad \\max \\{\\beta _{1},\\beta _{2}\\}\\le 0.00007,$$\n\nwhich implies that $$(H_2)$$ is satisfied for $$\\lambda _{1}=\\lambda _{2}=1$$. It is easy to verify that $$(H_1)$$$$(H_2)$$ are satisfied and the result follows from Theorems 1 and 2 (see Figs. 1, 2, 3). This completes the proof. $$\\square$$\n\n### Remark 10\n\nIn Example 1, the impulse coefficients ($$h_{1k}=0.5$$ and $$h_{2k}=0.6$$) do not satisfy (F). So it is impossible to obtain the permanence and existence of a unique uniformly asymptotically stable positive almost periodic solution of system (1) by the result in Zhang et al. (2012). Therefore, the work in this paper improves the results in paper Zhang et al. (2012).\n\n## Conclusion\n\nBy using the comparison theorem and the Lyapunov method of the impulsive differential equations, sufficient conditions are obtained for the permanence and existence of a unique uniformly asymptotically stable positive almost periodic solution in a class of impulsive Schoener’s competition model with pure-delays. Proposition 2 and Theorem 2 imply that the values of impulse coefficients $$h_{ik}\\,(i=1,2,3)$$ and the number of the impulse points $$\\tau _k$$ in each interval of length 1 are harm for the permanence and existence of a unique uniformly asymptotically stable positive almost periodic solution for a class of impulsive Schoener’s competition model with pure-delays. The main results obtained in this paper are completely new and the method used in this paper provides a possible method to study the permanence and existence of a unique uniformly asymptotically stable positive almost periodic solution of the models with impulsive perturbations and pure-delays in biological populations.\n\n## References\n\n• Bainov DD, Simeonov PS (1993) Impulsive differential equations: periodic solutions and applications. Longman Scientific and Technical, New York\n\n• Chen LS (1988) Mathematical ecology modelling and research methods. Science Press, Beijing (in Chinese)\n\n• Du ZJ, Lv YS (2013) Permanence and almost periodic solution of a Lotka–Volterra model with mutual interference and time delays. Appl Math Model 37:1054–1068\n\n• Gan WZ, Lin ZG (2012) The asymptotic periodicity in a Schoener’s competitive model. Appl Math Model 36:989–996\n\n• He MX, Chen FD, Li Z (2010) Almost periodic solution of an impulsive differential equation model of plankton allelopathy. Nonlinear Anal Real World Appl 11:2296–2301\n\n• Jin Z, Han MA, Li GH (2005) The persistence in a Lotka–Volterra competition systems with impulsive. Chaos Solitons Fractals 24:1105–1117\n\n• Lakshmikantham V, Bainov DD, Simeonov PS (1989) Theory of impulsive differential equations. World Scientific, Singapore\n\n• Li XP, Yang WS (2009) Permanence of a discrete $$n$$-species schoener competition system with time delays and feedback controls. Adv Differ Equ 2009:1–10\n\n• Liao YZ, Zhang TW (2012) Almost periodic solutions of a discrete mutualism model with variable delays. Discrete Dyn Nat Soc 2012:1–27\n\n• Liao YZ, Xu LJ (2014) Almost periodic solution for a delayed Lotka-Volterra system on time scales. Adv Differ Equ 2014:1–19\n\n• Lin X, Chen FD (2009) Almost periodic solution for a Volterra model with mutual interference and Beddington–DeAngelis functional response. Appl Math Comput 214:548–556\n\n• Liu ZJ, Chen LS (2007) Periodic solution of a two-species competitive system with toxicant and birth pulse. Chaos Solitons Fractals 32:1703–1712\n\n• Liu QM, Xu R (2004) Periodic solutions of Schoener’s competitive model with delays. J Biomath 19:385–394 (in Chinese)\n\n• Liu QM, Xu R, Wang WG (2006) Global asymptotic stability of Schoener’s competitive model with time delays. J Biomath 21:147–152\n\n• Liu QM, Xu R, Wang WG (2006) Global asymptotic stability of Schoener’s competitive model with delays. J Biomath 21:147–152 (in Chinese)\n\n• Meng XZ, Chen LS (2006) Almost periodic solution of non-autonomous Lotka–Volterra predator-prey dispersal system with delays. J Theor Biol 243:562–574\n\n• Nakata Y, Muroya Y (2010) Permanence for nonautonomous Lotka–Volterra cooperative systems with delays. Nonlinear Anal Real World Appl 11:528–534\n\n• Samoilenko AM, Perestyuk NA (1995) Impulsive differential equations. World Scientific, Singapore\n\n• Shang YL (2014) Group pinning consensus under fixed and randomly switching topologies with acyclic partition. Netw Heterog Media 9:553–573\n\n• Shang YL (2015) Global stability of disease-free equilibria in a two-group SI model with feedback control. Nonlinear Anal Model Control 20:501–508\n\n• Shang YL (2016) Consensus seeking over Markovian switching networks with time-varying delays and uncertain topologies. Appl Math Comput 273:1234–1245\n\n• Stamov GT (2009) On the existence of almost periodic solutions for the impulsive Lasota–Wazewska model. Appl Math Lett 22:516–520\n\n• Stamov GT (2012) Almost periodic solutions of impulsive differential equations. In: Lecture notes in mathematics. Springer, Berlin\n\n• Tian BD, Chen N, Qiu YH (2015) Almost periodic solution for an epidemic prey-predator system with impulsive effects and multiple delays. Discrete Dyn Nat Soc 2015:1–15\n\n• Wu LP, Chen FD, Li Z (2009) Permanence and global attractivity of a discrete Schoener’s competition model with delays. Math Comput Model 49:1607–1617\n\n• Xia YH, Han MA, Huang ZK (2008) Global attractivity of an almost periodic $$N$$-species nonlinear ecological competitive model. J Math Anal Appl 337:144–168\n\n• Xia YH (2013) Theory and applications of periodic solutions and almost periodic solutions. Discrete Dyn Nat Soc 2013:1–2\n\n• Xue YL, Xie XD, Chen FD, Han RY (2015) Almost periodic solution of a discrete commensalism system. Discrete Dyn Nat Soc 2015:1–11\n\n• Zhang TW (2013) Multiplicity of positive almost periodic solutions in a delayed Hassell–Varleytype predator–prey model with harvesting on prey. Math Methods Appl Sci 37:686–697\n\n• Zhang TW (2014) Almost periodic oscillations in a generalized Mackey–Glass model of respiratory dynamics with several delays. Int J Biomath 7:1–22\n\n• Zhang TW, Li YK (2011) Positive periodic solutions for a generalized impulsive $$n$$-species Gilpin–Ayala competition system with continuously distributed delays on time scales. Int J Biomath 4:23–34\n\n• Zhang H, Li YQ, Jing B (2014) Global attractivity and almost periodic solution of a discrete mutualism model with delays. Math Methods Appl Sci 37:3013–3025\n\n• Zhang H, Li YQ, Jing B, Zhao WZ (2014) Global stability of almost periodic solution of multispecies mutualism system with time delays and impulsive effects. Appl Math Comput 232:1138–1150\n\n• Zhang TW, Li YK, Ye Y (2012) On the existence and stability of a unique almost periodic solution of Schoener’s competition model with pure-delays and impulsive effects. Commun Nonlinear Sci Numer Simul 17:1408–1422\n\n• Zhang H, Feng F, Jing B, Li YQ (2015) Almost periodic solution of a multispecies discrete mutualism system with feedback controls. Discrete Dyn Nat Soc 2015:1–14\n\n• Zhou H, Wang J, Zhou Z (2013) Positive almost periodic solution for impulsive Nicholson’s blowflies model with multiple linear harvesting terms. Math Methods Appl Sci 36:456–461\n\n## Authors' contributions\n\nCL, ZGG and ZYZ collectively carried out this study, collected data, and analyzed them. CL and ZGG wrote the manuscript. ZYZ helped to draft the manuscript. All authors read and approved the final manuscript.\n\n### Acknowledgements\n\nThis work are supported by the Natural Science Foundation of Shanxi Province under Grant 2011011002-3, the Project of University Educational Reform in Shanxi Province under Grant J2015118 and the Education Science Foundation of Shanxi Province under Grant GH-13118. The authors thank the referee for his/her careful reading of the original manuscript and valuable comments and suggestions that greatly improved the presentation of this work.\n\n### Competing interests\n\nThe authors declare that they have no competing interests.\n\n## Author information\n\nAuthors\n\n### Corresponding author\n\nCorrespondence to Zhiyu Zhang.\n\n## Rights and permissions", null, "" ]	[ null, "https://springerplus.springeropen.com/track/article/10.1186/s40064-016-2068-x", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8133757,"math_prob":0.9999627,"size":36842,"snap":"2022-27-2022-33","text_gpt3_token_len":11253,"char_repetition_ratio":0.16724578,"word_repetition_ratio":0.18001525,"special_character_ratio":0.3311438,"punctuation_ratio":0.11867267,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000042,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-01T20:06:05Z\",\"WARC-Record-ID\":\"<urn:uuid:358ae06a-0dbf-44ec-a456-efd2b225faad>\",\"Content-Length\":\"353439\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5b908ca5-1360-4559-820b-37bbe73d5f09>\",\"WARC-Concurrent-To\":\"<urn:uuid:4a00f942-e5ee-48ff-a3bd-96994f25868d>\",\"WARC-IP-Address\":\"146.75.36.95\",\"WARC-Target-URI\":\"https://springerplus.springeropen.com/articles/10.1186/s40064-016-2068-x\",\"WARC-Payload-Digest\":\"sha1:HXTRR7T5NXD2PP5BFYVDQNUJJ762MD74\",\"WARC-Block-Digest\":\"sha1:WW3XNCDNN526N4KXJOPSGZJ2KRKC5DSY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103945490.54_warc_CC-MAIN-20220701185955-20220701215955-00235.warc.gz\"}"}
https://math.stackexchange.com/questions/2946510/solving-simple-equations-for-two-variables	[ "# Solving simple equations for two variables\n\n$$\\begin{cases} x + y = 0\\\\ \\displaystyle\\left(\\frac{1+\\sqrt{5}}{2}\\right)x + \\left(\\frac{1 - \\sqrt{5}}{2}\\right)y = 1 \\end{cases}$$\n\nHow can I solve for $$x$$ and $$y$$?\n\nI can't seem to solve this so that when I plug the values back in, they satisfy both equations. Seems like elementary math but...does anyone have any tips on how to solve this?\n\n• Do you happen to know how to use a coefficient matrix to solve a system of linear equations? – 高田航 Oct 8 '18 at 0:47\n• Let $x=-y$, therefore $1=\\frac {1+\\sqrt 5}2x+\\frac {-1+\\sqrt 5}2x \\implies \\sqrt 5x=1 \\implies x=\\frac {\\sqrt 5}5$ – Mohammad Zuhair Khan Oct 8 '18 at 0:48\n• It would help a lot if you showed us what you have so far. We are much more willing to show you what you did wrong than we are to tell you what the answer is. – steven gregory Oct 8 '18 at 1:07\n\nLet $$\\phi=\\frac{1+\\sqrt{5}}{2}$$, $$\\psi=\\frac{1-\\sqrt{5}}{2}$$. Thus the system becomes $$S=\\{(x,y)\\in \\Bbb R^2\|(x+y=0),(\\phi x+\\psi y=1) \\}$$ To find this, do the following\nGiven: $$x+y=0$$ $$\\phi x+\\psi y=1$$ Then: $$y=-x$$ $$\\phi x+\\psi y-1=0$$ Note: $$0=0$$ Therefore: $$\\phi x+\\psi y-1=x+y$$ $$\\phi x-x +\\psi y-y=1$$ Recall: $$y=-x$$\nTherefore: $$-x(\\psi-1)+x(\\phi-1)=1$$ $$x(\\phi-1+1-\\psi)=1$$ $$x=\\frac{1}{\\phi-\\psi}$$ $$x=\\frac{\\sqrt{5}}{5}$$ $$y=-\\frac{\\sqrt{5}}{5}$$ $$S=\\{(\\frac{\\sqrt{5}}{5},-\\frac{\\sqrt{5}}{5})\\}$$" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8062784,"math_prob":1.000009,"size":338,"snap":"2021-21-2021-25","text_gpt3_token_len":109,"char_repetition_ratio":0.11077844,"word_repetition_ratio":0.0,"special_character_ratio":0.33431953,"punctuation_ratio":0.0945946,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000099,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-14T23:19:15Z\",\"WARC-Record-ID\":\"<urn:uuid:2609eeb2-1f32-4708-85c7-e9fd426e66ba>\",\"Content-Length\":\"165790\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e3c0fbb2-517f-4f96-bda5-0091f5b6a3dd>\",\"WARC-Concurrent-To\":\"<urn:uuid:612f9b60-1401-491c-ab31-4755519a38d7>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/2946510/solving-simple-equations-for-two-variables\",\"WARC-Payload-Digest\":\"sha1:Y5RIJHT3SYF5NB6CL5O5JRD5DNXG5RGF\",\"WARC-Block-Digest\":\"sha1:KSWGJZPDQZIUXGLSJDLZ4MVU7FCQSUEN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991829.45_warc_CC-MAIN-20210514214157-20210515004157-00463.warc.gz\"}"}
https://forums.wolfram.com/mathgroup/archive/2000/Oct/msg00275.html	[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Re: Bugs in Abs and Sign !?\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg25705] Re: [mg25674] Bugs in Abs and Sign !?\n• From: Tomas Garza <tgarza01 at prodigy.net.mx>\n• Date: Thu, 19 Oct 2000 04:35:30 -0400 (EDT)\n• Sender: owner-wri-mathgroup at wolfram.com\n\n```I guess it's just a matter of how Abs[] works. Take\n\nIn:=\nAbs[x^2] // FullForm\nOut//FullForm=\nPower[Abs[x], 2]\n\nThis says that the calculation of Abs[x^2] proceeds by taking first the\nabsolute value of x, and then squaring it. This procedure gives the\nright result. However, for an arbitrary function g,\n\nIn:=\ng[x^2] // FullForm\nOut//FullForm=\ng[Power[x, 2]]\n\ni.e., first square the argument and then apply g, which is what one\nmight expect. Why does Abs[] goes in a different way? That is why one\ngets\n\nIn:=\nAbs[x^2] /. Abs[u_] -> fnt[u]\nOut=\nfnt[x]^2\n\ninstead of fnt[x^2]. But I wouldn't call it a bug - it's just the way\nAbs[] calculates, and it gives the right result.\n\nTomas Garza\nMexico City\n\nJack Goldberg <jackgold at umich.edu> wrote:\n\n> Try,\n>\n> Abs[x^2] /. Abs[u_] -> fnt[u]\n>\n> and\n>\n> Sign[x^2] /. Sign[u_] -> fnt[u]\n>\n> In each case I get fnt[x]^2. Shouldn't I get fnt[x^2]? After all,\n> g[x^2] /. g[u_] -> fnt[u] returns fnt[x^2] which is what I have been\nled\n> to expect from Mathematica. I am using ver 4.01 on a PowerMac.\n>\n> These peculiarities are driving me crazy. This is the fourth \"bug\" I\nhave\n> found in just 2 weeks. Please tell me I'm wrong.\n\n```\n\n• Prev by Date: Re: Mathematica -> \"AI\"\n• Next by Date: Re: Associated Legendre Definition\n• Previous by thread: Re: Bugs in Abs and Sign !?\n• Next by thread: Re: Bugs in Abs and Sign !?" ]	[ null, "https://forums.wolfram.com/mathgroup/images/head_mathgroup.gif", null, "https://forums.wolfram.com/mathgroup/images/head_archive.gif", null, "https://forums.wolfram.com/mathgroup/images/numbers/2.gif", null, "https://forums.wolfram.com/mathgroup/images/numbers/0.gif", null, "https://forums.wolfram.com/mathgroup/images/numbers/0.gif", null, "https://forums.wolfram.com/mathgroup/images/numbers/0.gif", null, "https://forums.wolfram.com/mathgroup/images/search_archive.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8759604,"math_prob":0.60019475,"size":1436,"snap":"2022-27-2022-33","text_gpt3_token_len":469,"char_repetition_ratio":0.11033519,"word_repetition_ratio":0.031620555,"special_character_ratio":0.3558496,"punctuation_ratio":0.16369048,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98197275,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-05T03:13:42Z\",\"WARC-Record-ID\":\"<urn:uuid:d5378896-0e8b-4603-9810-e1892e5ae6af>\",\"Content-Length\":\"44974\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e58841fe-8be1-4bbb-bbf5-2fef417acb97>\",\"WARC-Concurrent-To\":\"<urn:uuid:72b1ddc9-0824-483e-9600-3055c2ce6946>\",\"WARC-IP-Address\":\"140.177.205.73\",\"WARC-Target-URI\":\"https://forums.wolfram.com/mathgroup/archive/2000/Oct/msg00275.html\",\"WARC-Payload-Digest\":\"sha1:EDSRDP4ZRM7G45QGXRHMRKFCKSAPKYCG\",\"WARC-Block-Digest\":\"sha1:3Z4SWU63GJMLDT5S32LUH367BBUOP7DT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104512702.80_warc_CC-MAIN-20220705022909-20220705052909-00235.warc.gz\"}"}
https://convertwizard.com/186-carats-to-kilograms	[ "# 186 Carats to Kilograms (186 ct to kg)\n\nConvert 186 Carats to Kilograms (ct to kg) with our conversion calculator and conversion tables. To convert 186 ct to kg use direct conversion formula below.\n186 ct = 0.0372 kg.\nYou also can convert 186 Carats to other Weight (popular) units.\n\n186 CARATS\n\n=\n\n0.0372 KILOGRAMS\n\nDirect conversion formula: 1 Carats * 5000 = 1 Kilograms\n\nOpposite conversion: 186 Kilograms to Carats\n\nCheck out conversion of 186 ct to most popular weight units:\n\n186 ct to Pounds\n186 ct to Grams\n186 ct to Ounces\n186 ct to Milligrams\n186 ct to Tons (metric)\n\n## 186 ct from similar units:\n\nCarat 0.0381665579838\n\n## Conversion table: Carats to Kilograms\n\nCARATS KILOGRAMS\n1 = 0.0002\n2 = 0.0004\n3 = 0.0006\n4 = 0.0008\n5 = 0.001\n7 = 0.0014\n8 = 0.0016\n9 = 0.0018\n10 = 0.002\nKILOGRAMS CARATS\n1 = 5000\n2 = 10000\n3 = 15000\n4 = 20000\n5 = 25000\n7 = 35000\n8 = 40000\n9 = 45000\n10 = 50000\n\n## Nearest numbers for 186 Carats\n\nCARATS KILOGRAMS\n186.66 ct = 0.037332 kg\n187.5 ct = 0.0375 kg\n192.6 ct = 0.03852 kg\n196 ct = 0.0392 kg\n200 ct = 0.04 kg\n210 ct = 0.042 kg\n217.45 ct = 0.04349 kg\n222.95 ct = 0.04459 kg\n223.5 ct = 0.0447 kg\n235 ct = 0.047 kg\n238 ct = 0.0476 kg\n240 ct = 0.048 kg\n250 ct = 0.05 kg\n255.6 ct = 0.05112 kg\n260 ct = 0.052 kg\n280.2 ct = 0.05604 kg\n282 ct = 0.0564 kg\n297.3 ct = 0.05946 kg\n300 ct = 0.06 kg\n305 ct = 0.061 kg" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5104304,"math_prob":0.99987304,"size":1241,"snap":"2021-31-2021-39","text_gpt3_token_len":529,"char_repetition_ratio":0.26919967,"word_repetition_ratio":0.0,"special_character_ratio":0.6253022,"punctuation_ratio":0.16724738,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9989454,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-04T21:36:22Z\",\"WARC-Record-ID\":\"<urn:uuid:44e3c984-10fe-45f0-b73b-3f12d65c3766>\",\"Content-Length\":\"107861\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:653c4a6f-0db0-46f1-bf64-78578ab163ef>\",\"WARC-Concurrent-To\":\"<urn:uuid:1fc2dd32-e58e-44e4-b974-a52f1e17c1b6>\",\"WARC-IP-Address\":\"18.232.245.187\",\"WARC-Target-URI\":\"https://convertwizard.com/186-carats-to-kilograms\",\"WARC-Payload-Digest\":\"sha1:PDZD7PGIN3RLVKSV5USWTPQJ52ERA3QK\",\"WARC-Block-Digest\":\"sha1:NZXGQPS7I66FAZWBTWRNYAJPITXPOAYF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046155188.79_warc_CC-MAIN-20210804205700-20210804235700-00366.warc.gz\"}"}
https://bazargoods.com/what-is-electric-current-what-are-the-types-of-electric-current/	[ "1. Home\n2. /\n3. ITI Course\n4. /\n5. What is electric current? What are the types of electric current?\n\n# What is electric current? What are the types of electric current?\n\nFriends, welcome to my website, today I will show you Electric current What is? And what are its types? What is its unit, I will tell you about it. So let’s start\n\n### What is electric current?\n\nflow of electrons in an electric circuit electric current it is said.\n\nthis one scalar is, its units ampere it happens. english alphabets ‘i’ is displayed by\n\nto measure electric current ammeter is used.\n\nto measure electric current ammeter always in the circuit hierarchy is added to.\n\nThe direction of flow of electric current and electron are always opposite to each other. Where electric current flows from positive charge to negative charge, electrons flow from negative charge to positive charge.\n\nThe value of current flow in any electric circuit always depends on the resistance of the load to be added to the circuit and the voltage to be supplied in the circuit.\n\nThe value of electric current is proportional to voltage. When the value of voltage is high in the circuit, then the value of current is also high. When the value of voltage in the circuit is small, then the value of current is also low.\n\n### Types of electric current\n\nIt is of two types-\n\n#### 1. AC Current\n\nSuch an electric current whose direction and value keeps on changing or changing is called AC Current it is said. of this electric current Frequency 50Hz it occurs.\nFriends, the light coming from the headlight in our homes is only alternating current. this stream us alternator, oscillator etc. is obtained.\n\n#### 2.DC Current\n\nSuch an electric current whose direction and value does not change or change is called DC Current it is said. of this electric current Frequency 0 (zero) Hz it occurs.\nthis stream us Mobile Battery, Inverter, Cell, DC Generator and Rectifier etc. is obtained.\n\n### How do you measure electric current?\n\nto electric current ammeter measure from.\n\n### What are the types of electric current?\n\nIt is of two types-\n1.alternating current\n2.Direct current\n\n### What is the frequency of alternating current?\n\nhis Frequency 50 Hz it occurs.\n\n### What is the frequency of direct current?\n\nFriends, if you like this post of mine, then comment and share. post reading Of Thank you!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91929466,"math_prob":0.9792574,"size":2990,"snap":"2022-40-2023-06","text_gpt3_token_len":646,"char_repetition_ratio":0.21232417,"word_repetition_ratio":0.11023622,"special_character_ratio":0.20468228,"punctuation_ratio":0.12627986,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9568167,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-08T17:55:14Z\",\"WARC-Record-ID\":\"<urn:uuid:3eb5f902-d1b5-4167-b928-2de0b5473839>\",\"Content-Length\":\"83472\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:844c86cd-5f13-4952-a6c4-e85b6fb812da>\",\"WARC-Concurrent-To\":\"<urn:uuid:1c72fff4-f17e-4266-b169-c3121fc47346>\",\"WARC-IP-Address\":\"82.180.166.138\",\"WARC-Target-URI\":\"https://bazargoods.com/what-is-electric-current-what-are-the-types-of-electric-current/\",\"WARC-Payload-Digest\":\"sha1:UV4UKHP7JU3QD7ZLNKXYSTYNHFE4RTHI\",\"WARC-Block-Digest\":\"sha1:MVDJ2V6QBEIC77PUVXK7QE2XVLOPVUG5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500837.65_warc_CC-MAIN-20230208155417-20230208185417-00234.warc.gz\"}"}
https://patents.justia.com/patent/10834368	[ "# Kind of partially reliable transmission method based on hidden Markov model\n\n- SICHUAN UNIVERSITY\n\nThis invention discloses a kind of reliable data transmission method based on hidden Markov model that specific to video transmission quality enhancement over lossy satellite links. It uses HMM model to forecast retransmission rate and adopts the network packet loss rate to reflect the network performance. The underlying network packet loss rate is used as the hidden state of HMM model and the retransmission rate is used as the observation state. By analyzing the historically transmitted data sessions, APRT establishes the relationship between the hidden state transition probability and the emission probability. Then, a robust and efficient retransmission rate prediction is obtained. This invention takes into full consideration of high BER in satellite network channels and adjusts the reliable level according to the satellite network channel characteristics so as to make tradeoff between instantaneity and quality. Consequently, it guarantee the QoS and QoE of video stream.\n\n## Latest SICHUAN UNIVERSITY Patents:\n\nDescription\nFIELD OF TECHNOLOGY\n\nThis invention involves the end-to-end partially reliable transmission field, which involves a kind of transmission method based on hidden Markov model that specific to video transmission quality enhancement over lossy satellite links.\n\nBACKGROUND TECHNOLOGY\n\nMultimedia applications play an important role in satellite networks and High-Definition (HD) video dominated streaming applications increasingly, providing these applications with good QoS and QoE continues to be a critical problem. For the delay-sensitive type of applications, the conventional transmission policy cannot be adopted with the limited bandwidth and high bit error rate (BER) of satellite channel. Therefore, how to design efficient transfer protocol with guaranteed QoS and QoE for video transfer in satellite network is one of the key issues.\n\nReliable transmission control protocol (e.g., TCP and its variants) and unreliable user datagram protocol (e.g., UDP) are widely used for video transmission at the transport layer. TCP is widely used as the primary data transmission protocol, which can enable firewall penetration throughout the network and keep friendliness between different flows. In many papers, video is transferred with TCP protocol at the transport layer. For example, the existing work proposed the CS2P protocol by analyzing the throughput characteristics of different session data. CS2P can predict the initial throughput, and then use the hidden Markov model (HMM) to establish the medium flow forecasting model. Finally, using throughput rate control, CS2P can achieve adaptable for video transmission control. PATON improves the quality of real-time video streams by using the technologies, i.e., forward error correction (FEC), video frame priority selection, and adaptive redundancy. However, because satellite network has a high bit error rate environment, video applications directly use TCP protocol as the underlying protocol for data transmission may cause many problems. First, the reliable transport mechanism of TCP protocol will cause multiple retransmits until the lost packets are correctly received. For delay-sensitive applications, the retransmitted packets are likely to be invalid for video playout before they are received. Furthermore, when packet loss events occur, the congestion control and congestion avoidance mechanism of TCP protocol can result in low bandwidth occupation of video streams and cannot obtain higher bandwidth occupation when competing with UDP flow. In summary, reliable transmission strategies cannot guarantee QoS of video flows over satellite networks.\n\nNowadays, most multimedia applications are based on unreliable transmission protocols, such as UDP protocol and RTP protocol. Many well-known applications are based on UDP protocol (i.e., Skype and WebRTC). For example, GCC performs rate control based on RTP/RTCP protocol, and then uses the Kalman filter algorithm to estimate the delay gradient of packets in the networks, which can indicate the network congestion level. GCC is used for Google Hangouts and Chrome WebRTC protocol stack in reality. An existing work proposed the DCCP protocol which uses the principle of incremental sequence number. Moreover, by assessing whether the packet exceeds its scheduled life cycle, DCCP determines whether to retransmit the lost packet. However, transmission method of unreliable transmission protocol also has some disadvantages. First, UDP and RTP protocol without congestion control mechanism flood data packets into network, which may deteriorate the overall network performance. In addition, unreliable transmission protocol does not guarantee data transmission reliability. When the satellite network is in good condition but has large amount of stochastic packet losses, unreliable transfer methods will greatly influence the QoS of HD video.\n\nCONTENTS OF THE INVENTION\n\nThis invention aims to solve the technical problem which offers a kind of reliable transmission method based on hidden Markov model (hereafter referred to as APRT protocol) to handle the quality and real-time problem of delay-sensitive video service over lossy satellite links.\n\nTo solve the technical problem mentioned above, the technical program adopted by this invention is:\n\nStep 1: Present the specification definition of HMM model.\n\nThis method uses HMM model to forecast retransmission rate. The underlying network packet loss rate is used as the hidden state of HMM model and the retransmission rate is used as the observation state. By analyzing the historically transmitted data sessions, it establishes the relationship between the hidden state transition probability and the emission probability. Then, a robust and efficient retransmission rate prediction is obtained.\n\nRandom variables Wt is defined as the retransmission rate of APRT protocol in time period t. wt is defined as the actual retransmission rate measured in the network. Ŵt indicates the predicted value of random variable Wt. Assume that the retransmission rate of the strategy Wt is changing with the variation of some hidden state variables Xt∈χ, where χ∈{x1, x2, . . . , xN} represents a set of possible discrete state sequences and N=\|χ\| indicates the number of states. State variables Xt is a random variable and the probability distribution is defined as the vector π=(P(Xt=x1), . . . , P(Xt=xN)).\n\nThe HMM model assumes that the state varies with the Markov process, where the probability distribution of the current state is determined by the state of its previous period, that is P(Xt\|Xt−1, . . . , X1)=P(Xt\|Xt−1). The defined transition probability matrix is P={Pij}, where Pij=P(Xt=xi\|Xt−1=xj). According to Markov properties, it can be concluded that πt+τ tPτ. When a hidden state Xt is given, we assume that the probability distribution function (pdf) of network packet loss rate follows the following Gaussian distribution function: Wt\|Xt=x˜N (μx, σx2).\n\nStep 2: Offline training phase.\n\nWhen the number of states N is given, and the training data set is Set(Ms, s), utilize the maximum likelihood estimation algorithm to calculate the parameters of HMM model, which can be expressed by the equation θHMM={π0, P, {(μx, σx2), x∈χ}}.\n\nStep 3: Online prediction stage.\n\n• 1) Prediction initialization: Set(Ms, s) is the most matching feature for the transmission of flow s, which is defined as Ŵ1=Median(Set(Ms, s)), where Set(Ms, s) is the mean value of the data set and as the initial network packet loss rate.\n• 2) Forecast phase. In the period t, the updated pdf of the HMM model is given. Then, according to Markov feature, the pdf of current time period can be calculated by the equation πt\|1:t−1t−1\|1:t−1P. The predicted retransmission rate Ŵt can also be calculated.\n• 3) Parameter update of HMM model. When the actual network packet loss rate Wt=wt and πt\|1:t−1 are given, Baum-Welch algorithm is used to update the parameters of HMM model. πt\|1:t−1 is updated as πt\|1:t−1, and Pij is updated as Pij.\n\nStep 4: Congestion Control.\n\n• 1) Rate control for video flow. When the receiver uses the HMM model to determine the retransmission rate of the video, it returns the adaptive ACK feedback packet. When the sender receives the ACK packets, the APRT protocol congestion window control adopts the loss based congestion indicators, and the cubic window growth function, which is defined in Cubic TCP, so as to update the congestion window increase quickly and discreetly.\n• 2) Fast retransmission. When the network is in a non-congested state, the receiver side returns almost all the ACK feedbacks of all received data, and the sender performs fast retransmission after receiving duplicated ACKs or timeout as traditional transport protocol. When the network is predicted in a poor or a congested state, the receiver retransmits the acknowledgment feedback partially, and the sender only performs fast retransmission for the confirmed lost packets. For extreme conditions, such as if the network is in idle time, the APRT protocol is regarded as a completely reliable transmission protocol as the TCP protocol.\n\nAccording to the program mentioned above, in Step 3, the formula to calculate in Step 3 is defined as follows:\n\n$W ^ t = μ x$ $x = arg ⁢ ⁢ max x ∈ χ ⁢ ⁢ P ⁡ ( ∑ j n ⁢ X N ⁡ ( j ) )$ $X t + 1 ⁡ ( j ) = ∑ i = 1 , ⁢ … ⁢ , N ⁢ X t ⁡ ( i ) ⁢ P ij · W t + 1 ❘ X j$\n\nAccording to the program mentioned above, in Step 3, the formula to calculate πt\|1:t−1 and Pij in Step 3 is defined as follows:\n\n$π _ t ❘ 1 ⁢ : ⁢ t - 1 = P ⁡ ( W 1 , W 2 , … ⁢ , W t , X t = x i ❘ W 1 ⁢ : ⁢ t ) · P ⁡ ( W t + 1 ❘ X t = x i , W 1 ⁢ : ⁢ t ) ∑ i = 1 N ⁢ ⁢ P ⁡ ( W 1 , W 2 , … ⁢ , W t , X t = x i ❘ W 1 ⁢ : ⁢ t ) · P ⁡ ( W t + 1 ❘ X t = x i , W 1 ⁢ : ⁢ t )$ $⁢ P _ ij = P ⁡ ( W 1 , W 2 , … ⁢ , W t , X t = x i ❘ W 1 ⁢ : ⁢ t ) ⁢ P ij · W t + 1 ❘ X j · P ⁡ ( W t + 1 , W t + 2 ❘ X t + 1 = x i + 1 , W 1 ⁢ : ⁢ t + 1 ) ∑ i = 1 N ⁢ ⁢ P ⁡ ( W 1 , W 2 , ⁢ … ⁢ , W t , X t = x i ❘ W 1 ⁢ : ⁢ t ) · P ⁡ ( W t + 1 ❘ X t = x i , W 1 ⁢ : ⁢ t )$\n\nBeing compared with the current technology, the beneficial effects of this invention are: 1) It employ HMM model to depict reliable level of the transfer strategy, which can make tradeoffs between the quality of the video application and the real-time performance of video transfer. Through offline training initialization and online prediction, the relationship between the packet loss and the reliability of the transmission is established, which can represent good video quality and minimum packet delay. 2) It can retransmit the lost packets as the network changing without losing of key frames. It is compatible with the existing TCP congestion control scheme for rate control, and takes the video frame priority into consideration for partial reliability transmission. Comparing with other video transfer strategies, APRT achieves higher throughput while ensuring lower packet delay. Consequently, the proposed protocol can guarantee the video stream QoS for satellite network.\n\nSPECIFICATION OF THE ATTACHED FIGURES\n\nFIG. 1 shows the overview of HMM model.\n\nFIG. 2 shows the simulation topology.\n\nFIG. 3 shows the average throughput of each protocol under different link error rates when the bottleneck link bandwidth is 10 Mbps.\n\nFIG. 4 shows the average throughput of each protocol under different link error rates when the bottleneck link bandwidth is 200 Kbps.\n\nFIG. 5 shows the changing of RTT of different single flow over time when the link error rate is 10−5 in scenario A.\n\nFIG. 6 shows the changing of RTT of different single flow over time when the link error rate is 10−5 in scenario B.\n\nFIG. 7 shows the changing of delay jitter of different single flows over time in scenario A and scenario B.\n\nFIG. 8 shows the changing of delay jitter of different single flows over time in scenario B.\n\nFIG. 9 shows the subjective video quality of different single protocols using different video test sequence, when the bandwidth is 10 Mb/s and BER is 10−7 (1).\n\nFIG. 10 shows the subjective video quality of different single protocols using different video test sequence, when the bandwidth is 10 Mb/s and BER is 10−7 (2).\n\nFIG. 11 shows the subjective video quality of different single protocols using different video test sequence, when the bandwidth is 10 Mb/s and BER is 10−7 (3).\n\nFIG. 12 shows the subjective video quality of different single protocols using different video test sequence, when the bandwidth is 10 Mb/s and BER is 10−7 (4).\n\nFIG. 13 shows the subjective video quality of different single protocols using different video test sequence, when the bandwidth is 200 Kb/s and BER is 10−6 (1).\n\nFIG. 14 shows the subjective video quality of different single protocols using different video test sequence, when the bandwidth is 200 Kb/s and BER is 10−6 (2).\n\nFIG. 15 shows the subjective video quality of different single protocols using different video test sequence, when the bandwidth is 200 Kb/s and BER is 10−6 (3).\n\nFIG. 16 shows the impact of different playout buffer sizes on the quality of APRT protocol.\n\nFIG. 17 shows the average throughput of three video flows under different link BERs when the bottleneck link bandwidth is 10 Mbps.\n\nFIG. 18 shows the average throughput of three video flows under different link BERs when the bottleneck link bandwidth is 200 Kbps.\n\nFIG. 19 shows the changing of RTT of different concurrent flows over time in scenario A.\n\nFIG. 20 shows the changing of RTT of different concurrent flows over time in scenario B.\n\nFIG. 21 shows the changing of delay jitter of different concurrent flows over time in scenario A.\n\nFIG. 22 shows the changing of delay jitter of different concurrent flows over time in scenario B.\n\nFIG. 23 shows subjective video quality of different concurrent protocols using different video test sequence, when the bandwidth is 10 Mb/s and BER is 10−7 (1).\n\nFIG. 24 shows subjective video quality of different concurrent protocols using different video test sequence, when the bandwidth is 10 Mb/s and BER is 10−7 (2).\n\nFIG. 25 shows subjective video quality of different concurrent protocols using different video test sequence, when the bandwidth is 200 Kb/s and BER is 10−6 (1).\n\nFIG. 26 shows subjective video quality of different concurrent protocols using different video test sequence, when the bandwidth is 200 Kb/s and BER is 10−6 (2).\n\nSPECIFIC IMPLEMENTATION METHOD\n\nFurther specific specification is given as follows by combining the attached figures and the implementation case in detail.\n\nSimulation topology is shown in FIG. 2, which includes the senders, the receivers and the satellite networks. The following two main scenarios are simulated:\n\n• Scenario A: bandwidth of the bottleneck link is 10 Mb/s with varied round trip time (RTT), which represents broadband MEO satellite networks;\n• Scenario B: bandwidth of the bottleneck link is 200 Kb/s with varied RTT, which represents narrowband LEO satellite networks.\n\nThe APRT protocol is compared with different types of protocols, which include various reliability and order. Therein, variable ξ, denote the reliability degree of the transport service provided. A reliable service is 100% reliable. A partially reliable service is the service with ξ∈[0%, 100%], variable Ω defines the orderliness of the strategy, where Ω∈[0%, 100%]. P1 represents APRT protocol, P2 represents the protocol with ξ=1, Ω=1, P3 represents the protocol with ξ≤1, Ω=1, P4 represents the protocol with ξ≤1, Ω=1∘\n\nIt can be concluded from both FIGS. 3 and 4 that the average throughput of all protocols reduces with the increase of link error rate. As shown in FIG. 3, when the bottleneck link bandwidth is 10 Mbps, all protocols cannot take full advantage of the link bandwidth, so that the network is in a state of no congestion. When the BER is from 10−9 to 10−6, the network is in good state. The average throughput of APRT and p2 strategies are higher than p3 and p4 strategies. This is because the throughput of unreliable strategies (i.e., p3 and p4) is limited by the transfer speed of video encoded frames from the application layer to the transport layer at the sender side. Nevertheless, there is a reliable confirmation component in p2 strategy and APRT protocol, which produces part of the retransmission packets. When the link error rate is greater than 2.0×10−6, the average throughput of p3 and p4 strategies start to larger than that of reliable p2 strategy, while the APRT protocol begins to perform a similar performance to p4 strategy. The reason for this is that p2 strategy is based on the ACK acknowledgment to assure reliability while APRT adopts partially reliable transfer mechanism.\n\nAs shown in FIG. 4, when the bottleneck link bandwidth is 200 Kbps, the bottleneck link has been saturated, and is in a moderate congestion state. For different BER, p4 strategy always maintains the highest average throughput of about 200 Kbps, followed by p3 strategy. When the link error rate is from 10−9 to 10−6, average throughput of APRT protocol is slightly higher than of p2 strategy. When the BER is larger than 10−6, p2 strategy has an obvious performance degradation, the reason is that p3 and p4 strategies have no packet transmission control mechanism, so that the throughput is large. Therein, p3 strategy applies a packet in-order guarantee mechanism, so that the throughput of p3 is slightly smaller than that of p4 strategy, in which Ω=1. For APRT protocol, it will determine the state of the network according to the HMM model. When approaching the saturation bandwidth occupancy, APRT utilizes partially reliable transmission mechanism. When the network packet loss rate is large, APRT reduces the reliability of the protocol for real-time transmission. Thus, the proposed protocol can make trade-offs between high-quality and playback continuity.\n\nIt can be seen from FIG. 5 that the RTT of APRT protocol changes between [52.5 ms, 53 ms], the RTT of p2 strategy varies between [52.5 ms, 53.5 ms], and that of p3 and p4 strategies changes between [50 ms, 53 ms]. This is because that when the network bandwidth is not fully utilized, the reliable mechanism of p2 strategy will cause network routers cache a certain amount of data packets, so that the packet has a queuing delay. This will result in RTT increase. While for protocols without guaranteeing reliability, the more the protocol is unreliable, the lower the RTT is. Combining the jitter result shown in FIG. 7, p2 strategy has a high delay jitter. The larger the delay jitter is, the less stable queueing produced in the bottleneck routers.\n\nIt can be seen from FIG. 6 that RTT of different protocols has big difference. Therein, the RTT of p2 is much higher than the other three protocols. This is caused by the congestion control of p2 strategy; whose window changes regularly. At this time, the network utilization is high with high random BER in satellite networks, which results in a large number of packets buffered in routers. When the network has free bandwidth, the growth of the window is slow. Once packet loss events occur, congestion window of p2 turns into the congestion avoidance stage. The RTT of p4 packet is less than that of p2 strategy packet. Because of the high utilization of the network at this time, and that p4 strategy does not have congestion control mechanism, the routers of bottleneck link also buffer a large number of data packets. Because of p3 strategy guarantees the packet order and will discard the out-of-order packet at the receiver side, most RTT of p3 packets are less than 150 ms delay. APRT protocol also has a smaller round-trip delay APRT also represents a ladder curve characteristics, which is doe to the HMM model network prediction. APRT carries out an adaptive reliability and avoids unnecessary congestion window reduction so as to maintain the congestion window in a relatively stable state. According to the corresponding delay jitter in FIG. 8, it can be seen that the delay jitter range of p2 strategy is between [0,0.5], which is violent, and the APRT protocol is around 0.1, except that some individual points reach 0.3. p4 and p3 strategies achieve minimal delay jitter, which are all less than 0.06.\n\nTable 1 examines the packet loss results in different network states. Overall, it can be seen that for different BERs, packet loss rate of different protocols is: packet loss rate of p4>packet loss rate of p3>packet loss rate of APRT>packet loss rate of p2. When the bandwidth of 200 Kbps, the performance of p4 strategy and p3 strategy are affected greatly, and the total number of dropped frames are 452 and 314, respectively. The total number of dropped frames of APRT protocol is 178 at most.\n\nTABLE 1 Lost Frames of Different Protocols under Single Flow Scenarios with Different Bandwidths and BERs. Bandwidth protocol BER Packet loss rate Lost frame 10 Mb/s P1 10−7 0.0065 17 10−5 0.1215 52 P2 10−7 0.0062 16 10−5 0.1375 40 P3 10−7 0.068 18 10−5 0.1198 286 P4 10−7 0.0074 19 10−5 0.1166 299 200 Kb/s P1 10−7 0.0471 83 10−5 0.1409 178 P2 10−7 0.0285 24 10−5 0.1341 24 P3 10−7 0.1638 303 10−5 0.1364 314 P4 10−7 0.1762 452 10−5 0.1567 402\n\nThe PSNR results are shown in Table 2. The upper part of Table 2 shows a good state of the network, where the bandwidth is not fully utilized and the link BER is 10−8. The four protocols achieve a higher. PSNR value and the corresponding MOS value is 5, while the reliable protocol has higher video transmission quality than the unreliability protocol. The PSNR of the APRT protocol is 11.62% higher than that of the p4 strategy. The lower part of Table 2 shows a poor network status, where the bandwidth low and link BER of 10−5. Therein, PSNR of p2 strategy is 0, indicating that the video cannot be decoded and reconstruction. The video quality based on p3 strategy and p4 strategy is fair and poor, respectively. The APRT protocol with a certain reliability guarantee can reach 32.89, which ensures a better video transmission quality.\n\nTABLE 2 Video Quality of Different Protocols under Single Flow Scenarios with Different Bandwidths and BERs. Bandwidth MOS protocol BER PSNR MOS evaluation 10 Mb/s P1 10−8 41.525381 5 Excellent P2 41.623857 5 Excellent P3 37.296478 5 Excellent P4 37.202516 5 Excellent 200 Kb/s P1 10−5 32.889131 4 Good P2 0 1 Bad P3 29.525154 3 Fair P4 24.092692 2 Poor\n\nFIG. 9-12 and FIG. 13-15 show the video quality comparison of different video test sequences in different network states. By comparing with the source video image, it can be seen that the video quality of the APRT protocol and p2 strategy of the second and third columns is almost as clear as the source video image, and the video quality of p4 strategy is slightly worse, especially for dynamic image. In FIG. 13-15, due to this network state, p2 strategy is unable to carry out video decoding and reconstruction work, so we only show the quality of two compared video transfer protocols, i.e., p3 and p4. It can be seen that the quality of APRT protocol video is significantly better than that of p4 protocol.\n\nFIG. 16 shows that the bigger the playout buffer is, the better the video quality are. When the link error rate is 10−7 and the playout buffer size is less than 150, video quality only achieve fair quality. When the playout buffer size is higher than 150, the video quality is good. When the link error rate of 10−5 and the playout buffer size is less than 110, the video quality is poor, while the playout buffer size is higher than 240, the video quality is good. After the playout buffer size is higher than 350, continue to increase the size of the playout buffer does not improve the quality of the video any more. Overall, we can see that for a playout buffer with 150 to 300 packets, APRT can achieve a good quality of video playback. Moreover, because the poor satellite networks environment, larger playout buffer size is not necessary.\n\nIt can be seen from FIGS. 17 and 18 that the results are almost the same as the single flow experiments, that is, the average throughput of all protocols decrease as the link BER increases. As shown in FIG. 17, the bottleneck link bandwidth is 10 Mbps and the link error rate is 4.5×10−6, average throughput of APRT, p2 and p4 strategy are around 560 Kbps. The results of FIG. 18 differ from the single flow scenario, the average throughput of APRT protocol holds between the performance of p2 and p4 strategy. When the BER is 10−5, the average throughput of APRT protocol is higher than that of p2 strategy.\n\nThe evaluation of round trip delay and delay jitter is also similar to the result of a single flow scenario. FIG. 19 shows the delay curve changing over time when the link BER is 10−5 in scenario A. It can be seen from the figure that RTT of APRT protocol packet changes in the interval of [52.5 ms, 55 ms]. When p2 becomes stable, its RTT changes in the interval of [52 ms, 65 ms]. The RTT of p3 strategy and p4 strategy are between the interval of [52.5 ms, 60 ms]. The corresponding delay jitter results are given in FIG. 21, which shows that the APRT protocol has a larger round trip delay and can maintain a smaller queuing delay in the router:\n\nFIG. 20 shows the RTT curve varying over time when link BER is 10−5 in scenario B. It can be seen from the figure that when the network data flow increases, the round trip delay increases from 150 ms to 400 ms comparing with when there is only a single flow in the network. The overall round trip delay of APRT protocol is the smallest. However, as shown in FIG. 22, APRT has a large delay jitter like p2.\n\nTable 3 is used to check the coexistence of multiple data strip in different network state under different packet loss situation. It can be seen that in the case of multiple flow coexisting situations, all of the protocols have a high packet loss rate, especially when the bandwidth is 200 Kbps. The packet loss rate of APRT reduced by 77.69% and 55.05% comparing with the p4 strategy.\n\nTABLE 3 Lost Frames of Different Protocols under Concurrent Flow Scenarios with Different Bandwidths and BERs. Bandwidth Protocol BER Packet loss rate Lost frame 10 Mb/s P1 10−7 0.0173 48 10−5 0.1475 71 P2 10−7 0.0158 41 10−5 0.2198 87 P3 10−7 0.0672 54 10−5 0.1340 69 P4 10−7 0.0677 55 10−5 0.1093 74 200 Kb/s P1 10−7 0.1982 54 10−5 0.2981 64 P2 10−7 0.2068 55 10−5 0.3301 69 P3 10−7 0.8885 6814 10−5 0.8885 6791 P4 10−7 0.8885 6814 10−5 0.8867 6800\n\nFor the upper part of Table 4, when the network state is better, the four protocols can reach a higher PSNR value, and the MOS value is 4. The PSNR of difference protocol is not significant. However, when the network gets worse, the performance of APRT protocol is significantly better than the other three protocols. It can be concluded that APRT ensures a better video transmission quality.\n\nTABLE 4 Video Quality of Different Protocols under Concurrent Flow Scenarios with Different Bandwidths and BERs. Bandwidth MOS Protocol BER PSNR MOS evaluation 10 Mb/s P1 10−8 33.032264 4 Good P2 33.146254 4 Good P3 32.484776 4 Good P4 32.440097 4 Good 200 Kb/s P1 10−6 30.652149 3 Fair P2 0 1 Bad P3 27.088844 3 Fair P4 22.441292 2 Poor\n\nIt can be seen from FIGS. 23 and 24 that the video quality of the APRT protocol and the p2 strategy is almost as well as the source video image, while the video quality transmitted by the p4 strategy is slightly worse, especially for the dynamic images. For FIGS. 25 and 26, because in this network state, p2 strategy transmission of data has been unable to carry out video decoding and reconstruction work, so only shows the quality of other two comparing transmission protocol. It can be seen that video quality based on APRT protocol is significantly better than that of p4 strategy. Because the first sequence images are relatively static, so the quality is better than that of the second row. For second row, the transfer of p4 strategy represents a serious discontinuous phenomenon.\n\n## Claims\n\n1. A reliable transmission method based on a hidden Markov model (HMM model) to handle a quality and a real-time problem of a delay-sensitive video service over lossy satellite links, the method comprising:\n\nStep 1: Presenting a specification definition of the HMM model, including\nusing the HMM model to forecast a retransmission rate; using an underlying network packet loss rate as a hidden state of the HMM model; using the retransmission rate as an observation state; by analyzing historically transmitted data sessions, establishing a relationship between a hidden state transition probability and an emission probability, and obtaining a robust and efficient retransmission rate prediction, such that: a random variable Wt is defined as the retransmission rate in a time period t; wt is defined as an actual retransmission rate measured in a network; Ŵt indicates a predicted value of the random variable Wt; a retransmission rate of a strategy Wt is changing with a variation of hidden state variables Xt∈χ, where χ∈{x1, x2,..., xN} represents a set of possible discrete state sequences and N=\|χ\| indicates a number of states; the hidden state variables Xt are a random variable and a probability distribution is defined as a vector π=(P(Xt=x1),..., P(Xt=xN));\na current state varies with a Markov process, where a probability distribution of the current state is determined by a state of a previous period, expressed as P(Xt\|Xt−1,..., X1)=P(Xt\|Xt−1); a defined transition probability matrix is P={Pij}, where Pij=P(Xt=xi\|Xt−1=xj), and πt+τ =πtPτ; and when a hidden state Xt is given, a probability distribution function (pdf) of the network packet loss rate follows the Gaussian distribution function: Wt\|Xt=x˜N(μx, σx2);\nStep 2: Executing an offline training phase, including, when the number of states N is given and a training data set is Set (Ms, s), utilizing a maximum likelihood estimation algorithm to calculate the parameters of the HMM model, which can be expressed by the equation θHMM={π0, P, {(μx, σx2), x∈χ}};\nStep 3: Executing an online prediction stage, including a prediction initialization including: providing a Set (Ms, s) as a most matching feature for a transmission of flow s, which is defined as Ŵ1=Median (Set (Ms, s)), where Set(Ms, s) is a mean value of a data set and an initial network packet loss rate; a forecast phase including: in the time period t, providing an updated pdf of the HMM model, then according to a Markov feature, calculating the pdf of a current time period using the equation πt\|1:t−1=πt−1\|1:t−1P, and calculating the predicted retransmission rate Ŵt; and a parameter update of the HMM model including: when the network packet loss rate Wt=wt, and πt\|1:t−1 are provided, using a Baum-Welch algorithm to update the parameters of the HMM model, such that πt\|1:t−1 is updated as πt\|1:t−1, and Pij is updated as Pij;\n\n2. The method of claim 1, wherein the formula to calculate Ŵt in Step 3 is defined as follows: W ^ t = μ x x = arg ⁢ ⁢ max x ∈ χ ⁢ ⁢ P ⁡ ( ∑ j n ⁢ X N ⁡ ( j ) ) X t + 1 ⁡ ( j ) = ∑ i = 1, ⁢ … ⁢, N ⁢ X t ⁡ ( i ) ⁢ P ij · W t + 1 ❘ X j.\n\n3. The method of claim 1, wherein the formulas to calculate πt\|1:t−1 and Pij in Step 3 are defined as follows: π _ t ❘ 1 ⁢: ⁢ t - 1 = P ⁡ ( W 1, W 2, … ⁢, W t, X t = x i ❘ W 1 ⁢: ⁢ t ) · P ⁡ ( W t + 1 ❘ X t = x i, W 1 ⁢: ⁢ t ) ∑ i = 1 N ⁢ ⁢ P ⁡ ( W 1, W 2, … ⁢, W t, X t = x i ❘ W 1 ⁢: ⁢ t ) · P ⁡ ( W t + 1 ❘ X t = x i, W 1 ⁢: ⁢ t ) ⁢ P _ ij = P ⁡ ( W 1, W 2, … ⁢, W t, X t = x i ❘ W 1 ⁢: ⁢ t ) ⁢ P ij · W t + 1 ❘ X j · P ⁡ ( W t + 1, W t + 2 ❘ X t + 1 = x i + 1, W 1 ⁢: ⁢ t + 1 ) ∑ i = 1 N ⁢ ⁢ P ⁡ ( W 1, W 2, … ⁢, W t, X t = x i ❘ W 1 ⁢: ⁢ t ) · P ⁡ ( W t + 1 ❘ X t = x i, W 1 ⁢: ⁢ t ).\n\nPatent History\nPatent number: 10834368\nType: Grant\nFiled: Dec 13, 2018\nDate of Patent: Nov 10, 2020\nPatent Publication Number: 20190191130\nAssignee: SICHUAN UNIVERSITY (Chengdu)\nInventors: Junfeng Wang (Chengdu), Qin Luo (Chengdu), Ying Liu (Chengdu), Hao Niu (Chengdu), Xiaohui Li (Chengdu)\nPrimary Examiner: Jeffrey M Rutkowski\nAssistant Examiner: Mehedi S Aley\nApplication Number: 16/219,844\nClassifications\nCurrent U.S. Class: Determination Of Communication Parameters (370/252)\nInternational Classification: H04N 7/20 (20060101); H04B 17/391 (20150101); H04N 21/647 (20110101); H04N 21/643 (20110101); H04L 25/03 (20060101); H04N 21/24 (20110101); H04N 21/6375 (20110101); H04N 21/238 (20110101); H04N 21/2343 (20110101); H04N 21/2381 (20110101); H04N 21/61 (20110101);" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.884672,"math_prob":0.94441795,"size":19709,"snap":"2021-04-2021-17","text_gpt3_token_len":4626,"char_repetition_ratio":0.15833545,"word_repetition_ratio":0.17048268,"special_character_ratio":0.22918464,"punctuation_ratio":0.11962365,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96008277,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-17T20:09:56Z\",\"WARC-Record-ID\":\"<urn:uuid:add67229-249e-4db9-aab8-8747d0d4ef88>\",\"Content-Length\":\"103481\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:abbac678-b072-48d1-ab1c-f017a762233e>\",\"WARC-Concurrent-To\":\"<urn:uuid:db753bf0-6580-493b-b225-0da80861120a>\",\"WARC-IP-Address\":\"3.223.160.239\",\"WARC-Target-URI\":\"https://patents.justia.com/patent/10834368\",\"WARC-Payload-Digest\":\"sha1:MTFJYDEKW2AWAM35CQTKU2O2MDLDTKP5\",\"WARC-Block-Digest\":\"sha1:NWJI7VT3XHKGTGUFWTQUES66JOGIOVXJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703513144.48_warc_CC-MAIN-20210117174558-20210117204558-00573.warc.gz\"}"}
https://calculator.name/area/square-rod/square-meter/0.99	[ "Amount\nFrom\nTo\n\n# 0.99 square rods to square meters\n\nHow many square meters in 0.99 square rods? 0.99 square rods is equal to 25.039924114 square meters.\n\nThis page provides you how to convert between square rods and square meters with conversion factor.\n\n# How to convert 0.99 square rods to m2?\n\nTo convert 0.99 square rods into m2, follow these steps:\n\nWe know that, 1 square rods = 25.29285264 m2\n\nHence, to convert the value of 0.99 square rods into square meters, multiply both sides by 0.99 and we get:\n\n1 square rods = 25.29285264 m2\n\n(1 * 0.99) square rods = (25.29285264 * 0.99) m2\n\n0.99 square rods = 25.039924114 m2\n\n### Thus, 0.99 square rods equals to 25.039924114 m2\n\nSquare Rods Conversion of Square Rods to Square Meters\n0.98 square rods 0.98 square rods = 24.786995587 m2\n0.89 square rods 0.89 square rods = 22.51063885 m2\n0.99 square rods 0.99 square rods = 25.039924114 m2\n1.99 square rods 1.99 square rods = 50.332776754 m2" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7631905,"math_prob":0.999899,"size":797,"snap":"2023-40-2023-50","text_gpt3_token_len":272,"char_repetition_ratio":0.31651956,"word_repetition_ratio":0.044444446,"special_character_ratio":0.4404015,"punctuation_ratio":0.18085106,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997602,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-28T12:31:35Z\",\"WARC-Record-ID\":\"<urn:uuid:cc76e82c-67ff-47d9-b5ba-e64a434e40e4>\",\"Content-Length\":\"132236\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c51e1667-9564-4045-b7e3-a4ddf670adfb>\",\"WARC-Concurrent-To\":\"<urn:uuid:2d83f011-8d35-4e15-bbfd-ffe08604b72a>\",\"WARC-IP-Address\":\"68.178.147.160\",\"WARC-Target-URI\":\"https://calculator.name/area/square-rod/square-meter/0.99\",\"WARC-Payload-Digest\":\"sha1:EXWM22KEVEVLXAY5AHZH7SAM7HEYJLG7\",\"WARC-Block-Digest\":\"sha1:FGMAJA2WXAKEQI6FQ2LPCCZMZCWHSCAT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679099514.72_warc_CC-MAIN-20231128115347-20231128145347-00294.warc.gz\"}"}
http://ntour.ntou.edu.tw:8080/ir/handle/987654321/50482	[ "", null, "English \| 正體中文 \| 简体中文 \| Items with full text/Total items : 27314/39158 Visitors : 2473805 Online Users : 94\n Scope All of NTOUR 工學院河海工程學系 --期刊論文 Adv. Search\n National Taiwan Ocean University Institutional Repository > 工學院 > 河海工程學系 > 期刊論文 > Item 987654321/50482\n\n Please use this identifier to cite or link to this item: `http://ntour.ntou.edu.tw:8080/ir/handle/987654321/50482`\n\n Title: A Scalar Homotopy Method for Solving an Over/Under-Determined System of Non-Linear Algebraic Equations Authors: Chein-Shan LiuWeichung YeihChung-Lun KuoSatya N. Atluri Contributors: 國立臺灣海洋大學:河海工程學系 Keywords: Nonlinear algebraic equationsIterative methodOrdinary differential equationsScalar homotopy method (SHM) Date: 2009-12 Issue Date: 2018-10-12T03:24:20Z Publisher: CMES: Computer Modeling in Engineering & Science Abstract: Abstract: Iterative algorithms for solving a system of nonlinear algebraic equations (NAEs): Fi(xj) = 0, i, j= 1,. . . ,n date back to the seminal work of Issac Newton. Nowadays a Newton-like algorithm is still the most popular one to solve the NAEs, due to the ease of its numerical implementation. However, this type of algorithm is sensitive to the initial guess of solution, and is expensive in terms of the computations of the Jacobian matrix ∂Fi/∂ xj and its inverse at each iterative step. In addition, the Newton-like methods restrict one to construct an iteration procedure for n-variables by using n-equations, which is not a necessary condition for the existence of a solution for underdetermined or overdetermined system of equations. In this paper, a natural system of first-order nonlinear Ordinary Differential Equations (ODEs) is derived from the given system of Nonlinear Algebraic Equations (NAEs), by introducing a scalar homotopy function gauging the total residual error of the system of equations. The iterative equations are obtained by numerically integrating the resultant ODEs, which does not need the inverse of ∂Fi/∂ xj . The new method keeps the merit of homotopy method, such as the global convergence, but it does not involve the complicated computation of the inverse of the Jacobian matrix. Numerical examples given confirm that this Scalar Homotopy Method (SHM) is highly efficient to find the true solutions with residual errors being much smaller. Relation: 53(1) pp.47-71 URI: http://ntour.ntou.edu.tw:8080/ir/handle/987654321/50482 Appears in Collections: [河海工程學系] 期刊論文\n\nFiles in This Item:\n\nFile Description SizeFormat\nindex.html0KbHTML25View/Open" ]	[ null, "http://ntour.ntou.edu.tw:8080/ir/image/logo.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.80542964,"math_prob":0.773238,"size":2378,"snap":"2019-51-2020-05","text_gpt3_token_len":638,"char_repetition_ratio":0.10488627,"word_repetition_ratio":0.0,"special_character_ratio":0.23675357,"punctuation_ratio":0.14285715,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9684252,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-13T00:24:56Z\",\"WARC-Record-ID\":\"<urn:uuid:ec4f2537-496f-4aac-8677-23d1469835a3>\",\"Content-Length\":\"27440\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:22083248-54c0-4029-8a76-25b82fae9f69>\",\"WARC-Concurrent-To\":\"<urn:uuid:e0b2012e-889e-48a1-b194-3efeee313bd5>\",\"WARC-IP-Address\":\"140.121.140.13\",\"WARC-Target-URI\":\"http://ntour.ntou.edu.tw:8080/ir/handle/987654321/50482\",\"WARC-Payload-Digest\":\"sha1:L5IOFLURN4F7KHYTCZJQUKVSIB7ZSGXJ\",\"WARC-Block-Digest\":\"sha1:WDGNNULCJNQ3NWBRSOICM7A64K5AEYT4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540547536.49_warc_CC-MAIN-20191212232450-20191213020450-00047.warc.gz\"}"}
https://www.geeksforgeeks.org/circles/	[ "Open In App\nRelated Articles\n• CBSE Class 9 Maths Revision Notes\n• CBSE Class 9 Maths Formulas\n• NCERT Solutions for Class 9 Maths\n• RD Sharma Class 9 Solutions\n\n# Circle\n\nIn maths, a circle is a 2-D closed shape. It is the basic shape that we observed a lot in our daily life. In maths, a circle is defined as a curve in which all the points on the curve are equal distances from a fixed point. The fixed point is called the centre of the circle and the fixed distance is the radius of the circle. We came across various objects that resemble the circle in real life such as coins, CDs, DVDs, Bangles, etc all resemble circular shapes.\n\nThe circle has various formulas including the area of circle and the circumference of circle. Area of the circle is defined as the space occupied inside the boundary of the circle that the circumference of the circle is the length of the boundary of the circle.\n\nLet us more about what is a circle, circle formulas, and important terms along with examples and FAQs in detail in this article.", null, "Table of Content\n\n## What is Circle in maths?\n\nA circle is a two-dimensional closed figure formed by joining all the points which are at a fixed distance from a fixed point. It is also defined as the locus of the point which is always at a constant distance from a fixed point. The fixed point is called the origin or the centre of the circle and the fixed distance of the points from the origin is called the radius of the circle. We can have an infinite radius of a circle. All the lines passing through the centre of the circle divide the circle into two equal parts that are mirror images of each other. The equation of the circle in the cartesian plane with centre at (h, k) and radius r is,\n\n(x-h)2 + (y-k)2 = r2\n\n### Definition of Circle:\n\nA circle is a two-dimensional geometric shape that consists of all points in a plane that are equidistant (at the same distance) from a fixed point called the center. The distance from the center to any point on the circle is called the radius of the circle\n\n## Circle Shaped Objects\n\nVarious objects that we observe in real life are circular shapes. Some examples of circular-shaped objects are,\n\n• Chapattis\n• Coins\n• Wheels\n• Rings\n• Buttons\n• CDs / DVDs\n• Bangles\n• Plates\n\nApart from these many more objects from day-to-day life resembles circular objects. The image added below show some real example of the circle,", null, "## How to Draw a Circle?\n\nDrawing a circle is the most basic construction problem that is taught to students. Construction a circle requires some basic tools such as,\n\n• Pencil\n• Compass\n• Scale or Ruler\n\nTo draw a circle follow the steps below,\n\nStep 1: Attach the pencil to the compass and make sure the tip of the compass and pencil are aligned properly.\n\nStep 2: Mark a point on the paper and call it the centre of the circle.\n\nStep 3: Measure the radius of the circle using the ruler and adjust the compass accordingly.\n\nStep 4: Draw the circle using the compass from the centre point marked in Step 2.\n\n## Parts of a Circle\n\nA circle has various parts and some of the important parts of the circle that are used to study the properties and theorems of the circle are,\n\nCentre of Circle: It is defined as the point from which any point on the circumference of the circle is at a fixed distance.\n\nRadius of Circle: The distance between any point on the circumference of the circle and the fixed point called the centre is called the radius of the circle. It is denoted using the letter ‘r’. A circle can have an infinite radius.\n\nDiameter of Circle: We define the diameter of the circle as the straight line passing through the centre of the circle and connecting any two points on the boundary of the circle. It is denoted by the letter ‘D’. The relation between the radius and the centre is,\n\nD = 2r\n\nChord of a Circle: Any line which touches the circle at two points on its circumference is the chord of the circle. The diameter is also a chord of the circle. The longest chord of the circle is the diameter of the circle.\n\nTangent: A line that touches the circle only at one point on its circumference is called the tangent of the circle. It is always perpendicular to the radius of the circle at the point of tangency.\n\nSecant: A line intersecting the circumference of the circle is called the secant on the circle. We also call it the extended chord.\n\nArc of a Circle: Any portion of the circumference of the circle is called the arc of the circle.\n\nSegment in a Circle: The chord divides the circle into two parts and each part is called the segment of the circle. There are two segments formed by a chord that is,\n\n• Major Segment.\n• Minor Segment\n\nSector of a Circle: The area between two radii and the corresponding arc in a circle is called the sector of the circle. There are two types of sectors\n\n• Major Sector\n• Minor Sector\n\nThe image added below shows the parts of a circle.", null, "The fixed point is named the centre of the circle and therefore the fixed distance is named the radius of the circle. The constant distance between any point on the circle and its centre is called the radius. Note that the line segment joining the centre and any point on the circle is additionally called a radius of the circle.\n\nr = D/2", null, "### Diameter of Circle (D)\n\nA chord passing through the centre of the circle is called the diameter of the circle. Two radii at 180 degrees angle are the diameter of the circle. Diameter is the largest chord of the circle and every diameter has an equivalent length, which is adequate to twice the radius.\n\nD = 2r", null, "## Interior and Exterior of Circle\n\nIf we draw a circle it divides the 2-D plane into three parts which are,\n\n• Inside the Circle: A point lying in the plane of the circle such that its distance from its centre is less than the radius of the circle is known as the interior point.\n• On the Circle: Points lying in the plane of the circle such that its distance from its centre is equal to the radius of a circle. In simple words, a set of points lying on the circle are points on the circumference of a circle.\n• Outside the Circle: Points lying in the plane of the circle such that its distance from its centre is greater than the radius of the circle are exterior points.\n\n## Circle Formulas\n\nThere are various formulas related to the circle. Let the radius of the circle is ‘r’ then some of the important formulas related to the circle are,\n\nParts\n\nFormulae\n\nArea of Circleπr2\nCircumference of Circle2πr\nLength of Arc\n\nθ × r\n\nwhere,\n‘θ’ is the angle subtended by the arc in radian\n‘r’ is the radius of the arc\n\nArea of Sector of Circle\n\n(θ × r2) / 2\n\nwhere,\n‘θ’ is the angle subtended by the arc in radian\n‘r’ is the radius of the arc\n\nLength of Chord\n\n2 r sin(θ/2)\n\nwhere,\n\nArea of Segment\n\nr2(θ − sinθ)/2\n\nwhere,\n\nThe value of π is taken to be 3.14 or 22/7.\n\n## Properties of Circle\n\nA circle is the most common 2-D shape and some of the properties of the circle are,\n\n• Circle with the same radii are called the congruent circle.\n• Equal chords are equidistant from the centre of the circle.\n• Equidistant chords from the centre of the circle are always equal.\n• The perpendicular drawn from the centre of the circle to the chord always bisects the chord.\n• We can draw two tangents from an external point to a circle.\n• Tangents drawn from the endpoints of the diameter are always parallel to each other.\n\n## Area of a Circle Proof\n\nWe can easily prove the formula for the area of the circle using the area of the triangle formula. For this first, we have to draw various concentric circles inside the given circle. Then open all the concentric circles to form a right-angled triangle.", null, "If the radius of the given circle is r, then the outer circle would form the base of the right triangle having length 2πr.\n\nThe height of the triangle is ‘r’\n\nThe area of the right-angled triangle so formed is equal to the area of a circle.\n\nArea of a circle = Area of triangle = (1/2) × base × height\n\n= (1/2) × 2π r × r\n\nTherefore,\n\nArea of a Circle = πr2\n\nRelated Resources:\n\n## Solved Examples on Circles\n\nExample 1: If the diameter of a circle is 142.8 mm, then what is the radius of the circle?\n\nSolution:\n\nDiameter = 142.8 mm\n\nBy Formula,\n\n= 71.4 mm\n\nThus, the radius of the circle is 71.4 mm\n\nExample 2: The distance around a park is 21.98 yd. What is the radius of the park?\n\nSolution:\n\nCircumference of the Park = 21.98 yd\n\nWe know that,\n\n= 3.5\n\nThus, the radius of the circle is 3.5 yd\n\nExample 3: The inner circumference of a circular track is 440 m, and the track is 14 m wide. Calculate the cost of levelling the track at 25 rupees/m2.\n\nSolution:\n\nLet the radius of the inner circle be r m.\n\nNow,\n\nInner Circumference = 440 m\n\n2πr = 440\n\n2 × 22/7 × r = 440\n\nr = 440 × 744\n\nWidth of track = 14 m\n\n= (70 + 14) = 84 m\n\nArea of the Track = π(R2− r2)\n\n= π (842 – 702)\n\n= 22/7 × (7056 – 4900)\n\n= 6776 m2\n\nCost of levelling at 25 rupees per square meter = 6676 × 25\n= 169400 Rupees\n\nThus, the cost of leveling the track is 169400 Rupees\n\nExample 4: Find the length of the chord of a circle where the radius is 7 cm and the perpendicular distance from the chord to the centre is 4 cm.\n\nSolution:\n\nGiven,\n\nDistance of the chord to the centre, d = 3 cm\n\nChord length = 2√(r2 – d2\n\n= 2√(82 – 32\n\n= 2√(64 – 9)\n\n= 2√55\n\n= 2 × 7.416\n\nChord length = 14.83 cm\n\nThus, the length of the chord is 14.83 cm\n\nExample 5: If the radius of a circle is 5 cm and the measure of the angle of the arc is 110˚, what is the length of the arc?\n\nSolution:\n\nArc Length = (2 × π × r) × angle / 360°\n\n= 2 × 3.14 × 5 × 110/360°\n\n= 9.6 cm\n\nThus, the length of the arc is 9.6 cm.\n\nExample 6: If the area of a sector with a radius of 6 cm is 35.4 cm2. Calculate the angle of the sector.\n\nSolution:\n\nArea of Sector = Angle/360° × (π × r × r)\n\nAngle/360° × (π × 6 × 6) = 35.4\n\nAngle = (35.4/36π) × 360°\n\n= 112.67°\n\nThus, the angle of the sector is, 112.67°\n\n## FAQs on Circles\n\n### 1. What is a Circle in Geometry?\n\nWe define a circle in geometry as the locus of the point which is always at a fixed distance from a fixed point, called the centre of the circle.\n\n### 2. What are the formulas of a Circle?\n\nA circle has various formulas but the important formulas for the circle are,\n\n• Area of Circle(A) = πr2\n• Circumference of Circle(C) = 2πr\n\n### 3. What is the Chord in a Circle?\n\nA line segment joining two points of the circumference of the circle is called the chord of the circle. A chord passing through the centre of the circle is called the radius of the circle.\n\n### 4. What is the Difference between Chord and Secant in a Circle?\n\nA chord is a line segment which joints any two points on the circumference of the circle, whereas the secant is a line joining any two points on the circumference of the circle, it is also called as extended chord.\n\n### 5. What is the Circumference of a Circle?\n\nThe length of the boundary of the circle is called the circumference of the circle. It is calculated using the formula,\n\nC = 2πr\n\n### 6. What is the Area of a Circle?\n\nThe space occupied inside the circumference of the circle is called Area of the area of the circle. It is calculated using the formula,\n\nA = πr2\n\n### 7. What are the Properties of Circles?\n\nThe properties of the circle are,\n\n• It has the infinite number of radii.\n• The radius of the circle is always perpendicular to the tangent of the circle.\n• Diameter is the longest chord of the circle\n\n### 8. What are Types of Circles in maths?\n\nWe can categorize circles into various categories but the important types of circles are\n\n• Concentric Circles: Circles having different radii but the same centre are called concentric circles.\n• Congruent Circles: Circles having the same radius but different centres are called congruent circles." ]	[ null, "https://media.geeksforgeeks.org/wp-content/uploads/20230822111030/Cicle.png", null, "https://media.geeksforgeeks.org/wp-content/uploads/20230427100511/Circumference-of-Circle-3.jpg", null, "https://media.geeksforgeeks.org/wp-content/uploads/20230606132937/Area-Related-to-Circle-Extra-1.png", null, "https://media.geeksforgeeks.org/wp-content/uploads/20230606135050/Rdius.PNG", null, "https://media.geeksforgeeks.org/wp-content/uploads/20230606134955/Diameter.PNG", null, "https://media.geeksforgeeks.org/wp-content/uploads/20230606131601/Circle-area-using-triangle.PNG", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91572946,"math_prob":0.9887497,"size":11661,"snap":"2023-40-2023-50","text_gpt3_token_len":2960,"char_repetition_ratio":0.25349575,"word_repetition_ratio":0.13442768,"special_character_ratio":0.2575251,"punctuation_ratio":0.08728179,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99956065,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-27T15:33:33Z\",\"WARC-Record-ID\":\"<urn:uuid:b492f098-3c40-4874-b778-b9e0993060c3>\",\"Content-Length\":\"291817\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c99ac03c-bae6-4672-a9d9-043fee3e1782>\",\"WARC-Concurrent-To\":\"<urn:uuid:be4c11bb-637c-493c-8109-030ddab1b984>\",\"WARC-IP-Address\":\"23.218.218.77\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/circles/\",\"WARC-Payload-Digest\":\"sha1:OH76U6JANJVLFPD6S336AVRGRQNMKRJB\",\"WARC-Block-Digest\":\"sha1:E7OFQCJTRBXP6O5OKJBD666XZ4G7TKZC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510300.41_warc_CC-MAIN-20230927135227-20230927165227-00039.warc.gz\"}"}
https://au.mathworks.com/matlabcentral/cody/problems/42940-modulus-of-a-number/solutions/2202564	[ "Cody\n\n# Problem 42940. modulus of a number\n\nSolution 2202564\n\nSubmitted on 9 Apr 2020 by Jamal Nasir\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\nx = 1; y_correct = 1; assert(isequal(your_fcn_name(x),y_correct))\n\n2 Pass\nx = -1; y_correct = 1; assert(isequal(your_fcn_name(x),y_correct))\n\n3 Pass\nx = -3; y_correct = 3; assert(isequal(your_fcn_name(x),y_correct))\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7092015,"math_prob":0.94997084,"size":688,"snap":"2020-45-2020-50","text_gpt3_token_len":193,"char_repetition_ratio":0.14035088,"word_repetition_ratio":0.018867925,"special_character_ratio":0.2863372,"punctuation_ratio":0.103174604,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9741278,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-02T19:54:38Z\",\"WARC-Record-ID\":\"<urn:uuid:ed875355-1865-4c7f-b220-6f06dbd9b96e>\",\"Content-Length\":\"80302\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cfdf56ae-0555-4495-a0ee-73ac2cd982a2>\",\"WARC-Concurrent-To\":\"<urn:uuid:2e75f595-3897-4c0a-b2af-6004d517e649>\",\"WARC-IP-Address\":\"23.32.68.178\",\"WARC-Target-URI\":\"https://au.mathworks.com/matlabcentral/cody/problems/42940-modulus-of-a-number/solutions/2202564\",\"WARC-Payload-Digest\":\"sha1:OESRSRKK5ZLQCQHYUEI7XTRQBTZDCOIL\",\"WARC-Block-Digest\":\"sha1:GIA3EPYWPPNGUJGDSCDXBJWZEDTU6WH7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141715252.96_warc_CC-MAIN-20201202175113-20201202205113-00028.warc.gz\"}"}
https://samritchie.io/optics-and-the-principle-of-least-time/	[ "# Optics and the Principle of Least Time\n\n(This is a writeup of Exercise 1.3 from Sussman and Wisdom's \"Structure and Interpretation of Classical Mechanics\". See the solutions repository for more.)\n\nThe problem explores some consequences for optics of the principle of least time. The exercise states:\n\nFermat observed that the laws of reflection and refraction could be accounted for by the following facts: Light travels in a straight line in any particular medium with a velocity that depends upon the medium. The path taken by a ray from a source to a destination through any sequence of media is a path of least total time, compared to neighboring paths. Show that these facts imply the laws of reflection and refraction.\n\n## Law of Reflection\n\nThe law of reflection is described in the footnote:\n\nFor reflection the angle of incidence is equal to the angle of reflection.\n\nHere's the setup. The horizontal line is a mirror. The law states that $$\\theta_1 = \\theta_2$$.", null, "We have to show that if we consider all possible paths from a given starting point to a given endpoint, the path of minimum time will give us the law of reflection.\n\nThe actual path of minimum time is the straight line that avoids the mirror, of course. If we force the light to bounce off of the mirror, then we have to figure out where it will hit, where $$x_p$$ is, to minimize the time between the start and end points.\n\nThere are two ways to solve this problem. We can use geometry and visual intuition, or we can use calculus.\n\n### Geometry\n\nFirst, recall this fact from the problem text:\n\nLight travels in a straight line in any particular medium with a velocity that depends upon the medium.\n\nThere's no medium change, so if there were no mirror in its path, the light beam would continue in a straight line. Instead of figuring out what the beam will do when it hits the mirror, reflect the endpoint across the mirror and draw a straight line between the start and \"end\" points:", null, "The angle that the beam makes with the plane of the mirror is the same on both sides of the mirror.\n\nNow reflect the the \"end\" point and the segment of the beam that's crossed the mirror back up. By symmetry, $$\\theta_1 = \\theta_2$$, and we've proved the law of reflection.\n\n### Calculus\n\nWe can also solve this with calculus. Because the beam doesn't change media, its speed $$v$$ stays constant, so minimizing the total distance $$d$$ is equivalent to minimizing the time $$t = {d \\over v}$$.\n\nSet $$x_1 = 0$$ for convenience, and write the total distance the light travels as a function of $$x_p$$:\n\n\\begin{equation}\nd(x_p) = \\sqrt{y_1^2 + x_p^2} + \\sqrt{(x_2 - x_p)^2 + y_2^2}\n\\end{equation}\n\nFor practice, we can also define this function in Scheme.\n\n(define ((total-distance x1 y1 x2 y2) xp)\n(+ (sqrt (+ (square (+ x1 xp))\n(square y1)))\n(sqrt (+ (square (- x2 (+ x1 xp)))\n(square y2)))))\n\n\nHere's the function again, generated from code, with general $$t_1$$:\n\n(->tex-equation\n((total-distance 'x_1 'y_1 'x_2 'y_2) 'x_p))\n\n\n\\begin{equation}\n\\sqrt{{{x}_{1}}^{2} + 2 {x}_{1} {x}_{p} + {{x}_{p}}^{2} + {{y}_{1}}^{2}} + \\sqrt{{{x}_{1}}^{2} - 2 {x}_{1} {x}_{2} + 2 {x}_{1} {x}_{p} + {{x}_{2}}^{2} - 2 {x}_{2} {x}_{p} + {{x}_{p}}^{2} + {{y}_{2}}^{2}}\n\\end{equation}\n\nTo find the $$x_p$$ that minimizes the total distance,\n\n• take the derivative with respect to $$x_p$$,\n• set it equal to 0 and\n• solve for $$x_p$$.\n\nThe derivative will look cleaner in code if we keep the components of the sum separate and prevent Scheme from \"simplifying\". Redefine the function to return a tuple:\n\n(define ((total-distance* x1 y1 x2 y2) xp)\n(up (sqrt (+ (square (+ x1 xp))\n(square y1)))\n(sqrt (+ (square (- x2 (+ x1 xp)))\n(square y2)))))\n\n\nHere are the sum components:\n\n(->tex-equation\n((total-distance* 0 'y_1 'x_2 'y_2) 'x_p))\n\n\n\\begin{equation}\n\\begin{pmatrix} \\displaystyle{ \\sqrt{{{x}_{p}}^{2} + {{y}_{1}}^{2}}} \\cr \\cr \\displaystyle{ \\sqrt{{{x}_{2}}^{2} - 2 {x}_{2} {x}_{p} + {{x}_{p}}^{2} + {{y}_{2}}^{2}}}\\end{pmatrix}\n\\end{equation}\n\nTaking a derivative is easy with scmutils. Just wrap the function in D:\n\n(let* ((distance-fn (total-distance* 0 'y_1 'x_2 'y_2))\n(derivative (D distance-fn)))\n(->tex-equation\n(derivative 'x_p)))\n\n\n\\begin{equation}\n\\begin{pmatrix} \\displaystyle{ {{{x}_{p}}\\over {\\sqrt{{{x}_{p}}^{2} + {{y}_{1}}^{2}}}}} \\cr \\cr \\displaystyle{ {{ - {x}_{2} + {x}_{p}}\\over {\\sqrt{{{x}_{2}}^{2} - 2 {x}_{2} {x}_{p} + {{x}_{p}}^{2} + {{y}_{2}}^{2}}}}}\\end{pmatrix}\n\\end{equation}\n\nThe first component is the base of base $$x_p$$ of the left triangle over the total length. This ratio is equal to $$\\cos \\theta_1$$:", null, "The bottom component is $$-\\cos \\theta_2$$, or $${- (x_2 - x_p)}$$ over the length of the right segment. Add these terms together, set them equal to 0 and rearrange:\n\n\\begin{equation}\n\\label{eq:reflect-laws}\n\\cos \\theta_1 = \\cos \\theta_2 \\implies \\theta_1 = \\theta_2\n\\end{equation}\n\nThis description in terms of the two incident angles isn't so obvious from the Scheme code. Still, you can use Scheme to check this result.\n\nIf the two angles are equal, then the left and right triangles are similar, and the ratio of each base to height is equal:\n\n\\begin{equation}\n\\label{eq:reflect-ratio}\n{x_p \\over y_1} = {{x_2 - x_p} \\over y_2}\n\\end{equation}\n\nSolve for $$x_p$$ and rearrange:\n\n\\begin{equation}\n\\label{eq:reflect-ratio2}\nx_p = {{y_1 x_2} \\over {y_1 + y_2}}\n\\end{equation}\n\nPlug this in to the derivative of the original total-distance function, and we find that the derivative equals 0, as expected:\n\n(let* ((distance-fn (total-distance 0 'y_1 'x_2 'y_2))\n(derivative (D distance-fn)))\n(->tex-equation\n(derivative (/ (* 'y_1 'x_2) (+ 'y_1 'y_2)))))\n\n\n\\begin{equation}\n0\n\\end{equation}\n\nIf a beam of light travels in a way that minimizes total distance (and therefore time in a constant medium), then it will reflect off of a mirror with the same angle at which it arrived. The law of reflection holds.\n\n## Law of Refraction\n\nThe law of refraction is also called Snell's law. Here's the description from the footnote:\n\nRefraction is described by Snell's law: when light passes from one medium to another, the ratio of the sines of the angles made to the normal to the interface is the inverse of the ratio of the refractive indices of the media. The refractive index is the ratio of the speed of light in the vacuum to the speed of light in the medium.\n\nFirst we'll tackle this with calculus.\n\n### Calculus\n\nThe setup here is slightly different. We have a light beam traveling from one medium to another and changing speeds at a boundary located $$a$$ to the right of the starting point. The goal is to figure out the point where the light will hit the boundary, if we assume that the light will take the path of least time.", null, "The refractive index $$n_i = {c \\over v_i}$$, the speed of light $$c$$ in a vacuum over the speed in the material. Rearranging, $$v_i = {c \\over n_i}$$.\n\nTime is distance over speed, so the total time that the beam spends between the start and end points as a function of $$y_p$$, the point of contact with the boundary, is:\n\n\\begin{equation}\n\\begin{split}\nt(y_p) & = {c \\sqrt{a^2 + y_p^2}\\over v_1} + {c \\sqrt{(x_2 - x_p)^2 + y_2^2} \\over v_2} \\\n& = {n_1 \\over c} \\sqrt{a^2 + y_p^2} + {n_2 \\over c} \\sqrt{(x_2 - x_p)^2 + y_2^2}\n\\end{split}\n\\end{equation}\n\nTake the derivative:\n\n\\begin{equation}\nDt(y_p) = {1 \\over c} \\left({n_1 y_p \\over \\sqrt{a^2 + y_p^2}} - {n_2 (x_2 - x_p) \\over \\sqrt{(x_2 - x_p)^2 + y_2^2}}\\right)\n\\end{equation}\n\nSet the derivative equal to 0 and split terms:\n\n\\begin{equation}\n\\label{eq:almost-snell}\n{n_1 y_p \\over \\sqrt{a^2 + y_p^2}} = {n_2 (x_2 - x_p) \\over \\sqrt{(x_2 - x_p)^2 + y_2^2}}\n\\end{equation}\n\nSimilar to the law of reflection's result, each term (up to its $$n_i$$ multiple) is equal to the height of the left or right triangle over the length of the beam's path on the left or right of the boundary.\n\nEquation \\eqref{eq:almost-snell} simplifies to:\n\n\\begin{equation}\nn_1 \\sin \\theta_1 = n_2 \\sin \\theta_2\n\\end{equation}\n\nRearranging yields Snell's law:\n\n\\begin{equation}\n{n_1 \\over n_2} = {\\sin \\theta_2 \\over \\sin \\theta_1}\n\\end{equation}\n\n### Geometry\n\nI won't recreate this here, but the Feynman Lectures on Physics, in Lecture 26, has a fantastic discussion about, and derivation of, the law of refraction using no calculus, just geometry. I highly recommend you check out that lecture. Feynman lays out a number of examples of how the principle of least time is not just a restatement of the optical rules we already knew.\n\nYou can use the idea to guess what shape of mirror you'd want to build to focus many light rays on a single point (a parabola), or how you might force all light rays coming out of a single point to meet up again at another point (build a converging lens).\n\nThis whole area of optics and least time has obsessed scientists for hundreds of years. Spend a few minutes poking around and see what you find." ]	[ null, "https://github.com/sritchie/sicm/raw/master/images/Lagrangian_Mechanics/2020-06-10_10-31-24_screenshot.png", null, "https://github.com/sritchie/sicm/raw/master/images/Lagrangian_Mechanics/2020-06-10_10-36-53_screenshot.png", null, "https://github.com/sritchie/sicm/raw/master/images/Lagrangian_Mechanics/2020-06-10_10-36-53_screenshot.png", null, "https://github.com/sritchie/sicm/raw/master/images/Lagrangian_Mechanics/2020-06-10_12-03-11_screenshot.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8281164,"math_prob":0.9997863,"size":8785,"snap":"2023-14-2023-23","text_gpt3_token_len":2591,"char_repetition_ratio":0.13939187,"word_repetition_ratio":0.0734072,"special_character_ratio":0.31485486,"punctuation_ratio":0.07164179,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999833,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,4,null,8,null,8,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-06T13:36:10Z\",\"WARC-Record-ID\":\"<urn:uuid:f1c8ed0c-c512-4cef-a0da-342e40317ae7>\",\"Content-Length\":\"29041\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2ca1aee7-5d27-4d83-a554-17ac6d4d4e70>\",\"WARC-Concurrent-To\":\"<urn:uuid:c276e72b-ee8b-4fe1-8b9d-25981d0a1efa>\",\"WARC-IP-Address\":\"151.101.195.7\",\"WARC-Target-URI\":\"https://samritchie.io/optics-and-the-principle-of-least-time/\",\"WARC-Payload-Digest\":\"sha1:2EWK6Q2W2T6YTDIPBMWDLXVO7JQZQTKA\",\"WARC-Block-Digest\":\"sha1:UY4KP4GVIA7A3DOJSMJAATRVDIEBQCJQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224652569.73_warc_CC-MAIN-20230606114156-20230606144156-00296.warc.gz\"}"}
https://instrumentationtools.com/how-to-read-gauge-analog-scales/	[ " How to Read Gauge Analog Scales - Inst Tools\n\n# How to Read Gauge Analog Scales\n\n#### 1. Interpret the pressure measurement displayed by this gauge mechanism, assuming a gauge accuracy of ± 1% full scale:", null, "#### How low and how high could this pressure actually be, given the stated accuracy of this gauge?\n\nPressure = 3750 PSI ± 60 PSI\n\nThis means the actual pressure could be as low as 3690 PSI or as high as 3810 PSI\n\n#### 2. Interpret the pressure measurement displayed by this gauge mechanism, assuming a gauge accuracy of ± 2% full scale:", null, "#### How low and how high could this pressure actually be, given the stated accuracy of this gauge?\n\nPressure = 7800 PSI ± 400 PSI\n\nThis means the actual pressure could be as low as 7400 PSI or as high as 8200 PSI\n\n#### 3. Interpret the pressure measurement displayed by this gauge mechanism, assuming a gauge accuracy of ± 0.5% full scale:", null, "#### How low and how high could this pressure actually be, given the stated accuracy of this gauge?\n\nPressure = 560 PSI ± 5 PSI\n\nThis means the actual pressure could be as low as 555 PSI or as high as 565 PSI.\n\n#### 4. Interpret the pressure measurement displayed by this gauge mechanism, assuming a gauge accuracy of ± 2% full scale:", null, "#### How low and how high could this pressure actually be, given the stated accuracy of this gauge?\n\nPressure = 2.5 ”H2O ± 0.6 ”H2O. This means the actual pressure could be as low as 1.9 ”H2O or as high as 3.1 ”H2O.\n\nIf you look closely at the photograph, you can see that the camera’s angle to the gauge face is not straight-on, and therefore there will be some parallax error in reading this gauge’s face.\n\nIf we were to lower the camera’s view to get a more direct look at the gauge, we might see the needle pointing between the 2.5 and 3 divisions, which would mean a pressure of 2.75 ”H2O ±0.6 ”H2O.\n\n#### 5. Interpret the pressure measurement displayed by this gauge mechanism, assuming a gauge accuracy of ± 2% full scale:", null, "#### How low and how high could this pressure actually be, given the stated accuracy of this gauge?\n\nPressure = 14 PSI ± 2 PSI\n\nThis means the actual pressure could be as low as 12 PSI or as high as 16 PSI.\n\n#### 6. Interpret the temperature measurement displayed by this gauge mechanism, and also identify the meaning of the other pointer:", null, "The current temperature is 40 degrees Celsius (red pointer), and the other pointer is a low-temperature capture. In this case, the “capture” pointer shows that the temperature went down as low as 20 degrees Celsius (or perhaps a bit lower, since parallax error is making that pointer appear to read higher than it actually is).\n\n#### 7. Interpret the measurement displayed by this Level Indicator :", null, "The red pointer shows 61% process level. The setpoint is a few percent less than that.\n\n#### 8. Interpret the air flow measurement displayed by this rotameter:", null, "Rounding to the nearest half-division, we see the center of the ball-plummet nearest the “1” mark. No units of measurement are specified on the scale (from this viewing angle, anyway), and so we cannot say if this is 1 SCFH, 1 SCFM, or anything else.\n\n#### 9. Interpret the pressure measurement displayed by this U-tube manometer, recalling that the pressure will be equal to the difference in heights between the two liquid columns:", null, "" ]	[ null, "https://cdn.instrumentationtools.com/wp-content/uploads/2017/09/instrumentationtools.com_pressure-gauge-analog-scale.png", null, "https://cdn.instrumentationtools.com/wp-content/uploads/2017/09/instrumentationtools.com_pressure-gauge-scale.png", null, "https://cdn.instrumentationtools.com/wp-content/uploads/2017/09/instrumentationtools.com_pressure-gauge-measurement.png", null, "https://cdn.instrumentationtools.com/wp-content/uploads/2017/09/instrumentationtools.com_pressure-gauge-reading.png", null, "https://cdn.instrumentationtools.com/wp-content/uploads/2017/09/instrumentationtools.com_gauge-display-scale.png", null, "https://cdn.instrumentationtools.com/wp-content/uploads/2017/09/instrumentationtools.com_temperature-gauge-reading.png", null, "https://cdn.instrumentationtools.com/wp-content/uploads/2017/09/instrumentationtools.com_level-indicator-display-scale.png", null, "https://cdn.instrumentationtools.com/wp-content/uploads/2017/09/instrumentationtools.com_rotameter-scale-reading.png", null, "https://cdn.instrumentationtools.com/wp-content/uploads/2017/09/instrumentationtools.com_level-gauge-using-manometer.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92544234,"math_prob":0.92587674,"size":3402,"snap":"2021-31-2021-39","text_gpt3_token_len":812,"char_repetition_ratio":0.16421424,"word_repetition_ratio":0.33496732,"special_character_ratio":0.2489712,"punctuation_ratio":0.1125731,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9896863,"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,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-03T03:49:25Z\",\"WARC-Record-ID\":\"<urn:uuid:bb6af3e9-fc98-4f5d-91f5-d41c8ebd5694>\",\"Content-Length\":\"136302\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8195bed6-81c7-4555-8363-98b2631232f5>\",\"WARC-Concurrent-To\":\"<urn:uuid:005407dc-614d-49f1-9216-c6e5265f52b7>\",\"WARC-IP-Address\":\"172.67.218.42\",\"WARC-Target-URI\":\"https://instrumentationtools.com/how-to-read-gauge-analog-scales/\",\"WARC-Payload-Digest\":\"sha1:MHIFQ3XPNUP5ZAMTYKTBF6H2DXMTGA5H\",\"WARC-Block-Digest\":\"sha1:GVIIBYD7QTSQHTPD6ZYRFOUGZCVGYQRA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154420.77_warc_CC-MAIN-20210803030201-20210803060201-00686.warc.gz\"}"}
https://www.scribd.com/document/317792191/20PCoordFull-pdf	[ "You are on page 1of 27\n\n# CoordinateGeometry\n\nNotes\nFullSet\n\nstandard form is\n\ny = mx + b\n\n## 3. Complete the square and find the vertex.\n\nTo 1 decimal place, find the coordinates of the point(s) of intersection of the graphs\nof\n\n## a) what point is the center of the circle? _________________\n\nb) what is the radius of the circle? ____________________\nc) use the distance formula to find the equation of the circle.\n\n## a) what point is the center of the circle? _________________\n\nb) what is the radius of the circle? ____________________\nc) use the distance formula to find the equation of the circle.\n\n(3,-2)\n\n(-5,3)\n\n(5,-2)\n\n(0,3)\n\nstandard form\n\ngeneral form\n\n## Put the equation of the circle in general form:\n\n10\n\nWrite these circles in standard form by completing the squares in both x and y. Give the\n\n11\n\n12\n\n## Equation of the Tangent Line to a Circle\n\n1. Give the equation of the circle. Then find the equations of the vertical and horizontal\ntangent lines to the cirle graphed below.\n\n2. Find the equations of the horizontal and vertical tangents to the circle given by\n\n13\n\n## For all \"slanted\" tangents follow these steps:\n\nsketch\nfind slope of tangent line\nuse y = mx + b to find the equation of the tangent line\n3. Find the equation of the tangent to\n\nat (4,-5).\n\n## For all \"slanted\" tangents follow these steps:\n\nsketch\nfind slope of tangent line\nuse y = mx + b to find the equation of the tangent line\n4. Find the equation of the tangent to\n\nat (2,-1).\n\n14\n\n15\n\n16\n\n17\n\n## Splitting a Segment into Equal Smaller Segments\n\nFind the points that split (-10,5) to (6,1) into four equal smaller segments.\n\n18\n\n## Calculating the Distance between Points and Lines\n\n1. Calculate the shortest distance between the point and each of the lines pictured in\nthe following graph.\n\n2. Calculate the shortest distance between the point and the line.\n\n## The line y = -2x + 3 and the point (-5,1).\n\nThe steps.\nfind the slope of the line\nfind the slope of the\nperpendicular line through\nthe point\nfind the equation of the\nperpendicular line\nsolve the system to get\nthe closest point\nfind the distance\n\n19\n\n3. Calculate the shortest distance between the point and the line.\n\n## The line 2x + 3y - 6 = 0 and the point (5,4).\n\nThe steps.\nfind the slope of the line\nfind the slope of the\nperpendicular line through\nthe point\nfind the equation of the\nperpendicular line\nsolve the system to get\nthe closest point\nfind the distance\n\n20\n\n## 2. Find the distance between the two vertical lines.\n\n3. Find the horizontal distance, the vertical distance and the shortest distance between\nthe two parallel lines. y = 2x - 6 and y = 2x + 4\n\n21\n\n4. Find the horizontal distance, the vertical distance and the shortest distance between\nthe two parallel lines. 2x + 3y -6 = 0 and 2x + 3y + 12 = 0.\n\nVerifying Conjectures\n\n## Consider triangle A(-1,3), B(1,7) and\n\nC(5,5).\na) verify that triangle ABC is a right\ntriangle\n\nparallel to BC.\n\n22\n\n## A line segment has endpoints M(8,6) and N(-6,8).\n\na) verify that the endpoints are on the circle.\n\nb) determine the equation of the line through the center of the circle and the midpoint\nof the chord MN\n\n## c) verify that the new line\n\nand MN are perpendicular\n\nProving Conjectures\nProve that the diagonals of a square are perpendicular.\n\n23\n\n## Coordinate Geometry - unit review\n\nCalculate the length, the slope and the midpoint of the segment below.\n\nSeparate the segment from (-8,6) to (22,-6) into six smaller segments of equal\nlength.\n\n24\n\nFind\n\ncenter\ndomain\nrange\nx intercepts\nvertical tangents\n\n## Calculate the equation of the tangent to\n\nat (4,-1) .\n\n25\n\nFind the shortest distance from the point (1,3) to the line 4x - 3y - 12 = 0.\n\n26\n\n27" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86159956,"math_prob":0.99915636,"size":1951,"snap":"2019-43-2019-47","text_gpt3_token_len":509,"char_repetition_ratio":0.16692348,"word_repetition_ratio":0.16766468,"special_character_ratio":0.26089185,"punctuation_ratio":0.11640212,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99990916,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-13T04:48:12Z\",\"WARC-Record-ID\":\"<urn:uuid:99d63be3-3b2d-4a9a-b9e0-438e392f1ee7>\",\"Content-Length\":\"283983\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:98f46fff-b785-4b34-920d-0b35771602da>\",\"WARC-Concurrent-To\":\"<urn:uuid:9459390d-48cf-471b-a7bb-e967d9c343f9>\",\"WARC-IP-Address\":\"151.101.202.152\",\"WARC-Target-URI\":\"https://www.scribd.com/document/317792191/20PCoordFull-pdf\",\"WARC-Payload-Digest\":\"sha1:4R6VRDX73UGWU65CMJ2OLT55UY4ZVH33\",\"WARC-Block-Digest\":\"sha1:DCI5CWBT3ZYRVM5KOA46E4ZRPWBAC3MV\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496665985.40_warc_CC-MAIN-20191113035916-20191113063916-00172.warc.gz\"}"}
https://pdglive.lbl.gov/DataBlock.action?node=B024RS3&home=BXXX005	[ "# ${{\\boldsymbol N}{(2120)}}$ INELASTIC POLE RESIDUE\n\nThe normalized residue'' is the residue divided by $\\Gamma _{pole}$/2.\n\n# Normalized residue in ${{\\boldsymbol N}}$ ${{\\boldsymbol \\pi}}$ $\\rightarrow$ ${{\\boldsymbol N}{(2120)}}$ $\\rightarrow$ ${{\\boldsymbol N}{(1535)}}{{\\boldsymbol \\pi}}$ INSPIRE search\n\nMODULUS PHASE ($^\\circ{}$) DOCUMENT ID TECN COMMENT\n$0.15 \\pm0.08$ -90+-40\n 2014\nDPWA Multichannel\nReferences:\n GUTZ 2014\nEPJ A50 74 High Statistics Study of the Reaction" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5683319,"math_prob":0.9985361,"size":415,"snap":"2020-45-2020-50","text_gpt3_token_len":144,"char_repetition_ratio":0.20924574,"word_repetition_ratio":0.0,"special_character_ratio":0.41686746,"punctuation_ratio":0.06896552,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9950774,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-22T18:50:44Z\",\"WARC-Record-ID\":\"<urn:uuid:3ed0b6ed-cc95-4efa-a60e-1bce3877dc83>\",\"Content-Length\":\"17985\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5b67261b-95fa-4ace-ad64-fc77a4716588>\",\"WARC-Concurrent-To\":\"<urn:uuid:abde4034-b273-46b2-aa61-79e1edc07ed2>\",\"WARC-IP-Address\":\"54.212.177.198\",\"WARC-Target-URI\":\"https://pdglive.lbl.gov/DataBlock.action?node=B024RS3&home=BXXX005\",\"WARC-Payload-Digest\":\"sha1:E6JCDCWLLE47OGVZ3GZRWWZK4BOPT5VL\",\"WARC-Block-Digest\":\"sha1:Z3V6ZPVQQVBKK2N2E5UE6VXEBHC4M4GE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107880014.26_warc_CC-MAIN-20201022170349-20201022200349-00170.warc.gz\"}"}
https://www.weegy.com/?ConversationId=255EKXJH&Link=i&ModeType=2	[ "Solve for c. ab - c = 9\nUpdated 4/30/2015 6:59:20 PM\ns\nOriginal conversation\nUser: Solve for c. ab - c = 9\n\nWeegy: Q - 7 5/6 = 6 1/2\n\nQ = 6 1/2 + 7 5/6\n\n= ((13 X 6) + (47 X 2)) / (2 X 6)\n\n= (78 + 94) / 12\n\n= 172/12\n\n= 43/3\n\n= 14 1/3\n\nUser: Use the listing method to write the following set. {x \| x N, x < 3} {1, 2, 3} {0, 1, 2, 3} {1, 2}\n\nWeegy: I assume you mean divided by and the answer would be 1\ndebnjerry\|Points 68770\|\n\nUser: The length of a rectangle is 3 inches more than its width, and its perimeter is 22 inches. Find the width of the rectangle.\n\nWeegy: Let x be its width, then its length is (x + 3). perimeter = 2x + 2(x + 3) = 22, solve the equation we get x = 4. Its width is 4 inches.\nyeswey\|Points 29277\|\n\nUser: Given the line 2x - 3y - 5 = 0, find the slope of a line that is perpendicular to this line.\n\nWeegy: Given the line 2x - 3y - 5 = 0, the slope of a line that is perpendicular to this line is -3/2.\n\nUser: A man invests a certain amount of money at 2% interest and \\$800 more than that amount in another account at 4% interest. At the end of one year, he earned \\$92 in interest. How much money was invested in each account?\n\nWeegy: Let the man invest an amount x at 2% interest. Then, [ he invests an amount (x+800) at 4% interest. Total simple interest for one year is x2/100 + (x+800)4/100 = 92 2x+4x+3200 = 9200 6x = 9200-3200 6x = 6000 x = 1000 So the man invested \\$ 1000 at 2% interest and \\$ 1800 at 4% interest. ]\nsweet25\|Points 90\|\n\nUser: What is the algebraic expression for \"the difference between seven times a number and three times that number\"?\n\nUser: Describe the location of the point that has the following coordinates. positive abscissa, positive ordinate\n\na_dgzmn\|Points 96\|\n\nUpdated 4/30/2015 6:59:20 PM\nRating\n3\nab - c = 9\n- c = 9 - ab\nc = -9 + ab\n\nSolve for c. ab - c = 9\nWeegy: ab - c = 9; c = ab - 9 User: Solve 2(4x + 3) Weegy: 2(4x + 3) User: What is the equation of the line, in standard form, that passes through (4, -3) and is parallel to the line whose equation is 4x + y - 2 = 0? (More)\nUpdated 5/4/2015 12:44:56 PM\nThe equation of the line, in standard form, that passes through (4, -3) and is parallel to the line whose equation is 4x + y - 2 = 0 is 4x + y = 13\nConfirmed by Andrew. [5/4/2015 12:45:19 PM]\nWhat is the purpose of a thesis? It protects you against plagiarism. It completes your outline. It states what your paper will prove. It organizes your first paragraph.\nWeegy: The purpose of a thesis is that it states what your paper will prove. User: Select whether the following citation is done correctly: . . . discovered that cheetahs were the fastest animals on earth (J. Bogart, p.33). (More)\nUpdated 4/24/2015 4:08:09 PM\nSimplify 18 - 2[x + (x - 5)]. 8 - 4x 28 - 4x 28 - 2x\nWeegy: 18 - 2[x + (x - 5)] =(-2x + -2x) + (18 + 10) = -4x + 28 (More)\nUpdated 4/30/2015 12:56:14 PM\nA speech given at a graduation ceremony attended by the city mayor should be _____ in style. semi-formal formal informal\nWeegy: A speech given at a graduation ceremony attended by the city mayor should be formal in style. (More)\nUpdated 11/8/2015 1:46:00 AM\nSolve for x. 3(2x - 3) + 4(2 - x) = 0\nWeegy: 3(2x - 3) + 4(2 - x) = 0 6x - 9 + 8 - 4x = 0 2x - 1 = 0 2x = 1 x = 1/2 User: Given: y varies directly with x and has a constant rate of change of 7. When the value of y is 12, then the value of x would be _____. (More)\nUpdated 5/4/2015 10:14:47 AM\nGiven: y varies directly with x and has a constant rate of change of 7. When the value of y is 12, then the value of x would be 12/7 .\nIf y = 7x\nwhen y = 12\nx = 12/7\nConfirmed by Andrew. [5/4/2015 10:15:15 AM]\n32,009,014\nGET\nGET THE APP.\nweegy", null, "\nGet answers from Weegy and a team of really smart live experts.\nPopular Conversations\nWhich NIMS Management Characteristic helps to eliminate confusion ...\nWeegy: The National Incident Management System (NIMS) is a standardized approach to incident management developed by ...\nCharts can either be embedded in a worksheet or placed in a separate ...\nWeegy: Charts can either be embedded in a worksheet or placed in a separate chart sheet. User: A relative cell ...\nThe Incident Command System (ICS) is only applicable to large, ...\nWeegy: The incident command system ICS is only applicable to large complex incidents. FALSE. User: . Which type of ...\nThe Eighteenth Amendment, \"Prohibition,\" was reversed by the _______ ...\nWeegy: The Eighteenth Amendment, \"Prohibition,\" was reversed by the TWENTY-FIRST Amendment. User: Under Article ...\nWhich one of the following is a requirement when becoming a US ...\nWeegy: A requirement when becoming a US citizen is You must live in the United States for at least one year.\nS\nL\nP\nL\nP\nPoints 1013 [Total 7036] Ratings 1 Comments 1003 Invitations 0 Online\nS\nL\nPoints 926 [Total 3923] Ratings 2 Comments 906 Invitations 0 Offline\nS\nL\nP\nR\nP\nR\nL\nP\nP\nC\nR\nP\nR\nL\nP\nR\nPoints 850 [Total 12941] Ratings 1 Comments 580 Invitations 26 Online\nS\nL\nP\n1\nPoints 729 [Total 3585] Ratings 3 Comments 699 Invitations 0 Offline\nS\nL\nPoints 413 [Total 1906] Ratings 1 Comments 403 Invitations 0 Offline\nS\nL\nPoints 318 [Total 4702] Ratings 1 Comments 308 Invitations 0 Offline\nS\nL\n1\nPoints 187 [Total 3498] Ratings 3 Comments 157 Invitations 0 Online\nS\nL\nPoints 172 [Total 323] Ratings 4 Comments 132 Invitations 0 Offline\nS\nL\nL\nPoints 111 [Total 5399] Ratings 2 Comments 91 Invitations 0 Offline\nS\nL\nPoints 85 [Total 244] Ratings 0 Comments 85 Invitations 0 Offline\n* Excludes moderators and previous\nwinners (Include)\nHome \| Contact \| Blog \| About \| Terms \| Privacy \| © Purple Inc." ]	[ null, "https://www.weegy.com/img/ad-appstore-4.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8554591,"math_prob":0.98718834,"size":8042,"snap":"2020-34-2020-40","text_gpt3_token_len":2791,"char_repetition_ratio":0.118188605,"word_repetition_ratio":0.17882971,"special_character_ratio":0.3582442,"punctuation_ratio":0.13587873,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96965003,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-12T03:36:11Z\",\"WARC-Record-ID\":\"<urn:uuid:c4dc8396-1f55-4481-9e2a-6bcebb3197b2>\",\"Content-Length\":\"52186\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b6c51835-8615-419c-a4b8-846cd404d993>\",\"WARC-Concurrent-To\":\"<urn:uuid:34667961-6ebe-4927-83f4-ee6c09467341>\",\"WARC-IP-Address\":\"107.155.67.98\",\"WARC-Target-URI\":\"https://www.weegy.com/?ConversationId=255EKXJH&Link=i&ModeType=2\",\"WARC-Payload-Digest\":\"sha1:AHOHAGZLIEQPHHKYTFCOKSDRG6UNMCM6\",\"WARC-Block-Digest\":\"sha1:SGIXDXOBUMLI4DRJWVYVF5CPTA6V4I7X\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738864.9_warc_CC-MAIN-20200812024530-20200812054530-00052.warc.gz\"}"}
http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-550.html	[ "Microsoft Office Tutorials and References\nIn Depth Information\nSummary\nRANDBETWEEN function to create a set of random numbers and you don’t want Excel to\nrecalculate those random numbers each time you press Enter, you can convert the\nformulas to values. Just follow these steps:\n1. Select A1:A20.\n2. Choose Home\nClipboard\nCopy (or press Ctrl+C).\n3. Choose Home Clipboard Paste arrow Paste Values (V).\n4. Press Esc to cancel Copy mode.\nSummary\nThis chapter taught you the key details about entering formulas to perform calculations in\ncells. The chapter taught you how to:\nEnter formulas using operators.\nUse the order of precedence along with parentheses to ensure a formula calculates\ncorrectly.\nInclude functions in formulas to perform more sophisticated calculations.\nUse various methods to include ranges and functions in formulas.\nMake changes to formulas.\nUse relative and absolute references.\nUse formulas with a table.\nSee and ﬁ x formula errors.", 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, "Search JabSto ::\n\nCustom Search" ]	[ null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17199.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17200.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17201.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17202.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17203.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17204.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17205.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17206.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17207.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17208.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17209.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17210.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17211.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17212.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17213.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17214.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17215.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17216.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17217.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17218.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17219.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17220.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17221.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17222.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17223.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17224.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17225.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17226.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17227.png", null, "http://jabsto.com/Tutorial/topic-98/Microsoft-Office-2013-Bible_images/img-gen17228.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.66048855,"math_prob":0.8879731,"size":1011,"snap":"2019-13-2019-22","text_gpt3_token_len":239,"char_repetition_ratio":0.12810327,"word_repetition_ratio":0.0121951215,"special_character_ratio":0.20969337,"punctuation_ratio":0.13259669,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96819496,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-19T21:53:22Z\",\"WARC-Record-ID\":\"<urn:uuid:85791c0c-4d6c-45a9-8600-21c6ec38ceeb>\",\"Content-Length\":\"28581\",\"Content-Type\":\"application/http; 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https://useful_english.enacademic.com/133230/theory_of_gravitation	[ "\n\n# theory of gravitation\n\nnoun\n(physics) the theory that any two particles of matter attract one another with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them\nTopics: ↑physics, ↑natural philosophy\nHypernyms: ↑scientific theory\n\nUseful english dictionary. 2012.\n\n### Look at other dictionaries:\n\n• Nordström's theory of gravitation — In theoretical physics, Nordström s theory of gravitation was a predecessor of general relativity. Strictly speaking, there were actually two distinct theories proposed by the Finnish theoretical physicist Gunnar Nordström, in 1912 and 1913… … Wikipedia\n\n• Le Sage's theory of gravitation — is the most common name for the kinetic theory of gravity originally proposed by Nicolas Fatio de Duillier in 1690 and later by Georges Louis Le Sage in 1748. The theory proposed a mechanical explanation for Newton s gravitational force in terms… … Wikipedia\n\n• Yilmaz theory of gravitation — The Yilmaz theory of gravitation is an attempt by Huseyin Yilmaz (Turkish: Hüseyin Yılmaz ) and a handful of coworkers to formulate a classical field theory of gravitation which closely mimics general relativity in weak field conditions, but in… … Wikipedia\n\n• Whitehead's theory of gravitation — In theoretical physics, Whitehead s theory of gravitation was introduced by the distinguished mathematician and philosopher Alfred North Whitehead in 1922. Principal features of the theoryWhitehead s theory is said to feature a prior geometry .… … Wikipedia\n\n• Newton's theory of gravitation — noun (physics) the theory that any two particles of matter attract one another with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them • Syn: ↑theory of gravitation,… … Useful english dictionary\n\n• quantum theory of gravitation — kvantinė gravitacijos teorija statusas T sritis fizika atitikmenys: angl. quantum theory of gravitation vok. Quantentheorie der Gravitation, f rus. квантовая теория гравитации, f pranc. théorie quantique de la gravitation, f … Fizikos terminų žodynas\n\n• Gravitation — is a natural phenomenon by which objects with mass attract one another [http://math.ucr.edu/home/baez/physics/Relativity/GR/grav speed.html Does Gravity Travel at the Speed of Light?] , UCR Mathematics . 1998. Retrieved 3 July 2008] . In everyday … Wikipedia\n\n• Gravitation — hält die Planeten auf ihren Bahnen um die Sonne (nicht maßstabsgetreu) Die Gravitation (von lateinisch gravitas, Schwere), ist eine der vier Grundkräfte der Physik. Sie bewirkt die gegenseitige Anziehung von Massen. Gravitation besitzt… … Deutsch Wikipedia\n\n• gravitation — gravitational, adj. gravitationally, adv. /grav i tay sheuhn/, n. 1. Physics. a. the force of attraction between any two masses. Cf. law of gravitation. b. an act or process caused by this force. 2. a sinking or falling … Universalium\n\n• Theory of relativity — This article is about the scientific concept. For philosophical or sociological theories about relativity, see Relativism. For the silent film, see The Einstein Theory of Relativity. Two dimensional projection of a three dimensional analogy of… … Wikipedia" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.70839095,"math_prob":0.82825494,"size":3264,"snap":"2019-51-2020-05","text_gpt3_token_len":776,"char_repetition_ratio":0.1898773,"word_repetition_ratio":0.13195877,"special_character_ratio":0.19944853,"punctuation_ratio":0.11423221,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9827072,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-21T18:37:44Z\",\"WARC-Record-ID\":\"<urn:uuid:5868d16f-3b73-4b3f-9425-5d564f3f198f>\",\"Content-Length\":\"41454\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:91eb3ba0-f862-403b-b24f-31b93488e5f5>\",\"WARC-Concurrent-To\":\"<urn:uuid:0888d171-854e-4ac0-a203-25dd3c2e12e2>\",\"WARC-IP-Address\":\"84.16.246.89\",\"WARC-Target-URI\":\"https://useful_english.enacademic.com/133230/theory_of_gravitation\",\"WARC-Payload-Digest\":\"sha1:FBI7VH72QLSBLJ7QD6JIUFVUZ2VIZLSU\",\"WARC-Block-Digest\":\"sha1:Y2WFOLOGYRDSAGSDBHJGYOCQLNCRRW5J\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250604849.31_warc_CC-MAIN-20200121162615-20200121191615-00528.warc.gz\"}"}
https://scholars.hkbu.edu.hk/en/publications/a-volumetric-integral-radial-basis-function-method-for-time-depen-2	[ "# A volumetric integral radial basis function method for time-dependent partial differential equations. I. Formulation\n\nE. J. Kansa, H. Power, G. E. Fasshauer, Leevan LING\n\nCorresponding author for this work\n\nResearch output: Contribution to journalArticlepeer-review\n\n62 Citations (Scopus)\n\n## Abstract\n\nA strictly conservative volume integral formulation of the time dependent conservation equations in terms of meshless radial basis functions (RBFs) is presented. Rotational and translational transformations are considered that simplify the partial differential equations (PDEs) to be solved. As a result, the solutions that are represented at a finite sample of knots, x ∈ Ω/∂Ω ⊂ ℝd, are permitted to move as the system of equations evolves in time. Knots are inserted, deleted, or rearranged in such a manner to conserve the extensive physical quantities of mass, momentum components, and total energy. Our study consists of the following parts: (A) Local rotational and Galilean translational transformations can be obtained to reduce the conservation equations into steady-state forms for the inviscid Euler equations or Navier-Stokes equations. (B) The entire set of PDEs are transformed into the method of lines approach yielding a set of coupled ordinary differential equations whose homogeneous solution is exact in time. (C) The spatial components are approximated by expansions of meshless RBFs; each individual RBF is volumetrically integrated at one of the sampling knots x i, yielding a collocation formulation of the method of lines structure of the ODEs. (D) Because the volume integrated RBFs increase more rapidly away from the data center than the commonly used RBFs, we use a higher order preconditioner to counter-act the ill-conditioning problem. Domain decomposition is used over each piecewise continuous subdomain.\n\nOriginal language English 1191-1206 16 Engineering Analysis with Boundary Elements 28 10 SPEC. ISS. https://doi.org/10.1016/j.enganabound.2004.01.004 Published - Oct 2004\n\n## Scopus Subject Areas\n\n• Analysis\n• Engineering(all)\n• Computational Mathematics\n• Applied Mathematics\n\n## User-Defined Keywords\n\n• Galilean transformations\n• Momentum component transformations\n• Shock waves\n• Volumetric formulation\n\n## Fingerprint\n\nDive into the research topics of 'A volumetric integral radial basis function method for time-dependent partial differential equations. I. Formulation'. Together they form a unique fingerprint." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8726882,"math_prob":0.85946465,"size":1992,"snap":"2022-40-2023-06","text_gpt3_token_len":429,"char_repetition_ratio":0.120221324,"word_repetition_ratio":0.0,"special_character_ratio":0.18975903,"punctuation_ratio":0.07096774,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97074753,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-31T10:46:16Z\",\"WARC-Record-ID\":\"<urn:uuid:497f77e5-1e71-4e11-9702-ec6c37054e09>\",\"Content-Length\":\"56994\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b3e586b7-491d-4e1f-a1fd-bb7d6384771e>\",\"WARC-Concurrent-To\":\"<urn:uuid:73ea7017-e968-4a35-bd9e-481e1e63c812>\",\"WARC-IP-Address\":\"18.139.148.124\",\"WARC-Target-URI\":\"https://scholars.hkbu.edu.hk/en/publications/a-volumetric-integral-radial-basis-function-method-for-time-depen-2\",\"WARC-Payload-Digest\":\"sha1:PZZLM63SG3G3DFTSENNIP7ZUUIRI7BDF\",\"WARC-Block-Digest\":\"sha1:6YDKFAL3LH7GFYSB4WDXI6QHABIWNNXG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499857.57_warc_CC-MAIN-20230131091122-20230131121122-00708.warc.gz\"}"}
https://ashwinnarayan.blogspot.com/2014/07/inverse-kinematics-for-2dof-arm.html	[ "## Friday, 11 July 2014\n\n### Inverse Kinematics for 2DOF Arm\n\nWhen I first came across the problem of inverse kinematics I thought - quite naively - that it would be a simple matter to find a solution to the problem because the forward kinematics problem was so simple to solve. I decided to start out with a 2 link arm and see if I could work out the solution on my own from scratch without having any initial knowledge about inverse kinematics. After filling pages and pages with calculations that didn't quite seem to be going anywhere I realized why it was such a difficult problem.\n\nOf course a 2 link arm had a complete close form solution. After a bit of online research I found the solution. However, the solution for arms with more than 2 degrees of freedom turns out to be not so simple. It requires using numerical methods to slowly converge on a solution. This approach doesn't solve everything either. There's still the problem of choosing the right set of angles because there are often multiple solutions for a particular location of the end of the arm and some of these solutions are invalid because they are outside the range of motion of the motors at the joints of the arm.\n\nI've decided to slowly familiarize myself with all the different ways of finding inverse kinematics solutions. I think I'll make blogposts each with a higher degree of freedom and eventually I'll talk about the solution for the 6DOF arm. Why stop at 6? Because it turns out that in 3D space, you don't need more than 6DOF to reach a particular location/orientation. Any more than that and your arm becomes redundant. 3DOF for location in space plus 3DOF for orientation in space are all we need from an arm.\n\nSo in this post I'll talk about the solution to 2 link robotic arm. And this arm will have hinge type joints only. I think that the technical term used is \"revolute joint\".\n\nA 2 link planar robotic arm looks like this:\n\nThe forward kinematics equations for this arrangement are simple enough.\n$x = l_1 cos(\\theta_1) + l_2 cos(\\theta_1 + \\theta_2)$\n$y = l_1 sin(\\theta_1) + l_2 sin(\\theta_1 + \\theta_2)$\n\nThe inverse kinematics equations are quite monstrous. The full derivation is given in this fantastic pdf. I'll just write down the final equations here.\n\nGiven a value of x and y, the inverse kinematics equations are:\nLet\n$k_1 = l_1 + l_2 cos(\\theta_2)$\n$k_2 = l_2 sin(\\theta_2)$\n$\\gamma = atan2(k_2, k_1)$\n\n$\\theta_2 = atan2(\\sqrt{1 - (\\frac{x^2 + y^2 - l_1^2 - l_2^2}{2l_1 l_2})^2}, \\frac{x^2 + y^2 - l_1^2 - l_2^2}{2l_1 l_2})$\n$\\theta_1 = atan2(y, x) - \\gamma$\n\nI wrote a simple python script to calculate the values of theta. You can get the code here. The invkin2() function calculates the values of $\\theta_1$ and $\\theta_2$ for an arm with 2 links.\n\nIt was a bit difficult to test the code this way though. Can't see if I'm right or wrong in most cases. So I wrote this processing sketch to visualize the calculations so I can easily verify their correctness.\n\nYou can view the processing simulation here: http://rationalash.github.io/invkin/" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8951896,"math_prob":0.98972076,"size":4111,"snap":"2019-43-2019-47","text_gpt3_token_len":1128,"char_repetition_ratio":0.12247383,"word_repetition_ratio":0.50282484,"special_character_ratio":0.2821698,"punctuation_ratio":0.07643312,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99689114,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-16T01:17:39Z\",\"WARC-Record-ID\":\"<urn:uuid:f1b228a7-42c0-47f3-ba81-5b89098a1f30>\",\"Content-Length\":\"98697\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:481d03eb-734a-4e83-917a-6ac673cea33c>\",\"WARC-Concurrent-To\":\"<urn:uuid:11ea0207-88fe-4b9a-9f16-af553f05d4ce>\",\"WARC-IP-Address\":\"172.217.13.225\",\"WARC-Target-URI\":\"https://ashwinnarayan.blogspot.com/2014/07/inverse-kinematics-for-2dof-arm.html\",\"WARC-Payload-Digest\":\"sha1:K7TEOBNOVOLGVBSENVJWJF4FWU6SP7G7\",\"WARC-Block-Digest\":\"sha1:PNTXER3ARBYQNRE23DDDEMLHPC6LGNR6\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668716.69_warc_CC-MAIN-20191116005339-20191116033339-00221.warc.gz\"}"}
https://zims-en.kiwix.campusafrica.gos.orange.com/wikipedia_en_all_nopic/A/Maximal_torus	[ "# Maximal torus\n\nIn the mathematical theory of compact Lie groups a special role is played by torus subgroups, in particular by the maximal torus subgroups.\n\nA torus in a compact Lie group G is a compact, connected, abelian Lie subgroup of G (and therefore isomorphic to the standard torus Tn). A maximal torus is one which is maximal among such subgroups. That is, T is a maximal torus if for any torus T containing T we have T = T. Every torus is contained in a maximal torus simply by dimensional considerations. A noncompact Lie group need not have any nontrivial tori (e.g. Rn).\n\nThe dimension of a maximal torus in G is called the rank of G. The rank is well-defined since all maximal tori turn out to be conjugate. For semisimple groups the rank is equal to the number of nodes in the associated Dynkin diagram.\n\n## Examples\n\nThe unitary group U(n) has as a maximal torus the subgroup of all diagonal matrices. That is,\n\n$T=\\left\\{\\operatorname {diag} \\left(e^{i\\theta _{1}},e^{i\\theta _{2}},\\dots ,e^{i\\theta _{n}}\\right):\\forall j,\\theta _{j}\\in \\mathbb {R} \\right\\}.$", null, "T is clearly isomorphic to the product of n circles, so the unitary group U(n) has rank n. A maximal torus in the special unitary group SU(n) ⊂ U(n) is just the intersection of T and SU(n) which is a torus of dimension n 1.\n\nA maximal torus in the special orthogonal group SO(2n) is given by the set of all simultaneous rotations in any fixed choice of n pairwise orthogonal planes (i.e., two dimensional vector spaces). Concretely, the maximal torus consists of all block-diagonal matrices with $2\\times 2$", null, "diagonal blocks, where each diagonal block is a rotation matrix. This is also a maximal torus in the group SO(2n+1) where the action fixes the remaining direction. Thus both SO(2n) and SO(2n+1) have rank n. For example, in the rotation group SO(3) the maximal tori are given by rotations about a fixed axis.\n\nThe symplectic group Sp(n) has rank n. A maximal torus is given by the set of all diagonal matrices whose entries all lie in a fixed complex subalgebra of H.\n\n## Properties\n\nLet G be a compact, connected Lie group and let ${\\mathfrak {g}}$", null, "be the Lie algebra of G. The first main result is the torus theorem, which may be formulated as follows:\n\nTorus theorem: If T is one fixed maximal torus in G, then every element of G is conjugate to an element of T.\n\nThis theorem has the following consequences:\n\n• All maximal tori in G are conjugate.\n• All maximal tori have the same dimension, known as the rank of G.\n• A maximal torus in G is a maximal abelian subgroup, but the converse need not hold.\n• The maximal tori in G are exactly the Lie subgroups corresponding to the maximal abelian subalgebras of ${\\mathfrak {g}}$", null, "(cf. Cartan subalgebra)\n• Every element of G lies in some maximal torus; thus, the exponential map for G is surjective.\n• If G has dimension n and rank r then n r is even.\n\n## Root system\n\nIf T is a maximal torus in a compact Lie group G, one can define a root system as follows. The roots are the weights for the adjoint action of T on the complexified Lie algebra of G. To be more explicit, let ${\\mathfrak {t}}$", null, "denote the Lie algebra of T, let ${\\mathfrak {g}}$", null, "denote the Lie algebra of $G$", null, ", and let ${\\mathfrak {g}}_{\\mathbb {C} }:={\\mathfrak {g}}\\oplus i{\\mathfrak {g}}$", null, "denote the complexification of ${\\mathfrak {g}}$", null, ". Then we say that an element $\\alpha \\in {\\mathfrak {t}}$", null, "is a root for G relative to T if $\\alpha \\neq 0$", null, "and there exists a nonzero $X\\in {\\mathfrak {g}}_{\\mathbb {C} }$", null, "such that\n\n$\\mathrm {Ad} _{e^{H}}(X)=e^{i\\langle \\alpha ,H\\rangle }X$", null, "for all $H\\in {\\mathfrak {t}}$", null, ". Here $\\langle \\cdot ,\\cdot \\rangle$", null, "is a fixed inner product on ${\\mathfrak {g}}$", null, "that is invariant under the adjoint action of connected compact Lie groups.\n\nThe root system, as a subset of the Lie algebra ${\\mathfrak {t}}$", null, "of T, has all the usual properties of a root system, except that the roots may not span ${\\mathfrak {t}}$", null, ". The root system is a key tool in understanding the classification and representation theory of G.\n\n## Weyl group\n\nGiven a torus T (not necessarily maximal), the Weyl group of G with respect to T can be defined as the normalizer of T modulo the centralizer of T. That is,\n\n$W(T,G):=N_{G}(T)/C_{G}(T).$", null, "Fix a maximal torus $T=T_{0}$", null, "in G; then the corresponding Weyl group is called the Weyl group of G (it depends up to isomorphism on the choice of T).\n\nThe first two major results about the Weyl group are as follows.\n\n• The centralizer of T in G is equal to T, so the Weyl group is equal to N(T)/T.\n• The Weyl group is generated by reflections about the roots of the associated Lie algebra. Thus, the Weyl group of T is isomorphic to the Weyl group of the root system of the Lie algebra of G.\n\nWe now list some consequences of these main results.\n\n• Two elements in T are conjugate if and only if they are conjugate by an element of W. That is, each conjugacy class of G intersects T in exactly one Weyl orbit. In fact, the space of conjugacy classes in G is homeomorphic to the orbit space T/W.\n• The Weyl group acts by (outer) automorphisms on T (and its Lie algebra).\n• The identity component of the normalizer of T is also equal to T. The Weyl group is therefore equal to the component group of N(T).\n• The Weyl group is finite.\n\nThe representation theory of G is essentially determined by T and W.\n\nAs an example, consider the case $G=SU(n)$", null, "with $T$", null, "being the diagonal subgroup of $G$", null, ". Then $x\\in G$", null, "belongs to $N(T)$", null, "if and only if $x$", null, "maps each standard basis element $e_{i}$", null, "to a multiple of some other standard basis element $e_{j}$", null, ", that is, if and only if $x$", null, "permutes the standard basis elements, up to multiplication by some constants. The Weyl group in this case is then the permutation group on $n$", null, "elements.\n\n## Weyl integral formula\n\nSuppose f is a continuous function on G. Then the integral over G of f with respect to the normalized Haar measure dg may be computed as follows:\n\n$\\displaystyle {\\int _{G}f(g)\\,dg=\|W\|^{-1}\\int _{T}\|\\Delta (t)\|^{2}\\int _{G/T}f\\left(yty^{-1}\\right)\\,d[y]\\,dt,}$", null, "where $d[y]$", null, "is the normalized volume measure on the quotient manifold $G/T$", null, "and $dt$", null, "is the normalized Haar measure on T. Here Δ is given by the Weyl denominator formula and $\|W\|$", null, "is the order of the Weyl group. An important special case of this result occurs when f is a class function, that is, a function invariant under conjugation. In that case, we have\n\n$\\displaystyle {\\int _{G}f(g)\\,dg=\|W\|^{-1}\\int _{T}f(t)\|\\Delta (t)\|^{2}\\,dt.}$", null, "Consider as an example the case $G=SU(2)$", null, ", with $T$", null, "being the diagonal subgroup. Then the Weyl integral formula for class functions takes the following explicit form:\n\n$\\displaystyle {\\int _{SU(2)}f(g)\\,dg={\\frac {1}{2}}\\int _{0}^{2\\pi }f\\left(\\mathrm {diag} \\left(e^{i\\theta },e^{-i\\theta }\\right)\\right)\\,4\\,\\mathrm {sin} ^{2}(\\theta )\\,{\\frac {d\\theta }{2\\pi }}.}$", null, "Here $\|W\|=2$", null, ", the normalized Haar measure on $T$", null, "is ${\\frac {d\\theta }{2\\pi }}$", null, ", and $\\mathrm {diag} \\left(e^{i\\theta },e^{-i\\theta }\\right)$", null, "denotes the diagonal matrix with diagonal entries $e^{i\\theta }$", null, "and $e^{-i\\theta }$", null, ".\n\n## See also\n\nThis article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. 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https://discourse.elm-lang.org/t/json-encode-value-equality/898	[ "# Json.Encode.Value equality\n\nI was surprised by the following test results. I had expected all five tests to fail. Instead, `{a:1} == {a:1, b:2}` is `True`, and `Expect.equal {a:1, b:2} {a:1}` passes; where `{a:1}` stands for a `Json.Encode.Value` created by `Json.Encode.object`. (Note that there only three test failures, instead of five.)\n\nIs this expected? I don’t see anything in the docs for `Json.Encode.Value` or `Expect.Equal` that indicate this. The docs for `(==)` do list `Json.Encode.Value` as a “problematic type”, but the context suggests that this is because a `Value` might contain a function, not that `(==)` also returns false positives even when the properties have non-function types.\n\n(This came up while writing tests for my Encoder, to verify that it didn’t create any unexpected object properties in a case where a programming error could lead it to do so.)\n\n``````module JsonValueTests exposing (..)\n\nimport Expect exposing (Expectation)\nimport Json.Encode as Encode\nimport Test exposing (..)\n\nsuite : Test\nsuite =\nlet\nobj1 =\nEncode.object [ ( \"a\", Encode.int 1 ) ]\n\nobj2 =\nEncode.object [ ( \"a\", Encode.int 1 ), ( \"b\", Encode.int 2 ) ]\nin\ndescribe \"Json.Encode.Value\"\n[ test \"Expect.equal {a:1} {a:1, b:2}\" <\|\n\\_ ->\nExpect.equal obj1 obj2\n, test \"Expect.equal {a:1, b:2} {a:1}\" <\|\n\\_ ->\nExpect.equal obj2 obj1\n, test \"{a:1} == {a:1, b:2}\" <\|\n\\_ ->\nExpect.equal (obj1 == obj2) True\n, test \"{a:1, b:2} == {a:1}\" <\|\n\\_ ->\nExpect.equal (obj2 == obj1) True\n, test \"encode {a:1} == encode {a:1,b:2}\" <\|\n\\_ ->\nEncode.encode 0 obj1\n\|> Expect.equal (Encode.encode 0 obj2)\n]\n``````\n``````❯ elm-test tests/JsonValueTests.elm\nSuccess! Compiled 0 modules.\nSuccessfully generated /dev/null\nSuccess! Compiled 1 module.\nSuccessfully generated /Users/osteele/code/banyan/elm-stuff/generated-code/elm-community/elm-test/elmTestOutput.js\n\nelm-test 0.18.12\n----------------\n\nRunning 5 tests. To reproduce these results, run: elm-test --fuzz 100 --seed 2079329333 tests/JsonValueTests.elm\n\n↓ JsonValueTests\n↓ Json.Encode.Value\n✗ Expect.equal {a:1} {a:1, b:2}\n\n{ a = 1, b = 2 }\n╷\n│ Expect.equal\n╵\n{ a = 1 }\n\n↓ JsonValueTests\n↓ Json.Encode.Value\n✗ {a:1, b:2} == {a:1}\n\nTrue\n╷\n│ Expect.equal\n╵\nFalse\n\n↓ JsonValueTests\n↓ Json.Encode.Value\n✗ encode {a:1} == encode {a:1,b:2}\n\n\"{\\\"a\\\":1}\"\n╷\n│ Expect.equal\n╵\n\"{\\\"a\\\":1,\\\"b\\\":2}\"\n\nTEST RUN FAILED\n\nDuration: 285 ms\nPassed: 2\nFailed: 3\n``````\n\nThere are more cases than just function values where `Value` is a problematic type for equality, and this happens to be one of them. There’s no way for the equality kernel function to distinguish between the JavaScript representations of the Elm types it has to work on and anything you can represent as a `Value`. Eventually it falls through to the comparison case for record types, where it tries to check every key in the lefthand value against the key in the righthand value. This is why it only fails in the cases where the record with two keys is the lefthand value. It tries to find `b` in the object that has no `b`. In the reversed situation, it stops checking after `a` because it’s making an assumption that it’s comparing records and the keys will always match.\n\n2 Likes\n\nI ended up using the following in my tests:\n\n``````expectJsonEqual : Encode.Value -> Encode.Value -> Expectation\nexpectJsonEqual a =\nExpect.all\n[ Expect.equal a\n, flip Expect.equal a\n]\n``````\n\nWhen it reports an error, it’s not consistent about the order of the value and the expectation (it depends on which side has the extra key), but it’s good enough for my purposes for now.\n\nThis PR attempts to improve the documentation." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.60587883,"math_prob":0.87094563,"size":2294,"snap":"2022-05-2022-21","text_gpt3_token_len":733,"char_repetition_ratio":0.17510916,"word_repetition_ratio":0.042372882,"special_character_ratio":0.3435048,"punctuation_ratio":0.24513619,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9716361,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-20T03:42:37Z\",\"WARC-Record-ID\":\"<urn:uuid:84eb7835-c6fb-42c8-baee-f32655cebea4>\",\"Content-Length\":\"23739\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1bd4ef2c-a093-49c1-a4e4-bc409e557b7a>\",\"WARC-Concurrent-To\":\"<urn:uuid:55f9c8f2-b049-4d82-8c8c-cba3dfc74cc4>\",\"WARC-IP-Address\":\"64.62.250.111\",\"WARC-Target-URI\":\"https://discourse.elm-lang.org/t/json-encode-value-equality/898\",\"WARC-Payload-Digest\":\"sha1:WLG3KU5RAIG65QFMHQSSKHHQRKZK3GBN\",\"WARC-Block-Digest\":\"sha1:FKZ4LBDRYW3IONERYR2WTIFZ4NTFCT32\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662531352.50_warc_CC-MAIN-20220520030533-20220520060533-00602.warc.gz\"}"}
https://www.datasciencesociety.net/the-curious-case-of-household-budget-surveyhbs/	[ "# Datathon NSI Solution – The curious case of ‘Household Budget Survey(HBS)’\n\nThe National Statistical Institute of Bulgaria (NSI) conducts annually a Household Budget Survey (HBS) with an objective to get reliable and scientifically founded data on the income, expenditure, consumption and other elements of the living standard of the population as well as changes, which have occurred during the years. NSI is considering a change in the periodicity of the Household Budget Survey from yearly to once on every five years,In order to optimize the cost of carrying out the survey. Hence We are creating a model which will predict household expenditure for the next four years using linear regression model and time series. The algorithms that we will be taking help from are linear regression model & Autoregressive integrated moving average(ARIMA). So lets not waste any time and move on with it !\n\n2", null, "### The Curious Case Of ‘House Budget Survey’\n\n#### STEP I : Understanding the data.\n\nThe Household Budget Survey is a sample survey by implementation of two-stage cluster. The general population from which the sample for the survey is formed comprises all the households in the country. Institutional households are not studied by the household budget survey. The unit of observation is every randomly chosen ordinary household irrespective of the number of members and their material and personal status. The sample size since 2010 is 3060 households each quarter spread into three sub-samples for 1 020 households. Each sub-sample is monitored for one month from each quarter, based on the rotation sample model. In this way, each household in the sample participates in the survey four months during a twelve-month period. The data provided is from 2010-2017.\n\nDatasets provided :\n\n• Data for total expenditure average per capita by COICOP.\n• Monetary income by source of income.\n• Population by 5 years for age, group & sex.\n• Annual average wages and salaries of employees under labor contract.\n• Population, employed & unemployed count of labor force.\n\nThe following packages have been used-\n\n• dplyr\n• ggplot2\n• psych\n• tidyr\n• plotly\n• forecast\n• tseries\n• lubridate\n\nThe Algorithms that have been used are-\n\nApproach 1:\n\n• Linear regression : Linear regression is used to explain the relationship between one dependent variable and one or more independent variables.\n• Time Series Linear Model : To fit the linear model with multiple independent variables, in time series.\n\nApproach 2:\n\n• Time Series Linear Model : To fit the linear model with multiple independent variables, in time series.\n\nApproach 3:\n\n• Time series :Time series is a sequence of numerical data points in successive order. Time series tracks the movement of the chosen data points, such as a expenditure, over a specified period of time with data points recorded at regular intervals.\n\n#### STEP III: Exploratory Data Analysis (EDA)\n\nWe have used linear regression to find the significant relation between expenditure and other parameters such as income, age,sex,population,pension etc. We have also taken the help of various charts to dig deep into the data.\n\n• Some main observation are as follows:\n• The income of a person highly affects the expenditure.\n• Non-labor force affects the expenditure significantly.\n• Expenditure does not depend greatly over unemployment benefits.\n• People spend twice as much money on food and non alcoholic beverages as they do on their housing.\n• On an average citizens are earning almost 60% more than what they were earning in 2010 yet somehow the expenditure rate has not relatively increased.\n• There is no significant rise in earnings unless you are an employee under labor contract.\n\nWe have used three different approaches to predict the household values.\n\nApproach I: Using time series linear model\n\nThe provided data is in quarters but the expected prediction is supposed to be in years. Hence the first step is to transform data into years.\n\n1: Taken all the independent columns which highly affect the total expenditure(Consumer Expenditure, Non-Consumer Expenditure), into a dataset.\n\n2: Divided the data into test and train data.\n\n3: Used Time Series Linear Model to train the data based on the multiple significant columns identified and predict total expenditure.\n\n4: Then used test data to forecast data.\n\n[Note: Given values for the significant independent columns , we can predict the expenditure for future years]\n\n#### Codes and Graphs for prediction on yearly basis\n\nTrain Dataset: 2010-2015\n\nTest Dataset: 2016-2017\n\nPredicted Features:\n\n• Total Expenditure\n• Consumer Expenditure\n• Non-Consumer Expenditure\n\nSignificant Features:\n\n• Average pension of one pensioner-BGN\n• Wages and salaries\n• Self-employment income\n• Pensions\n• Other social benefits\n• Regular transfers from other households\n• Persons not in the labour force\n• Annual Average wages and salaries of the employees under labour contract\n\n#### Approach 2: Prediction using Linear model time series on quarterly basis\n\nHere are some predicted graphs that we performed on our test and train data.\n(Train data from 2010-2015 ; Test data 2016-2017).\nFrom this here are some following observations which is given below.\n\n1. The following graph represents Total expenditure against yearly Quarters.\na. The Q1 (Quarter-1) of every year shows less expenditure compared to Q4 (Quarter-4) which represents seasonality.\nb. This is because of the Christmas and New year’s eve that comes under Q4, where people celebrate on a larger scale.\n\nTrain Dataset: 2010-2015\n\nTest Dataset: 2016-2017\n\nPredicted Features:\n\n• Total Expenditure\n• Consumer Expenditure\n• Non-Consumer Expenditure\n• Food and non-alcoholic beverages\n• Housing, water, electricity, gas and other fuels\n• Taxes_and_social_insurance_contributions\n\nSignificant Features:\n\n• Average pension of one pensioner-BGN\n• Wages and salaries\n• Self-employment income\n• Pensions\n• Other social benefits\n• Regular transfers from other households\n\n2. The following graph represents Total consumer expenditure against yearly Quarters.\na. It represents the linearity in trend that people’s expenditure is always high during the 4th Quarter of the year w.r.t to first 3 Quarters.\nb. The expenditure in 3rd Quarter is comparatively high because of the SOFIA Restaurant week in September, as the expenditure on food and non – beverages is more as compared to Alcoholic beverages.\n\n3. The following graph represents Total Non-Consumer expenditure against yearly Quarters.\na. Non-consumer expenditure includes expenditures such as Taxes and social insurance contributions, which is linear and proportional.\nb. It also shows that there is not much increment in taxes and social insurance expenditure.\n\n• 4. The following graph represents Total Food and non-alcoholic expenditure against yearly Quarters.\na. The following graph represents that the expenditure on food and non-alcoholic beverages increases gradually from Q1 (Quarter-1) to Q4 (Quarter -4).\nb. And the alcoholic beverages expenditure is less as compared to food and non-alcoholic beverages.\n\n6. The following graph represents Total Taxes expenditure against yearly Quarters.\na. It shows the linearity in the graph that with every year the taxes are increased by the government , which also contributes in the non-consumer expenditure\n\nApproach 3: Prediction using AUTO ARIMA.\n\nWe have used auto-arima to forcast the household expenditure for next four years with reference to the historical household data from 2010.\n\n## STEP V: Conclusion\n\nWe would recommend to go with approach one and two which is ‘Time series linear model’ to predict expenditure to build on few significant factors that are mentioned in above approach descriptions.\n\nThank-you. 🙂\n\n#### 6 thoughts on “Datathon NSI Solution – The curious case of ‘Household Budget Survey(HBS)’”\n\n1.", null, "laura says:\n0\n\nIn the case of the linear regression model that you described: can you make it more clear, which were the observations in the train set, which were the observations in the test set? Which were the features and which were the predicted variables?\nCan you also comment on the main difference between using a classical linear Regression and the ARIMA model? Which is more appropriate?\n\n2.", null, "svetro says:\n0\n\nThank you for working on the NSI case!\nYou are saying that the TS linear model is better but for example from the “food and non-alcoholic expenditures” graph it can be seen a lost of seasonality which can be crucial when calculating consumer price indicies. It would be better if you provided predicted vs expected values for some kind of error estimation. Otherwise the article is readable, friendly and shows dedication and understanding of a certain level of the subject.\n\n1.", null, "omki says:\n0\n\nWe have thought of seasonality in the data and found also that.\nHowever, we tried to incorporate as many variables for prediction in the model to get as good fit as possible. As the time was less, we could not deal with the seasonality in an effective manner. We will further work on this to get a better model.\nWe have seen the R-square value which seemed to be good comparing the original and predicted and so for the time-constraints we have omitted presenting that however those were plotted in the graph. We will further update our article.\n\n3.", null, "apoorvakesarwani says:\n0\n\nEDA: We used linear regression model to identify the most significant factors on which the change in expenditure is dependent.\nModel: We used time series linear regression model to predict the household expenditure based on the above identified significant factors. Made model on both yearly basis and quarterly basis and predicted.\nUsed Auto Arima as another approach to make predictions.\nDetails of Training data, test data, predicted variables and features have been mentioned in the article now.\nIn Linear model we identify the linearity between the variables and in ARIMA model we use historical data to predict the future values.\n\n4.", null, "junior says:\n0\n1.", null, "omki says:" ]	[ null, "https://dss-www-production.s3.amazonaws.com/uploads/2018/09/Word-Art.jpeg", null, "https://dss-www-production.s3.amazonaws.com/uploads/avatars/30/5a79914a6e164-bpfull.png", null, "https://www.gravatar.com/avatar/c9d1a094eb0e656f14773f6a450aa088", null, "https://www.gravatar.com/avatar/518be455f39e987010cc409b66b4ce80", null, "https://www.gravatar.com/avatar/8aa922e38b9f4dd0bce1df273fec9e65", null, "https://dss-www-production.s3.amazonaws.com/uploads/avatars/11/5a7b646b1153c-bpfull.png", null, "https://www.gravatar.com/avatar/518be455f39e987010cc409b66b4ce80", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.94066876,"math_prob":0.83000785,"size":8762,"snap":"2022-05-2022-21","text_gpt3_token_len":1809,"char_repetition_ratio":0.13245033,"word_repetition_ratio":0.08250356,"special_character_ratio":0.20235106,"punctuation_ratio":0.09915748,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95155936,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,3,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-20T10:33:09Z\",\"WARC-Record-ID\":\"<urn:uuid:6d5fa42a-51b9-4c75-b232-59cf2a2f4f3d>\",\"Content-Length\":\"127566\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:34a41d90-50a3-4785-811f-8ac982e48211>\",\"WARC-Concurrent-To\":\"<urn:uuid:7f384182-b963-4e9d-92d4-729274e2271d>\",\"WARC-IP-Address\":\"104.21.26.14\",\"WARC-Target-URI\":\"https://www.datasciencesociety.net/the-curious-case-of-household-budget-surveyhbs/\",\"WARC-Payload-Digest\":\"sha1:IQUI6L272ICWKULOD5DSU7OAFDOSRHVI\",\"WARC-Block-Digest\":\"sha1:67OXZEII4DHPJAK7Q2UAOJ4QCULWNLAT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662531779.10_warc_CC-MAIN-20220520093441-20220520123441-00481.warc.gz\"}"}
https://pyautolens.readthedocs.io/en/latest/advanced/database.html	[ "# Database#\n\nThe default behaviour of PyAutoLens is for model-fitting results to be output to hard-disc in folders, which are straight forward to navigate and manually check the lens modeling results. For small samples of lenses this is sufficient, however many users have a need to perform many model fits to large lens datasets, making the manual inspection of results time consuming.\n\nPyAutoLens’s database feature outputs all model-fitting results as a sqlite3 (https://docs.python.org/3/library/sqlite3.html) relational database, such that all results can be efficiently loaded into a Jupyter notebook or Python script for inspection, analysis and interpretation. This database supports advanced querying, so that specific model-fits (e.g., which fit a certain lens model or dataset) can be loaded.\n\nTo make it so that results are output to an .sqlite database we simply open a database session and pass this session to the non-linear search:\n\n```session = af.db.open_database(\"database.sqlite\")\n\nemcee = af.Emcee(\nsession=session, # This instructs the search to write to the .sqlite database.\n)\n```\n\nWhen a model-fit is performed, a unique identifier is generated based on the model and non-linear search. However, if we were to fit many different lens datasets with the same model and non-linear search, they would all use the same unique identifier and not be distinguishable by the database.\n\nWe can overcome this by using the name of the dataset as the `unique_tag` passed to the search, which is used alongside the model and search to create the unique identifier:\n\n```session = af.db.open_database(\"database.sqlite\")\n\ndataset_name = \"lens_dataset_0\"\n\nemcee = af.Emcee(\npath_prefix=path.join(\"features\", \"database\"),\nunique_tag=dataset_name, # This makes the unique identifier use the dataset name\nsession=session, # This instructs the search to write to the .sqlite database.\n)\n```\n\nLets suppose that we have performed 100 model-fits to 100 strong lenses, and when we ran PyAutoLens we told it to write to the `.sqlite` database file. We can load these results in a Python script or Jupyter notebook using the `Aggregator`:\n\n```agg = Aggregator.from_database(\"path/to/output/database.sqlite\")\n```\n\nWe can now use the `Aggregator` to inspect the results of all model-fits. For example, we can load the `Samples` object of all 100 model-fits, which contains information on the best-fit model, posterior, Bayesian evidence, etc.\n\nBelow, we use the samples generator to create a list of the maximum log likelihood of every model-fit and print it:\n\n```for samples in agg.values(\"samples\"):\n\nprint(max(samples.log_likelihood))\n```\n\nThis object (and all objects loaded by the `Aggregator`) are returned as a generator (as opposed to a list, dictionary or other Python type). This is because generators do not store large arrays or classes in memory until they are used, ensuring that when we are manipulating large sets of results we do not run out of memory!\n\nWe can iterate over the samples to print the maximum log likelihood lens model of every fit:\n\n```for samps in agg.values(\"samples\"):\n\ninstance = samps.max_log_likelihood()\n\nprint(\"Maximum Likelihood Model-fit \\n\")\nprint(instance)\n```\n\nThe `Aggregator` contains tools for querying the database for certain results, for example to load subsets of model-fits. This can be done in many different ways, depending on what information you want.\n\nBelow, we query based on the model fitted. For example, we can load all results which fitted an `Isothermal` mass model (note that when we performed the model fit, we composed the lens model using a lens galaxy name `lens` and mass component named `mass`):\n\n```lens = agg.model.galaxies.lens\nagg_query = agg.query(lens.mass == al.Isothermal)\n\nsamples_gen = agg_query.values(\"samples\")\n```\n\nQueries using the results of model-fitting are also supported. Below, we query the database to find all fits where the inferred value of the `einstein_radius` for the `Isothermal` is above 1.0:\n\n```lens = agg.model.galaxies.lens\n```\n\nAdvanced queries can be constructed using logic, for example we below we combine the two queries above to find all results which fitted an `Isothermal` AND (using the & symbol) inferred a value of sigma less than 3.0.\n\nThe OR logical clause is also supported via the symbol \|.\n\n```agg_query = agg.query((lens.mass == al.Isothermal) & (lens.mass.einstein_radius > 1.0))\n```\n\nWe can query using the `unique_tag` to load the model-fit to a specific dataset:\n\n```agg_query = agg.query(agg.unique_tag == \"example_dataset_0\")\n```\n\nAn `info` dictionary can be passed into a model-fit, which contains information on the model-fit. The example below creates an `info` dictionary which is passed to the model-fit, which is then loaded via the database.\n\n```info = {\"example_key\": \"example_value\"}\n\nemcee.fit(model=model, analysis=analysis, info=info)\n\nagg = Aggregator.from_database(\"path/to/output/database.sqlite\")\n\ninfo_gen = agg.values(\"info\")\n```\n\nDatabases are an extremely powerful feature for users tasked with fitting extremely large datasets as well as fitting many different models, where the scale of the problem can make the management of the large quantity of results produced prohibitive. This is especially true on high performance computing facilities, which often have restrictions on the number of files that a user can store on the machine.\n\nFurthermore, if you are using the SLaM pipelines, non-linear grid searches or sensitivity mapping features you’ll be aware that one can easily start generating thousands or millions of model-fits. There is no way to manage this large library of results other than the database!\n\nA complete description of the database is given in the database folder on the autolens_workspace." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8287063,"math_prob":0.8509381,"size":5668,"snap":"2022-40-2023-06","text_gpt3_token_len":1231,"char_repetition_ratio":0.13259181,"word_repetition_ratio":0.03448276,"special_character_ratio":0.21030346,"punctuation_ratio":0.12369172,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96450585,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-29T13:31:55Z\",\"WARC-Record-ID\":\"<urn:uuid:490c4b42-f0e3-48bd-99f1-f963bdd2a599>\",\"Content-Length\":\"69995\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ab11746f-0d1a-49e4-8923-02ebc24d43cf>\",\"WARC-Concurrent-To\":\"<urn:uuid:67d19ded-1ac3-4dcd-be6d-54cce6e8fba0>\",\"WARC-IP-Address\":\"104.17.33.82\",\"WARC-Target-URI\":\"https://pyautolens.readthedocs.io/en/latest/advanced/database.html\",\"WARC-Payload-Digest\":\"sha1:H7KAJV6CFOAT76VCVCDPYLEGY6UQOG76\",\"WARC-Block-Digest\":\"sha1:5LGU2ANGEKPHUOLXDGS6D2GUFR7FXGQK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499713.50_warc_CC-MAIN-20230129112153-20230129142153-00332.warc.gz\"}"}
https://tex.stackexchange.com/questions/181141/automatic-tikz-coordinate-calculations	[ "# automatic tikz coordinate calculations\n\nRight now I have these defined:\n\n\\coordinate (Z1) at (0.9,0.5);\n\\coordinate (Z2) at (0.816,-0.112);\n\\coordinate (Z3) at (0.46848,-0.46336);\n\n\nBut I would really like to automate this to be:", null, "so that I can just specify Z1 and then vary the α factor easily to get Z2 and Z3. (Right now α = 0.8)\n\nWhat's the simplest way to do this? I know how to do some coordinate calculations but this is beyond my skillset right now.\n\nedit: FWIW(For what it's worth) this can also be expressed as a coordinate transformation H = [1 alpha; -alpha 1]/(1+alpha^2) so maybe that's something worth using. I don't know how to force tikz to compute transformed coordinates.\n\n• Do you need to generalise this to any k, or will k be no larger than 3? – Torbjørn T. May 29 '14 at 22:43\n• I don't need to generalize, just get Z2 and Z3. – Jason S May 29 '14 at 22:57\n• It looks like a complex numbers calculation. Does tikz can do calculations directly with complex numbers ? – Tarass May 31 '14 at 11:05\n\n## 5 Answers\n\nYou could do something like this:\n\n\\documentclass[border=3mm,tikz]{standalone}\n\\begin{document}\n\\begin{tikzpicture}\n\\pgfmathsetmacro{\\Xi}{0.9}\n\\pgfmathsetmacro{\\Yi}{0.5}\n\\pgfmathsetmacro{\\Coeff}{0.8}\n\n\\pgfmathsetmacro{\\Xii}{(\\Xi+\\Coeff\\Yi)/(1+\\Coeff^2)}\n\\pgfmathsetmacro{\\Yii}{(\\Yi-\\Coeff\\Xi)/(1+\\Coeff^2)}\n\\pgfmathsetmacro{\\Xiii}{(\\Xii+\\Coeff\\Yii)/(1+\\Coeff^2)}\n\\pgfmathsetmacro{\\Yiii}{(\\Yii-\\Coeff\\Xii)/(1+\\Coeff^2)}\n\n\\coordinate (Z1) at (\\Xi,\\Yi);\n\\coordinate (Z2) at (\\Xii,\\Yii);\n\\coordinate (Z3) at (\\Xiii,\\Yiii);\n\n\\draw (Z1) -- (Z2) -- (Z3);\n\n\\end{tikzpicture}\n\\end{document}\n\n\nVersion with 4-argument macro:\n\n\\documentclass[tikz,border=2mm]{standalone}\n\\newcommand{\\defcoords}[Z]{%\n\\pgfmathsetmacro{\\Xi}{#2}\n\\pgfmathsetmacro{\\Yi}{#3}\n\\pgfmathsetmacro{\\Coeff}{#4}\n\n\\pgfmathsetmacro{\\Xii}{(\\Xi+\\Coeff\\Yi)/(1+\\Coeff^2)}\n\\pgfmathsetmacro{\\Yii}{(\\Yi-\\Coeff\\Xi)/(1+\\Coeff^2)}\n\\pgfmathsetmacro{\\Xiii}{(\\Xii+\\Coeff\\Yii)/(1+\\Coeff^2)}\n\\pgfmathsetmacro{\\Yiii}{(\\Yii-\\Coeff\\Xii)/(1+\\Coeff^2)}\n\n\\coordinate (#11) at (\\Xi,\\Yi);\n\\coordinate (#12) at (\\Xii,\\Yii);\n\\coordinate (#13) at (\\Xiii,\\Yiii);\n}\n\n\\begin{document}\n\\begin{tikzpicture}\n\\defcoords{0.9}{0.5}{0.8}\n\\draw (Z1) -- (Z2) -- (Z3);\n\n\\defcoords[C]{0.9}{0.5}{2}\n\\draw (C1) -- (C2) -- (C3);\n\\end{tikzpicture}\n\\end{document}\n\n• is there any way to do this, given Z1 as an input? (get its x and y coordinates and compute the Z2 and Z3 coordinates) – Jason S May 30 '14 at 22:04\n• @JasonS I updated my answer, was it something like that you had in mind? – Torbjørn T. May 30 '14 at 22:18\n\nA solution using TikZ powerful let...in syntax, which produces a code short and readable:\n\n\\documentclass{standalone}\n\\usepackage{tikz}\n\\usetikzlibrary{calc}\n\n\\begin{document}\n\\def\\nextZ#1{\n\\path let\n\\p1 = (Z#1),\n\\n1 = {(\\x1+\\coeff\\y1)/(1+\\coeff^2)},\n\\n2 = {(\\y1-\\coeff\\x1)/(1+\\coeff^2)},\n\\n3 = {int(#1+1)},\n\\p2 = (\\n1,\\n2)\nin\n(\\p2) coordinate (Z\\n3);\n}\n\n\\begin{tikzpicture}[scale=2]\n\\coordinate (Z1) at (0.9,0.5);\n\\def\\coeff{0.8}\n\n\\foreach \\z in {1,2,3,4,5} {\n\\fill[red] (Z\\z) circle(2pt) node[black,above]{$Z_\\z$};\n\\nextZ\\z\n}\n\n\\end{tikzpicture}\n\\end{document}", null, "A solution with Lua and LuaLaTeX:\n\n\\documentclass{article}\n\\usepackage{luacode}\n\\usepackage{tikz}\n\n\\begin{document}\n\\luaexec{\ntp = tex.print\nlocal alfa = 0.8\nlocal k = 1 + (0.8 * 0.8)\nx = {}\ny = {}\nx = 0.9\ny = 0.5\nfor i = 2, 3 do\nx[i] = (x[i-1] + alfa * y[i-1]) / k\ny[i] = (y[i-1] - alfa * x[i-1]) / k\nend\ntp(\"\\\\begin{tikzpicture}[scale=3]\")\ntp(\"\\\\coordinate (Z1) at (\"..x..\",\"..y..\");\")\ntp(\"\\\\coordinate (Z2) at (\"..x..\",\"..y..\");\")\ntp(\"\\\\coordinate (Z3) at (\"..x..\",\"..y..\");\")\ntp(\"\\\\draw[-latex] (Z1)node[left]{$Z1$}--(Z2) node[right]{$Z2$};\")\ntp(\"\\\\draw[-latex] (Z2)--(Z3) node[right]{$Z3$};\")\ntp(\"\\\\end{tikzpicture}\")\n}\n\n\\end{document}", null, "For more then 3 coordinate calculations, say 10:\n\n\\documentclass{article}\n\\usepackage{luacode}\n\\usepackage{tikz}\n\n\\begin{document}\n\\luaexec{\ntp = tex.print\ntp(\"\\\\begin{tikzpicture}[scale=3]\")\nlocal n = 10\nlocal alfa = 0.8\nlocal k = 1 + (0.8 * 0.8)\nx = {}\ny = {}\nx = 0.9\ny = 0.5\nfor i = 2, n do\nx[i] = (x[i-1] + alfa * y[i-1]) / k\ny[i] = (y[i-1] - alfa * x[i-1]) / k\ntp(\"\\\\draw[-latex] (\"..x[i-1]..\",\"..y[i-1]..\"])--(\"..x[i]..\",\"..y[i]..\");\")\nend\ntp(\"\\\\end{tikzpicture}\")\n}\n\n\\end{document}", null, "Another possibility:\n\n\\documentclass[tikz,margin=5mm,convert=false]{standalone}\n\\newlength\\px%\n\\newlength\\py%\n\n\\newcommand\\morecoords{%\n\\foreach[count=\\j] \\i in {2,...,#1}{%\n% get the coordinates of the previous point (unit: pt)\n\\pgfextractx{\\px}{\\pgfpointanchor{#2\\j}{center}}\n\\pgfextracty{\\py}{\\pgfpointanchor{#2\\j}{center}}\n% calculate the coordinates of the next point (unit: pt)\n\\pgfmathsetmacro\\nx{(\\px+#3\\py)/(1+#3#3)}\n\\pgfmathsetmacro\\ny{(\\py-#3\\px)/(1+#3#3)}\n% define the new coordinate\n\\coordinate(#2\\i)at(\\nx pt,\\ny pt);\n}\n}\n\\begin{document}\n\\begin{tikzpicture}\n\\draw[help lines,step=0.5](-1,-1) grid (1,1);\n\\draw(0,-1)--(0,1)(-1,0)--(1,0);\n%\n\\coordinate(Z1)at(.9,.5);% define Z1\n\\morecoords{Z}{.8}% calculate Z2 and Z3\n\\foreach[count=\\i start=2] \\j in {2,3}\\draw[red,-latex](Z\\i)--(Z\\j);\n%\n\\coordinate(P1)at(-.8,.8);% define P1\n\\morecoords{P}{.7}% calculate P2,...,P10\n\\foreach[count=\\i start=2] \\j in {2,...,10}\\draw[blue,-latex](P\\i)--(P\\j);\n%\n\\coordinate(C1)at(-60:0.9);% define C1\n\\morecoords{C}{.9}% calculate C2,...,C6\n\\foreach[count=\\i start=2] \\j in {2,...,6}\\draw[green,-latex](C\\i)--(C\\j);\n\\end{tikzpicture}\n\\end{document}", null, "I think a recursive formula needs a recursion :P\n\n\\documentclass[tikz]{standalone}\n\n\\def\\myrecursion#1#2#3#4{%\n\\ifnum#4>0\\relax%\n\\pgfmathsetmacro{\\tempx}{(#1+#3#2)/(1+(#3)^2)}\n\\pgfmathsetmacro{\\tempy}{(#2-#3#1)/(1+(#3)^2)}\n\\pgfmathtruncatemacro{\\tempc}{(#4-1)}\n\\draw[line cap=round,->,very thin] (#1 cm,#2 cm)--(\\tempx cm,\\tempy cm);\n\\edef\\temp{\\noexpand\\myrecursion{\\tempx}{\\tempy}{#3}{\\tempc}}\\temp%\n\\fi%\n}\n\n\\begin{document}\n\\begin{tikzpicture}\n\\myrecursion{0.9}{0.5}{0.8}{10}\n\\end{tikzpicture}\n\\end{document}", null, "" ]	[ null, "https://i.stack.imgur.com/5c1ak.png", null, "https://i.stack.imgur.com/sGogq.png", null, "https://i.stack.imgur.com/q8Wcd.png", null, "https://i.stack.imgur.com/CMZoO.png", null, "https://i.stack.imgur.com/HE9Re.png", null, "https://i.stack.imgur.com/7VmXW.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9373327,"math_prob":0.99866736,"size":644,"snap":"2019-35-2019-39","text_gpt3_token_len":188,"char_repetition_ratio":0.140625,"word_repetition_ratio":0.0,"special_character_ratio":0.3167702,"punctuation_ratio":0.15068494,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997844,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-22T11:13:25Z\",\"WARC-Record-ID\":\"<urn:uuid:55f34e72-c9a5-4a27-9548-9d80c98ee599>\",\"Content-Length\":\"165199\",\"Content-Type\":\"application/http; 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https://www.delftstack.com/zh/howto/arduino/arduino-char-to-int/	[ "# Arduino 的字符转换为整型\n\n## 使用 Arduino 中的简单方法将 `char` 转换为 `int`\n\n``````void loop{\nchar someChar = '2'; // variable to store char\nint someInt = someChar - '0';\n}\n``````\n\n## 在 Arduino 中使用 `toInt()` 函数将 `char` 转换为 `int`\n\n``````void loop(){\nchar someChar = '123';\nString stringOne = String('a');// converting a constant char into a String\nstringOne.toInt();\n}\n``````\n\n## 使用 Arduino 中的 `Serial.parseInt()` 函数将 `char` 转换为 `int`\n\n``````void loop(){\n\nif (Serial.available()>0){\nint valA = Serial.parseInt();\n}\n}\n``````\n\n## 相关文章 - Arduino Integer\n\n• 在 Arduino 中将整数转换为字符串" ]	[ null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.9223705,"math_prob":0.8041455,"size":1193,"snap":"2021-21-2021-25","text_gpt3_token_len":583,"char_repetition_ratio":0.18923464,"word_repetition_ratio":0.0,"special_character_ratio":0.23470244,"punctuation_ratio":0.07446808,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98481244,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-11T03:59:15Z\",\"WARC-Record-ID\":\"<urn:uuid:c8b9b8ca-1d0b-48b3-9605-d72b727ec14d>\",\"Content-Length\":\"38186\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7b9f0417-69dc-4eb8-ba98-5081a8d44ab1>\",\"WARC-Concurrent-To\":\"<urn:uuid:c19e694d-50d9-461f-9d30-f6edfb341e65>\",\"WARC-IP-Address\":\"50.16.49.81\",\"WARC-Target-URI\":\"https://www.delftstack.com/zh/howto/arduino/arduino-char-to-int/\",\"WARC-Payload-Digest\":\"sha1:H3GVEXJUFLJQVKZYG5HV4FIJ2XYXUPJD\",\"WARC-Block-Digest\":\"sha1:RK7UL7FETYUKZWFYHYNYWQBMIUTEEQGD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991641.5_warc_CC-MAIN-20210511025739-20210511055739-00550.warc.gz\"}"}
https://www.slideserve.com/zeph-jensen/anova-overview-of-major-designs	[ "# ANOVA Overview of Major Designs - PowerPoint PPT Presentation", null, "Download Presentation", null, "ANOVA Overview of Major Designs\n\nANOVA Overview of Major Designs", null, "Download Presentation", null, "## ANOVA Overview of Major Designs\n\n- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -\n##### Presentation Transcript\n\n1. ANOVA Overview of Major Designs\n\n2. Between or Within Subjects • Between-subjects (completely randomized) designs • Subjects are nested within treatment conditions • By nested we mean that subjects are observed under only a single condition of the study • Within-subjects (randomized block) designs • Subjects are crossed by treatment conditions • By crossed we mean that subjects are observed under two or more conditions of the study\n\n3. When to Use Repeated Measures • Within-subjects designs are an advantage when • Scores under one condition are correlated with scores under another condition • When examining the effects of practice on performance of a learning task, or the effects of age in a longitudinal study of development • That in which a series of tests or subtests is to be administered to a group of subjects\n\n4. Randomized Block ANOVA\n\n5. Fixed Effects • Fixed factors are those in which we have selected particular levels of the factor in question not by random sampling but on the basis of our interest in those particular effects. • Cannot view these levels as representative • Cannot generalize to other levels • Examples: most manipulated variables, organismic variables, time, sessions, subtests\n\n6. Random Effects • Random factors are those in which we view the levels of the factor as having been randomly sampled from a larger population of such levels. • The most common random factor is subjects. • If subjects are not randomized we cannot generalize to others\n\n7. Error Terms in Four Designs • The appropriate choice of an error term in a repeated measures design depends on the fixed and random effects of within sampling units and between sampling units. • The effects (fixed or random) we want to test are properly tested by dividing the MS for that effect by the MS for a random source of variation.\n\n8. Aggregating Error Terms • When the number of df per error term is small, insignificant interactions can be aggregated with the error term to produce a pooled error term with more df. • Once we compute an aggregated (pooled) error term, it replaces all the individual error terms that contributed to its computation.\n\n9. Assumptions • Independence of errors • Normality • Homogeneity of variance including the sphericity assumption (homogeneity-of-variance-of-differences)\n\n10. What’s in a Name? • Choosing the appropriate statistic or design involves an understanding of • The number of independent variables and levels • The nature of assignment of subjects to treatment levels • The number of dependent variables • The source table for an analysis of variance describes the partition of the total sum of squares\n\n11. Between Subjects Completely Randomized ANOVA • One independent variable with two or more levels • Subjects completely randomly assigned to treatment levels • Also called • One-Way ANOVA\n\n12. Completely Randomized Analysis of Variance\n\n13. Within SubjectsRandomized Block ANOVA • One independent variable with two or more levels • Uses repeated measures of matching • Also called • One-Way with Repeated Measures ANOVA\n\n14. Randomized BlockAnalysis of Variance\n\n15. Between SubjectsCompletely Randomized Factorial ANOVA • Two or more independent variables each with two or more levels • Subjects are completely randomly assigned to all treatment combinations • Also called • Two-Way or Higher Order Analysis of Variance\n\n16. Two-Way Analysis of Variance\n\n17. Within Subjects Randomized Block Factorial Analysis of Variance • Two or more independent variables each with two or more levels • All treatment combinations use repeated measures or matching. • Also called • Two-Way or Higher Order Repeated Measures Analysis of Variance\n\n18. Randomized Block Factorial Analysis of Variance\n\n19. Mixed Analysis of Variance Split-Plot Factorial • Two or more independent variables each with two or more levels • At least one variable is completely randomized (between subjects) • At least one variable is randomized block (within subjects).\n\n20. Split Plot Factorials - Mixed Designs" ]	[ null, "https://www.slideserve.com/img/replay.png", null, "https://thumbs.slideserve.com/1_6556445.jpg", null, "https://www.slideserve.com/photo/205759.jpeg", null, "https://www.slideserve.com/img/output_cBjjdt.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8946704,"math_prob":0.7693415,"size":3694,"snap":"2021-04-2021-17","text_gpt3_token_len":703,"char_repetition_ratio":0.13604335,"word_repetition_ratio":0.11336718,"special_character_ratio":0.18597727,"punctuation_ratio":0.031141868,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9733473,"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\":\"2021-04-23T17:10:03Z\",\"WARC-Record-ID\":\"<urn:uuid:e154f04d-ca6e-4bce-b9aa-5f576802857a>\",\"Content-Length\":\"90801\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b6cda4e0-a08b-477a-b459-b399dc08ac04>\",\"WARC-Concurrent-To\":\"<urn:uuid:64659afa-9e94-4744-becc-64e9d9c31e5d>\",\"WARC-IP-Address\":\"44.236.52.34\",\"WARC-Target-URI\":\"https://www.slideserve.com/zeph-jensen/anova-overview-of-major-designs\",\"WARC-Payload-Digest\":\"sha1:GJGHIO3UB7U4KVHRULDGJ3DW4NWWDWW5\",\"WARC-Block-Digest\":\"sha1:D4KL7HNMILZJLIJM6JUBEIQ6MG3C22D4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618039596883.98_warc_CC-MAIN-20210423161713-20210423191713-00140.warc.gz\"}"}
https://globaltest.com/shop/sensors/vibration-transducer/acceleration/with-voltage-output/industrial-with-voltage-output/one-axial-industrial-with-voltage-output/	[ "# One-axial\n\n• Conversion factor: 1 mV/m/s²\n• Relative coefficient of transverse conversion (less than): 0,05\n• Operating frequency range: 0,5 ... 8000 Hz\n...\nGet price\n• Conversion factor: 1 mV/m/s²\n• Relative coefficient of transverse conversion (less than): 0,05\n• Operating frequency range: 0,5 ... 8000 Hz\n...\nGet price\n• Conversion factor: 1 mV/m/s²\n• Relative coefficient of transverse conversion (less than): 0,05\n• Operating frequency range: 0,5 ... 5000 Hz\n...\nGet price\n• Conversion factor: 10 mV/m/s²\n• Relative coefficient of transverse conversion (less than): 0,05\n• Operating frequency range: 0,5 ... 8000 Hz\n...\nGet price\n• Conversion factor: 10 mV/m/s²\n• Relative coefficient of transverse conversion (less than): 0,05\n• Operating frequency range: 0,5 ... 8000 Hz\n...\nGet price\n• Conversion factor: 10 mV/m/s²\n• Relative coefficient of transverse conversion (less than): 0,05\n• Operating frequency range: 0,5 ... 5000 Hz\n...\nGet price\n• Conversion factor: 3 mV/m/s²\n• Relative coefficient of transverse conversion (less than): 0,05\n• Operating frequency range: 0,5 ... 8000 Hz\n...\nGet price\n• Conversion factor: 3 mV/m/s²\n• Relative coefficient of transverse conversion (less than): 0,05\n• Operating frequency range: 0,5 ... 8000 Hz\n...\nGet price\n• Conversion factor: 3 mV/m/s²\n• Relative coefficient of transverse conversion (less than): 0,05\n• Operating frequency range: 0,5 ... 5000 Hz\n...\nGet price\n• Conversion factor: 5 mV/m/s²\n• Relative coefficient of transverse conversion (less than): 0,05\n• Operating frequency range: 0,5 ... 8000 Hz\n...\nGet price\nSUBMIT A REQUEST" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.54392105,"math_prob":0.9743512,"size":3255,"snap":"2020-10-2020-16","text_gpt3_token_len":928,"char_repetition_ratio":0.16364196,"word_repetition_ratio":0.5154867,"special_character_ratio":0.2596006,"punctuation_ratio":0.18438539,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9892479,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-08T11:51:05Z\",\"WARC-Record-ID\":\"<urn:uuid:f9562fc5-61dd-443a-931e-97b661c6968e>\",\"Content-Length\":\"150640\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cba39b1b-a64f-4a63-8356-b36aa3b8175e>\",\"WARC-Concurrent-To\":\"<urn:uuid:58ce5c24-6290-4a8c-a2f4-a8c14d16f3d8>\",\"WARC-IP-Address\":\"82.146.58.29\",\"WARC-Target-URI\":\"https://globaltest.com/shop/sensors/vibration-transducer/acceleration/with-voltage-output/industrial-with-voltage-output/one-axial-industrial-with-voltage-output/\",\"WARC-Payload-Digest\":\"sha1:Q43CVYDJMLQDZJRSLOBD7CWY2A7G6MNV\",\"WARC-Block-Digest\":\"sha1:DBWUTCFPWJBVVEJK7FTP7CZYVIVMP7HT\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585371813538.73_warc_CC-MAIN-20200408104113-20200408134613-00299.warc.gz\"}"}
https://www.mathworks.com/matlabcentral/cody/problems/43-subset-sum/solutions/180199	[ "Cody\n\nProblem 43. Subset Sum\n\nSolution 180199\n\nSubmitted on 20 Dec 2012\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\nTest Suite\n\nTest Status Code Input and Output\n1 Fail\n%% v = [2, 3, 5]; n = 8; correct = [2, 3]; actual = subset_sum(v, n); assert(isequal(actual, correct))\n\nError: You may not use the command(s) str2func in your code\n\n2 Fail\n%% v = [5, 3, 2]; n = 2; correct = 3; actual = subset_sum(v, n); assert(isequal(actual, correct))\n\nError: You may not use the command(s) str2func in your code\n\n3 Fail\n%% v = [2, 3, 5]; n = 4; correct = []; actual = subset_sum(v, n); assert(isequal(actual, correct))\n\nError: You may not use the command(s) str2func in your code\n\n4 Fail\n%% v = [1, 1, 1, 1, 1]; n = 5; correct = [1, 2, 3, 4, 5]; actual = subset_sum(v, n); assert(isequal(actual, correct))\n\nError: You may not use the command(s) str2func in your code\n\n5 Fail\n%% v = [1, 2, 3, 4, 100]; n = 100; correct = 5; actual = subset_sum(v, n); assert(isequal(actual, correct))\n\nError: You may not use the command(s) str2func in your code\n\n6 Fail\n%% v = [-7, -3, -2, 8, 5]; n = 0; correct = [2, 3, 5]; actual = subset_sum(v, n); assert(isequal(actual, correct))\n\nError: You may not use the command(s) str2func in your code" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5392727,"math_prob":0.9978827,"size":1383,"snap":"2019-43-2019-47","text_gpt3_token_len":479,"char_repetition_ratio":0.12255257,"word_repetition_ratio":0.39215687,"special_character_ratio":0.3933478,"punctuation_ratio":0.22712934,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.997011,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-20T13:59:29Z\",\"WARC-Record-ID\":\"<urn:uuid:241883d0-fc05-4a4b-96c5-999d5ee0cb06>\",\"Content-Length\":\"73424\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0e618b92-128f-455a-b79b-fc3a2612e8e0>\",\"WARC-Concurrent-To\":\"<urn:uuid:7a872387-ae5e-4651-abd4-1233acc1a121>\",\"WARC-IP-Address\":\"104.110.193.39\",\"WARC-Target-URI\":\"https://www.mathworks.com/matlabcentral/cody/problems/43-subset-sum/solutions/180199\",\"WARC-Payload-Digest\":\"sha1:FE2255OW3MAXUMV4DBMH4YF2UIWXMVIJ\",\"WARC-Block-Digest\":\"sha1:6Q5CU226KPI23IU2HZUFTEUBJMUKMAF5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986710773.68_warc_CC-MAIN-20191020132840-20191020160340-00379.warc.gz\"}"}
https://noteshunt.com/downloads/me2202-engineering-thermodynamics-important-questions/	[ "# ME2202 Engineering Thermodynamics Important Questions\n\nAnna University, Chennai\nDepartment of Mechanical Engineering\nME2202 Engineering Thermodynamics Important Questions\n\n1. A paddle wheel fixed to a shaft of an engine revolves in a closed hollow vessel containing water. This closed vessel is connected freely on the shaft and restraint to its turning moment is proved by mass attached to its side. Find the temperature rise for the following observations.\nEngine rpm = 650,\nLoad applied = 60 kg at a leverage of 1.2m,\nQuantity of water = 200 kg\nDuration of test\n\n2. A cylinder contains 0.45 m3 of gas at 1 x 105 N/m2 and 80oC. The gas is compressed to a volume of 0.13 m3, the final pressure being 5 X 105 N/m2 Determine i) The mass of gas ii) The value of index ‘n’ for compression The increase in internal energy of the gas and iv) The heat received or rejected by the gas during compression. Take γ = 1.4 R = 294.2 J/kg K\n\n3. A turbine operates under steady flow condition receives steam at the following state pressure = 1.2Mpa, Temperature =1880C, Enthalpy = 2785 kJ/kg, velocity = 33.3m/sec and elevation = 3m. The steam leaves the turbine at the following state: pressure = 20Kpa, Enthalpy = 2512 kJ/kg, velocity = 100m/sec and elevation = 0m. Heat is lost to the surrounding at the rate of 0.29kJ/sec. If the rate steam flow to the turbine is 0.42kg/sec what is the power output of the turbine in KW\n\n4. Three grams of nitrogen gas at 6atm and 160°c is expanded adiabatically to double its initial volume, then compressed at constant pressure to its initial volume and then compressed again at constant volume to its initial state. Calculate the net work done on the gas. Draw P-V diagram for the process. Specific heat ratio of nitrogen is 1.4.\n\n5. One litre of hydrogen at 273 K is adiabatically compressed to one-half of its initial volume. Find the change in temperature of the gas, if the ratio of two specific heats for hydrogen is 1.4\n\n6. An Engine-working on Otto cycle has a volume of 0.45 m3 , pressure 1 bar and temperature 30o,C at the beginning of compression stroke. At the end of compression stroke, the pressure is 11 bar and 210 KJ of heat is added at constant volume. Determine (i) Pressure, temperature and volumes at salient points in the cycle.’ (ii) Efficiency\n\n7. Derive the expression for thermal efficiency of a Carnot cycle with its p-V and Ts Diagram?\n\n8. A Carnot engine takes heat from an infinite reservoir at 5500C and rejects it to a sink 2750C. Half of the work delivered by the engine is used to run generator and the other half is used to run heat pump which takes heat at 2750C and rejects it at 4400C. Express the heat rejected at 4400C by the heat pump as {7bda82c1e6998b9ed8f6e0f52f1ca5ed9d6da876adcfbd0d890c6e0d88340190} of heat supplied to the engine at 5500C. If the operation of the generator is 500W, Find the heat rejected/hour by the heat pump at 4400C?\n\n9. Establish the inequality of Clausius and express Entropy change in irreversible process.\n\n10. 1 kg of steam initially dry saturated at 1.1 MPa expands in a cylinder following the law PV113 = C. The pressure at the end of expansion is 0.1 MPa. Determine (i) The final volume (ii) Final dryness fraction (hi) Work done (iv) The change in internal energy (v) The heat transferred\n\n11. Draw the p-V diagram of pure substance and explain its various regions of the diagram in details?\n\n12. with help of T-s diagram explain the various process of ideal Rankine’s cycle and derive the efficiency in terms of enthalpy?\n\n13. Steam at a pressure of 15 bar and 250°C expands according to the law PV1.25 = C to a pressure of 1.5 bar. Find the final conditions, work done, heat transfer and change in entropy. The mass is 0.8 kg\n\n14. Derive clausius-claperyon equation\n\n15. Deduce the Maxwell relations from thermodynamic property relations?\n\n16. Using Clausius Cleperyon’s equation, Estimate the enthalpy of vaporization at 2200C saturation temperature. Take the following data Ts = 2200C; vg = 0.08603m3/kg; vf =0.00119m3/kg.\n\n17. 2kg of air at 300C, 65{7bda82c1e6998b9ed8f6e0f52f1ca5ed9d6da876adcfbd0d890c6e0d88340190} RH is mixed adiabatically with 5kg of air at 200C, 10{7bda82c1e6998b9ed8f6e0f52f1ca5ed9d6da876adcfbd0d890c6e0d88340190}RH. Determine final condition of the mixture?\n\n18. Give short notes on following (a) Specific humidity (b) Relative humidity (c) Dew point temperature (d) Wet bulb depression\n\n19. Air has a dry bulb temperature of 250C and wet bulb temperature of 150C. If\n(a) Vapour pressure\n(b) Specific humidity\n(c) Saturation ratio\n(d) Relative humidity\n(e) Dew point temperature\n(f) Vapour density\n(g) Enthalpy of mixture\n\n0 reviews\n5 stars\n0\n4 stars\n0\n3 stars\n0\n2 stars\n0\n1 star\n0" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86677694,"math_prob":0.9252631,"size":4661,"snap":"2021-31-2021-39","text_gpt3_token_len":1304,"char_repetition_ratio":0.1112304,"word_repetition_ratio":0.0,"special_character_ratio":0.28041193,"punctuation_ratio":0.105320305,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.98372084,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-21T13:38:54Z\",\"WARC-Record-ID\":\"<urn:uuid:0bd85e6e-ef9b-4175-9bc1-f85a0ebba8ab>\",\"Content-Length\":\"135564\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aa1084b2-1430-40d2-9dfe-7eb815def3b6>\",\"WARC-Concurrent-To\":\"<urn:uuid:1a468557-27ea-4891-998f-76fad424655b>\",\"WARC-IP-Address\":\"172.67.132.58\",\"WARC-Target-URI\":\"https://noteshunt.com/downloads/me2202-engineering-thermodynamics-important-questions/\",\"WARC-Payload-Digest\":\"sha1:GEJ6PBQYMTRZ6FXPCQUY465SW7Z5UALB\",\"WARC-Block-Digest\":\"sha1:EVZVU3NP55TBHLPRNM3FB55NPMOUISKL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057225.38_warc_CC-MAIN-20210921131252-20210921161252-00674.warc.gz\"}"}