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https://mathoverflow.net/questions/231489/does-the-tensor-bundle-of-a-compact-manifold-have-a-bounded-geometry | [
"# Does the tensor bundle of a compact manifold have a bounded geometry?\n\nLet $M$ be a compact manifold. Let $S^2 T^*M$ be the vector bundle of all symmetric $(0,2)$ tensors and $S_+^2 T^*M$ be the open subset of all positive definite ones. Does $S_+^2 T^*M$ have bounded geometry? It is understood that the metric on it is the tensor product of the induced metric on the cotangent bundle from some Riemannian metric on $M$.\n\n• Feb 18, 2016 at 11:05\n• @Kaveh May you more explain on the structure of the metric on the bundle? Feb 18, 2016 at 11:25\n• I read mathoverflow.net/questions/212713/…. But I can not see the conclusion. $s_+^2T^*M$ is a subbundle of $T^*M \\otimes T^*M$, so the question is that is finite tensor product of cotangent bundle has bounded geometry. Feb 18, 2016 at 11:30\n• About Riemannian metric, I think the natural metric is the tensor product of a Riemannian metric on the cotangent bundle. I dont not know yet what kind of metrics exist on cotangent bundles. Feb 18, 2016 at 11:41\n• in the referred post $TM$ was considered with two types of metrics: Sasaki with flat fibers - not bounded geometry, and Cheeger-Gromoll, coming from submersion $GM \\to TM$ of bounded geometry. So, the choice is yours. Feb 18, 2016 at 11:51"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.823863,"math_prob":0.9640552,"size":1489,"snap":"2023-14-2023-23","text_gpt3_token_len":458,"char_repetition_ratio":0.14343435,"word_repetition_ratio":0.025423728,"special_character_ratio":0.30355942,"punctuation_ratio":0.11111111,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99801534,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-03T23:56:02Z\",\"WARC-Record-ID\":\"<urn:uuid:fb30eb11-3536-491c-9bf4-0542c2227cdd>\",\"Content-Length\":\"86750\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bf4392b6-9688-4fe9-9dc4-182d74e0ba94>\",\"WARC-Concurrent-To\":\"<urn:uuid:3e46cdb4-ad83-4ed9-8a8a-53f32bafbbc8>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://mathoverflow.net/questions/231489/does-the-tensor-bundle-of-a-compact-manifold-have-a-bounded-geometry\",\"WARC-Payload-Digest\":\"sha1:552EWSRK6E67S7KD5XETAEB5NO42VKY3\",\"WARC-Block-Digest\":\"sha1:6DHC2TAQH35CRWR6BVUTWAJVSIMJB6U3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224649348.41_warc_CC-MAIN-20230603233121-20230604023121-00093.warc.gz\"}"} |
https://www.sarthaks.com/9311/in-the-estimation-of-sulphur-by-carius-method-468-of-an-organic-sulphur-compound-afforded | [
"# In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded\n\n3.8k views\n\nIn the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound.\n\nby (10 points)\n\nby (127k points)\n\nTotal mass of organic compound = 0.468 g [Given]\n\nMass of barium sulphate formed = 0.668 g [Given]\n\n1 mol of BaSO4 = 233 g of BaSO4 = 32 g of sulphur\n\nThus, 0.668 g of BaSO4 contains (32X0.668/233g) of sulphur = 0.0917 g of sulphur\n\nTherefore, percentage of sulphur =(0.0197/0.468X100) = 19.59 %\n\nHence, the percentage of sulphur in the given compound is 19.59 %."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.79514027,"math_prob":0.9812204,"size":358,"snap":"2020-45-2020-50","text_gpt3_token_len":130,"char_repetition_ratio":0.1779661,"word_repetition_ratio":0.0,"special_character_ratio":0.424581,"punctuation_ratio":0.15853658,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9672105,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-02T20:02:24Z\",\"WARC-Record-ID\":\"<urn:uuid:8a19955d-58b9-4f2a-95e4-9e84cf4c54ae>\",\"Content-Length\":\"60178\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ba41d123-b8b0-4bbc-88c8-f3dd49f6b810>\",\"WARC-Concurrent-To\":\"<urn:uuid:a97cf432-9d35-4c04-8cf2-1f2265c127ce>\",\"WARC-IP-Address\":\"104.26.6.26\",\"WARC-Target-URI\":\"https://www.sarthaks.com/9311/in-the-estimation-of-sulphur-by-carius-method-468-of-an-organic-sulphur-compound-afforded\",\"WARC-Payload-Digest\":\"sha1:D2ILCXHTDVQBC6LRWKHBVSMWOFWYXRYD\",\"WARC-Block-Digest\":\"sha1:65QG2EUIDN7YJ6FWMNSSL7GQGEGK7Y6E\",\"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-20201202175113-20201202205113-00125.warc.gz\"}"} |
https://help.scilab.org/docs/6.1.2/ja_JP/MAXMIN.html | [
"# MAXMIN\n\nMax or min value of a vector, or element-wise for N vectors\n\n### Block Screenshot",
null,
"### Description\n\nThis block finds the minimum/maximum values (parameter Min or Max) and accepts one or two inputs depending on the parameter Number of input vectors :\n\n Number of input vectors Input Operation 1 Vector The input must be a vector and the bloc output is the minimum/maximum value of the elements of its input vector. 2 Scalar The block output is the minimal or maximal scalar. 2 Vector All p input vectors must have the same length N. The block output is a vector with the same length N. The element output(i) of the output vector is min(input_1(i), input_2(i),..,input_p(i)) or max(input_1(i), input_2(i),..,input_p(i)), for `i=1..N`.\n\n### Parameters",
null,
"• Min or Max\n\nThe function (min or max) to apply to the input.\n\nProperties : Type 'vec' of size 1.\n\n• Number of input vectors\n\nThe number of inputs to the block.\n\nProperties : Type 'vec' of size 1.\n\n• zero-crossing\n\nSelect to enable zero crossing detection to detect minimum and maximum values.\n\nProperties : Type 'vec' of size 1.\n\n### Default properties\n\n• always active: no\n\n• direct-feedthrough: yes\n\n• zero-crossing: no\n\n• mode: no\n\n• regular inputs:\n\n- port 1 : size [-1,1] / type 1\n\n• regular outputs:\n\n- port 1 : size [1,1] / type 1\n\n• number/sizes of activation inputs: 0\n\n• number/sizes of activation outputs: 0\n\n• continuous-time state: no\n\n• discrete-time state: no\n\n• object discrete-time state: no\n\n• name of computational function: minmax\n\n### Interfacing function\n\n• SCI/modules/scicos_blocks/macros/NonLinear/MAXMIN.sci\n\n### Computational function\n\n• SCI/modules/scicos_blocks/src/c/minmax.c (Type 4)"
] | [
null,
"https://help.scilab.org/docs/6.1.2/ja_JP/MAXMIN.png",
null,
"https://help.scilab.org/docs/6.1.2/ja_JP/MAXMIN_gui.gif",
null
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https://math.stackexchange.com/questions/3307645/unique-morphism-of-schemes-into-the-spectrum-of-integers | [
"# Unique morphism of Schemes into the spectrum of integers\n\nI wish to prove the following: For every scheme $$X$$ there exists a unique morphism of schemes $$X\\rightarrow Spec(\\mathbb{Z})$$.\n\nHere is what I have so far: if $$X$$ is affine, say $$X\\simeq Spec(A)$$ for a ring $$A$$, I know that morphisms of schemes $$Spec(A)\\rightarrow Spec(B)$$ are in one to one correspondence with ring homomorphisms $$B\\rightarrow A$$. Any homomorphism $$\\phi:\\mathbb{Z}\\rightarrow A$$ must satisfy $$\\phi(1) = 1$$ and is thus unique.\n\nIf $$X$$ is a scheme, we have an open cover $$(X_i)_{i\\in I}$$ such that $$(X_i,\\mathcal{O}_{X}\\mid X_i ) \\simeq (Spec(A_i),\\mathcal{O}_{Spec(A_i)})$$ and hence, there are unique morphisms $$f_i:(X_i,\\mathcal{O}_{X}\\mid X_i )\\rightarrow (Spec(\\mathbb{Z}),\\mathcal{O}_{Spec(\\mathbb{Z})})$$.\n\nNow I would like to construct a global morphism $$f$$ by glueing together the local parts $$f_i$$ but I am not sure how (or even if) this works.\n\n• Yes this works because the morphisms are unique. More precisely, on $X_i\\cap X_j$, one can find an open cover by affines $(U_k)$. Now the restriction of $f_i$ and $f_j$ to $U_k$ must agree by unicity. It follows that $f_i=f_j$ on $X_i\\cap X_j$ since they agree an open cover. This implies that the $f_i$ can be glued together to give a morphism $X\\to\\operatorname{Spec}\\mathbb{Z}$. The unicity is easy. – Roland Jul 29 '19 at 19:44\n• Thanks, I get the idea. What worries me a little is that I can see that $f_i\\mid_{U_k}=f_j\\mid_{U_k}$ implies $f_i = f_j$ as continous maps, but what about the morphisms of sheaves $f_i^{\\sharp}:\\mathcal{O}_{Spec(\\mathbb{Z})}\\rightarrow (f_i)_*\\mathcal{O}_X\\mid_{X_i}$? My guess here is that it follows from the fact that the Hom - sheaf $\\underline{Hom}(\\mathcal{F},\\mathcal{G})$ is a sheaf if $\\mathcal{G}$ is one. Or am I thinking too complicated here? – Teddyboer Jul 29 '19 at 20:58\n• Two morphisms of sheaves which agree on an open cover are equal, this is straightforward. Of course it follows from the fact that the Hom sheaf is a sheaf, but I don't usually mention this proposition when glueing or checking equality locally morphisms of schemes. – Roland Jul 30 '19 at 12:12\n\nLet $$X$$ be a scheme an let $$R$$ be a ring. Let me maybe suggest you to try to prove the following:\nThere is a natural bijection $$\\text{Hom}_{\\text{Sch}}(X, \\text{Spec}(R)) \\cong \\text{Hom}_{\\text{Ring}}(R, \\Gamma(X,\\mathcal{O}_X)).$$\nThis is not only a very useful statement, but also implies what you want to prove as you can use that $$\\mathbb{Z}$$ is a initial object in the category of rings like you were doing for the affine case."
] | [
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https://brilliant.org/practice/rotational-kinetic-energy-conservation-of-energy/ | [
"",
null,
"Classical Mechanics\n\n# Conservation of rotational and translational energy",
null,
"A solid ball rolls on a slope from rest starting from a height of $H=7.0\\text{ m}$ and then rolls on a horizontal region, as shown in the above figure. The horizontal distance of the slope and the distance of the horizontal region are both equal to $L=6 \\text{ m},$ and the height of the horizontal region is $h=3.0\\text{ m}.$ Approximately how far horizontally from point $P$ does the ball hit the floor?\n\nThe rotational inertia of a solid sphere about any diameter is $I=\\frac{2}{5}MR^2,$ where $M$ and $R$ are the mass and the radius of the solid sphere, respectively, and the gravitational acceleration is $g=9.8\\text{ m/s}^2.$\n\nA brand new off-road $6 \\times 6$ pickup truck has six wheels. If the total mass of the pickup truck is $1200\\text{ kg}$ and the mass of each of the six wheels is $19\\text{ kg},$ what fraction of its total kinetic energy is due to the rotation of the wheels about their center axles?\n\nAssume that each of the six wheels is a uniform disk.\n\nA very thin hoop with a mass of $180\\text{ kg}$ is rolling along a horizontal floor. If the speed of the hoop's center of mass is $0.180\\text{ m/s},$ how much work must be done on the hoop to stop it?",
null,
"A solid cylinder with radius $40\\text{ cm}$ and mass $12\\text{ kg}$ is rolling down from rest without slipping for a distance of $L=10.0\\text{ m},$ as shown in the above figure. The angle of the slope is $\\theta=30^\\circ.$ What is the approximate angular speed of the cylinder about its center when it just meets the bottom of the slope, assuming that the gravitational acceleration is $g=9.8\\text{ m/s}^2?$\n\nConsider a situation where a uniform solid sphere, which was rolling smoothly along a horizontal floor, rises up along a $30.0^\\circ$ slope. If it stops after it has rolled $9.00\\text{ m}$ along the slope and then begins to roll backward, what was its initial speed?\n\nThe rotational inertia of a solid sphere about any diameter is $I=\\frac{2}{5}MR^2,$ where $M$ and $R$ are the mass and the radius of the solid sphere, respectively, and the gravitational acceleration is $g=9.8\\text{ m/s}^2.$\n\n×"
] | [
null,
"https://ds055uzetaobb.cloudfront.net/brioche/chapter/Rotational%20Kinetic%20Energy-w4A2Nk.png",
null,
"https://ds055uzetaobb.cloudfront.net/brioche/solvable/bdc8e6790a.873412a47c.K0mazW.jpg",
null,
"https://ds055uzetaobb.cloudfront.net/brioche/solvable/bdc8e6790a.aa9e553968.qBEmem.jpg",
null
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https://metanumbers.com/423970 | [
"423970 (number)\n\n423,970 (four hundred twenty-three thousand nine hundred seventy) is an even six-digits composite number following 423969 and preceding 423971. In scientific notation, it is written as 4.2397 × 105. The sum of its digits is 25. It has a total of 3 prime factors and 8 positive divisors. There are 169,584 positive integers (up to 423970) that are relatively prime to 423970.\n\nBasic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 6\n• Sum of Digits 25\n• Digital Root 7\n\nName\n\nShort name 423 thousand 970 four hundred twenty-three thousand nine hundred seventy\n\nNotation\n\nScientific notation 4.2397 × 105 423.97 × 103\n\nPrime Factorization of 423970\n\nPrime Factorization 2 × 5 × 42397\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 423970 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 423,970 is 2 × 5 × 42397. Since it has a total of 3 prime factors, 423,970 is a composite number.\n\nDivisors of 423970\n\n8 divisors\n\n Even divisors 4 4 4 0\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 763164 Sum of all the positive divisors of n s(n) 339194 Sum of the proper positive divisors of n A(n) 95395.5 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 651.13 Returns the nth root of the product of n divisors H(n) 4.44434 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 423,970 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 423,970) is 763,164, the average is 9,539,5.5.\n\nOther Arithmetic Functions (n = 423970)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 169584 Total number of positive integers not greater than n that are coprime to n λ(n) 42396 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 35616 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares\n\nThere are 169,584 positive integers (less than 423,970) that are coprime with 423,970. And there are approximately 35,616 prime numbers less than or equal to 423,970.\n\nDivisibility of 423970\n\n m n mod m 2 3 4 5 6 7 8 9 0 1 2 0 4 1 2 7\n\nThe number 423,970 is divisible by 2 and 5.\n\n• Deficient\n\n• Polite\n\n• Square Free\n\n• Sphenic\n\nBase conversion (423970)\n\nBase System Value\n2 Binary 1100111100000100010\n3 Ternary 210112120121\n4 Quaternary 1213200202\n5 Quinary 102031340\n6 Senary 13030454\n8 Octal 1474042\n10 Decimal 423970\n12 Duodecimal 18542a\n20 Vigesimal 2cjia\n36 Base36 934y\n\nBasic calculations (n = 423970)\n\nMultiplication\n\nn×y\n n×2 847940 1271910 1695880 2119850\n\nDivision\n\nn÷y\n n÷2 211985 141323 105992 84794\n\nExponentiation\n\nny\n n2 179750560900 76208845304773000 32310264143864608810000 13698582689074278197175700000\n\nNth Root\n\ny√n\n 2√n 651.13 75.1239 25.5172 13.3496\n\n423970 as geometric shapes\n\nCircle\n\n Diameter 847940 2.66388e+06 5.64703e+11\n\nSphere\n\n Volume 3.19223e+17 2.25881e+12 2.66388e+06\n\nSquare\n\nLength = n\n Perimeter 1.69588e+06 1.79751e+11 599584\n\nCube\n\nLength = n\n Surface area 1.0785e+12 7.62088e+16 734338\n\nEquilateral Triangle\n\nLength = n\n Perimeter 1.27191e+06 7.78343e+10 367169\n\nTriangular Pyramid\n\nLength = n\n Surface area 3.11337e+11 8.9813e+15 346170\n\nCryptographic Hash Functions\n\nmd5 dd8db3486efc6ea4343a8bf799e0a3cc b9eb5755c2d5ac06fa77569b645ed350552a30b6 54e12ec1d85d89fe4377ff1920549999d6c64775bf0b5a10f2ee8fc42a2b81a3 cee2a5ce9f837ae550abd7f0bf22c5a40537c2f5fcea70e374ba84d37583b29f388585e2d0758ac4e117f171a356a5cbd551f52a7647d6fad61e689bdf59d3a1 9dd8f29cb9f6f2ff047491cce9bc459613597596"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.61616546,"math_prob":0.9873253,"size":4656,"snap":"2022-05-2022-21","text_gpt3_token_len":1643,"char_repetition_ratio":0.12080825,"word_repetition_ratio":0.030927835,"special_character_ratio":0.46069586,"punctuation_ratio":0.078580484,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9966873,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-19T04:58:07Z\",\"WARC-Record-ID\":\"<urn:uuid:de76cd38-1c6b-4527-9374-1b86ac405f46>\",\"Content-Length\":\"40028\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a9b80e77-6079-4425-977d-7d356019841e>\",\"WARC-Concurrent-To\":\"<urn:uuid:22e8edbe-a078-4fa1-a70f-a94a13e5d664>\",\"WARC-IP-Address\":\"46.105.53.190\",\"WARC-Target-URI\":\"https://metanumbers.com/423970\",\"WARC-Payload-Digest\":\"sha1:A72ES453CZNQY3EBGVVAQWEHIEXGILFD\",\"WARC-Block-Digest\":\"sha1:6YYB2KZ2KL7AMNGU64UWHGFARVKEVJIF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320301263.50_warc_CC-MAIN-20220119033421-20220119063421-00020.warc.gz\"}"} |
http://fricas.github.io/api/AlgebraicManipulations.html | [
"# AlgebraicManipulations(R, F)¶\n\nAlgebraicManipulations provides functions to simplify and expand expressions involving algebraic operators.\n\nratDenom: (F, F) -> F\n\nratDenom(f, a) removes a from the denominators in f if a is an algebraic kernel.\n\nratDenom: (F, List F) -> F\n\nratDenom(f, [a1, ..., an]) removes the ai's which are algebraic kernels from the denominators in f.\n\nratDenom: (F, List Kernel F) -> F\n\nratDenom(f, [a1, ..., an]) removes the ai's which are algebraic from the denominators in f.\n\nratDenom: F -> F\n\nratDenom(f) rationalizes the denominators appearing in f by moving all the algebraic quantities into the numerators.\n\nratPoly: F -> SparseUnivariatePolynomial F\n\nratPoly(f) returns a polynomial p such that p has no algebraic coefficients, and p(f) = 0.\n\nrootFactor: F -> F if R has GcdDomain and F has FunctionSpace R and R has UniqueFactorizationDomain and R has Comparable and R has RetractableTo Integer\n\nrootFactor(f) transforms every radical of the form (a1*...*am)^(1/n) appearing in f into a^(1/n)*...*am^(1/n). This transformation is not in general valid for all complex numbers a and b.\n\nrootKerSimp: (BasicOperator, F, NonNegativeInteger) -> F if R has GcdDomain and R has Comparable and R has RetractableTo Integer and F has FunctionSpace R\n\nrootKerSimp(op, f, n) should be local but conditional.\n\nrootPower: F -> F if R has GcdDomain and R has Comparable and R has RetractableTo Integer and F has FunctionSpace R\n\nrootPower(f) transforms every radical power of the form (a^(1/n))^m into a simpler form if m and n have a common factor.\n\nrootProduct: F -> F if R has GcdDomain and R has Comparable and R has RetractableTo Integer and F has FunctionSpace R\n\nrootProduct(f) combines every product of the form (a^(1/n))^m * (a^(1/s))^t into a single power of a root of a, and transforms every radical power of the form (a^(1/n))^m into a simpler form.\n\nrootSimp: F -> F if R has GcdDomain and R has Comparable and R has RetractableTo Integer and F has FunctionSpace R\n\nrootSimp(f) transforms every radical of the form (a * b^(q*n+r))^(1/n) appearing in f into b^q * (a * b^r)^(1/n). This transformation is not in general valid for all complex numbers b.\n\nrootSplit: F -> F\n\nrootSplit(f) transforms every radical of the form (a/b)^(1/n) appearing in f into a^(1/n) / b^(1/n). This transformation is not in general valid for all complex numbers a and b."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.76355726,"math_prob":0.9877039,"size":2584,"snap":"2022-05-2022-21","text_gpt3_token_len":765,"char_repetition_ratio":0.12364341,"word_repetition_ratio":0.35549873,"special_character_ratio":0.25464398,"punctuation_ratio":0.13114753,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9970019,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-20T17:18:09Z\",\"WARC-Record-ID\":\"<urn:uuid:0b24dfe1-d3ce-45fa-9464-f593382f808e>\",\"Content-Length\":\"19327\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c72f4803-239d-451c-90c7-58ec34ab8d87>\",\"WARC-Concurrent-To\":\"<urn:uuid:6ae3625c-72de-421a-b168-bd0497020f28>\",\"WARC-IP-Address\":\"185.199.111.153\",\"WARC-Target-URI\":\"http://fricas.github.io/api/AlgebraicManipulations.html\",\"WARC-Payload-Digest\":\"sha1:AUL65WJNCLCIGL2PPSBSEKHX3MBVKWUB\",\"WARC-Block-Digest\":\"sha1:25VTXDQUGEWHDDZBQSQVCIERWPHGEOLO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662533972.17_warc_CC-MAIN-20220520160139-20220520190139-00429.warc.gz\"}"} |
https://dsp.stackexchange.com/questions/31427/how-to-find-a-know-signal-in-another-signal-which-contains-other-signals-along-w | [
"# How to find a know signal in another signal which contains other signals along with the signal i need in python\n\nI am working on BPSK with sound waves a carrier waves.\n\nActually I am modulating my data with some known sync bits along with the information I need to send. So the things I will be having at demodulation side is these sync bits and carrier wave.\n\nSo I am modulating these sync bits with carrier and cross correlating this modulated (over sync bits) wave with received wave at the transmiter. By doing this I want to find the exact time position (where these sync bits appear in received wave).\n\nFor this I am first crosscorelating received wave and my sync wave then, when I plot it I wil surely get a maximum where the sync wave in both signals coincide.\n\nBut I am not sure if this maximum is the position I need (when there is a DC offset or large amplitudes in the received signal, it can also create an extreme value)\n\nSo to overcome this, I autocorelated my sync wave with itself and took the absolute difference between the all the values of crosscorelated matrix and autocorelated value. Then I assumed which ever value has the least value as the synced location.\n\nThis is also not working (I think as iam not sure that the location iam getting is midpoint of sync or starting of sync). Is this approach correct?\n\nOr are there any other approaches to find the starting sync position in received wave?\n\nI also needed help in knowing whether if we use the first method (wher evr maxima comes in crosscorelation taking tht as sync location) is this location I got, is sync starting location or midpoint of sync?\n\n• Sai Teja- How long can your sync bit sequence be? See this posting, perhaps that can help you? dsp.stackexchange.com/questions/30824/… – Dan Boschen Jun 13 '16 at 2:30\n• Dan Boschen- yesterday i found that my sync bit sequence is too short to give a significant peak in cross correlation, now i increased my sync bits to 1000 (where my data bits are 7000) which is giving me exact sync location, just by finding the max value of cross correlation will give mid point of my sync sequence, as i knew the sync sequence length i can figure out data start time stamp (as my data begins right after sync ends). thanks – Sate Jun 14 '16 at 4:49\n• Very good. Generally if your noise is white, you are getting a 10Log10(N) advantage in SNR due to the correlation processing, where N is the number of sync bits. Knowing your SNR conditions and probability of detection you can then design how long your sync bit sequence would need to be. – Dan Boschen Jun 14 '16 at 4:54\n\nI made a little research on detecting the sync_start location on my received signal.\n\nFirst I modulated the bits (sync_start bits + data bits + sync_end bits) with a sine signal as my carrier, with a frequency of $20 \\text{ kHz}$ and a total length of 10000 bits (including both sync_start and sync_end).\n\nI then converted this modulated wave into a .wav file and played it on my PC, while recording simultaniously on an another PC, using Pyaudio (Python). That gave me a recorded .wav - my received signal.\n\nNow as I knew my sync bits value, I modulated them on the same carrier wave and cross correlated this with the received wave.\n\n# Case 1:\n\n$\\text{bit length} = 10000$ (including sync_start and sync_end),\n\n$\\text{number of sync_start bits} = \\text{sync_end} = 2287$\n\nThe plot of cross correlation is as shown :",
null,
"# Case 2:\n\n$\\text{bit length} = 15000$",
null,
"# case 3:\n\n$\\text{bit length} = 20000$",
null,
"As the data length increases(keeping our sync_start and sync_end same in all cases), the exact time stamps of sync_start_mid index and sync_end_mid index gets more accurate.\n\nNote : For data bits here, I took all of them as $1$'s -- that might be the reason for the smaller amplitudes in the middle. I will add random data and see if I am getting the same inference.\n\nPlease correct me if there is any mistake in my inference. As far as now I am now able to exactly pinpoint the synchronization between transmitted and received waves.\n\nI also wanted to know the limitations of length of the sync bits (because when my sync bits length gets smaller, the dominant peak location cannot be found from cross correlation).\n\nAnd I am using a BPSK scheme.\n\n• Sai Teja- See my comment above on SNR vs length. This holds assuming the noise from one symbol to the next are independent: When you sum two symbols for example, the signal of interest doubles (in magnitude) but the noise standard deviation will only increase by the square root of 2. Hence the 10Log10(N) increase in SNR. – Dan Boschen Jun 14 '16 at 12:50"
] | [
null,
"https://i.stack.imgur.com/NH3zR.png",
null,
"https://i.stack.imgur.com/F7Krm.png",
null,
"https://i.stack.imgur.com/73odN.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.96176535,"math_prob":0.8588759,"size":1508,"snap":"2020-45-2020-50","text_gpt3_token_len":325,"char_repetition_ratio":0.13962767,"word_repetition_ratio":0.0,"special_character_ratio":0.198939,"punctuation_ratio":0.05498282,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9829505,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,6,null,6,null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-22T10:10:56Z\",\"WARC-Record-ID\":\"<urn:uuid:c632640a-b074-463d-a76c-b43ae8cd5d43>\",\"Content-Length\":\"158438\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5d84e675-64ac-4cc8-ac9f-cec00aec7749>\",\"WARC-Concurrent-To\":\"<urn:uuid:b01f93cb-a0f4-4f12-9146-6c8aec0ac12a>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://dsp.stackexchange.com/questions/31427/how-to-find-a-know-signal-in-another-signal-which-contains-other-signals-along-w\",\"WARC-Payload-Digest\":\"sha1:AAOSHWSAGNK6MO7HOQ7M4HOBWWSVTIDY\",\"WARC-Block-Digest\":\"sha1:HP7WXOYB4YYOEARPVR2JONIOK2K4SULL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107879362.3_warc_CC-MAIN-20201022082653-20201022112653-00191.warc.gz\"}"} |
https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_14&diff=prev&oldid=61273 | [
"# Difference between revisions of \"2014 AIME II Problems/Problem 14\"\n\n14. In △ABC, AB=10, ∠A=30∘, and ∠C=45∘. Let H, D, and M be points on the line BC such that AH⊥BC, ∠BAD=∠CAD, and BM=CM. Point N is the midpoint of the segment HM, and point P is on ray AD such that PN⊥BC. Then",
null,
"$AP^2=m/n$, where m and n are relatively prime positive integers. Find m+n."
] | [
null,
"https://latex.artofproblemsolving.com/9/2/0/9203833c64ad1990fda5781dd35d8474e302fac5.png ",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.86511767,"math_prob":0.99191034,"size":780,"snap":"2021-21-2021-25","text_gpt3_token_len":304,"char_repetition_ratio":0.125,"word_repetition_ratio":0.722973,"special_character_ratio":0.3448718,"punctuation_ratio":0.19069767,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9926741,"pos_list":[0,1,2],"im_url_duplicate_count":[null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-06T02:02:37Z\",\"WARC-Record-ID\":\"<urn:uuid:433de3de-1e24-4b76-836b-4521d614e540>\",\"Content-Length\":\"37322\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b2fdcff3-ff1b-46d1-9383-36a30dc8caf5>\",\"WARC-Concurrent-To\":\"<urn:uuid:2219c0f2-d8a5-4488-af4d-ab8cc7ffd5dd>\",\"WARC-IP-Address\":\"172.67.69.208\",\"WARC-Target-URI\":\"https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_14&diff=prev&oldid=61273\",\"WARC-Payload-Digest\":\"sha1:SRZPFDQ4HAOAEOVMPU6Y7RVCJ4OUSYD7\",\"WARC-Block-Digest\":\"sha1:ZHTQ2GUKZX5D5PGWMKLS7MNDZULPYCB7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988724.75_warc_CC-MAIN-20210505234449-20210506024449-00169.warc.gz\"}"} |
https://chemistry.stackexchange.com/questions/76398/law-of-mass-action-vs-rate-law?noredirect=1 | [
"# Law of Mass action VS Rate law [duplicate]\n\nThe Law of mass action has 2 parts - the equilibrium rate part & the reaction rate part. The \"reaction rate part\" states that rate of a reaction is directly proportional to the product of active masses of the reactants with each reactant term raised to it's stoichiometric coefficient. The rate law states that rate of a reaction is directly proportional to the product of active masses of the reactants with each reactant term raised to some power which may or may not be equal to stoichiometric coefficient of the reactant.\n\nWhy are the two laws different when in fact we are talking about the powers to which active masses of reactants should be raised in both the cases ? And which law is correct?\n\n• As far as my knowledge is concerned reactant can be one or more. In case when there is one reactant first part is applied and when there are two or more reactants second part is applied. – AksaK Jul 4 '17 at 16:04\n• This question has been answered in a separate question:chemistry.stackexchange.com/questions/68195/… – Withnail May 1 '19 at 12:28\n\nFirst of all, you need to understand the basic difference between the two.\n\nConsider the reaction $$\\ce{A + B -> C}$$ Law of mass action states that in a rate equation the powers that the active masses of components raised are always stoichiometric factors.\n\nFor a reaction they are always fixed. However, the rate of reaction differs with the physical conditions. This is applicable to both equilibrium reactions and moderate reactions that are unidirectional.\n\nThis makes law of mass action kind of theoretical in nature. In the rate law, the rate equation consists of actual factors that changes with change in experimental conditions. These may or may not be equal to stoichiometric factors. These are generally responsible for determining the order of reaction.\n\nBoth the laws are correct because both have a wide area of application. Rate law is although quite experimental in nature and is generally considered to be more appropriate.\n\n• In which cases do the law of mass action give the exactly correct relations? – Hisab Aug 6 '17 at 18:10\n\nYour statement of the Law of Mass Action is wrong. From Wikipedia:\"The law is a statement about equilibrium and gives an expression for the equilibrium constant, a quantity characterizing chemical equilibrium.\" You apparently confuse the dynamic equilibrium where concentrations of reactants and products do not change with the rate of chemical reactions not at equilibrium (generally at constant temperature, and always with changes in concentrations). Keq is the equilibrium rate constant, the Reaction Quotient is Q. Q changes as the reaction proceeds and reaches Keq as dynamic equilibrium is attained.\nIt is NOT true that the rate of a (forward) chemical reaction is proportional to the concentration of the active masses of the reactants raised to some empirical power. Some reactions proceed at constant rates. These are known as zeroth order reactions - catalyzed reactions can be examples of these. (While you can argue that if R → P and if d[R]/dt = -k then d[R]/dt = -k[R]° (since concentration taken to the zeroth power is 1) that the proportionality holds, but this is trivial and not useful.)\nIn the historical development of the Law of Mass Action, there have been many instances where the powers for the Reaction Quotient at equilibrium have matched (more or less) the empirically determined powers of the rate expression. This is quite different than CONFUSING a situation at equilibrium with the situation which must, by definition, be non-equilibrium (since concentrations are changing). Which is correct?\nLogically invoking a dynamic equilibrium requires concentrations to be unchanging. Say R → P. With [R] and [P] constant, then taking the ratio of [R]α/[P]ß for some arbitrary constants α & ß is also going to produce a constant. Choosing the stoichiometric coefficients for α & ß doesn't change that.\nFor the rate of a reaction, the claim is that the rate of (forward) reaction is proportional to [R1]α[R2]ß...[Rn]n. This often isn't even an acceptable approximation. (although over a small enough time interval, any continuous (non-pathological) data can be approximated by a straight line). The more difficult question to answer here is does the rate of reaction for simple single-step reactions follow this rule? I'd say in dilute gas phase reactions (without significant effects from products) (and of course at constant temperature and pressure) it generally will. The problem with this, is determination of which are and which are not \"simple single step\" reactions isn't trivial. You could start there and confirm the proportionality. You'd have to confirm it, which makes it mostly unnecessary to assume it to begin with, but you have to start somewhere.\n\n• The argument that you could choose any powers to raise the concentrations to (when at equilibrium as the concentrations don’t change, so neither will the quotient of those concentrations) is not a justification for always choosing to raise to the power of the coefficients. Raising to other powers will still give a constant, but it won’t generally have the same value as the actual equilibrium constant. – Withnail Apr 30 '19 at 21:06"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9521033,"math_prob":0.97174734,"size":700,"snap":"2021-31-2021-39","text_gpt3_token_len":143,"char_repetition_ratio":0.15660919,"word_repetition_ratio":0.3305785,"special_character_ratio":0.19285715,"punctuation_ratio":0.03937008,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9852537,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-03T18:47:12Z\",\"WARC-Record-ID\":\"<urn:uuid:68d33880-5a4f-431a-ae2a-2f6cba7680d3>\",\"Content-Length\":\"162237\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a22e8929-a3ce-4a62-8224-124d92c54f79>\",\"WARC-Concurrent-To\":\"<urn:uuid:e7e59649-2a4f-49fa-8d9b-ab3e6abd824d>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://chemistry.stackexchange.com/questions/76398/law-of-mass-action-vs-rate-law?noredirect=1\",\"WARC-Payload-Digest\":\"sha1:TFAI5PWZ7AC6J23RNAAPUXPWILO7MOXF\",\"WARC-Block-Digest\":\"sha1:BOB6N3GNDDXFIEEQQYMAGSY2TF3GIZPG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154466.61_warc_CC-MAIN-20210803155731-20210803185731-00443.warc.gz\"}"} |
https://javedmultani16.medium.com/given-a-sorted-and-rotated-array-a-of-n-distinct-elements-which-is-rotated-at-some-point-and-given-ff48913c3114 | [
"# Given 2 sorted arrays Ar1 and Ar2 of size N each. Merge the given arrays and find the sum of the two middle elements of the merged array.\n\nGiven 2 sorted arrays Ar1 and Ar2 of size N each. Merge the given arrays and find the sum of the two middle elements of the merged array.\n\nExample 1:\n\n`Input:N = 5Ar1[] = {1, 2, 4, 6, 10}Ar2[] = {4, 5, 6, 9, 12}Output: 11Explanation: The merged array looks like{1,2,4,4,5,6,6,9,10,12}. Sum of middleelements is 11 (5 + 6).`\n\nExample 2:\n\n`Input:N = 5Ar1[] = {1, 12, 15, 26, 38}Ar2[] = {2, 13, 17, 30, 45}Output: 32Explanation: The merged array looks like{1, 2, 12, 13, 15, 17, 26, 30, 38, 45} sum of middle elements is 32 (15 + 17).`\n\n`//Given 2 sorted arrays Ar1 and Ar2 of size N each. Merge the given arrays and find the sum of the two middle elements of the merged array.func sumOfSortedArray(Ar1:[Int], Ar2:[Int],N:Int)->Int{var newArray = [Int]()newArray.append(contentsOf: Ar1)newArray.append(contentsOf: Ar2)newArray = newArray.sorted()//this is how we can get middle index by total of both array divided by 2 and need to minus 1 because array always start with 0 not 1.let middleElementIndex = ((N+N)/2) - 1let sum = newArray[middleElementIndex] + newArray[middleElementIndex + 1]print(sum)return sum}`"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.73415715,"math_prob":0.9932999,"size":1588,"snap":"2023-40-2023-50","text_gpt3_token_len":515,"char_repetition_ratio":0.14709596,"word_repetition_ratio":0.28985506,"special_character_ratio":0.35642317,"punctuation_ratio":0.19021739,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9885626,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-02T22:11:51Z\",\"WARC-Record-ID\":\"<urn:uuid:e226b90a-d6e8-4b6d-a136-b3d1aeb27169>\",\"Content-Length\":\"187241\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:db34af80-89c5-41a8-b0f7-e5720b0aeb05>\",\"WARC-Concurrent-To\":\"<urn:uuid:9265c4e6-0287-4a4c-a14d-10f0543917b7>\",\"WARC-IP-Address\":\"162.159.153.4\",\"WARC-Target-URI\":\"https://javedmultani16.medium.com/given-a-sorted-and-rotated-array-a-of-n-distinct-elements-which-is-rotated-at-some-point-and-given-ff48913c3114\",\"WARC-Payload-Digest\":\"sha1:ZC5PZA55M3DD7A5MZXQS64QDZMUCOOLD\",\"WARC-Block-Digest\":\"sha1:3CDZ4CA3YZXATVQYG3TDVW4642QIMA4J\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511021.4_warc_CC-MAIN-20231002200740-20231002230740-00414.warc.gz\"}"} |
https://www.wildcatartists.net/2013/03/ah-electronics.html | [
"## Thursday, March 14, 2013\n\n### Ah! Electronics!\n\nToday we lots of discussions on how the electronics were going to be put into the Alebrijes. There were many discussions and questions that were asked by students. We looked at their original concept sketches and then students quickly found interesting solutions to their design problem.\n\nA review of how to read resistors and of what the equation and schematic looked like was discussed today.\n\nVs = Batter Voltage\nVf = LED Voltage\nI = Current of LED\n\nIn general students have been using a 9V battery for Vs. Where it gets a little more difficult is when students needed to select their LED's.\n\nThe data sheet on the package was a little difficult to read but we eventually found the maximum forward voltage for the LED's that the students selected.\n\nI modified the original equation a little since LED1 and LED2's voltages should be the same we can just simply multiply by 2:\n\n$\\large R_1=\\frac{V_s-2V_f}{I}$\n\nSo if I have a red LED's whose maximum voltage is 2.3V I can simply set up my problem like so.\nFirst I define my variables:\n• $V_s=9$\n• $V_f=2.3$\n• I=20mA or since we need amps I= .02A\nNow we just substitute the numbers into the variables of our equation:\n\n$\\large R_1=\\frac{V_s-2V_f}{I}$\n\n$\\large R_1=\\frac{9-2(2.3)}{.02}$\n\n$\\large R_1=\\frac{9-4.6}{.02}$\nNote: We must follow the order of operations here!\n\n$\\large R_1=\\frac{4.4}{.02}$\nNote: This looks daunting but since the denominator is in hundredths lets multiply the denominator and numerator by 100!\n\n$\\large R_1=\\frac{4.4 * 100}{.02 * 100}$\nNote: Now just simply move the decimal two places to the right in both the numerator and denominator!\n\n$\\large R_1=\\frac{440}{2}$\nNote: Ahhh much better!\n\n$R_1=220 \\Omega$\n\nNow we have found our resistor value! We will need to see the chart below to determine the color bands of our resistor.\n\nSince we need a resistor that is $220 \\Omega$ we know that the first number relates to the first band of our resistor which 2 = red. Next we look at the second number and this is our second band of our resistor which is 2 which is red.\nRecall that the third band of our resistor is reserved for our multiplier.\n22 multiplied by what will give us 220?\nI hope you said 10!\nIf you look at your chart a multiplier of 10 gives us brown for our third color.\nTherefore the resistor in this example would be: Red Red Brown!\n\n Color First Band Second Band Multiplier Tolerance Black 0 0 1 Brown 1 1 10 Red 2 2 100 Orange 3 3 1000 Yellow 4 4 10000 Green 5 5 100000 Blue 6 6 1000000 Violet 7 7 10000000 Gray 8 8 100000000 White 9 9 1000000000 Gold ±5% Silver ±10%"
] | [
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https://s127041.gridserver.com/.tmb/journal/6nmqq7b.php?e5b311=mangalyaan-2 | [
"# mangalyaan 2\n\nmangalyaan 2\nOctober 28, 2020\n\nFormula works only when wood is between 6 percent and 14 percent moisture, but this is a fair range for furniture. Here is a series of \"Wood Doctor Approved\" calculators to make your lumber and timber related estimating a little easier. Recently, one of the users asked to calculate \"How many cubic meters in a single board, the calculation of the lumber/timber's cubic capacity, in general.\" See below for two full examples working with different initial criteria. If you have multiple planks of wood lying end to end, then the calculator can also work out the total board feet of the entire construction: $$Total\\,Board\\,Feet = Board\\,Feet \\times Number\\,of\\,Pieces$$. I shall show you two types of formulas to calculate the volume of wooden log in this video. Your Attention, Please: 1 Bd. AMERICAN WOOD … Construction Calculations is a manual that provides end users with a comprehensive guide for many of the formulas, mathematical vectors and conversion factors that are commonly encountered during the design and construction stages of a construction project. I would enter these measurements into the calculator. It would then perform the following calculations to work out total board feet and linear feet: $$Board\\,Feet\\,(bf) = {Thickness\\,(in) \\times Width\\,(in) \\times Length\\,(ft) \\over 12} = {1\\,in \\times 3\\,in \\times 10\\,ft \\over 12} = 2.5\\,bf$$, $$Total\\,Board\\,Feet = Board\\,Feet \\times Number\\,of\\,Pieces = 2.5\\,bf \\times 5 = 12.5\\,bf$$, $$Linear\\,Feet = Length \\times Number\\,of\\,Pieces = 10\\,ft \\times 5 = 50\\,ft$$. Formula: Estimated Log Weight = A x (A x (B x 0.5454)) / 100 x C Where, A = Log Diameter in Inches B = Log Length in Feet C = Type of Wood Value Related Calculator: I was also thrilled to find … Multiply the obtained value by log diameter (inches). I would then enter the measurements into the calculator to calculate the total board feet and linear feet. Add a Free Cord Calculator Widget to Your Site! At this stage of the design process, the combined live load and dead load on the beam should already have been calculated. This was part of a multi-sheet project for the complete design of an A-frame wood structure. At present, the Wood Shear Wall module implements the Individual Full-Height Wall Segment Shear Walls method. There are several ways to solve this problem and several \"correct\" answers. Finally, the calculator would work out the total cost of the wood: $$Cost = Price\\,Per\\,Linear\\,Foot \\times Linear\\,Feet = 12\\,/ft \\times 50\\,ft = 600$$, ✔️ Two planks of wood calculating price per board foot. Maximum Density kg/m 3. Otherwise, make your measurements accurately and enter the values into our online lumber calculator to get started! Surprisingly, the calculations are extremely easy! Cft or Cubic feet or ft 3 is a unit of measurement of volume in imperial unit. You’ll want to measure the width in inches for this calculation … This calculator attempts to generate the most efficient cut list for a given set of pieces. ByExample.com provides an instant geodesic dome calculator to simplify dome formulas. Find here the maximum and minimum density. The volume calculations for these individual frustums of trunk segments can be further refined by considering the overall shape of the trunk. Since your density figure is metric, inches are the wrong unit. Kerf Spacing Calculator for Bending Wood. Wood Density The common definition of wood moisture content on a dry basis can be written as (1) where Wg is the green weight of wood (pounds or kilograms) at moisture content M (percent), and Wd is the oven-dry weight of wood. This calculator finds the spacing of the cuts (at entered dimensions), so when the piece is bent, the inside edges of each cut touch together to form the correct curve. Ft. = 144 cubic inches. Values of K F and phi are automatically determined and applied for the LRFD method. Both lateral (single and double shear) and withdrawal capacities can be determined. The beam calculator uses these equations to generate bending moment, shear force, slope and defelction diagrams. If you know the price per linear foot or board foot, the calculator can work out the total cost of the wood using the formula: $$Cost = Price\\,Per\\,Board\\,Foot \\times Total\\,Board\\,Feet$$. the carpenter is confusing me totally. It's based on linear board-feet and so works only in the one dimension. Sample Design Calculations. Enter the length, breadth and depth of your requirement, and the volume of wood required will be presented to you. Calculate the perpendicular to grain compressive strength for a wood member. Use our dome calculators to make a plan for your own geodesic dome and read about the disadvantages of living in a dome home. Calculator - Log volume calculator - Wood log volume calculation - Logs - Volume m³ BF - - Wood calculators However, Do not enter decimals or fractions, and do not enter the percentage sign after the value. On this page, you can calculate the volume of wood/ lumber in cubic feet (Cft). We sell most of our lumber by the board foot. All calculations are according to the NDS code. I want to estimate the total board feet and cost for this landscaping project. The calculators \"remember\" previous calculations, and generate a running report of entries and subtotals. P = total concentrated load, lbs. Share. Example: Consider that the density of wood with weight 2 kg, minimum and … Thanks eBay. In terms of quantity and quality, the tree taper (stem form) is useful for maximizing the value of primary wood processing. The length of the wooden board should be expressed in feet, while the width and thickness - … T\" x Sq. The calculator would perform the following operations: $$Board\\,Feet\\,(bf) = {Thickness\\,(in) \\times Width\\,(in) \\times Length\\,(ft) \\over 12} = {2\\,in \\times 5\\,in \\times 12\\,ft \\over 12} = 10\\,bf$$, $$Total\\,Board\\,Feet = Board\\,Feet \\times Number\\,of\\,Pieces = 10\\,bf \\times 2 = 20\\,bf$$, $$Linear\\,Feet = Length \\times Number\\,of\\,Pieces = 12\\,ft \\times 2 = 24\\,ft$$. The calculators \"remember\" previous calculations, and generate a running report of entries and subtotals. Result : Minimum Density kg/m 3. Thickness uses the rough sawn dimension, not your final net dimension. Wood Angled Bearing: Calculate the compressive strength or bearing strength for a wood member at an angle to the grain. You’ll agree with us when we say that working out how much wood your next landscaping project will require is tricky. The first wood I'm working with has a density of 710 - 770 kg/m3. Minimum Volume volume. See items below for descriptions of items that are specific to the Wood Column module. A formula developed by Hiram Hallock, retired, U.S. Forest Service, can be used to determine horsepower requirements. Case in point: if your timber grows near several mills, it may command a higher price than if it were located further away from those mills. This unit is very commonly used in wood industry for pricing wooden planks/ lumbers. This appendix presents design examples of the retrofitting techniques for elevation, dry floodproofing, wet . A key “feature” of a tree is that trees sequester carbon – the process of removal and long-term storage of carbon dioxide (CO2) from our atmosphere. Calculator - Log volume calculator - Wood log volume calculation - Logs - Volume m³ BF - - Wood calculators Our calculator works out the board feet of standard three dimensional wood material based on the measurements you provide. Underestimates will compromise the safety of the structure, and overestimates will lead to unnecessarily high costs. All calculations are according to the NDS code. Finally, the estimator would work out the total cost of the wood: $$Cost = Price\\,Per\\,Board\\,Foot \\times Total\\,Board\\,Feet = 3\\,/bf \\times 20\\,bf = 60$$. Accurate calculations of wood beam strength are essential in construction. American Wood Council American Forest & Paper Association NDS 2005 EDITION ® ANSI/AF&PA NDS-2005 Approval Date: JANUARY 6, 2005 WITH COMMENTARY AND SUPPLEMENT: DESIGN VALUES FOR WOOD CONSTRUCTION ASD/LRFD American W ood Council BEAM DESIGN FORMULAS WITH SHEAR AND MOMENT DIAGRAMS American Forest & Paper Association w R V V 2 2 Shear M max Moment x DESIGN AID No. Loads, and construction of a floodwall in a residential setting and 14 moisture. Unit of measurement of volume cord calculator Widget to your Site two full examples working with initial... ( cft ) s why we built our Free online lumber calculator,... Overestimates will lead to unnecessarily high costs is expressed in units of.! Spacing of the way — from explaining the differences between wood flooring types to installing them for.... And contracts the most straightforward method from the projected results lumber in cubic feet show you types. Read on to learn how to apply the heat capacity formula correctly to obtain a valid.. Three times, because you have cubic inches by 1,728 to find the volume of wood/ lumber cubic... In imperial units ( inches, feet, yards ) or metric units ( inches, feet, yards or... Inches to meters ( three times, because you have cubic inches.! And height unit of measurement of volume and is known as the formula '' box above cost for landscaping. Occupied by a cube with 1 foot width, length and height calculate wood for door {... Ll be constructing a rather large white oak built-in at my home and timber related estimating a little.! After the value of primary wood processing ( based on the beam measuring feet! Free online lumber calculator to simplify dome formulas Hiram Hallock, retired, U.S. Forest Service, can used! The one dimension, our dome calculators to make your lumber and timber related estimating a easier... Walls method dead load on the beam calculator uses these equations to generate bending moment, shear force slope... Wood-To-Concrete, and overestimates will lead to unnecessarily high costs load of any roofing other! Next landscaping project will require is tricky and quality, the wood,... That reflects how much a certain species of wood to obtain the curve volume in cubic from. Of your wood in board feet and linear feet an Excel file that! Make your lumber and timber related estimating a little easier measurements into the calculator to simplify formulas! You want to estimate the total board feet and cost for this landscaping project will require is tricky wood calculation formula 60! Say I need to figure out the weight of the cut, and construction of a floodwall in dome. Tools will provide quick answers to your Site combined live load and dead on. Of measurement of volume and is known as the formula '' box above also thrilled to find volume... Log weight formula mentioned below to calculate wood for door frames { }. Meters ( three times, because you have cubic inches ) feet =! For these individual frustums of trunk segments can be used to determine horsepower requirements use of the working drawings quickly. I 'm working with different initial criteria member at an angle to the Forest Products Laboratory end, multiply obtained... Contract more than narrower ones below for two full examples working with has a density of 710 770... Cubic meters from unit sizes and number of units and vice versa single bolts, nails, screws. Bolts, nails, lag screws and wood screws per the 2015 NDS, multiply the wood log mentioned! And read about the disadvantages of living in a dome home defelction diagrams one way and gives you a answer! Little easier ( inches, feet cost = Price\\, Per\\, Linear\\ foot... Log length ( feet ) with 100 vice versa strength is expressed units... With shelves density of 710 - 770 kg/m3 feet in length, 3 inches wide 1! To make a plan for your own geodesic dome formulas as accurate as possible to limit any waste! Measurement of volume in cubic meters from unit sizes and number of units and vice versa load takes account. Units ( inches, feet in ft3 ) 12 = Bd ( )! Moisture meter that 's built like a tank into account the entire load any! This appendix presents design examples of the design process, the inside edges of working. While prices in the one dimension conception of determining equivalent moment is authored by wood and concrete beams under loading! Module, end fixity, loads, and the volume in imperial unit in width,. Frames { chowkath } A-frame wood structure box above to use, what it calculates and how does. Pricing wooden planks/ lumbers a great tool to quickly validate forces in beams a certain species of in... Formula works only in the one dimension in cu ft loads, and load combinations click here double!"
] | [
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https://codepen.io/jaehee/pen/grVmXp/ | [
"## Pen Settings\n\n### CSS\n\n#### Vendor Prefixing\n\nAny URL's added here will be added as `<link>`s in order, and before the CSS in the editor. If you link to another Pen, it will include the CSS from that Pen. If the preprocessor matches, it will attempt to combine them before processing.\n\n### JavaScript\n\n#### JavaScript Preprocessor\n\nBabel includes JSX processing.\n\nAny URL's added here will be added as `<script>`s in order, and run before the JavaScript in the editor. You can use the URL of any other Pen and it will include the JavaScript from that Pen.\n\n### Packages\n\nSearch for and use JavaScript packages from npm here. By selecting a package, an `import` statement will be added to the top of the JavaScript editor for this package.\n\n### Behavior\n\n#### Save Automatically?\n\nIf active, Pens will autosave every 30 seconds after being saved once.\n\n#### Auto-Updating Preview\n\nIf enabled, the preview panel updates automatically as you code. If disabled, use the \"Run\" button to update.\n\n#### Format on Save\n\nIf enabled, your code will be formatted when you actively save your Pen. Note: your code becomes un-folded during formatting.\n\n## HTML\n\n``` ```\n<div class=\"box box--fixed overflow\">\nNo ellip. Too short.\n</div>\n\n<div class=\"box box--fixed one-line\">\nForcing one line regardless\n</div>\n\n<div class=\"box box--fixed two-lines\">\nForcing two lines of text\nregardless of overflow\n</div>\n\n<div class=\"box box--fixed overflow\">\nTrying to ellipsis any overflowed content.\nThe quick brown fox jumped over the lazy dogs.\nThe quick brown fox jumped over the lazy dogs.\n</div>\n\n<div class=\"box box--responsive\">\nThis is a responsive box that will update it's\nellipsis when the screen resizes. The quick brown\nfox jumped over the lazy dogs. The quick brown fox\njumped over the lazy dogs.\n</div>\n```\n```\n!\n\n## CSS\n\n``` ```\nbody {\n}\n\n.box {\nheight: 4.5em;\nbackground: #eee;\nborder: 1px solid #ccc;\nmargin-right: 30px;\nmargin-bottom: 30px;\noverflow: hidden;\nfloat: left;\n}\n\n.box--fixed {\nwidth: 130px;\n}\n\n.box--responsive {\nwidth: 30%;\n}\n\n/* necessary plugin styles */\n.ellip {\ndisplay: block;\nheight: 100%;\n}\n\n.ellip-line {\ndisplay: inline-block;\ntext-overflow: ellipsis;\nwhite-space: nowrap;\nword-wrap: normal;\n}\n\n.ellip,\n.ellip-line {\nposition: relative;\noverflow: hidden;\nmax-width: 100%;\n}\n```\n```\n!\n\n## JS\n\n``` ```\n/*!\n* jQuery.ellipsis\n* https://github.com/jjenzz/jquery.ellipsis\n* --------------------------------------------------------------------------\n* Copyright (c) 2013 J. Smith (@jjenzz)\n*\n* adds a class to the last 'allowed' line of text so you can apply\n* text-overflow: ellipsis;\n*/\n(function(a){if(typeof define===\"function\"&&define.amd){define([\"jquery\"],a)}else{a(jQuery)}}(function(d){var c=\"ellipsis\",b='<span style=\"white-space: nowrap;\">',e={lines:\"auto\",ellipClass:\"ellip\",responsive:false};function a(h,q){var m=this,w=0,g=[],k,p,i,f,j,n,s;m.\\$cont=d(h);m.opts=d.extend({},e,q);function o(){m.text=m.\\$cont.text();m.opts.ellipLineClass=m.opts.ellipClass+\"-line\";m.\\$el=d('<span class=\"'+m.opts.ellipClass+'\" />');m.\\$el.text(m.text);m.\\$cont.empty().append(m.\\$el);t()}function t(){if(typeof m.opts.lines===\"number\"&&m.opts.lines<2){m.\\$el.addClass(m.opts.ellipLineClass);return}n=m.\\$cont.height();if(m.opts.lines===\"auto\"&&m.\\$el.prop(\"scrollHeight\")<=n){return}if(!k){return}s=d.trim(m.text).split(/\\s+/);m.\\$el.html(b+s.join(\"</span> \"+b)+\"</span>\");m.\\$el.find(\"span\").each(k);if(p!=null){u(p)}}function u(x){s[x]='<span class=\"'+m.opts.ellipLineClass+'\">'+s[x];s.push(\"</span>\");m.\\$el.html(s.join(\" \"))}if(m.opts.lines===\"auto\"){var r=function(y,A){var x=d(A),z=x.position().top;j=j||x.height();if(z===f){g[w].push(x)}else{f=z;w+=1;g[w]=[x]}if(z+j>n){p=y-g[w-1].length;return false}};k=r}if(typeof m.opts.lines===\"number\"&&m.opts.lines>1){var l=function(y,A){var x=d(A),z=x.position().top;if(z!==f){f=z;w+=1}if(w===m.opts.lines){p=y;return false}};k=l}if(m.opts.responsive){var v=function(){g=[];w=0;f=null;p=null;m.\\$el.html(m.text);clearTimeout(i);i=setTimeout(t,100)};d(window).on(\"resize.\"+c,v)}o()}d.fn[c]=function(f){return this.each(function(){try{d(this).data(c,(new a(this,f)))}catch(g){if(window.console){console.error(c+\": \"+g)}}})}}));\n\n\\$('.overflow').ellipsis();\n\\$('.one-line').ellipsis({ lines: 1 });\n\\$('.two-lines').ellipsis({ lines: 2 });\n\\$('.box--responsive').ellipsis({ responsive: true });\n```\n```\n!\n999px"
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https://proceedings.neurips.cc/paper/2020/hash/440924c5948e05070663f88e69e8242b-Abstract.html | [
"#### Reinforcement Learning with General Value Function Approximation: Provably Efficient Approach via Bounded Eluder Dimension\n\nValue function approximation has demonstrated phenomenal empirical success in reinforcement learning (RL). Nevertheless, despite a handful of recent progress on developing theory for RL with linear function approximation, the understanding of \\emph{general} function approximation schemes largely remains missing. In this paper, we establish the first provably efficient RL algorithm with general value function approximation. We show that if the value functions admit an approximation with a function class $\\mathcal{F}$, our algorithm achieves a regret bound of $\\widetilde{O}(\\mathrm{poly}(dH)\\sqrt{T})$ where $d$ is a complexity measure of $\\mathcal{F}$ that depends on the eluder dimension~[Russo and Van Roy, 2013] and log-covering numbers, $H$ is the planning horizon, and $T$ is the number interactions with the environment. Our theory generalizes the linear MDP assumption to general function classes. Moreover, our algorithm is model-free and provides a framework to justify the effectiveness of algorithms used in practice."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7980492,"math_prob":0.99202466,"size":1226,"snap":"2020-45-2020-50","text_gpt3_token_len":264,"char_repetition_ratio":0.13584287,"word_repetition_ratio":0.0,"special_character_ratio":0.1949429,"punctuation_ratio":0.085427135,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99578136,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-25T22:29:53Z\",\"WARC-Record-ID\":\"<urn:uuid:613d0dd3-414c-4114-9d76-0b4f3ea49177>\",\"Content-Length\":\"7821\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b9f2408c-7a00-44f3-b9ee-4552bec49fee>\",\"WARC-Concurrent-To\":\"<urn:uuid:119ada25-bf35-4a3a-8b76-247f36d91d11>\",\"WARC-IP-Address\":\"198.202.70.94\",\"WARC-Target-URI\":\"https://proceedings.neurips.cc/paper/2020/hash/440924c5948e05070663f88e69e8242b-Abstract.html\",\"WARC-Payload-Digest\":\"sha1:6JYSYM2LFWOEM44OXTIP6QIKPFYX4JL5\",\"WARC-Block-Digest\":\"sha1:3PNC5CDTQSZ5SZ5KUSG7GD7ZBV7C65DH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141184870.26_warc_CC-MAIN-20201125213038-20201126003038-00016.warc.gz\"}"} |
https://en.wikipedia.org/wiki/Eigenvector_centrality | [
"# Eigenvector centrality\n\nIn graph theory, eigenvector centrality (also called eigencentrality or prestige score) is a measure of the influence of a node in a network. Relative scores are assigned to all nodes in the network based on the concept that connections to high-scoring nodes contribute more to the score of the node in question than equal connections to low-scoring nodes. A high eigenvector score means that a node is connected to many nodes who themselves have high scores. \n\nGoogle's PageRank and the Katz centrality are variants of the eigenvector centrality.\n\n## Using the adjacency matrix to find eigenvector centrality\n\nFor a given graph $G:=(V,E)$",
null,
"with $|V|$",
null,
"vertices let $A=(a_{v,t})$",
null,
"be the adjacency matrix, i.e. $a_{v,t}=1$",
null,
"if vertex $v$",
null,
"is linked to vertex $t$",
null,
", and $a_{v,t}=0$",
null,
"otherwise. The relative centrality, $x$",
null,
", score of vertex $v$",
null,
"can be defined as:\n\n$x_{v}={\\frac {1}{\\lambda }}\\sum _{t\\in M(v)}x_{t}={\\frac {1}{\\lambda }}\\sum _{t\\in G}a_{v,t}x_{t}$",
null,
"where $M(v)$",
null,
"is a set of the neighbors of $v$",
null,
"and $\\lambda$",
null,
"is a constant. With a small rearrangement this can be rewritten in vector notation as the eigenvector equation\n\n$\\mathbf {Ax} =\\lambda \\mathbf {x}$",
null,
"In general, there will be many different eigenvalues $\\lambda$",
null,
"for which a non-zero eigenvector solution exists. However, the additional requirement that all the entries in the eigenvector be non-negative implies (by the Perron–Frobenius theorem) that only the greatest eigenvalue results in the desired centrality measure. The $v^{\\text{th}}$",
null,
"component of the related eigenvector then gives the relative centrality score of the vertex $v$",
null,
"in the network. The eigenvector is only defined up to a common factor, so only the ratios of the centralities of the vertices are well defined. To define an absolute score one must normalise the eigen vector e.g. such that the sum over all vertices is 1 or the total number of vertices n. Power iteration is one of many eigenvalue algorithms that may be used to find this dominant eigenvector. Furthermore, this can be generalized so that the entries in A can be real numbers representing connection strengths, as in a stochastic matrix.\n\n## Normalized eigenvector centrality scoring\n\nGoogle's PageRank is based on the normalized eigenvector centrality, or normalized prestige, combined with a random jump assumption. The PageRank of a node $v$",
null,
"has recursive dependence on the PageRank of other nodes that point to it. The normalized adjacency matrix $N$",
null,
"is defined as:\n\n$N(u,v)={\\begin{cases}{1 \\over \\operatorname {od} (u)},&{\\text{if }}(u,v)\\in E\\\\0,&{\\text{if }}(u,v)\\not \\in E\\end{cases}}$",
null,
"where $od(u)$",
null,
"is the out-degree of node $u$",
null,
".\n\nThe normalized eigenvector centrality score is defined as:\n\n$p(v)=\\sum _{u}{N^{T}(v,u)\\cdot p(u)}$",
null,
"## Applications\n\nEigenvector centrality is a measure of the influence a node has on a network. If a node is pointed to by many nodes (which also have high eigenvector centrality) then that node will have high eigenvector centrality.\n\nThe earliest use of eigenvector centrality is by Edmund Landau in an 1895 paper on scoring chess tournaments.\n\nMore recently, researchers across many fields have analyzed applications, manifestations, and extensions of eigenvector centrality in a variety of domains:\n\n• Eigenvector centrality is the unique measure satisfying certain natural axioms for a ranking system.\n• In neuroscience, the eigenvector centrality of a neuron in a model neural network has been found to correlate with its relative firing rate.\n• In a standard class of models of opinion updating or learning (sometimes called DeGroot learning models), the social influence of a node over eventual opinions is equal to its eigenvector centrality.\n• The definition of eigenvector centrality has been extended to multiplex or multilayer networks.\n• In a study using data from the Philippines, the authors showed how political candidates' families had disproportionately high eigenvector centrality in local intermarriage networks.\n• In a economic public goods problems, a person's eigenvector centrality can be interpreted as how much that person's preferences influence an efficient social outcome (formally, a Pareto weight in a Pareto efficient social outcome)."
] | [
null,
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null,
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null,
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null,
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null,
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null,
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null,
"https://wikimedia.org/api/rest_v1/media/math/render/svg/7b273e99e678ece858ca3873d6ae2b89bfa343d2",
null,
"https://wikimedia.org/api/rest_v1/media/math/render/svg/e07b00e7fc0847fbd16391c778d65bc25c452597",
null,
"https://wikimedia.org/api/rest_v1/media/math/render/svg/e07b00e7fc0847fbd16391c778d65bc25c452597",
null,
"https://wikimedia.org/api/rest_v1/media/math/render/svg/f5e3890c981ae85503089652feb48b191b57aae3",
null,
"https://wikimedia.org/api/rest_v1/media/math/render/svg/a998ae74c021370f0dceb81d4b42b089f18c5244",
null,
"https://wikimedia.org/api/rest_v1/media/math/render/svg/00dc1fb84f3b724914e4cead47be71d0b1660b4c",
null,
"https://wikimedia.org/api/rest_v1/media/math/render/svg/c3e6bb763d22c20916ed4f0bb6bd49d7470cffd8",
null,
"https://wikimedia.org/api/rest_v1/media/math/render/svg/d9705bce3cf44f312e2efbc0619e7da9cec9ea09",
null
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https://sicp.hexlet.io/exercises/83 | [
"2.2.3. Sequences as Conventional Interfaces\nExercise 2.37\n\n# Matrix operations\n\nSuppose we represent vectors v = (vᵢ) as sequences of numbers, and matrices m = (mᵢⱼ) as sequences of vectors (the rows of the matrix). For example, the matrix",
null,
"is represented as the sequence ((1 2 3 4) (4 5 6 6) (6 7 8 9)). With this representation, we can use sequence operations to concisely express the basic matrix and vector operations. These operations (which are described in any book on matrix algebra) are the following:\n\n(dot-product v w) returns the sum ∑ᵢvᵢwᵢ;\n\n(matrix-*-vector m v) returns the vector t, where tᵢ = ∑ⱼmᵢⱼvᵢ;\n\n(matrix-*-matrix m n) returns the matrix p, where pᵢⱼ = ∑ₖmᵢₖnₖⱼ\n\n(transpose m) returns the matrix n, where nᵢⱼ = mⱼᵢ\n\nWe can define the dot product as\n\n``````(define (dot-product v w)\n(accumulate + 0 (map * v w)))\n``````\n\nFill in the missing expressions in the following procedures for computing the other matrix operations. (The procedure accumulate-n is defined in exercise 2.36.)\n\n``````(define (matrix-*-vector m v)\n(map <??> m))\n(define (transpose mat)\n(accumulate-n <??> <??> mat))\n(define (matrix-*-matrix m n)\n(let ((cols (transpose n)))\n(map <??> m)))\n``````"
] | [
null,
"https://sicp.hexlet.io/img/exercises/2_37.gif",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.81048316,"math_prob":0.998754,"size":1183,"snap":"2020-24-2020-29","text_gpt3_token_len":362,"char_repetition_ratio":0.14503817,"word_repetition_ratio":0.0,"special_character_ratio":0.27979714,"punctuation_ratio":0.11210762,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99992406,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-04T21:45:39Z\",\"WARC-Record-ID\":\"<urn:uuid:c3ef0bf0-ac62-4b8a-a28e-d7acfda5699f>\",\"Content-Length\":\"16941\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1bd167aa-5b57-4536-ad73-2c4caf8414c8>\",\"WARC-Concurrent-To\":\"<urn:uuid:a9f6326a-5cf5-462f-a923-1b1c12815e9a>\",\"WARC-IP-Address\":\"52.48.90.17\",\"WARC-Target-URI\":\"https://sicp.hexlet.io/exercises/83\",\"WARC-Payload-Digest\":\"sha1:XIRUGIQ3QARPDPWRODH6WQRRAF6B22DB\",\"WARC-Block-Digest\":\"sha1:R7LN3SUP3ILWAFDDL3WTUZEXAYGVNY4T\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655886706.29_warc_CC-MAIN-20200704201650-20200704231650-00413.warc.gz\"}"} |
https://www.gradesaver.com/textbooks/math/algebra/algebra-2-1st-edition/chapter-7-exponential-and-logarithmic-functions-7-5-apply-properties-of-logarithms-7-5-exercises-mixed-review-page-513/80 | [
"## Algebra 2 (1st Edition)\n\n$$x=10$$\nSimplifying using the rules of radicals, we find: $$\\left(\\sqrt{x+6}\\right)^2=\\left(\\sqrt{3x-14}\\right)^2 \\\\ x+6=3x-14 \\\\ x=10$$"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7935112,"math_prob":1.0000082,"size":409,"snap":"2022-27-2022-33","text_gpt3_token_len":115,"char_repetition_ratio":0.09876543,"word_repetition_ratio":0.0,"special_character_ratio":0.29095355,"punctuation_ratio":0.072289154,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998802,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-03T08:55:02Z\",\"WARC-Record-ID\":\"<urn:uuid:d77d6a39-ca7e-4e4e-80d7-a44b8a3fe824>\",\"Content-Length\":\"101821\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:72303cb3-9620-4469-900e-b1cc2c4e0d81>\",\"WARC-Concurrent-To\":\"<urn:uuid:8de6613a-5e53-4858-b71f-2e50131ae509>\",\"WARC-IP-Address\":\"54.166.60.202\",\"WARC-Target-URI\":\"https://www.gradesaver.com/textbooks/math/algebra/algebra-2-1st-edition/chapter-7-exponential-and-logarithmic-functions-7-5-apply-properties-of-logarithms-7-5-exercises-mixed-review-page-513/80\",\"WARC-Payload-Digest\":\"sha1:GJH5ONBFS6THPQPEGVUZUZSTG4HEURW2\",\"WARC-Block-Digest\":\"sha1:A5ARDGQBMWV5H6NHS3NAX6MIC3XXP3F6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104215805.66_warc_CC-MAIN-20220703073750-20220703103750-00796.warc.gz\"}"} |
https://www.enotes.com/homework-help/can-someone-please-explain-how-do-17-thanks-458320 | [
"# Can someone please explain how to do #17? Thanks!\n\nImages:\nThis image has been Flagged as inappropriate Click to unflag",
null,
"Since m||n then we know that their are angle pairs that exist to make equality statements.\n\n`/_3 and/_5` are known as either same-side interior angles or consecutive interior angles. When lines are parallel, same-side interior angles together = 180˚. This makes the equation:\n\n`(6x + y) + (8x + 2y) = 180`\n\n`14x + 3y = 180`\n\nWe also know that `/_5 and/_6` form a linear pair which, together, must equal 180˚. This makes the equation:\n\n`(8x + 2y) + (4x + 7y) = 180`\n\n`12x + 9y = 180`\n\nNow that we have a system of equations, we can solve for x and y. This case I will use the elimination method.\n\ni.) 14x + 3y = 180\n\nii.) 12x + 9y = 180\n\nMultiply equation (i.) by -3 to obtain the coefficient -9 which will be opposite of 9 in eqution ii.\n\n`-3(14x + 3y=180) rArr -42x -9y = -540`\n\nAdd the new equation to equation ii.\n\n` -42x - 9y = -540`\n\n` 12x + 9y = 180 `\n\n`-30x = -360`\n\n`x = 12`\n\nSubstitute to find y.\n\n`12(12) + 9y = 180 `\n\n`144 + 9y = 180 `\n\n`9y = 36 `\n\n`y = 4`\n\nx = 12 and y = 4\n\nTo find `/_7` , we know that `/_6 and/_7` are equal because all vetical angles are equal.\n\nTherefore,`/_6` is 4x + 7y = 4(12) + 7(4) = 48 + 28 = 76.\n\n`/_7` is also 76˚\n\nThe solution for the given problem is x = 12, y = 4, and `/_7` = 76˚.\n\nApproved by eNotes Editorial Team"
] | [
null,
"https://static.enotescdn.net/images/main/illustrations/illo-answer.svg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8691183,"math_prob":0.99989474,"size":1533,"snap":"2021-31-2021-39","text_gpt3_token_len":542,"char_repetition_ratio":0.120340094,"word_repetition_ratio":0.02208202,"special_character_ratio":0.4266145,"punctuation_ratio":0.10377359,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998264,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-27T13:11:51Z\",\"WARC-Record-ID\":\"<urn:uuid:c33f952a-572f-4601-8f6a-7487ecc386fd>\",\"Content-Length\":\"73904\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:155c6e4c-e96a-4a33-86fd-2f4b2b11166a>\",\"WARC-Concurrent-To\":\"<urn:uuid:dfce116b-a4c0-4357-94a7-24f6ccfb225f>\",\"WARC-IP-Address\":\"104.26.5.75\",\"WARC-Target-URI\":\"https://www.enotes.com/homework-help/can-someone-please-explain-how-do-17-thanks-458320\",\"WARC-Payload-Digest\":\"sha1:FVSTSBOBAQJZDRWDFMMEFYWPDNDHP3TL\",\"WARC-Block-Digest\":\"sha1:SPHRBXI73VNFTLDTC2HUBRXW7L3SW2KT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153391.5_warc_CC-MAIN-20210727103626-20210727133626-00293.warc.gz\"}"} |
https://blog.maxkit.com.tw/2014/04/erlang-interfacing-technology.html | [
"## 2014/04/01\n\n### erlang - interfacing technology\n\nerlang 利用 port 控制通訊，「負責建立一個 port」的行程，稱為該 port 的 connected process，所有送往外部的訊息都必須貼上connected process的 PID，而外部程式的所有訊息都會被送到 connected process。\n\nport 的作用就像是一個 erlang process，可送訊息給它，可以註冊，如果外部程式 crash，離開訊號就會送到 connected process，如果 connected process 死亡，外部程式就會被 kill。\n\n# Port\n\nPort = open_port(PortName, PortSettings)\n\nPort ! {PidC, {command, Data}}\n\nPort ! {PidC, {connect, Pid1}}\n\nPort ! {PidC, close}\n\nconnected process 可以收到外部程式送來的訊息，類似這樣：\n\nreceive\n{Port, {data, Data}} ->\n...\n\n### open_port\n\nopen_port可接受很多Opt設定值，以下列出常見的設定：\n\n@spec open_port(PortName, [Opt]) -> Port\n\nPortName 可以是下列其中之一\n\n1. {spawn, Command}\n啟動一個外部程式，Command 是外部程式的名稱，除非是 linked-in driver，否則它會是在 erlang 工作空間外部的地方執行\n2. {fd, In, Out}\n\nOpt 可以是下列其中之一\n\n1. {packet, N}\n2. stream\n3. {line, Max}\n4. {cd, Dir}\n5. {env, Env}\n\n# 連接外部的 C 程式\n\n// example1.c\nint twice(int x){\nreturn 2*x;\n}\n\nint sum(int x, int y){\nreturn x+y;\n}\n\nX1 = example1:twice(23),\nY1 = example1:sum(45, 32),\n\n### 定義 port 與 外部程式 之間的協定\n\n1. 所有封包一開始都是 2 bytes 的 Len，後面接著 Len 個 bytes 資料\n2. 呼叫 twice(N)，在協定中轉換為 [1,N]，1代表 twice，N為參數\n3. 呼叫 sum(N,M), 在協定中轉換為 [2,N,M]\n4. 回傳值的長度為 1 byte\n\n1. port 送出 [0,3,2,45,32] 給外部程式，0,3 表示封包長度為 3，2 表示呼叫 sum，45 與 32是 sum 的參數\n2. 外部程式從 stdin 讀取這五個位元，呼叫 sum 函數，寫出位元組序列「0,1,77」到 stdout，0,1 表示封包長度為 1，資料內容為 77\n\n### C 語言外部程式，實作協定\n\n1. example1.c: 包含了 twice 與 sum 兩個函式\n2. example1_driver.c: 會終結 byte 串流協定，且呼叫 example1.c 內的函式\n3. erl_comm.c: 具有讀寫記憶體緩衝區的函式\n\nexample1_drive.c 執行一個無窮迴圈，持續從 stdin 讀取資料，並把結果寫入 stdout\n\n// example1_drive.c\n#include <stdio.h>\ntypedef unsigned char byte;\n\nint write_cmd(byte *buff, int len);\n\nint main() {\nint fn, arg1, arg2, result;\nbyte buff;\n\nfn = buff;\n\nif (fn == 1) {\narg1 = buff;\nresult = twice(arg1);\n} else if (fn == 2) {\narg1 = buff;\narg2 = buff;\n/* debug -- you can print to stderr to debug\nfprintf(stderr,\"calling sum %i %i\\n\",arg1,arg2); */\nresult = sum(arg1, arg2);\n}\n\nbuff = result;\nwrite_cmd(buff, 1);\n}\n}\n\nerl_comm.c，負責在 stdin/stdout 讀寫 2 bytes 開頭的封包。\n\n/* erl_comm.c */\n#include <unistd.h>\n\ntypedef unsigned char byte;\n\nint write_cmd(byte *buf, int len);\nint write_exact(byte *buf, int len);\n\n{\nint len;\n\nreturn(-1);\nlen = (buf << 8) | buf;\n}\n\nint write_cmd(byte *buf, int len)\n{\nbyte li;\n\nli = (len >> 8) & 0xff;\nwrite_exact(&li, 1);\n\nli = len & 0xff;\nwrite_exact(&li, 1);\n\nreturn write_exact(buf, len);\n}\n\n{\nint i, got=0;\n\ndo {\nif ((i = read(0, buf+got, len-got)) <= 0)\nreturn(i);\ngot += i;\n} while (got<len);\n\nreturn(len);\n}\n\nint write_exact(byte *buf, int len)\n{\nint i, wrote = 0;\n\ndo {\nif ((i = write(1, buf+wrote, len-wrote)) <= 0)\nreturn (i);\nwrote += i;\n} while (wrote<len);\n\nreturn (len);\n}\n\n### erlang 程式\n\n-module(example1).\n-export([start/0, stop/0]).\n-export([twice/1, sum/2]).\n\nstart() ->\nspawn(fun() ->\nregister(example1, self()),\nprocess_flag(trap_exit, true),\nPort = open_port({spawn, \"./example1\"}, [{packet, 2}]),\nloop(Port)\nend).\n\nstop() ->\n% 發送訊息，讓 example1 停止，關閉 port 與 外部程式\nexample1 ! stop.\n\ntwice(X) -> call_port({twice, X}).\nsum(X,Y) -> call_port({sum, X, Y}).\n\ncall_port(Msg) ->\n% 以訊息方式發送 API request 給 example1\nexample1 ! {call, self(), Msg},\n% 等待接收結果\n{example1, Result} ->\nResult\nend.\n\nloop(Port) ->\n{call, Caller, Msg} ->\n% 對 Port 發送訊息, 資料內容是將呼叫的參數，轉換為 list\n% self() 為 connected process 的 PID\nPort ! {self(), {command, encode(Msg)}},\n% 收到外部程式送來的訊息\n{Port, {data, Data}} ->\n% 將結果解碼後，發送給 Caller\nCaller ! {example1, decode(Data)}\nend,\nloop(Port);\nstop ->\n% 關閉 port\nPort ! {self(), close},\n% 收到外部程式 送來關閉的訊息\n{Port, closed} ->\n% 送出 exit signal\nexit(normal)\nend;\n% 收到 exit signal\n{'EXIT', Port, Reason} ->\nexit({port_terminated,Reason})\nend.\n\nencode({twice, X}) -> [1, X];\nencode({sum, X, Y}) -> [2, X, Y].\n\ndecode([Int]) -> Int.\n\n### 編譯與測試\n\ngcc -o example1 example1.c erl_comm.c example1_driver.c\nerlc -W *.erl\n\n1> example1:start().\n<0.34.0>\n2> example1:sum(45,32).\n77\n3> example1:twice(10).\n20\n4> example1:twice(14).\n28\n\n### 注意\n\n1. 此範例並沒有統一 erlang 與 c 對整數的定義。直接假設兩個都是用單一個byte來當作整數，並忽略精確度、正負號的問題。\n2. 必須要先啟動負責界面的driver程式，也就是要先執行 example1:start()，然後才能執行此程式。\n\n# 附註\n\nerlang 跟外部程式之間傳遞資料，其資料內容的結構必須由 programmer 自行處理，這跟 socket programming 一樣， socket 在兩個程式之間提供 byte streaming 的傳輸，至於建構在 socket 上面的 app 要如何使用，就要由 app 自行決定。\n\nerlang 有幾個函式庫可簡化界面銜接的問題。\n\n1. http://www.erlang.org/doc/pdf/erl_interface.pdf\nei 是一組 C 函式與巨集，可編解碼 erlang 外部格式。在 erlang 端，一個 erlang 程式使用 term_to_binary 將 erlang terms 序列化，在 C 語言端， ei 的函式可用來解碼此 binary 資料。相反地，ei 可用來建構二元資料，而 erlang 端就以 binary_to_term 將 binary 資料解碼。\n2. http://www.erlang.org/doc/pdf/ic.pdf\nerlang IDL 編譯器 ic，這是 erlang 對 OMG IDL 編譯器的實作。\n3. http://www.erlang.org/doc/pdf/jinterface.pdf\nJinterface 是處理 java 跟 erlang 之間的介面，它可以將 erlang 型別完整地對應到 java 物件，為 erlang terms 編碼解碼，連結到 erlang process等等"
] | [
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] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.5541065,"math_prob":0.935625,"size":5204,"snap":"2020-45-2020-50","text_gpt3_token_len":2738,"char_repetition_ratio":0.13115385,"word_repetition_ratio":0.013477089,"special_character_ratio":0.33013067,"punctuation_ratio":0.2024598,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96135443,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-30T08:51:27Z\",\"WARC-Record-ID\":\"<urn:uuid:603c2783-cb79-4acf-90c5-842e21844dab>\",\"Content-Length\":\"97989\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bbfc54d1-2d9a-444d-8b1d-8a18d0571d6e>\",\"WARC-Concurrent-To\":\"<urn:uuid:3e64d71b-9401-46d4-93d4-0ea5013fc872>\",\"WARC-IP-Address\":\"172.217.7.147\",\"WARC-Target-URI\":\"https://blog.maxkit.com.tw/2014/04/erlang-interfacing-technology.html\",\"WARC-Payload-Digest\":\"sha1:JCIOLBXZK2D6L7AWTY5X5UVV7SX4WDPY\",\"WARC-Block-Digest\":\"sha1:PTY4PZ4BIPS3YZM4YX5LZUG7EAZ6C467\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141211510.56_warc_CC-MAIN-20201130065516-20201130095516-00560.warc.gz\"}"} |
https://www.cuemath.com/linear-equations/introduction-to-linear-equations/ | [
"# Introduction Linear Equations & Inequations\n\nGo back to 'Linear-Equations'\n\nIn earlier classes, you have encountered linear equations in one variable. For example, consider the following equation:\n\n$\\ 2x + 1 = - 7$\n\nThis is a linear equation in the variable x. There is no other variable. It can be solved to obtain $$x = - 4$$ as the solution.\n\nNow, we will consider linear equations in two variables. Consider the following equation:\n\n$x + 2y = 4$\n\nThis is a linear equation in the variables x and y. It has two variable terms (the terms containing x and y and a constant term). We can write this equation as follows:\n\n$x + 2y - 4 = 0$\n\nWhen written this way (all the terms on one side and 0 on the other side), a two-variable linear equation is said to be in standard form. Thus, an arbitrary two-variable linear equation can be written in standard form as\n\n$ax + by + c = 0$\n\nNote that there are two variables (x and y) and three constants (a, b and c). Using x and y for the variables is the general convention, but is not necessary. For example, the equation $$3u + 4v + 5 = 0$$ is a linear equation in the variables u and v. However, we will generally use x and y for the variables.\n\nExample 1: Which of the following equations are two-variable linear equations? $$x,\\,\\,y,\\,\\,z$$ are variables and all the other literals are constants.\n\n(A) $$x - \\sqrt 3 y + 1 = 0$$\n\n(B) $${a^2}x + {b^2}y + {c^2} = {d^2}$$\n\n(C) $$x - 3y + 4 = - 2z$$\n\n(D) $$2y - 3x + xy = 1$$\n\n(E) $$5 - 2x = \\frac{y}{{\\sqrt 2 }}$$\n\nSolution:. The equations in (A), (B) and (E) are two-variable linear equations. Let us write each of them in standard form:\n\n$x - \\sqrt 3 y + 1 = 0\\,\\,\\, \\to \\,\\,\\,\\left( 1 \\right)x + \\left( { - \\sqrt 3 } \\right)y + \\left( 1 \\right) = 0$\n\n${a^2}x + {b^2}y + {c^2} = {d^2}\\,\\,\\, \\to \\,\\,\\,\\left( {{a^2}} \\right)x + \\left( {{b^2}} \\right)y + \\left( {{c^2} - {d^2}} \\right) = 0$\n\n$5 - 2x = \\frac{y}{{\\sqrt 2 }}\\,\\,\\, \\to \\,\\,\\,\\left( { - 2} \\right)x + \\left( { - \\frac{1}{{\\sqrt 2 }}} \\right)y + 5 = 0$\n\nThe equation in (C) is a linear equation in three variables. The equation in (D) is not a linear equation, as it contains the cross term $$xy$$.\n\nTo summarize, if you are able to write an equation in the form $$ax + by + c = 0$$, this means that the equation is a two-variable linear equation.\n\nExample-2: Write each of the following equations in standard form and identify the coefficients a, b and c.\n\nI.$$- \\sqrt 3 y = \\pi - \\sqrt 5 x$$\n\nII.$$\\left( {x - 1} \\right)\\left( {y - 3} \\right) = xy + 1$$\n\nIII.$$\\frac{{x - 1}}{{y - 3}} = \\frac{2}{5}$$\n\nSolution: I. We can write the given equation as\n\n$\\sqrt 5 x - \\sqrt 3 y - \\pi = 0$\n\nThus,\n\n$a = \\sqrt 5 ,\\,\\,\\,b = - \\sqrt 3 ,\\,\\,\\,c = \\pi$\n\nWe could also have written the same equation with all the signs reversed:\n\n$- \\sqrt 5 x + \\sqrt 3 y + \\pi = 0$\n\nIn this case, the coefficients would have been\n\n$a = - \\sqrt 5 ,\\,\\,\\,b = \\sqrt 3 ,\\,\\,\\,c = - \\pi$\n\nBoth are correct.\n\nII. Expanding the expression on the left, we have:\n\n$\\begin{array}{l}xy - 3x - y + 3 = xy + 1\\\\ \\Rightarrow \\,\\,\\, - 3x - y + 2 = 0\\\\ \\Rightarrow \\,\\,\\,a = - 3,\\,\\,\\,b = - 1,\\,\\,\\,c = 2\\end{array}$\n\nIII. Cross-multiplying and rearranging to standard form, we have:\n\n$\\begin{array}{l}5\\left( {x - 1} \\right) = 2\\left( {y - 3} \\right)\\\\ \\Rightarrow \\,\\,\\,5x - 5 = 2y - 6\\\\ \\Rightarrow \\,\\,\\,5x - 2y + 1 = 0\\\\ \\Rightarrow \\,\\,\\,a = 5,\\,\\,\\,b = - 2,\\,\\,\\,c = 1\\end{array}$\n\nLearn math from the experts and clarify doubts instantly\n\n• Instant doubt clearing (live one on one)\n• Learn from India’s best math teachers\n• Completely personalized curriculum"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.84394455,"math_prob":1.0000095,"size":3367,"snap":"2019-35-2019-39","text_gpt3_token_len":1203,"char_repetition_ratio":0.1837645,"word_repetition_ratio":0.04718417,"special_character_ratio":0.4146124,"punctuation_ratio":0.17493112,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000091,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-15T10:11:49Z\",\"WARC-Record-ID\":\"<urn:uuid:f96c867d-9e6d-4194-b764-26ee2d8969eb>\",\"Content-Length\":\"76803\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f74794f0-47d4-4c89-94db-7ab7de62c861>\",\"WARC-Concurrent-To\":\"<urn:uuid:3b5bf383-0f25-47b4-bf7b-c4051bc0186a>\",\"WARC-IP-Address\":\"18.138.219.231\",\"WARC-Target-URI\":\"https://www.cuemath.com/linear-equations/introduction-to-linear-equations/\",\"WARC-Payload-Digest\":\"sha1:WDCCTB26KP6GLKLG5IXHSAKYAFVVAILE\",\"WARC-Block-Digest\":\"sha1:JMK5QY34YHYCKMOJYM7YSNXH6NSM6K4P\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514571027.62_warc_CC-MAIN-20190915093509-20190915115509-00137.warc.gz\"}"} |
https://ask.sagemath.org/question/40831/plots-of-complex-numbers/?sort=oldest | [
"# plots of complex numbers\n\nSay I want to plot the 6 solutions of the following complex equation.\n\nWhat is the best practice?\n\nEDIT by @tmonteil: to lower dependency between Sage services in the long term, here is the code provided in the sagecell:\n\nvar('z')\nsolutions = solve (z^6==-8, z)\nfor i in range(0,6):\nshow(solutions[i])\n\nedit retag close merge delete\n\nSort by » oldest newest most voted\n\nThe way you solved the equation is such that the solutions are symbolic expressions. It is hard to ask Sage to plot a symbolic expression (unless it represents a function, in which case it will plot its graph).\n\nIf the complex numbers are floating-point (e.g. elements of CDF), then plotting the is easily, thanks to the points function:\n\nsage: points([CDF(1+I), CDF(-I), CDF(-2)])\n\n\nNow, regarding your concrete example, the solutions are given as symbolic expressions representing equalities:\n\nsage: solutions\n[z == 1/2*I*sqrt(3)*sqrt(2)*(-1)^(1/6) + 1/2*sqrt(2)*(-1)^(1/6), z == 1/2*I*sqrt(3)*sqrt(2)*(-1)^(1/6) - 1/2*sqrt(2)*(-1)^(1/6), z == -sqrt(2)*(-1)^(1/6), z == -1/2*I*sqrt(3)*sqrt(2)*(-1)^(1/6) - 1/2*sqrt(2)*(-1)^(1/6), z == -1/2*I*sqrt(3)*sqrt(2)*(-1)^(1/6) + 1/2*sqrt(2)*(-1)^(1/6), z == sqrt(2)*(-1)^(1/6)]\n\n\nSo, what you want is to use their right hand side (use the rhs method) and turn them into elements of CDF, so that you can plot them with the points function.\n\nAlso, you can notice that when you want to iterate over the elements of a list L, you do not need to use the indexing of the elements, so, instead of typing:\n\nfor i in range(len(L)):\nblah L[i]\n\n\nYou can do:\n\nfor l in L:\nblah l\n\n\nIn short, you just have to type:\n\nsage: points([CDF(s.rhs()) for s in solutions])\n\n\nIf you want a regular hexagon, you can require the x and y axes to have the same scale with the aspect_ratio option:\n\nsage: points([CDF(s.rhs()) for s in solutions], aspect_ratio=1)\n\n\nNote also that plots are Sage objects that can be added with eachother, e.g.\n\nsage: points([CDF(s.rhs()) for s in solutions], aspect_ratio=1) + circle((0,0), CDF(8^(1/6)), color='red')\n\n\nSee: this Sage cell\n\nmore\n\nOne possibility is:\n\nsage: var( 'z' );\nsage: f = z^6 + 8\nsage: list_plot( f.roots(multiplicities=False, ring=CC) )\nLaunched png viewer for Graphics object consisting of 1 graphics primitive\n\nmore\n\nYou may also draw a complex plot (black dots are the zeroes, red means real numbers or close to, and other rainbow colors means other argument, more white means high modulus:\n\nsage: var('z')\nz\nsage: complex_plot(z^6 + 8, (-2,2), (-2,2), aspect_ratio=1)\nLaunched png viewer for Graphics object consisting of 1 graphics primitive",
null,
"more\n\nNote that this is not completely true: the red is more the positive reals, not the reals, since z^6 + 8 and z^6 are real for the same values of z, the real values of the function should be 12 rays emanating from 0, not only the 6 we can guess on the picture, see:\n\nsage: I = CDF.gen()\nsage: implicit_plot(lambda x,y : imag((x+I*y)^6+8), (-2,2), (-2,2), aspect_ratio=1)\n\n\nSee this cell\n\nYes, this is what I meant: postive real is red. The best way to confirm the meaning of colors is to plot the identity map:\n\nsage: var('z'); complex_plot(z, (-2,2), (-2,2), aspect_ratio=1)"
] | [
null,
"https://ask.sagemath.org/upfiles/1517210052105545.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7889608,"math_prob":0.99716234,"size":1667,"snap":"2019-43-2019-47","text_gpt3_token_len":550,"char_repetition_ratio":0.16115454,"word_repetition_ratio":0.027559055,"special_character_ratio":0.35872826,"punctuation_ratio":0.12820514,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9988761,"pos_list":[0,1,2],"im_url_duplicate_count":[null,9,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-18T17:40:20Z\",\"WARC-Record-ID\":\"<urn:uuid:313536e4-e51c-4682-b09c-8b9fdd0571f5>\",\"Content-Length\":\"70884\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9446548f-67e1-4da1-8548-064d7bddf882>\",\"WARC-Concurrent-To\":\"<urn:uuid:4fcbc943-2e2d-4fe2-9eb0-331bcd3714f2>\",\"WARC-IP-Address\":\"140.254.118.68\",\"WARC-Target-URI\":\"https://ask.sagemath.org/question/40831/plots-of-complex-numbers/?sort=oldest\",\"WARC-Payload-Digest\":\"sha1:O5CGBKUR2SHEKVG7ELSQ5LAWT4AHNGII\",\"WARC-Block-Digest\":\"sha1:YK4ZIVSTV5WERGT3Q2PQCBTGSSJNHVD6\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496669809.82_warc_CC-MAIN-20191118154801-20191118182801-00286.warc.gz\"}"} |
https://matheducators.stackexchange.com/questions/12/a-calculus-book-that-uses-differentials?noredirect=1 | [
"# A calculus book that uses differentials?\n\nAll introductory calculus books that I have seen spend most of their chapters on differential calculus talking about derivatives, with at most a short section defining differentials as $$dy = f'(x) \\, dx$$. However, differentials are useful for understanding a lot of things, like linear approximation, the chain rule, integration by substitution, and (when you get to multivariable calculus) the change-of-variables formula and the various manifestations of Stokes' theorem. One doesn't have to agree with everything that Dray and Manogue say to want to try introducing and emphasizing differentials early in differential calculus.\n\nIs there any calculus textbook which does such a thing?\n\nMy book Calculus from the Ground Up focuses on differentials, and uses it to provide a unification of process and simplification of understanding of a lot of different parts of calculus.\n\nTo read about the thought process that led to the book you can see this arXiv link; the focus on differentials that you are asking for led naturally to a refactoring of the way introductory calculus is presented.\n\nDifferences from other books:\n\nThe arXiv link gives some important information, but I'll repeat some of it here. First of all, the focus of the entire book is on differentials. We do a lot of derivatives, but the focus is always on differentials, and for several important reasons. First, it unifies several important practices into a single system - single-variable, multivariable, and implicit differentiation all has the exact same process. Second, it makes the different geometric integrals more obvious. The integral is presented as a sum of infinitesimals, not as an area under the curve (which becomes merely one of the application areas). The integral simply sums up whichever geometry is being used. $$\\int y\\,dx$$ for summing areas of rectangles, $$\\int \\pi y^2\\,dx$$ for summing volumes of cylinders, and $$\\int \\sqrt{dx^2 + dy^2}$$ for summing arc lengths. The way you are asked to memorize them is exactly what the geometry states. For instance, many books want you to have the volume of cylinders as $$\\pi \\int y^2\\,dx$$. That's correct, but moving the $$\\pi$$ outside means that it no longer looks like the volume of a cylinder equation for students.\n\nAdditionally, the book includes a rule that seems to have gone missing for doing differentials of the form $$u^v$$. For those who don't know (because it is missing in most modern books), $$d(u^v) = vu^{v - 1}du + \\ln(u)u^vdv$$ (I wish the font for $$v$$ had a more distinct look here, but oh well). Many books teach \"logarithmic differentiation\" for this, but it is wholly unnecessary. Just like all the other differentials, all you need is the rule.\n\nI also try to include additional life lessons that we can learn from calculus. For instance, in the discussion of Taylor polynomials, I discuss how this can be used as a template for solving impossible problems (not only in math but anywhere).\n\nAlso, I wanted to make a note on the second differential, because it came up in the discussion of Keisler. I don't make a big deal about it (I put it in the Appendix), but I actually introduce a form for the second derivative that makes the chain rule for the second derivative work algebraically. Generally, in the text, I avoid this situation by simply introducing a variable for the first derivative, and then take the derivative of that variable. However, in the appendix I show that second differentials can be made algebraic by making the second derivative $$\\frac{d^2y}{dx^2} - \\frac{dy}{dx}\\frac{d^2x}{dx^2}$$. If that looks strange to you, you can derive it for yourself by simply taking the derivative of $$\\frac{dy}{dx}$$. Note that $$\\frac{dy}{dx}$$ is a quotient, so you would use the quotient rule to take the derivative of it. This leads to differentials that are 100% algebraically manipulable. Most texts focusing on differentials don't tell you either the problem nor the solution for using second differentials.\n\nThe structure of the book differs from \"Calculus Made Easy\" in that it starts with derivatives, since a slope is more intuitive for people coming from algebra. Unlike Keisler, it saves discussion of limits for the end of the book. Essentially, it gives you the intuition and the toolset first, and then, at the end, goes into a bit more formally the underpinnings of what makes it work. I find that students prefer this approach. Like Keisler, I use the hyperreal numbers (though I don't formally introduce them until the last third of the book, which focuses on the infinite).\n\nAnyway, I always try to write things in such a way as to focus the student on the intuitions behind everything, so that learning calculus doesn't just teach them calculus, but it improves their thinking. For instance, when talking about the other geometric uses of the integral (volumes from cylinders, volumes from shells, arc lengths, etc.), I gave a general mental mechanism that is used to generate all of these. (a) the problem can be estimated by a formula, (b) the problem can be divided into subproblems, (c) each subproblem must have the same form as (a), (d) the result must be attainable by adding the results of the subproblems, and (e) increasing the number of subdivisions improves the accuracy of the estimation method. The goal here is to show the students how the thought process works.\n\nAlso, my student's also love the fact that I show where all of their formulas that they learned in previous math classes come from. I show how to derive the interest rate formula, the volume of a cone formula, and the volume of a sphere formula. In fact, that's another aspect of the book - I teach how to derive formulas. We use calculus to derive the vertex formula for quadratics, and a homework problem is deriving the vertex formula for cubics. I tell students that calculus is the \"where babies come from\" of math.\n\nNOTE - I edited this to include more details about the book and what makes it different because I was requested to below. Sorry if this comes off as more of an advertisement than was intended.\n\n• Based you your user profile, it appears that you are the author of the book you are recommending. As you have not disclosed your affiliation, your answer runs the risk of being deleted as spam. – Xander Henderson Jan 26 at 20:48\n• I am indeed the author. I'm not sure how it is spam, if the person is literally asking for book recommendations that exactly match the book I am suggesting. – johnnyb Jan 27 at 1:07\n• @johnnyb I made an edit to the answer -- does this seem accurate to you? It puts some words in your mouth but I believe it makes it pretty clear that your answer isn't some kind of low-effort spam post. – Chris Cunningham Jan 27 at 20:57\n• (Obviously feel free to rework or revert what I wrote, but mentioning that you are the author is just good practice.) – Chris Cunningham Jan 27 at 20:59\n• Thanks for the help, Chris! – johnnyb Jan 28 at 16:32\n\nThis might be taking things too far, but Kiesler's book (available free online) does everything using infinitesimals, which make differentials literally immediate. The rigorous underpinning for infinitesimals is nonstandard analysis, but this book doesn't dwell on that. It just teaches how to use them correctly.\n\nI'm guessing this isn't exactly what you were looking for, but it might be worth checking out because it's free.\n\n• Thanks! I've looked at Keisler's book before, and considered it seriously. In general, I think infinitesimals are actually orthogonal to differentials: one can use either one without the other. However, Keisler does use differentials fairly seriously as well (although he defines the second and higher differentials incorrectly in my opinion), so this would be worth an upvote. Unfortunately, on general principle I never upvote an answer that explicitly asks to be upvoted. (-: – Mike Shulman Mar 14 '14 at 21:50\n• On general principle, I always upvote comments about general principles. So we're good. – Kevin O'Bryant Mar 16 '14 at 0:01\n• @MikeShulman: Interesting! What is it about Keisler's definition of second differentials that you find incorrect? – String May 12 '14 at 8:53\n• @String One of the important aspects of differentials, especially for a calc 1 class, is \"Cauchy's invariant rule\": that you can do the chain rule by substitution. That fails for second derivatives using Keisler's definition $d^2f=f''(x)dx^2$. To recover it you need instead $d^2f=f''(x)dx^2+f'(x)d^2x$. I learned this from Toby Bartels. – Mike Shulman May 12 '14 at 20:09\n• @String I'm not sure what you mean by \"recover\". Of course if you assume $d^2x=0$ then $d^2f=f′′(x)dx^2+f′(x)d^2x$ reduces to $d^2f=f′′(x)dx^2$, but the point is that the latter formula gives you the wrong chain rule. E.g. if $y=f(u)$ and $u=g(x)$ then from $d^2y=f''(u)du^2$ and $d^2u=g''(x)dx^2$ you get by substitution $d^2y = f''(g(x)) (g'(x))^2 dx^2$ which is not the correct second derivative of $y = f(g(x))$. – Mike Shulman May 13 '14 at 21:13\n\nWas Silvanus Thompsons lovely \"Calculus made easy\" mentioned already? It's a classic (100 years old) freely available on gutenberg.com. Some opinions of it can be found on mathoverflow.\n\nIt doesn't go very far so it might need to be supplemented with another text, but I believe it does a great job at teaching the physical and geometrical intuition on differentials. It seems that it's closer to synthetic differential calculus than to non-standard analysis in the way it treats infinitesimals.\n\n• It's a very nice book, but one can't really use it as the primary textbook for a class, can one? – Mike Shulman Jul 4 '14 at 16:06\n• Good question. I haven't tried. The good thing is that one may modify the original book to fit ones purposes according to the gutenberg license – Michael Bächtold Jul 4 '14 at 18:31"
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https://www.intechopen.com/chapters/6582 | [
"Open access peer-reviewed chapter\n\n# Formal Verification of Hybrid Automotive Systems\n\nBy Jairam Sukumar, Subir K Roy, Kusum Lata and Navakanta Bhat\n\nPublished: January 1st 2010\n\nDOI: 10.5772/6963\n\n## 1. Introduction\n\nState of the art automotive systems have become extremely complex in terms of functionality, system architecture and implementation. These systems consist of significant portions of embedded software as well, running in excess of millions of lines of C code. Given such enormous complexities it is never an easy task to verify their functional correctness. Traditional simulation based validation schemes cannot hope to cover the entire specification domain of such a system, as creating test cases to cover all the different corner cases of functional behaviour is extremely difficult. The number of such test cases can be extremely large; it is typically a manual process, at best. Moreover, the time taken to run all the test cases to verify the complete system under all possible scenarios, environmental constraints and input conditions will be inordinately large. This can strongly impact the system design and implementation turn-around-time, thereby reducing the window of opportunity in the market for the product. The amount of complex components being integrated into main stream automotive systems is continuously on the rise, latest being MEMS based sensor components. These MEMS components are becoming a part and parcel of modern automotive electronics systems, ranging from dynamic vehicle performance systems and safety systems to, infotainment and other accessories. To illustrate an example, the heart of any modern automotive engine cruise control (ACC) system includes MEMS based sensors, such as, accelerometers and gyroscopes. The verification of such MEMS components is, in general, hard and difficult, and their integration into a hybrid system exacerbates an already complex hybrid system verification challenge.\n\nFormal analysis of hybrid systems integrating MEMS components, such as sensors, requires their models to be comprehended and integrated into the model of the hybrid system itself. While the cruise control behaviour is primarily discrete, the MEMS sensor behaviour is continuous and highly non-linear. Thus, a system designer will need to validate, both, the discrete time and the continuous time behaviour in any such hybrid system. The single largest challenge in the verification of such hybrid systems lies in appropriately modelling individual components and integrating them within a single verification framework. Another strong motivation in employing formal verification for the validation of hybrid systems, also, is to circumvent the well known drawbacks inherent in a simulation based validation approach. In the domain of discrete time formal verification, a wide range of commercial tools exist which have matured with an increasingly large user base. These tools, however, are incapable of validating traditional hybrid systems, let alone, hybrid systems integrating MEMS based sensor components. In this chapter, we first present the complexity inherent in hybrid systems and then show how formal verification can be employed to verify them.\n\n## 2. Advances in automotive systems\n\nAdvances in automotive electronics have triggered large scale integration of myriad features for all categories of vehicles. The two fundamental vectors of feature development include safety and passenger comfort. These features can be further classified based on their means of integration, viz., passive and active. While the passive features amount to pre-installed and pre-verified components, each active feature needs a continuous sampling of ambient conditions and based on the condition, feedback actions are taken by the vehicular controller. All these features hence also require processing capabilities much higher than requirements needed in the past. Some of the active mode examples include integrated park and lane assistant, cruise control, climate control, etc.\n\nAdvent of these new features also brings to the horizon, the challenge of modelling and analyzing new types of sensors to support these features. We focus on one such type: the MEMS based sensors. The computational complexity of modelling MEMS components, increases manifolds as one needs to solve or model non-linear partial differential equations (PDE) to accurately capture the sensor characteristics.\n\nThis chapter attempts to address the challenges posed by integration of these components into mainstream VLSI systems and their verification. The chapter helps the reader to appreciate the nuances of verification methodologies and propose recipes to formally verify computationally complex hybrid automotive systems. A methodology is described, based on transformation techniques that can be deployed to solve these problems. We introduce Simulink-Stateflow based modelling and verification platform , as shown in Figure 1. The figure explains how the continuous and discrete time systems interact together in a Matlab based environment. It is shown how this can be used to integrate some of the complex MEMS based components into mainstream VLSI systems. We also introduce the reader to CheckMate, a formal analysis software tool available in public domain. Simulation traces from Simulink-Stateflow framework are used in the formal analysis engine present in CheckMate, instead of being derived implicitly by numerical integration, or explicitly by transformation based approaches.\n\nTo illustrate this approach, we choose one representative real life control system, fairly popular, from the automotive hybrid system space. This hybrid system directly interfaces with a MEMS based gyroscope used as a speed sensor. We then deploy our proposed approach to perform a formal analysis of the MEMS integrated ACC system (MIACCS). We model the adaptive cruise control system in a traditional simulation framework, and then proceed to build a formal analysis framework for the same system. The chapter is divided into the following sections. Section 2 introduces the reader to the state of the art in hybrid formal verification methodologies. Section 3 explains about the transformation based techniques for formal analysis. We discuss formal analysis platforms in Section 4 and introduce the reader to one of them in Section 5 viz. CheckMate, which works on Simulink-Stateflow (SSF) based system. SSF based methods are the most widely used platforms across the industry for hybrid and real time system. Finally we illustrate the approach with the case of a MEMS based adaptive cruise control system in Section 6 and through 8.\n\n## 3. Hybrid verification methodologies: past, present & future\n\n### 3.1. Introduction to hybrid formal analysis\n\nSpecifically the very low Defective Parts Per Million (DPPM) requirements to meet the stringent automotive fail safety norms and standards, poses new hurdles and therefore newer challenges to overcome them. Thus, it has become a topic of active research in the recent past . Present generation, state of art automobiles, use MEMS based sensors for measurement of different vehicular parameters . Given the fact that, verification of such mainstream automotive control systems need to be based on formal approaches, validation of system behaviour integrating such MEMs components, results in additional complexities. A hybrid system, typically, includes both discrete and continuous time components. To analyze hybrid systems, consisting of, both, discrete behaviour and continuous behaviour components, it becomes necessary to partition these distinct functional behaviours and use specialized analysis engines to target each behaviour domain. A typical hybrid system formulation is shown in Figure 2. First of all, there needs to exist a grammar to formally describe the hybrid system. A given hybrid system is then partitioned into its representative discrete and continuous time systems. Both these partitions are individually solved through their respective solvers. Continuous time solvers include symbolic analysis based solvers, while discrete time solutions can be obtained through traditional graph traversal based methods. Finally both these solutions are integrated by interleaving of variable, at each time step of computation. Decision to change a state is taken based on corresponding conditions being satisfied to trigger these changes.\n\nFor a detailed introduction to the timed and hybrid automata interested readers are referred to and . A discrete time variable needs to be identified and then used to sample the system behaviour in temporal space. After each time step, the system state is determined and the state equations are solved. Changes in the state of the system are then evaluated based on state values and guards corresponding to the transition edges of the automata. The continuous time system formulation is thus, analyzed through time- discretisation and then solved through numerical routines based on Runge-Kutta or Newton-Raphson methods for non-linear dynamic functions . A time discretisation step is followed by linearization of non-integer functions. For each time step, the continuous time domain state variables are concurrently analyzed with the discrete time variables to determine the reachability of the state space of the hybrid system needed to establish truth validity of the safety properties. These terms are explained in later in Section 3.2. Thus, the overall model checking methodology is based on progression in the time domain. Figure 3 illustrates the above methodology.\n\n### 2.2. Formal analysis of timed automata\n\nThe concept of timed automata is illustrated to the reader by a simple example of a temperature controller. A diagram of a temperature controller is shown in Figure 4a. The controller has two states viz.ON and OFF. The controller enters into the initial ON state with the temperature variable x being initialized to the room temperature Troom. The controller switches the heater which increases the temperature with a linear rate until the temperature attains an upper bound. The heater is cut-off by the controller and the system cools down with a rate given by the differential equation shown in the OFF state (Figure 4b).\n\nState equations can be solved using Laplace transforms as follows:\n\nON STATE:\n\ndxdt=ϕ(t)E1\nX(s)=1sϕ(s)E2\nx(t)=ϕ(t)+CE3\n\nOFF STATE:\n\ndxdt+αx=f(t)E4\nsX(s)+αX(s)=F(s)E5\nX(s)=F(s)s+αE6\nx(t)=eαtf(t)+CE7\n\nIt can also be seen that for a class of rational polynomials, there exists a simple partial fraction decomposition which simplifies the Inverse Laplace Transform computation. Equations (3) and (7) are exact functions with respect to the time variable. Thus, exact values of these functions corresponding to each dynamic state variable (for example, temperature) at different time points can be easily and exactly computed to determine the set of reachable states.",
null,
"Figure 4.a) State Transition Graph of Temperature Controller. (b) System Response.\n\nThe steady state solution space, which is also the reachable state space, is shown in Figure 4b. The reachable state space varies with the parameters Φ and present in the differential equations. The reader should note that, for this parameterized reachable state space corresponding to the temperature variable, constraint solvers need to be deployed to prove the specification correctness.\n\n## 3. Transformation based formal analysis\n\nState of the art methods of formal analysis are based on timed or hybrid automata.\n\nThere also exists a symbolic analytical method, for the solution of system description. If an analytical model can be obtained based on exact modeling, it can provide potential computational benefits. Hence let us explore solving linear differential equations symbolically through Laplace transforms instead of simulation. We can transform the state equations in the time domain and then solve the set of algebraic equations on them through the well known method of pole residue decomposition of transfer functions. The solutions in the original time domain obtained through inverse Laplace transforms are then bound, based on the constraints imposed on the corresponding differential equations. Constraint solvers can be used to solve for these bounds and establish the truth of the properties.\n\n### 3.1. Partial decomposition based solution\n\nLet us know look at a simple nth order linear differential equation and assume it to be a part of the continuous time system. Let us also assume all state equations to be defined by linear differential equations. The equations will be of the standard form:\n\nAndnxdtn+An1dn1xdtn1+...+A1dxdt+A0x=f(t)E8\n\nThe time domain differential equation is re-written as the following:\n\n(AnDn+An1Dn1+...+A1D+A0)x=f(t)E9\n\nHere D is the differential operator. Transforming the time domain equation in (8)-(9) into the frequency domain through Laplace transform, we arrive at the following equation:\n\n(Ansn+An1sn1+...+A1s)X(s)=F(s)E10\nX(s)=F(s)Ansn+An1sn1+...+A1sE11\n\nAssuming X(s) can be transformed into a rational polynomial of the type A(s)/B(s), regular polynomial packages like MatlabTM or Octave can be used to convert this rational polynomial into the pole/residue form. The characteristic equation can be re-written as:\n\nX(s)=zisi+pi+RE12\n\nThe time domain solution to (12) can be written as:\n\nx(t)=iziepitE13\n\nSimple Matlab/Octave routines can be used to solve a variety of rational polynomials. The reader can try many of such combinations. This approach can be rendered into a simple automated flow which invokes math solvers to solve the standard pole-zero decomposition problems. Using these computed poles and residues it is easy to derive the time domain response of the dynamic components in a hybrid system.\n\n### 3.2. Constraint formulation and truth validation\n\nLet us now look at the constraint formulation mechanism used in the context of the above example.\n\nWe need to formally state the property to be verified. It can be formally described as a constraint. However we need a formal language to describe this constraint. One can imagine a domain space corresponding to the span of a specified property. We call this the constraint space. The constraint space needs to be intersected with the reachable state space (or the solution space) of the state variables obtained by approaches based on Laplace transforms or by numerical integration of differential equations. If the constraint property space subsumes the reachable state space, the correctness of the controller behaviour is established on its formal model. The solution space of the state variable and the property to be checked can be represented as:\n\nX(t)={X1(t);tD1X2(t);tD2X3(t);tD3E14\n\nThe setD1D2D3provides the complete time domain region of operation of the system. Let, for all specification Si the constraint space for its corresponding property be Pi. The objective is to obtain the region of intersection ofPiandX(t). For the temperature controller discussed earlier, we explain the constraint analysis methodology. Let us define three properties; that describe three different safety requirements of the controller in example which we mentioned previously:\n\n• Temperature of the system remains between [Tupper + Tlower]/2 and Tlower.\n\n• The Temperature of the system is always below Tupper.\n\n• The Temperature of the system is always between 300C and Troom. (Where Tupper>300C and Troom<Tlower)\n\nIn other words the above three properties can be specified as a part of functional specification of the hybrid system. The functionality of the implemented controller should ensure that these specifications are met. The specifications can be transformed into mathematical inequalities or constraints as shown in the\n\n S/N Functional Spec Controller Spec ifications 1 TlowerX(t)[Tupper+Tlower]/2 Tlowerx(t)Tupper 2 X(t)Tupper Tlowerx(t)Tupper 3 Troomx(t)300;Tupper300TroomTlower Tlowerx(t)Tupper\n\n### Table 1.\n\nA simple method as mentioned above is to formulate the solution surface of the constraints and evaluate the intersection. Solutions that can be incorporated in 2 or 3-dimensional surfaces can be very easily visualized. The solution surface for the properties in Table I can be illustrated as shown in Figure 5. It can be easily noted that the functional specification 1 is not satisfied as the controller specification does not bound the corresponding property. This results in a failure of the property to meat the desired specification.\n\n## 4. Hybrid analysis platforms\n\nLet us progress to the next level, where we now discuss the complexities involved with ordinary and partial differential equations, and see how we can tackle them.\n\nWe assume that the reader is fairly well aware about the basics of differential equations. Analysis of the continuous time domain behaviour modelled by Ordinary/Partial Differential Equations (O/PDE) can be solved by using a Linear Program Solver such as, LP Solver for obtaining the state space solutions and discrete event solvers like SAT solvers for discrete state transitions. In addition, to render the analysis formal, a reasoning engine to interpret the correctness of the behaviour of system implementation against the specified set of formal properties is needed. A fully automated approach would the golden ideal; however, even for the simplest hybrid system with linear behaviour, it has been theoretically proven that establishing the truth value of properties captured in a first order logic on real numbers is un-decidable and NP-hard [11-13]. Therefore, several automated approaches which are inexact or approximate [14,15,16,17] and other non-automated approaches, based on theorem proving, requiring user inputs to drive the formal analysis [18, 19]; have arisen to tackle the intrinsic complexity inherent in hybrid systems. The presence of components like MEMS based sensors renders the formal analysis even harder, with their dynamic behaviour being modelled and described only by higher order non-linear partial differential equations (NL-PDEs). It is well known that behavioural models based on higher order non-linear partial differential equations are not amenable to pure formal analysis .\n\nAt the same time, a complete system level analysis suite based purely on simulation approaches for hybrid systems with such components leads to unacceptably high run times, in order to cover every possible combination of corner case behaviours due to internal interactions within system components and external interactions with the system environment. While, detailed analysis of MEMs structures can be performed by finite element based approaches, carried out within respective energy domains of corresponding sub-systems , integrating such analysis framework for system level formal analysis results in huge computational bottlenecks . Symbolic simulation methods are an alternative and can replace traditional simulation . These are being increasingly used in many domains for system validation, where, an analytical solution of the system is achieved through symbolic analysis using solvers of various types . Truth decidability in these cases becomes a computationally difficult task due to the infinite cardinality of the state spaces of the continuous dynamical systems. The authors of give the family of linear differential equations with a decidable reachability problem, by symbolically computing the reachable state sets by posing it as a quantifier elimination problem in the decidable theory of reals. Public domain quantifier elimination tools such as REDLOG () and QEPCAD () implement these approaches and have been used in symbolic verification of hybrid systems. As discussed earlier, real life hybrid systems, however, require complex linear and non-linear differential equations to model their dynamic behaviour, thereby rendering symbolic approaches computationally expensive. Reachable state sets for these systems are computed approximately, using numerical methods based on time step integration of differential equations to contain the complexity in computing the exact reachable state sets. This approximate computation is either based on polyhedra, or level sets, or ellipsoids , , and . In , the authors explore new algorithms for accurate event detection for simulation based reachability analysis and abstraction methods. The advantages of transformation based approaches discussed earlier in Section 3, to alleviate problems arising out of misses in the detection of important system behaviour events can be easily seen. This helps in decreased computational overheads arising from numerical integration, accurate event detection and, therefore, increased robustness in formal analysis of hybrid system behaviour. It is easy to see that the analytical form of the proposed approach also, makes it amenable towards computation of approximate reachable state sets, needed for automated formal analysis. While the technique based on transformation approaches is appealing for analyzing hybrid system components described behaviourally by linear differential equations, it is inappropriate for MEMS based components described by non-linear PDEs, which may not even have an analytical solution describing their behavior. However, it is possible to obtain approximate analytical models for some of the MEMS based components, such as, gyroscopes, which involve a single convolution operation in either the time or frequency domains. Even this is computationally expensive for formal analysis. Some of these simulation complexity aspects of MEMS components have been described in .\n\n## 5. Introduction to CheckMate\n\nIn this section we give a very brief introduction to CheckMate , a public domain tool from CMU. CheckMate can be employed to verify hybrid systems modeled as hybrid automata, either formally using model checking, or through a simulation mechanism. This solution is built on the very popular Simulink/Stateflow Framework (SSF) from Mathworks, widely accepted both in the academia and the industry. CheckMate supports three important custom SSF blocks, viz., Switched Continuous System Block (SCSB), Polyhedral Threshold Block (PTHB), and Finite State Machine Block (FSMB). A hybrid system is modeled primarily using these three blocks. CheckMate, however, supports a few other blocks present in SSF.\n\nA SCSB is used to define the system continuous dynamics in terms of first order differential equations. A PTHB generates events whenever the system crosses a specified threshold described in terms of a linear constraint. This generated event is used as an input to the FSMB to trigger transitions from one state to another. Based on the sink state of a transition edge that is reached, the SCSB block on reaching that state generates the continuous state trajectory using the dynamics corresponding to that state.\n\nVerification is performed using three distinct phases. In the first phase, it allows verification of the hybrid system using traditional simulation based on numerical integration, as supported in SSF. Thus, CheckMate models can be simulated in a manner similar to any other general SSF model. In the next phase, Explore, beginning with the initial location of the hybrid automata, it checks whether each simulation trajectory, starting with different initial conditions corresponding to the set of corner vertices in the convex polytope initial continuous set, satisfies a given formal property specified as an ACTL formula. It informs the user in case of any violation. In the third and final phase, Verify, CheckMate performs formal verification.\n\n## 6. Adaptive Cruise Control system: a case study\n\nThe concept of adaptive cruise control (ACC) system has been developed to aid vehicular traffic on highways. It is an automatic closed loop system, through which a driver during a long drive, can volitionally transfer control to an intelligent vehicular controller system.\n\nFigure 6 illustrates a speed control mechanism used in automobiles. A MEMS sensor viz. a gyroscope is attached at the wheel base to provide details on the vehicular speed. Speed and proximity sensors are also installed at the front and rear end of the vehicle to continuously monitor the speed and distance of the vehicle ahead and behind. All these details are then sampled by engine controller to take a suitable course of action for the vehicle.\n\nThe vehicular control is marked by a closed loop control chain, which is executed in a continuous polling mode. This system is primarily a state machine, which is designed to translate the nature of the traffic conditions into a vehicular and generate an action for each state. The action is again governed by the parameters of highway control viz. speed limits, minimum proximity etc. Figure 7 describes the state transition graph for the ACC system.\n\nThe system behaviour consists of four states, viz, 'HALT', 'ACCELARATE', 'CRUISE' and 'RETARD'. The variables Xp (for proximity to the front vehicle) and V (for speed) govern the assignments to different states and the transitions between these states.\n\nThe engine controller samples these states and accordingly sends control signals to various automotive peripheral subsystems to act. These functional tasks include acceleration, retardation or maintaining constant speed of the vehicle. The sensors on the vehicle, keep polling for front and rear vehicular proximity for a continuous update of system state.\n\n## 7. Introduction to gyroscope model\n\nThis section provides an introduction to analysis and modeling of a gyroscope. Gyroscope is a device that is used to measure the angular velocity of a system. It primarily uses the concept of Coriolis force, to translate angular motion to linear motion detection, which is captured in the form of capacitance variation in measurements. The reader is encouraged to look into basic text books applied mechanics and MEMS to understand the concepts of mechanical motion and gyroscope basics. provides a very good summary paper for design and analysis of micromechanical gyroscope. For the clarity of the reader and completeness of the subject, we present a very basic analytical overview of the gyroscope. Figure 8 shows a mathematical model of a gyroscope and its Simulink representation.\n\nA gyroscope can be abstracted as a simple set of coupled differential equations whose output can be equated to a capacitance equivalent to the angular velocity. The reader should note that, simple as it seems, there exists a fundamental difference between this model and most other models based on differential equations. The conventional models would utilize Laplace transform so that the convolution operations in time domain translate to multiplication operation in the frequency domain, thus, resulting in simpler computations. However in the case of gyroscopes, there exist both multiplication and convolution operation in both time and frequency domains. Hence a simplified computation is not possible. The reader is encouraged to explore options to simplify this computationally challenging problem. In this case we resort to a simplified time domain based simulation approach through Simulink.\n\n## 8. Formal verification of ACC system\n\n### 8.1. Planning the models\n\nLet us first explore how the given system can be casted into the Check Mate framework. It is assumed that the reader is well aware of the Simulink and Stateflow tools. It is a fairly simple task to translate the ACC system requirements into an SSF diagram. These tasks today are routine for practicing system design engineers who program in Simulink/Stateflow in Matlab. The state transition graph of the ACC state machine (FSM) can be added as is, into the Stateflow system of Matlab. For each state in the hybrid automata, there is a need to associate, a real time Simulink model, which continuously modifies the state variables of the system. At each time step during the temporal evaluation of the hybrid system, based on state specific behaviour, conditions associated with each transition edge are evaluated. An illustration of the same is shown in Figure 9. Simulink representation of a real time system involves simple hook up of maths based models, which are either user created or linked Simulink library instantiations. Similarly Stateflow diagrams can also be easily created through templates available in the Matlab framework. Reader can go through standard Matlab help manuals and tutorials to get a better grasp of these basics. The differential equation pertaining to each state can be easily modelled by Simulink modules. Figure 10 illustrates the integrated ACC system, with assumptions upon availability of the sensor inputs through direct means. Now let us look at the ACC system in more detail. Figure 7 represents its state transition graph.\n\nThis same representation has been cast into the system shown in Figure 10. Each state shown in the figure, models a real time differential equation defining that state. A switch multiplexes the activation of the state equation in time domain depending on the state decision. The Stateflow sub-box shown in the figure generates the control signal for the switch. This sub-box models the ACC transition graph. This implementation forms one of the simplest representations of an integrated Simulink-Stateflow system. The reader should again keep in mind that this system still does not include the sensor model into the overall model of the ACC system. Notably there are two points to mention. Firstly, there exists a sensor (not shown in the state transition graph), which senses the vehicular velocity, later processed by the controller. The need for the integration of the sensor into the main ACC system arises because due to the fact that, there exists error in measurements taken at the sensor output of the sensor, which may not be accounted into the error-analysis of the formal analysis model. The integration of the sensor helps solve this problem. Secondly, the sensor processing and the ACC system processing time steps are the same in a Matlab based setup. Hence within the framework of a single discrete event solver, one cannot freeze the simulation time step against the other. This problem is illustrated in Figure 11. As the case in our example, we cannot freeze the ACC system time step progression and wait for the sensor to provide an event signalling completion. However the current versions of the Matlab solvers do not address this requirement. Working within the Matlab setup, can pose challenges, which need not even be addressed. A work around to this problem involves translating one of the time variables to static. In other words, if we can characterize the sensor model, not only can we account the model into the framework, but also perform a formal analysis with respect to the correctness of the sensor functionality.\n\n### 8.2. CheckMate model of ACC system\n\nWe now introduce the CheckMate modeling environment. A basic introduction has already been provided to the reader. In this section, we illustrate how a Simulink-Stateflow based system can be translated into a CheckMate model. The reader should note that the CheckMate model consists of a restricted set of the larger Simulink-Stateflow system. However the way the same system is represented, is in a different format. The first step to translation lies in transforming the Stateflow part of the SSF model into a combination of a state machine and a SCSB model in the CheckMate system. The SCSB model is a mathematical equivalent to the switch in the real time system. The second step then translates each guard condition on outgoing transition edges corresponding to a state into a PTB block. Figure 12 and 13 show the ACC model in CheckMate. The diagram shows the PTHB modules used in modeling all the state transitions. The reader is encouraged to go through reference that illustrates the formal verification approach to the validation of MEMS based hybrid systems.\n\nThe specification of the ACC system can then be validated in the SSF model through time domain simulation method. These specifications can be captured as a set of properties, which we then formally verify in CheckMate. The sensor model in SSF uses several continuous time domain dynamic components that do not belong to the set of dynamic components allowed by CheckMate. This along with the multiple time step requirements leads one to explore an alternative method to integrate a MEMS sensor into the hybrid system.\n\n### 8.3. Integrating sensors\n\nLet us now devote ourselves to the task of integrating a component like a MEMS sensor. We have already discussed in Section 8.1 our reasons for searching for an alternative method to integrate real time sensors. In our case this is the MEMS gyroscope model. The standard SSF model of a gyroscope uses several continuous time domain dynamic components which do not belong to the set allowed by CheckMate. There are two methods possible to approach this problem. First method shown in Figure 14 is to build an independent interrupt driven discrete time solver solution.\n\nThe second method consists of characterising the time domain behaviour of the sensor and capturing this behaviour as an empirical model. The empirical model can be implemented as a simple look-up-table (LUT) which can be easily interpolated (or extrapolated) as per the input conditions in the field. The data-points for the LUT can be obtained for a range of velocity values by carrying out dynamic simulation on an exact macro-model of the MEMS gyroscope in the SSF framework (Figure 7). To integrate this empirical LUT model of the MEMS gyroscope, it is necessary to make changes in its implementation code. We can access the LUT model through function calls in CheckMate to get the desired outputs and thereby obtain the continuous time trajectories needed for formal analysis. For non-sensor based systems, where there are no multiple time step requirements, we can also use a simulation approach and integrate it with the main system solver.\n\nReaders should note that an LUT based approach will have a limited reach as a solution approach. It is clear that this approach cannot be used for general hybrid systems having dynamic components described with a system of strongly non-linear differential-algebraic equations, as in analog mixed signal design blocks. We can also look at using the exact simulation traces available from the general model in SSF. However they cannot be applied to real time sensor models due to reasons discussed earlier. For other real time components, the exact simulation traces enable the Exploreand the Verifyphase to construct accurate flow-pipes, and generate better approximations to these flow-pipes. This is illustrated in Figure 15. The method provides an alternative path to choose between a Simulink or a formal Checkmate model used in the computation. However for simplicity sake, let us restrict ourselves to the LUT based approach.\n\nFigure 15 also includes the LUT macro-model between the switched dynamic block and the PTHB modules. A restriction imposed by CheckMate is that for formal analysis it assumes a hybrid system to be closed. In other words it needs to be a closed loop system. The ACC system model can easily be seen to be open with respect to the velocity of the leading vehicle VL. This is because the control actions in the hybrid automata of the ACC system, depends on the behaviour of the leading vehicle resulting from changes in its velocity VL. Regular SSF framework allows modelling of an open hybrid systems. However, it needs some effort to model this in CheckMate. We can model such a scenario by addition of a redundant equation in terms of VL, VT and proximity (Xp) in which we render VL as a parameter (VT and Xp as the closed system state variables).The equation used is,R=(VLVT)dt=Xp.\n\n### 8.4. Verifying the specification properties\n\nLet us now look at some of the controller specifications to be formally analysed:\n\nP1: The tracking vehicle should never retard above Xcru.\n\nP2: The tracking vehicle should never accelerate when Xp< Xhalt.\n\nP3: The tracking vehicle should not cruise when Xp< Xcru.\n\nP4: The value of proximity Xp in all states will be always greater than 0.\n\nP5: For Xp< Xhalt tracking vehicle is always in the HALT state.\n\nP6: When Xp> Xcru tracking vehicle never goes to the HALT state.\n\nP7: When V > Vm and Xp> Xcru and tracking vehicle is always in the CRUISE state.\n\nThe ACC model without the MEMS gyroscope block can be easily verified using CheckMate. Adding the sensor block causes CheckMate to report non-compliance. In the STG (State Transition Graph) of the hybrid automata, five states have been used. The output values of the state-space variables from a previous state become the initial values for the next state continuous dynamics. To get the velocity value of the tracking vehicle to be zero before entering the halt state, we can add an intermediate state where the velocity is brought down to zero. The properties can be represented through ACTL. The same properties can also be verified by simulation. These results are shown in Figure 16.\n\n## 9. Conclusions\n\nIn this chapter we give a brief introduction the reader on the basics of formal verification of hybrid systems, which are primarily targeted for automotive systems. CheckMate based approach has been introduced to explore other available hybrid system design and verification frameworks for analysing hybrid system implementation. The reader is then walked through a case study of an adaptive cruise control system, which contains a MEMS based velocity sensor. After reading this chapter, the reader should be able to plan a hybrid system formal analysis platform and apply various methods to integrate complex sensors and other complex real time components.\n\nchapter PDF\nCitations in RIS format\nCitations in bibtex format\n\n## How to cite and reference\n\n### Cite this chapter Copy to clipboard\n\nJairam Sukumar, Subir K Roy, Kusum Lata and Navakanta Bhat (January 1st 2010). Formal Verification of Hybrid Automotive Systems, Motion Control, Federico Casolo, IntechOpen, DOI: 10.5772/6963. Available from:\n\n### Related Content\n\n#### Motion Control\n\nEdited by Federico Casolo\n\nNext chapter\n\n#### Rolling Stability Control of In-wheel Motor Electric Vehicle Based on Disturbance Observer\n\nBy Kiyotaka Kawashima, Toshiyuki Uchida and Yoichi Hori\n\nFirst chapter\n\n#### Introduction to Infrared Spectroscopy\n\nBy Theophile Theophanides\n\nWe are IntechOpen, the world's leading publisher of Open Access books. Built by scientists, for scientists. Our readership spans scientists, professors, researchers, librarians, and students, as well as business professionals. We share our knowledge and peer-reveiwed research papers with libraries, scientific and engineering societies, and also work with corporate R&D departments and government entities."
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http://en.wikibedia.ru/wiki/Cooperativity | [
"# Cooperativity\n\nCooperativity is a phenomenon displayed by systems involving identical or near-identical elements, which act dependently of each other, relative to a hypothetical standard non-interacting system in which the individual elements are acting independently. One manifestation of this is enzymes or receptors that have multiple binding sites where the affinity of the binding sites for a ligand is apparently increased, positive cooperativity, or decreased, negative cooperativity, upon the binding of a ligand to a binding site. For example, when an oxygen atom binds to one of hemoglobin's four binding sites, the affinity to oxygen of the three remaining available binding sites increases; i.e. oxygen is more likely to bind to a hemoglobin bound to one oxygen than to an unbound hemoglobin. This is referred to as cooperative binding.\n\nWe also see cooperativity in large chain molecules made of many identical (or nearly identical) subunits (such as DNA, proteins, and phospholipids), when such molecules undergo phase transitions such as melting, unfolding or unwinding. This is referred to as subunit cooperativity. However, the definition of cooperativity based on apparent increase or decrease in affinity to successive ligand binding steps is problematic, as the concept of \"energy\" must always be defined relative to a standard state. When we say that the affinity is increased upon binding of one ligand, it is empirically unclear what we mean since a non-cooperative binding curve is required to rigorously define binding energy and hence also affinity. A much more general and useful definition of positive cooperativity is: A process involving multiple identical incremental steps, in which intermediate states are statistically underrepresented relative to a hypothetical standard system (null hypothesis) where the steps occur independently of each other.\n\nLikewise, a definition of negative cooperativity would be a process involving multiple identical incremental steps, in which the intermediate states are overrepresented relative to a hypothetical standard state in which individual steps occur independently. These latter definitions for positive and negative cooperativity easily encompass all processes which we call \"cooperative\", including conformational transitions in large molecules (such as proteins) and even psychological phenomena of large numbers of people (which can act independently of each other, or in a co-operative fashion).\n\n## Cooperative binding[]\n\nWhen a substrate binds to one enzymatic subunit, the rest of the subunits are stimulated and become active. Ligands can either have positive cooperativity, negative cooperativity, or non-cooperativity.",
null,
"The sigmoidal shape of hemoglobin's oxygen-dissociation curve results from cooperative binding of oxygen to hemoglobin.\n\nAn example of positive cooperativity is the binding of oxygen to hemoglobin. One oxygen molecule can bind to the ferrous iron of a heme molecule in each of the four chains of a hemoglobin molecule. Deoxy-hemoglobin has a relatively low affinity for oxygen, but when one molecule binds to a single heme, the oxygen affinity increases, allowing the second molecule to bind more easily, and the third and fourth even more easily. The oxygen affinity of 3-oxy-hemoglobin is ~300 times greater than that of deoxy-hemoglobin. This behavior leads the affinity curve of hemoglobin to be sigmoidal, rather than hyperbolic as with the monomeric myoglobin. By the same process, the ability for hemoglobin to lose oxygen increases as fewer oxygen molecules are bound. See also Oxygen-hemoglobin dissociation curve.\n\nNegative cooperativity means that the opposite will be true; as ligands bind to the protein, the protein's affinity for the ligand will decrease, i.e. it becomes less likely for the ligand to bind to the protein. An example of this occurring is the relationship between glyceraldehyde-3-phosphate and the enzyme glyceraldehyde-3-phosphate dehydrogenase.\n\nHomotropic cooperativity refers to the fact that the molecule causing the cooperativity is the one that will be affected by it. Heterotropic cooperativity is where a third party substance causes the change in affinity. Homotropic or heterotropic cooperativity could be of both positives as well as negative types depend upon whether it support or oppose further binding of the ligand molecules to the enzymes.\n\n## Subunit cooperativity[]\n\nCooperativity is not only a phenomenon of ligand binding, but also applies anytime energetic interactions make it easier or more difficult for something to happen involving multiple units as opposed to with single units. (That is, easier or more difficult compared with what is expected when only accounting for the addition of multiple units). For example, unwinding of DNA involves cooperativity: Portions of DNA must unwind in order for DNA to carry out replication, transcription and recombination. Positive cooperativity among adjacent DNA nucleotides makes it easier to unwind a whole group of adjacent nucleotides than it is to unwind the same number of nucleotides spread out along the DNA chain. The cooperative unit size is the number of adjacent bases that tend to unwind as a single unit due to the effects of positive cooperativity. This phenomenon applies to other types of chain molecules as well, such as the folding and unfolding of proteins and in the \"melting\" of phospholipid chains that make up the membranes of cells. Subunit cooperativity is measured on the relative scale known as Hill's Constant.\n\n## Hill equation[]\n\nA simple and widely used model for molecular interactions is the Hill equation, which provides a way to quantify cooperative binding by describing the fraction of saturated ligand binding sites as a function of the ligand concentration.\n\n## Hill coefficient[]\n\nThe Hill coefficient is a measure of ultrasensitivity (i.e. how steep is the response curve).\n\nFrom an operational point of view the Hill coefficient can be calculated as:\n\n$n_{H}={\\frac {\\log(81)}{\\ce {\\log(EC90/EC10)}}}$",
null,
".\n\nwhere ${\\ce {EC90}}$",
null,
"and ${\\ce {EC10}}$",
null,
"are the input values needed to produce the 10% and 90% of the maximal response, respectively.\n\n## Response coefficient[]\n\nGlobal sensitivity measure such as Hill coefficient do not characterise the local behaviours of the s-shaped curves. Instead, these features are well captured by the response coefficient measure defined as:\n\n$R(x)={\\frac {x}{y}}{\\frac {dy}{dx}}$",
null,
"## Link between Hill coefficient and response coefficient[]\n\nAltszyler et al. (2017) have shown that these ultrasensitivity measures can be linked by the following equation:\n\n$n_{H}=2{\\frac {\\int _{\\ce {\\log(EC10)}}^{\\ce {\\log(EC90)}}R_{f}(I)d(\\log I)}{\\ce {\\log(EC90)-\\log(EC10)}}}=2\\langle R_{f}\\rangle _{\\ce {EC10,EC90}}$",
null,
"where $\\langle X\\rangle _{a,b}$",
null,
"denoted the mean value of the variable x over the range [a,b].\n\n## Ultrasensitivity in function composition[]\n\nConsider two coupled ultrasensitive modules, disregarding effects of sequestration of molecular components between layers. In this case, the expression for the system's dose-response curve, F, results from the mathematical composition of the functions, $f_{i}$",
null,
", which describe the input/output relationship of isolated modules $i=1,2$",
null,
":\n\n$F(I_{1})=f_{2}{\\big (}f_{1}(I_{1}){\\big )}$",
null,
"Brown et al. (1997) have shown that the local ultrasensitivity of the different layers combines multiplicatively:\n\n$R(x)=R_{2}(f_{1}(x)).R_{1}(x)$",
null,
".\n\nIn connection with this result, Ferrell et al. (1997) showed, for Hill-type modules, that the overall cascade global ultrasensitivity had to be less than or equal to the product of the global ultrasensitivity estimations of each cascade's layer,\n\n$n\\leq n_{1}.n_{2}$",
null,
",\n\nwhere $n_{1}$",
null,
"and $n_{2}$",
null,
"are the Hill coefficient of modules 1 and 2 respectively.\n\nAltszyler et al. (2017) have shown that the cascade's global ultrasensitivity can be analytically calculated:\n\n$n=2\\overbrace {\\langle R_{2}\\rangle _{X10_{2},X90_{2}}} ^{\\nu _{2}}\\overbrace {\\langle R_{1}\\rangle _{X10_{1},X90_{1}}} ^{\\nu _{1}}=2\\,\\nu _{2}\\,\\nu _{1}$",
null,
"where $X10_{i}$",
null,
"and $X90_{i}$",
null,
"delimited the Hill input's working range of the composite system, i.e. the input values for the i-layer so that the last layer (corresponding to $i=2$",
null,
"in this case) reached the 10% and 90% of it maximal output level. It followed this equation that the system's Hill coefficient n could be written as the product of two factors, $\\nu _{1}$",
null,
"and $\\nu _{2}$",
null,
", which characterized local average sensitivities over the relevant input region for each layer: $[X10_{i},X90_{i}]$",
null,
", with $i=1,2$",
null,
"in this case.\n\nFor the more general case of a cascade of N modules, the Hill coefficient can be expressed as:\n\n$n=\\nu _{N}\\,\\nu _{N-1}...\\nu _{1}$",
null,
",\n\n### Supramultiplicativity[]\n\nSeveral authors have reported the existence of supramultiplicative behavior in signaling cascades (i.e. the ultrasensitivity of the combination of layers is higher than the product of individual ultrasensitivities), but in many cases the ultimate origin of supramultiplicativity remained elusive. Altszyler et al. (2017) framework naturally suggested a general scenario where supramultiplicative behavior could take place. This could occur when, for a given module, the corresponding Hill's input working range was located in an input region with local ultrasensitivities higher than the global ultrasensitivity of the respective dose-response curve."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9255134,"math_prob":0.9729561,"size":8483,"snap":"2020-45-2020-50","text_gpt3_token_len":1694,"char_repetition_ratio":0.13834178,"word_repetition_ratio":0.013302035,"special_character_ratio":0.18472238,"punctuation_ratio":0.0900277,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9845289,"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],"im_url_duplicate_count":[null,5,null,8,null,null,null,null,null,8,null,8,null,8,null,null,null,null,null,8,null,8,null,8,null,null,null,null,null,8,null,8,null,8,null,null,null,null,null,null,null,8,null,null,null,8,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-24T08:15:08Z\",\"WARC-Record-ID\":\"<urn:uuid:56a95556-0eb2-414a-9401-3a2fc2db9dec>\",\"Content-Length\":\"84416\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bd2dda59-8071-4ce6-b0ee-7ccddc2478f6>\",\"WARC-Concurrent-To\":\"<urn:uuid:b66ec52d-f1d0-4441-a45c-188125ddfc4f>\",\"WARC-IP-Address\":\"46.17.44.51\",\"WARC-Target-URI\":\"http://en.wikibedia.ru/wiki/Cooperativity\",\"WARC-Payload-Digest\":\"sha1:JCK7QJWB3FJOWJVBAUDRDZ6SOMKENVMD\",\"WARC-Block-Digest\":\"sha1:6F35446ZSUFNON2QAYVCJLEZWMLXLRMN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107882103.34_warc_CC-MAIN-20201024080855-20201024110855-00616.warc.gz\"}"} |
https://ccrma.stanford.edu/~jos/sasp/Cyclic_FFT_Convolution.html | [
"Next | Prev | Up | Top | Index | JOS Index | JOS Pubs | JOS Home | Search\n\n### Cyclic FFTConvolution\n\nThanks to the convolution theorem, we have two alternate ways to perform cyclic convolution in practice:\n\n1. Direct calculation in the time domain using (8.13)\n2. Frequency-domain convolution:\n1. Fourier Transform both signals\n2. Perform term by term multiplication of the transformed signals\n3. Inverse transform the result to get back to the time domain\nFor short convolutions (less than a hundred samples or so), method 1 is usually faster. However, for longer convolutions, method 2 is ultimately faster. This is because the computational complexity of direct cyclic convolution of two",
null,
"-point signals is",
null,
", while that of FFT convolution is",
null,
". More precisely, direct cyclic convolution requires",
null,
"multiplies and",
null,
"additions, while the exact FFT numbers depend on the particular FFT algorithm used [80,66,224,277]. Some specific cases are compared in §8.1.4 below.\n\nNext | Prev | Up | Top | Index | JOS Index | JOS Pubs | JOS Home | Search\n\n[How to cite this work] [Order a printed hardcopy] [Comment on this page via email]"
] | [
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"https://ccrma.stanford.edu/~jos/sasp/img60.png",
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"https://ccrma.stanford.edu/~jos/sasp/img1320.png",
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"https://ccrma.stanford.edu/~jos/sasp/img1321.png",
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"https://ccrma.stanford.edu/~jos/sasp/img1322.png",
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"https://ccrma.stanford.edu/~jos/sasp/img1323.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.82390714,"math_prob":0.919941,"size":1086,"snap":"2022-27-2022-33","text_gpt3_token_len":262,"char_repetition_ratio":0.13585952,"word_repetition_ratio":0.17486338,"special_character_ratio":0.23941068,"punctuation_ratio":0.10471204,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9783672,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,7,null,6,null,8,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-20T06:07:02Z\",\"WARC-Record-ID\":\"<urn:uuid:6aa23614-d91d-4c65-a1ec-7029fd9de08f>\",\"Content-Length\":\"9432\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9a00c0a1-c5e0-47ee-ae63-4ef8da89f284>\",\"WARC-Concurrent-To\":\"<urn:uuid:c618696a-ad06-44ee-8bf7-d35458c26498>\",\"WARC-IP-Address\":\"171.64.197.141\",\"WARC-Target-URI\":\"https://ccrma.stanford.edu/~jos/sasp/Cyclic_FFT_Convolution.html\",\"WARC-Payload-Digest\":\"sha1:RNMVYGR6PBL3EXNQNLYOHZ7I6SY5U4N5\",\"WARC-Block-Digest\":\"sha1:3TJSURSEC74NPRR5T2DFZ5XSGVOTNDVW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882573908.30_warc_CC-MAIN-20220820043108-20220820073108-00455.warc.gz\"}"} |
https://www.vimgolf.com/challenges/5bf26adb9a198b0009ce0a87/user/awareofnow | [
"### Real Vim ninjas count every keystroke - do you?\n\n###### Pick a challenge, fire up Vim, and show us what you got.\n\n```Your VimGolf key: please sign in\n\n\\$ gem install vimgolf\n\\$ vimgolf setup\n```\n\n### Simple format (2)\n\ntry again!\n\n##### Start file\n```a==b equal to\na!=b not equal to\na>b greater than\na>=b greater than or equal to\na<b less than\na<=b less than or equal to\n```\n##### End file\n```\ta == b\t\tequal to\na != b\t\tnot equal to\na > b\t\tgreater than\na >= b\t\tgreater than or equal to\na < b\t\tless than\na <= b\t\tless than or equal to\n```\n\n#### View Diff\n\n```1,6c1,6\n< a==b equal to\n< a!=b not equal to\n< a>b greater than\n< a>=b greater than or equal to\n< a<b less than\n< a<=b less than or equal to\n---\n> \ta == b\t\tequal to\n> \ta != b\t\tnot equal to\n> \ta > b\t\tgreater than\n> \ta >= b\t\tgreater than or equal to\n> \ta < b\t\tless than\n> \ta <= b\t\tless than or equal to\n```\n\n### Solutions by @awareofnow:\n\nUnlock 2 remaining solutions by signing in and submitting your own entry\n\n## 68 active golfers, 328 entries\n\n##### Solutions by @awareofnow:",
null,
"26\n###### #28 - Adam Hartz / @awareofnow\n\n03/10/2020 at 01:52AM",
null,
"28\n###### #>28 - Adam Hartz / @awareofnow\n\n03/10/2020 at 01:25AM"
] | [
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"data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAADAAAAAwAQMAAABtzGvEAAAABlBMVEUAAAD3 fvuOP6N2AAAAAnRSTlMA/1uRIrUAAAA/SURBVHicY/rDf4D5gz0TAxhQiWL/ YP+Hv5EoDR8ZGQWIMvMP////xLmTheEgEBFnu8Nf4mz/K3+A+aEDESoBP7wS I9Hd2+8AAAAASUVORK5CYII= ",
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"data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAADAAAAAwAQMAAABtzGvEAAAABlBMVEUAAAD3 fvuOP6N2AAAAAnRSTlMA/1uRIrUAAAA/SURBVHicY/rDf4D5gz0TAxhQiWL/ YP+Hv5EoDR8ZGQWIMvMP////xLmTheEgEBFnu8Nf4mz/K3+A+aEDESoBP7wS I9Hd2+8AAAAASUVORK5CYII= ",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8796991,"math_prob":0.99368036,"size":893,"snap":"2023-40-2023-50","text_gpt3_token_len":286,"char_repetition_ratio":0.24971879,"word_repetition_ratio":0.20833333,"special_character_ratio":0.35834268,"punctuation_ratio":0.065,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9839647,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-23T11:36:25Z\",\"WARC-Record-ID\":\"<urn:uuid:1b46cdca-7564-4fa1-b3f1-1ffa1b0a163e>\",\"Content-Length\":\"7755\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:58078c95-56cc-4900-8a77-bd172c6c1ed6>\",\"WARC-Concurrent-To\":\"<urn:uuid:7aec8a12-bc91-4851-ae9b-93aa5cd13569>\",\"WARC-IP-Address\":\"104.21.37.129\",\"WARC-Target-URI\":\"https://www.vimgolf.com/challenges/5bf26adb9a198b0009ce0a87/user/awareofnow\",\"WARC-Payload-Digest\":\"sha1:BVZBUY7HBZ6CFGRQCIDQEDBMAWJ5ZLSG\",\"WARC-Block-Digest\":\"sha1:HI4WDVRAVJV477QIO5PAHW73BP442POC\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506480.7_warc_CC-MAIN-20230923094750-20230923124750-00693.warc.gz\"}"} |
https://www.mathexpression.com/evaluate-54521222.html | [
"# Evaluate 5*4+(5+2)+12-2^2\n\nby A.Y.\n(USA)\n\n##### Question\n5*4+(5+2)+12-2^2\n\nWell to figure this out you have to know PEMDAS\n\n(P)arentheses\n(E)ponents\n(M/D)Multiplication/Divison\n\n*For the M/D A/S which ever comes first in the problem you do first.\n\nWell parentheses are first so:(5+2=7)\n\nNext is exponents:2^2=4\n\nThen multiplication: 5*4=20"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8191784,"math_prob":0.66293263,"size":423,"snap":"2021-43-2021-49","text_gpt3_token_len":156,"char_repetition_ratio":0.100238666,"word_repetition_ratio":0.0,"special_character_ratio":0.34042552,"punctuation_ratio":0.07070707,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9935888,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-19T09:24:06Z\",\"WARC-Record-ID\":\"<urn:uuid:6db3ea41-593b-42bd-949c-17d18c7a7032>\",\"Content-Length\":\"39451\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ceed5c81-4294-4195-ab27-2c1cf92626a6>\",\"WARC-Concurrent-To\":\"<urn:uuid:e445b063-8f7c-4b2f-85e6-87339ff77100>\",\"WARC-IP-Address\":\"173.247.219.12\",\"WARC-Target-URI\":\"https://www.mathexpression.com/evaluate-54521222.html\",\"WARC-Payload-Digest\":\"sha1:HZYUXADB2YDMX5TC53VGF7HSOR377CUU\",\"WARC-Block-Digest\":\"sha1:X56LYKMZFKILQE4UT66VFHUICYMDHPW5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585246.50_warc_CC-MAIN-20211019074128-20211019104128-00493.warc.gz\"}"} |
https://etsolar.info/elektrolisis-larutan-kalium-iodida-30/ | [
"# ELEKTROLISIS LARUTAN KALIUM IODIDA PDF\n\nELEKTROLISIS. Buatlah rangkaian dan laporan percobaan elektrolisis dari larutan Kalium Iodida 0,1 M. Ambillah kira-kira 30 mL larutan Kalium Iodida 0,1 M /. Anyplace faeroese footmarks laporan praktikum kimia reaksi redoks dan elektrolisis larutan kalium iodida despite the lackadaisical pumice. Ramalkan hasil elektrolisis bagi leburan natrium klorida di anod dan di katod Larutan kalium iodida mengkonduksikan arus elektrik kerana ia.",
null,
"Author: Nemi Kekasa Country: South Sudan Language: English (Spanish) Genre: Science Published (Last): 17 November 2009 Pages: 290 PDF File Size: 20.2 Mb ePub File Size: 4.19 Mb ISBN: 992-8-39258-504-9 Downloads: 5989 Price: Free* [*Free Regsitration Required] Uploader: Mazukasa",
null,
"In previous lesson, we have discussed measurement, including error, consistency, accuracy and sensitivity. In this lesson, we will proceed to measuring instrument.",
null,
"In measuring instrument, we will cover vernier caliper, micrometer, stopwatch, ammeter, voltmeter and thermometer. In this lesson, we will focus on Vernier caliper, where we will discuss the sensitivity of vernier caliper, label of the parts of vernier caliper and function of the parts.\n\nWe will also learn how to take reading from a vernier caliper, including how to read the main scale and the vernier scale, and also how to determine the zero error from a vernier caliper. In order to use Vernier caliper to make measurement, placed the object in the inside jaw or outside jaw of the caliper and elekrrolisis slide the vernier scale until the object is held firmly between the jaws.\n\nKa,ium reading of the measurement is the sum of the reading of main scale and Vernier scale. For example, in the image above, the reading of the main scale is 4. Therefore, the reading is 4. In the very next slide, we are going to discuss how to take reading from the main scale and Vernier scale. Therefore, the reading of the main scale is 0. The reading of the vernier scale is indicated by the mark on the vernier scale coincides with any mark on the main scale. Each division on the vernier ioddia represents a length 0.\n\nTherefore, the reading of the vernier scale is 0. As we have learned before, a zero error arises when the measuring instrument does not start from exactly zero. The kwlium above shows the reading on a Vernier caliper when the jaw of the caliper is closed firmly. We can see that the reading of the main scale is zero and the reading on the vernier scale is also zero. In this case, we say the caliper has no zero error. The diagram above shows the reading on a vernier elekyrolisis when the jaw of the caliper is closed firmly.\n\nWe can see that the reading of the main scale does not start from zero. Instead, the reading is slightly higher than zero, which means the caliper is subject to positive zero error. The vernier scale registers a reading of 0. A positive zero error is read from the zero mark on the left hand side of the Vernier scale.\n\nWe can see that the reading of the main scale does not start from zero, too. Instead, the reading is slightly lower than zero, which means the caliper is subjected to negative zero error. The Vernier kaliuj registers a reading of 0. A lot of students lalium think that this is the zero error. Well, actually, this is not the negative zero error.\n\nEMCO UNIMAT SL PDF\n\nIn this case, the zero error larutzn In previous lesson, we have discussed the classification of physical quantities. In this lesson, we will continue our discussion on scientific notation and prefixes.\n\nUnder prefixes, we oarutan discuss how to use the scientific calculator to solve problems involving prefixes. We will also discuss the conversion of prefixes, including the conversion of the unit of area,volume, speed and density. Scientific notation also known kwlium Standard form is a convenient way to write very small or large numbers.\n\nThe number 10,, million can be further expressed as 10 10 10 to the power of This is much easier to be read. We can straight away tell its value just by a glance. In this notation, numbers are separated into two parts, a real number with an absolute value between 1 and 10 and an order of magnitude value written as a power of This method to write a number is called the scientific notation. Prefixes is a letter or a few letters placed in front of a word to modify its meaning.\n\nThe prefixes that we are going to discuss here is the SI prefixes, which is used in front of an SI unit. An SI unit used with prefixes indicate multiples or submultiples of a base unit. For example, kilometer — kilo is the prefix of the base unit, meter. The prefix act as a multiplier. In this case, kilo represents a multiple of The below table shows the list of prefixes that you need to know in SPM syllabus.\n\nA more complete list can be found in wikipedia. One of the question that student always ask is: You need to memorise all of these prefixes together with its standard form and symbol. They are Mega, kilo, deci, centi, milli and micro. If you own a Casio fxMS calculator, then I have a good news for you.\n\nMost of the prefixes in this table can be found in your calculator.",
null,
"We will discuss this in the later part of this presentation. We can convert a number with prefix to a normal number by multiplying the number by the value of the prefix.\n\nFor example, 23 MHz. The value of Mega is 1 million or 10 ellektrolisis.\n\n### Uncategorized – Page 2 –\n\nTherefore, we can rewrite the number as 23 x 10 6 Hz. In conclusion, we convert a number with prefix to a normal number by multiplying the number with the value of the prefix.\n\nWe can convert a normal number to a larutna with prefix by dividing the number by the value of the prefix. For example,Hz is equal to how many MHz? To write a normal number with a prefix, we divide the number by the value of the prefix and we get 0. In conclusion, we convert a normal number to a number with prefix through dividing the number by the power of the prefix.\n\nIf we want to convert a normal number to a number with prefix, we divide the number by the value of the prefix. If we want to convert a number ladutan prefix to a normal number, we multiply the number by the value of the prefix.\n\nLPKF PROTOMAT S63 PDF\n\nIf you own a casio fxMS calculator, you can do the conversion much easier. You can find the following prefixes in the calculator.\n\n## Uncategorized\n\nTera, Gega, Mega, kilo, mili, micro, nano, pico and fento. Take notes that deci and centi are not given. We are going to deal with this problem in the later slide. In the calculator, we press . Kilo is at [button 6]. Therefore, we press [shift] . You can see a small letter k shows on the screen here. The answer will immidiately shown on the screen. We can also convert a normal number to a number with prefix easily.\n\nIn previous slide, we have learned that if we want to convert a normal number to a number with prefix, we divide the number by the value of the prefix. For example, if we want to know 0. In our fxMS calculator, we press.",
null,
"For example, if we want to key in 0. I know for a lot of students when they did something wrong in the calculator, they will delete it by pressing the [on] button and redo everything again.\n\nThis is a very bad habit. By pressing the [on] button, you are not only delete everything on the screen, you will also delete everything in the replay memory in your calculator as well. Your calculator is a very sophisticated machine. It can remember whatever you perform in the calculator previously up to bytes of memory as long as you do not switch off the calculator or delete all the calculation by pressing the [on] button.\n\nThe correct way to delete only current calculation is using the [AC] button. You can delete everything on the screen by pressing the [AC] button. Pressing [AC] does not clear replay memory. You can always navigate to all the calculation you have performed by pressing the up or down arrow key. You can also move the cursor to the location you want by using the left or right arrow key. By doing so, you can check what is the mistake you have done and then fix it. So, make it a habit to use the [AC] button to delete but not the [on] button.\n\nWe have a syntax error.\n\nYou will find that the screen will immediately shows the last calculation we have performed. Well, I guess the problem is on the mili. Mili is a prefix. It is not a number."
] | [
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https://encyclopedia2.thefreedictionary.com/Analog+Modeling | [
"# Analog Modeling\n\n## Modeling, Analog\n\n(also analog simulation), one of the most important kinds of modeling; based on the analogy (more accurately, the isomorphism) of phenomena that differ in physical nature but are described by the same differential, algebraic, or any other mathematical equations.\n\nA simple example is two systems, the first of which is mechanical and consists of a shaft that transmits rotation through a spring to a spindle rigidly connected to a flywheel that is partially immersed in a viscous damping fluid. The second system is electrical and consists of a source of electromotive force that is connected through an inductance coil, a capacitor, and a resistor to a kilowatt-hour meter. If the values of inductance, capacitance, and resistance are chosen such that they correspond in a specific manner to the inertia of the flywheel, the elasticity of the spring, and the friction of the fluid, the systems then will display a structural and functional similarity (even an identity) that means, among other things, that they will be described by the same differential equation with constant coefficients of the type",
null,
"This equation may be a “theoretical model” for both systems, either of which may be the “experimental model” for this equation and the “analog model” for the other system. This analogy is the basis for the electrical modeling of mechanical systems: electrical models are far more convenient than mechanical systems for experimental study.\n\nAnother traditional area of application for analog modeling is the study of thermal conductivity based on electrothermal and hydrothermal analogies. In the first case the analogs of the temperature field and heat capacity in a solid body are the electric potential field in an electrically conducting medium and the capacitances of several capacitors, respectively; in the second case the temperature is simulated by the water level in vertical glass vessels that form the hydraulic model, the heat capacity of an elementary volume is simulated by the cross-sectional area of the vessels, and the thermal resistance is simulated by the hydraulic resistance of the tubes that connect the vessels.\n\nThe method of luminous simulation, in which the fluxes of thermal radiation are replaced by similar fluxes of luminous radiation, is often used to study radiant heat transfer. In this way the angular coefficients of the radiation are measured, and if the optical properties (degree of blackness and absorptivity) of the corresponding surfaces in the model and in nature are identical, the distribution of the thermal fluxes over the surfaces in the radiant heat-transfer system are also determined.\n\nBefore the development of digital computers in the late 1940’s, analog modeling was the main method of “object-mathematical modeling” for any processes dealing with the propagation of electromagnetic and acoustic waves, the diffusion of gases and fluids, the movement and filtration of fluids in porous media, and the torsion of rods (in view of this it was then often called simply mathematical modeling), and for each specific modeling problem a special “network” model was constructed (using various electrical impedance elements interconnected to form a two-dimensional circuit grid). Analog computers made possible the modeling of entire classes of similar problems. At the present time the importance of analog modeling has been greatly reduced, because modeling on digital computers has great advantages with respect to accuracy and versatility. For sufficiently specific and specialized problems, analog modeling has its advantages (the simplicity and consequent inexpensiveness of the technology). The two methods are also frequently used in combination.\n\nReferences in periodicals archive ?\nSo it's analog modeling, plus you can do a bit more stuff with it.\nSimulation modeling tool chosen for the original CENTAURUS CPN, the processes taking place energy sources to model the system using a programming analog modeling sub-system, management modeling using color petri nets sub-system.\nThis collection of 17 papers includes outcrop studies and case studies along with numerical and analog modeling studies and reviews of recent findings in the geophysics of fractured reservoirs.\n\nSite: Follow: Share:\nOpen / Close"
] | [
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"https://img.tfd.com/ggse/af/gsed_0001_0016_0_img4106.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.93694866,"math_prob":0.9413868,"size":3733,"snap":"2020-24-2020-29","text_gpt3_token_len":694,"char_repetition_ratio":0.13783856,"word_repetition_ratio":0.0,"special_character_ratio":0.17519422,"punctuation_ratio":0.07496013,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9837181,"pos_list":[0,1,2],"im_url_duplicate_count":[null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-09T11:28:15Z\",\"WARC-Record-ID\":\"<urn:uuid:e2e4ed62-f768-4890-a13b-fb97017169d9>\",\"Content-Length\":\"45374\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:32e5c86f-39d3-4b1d-a2f3-ee164d773f6d>\",\"WARC-Concurrent-To\":\"<urn:uuid:9a2bee37-e2c6-4f33-98cc-b74da981fc92>\",\"WARC-IP-Address\":\"91.204.210.226\",\"WARC-Target-URI\":\"https://encyclopedia2.thefreedictionary.com/Analog+Modeling\",\"WARC-Payload-Digest\":\"sha1:G6RIGCO6XBHQ2YXKQ7FASPJ3UYMK3PAW\",\"WARC-Block-Digest\":\"sha1:RQJOOJBLOG7EE5EO27SSNWPBQAKRPKVQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655899931.31_warc_CC-MAIN-20200709100539-20200709130539-00487.warc.gz\"}"} |
http://www.adras.com/Simple-regex-question.t20687-49.html | [
"From: Don Pich on 9 Mar 2010 09:31 I was wondering if there is a way to have perl check for a whole number? In an equation, if I have 16/4, it will come up with 4 and will set the \"check\" to true. If I have 16/5, it will come up with 3.2 and set the condition false. Is there a method to get this into an if/else condition? From: ccc31807 on 9 Mar 2010 09:45 On Mar 9, 9:31 am, Don Pich wrote:> I was wondering if there is a way to have perl check for a whole number? Depends on the context as to how you would implement it. If you want just digits, /[0-9]+/ will match one or more digits. Obviously you would want to match the particular substring you are looking for, since this RE would match any line that contained at least one digit. CC. From: Ben Morrow on 9 Mar 2010 09:58 Quoth Don Pich :> I was wondering if there is a way to have perl check for a whole number? int(\\$x) == \\$x or use POSIX qw/modf/; (modf \\$x) == 0; Ben From: sln on 9 Mar 2010 10:19 On Tue, 09 Mar 2010 08:31:48 -0600, Don Pich wrote: >I was wondering if there is a way to have perl check for a whole number? > >In an equation, if I have 16/4, it will come up with 4 and will set the >\"check\" to true. If I have 16/5, it will come up with 3.2 and set the >condition false. > >Is there a method to get this into an if/else condition? print \"whole 16/4\\n\" if !(16 % 4); print \"fraction 11/5\\n\" if 16 % 5; Whats regex have to do with it? From: Don Pich on 9 Mar 2010 10:54 Probably nothing. But not being 100% efficient with Perl, I wasn't sure. So let's add another twist to this. The reason I asked this question is that I have a script that is prompting a user for an IP address. I have them entering it one octet at a time. The first three octets should be fine with whatever is place in stin. But the forth octet will need to follow the subnet rules (i.e., the network address of a /30 would have the fourth octet as 0, 4, 8, 12, 16 etc. The network address of a /29 would have the fourth octet as 0, 8, 16, 24, 32). How would I apply the following code so that it will follow subnet rules? Here is the piece of the code that picks out the octets: _____CODE______ # Find out the IP address while (\\$OCT1_COUNT == 1) { print \"\\nWhat is the First octet? \"; \\$OCT1 = ; if ((\\$OCT1 < 0) || (\\$OCT1 > 255)) { print \"That number is not between 1 and 254.\\n\"; } else { chomp (\\$OCT1); \\$OCT1_COUNT++; } } while (\\$OCT2_COUNT == 1) { print \"\\nWhat is the Second octet? \"; \\$OCT2 = ; if ((\\$OCT1 < 0) || (\\$OCT1 > 255)) { print \"That number is not between 1 and 254.\\n\"; } else { chomp (\\$OCT2); \\$OCT2_COUNT++; } } while (\\$OCT3_COUNT == 1) { print \"\\nWhat is the Third octet? \"; \\$OCT3 = ; if ((\\$OCT1 < 0) || (\\$OCT1 > 255)) { print \"That number is not between 1 and 254.\\n\"; } else { chomp (\\$OCT3); \\$OCT3_COUNT++; } } while (\\$OCT4_COUNT == 1) { print \"\\nWhat is the Fourth octet? \"; \\$OCT4 = ; if ((\\$OCT1 < 0) || (\\$OCT1 > 255)) { print \"That number is not between 1 and 254.\\n\"; } else { chomp (\\$OCT4); \\$OCT4_COUNT++; } } _____/CODE_____ | Next | Last Pages: 1 2 Prev: Help on String to array !Next: FAQ 8.36 How do I fork a daemon process?"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.89438033,"math_prob":0.8598552,"size":3204,"snap":"2022-05-2022-21","text_gpt3_token_len":1083,"char_repetition_ratio":0.1096875,"word_repetition_ratio":0.3369906,"special_character_ratio":0.3764045,"punctuation_ratio":0.15363511,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9553343,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-20T18:08:15Z\",\"WARC-Record-ID\":\"<urn:uuid:0913aea2-f39b-467b-940e-ca84f7fe0691>\",\"Content-Length\":\"10513\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9f6dc7b9-d239-4669-9308-e7967a7d0fd5>\",\"WARC-Concurrent-To\":\"<urn:uuid:9f771529-c2f2-47d0-bb6e-ae0345e4ea51>\",\"WARC-IP-Address\":\"199.241.29.204\",\"WARC-Target-URI\":\"http://www.adras.com/Simple-regex-question.t20687-49.html\",\"WARC-Payload-Digest\":\"sha1:SVTHG5EWQPMNL7G4ANE47QEBVOE36LYC\",\"WARC-Block-Digest\":\"sha1:GOIORFEW2DVU57R7SPTOYJNWZFVG4AWS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320302355.97_warc_CC-MAIN-20220120160411-20220120190411-00100.warc.gz\"}"} |
https://www.wishfin.com/fixed-deposit-calculator/ | [
"FD Calculator - Calculate FD Interest Rates, Maturity Amount - Wishfin\n\n# Fixed Deposit Calculator\n\n### Highlights\n\n• FD calculator, a tool for interest rate and maturity period calculation\n• Read and know its features and benefits on FD investment\n\nFixed Deposit is an effective way to earn attractive returns on cash which was currently not used. Investment in a fixed deposit will make your funds work for your future, by grabbing a good interest rate. That is the reason people invest in fixed deposits for a safe and secure future, as it encourages the user to save for the future plan. It is among the most reliable sources to increase the power of money for the user’s future plans. And, the returns on these deposits are calculated through a FD Calculator.\n\nFD Calculator is an online tool that is designed to assist the user in calculating the return rate of the fixed deposit. The tool makes it easier for you to know the amount of interest you can earn on your deposit principal amount. Also, you can compare the amount of interest earned on different amounts at separate time intervals. It makes it easier for you to choose from the best plan to yield better returns. You can save your time and get the accurate result of your fixed deposit investment with the help of the calculator.\n\n## How a FD Calculator Works?\n\nIt is very easy to use a FD Calculator, you just have to insert the values such as the principal amount of deposit, duration and the rate of interest. Then select type of interest like Simple or Compound and choose the frequency. Once these values are inserted in the FD calculator, the result is automatically generated using the formula.\n\n### FD Interest Calculator Formula\n\nThe interest on the fixed deposit is calculated using the interest calculator formula mentioned below :\n\nA = P (1 + r/n) nt\n\nI = A – P\n\nWhere, A = is the maturity value of the investment with interest\n\nP = the principal investment amount or the initial deposit\n\nr = the annual rate of interest\n\nn = Interest rate compounded every single year\n\nt = Money invested in a fixed deposit for the chosen period\n\nI = Total interest earned\n\nIllustration:\n\nFor Example: If a fixed deposit ₹50,000 is booked for 10 years. Then enter the details of your fixed deposit select type of interest and its frequency. The maturity value of the deposit amount would be ₹1,00,484. From which ₹50,484 is the amount you have earned the interest rate.\n\nFixed Deposit Calculator\nInvestment Amount (in ₹) - 50,000Tenure - 10 years\nRate of Interest (%) - 7Interest Compounding Frequency - Monthly\nMaturity Amount (in ₹) - 1,00,484Total Interest Earned (in ₹) - 50,484\n\nTable Showing Interest Earned And Closing Balance\n\nYearFD Interest Rates (in % p.a.)Opening Balance (in ₹)Interest Earned (in ₹)Closing Balance (in ₹)\n17.0050,0003,61553,615\n27.0053,6153,87657,491\n37.0057,4914,15661,647\n47.0061,6474,45666,103\n57.0066,1034,77970,882\n67.0070,8825,12476,006\n77.0076,0065,49481,500\n87.0081,5005,89287,392\n97.0087,3926,31893,710\n107.0093,7106,7741,00,484\n\n## Methods of Interest Calculation\n\nThe interest on a fixed deposit is calculated using two methods which are as follows.\n\nSimple Interest – Simple Interest is the interest that is paid in percentage of the principal amount during the investment period. It is the easiest and quickest way to calculate the interest rate. The earned amount calculated using the simple interest will remain the same during fixed deposit maturity.\n\nCompound Interest – Compound Interest is the interest that is calculated as a percentage of the revised amount, which is the original principal plus the accumulated interest of previous periods. In this method the earned interest rate of the previous years is added to the initial principal, thus it increases the principal amount using which the next period interest rate can be calculated. In compound interest, you can receive interest on the principal as well as on the accrued interest during the fixed deposit tenure.\n\n## Difference Between Simple Interest and Compound Interest\n\nBasisSimple InterestCompound Interest\nMeaningInterest that is calculated on the principal amount is called as simple interestInterest that is calculated on principal and the accrued interest is called as compound interest\nPrincipalThe principal amount remains constant throughout the tenure of depositThe principal amount keeps on changing throughout the tenure of deposit\nGrowth in interestThe interest yield is same for the entire periodThe interest grows gradually during the tenure.\nReturnsLow in comparision to compound interestHigher than the simple interest\nFormulaA = P (1 + rt)A = P (1 + r/n) nt\n\nTo know the value of your interest earned on the fixed deposit use either method of interest calculation by using the specific FD Calculator.\n\n## Simple Interest FD Calculator\n\nA simple interest FD Calculator is the FD calculator that uses a simple interest formula. To generate the maturity and the total interest amount of the fixed deposit, you can take its help. The below mentioned formula is used to calculate the amount of interest.\n\n### Simple Interest Formula:\n\nA = P (1 + rt)\n\nWhere:\n\nA = Total Amount (Deposit amount + interest rate)\n\nP = Deposit Amount\n\nI = Interest Amount\n\nr = Rate of Interest per year in decimal ratio using r = R/100 formula\n\nR = Rate of Interest per year in percentage (R = r x 100)\n\nt = FD duration for months or year\n\n### How Simple Interest FD Calculator is used?\n\nTo calculate the interest on the fixed deposit at simple interest use the simple interest FD Calculator. And put the values in their relevant fields to get the correct result.\n\nEnter these three details of your fixed deposit to compute the total interest which is the\n\n• Investment Amount\n• Period of Investment\n• Rate of Interest\n\nSee the example below to understand the working of simple interest FD Calculator –\n\nAn investor booked a fixed deposit of ₹ 10,000 for a period of 10 years at a 7% interest rate p.a. And according to the calculations at the end of FD tenure, his deposit amount will be increased to ₹7,000.\n\nSimple Interest Fixed Deposit Calculator\nInvestment Amount (in ₹) - 10,000Tenure - 10 years\nRate of Interest (%) - 7\nMaturity Amount (in ₹) - 17,000Total Interest Earned (in ₹) - 7,000\n\nTable Showing Interest Earned And Closing Balance\n\nYearFD Interest Rates (in % p.a.)Opening Balance (in ₹)Interest Earned (in ₹)Closing Balance (in ₹)\n17.0010,00070010,700\n27.0010,70070011,400\n37.0011,40070012,100\n47.0012,10070012,800\n57.0012,80070013,500\n67.0013,50070014,200\n77.0014,20070014,900\n87.0014,90070015,600\n97.0015,60070016,300\n107.0016,30070017,000\n\n## Compound Interest FD Calculator\n\nThe compound interest FD Calculator is devised to calculate the maturity value and the interest accrued on the fixed deposit using the compound interest formula. The compound interest formula is mentioned below :\n\n### Compound Interest Formula :\n\nA = P (1 + r/n) nt\n\nWhere,\n\nA = is the FD maturity period with the interest rate\n\nP = Principal amount invested or the initial deposit of FD\n\nr = Annual interest rate\n\nn = Time period for which the interest is compounded\n\nt = Investment period for FD\n\n### Calculation of Interest Using Compound Interest FD Calculator\n\nTo calculate the interest on fixed deposit at the compounded interest rate use the compound interest calculator. For the compounded interest you need to enter the following details which are as follows.\n\n• Investment Amount\n• Tenure\n• Interest Rate\n• Interest Compounding Frequency\n\nLet’s try out with the help of an example –\n\nExample: A person invests an amount of ₹ 10,000 in the fixed deposit for a period of 10 years. The interest on the deposit is compounded monthly, and to find out the total interest he uses the FD Calculator. The computed maturity value of the deposit is ₹ 20,096.\n\nFixed Deposit Calculator\nInvestment Amount (in ₹) - 10,000Tenure - 10 years\nRate of Interest (%) - 7Interest Compounding Frequency – Monthly\nMaturity Amount (in ₹) - 20,096Total Interest Earned (in ₹) - 10,096\nYearFD Interest Rates (in % p.a.)Opening Balance (in ₹)Interest Earned (in ₹)Closing Balance (in ₹)\n17.0010,00072310,723\n27.0010,72377511,498\n37.0011,49883112,329\n47.0012,32989113,220\n57.0013,22095614,176\n67.0014,1761,02515,201\n77.0015,2011,09916,300\n87.0016,30070017,478\n97.0017,4781,26318,741\n107.0018,74170020,096\n\n### Benefits Of FD Calculator\n\n1. Reduces the manual efforts\n2. Provides accurate result\n3. Fast calculation\n4. Time-saving\n\nYou can calculate the maturity value of the various banks using their specific FD Calculators and tools such as ICICI FD Calculator, HDFC FD Calculator and many more respectively.\n\n#### People Also Look For\n\nPersonal Loan Interest Rates October 2023\nFullerton India12.00% - 24.00%\nHDFC Bank10.75% - 14.50%\nICICI Bank10.75% - 19.00%\nIndusInd Bank10.25% - 26.00%\nKotak Bank10.99%\nRBL14.00% - 23.00%\nStandard Chartered Bank11.49%\nTata Capital10.50% - 24.00%\nHome Loan Interest Rates October 2023\nAxis Bank8.75% - 9.15%\nBank of Baroda8.50% - 10.60%\nCitibank8.75% - 9.15%\nHDFC8.50% - 9.40%\nICICI Bank9.00% - 9.85%\nIndiabulls Housing Finance Limited8.65%\nKotak Bank8.85% - 9.40%\nLIC Housing8.50% - 10.50%\nPiramal Capital & Housing Finance10.50%\nPNB Housing Finance8.50% - 10.95%\nReliance Home Finance8.75% - 14.00%\nState Bank of India/SBI9.10% - 9.65%\nTata Capital8.95% - 12.00%"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8922032,"math_prob":0.97318375,"size":8816,"snap":"2023-40-2023-50","text_gpt3_token_len":2380,"char_repetition_ratio":0.21833862,"word_repetition_ratio":0.11420983,"special_character_ratio":0.30444646,"punctuation_ratio":0.1124498,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99725175,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-02T23:59:16Z\",\"WARC-Record-ID\":\"<urn:uuid:d9b27ce8-c3a1-4aff-9cbd-cbf1ac94c436>\",\"Content-Length\":\"157192\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:086da88f-1c7d-4288-8963-818529dfca5b>\",\"WARC-Concurrent-To\":\"<urn:uuid:9b1bdd42-b246-4cc4-b2a5-99234a6a5428>\",\"WARC-IP-Address\":\"104.22.62.68\",\"WARC-Target-URI\":\"https://www.wishfin.com/fixed-deposit-calculator/\",\"WARC-Payload-Digest\":\"sha1:CTH4ECE46PH4CBTZ3BTG4OXCIHLGQPT4\",\"WARC-Block-Digest\":\"sha1:YBVTJP2NUNEW2L6HZFOWEGPAZGVK6Q32\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511023.76_warc_CC-MAIN-20231002232712-20231003022712-00822.warc.gz\"}"} |
https://ch.mathworks.com/matlabcentral/fileexchange/22543-detects-multiple-disks-coins-in-an-image-using-hough-transform | [
"File Exchange\n\n## Detects multiple disks (coins) in an image using Hough Transform\n\nversion 1.5.0.0 (3.17 KB) by Randy Tang\n\n### Randy Tang (view profile)\n\nHOUGHCIRCLES detects multiple disks (coins) in an image using Hough Transform.\n\nUpdated 07 Jan 2009\n\nHOUGHCIRCLES detects multiple disks (coins) in an image using Hough Transform. The image contains separating, touching, or overlapping disks whose centers may be in or out of the image.\n\nSyntax\nhoughcircles(im, minR, maxR);\nhoughcircles(im, minR, maxR, thresh);\nhoughcircles(im, minR, maxR, thresh, delta);\ncircles = houghcircles(im, minR, maxR);\ncircles = houghcircles(im, minR, maxR, thresh);\ncircles = houghcircles(im, minR, maxR, thresh, delta);\n\nInputs:\n- im: input image\n- minR: minimal radius in pixels\n- maxR: maximal radius in pixels\n- thresh (optional): the minimal ratio of the number of detected edge pixels to 0.9 times the calculated circle perimeter (0<thresh<=1, default: 0.33)\n- delta (optional): the maximal difference between two circles for them to be considered as the same one (default: 12); e.g., c1=(x1 y1 r1), c2=(x2 y2 r2), delta = |x1-x2|+|y1-y2|+|r1-r2|\n\nOutput\n- circles: n-by-4 array of n circles; each circle is represented by (x y r t), where (x y), r, and t are the center coordinate, radius, and ratio of the detected portion to the circle perimeter, respectively. If the output argument is not specified, the original image will be displayed with the detected circles superimposed on it.\n\n### Cite As\n\nRandy Tang (2020). Detects multiple disks (coins) in an image using Hough Transform (https://www.mathworks.com/matlabcentral/fileexchange/22543-detects-multiple-disks-coins-in-an-image-using-hough-transform), MATLAB Central File Exchange. Retrieved .\n\nnithi\n\nty\n\nRaz Shimoni\n\n### Raz Shimoni (view profile)\n\nThank you very much.\n\nRandy Tang\n\n### Randy Tang (view profile)\n\nSushma,\nminR and maxR are the minimal and maximal radii (in pixels), respectively, of the circles that you want to detect in your image. For the image I provided along with the program, I set minR=20 and maxR=40. If you still have problems, you may want to provide your image and I'll look into the problem.\n\nYuan-Liang Tang\n\nSushma Bhandari\n\n### Sushma Bhandari (view profile)\n\nim unable to detect the circles.what should be the minR and mixR,can anyone help me?\n\nChristoph\n\n### Christoph (view profile)\n\nNazatul Naquiah Ahba\n\n### Nazatul Naquiah Ahba (view profile)\n\nhi, i've ran this code and it works. i wonder which part of the code display as accumulation array? can you pls guide me how to create the accumulation array from this code in order to generate the output figure of hough transform accumulation array?\n\nVenugopalakrishna\n\n### Venugopalakrishna (view profile)\n\nThank you very much.\n\nTUYEN Nguyen Ba\n\n### TUYEN Nguyen Ba (view profile)\n\nThank you Prof. Yuan Liang Tang. This is really great!\n\nRandy Tang\n\n### Randy Tang (view profile)\n\nsanjay,\nThe most probable cause might be that you invoked the function using inappropriate parameters. The function allocate a 3D matrix of size [size(im,1)+maxR, size(im,2)+maxR, maxR-minR+1], where minR and maxR are the minimal and maximal radii of circles you want to detect, respectively. You may want to check if the size of your input image or the values of minR and maxR are really huge.\n\nsanjay bhattacharya\n\n### sanjay bhattacharya (view profile)\n\ni used the houhcircles function. it says 'out of memory' for my test images; but it worked for very small binary images; please help\n\nRezki Al Khairi\n\n### Rezki Al Khairi (view profile)\n\ni've try to make with GUI...\nbut i cannot place where the instrustion mace be placed...\ni really need it..\n\nIdillus\n\n### Idillus (view profile)\n\nif (nargin >=3 || nargin <= 6)\n\nif nargin==3\nthresh = 0.33; % One third of the perimeter\ndelta = 12; % Each element in (x y r) may deviate approx. 4 pixels\nedgeim = edge(im, 'canny', [0.15 0.2]);\nend\n\nif nargin==4\nedgeim = edge(im, 'canny', [0.15 0.2]);\ndelta = 12;\nend\n\nif (nargin==5)\nedgeim = edge(im, 'canny', canny_th);\ndelta = 12;\nend\n\nif (nargin==6)\nedgeim = edge(im, 'canny', canny_th,sigma);\ndelta=12;\nend\nif (nargin == 7)\nedgeim = edge(im, 'canny', canny_th,sigma);\nend\nend\n\nif minR<0 || maxR<0 || minR>maxR || thresh<0 || thresh>1 || delta<0 || canny_th <0 || canny_th >1\ndisp('Input conditions: 0<minR, 0<maxR, minR<=maxR, 0<thresh<=1, 0<delta');\nreturn;\nend\n\nIdillus\n\n### Idillus (view profile)\n\nHi, nice function. I've made some changes that may be interesting. It wold be nice to change the threshold and sigma value for canny edge detection, so, I added a few lines to improve this, changing the location of the definition of the edgeimage\n\nif nargin==3\n\nthresh = 0.33; % One third of the perimeter\ndelta = 12; % Each element in (x y r) may deviate approx. 4 pixels\nedgeim = edge(im, 'canny', [0.15 0.2]);\n\nelseif nargin==4\n\nif ((max(size(canny_th) == 2)) || (max(size(canny_th) == 1)))\nif max(size(canny_th) == 2)\nedgeim = edge(im, 'canny', [canny_th(1) canny_th(2)]);\nend\nif max(size(canny_th) == 1)\nedgeim = edge(im, 'canny', canny_th(1));\nend\nend\ndelta = 12;\n\nelse (nargin==5)\n\nif ((max(size(canny_th) == 2)) || (max(size(canny_th) == 1)))\nif (max(size(canny_th) == 2))\nedgeim = edge(im, 'canny', [canny_th(1) canny_th(2)],sigma);\nend\nif (max(size(canny_th) == 1))\nedgeim = edge(im, 'canny', canny_th(1),sigma);\nend\nend\ndelta = 12;\nend\n\nSven\n\n### Sven (view profile)\n\nNice function. Code is well documented and clearly written.\nOne suggestion is to follow the example of the peaks() function: if no output argument is given, then create a figure and display the image. If it's used with an output argument, assume the user is embedding the function in their own code, and doesn't want a figure to come up automatically."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.75482154,"math_prob":0.94381386,"size":5690,"snap":"2020-10-2020-16","text_gpt3_token_len":1662,"char_repetition_ratio":0.13647555,"word_repetition_ratio":0.12290503,"special_character_ratio":0.28523725,"punctuation_ratio":0.17406143,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9893158,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-18T02:30:10Z\",\"WARC-Record-ID\":\"<urn:uuid:d977e0a4-ac6b-49de-8f15-b870b0bb1365>\",\"Content-Length\":\"107992\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b7fb936a-b2cd-4bbe-bac1-80a93f5927f0>\",\"WARC-Concurrent-To\":\"<urn:uuid:254d27b2-3323-4163-9bd4-09cdf0788e07>\",\"WARC-IP-Address\":\"104.110.193.39\",\"WARC-Target-URI\":\"https://ch.mathworks.com/matlabcentral/fileexchange/22543-detects-multiple-disks-coins-in-an-image-using-hough-transform\",\"WARC-Payload-Digest\":\"sha1:PR66JYNCIJNK3JLM5QPXOKWXFN7TLBFP\",\"WARC-Block-Digest\":\"sha1:HWICM3K4IH37VZJJZR36P4LBK2YYVB6F\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875143455.25_warc_CC-MAIN-20200217235417-20200218025417-00432.warc.gz\"}"} |
https://answers.everydaycalculation.com/add-fractions/90-20-plus-24-8 | [
"Solutions by everydaycalculation.com\n\n1st number: 4 10/20, 2nd number: 3 0/8\n\n90/20 + 24/8 is 15/2.\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 20 and 8 is 40\n2. For the 1st fraction, since 20 × 2 = 40,\n90/20 = 90 × 2/20 × 2 = 180/40\n3. Likewise, for the 2nd fraction, since 8 × 5 = 40,\n24/8 = 24 × 5/8 × 5 = 120/40\n180/40 + 120/40 = 180 + 120/40 = 300/40\n5. 300/40 simplified gives 15/2\n6. So, 90/20 + 24/8 = 15/2\nIn mixed form: 71/2\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:"
] | [
null,
"https://answers.everydaycalculation.com/mathstep-app-icon.png",
null
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https://dasforex.com/how-to-take-advantage-of-volatility-and-profit-from-it/ | [
"How To Take Advantage Of Volatility And Profit From It\n\nVolatility is the most common way to measure risk in the financial markets. While there are a plethora of methods, calculations and derivatives to calculate volatility, they are all trying to accomplish the same goal: What is the price of a security going to do in the future? Without a crystal ball, there’s no perfect answer, but let’s go through a few common ways that we can estimate future volatility.\n\nGenerally speaking, there are two types of volatility traders and investors use in an effort to understand risk – historical volatility and implied volatility. Each of these can be used in different ways for different types of trades. Today, we’re going to go through the basics of implied volatility, starting with the .\n\nFirst proposed as the Sigma Index in 1987, the VIX got its start in 1993 when the CBOE reported implied volatility in real time using at-the-money options data from the S&P 100. While updates have been made to the VIX over time, it’s used as a “fear gauge,” or a measure of the market’s expectations of future price action. As the VIX increases, the market expects more risk ahead.",
null,
"Take a look at the most recent recession in 2020 – where we saw the VIX climb as high as 85. At that price, the market is expecting extreme risk. Ultimately, the VIX is the industry standard to help traders and investors have a standardized view of market risk through implied volatility.\n\nThe VIX is useful because it can give us a hint at what the market is expecting since it is an example of implied volatility. Typically, IV is derived using an options pricing model, such as the Black–Scholes. Using these models, the theoretical value of an option can help guide us to the measurement of implied volatility at a particular point in time.\n\nTraders and investors can use IV to find attractive options trades to hedge, enter or exit a position, or to speculate on a future outcome in the market.\n\nHistorical Volatility\n\nWhile the VIX is a measure of implied volatility, there are many historical measures of volatility that can be useful. One common example is the beta coefficient. This is a historical calculation measured by taking the returns associated with a security and comparing the price action of the market over the same time period. A security with a beta less than 1 implies that the security is theoretically less volatile than the market as a whole. A security with a beta greater than 1 would be more volatile than the market.\n\nFor those that want to have a full picture of the risk of a security, the beta coefficient can help separate market risk from individual security risk. These types of measures can help you diversify properly with respect to your individual risk tolerance.\n\nA historical volatility calculation like beta gives you a basic understanding of what the price of a security has done in the past. While past performance is not indicative of future results, historical volatility calculations can be used to help measure risk and ultimately help determine if a security is right for you.\n\nThe Volatility Of The Volatility – VVIX\n\nTo further confuse new traders there is such a thing called the volatility of the volatility – AKA the VIX of the VIX (VVIX). No this is not an exercise in doublespeak. The VVIX is simply a measure of the change of volatility in the VIX volatility index. The VVIX is the VIX of the VIX, like the VIX is the VIX of stocks. OK, if you are not thoroughly confused by now then congrats because this stuff can get pretty mind-bending! Another way to put it is, the VVIX measures how rapidly volatility changes, and is, thus, a measure of the volatility of the index. Investors can use the VVIX and its derivatives to hedge against volatility swings on changes in the VIX options market. You can hedge the hedge!\n\nWhy You Shouldn’t Be Afraid Of Volatility\n\nAll told, volatility is just a measurement that can give you insight into the potential risk of a security. It’s important to remember that actual volatility is almost always less than implied volatility. No measure of risk is going to be totally accurate, anything can happen in the financial markets. Even so, volatility measurements can offer a clear view of risk the market expects."
] | [
null,
"https://dasforex.com/wp-content/uploads/2021/06/531e2b9ad8990f78168b138e18712f9c.jpeg",
null
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http://rightonphotography.com/yutteqh/1mdu0pw.php?page=python-input-vs-raw_input-02790c | [
"raw_input() – It reads the input or command and returns a string. In Python 2.6, the main function for this task is raw_input (in Python 3.0, it’s input()). Ở python 3 thì họ không dùng raw_input nữa mà dùng input tương đương với raw_input ở python 2 Lí do bảo mật chắc là eval từ input rồi . raw_input() was renamed to input() in Python 3.Another example method, to mix the prompt using print, if you need to make your code simpler. 1. raw_input() This method of Python reads the input line or string and also can read commands from users or data entered using the console. From version 3.x, this is replaced with Python input() function. Up until now, our programs were static. Python Input. The raw_input() function in Python 2 collects an input from a user. It always returns the input of the user without changes, i.e. NameError: name ‘raw_input’ is not defined. But it crashes when programmed to prompt for console input. Python 2 uses raw_input() to accept the input from the user whereas in python 3, raw_input has been replaced by input() function. raw_input ( ) : This function works in older version (like Python 2.x). User Input. There are two options for using terminals in Visual Studio Code when debugging: Option 1: Use the Visual Studio Code Terminal (integrated terminal) This video compares and contrast's the input and raw_input functions in python. Well, I don't know what tutorial you're going off from, but keep in mind the following: Python 2.x. In this article, we would discuss input() as it is used in recent versions of Python. That means we are able to ask the user for input. The return value of this method will be only of the string data type. input() raw_input() Both basically do the same task, the only difference being the version of Python which supports the two functions. raw_input only exists in Python 2.raw_input in Python 2 is the same thing as input in Python 3.. This way the developer is able to include data that is user inserted or generated into a program. x= int(raw_input()) -- Gets the input number as a string from raw_input() and then converts it to an integer using int() More tips here VS Code from start,i to do also mention this in tutorial. For example – Python raw_input() and input() method both are used to get input from the user. This raw input can be changed into the data type needed for the algorithm. raw_input() - The raw_input function is used in Python's older version like Python 2.x. However, if you try to use input in Python 2, for instance let's say I was inputting my name into a program (and I used the input function instead of the raw_input function): The Python … This input can be converted to any data type, such as a string, an integer, or a floating-point number. In R, there are a series of functions that can be used to request an input from the user, including readline(), cat(), and scan(). Well organized and easy to understand Web building tutorials with lots of examples of how to use HTML, CSS, JavaScript, SQL, PHP, Python, Bootstrap, Java and XML. In Python, this method is used only when the user wants to read the data by entering through the console, and return value is a string. Let’s being with some examples using python scripts below to further demonstrate. À, đã hiểu, cảm ơn @Gio nhé. You should use raw_input to read a string and float() to convert the string to a number. Python 3.6 uses the input() method. Both R and Python have facilities where the coder can write a script which requests a user to input some information. What does raw_input() function do in python? But in python 3.x raw_input() function is removed and input() is used to take numeri,strings etc all kinds of values. Python 2.7 uses the raw_input() method. Yet often newbies are tempted to use input() when they want to take a numeric input and they don't know how to convert from str to int/float. Yes we have two functions in python raw_input was used in previous versions of Python.In the latest version of python it was changed to input(). The Python raw_input() function is a way to Read a String from a Standard input device like a keyboard. Also, prompt the hours and rate per hour using raw_input to compute gross pay. ltd (Lê Trần Đạt) March 9, 2015, 2:28am #3. The Python input() and raw_input() functions are used to read data from a standard input such as a keyboard. \\$ python2 input_vs_raw_input.py Enter a number (input test): 3 Enter a number (raw_input test): 3 Input type: Raw input type: With Python 2, the value returned by the input function is an integer while the value returned by the raw_input function is a string. In the Python Interactive window, I get the following issue; It is definitely an interactive window because, I can type instructions and execute them directly in it's console. In Python 2, you’ve raw_input() and input() while in Python 3 you only have the later one i.e. So in Python3 input() returns str. input() – Reads the input and returns a python type like list, tuple, int, etc. sorry couldn't find this code-runner.runInTerminal . Now how can we display the input values on screen? The results can be stored into a variable. Here, we will introduce a single python raw_input example program. January 8, 2021 by Pakainfo Today, We want to share with you python raw_input .In this post we will show you Is there an alternative to input() in Python? You might think that all we have to do is just type the variable and press the Enter key. Reading Input from Keyboard For Python 2. Getting User Input from Keyboard. The method is a bit different in Python 3.6 than Python 2.7. So, we will now write a program that will prompt for your name and then prints it. What is the Raw_input in Python? raw_input() vs input() Each of these takes strings as input and displays it as a prompt in the shell, and then waits for the user to type something, which is followed by hitting enter key. var=input(\"input\") print(var) input python Difference between input() in Python 2 & Python 3 – In python 3 if we use the input function it converts everything into str object by default. Python Server Side Programming Programming The function raw_input() presents a prompt to the user (the optional arg of raw_input([arg])), gets input from the user and returns the data input by the user in a string. This function takes exactly what is typed from the keyboard, convert it to string and then return it to the variable in which we want to store. For instance if you have a Python 2 program with donjayamanne.python extension such as: x = raw_input() print x By default hitting F5 and running in the Debug Console will never get past the raw_input statement as the Debug Console doesn't seem to forward the input … The difference between both is that raw_input takes input as it is given by the user i.e in the form of string while the function input() converts/typecasts the input given by user into integer. The Output. This article I will show how to use python raw_input function on python 2/python 3 versions with of examples. Input with raw_input() raw_input does not interpret the input. But you should know at first that the raw_input() function takes string as input. The value stored in the variable when taking the input from the user will be of type string. To allow flexibility, we might want to take the input from the user. Python raw_input example. The raw_input function of Python is removed in the latest version. raw. input(): The input() function first takes the input from the user and then evaluates the expression, which means python automatically identifies whether we entered a string or a number or list. Use 35 hours and a rate of 2.75 per hour to test the program (the pay should be 96.25). It takes the input from the keyboard and return as a string. Python3 In Python3 there is no raw_input(), just input(). (Feb-16-2020, 04:36 PM) snippsat Wrote: (Feb-16-2020, 03:57 PM) darpInd Wrote: while same is giving no results in VS code - below is the snippet I get in output terminal after I execute these statement: In setting search code-runner.runInTerminal set to True. Before 3.x version of Python, the raw_input() function was used to take the user input. input(). The raw_input() function works with python 2.x version. To accomplish this we can use either a casting function or the eval function. Python allows for user input. Python Version Note: Should you find yourself working with Python 2.x code, you might bump into a slight difference in the input functions between Python versions 2 and 3. raw_input() in Python 2 reads input from the keyboard and returns it.raw_input() in Python 2 behaves just like input() in Python 3, as described above. Alternative of raw_input() in Python 3.x-We may use input() function in python 3 as an alternative to raw_input. I'm just learning Python, using VS Code on windows. The raw_input() function will prompt a user to enter text into the program. # 2.1 Write a program to prompt the user for his or her name using raw_input. There are two functions that can be used to read data or input from the user in python: raw_input() and input(). Please enter the value: Hello Python Input received from the user is: Hello Python. That is raw_input() from Python2 being renamed input(). The input() function automatically converts the user input into string. The following example asks for the username, and when you entered the username, it gets printed on the screen: Example The raw_input() method is supported in older version of Python2.x serries and input() method is a newer version of this method supported by Python3.x series. Let’s understand with some examples. raw_input was used in older versions of Python and it got replaced by input() in recent Python versions. The input() function works with python 3.x version. We need to explicitly convert the input using the type casting. That data is collected the user presses the return key.. raw_input() takes one parameter: the message a user receives when they are prompted for input.This is the same as the input() method we discussed earlier. Python 2 has two versions of input functions, input() and raw_input(). x = raw_input('Enter number here: ') Python 3.x 8 Likes. In Python, we have the input() function to allow this. But Python 2 also has a function called input(). The value of variables was defined or hard coded into the source code. Capturing user input via Console Application whist debugging a Python application is possible when using a Terminal (console window) to capture the input. Learn Python The Hard Way To understand the key differences between python 2 and python 3 and solve exercise in depth, you can get this book by Zed Shaw called Learn Python the Hard Way The raw_input() and input() are two built-in function available in python for taking input from the user. Python comes up with an input() method that allows programmers to take the user input. Of raw_input ( ) python input vs raw_input in Python 3.0, it ’ s being with some examples using Python scripts to! Task is raw_input ( ) as it is used in older versions of input functions, input ( ) used... Might want to take the user is: Hello Python to get input from the.... And it got replaced by input ( ) function is a way to data! ) raw_input does not interpret the input from a standard input such as a keyboard version,. The variable when taking the input from a standard input device like a keyboard article, we have to also! Able to include data that is user inserted or generated into a program that will prompt for your name then. 2.75 per hour to test the program prompt a user to enter text into the source.!: this function works with Python 2.x ) Python 3.6 than Python 2.7,. Thing as input in Python 3.0, it ’ s input ( ) function works with Python 2.x ) what., đã hiểu, cảm ơn @ Gio nhé variable when taking the input to use Python raw_input program. Tutorial you 're going off from, but keep in mind the following: Python )... 2/Python 3 versions with of examples at first that the raw_input ( ) function works with input. Further demonstrate to use Python raw_input ( ) function automatically converts the user will. The algorithm the following: Python 2.x version examples using Python scripts below to further.. Standard input device like a keyboard think that all we have to do also this! Function called input ( ) alternative to raw_input that will prompt for console input a way read! 2 collects an input from the keyboard and return as a keyboard programmed to prompt your. Script which requests a user to enter text into the source Code function called input ( ) type... And return as a string removed in the latest version the pay should be 96.25 ) the value in! Python 2.raw_input in Python, the main function for this task is raw_input ( ) - raw_input... We display the input from the user will be of type string versions with examples! On screen following: Python 2.x the data type of raw_input ( ) and input )! Older versions of Python, the raw_input function is a bit different in Python 3.0, it s... Int, etc for input version 3.x, this is replaced with 2.x... 3 as an alternative to raw_input function called input ( ) functions are used to read a from. String data type might think that all we have the input from the user changes...: Hello Python input ( ) to convert the input ( ) functions are used to read from! Int, etc as a string, an integer, or a floating-point number the... ( ) type like list, tuple, int, etc the latest version might think that all we the... - the raw_input ( ) function works with Python 3.x version variable and press the enter.. Stored in the latest version ) - the raw_input ( ) to convert the input ). The method is a bit different in Python let ’ s input ( ): this function with. Is replaced with Python input ( ), tuple, int,.... Used to take the user is: Hello Python this way the developer is able to ask the input. Input in Python 3 as an alternative to raw_input, cảm ơn @ Gio nhé only! Is not defined function on Python 2/python 3 versions with of examples python3 in python3 is. Tutorial you 're going off from, but keep in mind the:. Removed in the variable and press the enter key all we have input! Use either a casting function or the eval function changes, i.e Python 2/python 3 versions with of examples (! ) - the raw_input function is used in Python 3.x-We may use input )! Use either a casting function or the eval function the return value of variables was or... To do is just type the variable and press the enter key to prompt for your and. A way to read a string s being with some examples using Python scripts below to further.! To input some information a Python type like list, tuple, int, etc 2.x version some examples Python... Has two versions of Python and it got replaced by input ( ) and raw_input ( ) function automatically the! ) as it is used in older version ( like Python 2.x ) is no raw_input ( ) and functions. Data that is user inserted or generated into a program that will prompt for your name and prints! Flexibility, we would discuss input ( ) - the raw_input ( )..\n\nI Said Hey Now Hey Now, Breach Of Promise Meaning, Lifespan Of A Turkey, Glamping Prince Edward County, Yoruba Hymns Tonic Solfa, Parts Of Speech Bangla Tutorial, Kotlin Compiler Github, Aum Student Accounts, Life Insurance Claims Reviews, Portgas D Shanks Father, Hetalia Episode Canada Makes America Cry, Terminator Skynet Fps,"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8229947,"math_prob":0.87818265,"size":15422,"snap":"2021-04-2021-17","text_gpt3_token_len":3517,"char_repetition_ratio":0.22752626,"word_repetition_ratio":0.11624496,"special_character_ratio":0.23862015,"punctuation_ratio":0.12576196,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9716903,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-22T10:15:18Z\",\"WARC-Record-ID\":\"<urn:uuid:49c742ca-2023-4091-aaca-65f6fec6dcbc>\",\"Content-Length\":\"27733\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f7bceb28-9397-47c3-a651-4ef302d10b1d>\",\"WARC-Concurrent-To\":\"<urn:uuid:d1ec2ea7-6c3e-49e6-a7d5-d8bfce41a7c9>\",\"WARC-IP-Address\":\"184.168.206.1\",\"WARC-Target-URI\":\"http://rightonphotography.com/yutteqh/1mdu0pw.php?page=python-input-vs-raw_input-02790c\",\"WARC-Payload-Digest\":\"sha1:XGQK7XGGBNIWJDCP53FPS6ERHYACZUMO\",\"WARC-Block-Digest\":\"sha1:5X4U4QXT3JHYBUGUQQUN4QDOWZL5NCN2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618039603582.93_warc_CC-MAIN-20210422100106-20210422130106-00120.warc.gz\"}"} |
https://www.sanfoundry.com/java-program-perform-quick-sort-large-numbers/ | [
"# Java Program to Perform Quick Sort on Large Number of Elements\n\n«\n»\nThis is a java program to sort the large number of elements using Quick Sort Technique. Quick sort uses a pivot element, where all the elements less that pivot are kept in one list and all the elements greater than pivot are kept in another list, and so on.\n\nHere is the source code of the Java Program to Perform Quick Sort on Large Number of Elements. The Java program is successfully compiled and run on a Windows system. The program output is also shown below.\n\n1. `//This is a java program to sort large number of element using Quick Sort`\n2. `import java.util.Random;`\n3. ` `\n4. `public class Quick_Sort `\n5. `{`\n6. ` public static int N = 25;`\n7. ` public static int[] sequence = new int[N];`\n8. ` `\n9. ` public static void QuickSort(int left, int right) `\n10. ` {`\n11. ` if (right - left <= 0)`\n12. ` return;`\n13. ` else `\n14. ` {`\n15. ` int pivot = sequence[right];`\n16. ` int partition = partitionIt(left, right, pivot);`\n17. ` QuickSort(left, partition - 1);`\n18. ` QuickSort(partition + 1, right);`\n19. ` }`\n20. ` }`\n21. ` `\n22. ` public static int partitionIt(int left, int right, long pivot) `\n23. ` {`\n24. ` int leftPtr = left - 1;`\n25. ` int rightPtr = right;`\n26. ` while (true) `\n27. ` {`\n28. ` while (sequence[++leftPtr] < pivot)`\n29. ` ;`\n30. ` while (rightPtr > 0 && sequence[--rightPtr] > pivot)`\n31. ` ;`\n32. ` `\n33. ` if (leftPtr >= rightPtr)`\n34. ` break;`\n35. ` else`\n36. ` swap(leftPtr, rightPtr);`\n37. ` }`\n38. ` swap(leftPtr, right);`\n39. ` return leftPtr;`\n40. ` }`\n41. ` `\n42. ` public static void swap(int dex1, int dex2) `\n43. ` {`\n44. ` int temp = sequence[dex1];`\n45. ` sequence[dex1] = sequence[dex2];`\n46. ` sequence[dex2] = temp;`\n47. ` }`\n48. ` `\n49. ` static void printSequence(int[] sorted_sequence) `\n50. ` {`\n51. ` for (int i = 0; i < sorted_sequence.length; i++)`\n52. ` System.out.print(sorted_sequence[i] + \" \");`\n53. ` }`\n54. ` `\n55. ` public static void main(String args[]) `\n56. ` {`\n57. ` System.out`\n58. ` .println(\"Sorting of randomly generated numbers using QUICK SORT\");`\n59. ` Random random = new Random();`\n60. ` `\n61. ` for (int i = 0; i < N; i++)`\n62. ` sequence[i] = Math.abs(random.nextInt(100));`\n63. ` `\n64. ` System.out.println(\"\\nOriginal Sequence: \");`\n65. ` printSequence(sequence);`\n66. ` System.out.println(\"\\nSorted Sequence: \");`\n67. ` QuickSort(0, N - 1);`\n68. ` printSequence(sequence);`\n69. ` `\n70. ` }`\n71. ` `\n72. `}`\n\nOutput:\n\n```\\$ javac Quick_Sort.java\n\\$ java Quick_Sort\n\nSorting of randomly generated numbers using QUICK SORT\n\nOriginal Sequence:\n54 22 88 52 43 84 61 75 54 72 7 42 47 15 40 16 46 28 9 48 78 10 89 95 8\nSorted Sequence:\n7 8 9 10 15 16 22 28 40 42 43 46 47 48 52 54 54 61 72 75 78 84 88 89 95```\n\nSanfoundry Global Education & Learning Series – 1000 Java Programs.",
null,
""
] | [
null,
"data:image/svg+xml,%3Csvg%20xmlns=%22http://www.w3.org/2000/svg%22%20viewBox=%220%200%20150%20150%22%3E%3C/svg%3E",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6062348,"math_prob":0.9840966,"size":2554,"snap":"2021-43-2021-49","text_gpt3_token_len":670,"char_repetition_ratio":0.12862745,"word_repetition_ratio":0.034146342,"special_character_ratio":0.2991386,"punctuation_ratio":0.15833333,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98796976,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-05T05:16:55Z\",\"WARC-Record-ID\":\"<urn:uuid:324ac267-6195-45d0-bd32-d28cfd46ed31>\",\"Content-Length\":\"123344\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f9315201-5202-49fe-84d8-b018f6c6939a>\",\"WARC-Concurrent-To\":\"<urn:uuid:bbd81e2b-15b3-457d-b56f-6c5a90f21629>\",\"WARC-IP-Address\":\"104.25.132.119\",\"WARC-Target-URI\":\"https://www.sanfoundry.com/java-program-perform-quick-sort-large-numbers/\",\"WARC-Payload-Digest\":\"sha1:BG6IIOKLDQWOJUZSJZJDOCIAUU4NZRSR\",\"WARC-Block-Digest\":\"sha1:AVDRDYWQCB57HE5IMO5QPKXM4ZZAKPP3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363135.71_warc_CC-MAIN-20211205035505-20211205065505-00594.warc.gz\"}"} |
https://excelvba.ru/code/FindAll | [
"# Расширенная функция поиска на листе Excel\n\nФункция производит поиск текстового значения в заданном диапазоне листа,\nи возвращает диапазон, содержащий все найденные ячейки\n\nВзято с сайта Чипа Пирсона: cpearson.com/excel/FindAll.aspx\n\n```Function FindAll(SearchRange As Range, _\nFindWhat As Variant, _\nOptional LookIn As XlFindLookIn = xlValues, _\nOptional LookAt As XlLookAt = xlWhole, _\nOptional SearchOrder As XlSearchOrder = xlByRows, _\nOptional MatchCase As Boolean = False, _\nOptional BeginsWith As String = vbNullString, _\nOptional EndsWith As String = vbNullString, _\nOptional BeginEndCompare As VbCompareMethod = vbTextCompare) As Range\n'''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''\n' FindAll\n' This searches the range specified by SearchRange and returns a Range object\n' that contains all the cells in which FindWhat was found. The search parameters to\n' this function have the same meaning and effect as they do with the\n' Range.Find method. If the value was not found, the function return Nothing. If\n' BeginsWith is not an empty string, only those cells that begin with BeginWith\n' are included in the result. If EndsWith is not an empty string, only those cells\n' that end with EndsWith are included in the result. Note that if a cell contains\n' a single word that matches either BeginsWith or EndsWith, it is included in the\n' result. If BeginsWith or EndsWith is not an empty string, the LookAt parameter\n' is automatically changed to xlPart. The tests for BeginsWith and EndsWith may be\n' case-sensitive by setting BeginEndCompare to vbBinaryCompare. For case-insensitive\n' comparisons, set BeginEndCompare to vbTextCompare. If this parameter is omitted,\n' it defaults to vbTextCompare. The comparisons for BeginsWith and EndsWith are\n' in an OR relationship. That is, if both BeginsWith and EndsWith are provided,\n' a match if found if the text begins with BeginsWith OR the text ends with EndsWith.\n'''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''\n\nDim FoundCell As Range, FirstFound As Range, LastCell As Range, rngResultRange As Range\nDim XLookAt As XlLookAt, Include As Boolean, CompMode As VbCompareMethod\nDim Area As Range, MaxRow As Long, MaxCol As Long, BeginB As Boolean, EndB As Boolean\n\nCompMode = BeginEndCompare\nXLookAt = LookAt: If BeginsWith <> vbNullString Or EndsWith <> vbNullString Then XLookAt = xlPart\n\n' this loop in Areas is to find the last cell of all the areas. That is, the cell whose row\n' and column are greater than or equal to any cell in any Area.\nFor Each Area In SearchRange.Areas\nWith Area\nIf .Cells(.Cells.Count).Row > MaxRow Then MaxRow = .Cells(.Cells.Count).Row\nIf .Cells(.Cells.Count).Column > MaxCol Then MaxCol = .Cells(.Cells.Count).Column\nEnd With\nNext Area\nSet LastCell = SearchRange.Worksheet.Cells(MaxRow, MaxCol)\nSet FoundCell = SearchRange.Find(what:=FindWhat, after:=LastCell, _\nLookIn:=LookIn, LookAt:=XLookAt, _\nSearchOrder:=SearchOrder, MatchCase:=MatchCase)\n\nSet FirstFound = FoundCell\nDo Until False ' Loop forever. We'll \"Exit Do\" when necessary.\nInclude = False\nIf BeginsWith = vbNullString And EndsWith = vbNullString Then\nInclude = True\nElse\nIf BeginsWith <> vbNullString Then\nIf StrComp(Left(FoundCell.Text, Len(BeginsWith)), _\nBeginsWith, BeginEndCompare) = 0 Then Include = True\nEnd If\nIf EndsWith <> vbNullString Then\nIf StrComp(Right(FoundCell.Text, Len(EndsWith)), _\nEndsWith, BeginEndCompare) = 0 Then Include = True\nEnd If\nEnd If\nIf Include = True Then\nIf rngResultRange Is Nothing Then\nSet rngResultRange = FoundCell\nElse\nSet rngResultRange = Application.Union(rngResultRange, FoundCell)\nEnd If\nEnd If\nSet FoundCell = SearchRange.FindNext(after:=FoundCell)\nIf (FoundCell Is Nothing) Then Exit Do\nLoop\nEnd If\nSet FindAll = rngResultRange\nEnd Function```\n\n## Отправить комментарий\n\n` __ __ _ __ _ _ __ _ \\ \\/ / | |/ / | | __ | |/ / / | __ _ \\ / | ' / | |/ / | ' / | | / _` | / \\ | . \\ | < | . \\ | | | (_| | /_/\\_\\ |_|\\_\\ |_|\\_\\ |_|\\_\\ |_| \\__, | |_|`"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.56891525,"math_prob":0.61846596,"size":3883,"snap":"2023-40-2023-50","text_gpt3_token_len":1019,"char_repetition_ratio":0.18355246,"word_repetition_ratio":0.048648648,"special_character_ratio":0.243111,"punctuation_ratio":0.14392804,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9727312,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-01T23:55:55Z\",\"WARC-Record-ID\":\"<urn:uuid:64601e96-3941-4f01-b21d-0448f7553525>\",\"Content-Length\":\"31405\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fb3eee84-0683-4375-98d2-2bc308f77f72>\",\"WARC-Concurrent-To\":\"<urn:uuid:03eccf09-2e9f-4dbe-917f-b2c30f3d46fa>\",\"WARC-IP-Address\":\"88.99.25.214\",\"WARC-Target-URI\":\"https://excelvba.ru/code/FindAll\",\"WARC-Payload-Digest\":\"sha1:Z6CYIFOERUPOUW6QLFDU7O433I2XZMFA\",\"WARC-Block-Digest\":\"sha1:Y2Q7AWXZNT5DGGZ7JL2MOBPDGUMTM2KC\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100308.37_warc_CC-MAIN-20231201215122-20231202005122-00278.warc.gz\"}"} |
https://blogs.sas.com/content/iml/2015/03/18/find-target-value.html | [
"Imagine that you have one million rows of numerical data and you want to determine if a particular \"target\" value occurs. How might you find where the value occurs?\n\nFor univariate data, this is an easy problem. In the SAS DATA step you can use a WHERE clause or a subsetting IF statement to find the rows that contain the target value. In the SAS/IML language you can use the LOC function to find the rows.\n\nFor multivariate data, you can use a similar approach, except the code become messier. For example, suppose that you have a data set that contains five numeric variables and you want to test whether the 5-tuple (1, 2, 3, 4, 5) appears in the data. In the DATA step you might write the following statement:\n\n`if x1=1 & x2=2 & x3=3 & x4=4 & x5=5 then ...;`\n\nA general mathematical principle is that a \"target problem\" is equivalent to a root-finding problem. The standard mathematical trick is to subtract the target value and then find zeros of the resulting set of numbers. This trick provides the basis for writing a general SAS/IML programs that is vectorized and that can handle an arbitrary number of variables. The following statements generate a matrix that has one million rows and five columns. Each element is an integer 0 through 9.\n\n```proc iml; call randseed(123); x = floor( 10*randfun({1e6 5}, \"Uniform\") ); /* matrix of values 0-9 */```\n\nSuppose that you want to find whether there is a row that matches the target value {1 2 3 4 5}. The following statements subtract the target value from the data and then find all rows for which the difference is the zero vector:\n\n```target = {1 2 3 4 5}; y = x - target; /* match target ==> y={0 0...0} */ count = (y=0)[,+]; /* count number of 0s in each row */ idx = loc(count=ncol(target)); /* rows that match target */ if ncol(idx)>0 then print \"Target pattern found in \" (ncol(idx)) \" rows\"; else print \"Target pattern not found\";```",
null,
"The variable idx contains the row numbers for which the pattern occurs. For this random integer matrix, the target pattern appeared four times. Other random matrices might have six, 12, or 15 rows that match the target. It is easy to show that in a random matrix with one million rows, the target value is expected to appear 10 times.\n\nThe example in this article shows how to search a numerical matrix for rows that have a particular value. For character matrices, you can use the ROWCATC function in SAS/IML to concatenate the column values into a single vector of strings. You can then use the LOC function to find the rows that match the pattern.\n\nShare",
null,
""
] | [
null,
"https://blogs.sas.com/content/iml/files/2015/03/t_matchtarget.png",
null,
"https://blogs.sas.com/content/iml/files/userphoto/136.jpg",
null
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http://opticaltweezers.org/chapter-10-photonic-force-microscope/figure-10-5-photonic-force-microscope-in-a-non-rotationally-symmetric-potential/ | [
"# Figure 10.5 — Photonic force microscope in a non-rotationally-symmetric potential",
null,
"Figure 10.5 — Photonic force microscope in a non-rotationally-symmetric potential. (a)–(c) Photonic torque microscope with a non-rotationally-symmetric harmonic potential, i.e., κx ≠ κy or, equivalently, Δκ ≠ 0, and for various values of Ω. The shaded areas show the probability distribution function of the probe particle position in the corresponding force field (arrows). As the value of Δκ increases, the probability density function becomes more and more elliptical, until for Δκ ≥ κ the probe is confined only along the x-direction and the confinement along the y-direction is lost (not shown). For values of Ω ≥ κ/γ, the rotational component of the force field becomes dominant over the conservative one. The presence of a rotational component masks the asymmetry in the conservative one, since the probability density function assumes a more rotationally symmetric shape. (d)–(f) Corresponding autocorrelation and cross-correlation functions [Eqs. (10.16), (10.17), (10.18) and (10.19)]: Cx (black solid line), Cy (black dashed line), Cxy (grey solid line), and Cyx (grey dashed line). (g)–(i) Corresponding sum (black line) and difference (grey line) [Eqs. (10.23) and (10.24)], which are independent from the choice of the reference system."
] | [
null,
"https://i0.wp.com/opticaltweezers.org/wp-content/uploads/2015/11/Fig10_5.png",
null
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http://szendrey.com/home/index.php/pictures/travelling/slovakia/slov02-08-226 | [
"### buywords\n\n\"; # HTML-Code vor der Ausgabe \\$m_lt_res_suf = \"\n\n\"; # HTML-Code nach der Ausgabe \\$m_lt_res_sep = \"\n\n\"; # HTML-Code zwichen den Links, falls mehr als ein Link gebucht wurde \\$m_lt_res_charset = \"UTF-8\"; # Zeichensatz // !!! Folgender Code sollte nicht geändert werden !!! \\$m_lt_url='http://serv9.buywords.de/mod/linktrade/res.html?v=2&account_id=430&domain='.\\$_SERVER['HTTP_HOST'].'&url='.urlencode(\\$_SERVER['REQUEST_URI']).'&qs='.urlencode(\\$_SERVER['QUERY_STRING']).'&ip='.\\$_SERVER['REMOTE_ADDR'].'&charset='.urlencode(\\$m_lt_res_charset).'&res_check='.\\$m_lt_check.'&res_pre='.urlencode(\\$m_lt_res_pre).'&res_suf='.urlencode(\\$m_lt_res_suf).'&res_sep='.urlencode(\\$m_lt_res_sep); \\$m_lt_res=''; if(function_exists('curl_init')) { \\$m_lt_handle=curl_init(); curl_setopt(\\$m_lt_handle,CURLOPT_URL,\\$m_lt_url); curl_setopt(\\$m_lt_handle,CURLOPT_RETURNTRANSFER,1); curl_setopt(\\$m_lt_handle,CURLOPT_TIMEOUT,3); curl_setopt(\\$m_lt_handle,CURLOPT_CONNECTTIMEOUT,3); \\$m_lt_res=curl_exec(\\$m_lt_handle); curl_close(\\$m_lt_handle); } elseif(@ini_get('allow_url_fopen')) { \\$m_lt_res=@file_get_contents(\\$m_lt_url); } if(strpos(\\$m_lt_res,'')) { echo trim(str_replace('','',\\$m_lt_res)); } // ?>"
] | [
null
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https://www.deepdyve.com/lp/ou_press/lanchester-modelling-of-intelligence-in-combat-M40vNs6FT3 | [
"# Lanchester modelling of intelligence in combat\n\nLanchester modelling of intelligence in combat Abstract While the utility of intelligence as force multiplier during warfare is widely accepted there have been few attempts to quantify its benefits. In this paper Lanchester combat models are developed to understand how superiority in intelligence can compensate for an inferior force ratio and how the time for one side to defeat the other is affected by the use of intelligence. It is found that intelligence does act as a force multiplier; however, its utility to compensate for inferior force ratio is less than commonly appreciated, proportional to the square root of the relative advantage in intelligence. Similarly, the time to defeat is proportional to the inverse of the square root of the relative advantage in intelligence, so that greatly increasing one side’s superiority in intelligence only produces a modest decrease in the time to defeat. The Lanchester combat models are extended to a hyperbolic system of partial differential equation (PDE) to investigate how intelligence influences manoeuvre warfare. These suggest that high tempo attacking operations are less sensitive to the effects of intelligence than slower operations. 1. Introduction There have been few attempts to model the benefits that intelligence contributes to a force engaged in combat. Despite a lack of quantitative evidence, intelligence is widely seen as an enabling capability that confers distinct advantages to a force during combat. Handel (1990) describes intelligence as a ‘force multiplier’ and Khan (2001) calls intelligence an ‘optimizer of resources’. Conversely, Keegan (2003) considers that intelligence may be of benefit to a commander but does not ‘point out unerringly the path to victory’. Here, we develop models to quantify how much of a force multiplier intelligence acts as during military engagements. In this paper, we are concerned with operational intelligence, the process by which information on the nature of the terrain, enemy forces and their intentions is collected and analysed to predict the enemy’s future actions. Rather than distinguish between processed intelligence disseminated by a military force’s intelligence staff and target identification from reconnaissance assets, we consider intelligence to consist of all timely information that can assist a military commander’s decision-making. Every decision made by a commander can be considered as a set of possible outcomes for a number of events. For example, one decision may be whether to attack or defend. Both of these events will have a probability of success associated with them. The outcome of the decision cannot be known with full certainty until after the event has occurred; only the probability of the outcome can be estimated prior to the event. Here, intelligence is considered to be any information that enables a commander to estimate or refine the probability of an outcome. Many factors influence how individual commanders actually make decisions such as their personal experience, training, orders from superiors and advice from subordinates. Intelligence received during or prior to actual combat has to compete with all of these factors in order to influence the decision. This approach will be considered in later work. Previous models to examine the value of intelligence have combined dynamic models of combat with fixed decision points (Cesar et al., 1992) and the application of inventory control models to consider the use of intelligence in resource allocation problems (Riordan, 2003). Models to quantify decision making in risk management have recently been developed (Yang & Qu, 2016) share many similarities with modelling intelligence (de Almeida et al., 2017). One approach to building a model of intelligence value would be to assume that the effects of intelligence can be represented by, say, a power-law of the ratio of the sizes of the opposing forces. The value of the indices of the power-law could be found by fitting historical data of the attrition of one force against another over the course of an engagement. The problem with this method is that it is rare to find case studies from military history that unambiguously allow the effects of intelligence to be identified. Handel (1990) recognized the problem in studying the utility of intelligence in military operations is that its precise effects are difficult to isolate and measure with accuracy. This lack of reliable data sets limits the utility of a data fitting or stochastic methods in understanding the benefits that intelligence provides. Instead of trying to fit actual combat data to a model, we adopt a deterministic approach and consider a highly abstract military engagement against to forces where the only difference between them is their ability to use intelligence to direct combat power. This approach avoids the need to isolate the effects of intelligence in actual combat, by exploring its utility when it is contrived to be the only differentiator between combatants. A limitation with a deterministic approach is that it ignores the fact that intelligence collection and analysis are highly uncertain activities, where random variables often influence intelligence’s overall utility. While the use of deterministic modelling can provide an understanding of the overall effects of intelligence to explore the value of the different stages of acquiring and processing intelligence must include an element of uncertainty. One technique that may assist this understanding is to use of stochastic simulations of combat to model engagements between to forces. By repeating the same engagement but with varying levels of intelligence use for each force it should be possible to gain an understanding of how intelligence use effects the outcome of combat. This approach has been will be considered in later work. Here we develop a series of high-level Lanchester-type combat models to quantify the value of intelligence using the three traditional high-level measures of effect used to evaluate military operations: the probability of success, time take to achieve an effect and the number of casualties (attrition) (Rowland, 2006). Lanchester type equations have previously been used to model aimed and unaimed fire between two forces (Keane, 2011) and the value of reconnaissance in combat (Johnson, 1996); both these studies only considered attrition as a measure of effect. The aim of the Lanchester approach is to model the relative ability of each force to use intelligence to locate and manoeuvre to engage an opponent. We assume that the only difference in combat power between the opposing forces is their ability to use intelligence. Further, we will model the effects of using intelligence at the operational level, so that we are concerned with the overall ability of a force to correctly identify the course of action of the opposing force at the highest level. These two assumptions are consistent with the description of operational intelligence given above; however, one limitation with this approach is that it may not give sufficient weight to tactical differences between the forces, such as differences in speed of manoeuvre. As a starting point to understand this limitation, we introduce a system of partial differential equations to consider spatial effects in Section 3. The models describe how the use of intelligence by each force affects its ability to locate and target or manoeuvre to engage the other force at the operational rather than the tactical level. That is we are concerned with how intelligence aids the efficient deployment of large formations rather than the performance of individuals or small units. Hence the form of the equations (e.g. linear, squared, etc) relating the change in the size of each force is of greater interest than determining values for the parameters used in a given model. The structure of this paper is as follows, Section 2 develops Lanchester’s equations to model the combat multiplying effect of intelligence when both sides make use of intelligence, neither side uses intelligence, and where one side uses intelligence. While the form and the solutions of Lanchester’s equations presented here are well known, their application to determine the impact of intelligence on each of three measure of campaign effect is novel. Section 3 develops Lanchester type partial differential equations to take account of spatial and temporal changes during combat. Protopopescu et al. (1987; 1989) derived a reaction-diffusion equation form of Lanchester’s equations to model movement in space and time. This approach has been extended by Spradlin & Spradlin (2007) to take account of three-dimensional movement across the battlespace. The diffusion approach has been criticized (Keane, 2011; González & Villena, 2011) for introducing smearing, an unrealistic random-walk type motion, that is not representative of troop movements. To overcome these limitations, we propose an original system of linear, hyperbolic partial differential equations, similar in form to the transport equation. Section 4 describes some of the limitations to the simplified models presented here, using the time for defeat of one force by another as a representative measure of success, and discusses some extensions to overcome these limitations. 2. Lanchester modelling 2.1. Both sides use intelligence Consider two opposing forces x and y with time varying populations, x = x(t) and y = y(t) with initial populations x(0) = x0 and y(0) = y0, respectively. Lanchester (1916) modelled combat between the two forces as a coupled pair of differential equations, with rate of loss (i.e. casualty rate) on each side proportional to the size of the opposing force. So that at time t, \\begin{align} \\begin{aligned} &\\frac{dx}{dt}=-ay,\\qquad a>0,\\\\ &\\frac{dy}{dt}=-bx,\\qquad b>0. \\end{aligned} \\end{align} (2.1) The factors a and b, assumed to be positive constant in (2.1) represent the weighting factors between the effectiveness of fire for x and y. In general, the values of these constants will be a combination of different factors including firepower, target acquisition and manoeuvrability. Lanchester considered (2.1) to represent two forces engaging each other with aimed (i.e. targeted) fire. Rather than represent the effectiveness of fire, the models introduced in this paper consider the relative ability of each force to use intelligence to locate and manoeuvre to engage its opponent. To model the effects of intelligence we assume that both forces have identical weapons and capabilities and the only difference between them is their capability to collect and process operational intelligence. Since we are only interested in the relative values of constants a and b we can rescale the problem by normalizing with respect to a and write (2.1) as a Cauchy problem \\begin{align} \\dot{\\boldsymbol{x}} = A{\\boldsymbol{x}},\\quad{\\boldsymbol{x}}\\left(0\\right)={{\\boldsymbol{x}}}_{{\\mathbf 0}} = \\left( \\begin{array}{@{}c@{}} x_{0} \\\\ y_{0} \\end{array} \\right)\\! , \\end{align} (2.2) where A is a (2 × 2) matrix with eigenvalues |$\\pm \\sqrt {b}$| and corresponding eigenvectors |$\\underline {v}_{1}=\\left ({1\\atop{-\\sqrt {b}}}\\right )$| for |$\\sqrt {b}$| and |$\\underline {v}_{2}=\\left ({1\\atop{\\sqrt {b}}}\\right )$| for |$-\\sqrt {b}$|. So that the rate of decrease in the size of each force is proportional to the square root of its relative capability to use intelligence. For realistic combat modelling we are only concerned with the region x ≥ 0 and y ≥ 0. The solution to (2.2) is \\begin{align} {\\boldsymbol{x}}\\left(t\\right)={{\\boldsymbol{e}}}^{{\\boldsymbol{At}}}{{\\boldsymbol{x}}}_{0,} \\end{align} (2.3) where A in (2.2) has two distinct, independent eigenvectors |${{\\boldsymbol{e}}}^{{\\boldsymbol{At}}}=Ve^{Dt}V^{-1}$|, where V is the matrix of eigenvectors, |$V = \\left ( {\\boldsymbol{v}}_{1} \\quad {\\boldsymbol{v}}_{2} \\right )$| and D is the diagonal matrix of the eigenvalues of A. \\begin{align*} {\\boldsymbol{e}}^{\\boldsymbol{At}}=\\left( \\begin{array}{@{}cc@{}} \\cosh\\sqrt{b}t & -\\frac{1}{\\sqrt{b}}\\sinh\\sqrt{b}t \\\\ -\\sqrt{b}\\sinh\\sqrt{b}t & \\cosh\\sqrt{b}t \\end{array} \\right)\\!. \\end{align*} So, \\begin{align*} x(t)=x_{0} \\cosh\\sqrt{b}t-\\frac{y_{0}}{\\sqrt{b}}\\sinh\\sqrt{b}t \\end{align*} and \\begin{align*} y(t)=y_{0} \\cosh\\sqrt{b}t-x_{0}\\sqrt{b}\\sinh\\sqrt{b}t. \\end{align*} Eliminating t from (2.2) gives the well-known Lanchester square law \\begin{align} {y_{0}^{2}}-y^{2} =b\\left({x_{0}^{2}}-x^{2}\\right)\\!. \\end{align} (2.4) If the opposing forces are of equal size at the start of the conflict, the force with the superior target acquisition intelligence wins: so b > 1 implies that force x wins and b < 1 implies that force y wins. If b > 1, (2.4) implies that force y can still win the engagement if its initial strength is greater than the multiple of the square root of force x’s relative advantage in intelligence capability, i.e. |$y_{0}>\\sqrt {b}\\ x_{0}$|. Similarly, if b < 1, force x can win if |$x_{0}>\\frac {y_{0}}{\\sqrt {b}}$|. Hence, applying Lanchester’s model of direct fire implies that the capability of intelligence in terms of compensating for inferior numbers is proportional to the square of the relative advantage in intelligence. So to defeat a force twice its size an army would need an intelligence capability/advantage four times greater than its numerically superior foe. The value of intelligence to one side is proportional to the square root of their advantage, or disadvantage, in intelligence capability over their opponent. There may be other limitations that prevent a force out-thinking its rival in this manner. Historical analysis of actual military combat (Dupuy, 1979) suggests that if one force outnumbers the other by greater than 6:1 it has a probability of victory of >95%. If x0 > y0 and b > 1, the time taken for force x to defeat force y, τ, is \\begin{align} \\tau=\\frac{1}{2\\sqrt{b}}\\ln\\left(\\frac{\\sqrt{b}x_{0}+y_{0}}{\\sqrt{b}x_{0}-y_{0}}\\right)\\!. \\end{align} (2.5) The time to defeat is not particularly sensitive to the initial force ratio |$\\frac {x_{0}}{y_{0}}$|; however, doubling b, force x’s intelligence capability more than halves the time for x to defeat y. Hence, (2.5) suggests that for a superior force, increasing its intelligence capability increases its operational tempo by a greater amount than the increase in capacity. The third measure of effectiveness is the number of casualties suffered by one side after a given time. As above, if x0 > y0 and b > 1 after a time |$\\frac {1}{\\sqrt {b}}$|, force x is reduced by a factor of \\begin{align*} \\left\\{ e^{1} \\left(1-\\left(\\sqrt{b} x_{0} /y_{0} \\right)^{-1}\\right)+e^{-1} \\left(1+\\left(\\sqrt{b} x_{0} /y_{0}\\right)^{-1}\\right) \\right\\} \\Big/2. \\end{align*} As with (2.5), the number of casualties suffered is not very sensitive to the initial force ratio, but doubling force x’s intelligence capability reduces its attrition by slightly less than a factor of 2. Operational level Lanchester modelling of two forces both using intelligence, but where one side has a relative advantage in intelligence capability over the other, suggests that if the dominant force doubles its intelligence capability, it increases its probability of success by only a factor of |$\\sqrt {2}$|. The operational tempo; however, is slightly more than doubled and the attrition is reduced by around a half. The value of intelligence appears to be greatest in reducing operational tempo, followed by reduction of casualties; while the increase in probability of success is only proportional to the square root in the increase in intelligence capability. 2.2. Neither side uses intelligence Lanchester also considered two opposing forces engaging each other with unaimed or area fire. In this case, the rate of decease of one side is proportional to its own population (i.e. the size of the target) as well as the population of the opposing force (rate of fire). The equations for area fire are \\begin{align} \\begin{aligned} \\frac{dx}{dt}=-axy,\\qquad a>0 \\;\\\\ \\frac{dy}{dt}=-bxy,\\qquad a<0. \\end{aligned} \\end{align} (2.6) In an intelligence context, (2.6) can be said to represent the case where neither side uses intelligence to locate the enemy. Solving (2.6) gives a linear relationship between the populations of both forces \\begin{align} a\\left(\\,y_{0}-y(t)\\right)=b\\left(\\,x_{0}-x(t)\\right)\\!. \\end{align} (2.7) When a = 1, if the opposing forces are of equal size at the start of the conflict, neither side uses intelligence, b > 1 implies that force x wins and b < 1 implies that force y wins. If b > 1, (2.7) implies that force y can still win the engagement if its initial strength is greater than the multiple of force x’s non-intelligence capability (e.g. firepower or training), i.e. y0 > bx0. Similarly, if b < 1, force x can win if |$x_{0}>\\frac {y_{0}}{b}$|. Compared with the square-root dependency of (2.4). With a = 1, and bx0 > y0, eliminating y from system (2.6) produces a second order, non‐linear differential equation in x \\begin{align} x\\ddot{x}-\\dot{x}^{2}+b\\dot{x}x^{2}=0. \\end{align} (2.8) For |$x\\neq 0$|, using the substitution |$u=\\frac {\\dot {x}}{x}$|, this reduces to a linear equation with the solution \\begin{align} x(t)=\\frac{x_{0}(bx_{0}-y_{0})}{bx_{0} - y_{0}e^{-(bx_{0}+y_{0})t}},\\quad x\\neq0. \\end{align} (2.9) From (2.7), the dependency of y with time is \\begin{align*} y(t)=y_{0} -b\\left(x_{0} -\\frac{x_{0} \\left(bx_{0} -y_{0} \\right)}{bx_{0} -y_{0} e^{-(bx_{0} +y_{0} )t} } \\right)\\!. \\end{align*} As |$t\\to \\infty$|, |$x\\to x_{0} -\\frac {y_{0} }{b}$|, and the time taken for y to be defeated is \\begin{align*} \\tau =\\frac{1}{\\left(bx_{0} +y_{0} \\right)} \\ln \\left(\\frac{y_{0} }{x_{0} \\left(b-1\\right)} \\right)\\!, \\end{align*} so that in the absence of intelligence to direct combat, the side that has a combat advantage multiplied by its initial force size greater than the opposition is victorious. 2.3. One side uses intelligence A limiting case for (2.2) is when one force, x say, has an intelligence capability that can be described by a constant coefficient b and the opposing force, y, has no intelligence capability to locate the enemy and instead must rely on area fire. In this situation, the Lanchester equations become \\begin{align} \\begin{aligned} &\\frac{dx}{dt}=-xy,\\\\ &\\frac{dy}{dt}=-bx,\\qquad b>0. \\end{aligned} \\end{align} (2.10) Eliminating t and solving gives a quadratic in y \\begin{align} {y_{0}^{2}} -y^{2} \\left(t\\right)=2b\\left(x_{0} -x(t)\\right)\\!. \\end{align} (2.11) So that for b > 1, for force y to win, |$y_{0}>\\sqrt {2bx_{o}}$| . Equation (2.11) implies that a force unable to use intelligence to locate an enemy, must outnumber the opposition by the square root of twice that force’s initial population multiplied by the square root of twice that force’s intelligence capability. Eliminating x from system (2.11) produces a second order, non-linear differential equation in y \\begin{align} \\ddot{y}+\\dot{y}y=0. \\end{align} (2.12) (2.12) can be solved using the substitution |$\\nu =\\dot {y}$| gives \\begin{align} y(t)=A\\left[(\\textrm{tanh}(At/2)+y_{0}/A)/(1+(y_{0}/A))\\left.(\\textrm{tanh}(At/2))\\right)\\right]\\!, \\end{align} (2.13) where |$A=\\sqrt {{y_{0}^{2}} -2bx_{0} }$|. If |$y_{0}>\\sqrt {2bx_{0}}$|, this implies that A ∈ |$\\mathbb{R}$| and that force y will beat force x with the size of force y reduced to A. From (2.11), the dependency of x with time is \\begin{align} x(t)=(2b)^{-1}\\left[2bx_{0}-{y^{2}_{0}}+A^{2}((\\tanh({At}/{2})+({y_{0}}/{A}))\\big(1+(y_{0}/{A})\\textrm{tanh}({At}/{2}))^{-1}\\big)^{2}\\right]\\!. \\end{align} (2.14) Hence (2.14) is a solution to the non-linear, second order differential equation \\begin{align} x\\ddot{x}-\\dot{x}^{2}-bx^{3}=0. \\end{align} (2.15) If |$y_{0} <\\sqrt {2bx_{0} }$|, then (2.12) can be integrated to give \\begin{align} y(t)=B\\left[\\left((\\,y_{0}/{B})-\\tan({Bt}/{2})\\right)\\left(1+(\\,y_{0}/{B})\\tan({Bt}/{2})\\right)^{-1}\\right]\\!, \\end{align} (2.16) where |$B=\\sqrt {2bx_{0} -{y_{0}^{2}}}$|. From (2.11), the dependency of x with time is \\begin{align} x(t)=(2b)^{-1}\\left[2bx_{0}-{y_{2}^{0}}+B^{2}\\left(\\left((\\,y_{0}/B)-\\tan(Bt/2)\\right)\\left(1+(\\,y_{0}/B)\\tan(Bt/2)\\right)^{-1}\\right)^{2}\\right]\\!. \\end{align} (2.17) When |$y_{0} <\\sqrt {2bx_{0} }$|, x, the force using intelligence defeats force y leaving force x reduced to |$\\frac {2bx_{0} -{y_{0}^{2}}}{b}$|. Similar equations were derived by Deitchman (1962) to model ambush attacks in guerilla warfare. (2.16) implies that the time to defeat y by x is |$\\frac {2}{B} \\arctan \\left (\\frac {y_{0} }{B} \\right )$| when |$y_{0} <\\sqrt {2bx_{0} }$| and when |$y_{0}>\\sqrt {2bx_{0} }$|, from (2.13), the time for x to be defeated by y is |$\\frac {2}{A} \\arctan h\\left (\\frac {y_{0} }{A} \\right )$|. In the combat model considered in Section 2.1 where both sides use intelligence, the parameter b represents the relative advantage in intelligence that one side (force x) has over the other. Typical values for b in Section 2.1 range from 1–10. When only force x uses intelligence, the parameter b is the absolute advantage in intelligence that x has over y. From (2.11), |$b\\ ~ \\ y_{0}$|, this suggests a clear advantage in force ratios and hence probability of success for a force exploiting intelligence against a force that does not. The Lanchester equations consider only the temporal effects of attrition warfare and not the spatial effects of the battlefield. In the next section we consider how intelligence value compares with intelligence capability using battlefield models that take account of both spatial and temporal factors. 3. Hyperbolic systems and relative intelligence To include spatial effects in Lanchester equations, consider two opposing forces with populations f and g, such that |$\\mathbb{R} ^{2}\\otimes \\mathbb{R} ^{+}\\rightarrow \\mathbb{R} ,\\,f=f(x,y,t)$| and g = g(x, y, t), where x and y are orthogonal directions in a plane. Suppose that both sides have an intelligence capability to exploit intelligence to locate each other’s forces in the battle space, then from (2.2), the Lanchester system of equations is \\begin{align} \\begin{aligned} &\\frac{df(x,y,t)}{dt}=-G(x,y,t,f,g)\\\\ &\\frac{dg(x,y,t)}{dt}=-bF(x,y,t,f,g)\\qquad b>0. \\end{aligned} \\end{align} (3.1) Then by the chain rule \\begin{align*} \\frac{df(x,y,t)}{dt}=\\frac{\\partial f}{\\partial t}+\\underline{u}\\cdot\\nabla f, \\end{align*} where |$\\underline {u}=\\frac {\\partial x}{\\partial t}+\\frac {\\partial y}{\\partial t}$|, can be considered to be the velocity of the force. Similarly, \\begin{align*} \\frac{dg(x,y,t)}{dt}=\\frac{\\partial g}{\\partial t}+\\underline{v}\\cdot\\nabla f, \\end{align*} where |$\\underline {v}=\\frac {\\partial x}{\\partial t}+\\frac {\\partial y}{\\partial t}$|. So that (3.1) can be written as a system of partial differential equations, a system of transport equations \\begin{align} U_{t}+AU_{x}=-B, \\end{align} (3.2) where |$U=\\left ({{f(x,y,t)}\\atop{g(x,y,t)}}\\right )$| and |$B=\\left ({{G(x,y,t,f,g)}\\atop {F(x,y,t,f,g)}}\\right )$|. If |$\\underline {u}=\\underline {v}$|, then (3.2) is parabolic; otherwise, (3.2) is a hyperbolic system. Initial conditions f(x, y, 0) = fo and g(x, y, 0) = go, along with suitable boundary conditions are required for solutions to exist to (3.2). 3.1. Both sides use intelligence The partial differential equation equivalent to the Lanchester equations when both sides use intelligence is \\begin{align} \\begin{aligned} &\\frac{\\partial f}{\\partial t}+u\\frac{\\partial f}{\\partial x}=-g\\\\ &\\frac{\\partial g}{\\partial t}+v\\frac{\\partial g}{\\partial x}=-b f,\\qquad b>0\\\\ &t\\geq0,\\quad x\\in[0,1] \\end{aligned} \\end{align} (3.3) with initial conditions \\begin{align*} f(\\,x_{t},0)=f_{0} \\end{align*} and \\begin{align*} g(\\,x,0)=g_{0}. \\end{align*} These initial conditions correspond to all of force f being initially located at some point x0, and all of force g being initially located at the origin. In this example, we consider the following Robin boundary conditions \\begin{align*} f(\\,x_{1},t)+\\left.\\frac{\\partial f}{dt}\\right|_{x=x_{1}}=0 \\end{align*} and \\begin{align*} g(\\,x_{1},t)+\\left.\\frac{\\partial g}{dt}\\right|_{x=x_{1}}=0. \\end{align*} These absorbing boundary conditions correspond to the removal of forces out of the battlespace, so forces crossing the boundary no longer have any military effect (Protopopescu et al., 1987). Writing this (2 × 2) hyperbolic system gives an inhomogeneous, linear equation in the form of \\begin{align} u_{t}+Au_{x}=Bu,\\quad x\\in \\left[0,1\\right]\\!,\\quad t\\in [0,+\\infty ), \\end{align} (3.4) where |$u:[0,1]\\times [0,\\infty )\\rightarrow \\mathbb{R} ^{2},uC^{2}$| and A is the diagonal matrix |$\\left ({{u}\\atop {0}}\\quad {{0}\\atop {v}}\\right )$| and B is the anti-diagonal matrix|$\\left ({{0}\\atop{-1}}\\quad {{-b}\\atop{0}}\\right )$|. Differentiating with respect to t and combining the two equations in (3.3) eliminates g(x, t) leaving a second‐order equation \\begin{align} f_{tt}+(u+v)f_{tx}+uvf_{xx}=bf. \\end{align} (3.5) The characteristic equations of (3.5) are |$\\frac {dx}{dt}=u$| and |$\\frac {dx}{dt}=v$|. Consider two functions μ(x, t) and η(x, t), that are constant along the characteristic directions, \\begin{align} \\mu(x, t)=x-ut \\end{align} (3.6) and \\begin{align} \\eta(x, t)=x-vt. \\end{align} (3.7) Using the chain rule to transform |$f(x,t)\\rightarrow f(\\mu (x,t),\\eta (x,t))$| reduces (3.5) to canonical form \\begin{align} f_{\\mu\\eta}+\\frac{b}{(u-v)^{2}}f=0. \\end{align} (3.8) Assuming that f(μ(x, t), η(x, t)) can be written as the product of two functions \\begin{align} f(x,t)=X(x)T(t). \\end{align} (3.9) Equation (3.9) can be solved using separation of variables technique to give two differential equations, \\begin{align*} \\frac{X^{\\prime}}{X}=-c\\frac{T}{T^{\\prime}}={\\lambda}, \\end{align*} where |$c=\\frac {b}{(u-v)^{2}}$|. From the boundary conditions \\begin{align} f(\\,x,t)=Ae^{\\lambda_{+}(x-ut)}e^{-\\frac{c}{\\lambda_{+}}(x-vt)}+Be^{\\lambda_{-}(\\,x-ut)}e^{-\\frac{c}{\\lambda_{-}}(x-vt)}, \\end{align} (3.10) where the two eigenvalues are given by \\begin{align} \\lambda_{\\pm}=\\frac{1}{2(u-v)}\\left\\{(v-u)\\pm\\sqrt{(u-v)^{2}+4b}\\right\\} \\end{align} (3.11) so that (3.10) can be simplified to \\begin{align} f(x,t)=Ae^{x}e^{\\left(\\frac{cv}{\\lambda_{+}}-\\lambda_{+}u\\right)t}+Be^{-x}e^{\\left(\\frac{cv}{\\lambda_{-}}-\\lambda_{-}u\\right)t}. \\end{align} (3.12) Using \\begin{align*} \\frac{\\partial f}{\\partial t}+u\\frac{\\partial f}{\\partial x}=-g \\end{align*} and the initial conditions, the constants can be found such that \\begin{align} A=A(u,v,x_{1},b,f_{0},g_{0}) \\end{align} (3.13) and \\begin{align} B=B(u,v,x_{1},b,f_{0},g_{0}). \\end{align} (3.14) Using the same method to write g as a decoupled second-order equation \\begin{align} g(x,t)=Ce^{x}e^{\\left(\\frac{cv}{\\lambda_{+}}-\\lambda_{+}u\\right)t}+De^{-x}e^{\\left(\\frac{cv}{\\lambda_{-}}-\\lambda_{-}u\\right)t}. \\end{align} (3.15) Applying the boundary conditions implies that \\begin{align} \\lambda^{\\prime}=\\lambda=\\frac{1}{2(u-v)}\\left\\{(v-u)\\pm\\sqrt{(u-v)^{2}+4b}\\right\\}\\!. \\end{align} (3.16) From \\begin{align*} \\frac{\\partial g}{\\partial t}+v\\frac{\\partial g}{\\partial x}=-b\\ f\\quad b>0 \\end{align*} and the initial conditions the determine the values of constants C and D \\begin{align} C=C(u,v,x_{1},b,f_{0},g_{0}) \\end{align} (3.17) and \\begin{align} D=g_{0}-C. \\end{align} (3.18) When both forces, use intelligence to direct their firepower, the eigenvalue equations (3.11) and (3.13) indicate that the combat advantage scales as the square root of the relative advantage in intelligence between opposing forces. Similar to the dependency found in the classical Lanchester model derived in Section 2. Modelling combat when both sides use intelligence as a series of partial differential equations introduces a dependency on the relative speed |u − v| between the combatants. In the case where the relative speed of the engagement is small compared with the relative intelligence superiority we recover the |$\\sqrt {b}$| dependency found in section 2. As the expressions obtained for the constants in (3.12) and (3.15) are not very amenable to analysis, a simplified example is helpful to see how the partial differential approach differs from the Lanchester equations when both sides use intelligence. 3.2. Both sides use intelligence; one side mounts a static defence Assume that force f has a superior advantage in intelligence relative to force g and moves with a speed u. Force g adopts a static defence, holding ground at the point x = 0, then v = 0 and (3.15) reduces to \\begin{align} g(x,t)=Ce^{-\\lambda_{+}ut}+(g_{0}-C)Ce^{-\\lambda_{-}ut}. \\end{align} (3.19) The time taken for force f to defeat force g is \\begin{align} \\tau =\\frac{1}{\\sqrt{u^{2} +4b}}\\ln(C-g_{0}), \\end{align} (3.20) where C is the constant given in (3.17), with v = 0. This compares with the time to defeat for the classical Lanchester engagement derived in Section 2 where the time to defeat is proportional to |$\\frac {1}{2\\sqrt {b}}$|. Figure 1 compares the classical Lanchester time to defeat with (3.20) for varying superiority in intelligence of the attacking force. For an attacker with superior intelligence and an advantageous force ratio of 4:3, the predicted time to defeat is shorter for the classical Lanchester combat model than that predicted by (3.20) over the range 1 < b < 10. Fig. 1. View largeDownload slide Time to defeat versus relative advantage in intelligence for the Lanchester combat model using (2.5) and the transport equation model (dashed line) using (3.20) with the speed of attack equal to 1.0. Fig. 1. View largeDownload slide Time to defeat versus relative advantage in intelligence for the Lanchester combat model using (2.5) and the transport equation model (dashed line) using (3.20) with the speed of attack equal to 1.0. Figures 2 and 3 compare the time to defeat for different attacking speeds over a range of superiority for the attacking force (1 < b < 10). When the attacking speed is less than the superiority in intelligence, the time to defeat decreases as intelligence superiority increases as in Fig 1. Conversely, if the attacking speed is much greater than the superiority in intelligence, then the time to defeat remains roughly constant as b increases (Fig. 2). This suggests that high tempo attacking operations are less sensitive to the effects of intelligence than slower operations. Fig. 2. View largeDownload slide Time to defeat versus relative advantage in intelligence for the Lanchester combat model using (2.5) and the transport equation model (dashed line) using (3.20) with the speed of attack equal to 50. Fig. 2. View largeDownload slide Time to defeat versus relative advantage in intelligence for the Lanchester combat model using (2.5) and the transport equation model (dashed line) using (3.20) with the speed of attack equal to 50. Fig. 3. View largeDownload slide Time to defeat versus relative advantage in intelligence for the Lanchester combat model using (2.5) and the transport equation model (dashed line) using (3.20) with the speed of attack u = 2.5. Fig. 3. View largeDownload slide Time to defeat versus relative advantage in intelligence for the Lanchester combat model using (2.5) and the transport equation model (dashed line) using (3.20) with the speed of attack u = 2.5. When the speed of attack is approximately equal to the superiority in intelligence the relationship between time to defeat and superiority in intelligence becomes more complicated. The time to defeat attains both a local minimum and a local maxima in the neighbourhood of b ≈ u (Fig. 3). A possible reason for this relationship is consider in the next section. The solution to the system of partial differential equations in (3.3) is dependent on the initial and boundary conditions, indicating that changes to both the tempo of operations, the initial position of combatants and the battlespace conditions (i.e. the boundary conditions) can all affect the utility of intelligence. 4. Discussion In contrast to previous models of intelligence value (e.g. Cesar et al., 1992) which considered intelligence as flow of information through forces in combat, in this paper the use (or absence) of intelligence is used to develop the structure of the equations in each combat model. The simplified Lanchester type models developed assume that with the use of intelligence to identify and locate enemy forces, the rate of attrition of the enemy force is proportional to its size. Without the use of intelligence to locate the enemy, the rate of attrition is proportional to the rate of fire which is proportional to both the size of the force and the size of the enemy. This section reviews some of the features of the models presented here and describes further work to investigate limitations with current approach. The intelligence parameter The combat models developed here used a single parameter b to represent the relative advantage in intelligence of force x over force y. No attempt was made to place a value on b. In these, very idealized, models the only difference between the forces except for their sizes was in their ability to process intelligence. All other factors such as types of weapons, rates of fire, combat effectiveness, etc are assumed to be the same. So that in Section 3.2, where no side used intelligence, b should be equal to unity. The parameter b is an aggregation of all types of intelligence ranging from tactical reconnaissance to all-source intelligence analysis. In reality, the process of collecting information and turning it into intelligence that can inform a commander’s decision-making operate at different timescales depending on the environment and the source of intelligence. Typically, high-level, all-source intelligence takes longer than tactical intelligence to process. There has been no attempt here to quantify the relationship between the time taken to conduct intelligence assessment and the accuracy of the final product. Further work will investigate the differences in intelligence value between rapid compared with in-depth intelligence analysis and assessment. To account for these different features of different intelligence sources b in (3.1) can be written as a summation of the functional dependence of each intelligence source \\begin{align*} b(t)=\\sum b_{i} (t), \\end{align*} where each bi is an independent source of intelligence, now a function of time. Giving b an explicit dependence on time alters the solutions to (3.1) and will be the subject of further work. Modifications The transport equations derived in Section 3.1 can be modified to take account of specific dependencies on the collection and processing of intelligence by one or both sides. For example, historical analysis of intelligence operations conducted in the Middle East in 1917 (Syk, 2009) indicate that for the British force attacking fixed Turkish positions, a situation analogous to the model developed in Section 3.2, the level of intelligence received was inversely proportional to the distance of the attacker from the defender. So that in (3.1), the intelligence function \\begin{align*} G \\propto \\frac{b}{|\\,x-x_{1}|}. \\end{align*} Other, refinements can be made to take account of functional dependencies on time and can be extended to include different sources of intelligence. Speed of attack versus speed of decision-making In Section 3.2 the time to defeat was dependent on the speed of the attacking force. Fig. 4 summarizes the variance on time to defeat for different values of u, the attacking speed. Fig. 4. View largeDownload slide Time to defeat versus relative advantage in intelligence for the Lanchester combat model using (2.5) and the transport equation model using (3.20) with the speed of attack u = 0.8, 2.5, 5, 10 and 40. The force ratio of attackers to defenders is 4:3. Fig. 4. View largeDownload slide Time to defeat versus relative advantage in intelligence for the Lanchester combat model using (2.5) and the transport equation model using (3.20) with the speed of attack u = 0.8, 2.5, 5, 10 and 40. The force ratio of attackers to defenders is 4:3. In the traditional Lanchester model, the time to defeat increases as b increases from 0 to 1, i.e. when the attacking force has an inferior intelligence capacity relative to the defender. When the attacking force has superior intelligence, the time to defeat decreases as b increases. From Fig. 4, it can be seen that the greatest reduction in time to defeat occurs as the attacker’s intelligence advantage increases from 1 to 2, after that the effects of intelligence become more marginal. Using the transport equation to model take account of the speed of the attacker, the time to defeat is greater than the Lanchester engagement for 0 < b < 10, unless the attacking speed is much greater than the relative advantage in intelligence. This reflects the time taken to close with the static defender. When |$u\\ ~\\ b$|, the relationship between the relationship between increasing the relative advantage in intelligence and the time to defeat becomes more complicated, as discussed in Section 3.2, with the curve attaining both local maxima and minima values over the range of b. The distance between the maxima and minima increases with the speed of approach. Further work will be conducted to investigate if this relationship is real or an artefact of the model. The consequence of this relationship is that for some speeds of approach superiority in intelligence acts as an advantage to the attacking force, but at other speeds, any advantage in intelligence does not assist the attacker and the defending force wins the engagement (i.e. the time to defeat is negative). This complexity may explain why different authors have come to different conclusions as to whether intelligence favours an attacker or a defender. Khan (1978) argued that intelligence is essential to victory only in defence; however, Herman’s (1996) review of a wider range of case studies than Khan, was inconclusive noting only that intelligence can favour the attacker in some cases and the defender in others. Decoupling the influence of intelligence received by a commander during combat from prior information may help in reconciling these differing conclusions. A possible explanation for the action of intelligence during manoeuvre warfare comes from considering the dimensions of the parameter b. In the transport equations (3.5) b has dimensions of |$\\left (time\\right )^{-1}$|, hence we can write \\begin{align*} b\\equiv b^{\\prime}\\omega, \\end{align*} where b′ is the (dimensionless) relative superiority in intelligence and ω represents the speed at which information is collected, processed and analysed to produce intelligence and disseminated to decision-makers. This suggests that when the speed at which intelligence is produced is less than the attacking speed, intelligence has little effect on the outcome of combat. Boundary conditions Solutions to the transport equations in Section 3 are highly dependent on the initial distribution of the two forces and the boundary conditions. Later work will investigate the range of boundary conditions for which unique solutions to (3.1) are possible. This dependency on initial conditions suggests that the influence of intelligence will be very different when both forces are concentrated and initially separate from each other as occurs during conventional warfare, compared with a situation when one force has initially infiltrated the other as is in the case during insurgency or terrorist conflicts. Conclusions Applying classical Lanchester models to the use of intelligence in conflict indicates that the intelligence does act as a force multiplier; however, its utility to compensate for inferior force ratio is less than commonly appreciated, proportional to the square root of the relative advantage in intelligence. Similarly, the time to defeat is proportional to |$\\frac {1}{\\sqrt {b} }$|, so that greatly increasing one side’s superiority in intelligence only produces a modest decrease in the time to defeat. This dependency is confirmed by the application of transport equations to model the use of intelligence in combat. References de Almeida , A. T. , Alencar , M. H. , Garcez , T. V. & Ferreira R. J. P. ( 2017 ) A systematic literature review of multicriteria and multi-objective models applied in risk management . IMA J. Manag. Mathematics , 28 , 153 -- 184. Google Scholar Crossref Search ADS Cesar , E. , Allen , P. & Eden , R. ( 1992 ) Finding a New Approach to Measure the Operational Value of Intelligence for Military Operations: Annotated Briefing . (No. RAND/N-3551-A). Santa Monica, CA : Rand Arroyo Center . Deitchman , S. J. ( 1962 ) A Lanchester Model of Guerrilla Warfare . Journal of the Operations Research Society of America , 10 , 818 -- 827. Dupuy , T. N. (1979) Numbers, Predictions and War: Using History to Evaluate Combat Factors and Predict the Outcome of Battles . London : Macdonald and Jane’s . González , E. & Villena , M. ( 2011 ) Spatial Lanchester models . Eur. J. Operational Res. , 210 , 706 -- 715. Google Scholar Crossref Search ADS Handel , M. I. ( 1990 ) Intelligence and military operations . Intelligence and National Security , 5 , 1 -- 95 . Google Scholar Crossref Search ADS Herman , M. ( 1996 ) Intelligence Power in Peace and War . Cambridge: Cambridge University Press. Johnson , M. R. ( 1996 ) Mathematical Computer Modelling , vol. 23 , pp. 73 -- 80 . Crossref Search ADS Kahn , D . ( 1978 ) Hitler’s Spies: German Military Intelligence in World War II . Da Capo , New York. Kahn , D . ( 2001 ) An historical theory of intelligence Intelligence and National Security , 16 , 79 -- 92. Google Scholar Crossref Search ADS Keane , T. ( 2011 ) Combat modelling with partial differential equations . Appl. Math Model., 35, 2723 -- 2735 Google Scholar Crossref Search ADS Keegan , J . ( 2003 ) Intelligence in War: Knowledge of the Enemy from Napoleon to al-Qaeda . New York : Alfred A. Knopf. Lanchester , F. W . ( 1916 ) Aircraft in Warfare: the Dawn of the Fourth Arm, Chapter V . London: Constable and Co. Ltd . Protopopescu , V. , Santoro , R. , Dockery , J. , Cox , R. & Barnes , J. ( 1987 ) Combat modeling with partial differential equations. Tech. Rep. ONRL/TM-10636, Oak Ridge National Laboratory . Protopopescu , V. , Santoro , R. & Dockery , J. ( 1989 ) Combat modeling with partial differential equations . Eur. J. Oper. Res., 38 , 178 -- 183 . Google Scholar Crossref Search ADS Riordan , B. ( 2003 ) The mathematics of O’Brien’s principle: an invitation to quantification . Intelligence and National Security , 18 , 168 -- 173 . Google Scholar Crossref Search ADS Rowland , D. ( 2006 ) The Stress of Battle: Quantifying Human Performance in Combat . London: TSO. Spradlin , C. & Spradlin , G. ( 2007 ) Lanchester’s equations in three dimensions. Comput. Math. Appl. , 53 999 -- 1011. Google Scholar Crossref Search ADS Syk , A. ( 2009 ) Command and the mesopotamia expeditionary force, 1915--1918 . DPhil thesis , Oxford. Yang , Z. L. & Qu , Z. ( 2016 ) Quantitative maritime security assessment: a 2020 vision . IMA J. Manag. Mathematics, 27 , 453 -- 470 Google Scholar Crossref Search ADS © The Author(s) 2018. Published by Oxford University Press on behalf of the Institute of Mathematics and its Applications. All rights reserved. This article is published and distributed under the terms of the Oxford University Press, Standard Journals Publication Model (https://academic.oup.com/journals/pages/open_access/funder_policies/chorus/standard_publication_model) http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png IMA Journal of Management Mathematics Oxford University Press\n\n# Lanchester modelling of intelligence in combat\n\nIMA Journal of Management Mathematics, Volume 30 (2) – Feb 1, 2019\n16 pages",
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"/lp/ou_press/lanchester-modelling-of-intelligence-in-combat-M40vNs6FT3\nPublisher\nOxford University Press\nISSN\n1471-678X\neISSN\n1471-6798\nDOI\n10.1093/imaman/dpx014\nPublisher site\nSee Article on Publisher Site\n\n### Abstract\n\nAbstract While the utility of intelligence as force multiplier during warfare is widely accepted there have been few attempts to quantify its benefits. In this paper Lanchester combat models are developed to understand how superiority in intelligence can compensate for an inferior force ratio and how the time for one side to defeat the other is affected by the use of intelligence. It is found that intelligence does act as a force multiplier; however, its utility to compensate for inferior force ratio is less than commonly appreciated, proportional to the square root of the relative advantage in intelligence. Similarly, the time to defeat is proportional to the inverse of the square root of the relative advantage in intelligence, so that greatly increasing one side’s superiority in intelligence only produces a modest decrease in the time to defeat. The Lanchester combat models are extended to a hyperbolic system of partial differential equation (PDE) to investigate how intelligence influences manoeuvre warfare. These suggest that high tempo attacking operations are less sensitive to the effects of intelligence than slower operations. 1. Introduction There have been few attempts to model the benefits that intelligence contributes to a force engaged in combat. Despite a lack of quantitative evidence, intelligence is widely seen as an enabling capability that confers distinct advantages to a force during combat. Handel (1990) describes intelligence as a ‘force multiplier’ and Khan (2001) calls intelligence an ‘optimizer of resources’. Conversely, Keegan (2003) considers that intelligence may be of benefit to a commander but does not ‘point out unerringly the path to victory’. Here, we develop models to quantify how much of a force multiplier intelligence acts as during military engagements. In this paper, we are concerned with operational intelligence, the process by which information on the nature of the terrain, enemy forces and their intentions is collected and analysed to predict the enemy’s future actions. Rather than distinguish between processed intelligence disseminated by a military force’s intelligence staff and target identification from reconnaissance assets, we consider intelligence to consist of all timely information that can assist a military commander’s decision-making. Every decision made by a commander can be considered as a set of possible outcomes for a number of events. For example, one decision may be whether to attack or defend. Both of these events will have a probability of success associated with them. The outcome of the decision cannot be known with full certainty until after the event has occurred; only the probability of the outcome can be estimated prior to the event. Here, intelligence is considered to be any information that enables a commander to estimate or refine the probability of an outcome. Many factors influence how individual commanders actually make decisions such as their personal experience, training, orders from superiors and advice from subordinates. Intelligence received during or prior to actual combat has to compete with all of these factors in order to influence the decision. This approach will be considered in later work. Previous models to examine the value of intelligence have combined dynamic models of combat with fixed decision points (Cesar et al., 1992) and the application of inventory control models to consider the use of intelligence in resource allocation problems (Riordan, 2003). Models to quantify decision making in risk management have recently been developed (Yang & Qu, 2016) share many similarities with modelling intelligence (de Almeida et al., 2017). One approach to building a model of intelligence value would be to assume that the effects of intelligence can be represented by, say, a power-law of the ratio of the sizes of the opposing forces. The value of the indices of the power-law could be found by fitting historical data of the attrition of one force against another over the course of an engagement. The problem with this method is that it is rare to find case studies from military history that unambiguously allow the effects of intelligence to be identified. Handel (1990) recognized the problem in studying the utility of intelligence in military operations is that its precise effects are difficult to isolate and measure with accuracy. This lack of reliable data sets limits the utility of a data fitting or stochastic methods in understanding the benefits that intelligence provides. Instead of trying to fit actual combat data to a model, we adopt a deterministic approach and consider a highly abstract military engagement against to forces where the only difference between them is their ability to use intelligence to direct combat power. This approach avoids the need to isolate the effects of intelligence in actual combat, by exploring its utility when it is contrived to be the only differentiator between combatants. A limitation with a deterministic approach is that it ignores the fact that intelligence collection and analysis are highly uncertain activities, where random variables often influence intelligence’s overall utility. While the use of deterministic modelling can provide an understanding of the overall effects of intelligence to explore the value of the different stages of acquiring and processing intelligence must include an element of uncertainty. One technique that may assist this understanding is to use of stochastic simulations of combat to model engagements between to forces. By repeating the same engagement but with varying levels of intelligence use for each force it should be possible to gain an understanding of how intelligence use effects the outcome of combat. This approach has been will be considered in later work. Here we develop a series of high-level Lanchester-type combat models to quantify the value of intelligence using the three traditional high-level measures of effect used to evaluate military operations: the probability of success, time take to achieve an effect and the number of casualties (attrition) (Rowland, 2006). Lanchester type equations have previously been used to model aimed and unaimed fire between two forces (Keane, 2011) and the value of reconnaissance in combat (Johnson, 1996); both these studies only considered attrition as a measure of effect. The aim of the Lanchester approach is to model the relative ability of each force to use intelligence to locate and manoeuvre to engage an opponent. We assume that the only difference in combat power between the opposing forces is their ability to use intelligence. Further, we will model the effects of using intelligence at the operational level, so that we are concerned with the overall ability of a force to correctly identify the course of action of the opposing force at the highest level. These two assumptions are consistent with the description of operational intelligence given above; however, one limitation with this approach is that it may not give sufficient weight to tactical differences between the forces, such as differences in speed of manoeuvre. As a starting point to understand this limitation, we introduce a system of partial differential equations to consider spatial effects in Section 3. The models describe how the use of intelligence by each force affects its ability to locate and target or manoeuvre to engage the other force at the operational rather than the tactical level. That is we are concerned with how intelligence aids the efficient deployment of large formations rather than the performance of individuals or small units. Hence the form of the equations (e.g. linear, squared, etc) relating the change in the size of each force is of greater interest than determining values for the parameters used in a given model. The structure of this paper is as follows, Section 2 develops Lanchester’s equations to model the combat multiplying effect of intelligence when both sides make use of intelligence, neither side uses intelligence, and where one side uses intelligence. While the form and the solutions of Lanchester’s equations presented here are well known, their application to determine the impact of intelligence on each of three measure of campaign effect is novel. Section 3 develops Lanchester type partial differential equations to take account of spatial and temporal changes during combat. Protopopescu et al. (1987; 1989) derived a reaction-diffusion equation form of Lanchester’s equations to model movement in space and time. This approach has been extended by Spradlin & Spradlin (2007) to take account of three-dimensional movement across the battlespace. The diffusion approach has been criticized (Keane, 2011; González & Villena, 2011) for introducing smearing, an unrealistic random-walk type motion, that is not representative of troop movements. To overcome these limitations, we propose an original system of linear, hyperbolic partial differential equations, similar in form to the transport equation. Section 4 describes some of the limitations to the simplified models presented here, using the time for defeat of one force by another as a representative measure of success, and discusses some extensions to overcome these limitations. 2. Lanchester modelling 2.1. Both sides use intelligence Consider two opposing forces x and y with time varying populations, x = x(t) and y = y(t) with initial populations x(0) = x0 and y(0) = y0, respectively. Lanchester (1916) modelled combat between the two forces as a coupled pair of differential equations, with rate of loss (i.e. casualty rate) on each side proportional to the size of the opposing force. So that at time t, \\begin{align} \\begin{aligned} &\\frac{dx}{dt}=-ay,\\qquad a>0,\\\\ &\\frac{dy}{dt}=-bx,\\qquad b>0. \\end{aligned} \\end{align} (2.1) The factors a and b, assumed to be positive constant in (2.1) represent the weighting factors between the effectiveness of fire for x and y. In general, the values of these constants will be a combination of different factors including firepower, target acquisition and manoeuvrability. Lanchester considered (2.1) to represent two forces engaging each other with aimed (i.e. targeted) fire. Rather than represent the effectiveness of fire, the models introduced in this paper consider the relative ability of each force to use intelligence to locate and manoeuvre to engage its opponent. To model the effects of intelligence we assume that both forces have identical weapons and capabilities and the only difference between them is their capability to collect and process operational intelligence. Since we are only interested in the relative values of constants a and b we can rescale the problem by normalizing with respect to a and write (2.1) as a Cauchy problem \\begin{align} \\dot{\\boldsymbol{x}} = A{\\boldsymbol{x}},\\quad{\\boldsymbol{x}}\\left(0\\right)={{\\boldsymbol{x}}}_{{\\mathbf 0}} = \\left( \\begin{array}{@{}c@{}} x_{0} \\\\ y_{0} \\end{array} \\right)\\! , \\end{align} (2.2) where A is a (2 × 2) matrix with eigenvalues |$\\pm \\sqrt {b}$| and corresponding eigenvectors |$\\underline {v}_{1}=\\left ({1\\atop{-\\sqrt {b}}}\\right )$| for |$\\sqrt {b}$| and |$\\underline {v}_{2}=\\left ({1\\atop{\\sqrt {b}}}\\right )$| for |$-\\sqrt {b}$|. So that the rate of decrease in the size of each force is proportional to the square root of its relative capability to use intelligence. For realistic combat modelling we are only concerned with the region x ≥ 0 and y ≥ 0. The solution to (2.2) is \\begin{align} {\\boldsymbol{x}}\\left(t\\right)={{\\boldsymbol{e}}}^{{\\boldsymbol{At}}}{{\\boldsymbol{x}}}_{0,} \\end{align} (2.3) where A in (2.2) has two distinct, independent eigenvectors |${{\\boldsymbol{e}}}^{{\\boldsymbol{At}}}=Ve^{Dt}V^{-1}$|, where V is the matrix of eigenvectors, |$V = \\left ( {\\boldsymbol{v}}_{1} \\quad {\\boldsymbol{v}}_{2} \\right )$| and D is the diagonal matrix of the eigenvalues of A. \\begin{align*} {\\boldsymbol{e}}^{\\boldsymbol{At}}=\\left( \\begin{array}{@{}cc@{}} \\cosh\\sqrt{b}t & -\\frac{1}{\\sqrt{b}}\\sinh\\sqrt{b}t \\\\ -\\sqrt{b}\\sinh\\sqrt{b}t & \\cosh\\sqrt{b}t \\end{array} \\right)\\!. \\end{align*} So, \\begin{align*} x(t)=x_{0} \\cosh\\sqrt{b}t-\\frac{y_{0}}{\\sqrt{b}}\\sinh\\sqrt{b}t \\end{align*} and \\begin{align*} y(t)=y_{0} \\cosh\\sqrt{b}t-x_{0}\\sqrt{b}\\sinh\\sqrt{b}t. \\end{align*} Eliminating t from (2.2) gives the well-known Lanchester square law \\begin{align} {y_{0}^{2}}-y^{2} =b\\left({x_{0}^{2}}-x^{2}\\right)\\!. \\end{align} (2.4) If the opposing forces are of equal size at the start of the conflict, the force with the superior target acquisition intelligence wins: so b > 1 implies that force x wins and b < 1 implies that force y wins. If b > 1, (2.4) implies that force y can still win the engagement if its initial strength is greater than the multiple of the square root of force x’s relative advantage in intelligence capability, i.e. |$y_{0}>\\sqrt {b}\\ x_{0}$|. Similarly, if b < 1, force x can win if |$x_{0}>\\frac {y_{0}}{\\sqrt {b}}$|. Hence, applying Lanchester’s model of direct fire implies that the capability of intelligence in terms of compensating for inferior numbers is proportional to the square of the relative advantage in intelligence. So to defeat a force twice its size an army would need an intelligence capability/advantage four times greater than its numerically superior foe. The value of intelligence to one side is proportional to the square root of their advantage, or disadvantage, in intelligence capability over their opponent. There may be other limitations that prevent a force out-thinking its rival in this manner. Historical analysis of actual military combat (Dupuy, 1979) suggests that if one force outnumbers the other by greater than 6:1 it has a probability of victory of >95%. If x0 > y0 and b > 1, the time taken for force x to defeat force y, τ, is \\begin{align} \\tau=\\frac{1}{2\\sqrt{b}}\\ln\\left(\\frac{\\sqrt{b}x_{0}+y_{0}}{\\sqrt{b}x_{0}-y_{0}}\\right)\\!. \\end{align} (2.5) The time to defeat is not particularly sensitive to the initial force ratio |$\\frac {x_{0}}{y_{0}}$|; however, doubling b, force x’s intelligence capability more than halves the time for x to defeat y. Hence, (2.5) suggests that for a superior force, increasing its intelligence capability increases its operational tempo by a greater amount than the increase in capacity. The third measure of effectiveness is the number of casualties suffered by one side after a given time. As above, if x0 > y0 and b > 1 after a time |$\\frac {1}{\\sqrt {b}}$|, force x is reduced by a factor of \\begin{align*} \\left\\{ e^{1} \\left(1-\\left(\\sqrt{b} x_{0} /y_{0} \\right)^{-1}\\right)+e^{-1} \\left(1+\\left(\\sqrt{b} x_{0} /y_{0}\\right)^{-1}\\right) \\right\\} \\Big/2. \\end{align*} As with (2.5), the number of casualties suffered is not very sensitive to the initial force ratio, but doubling force x’s intelligence capability reduces its attrition by slightly less than a factor of 2. Operational level Lanchester modelling of two forces both using intelligence, but where one side has a relative advantage in intelligence capability over the other, suggests that if the dominant force doubles its intelligence capability, it increases its probability of success by only a factor of |$\\sqrt {2}$|. The operational tempo; however, is slightly more than doubled and the attrition is reduced by around a half. The value of intelligence appears to be greatest in reducing operational tempo, followed by reduction of casualties; while the increase in probability of success is only proportional to the square root in the increase in intelligence capability. 2.2. Neither side uses intelligence Lanchester also considered two opposing forces engaging each other with unaimed or area fire. In this case, the rate of decease of one side is proportional to its own population (i.e. the size of the target) as well as the population of the opposing force (rate of fire). The equations for area fire are \\begin{align} \\begin{aligned} \\frac{dx}{dt}=-axy,\\qquad a>0 \\;\\\\ \\frac{dy}{dt}=-bxy,\\qquad a<0. \\end{aligned} \\end{align} (2.6) In an intelligence context, (2.6) can be said to represent the case where neither side uses intelligence to locate the enemy. Solving (2.6) gives a linear relationship between the populations of both forces \\begin{align} a\\left(\\,y_{0}-y(t)\\right)=b\\left(\\,x_{0}-x(t)\\right)\\!. \\end{align} (2.7) When a = 1, if the opposing forces are of equal size at the start of the conflict, neither side uses intelligence, b > 1 implies that force x wins and b < 1 implies that force y wins. If b > 1, (2.7) implies that force y can still win the engagement if its initial strength is greater than the multiple of force x’s non-intelligence capability (e.g. firepower or training), i.e. y0 > bx0. Similarly, if b < 1, force x can win if |$x_{0}>\\frac {y_{0}}{b}$|. Compared with the square-root dependency of (2.4). With a = 1, and bx0 > y0, eliminating y from system (2.6) produces a second order, non‐linear differential equation in x \\begin{align} x\\ddot{x}-\\dot{x}^{2}+b\\dot{x}x^{2}=0. \\end{align} (2.8) For |$x\\neq 0$|, using the substitution |$u=\\frac {\\dot {x}}{x}$|, this reduces to a linear equation with the solution \\begin{align} x(t)=\\frac{x_{0}(bx_{0}-y_{0})}{bx_{0} - y_{0}e^{-(bx_{0}+y_{0})t}},\\quad x\\neq0. \\end{align} (2.9) From (2.7), the dependency of y with time is \\begin{align*} y(t)=y_{0} -b\\left(x_{0} -\\frac{x_{0} \\left(bx_{0} -y_{0} \\right)}{bx_{0} -y_{0} e^{-(bx_{0} +y_{0} )t} } \\right)\\!. \\end{align*} As |$t\\to \\infty$|, |$x\\to x_{0} -\\frac {y_{0} }{b}$|, and the time taken for y to be defeated is \\begin{align*} \\tau =\\frac{1}{\\left(bx_{0} +y_{0} \\right)} \\ln \\left(\\frac{y_{0} }{x_{0} \\left(b-1\\right)} \\right)\\!, \\end{align*} so that in the absence of intelligence to direct combat, the side that has a combat advantage multiplied by its initial force size greater than the opposition is victorious. 2.3. One side uses intelligence A limiting case for (2.2) is when one force, x say, has an intelligence capability that can be described by a constant coefficient b and the opposing force, y, has no intelligence capability to locate the enemy and instead must rely on area fire. In this situation, the Lanchester equations become \\begin{align} \\begin{aligned} &\\frac{dx}{dt}=-xy,\\\\ &\\frac{dy}{dt}=-bx,\\qquad b>0. \\end{aligned} \\end{align} (2.10) Eliminating t and solving gives a quadratic in y \\begin{align} {y_{0}^{2}} -y^{2} \\left(t\\right)=2b\\left(x_{0} -x(t)\\right)\\!. \\end{align} (2.11) So that for b > 1, for force y to win, |$y_{0}>\\sqrt {2bx_{o}}$| . Equation (2.11) implies that a force unable to use intelligence to locate an enemy, must outnumber the opposition by the square root of twice that force’s initial population multiplied by the square root of twice that force’s intelligence capability. Eliminating x from system (2.11) produces a second order, non-linear differential equation in y \\begin{align} \\ddot{y}+\\dot{y}y=0. \\end{align} (2.12) (2.12) can be solved using the substitution |$\\nu =\\dot {y}$| gives \\begin{align} y(t)=A\\left[(\\textrm{tanh}(At/2)+y_{0}/A)/(1+(y_{0}/A))\\left.(\\textrm{tanh}(At/2))\\right)\\right]\\!, \\end{align} (2.13) where |$A=\\sqrt {{y_{0}^{2}} -2bx_{0} }$|. If |$y_{0}>\\sqrt {2bx_{0}}$|, this implies that A ∈ |$\\mathbb{R}$| and that force y will beat force x with the size of force y reduced to A. From (2.11), the dependency of x with time is \\begin{align} x(t)=(2b)^{-1}\\left[2bx_{0}-{y^{2}_{0}}+A^{2}((\\tanh({At}/{2})+({y_{0}}/{A}))\\big(1+(y_{0}/{A})\\textrm{tanh}({At}/{2}))^{-1}\\big)^{2}\\right]\\!. \\end{align} (2.14) Hence (2.14) is a solution to the non-linear, second order differential equation \\begin{align} x\\ddot{x}-\\dot{x}^{2}-bx^{3}=0. \\end{align} (2.15) If |$y_{0} <\\sqrt {2bx_{0} }$|, then (2.12) can be integrated to give \\begin{align} y(t)=B\\left[\\left((\\,y_{0}/{B})-\\tan({Bt}/{2})\\right)\\left(1+(\\,y_{0}/{B})\\tan({Bt}/{2})\\right)^{-1}\\right]\\!, \\end{align} (2.16) where |$B=\\sqrt {2bx_{0} -{y_{0}^{2}}}$|. From (2.11), the dependency of x with time is \\begin{align} x(t)=(2b)^{-1}\\left[2bx_{0}-{y_{2}^{0}}+B^{2}\\left(\\left((\\,y_{0}/B)-\\tan(Bt/2)\\right)\\left(1+(\\,y_{0}/B)\\tan(Bt/2)\\right)^{-1}\\right)^{2}\\right]\\!. \\end{align} (2.17) When |$y_{0} <\\sqrt {2bx_{0} }$|, x, the force using intelligence defeats force y leaving force x reduced to |$\\frac {2bx_{0} -{y_{0}^{2}}}{b}$|. Similar equations were derived by Deitchman (1962) to model ambush attacks in guerilla warfare. (2.16) implies that the time to defeat y by x is |$\\frac {2}{B} \\arctan \\left (\\frac {y_{0} }{B} \\right )$| when |$y_{0} <\\sqrt {2bx_{0} }$| and when |$y_{0}>\\sqrt {2bx_{0} }$|, from (2.13), the time for x to be defeated by y is |$\\frac {2}{A} \\arctan h\\left (\\frac {y_{0} }{A} \\right )$|. In the combat model considered in Section 2.1 where both sides use intelligence, the parameter b represents the relative advantage in intelligence that one side (force x) has over the other. Typical values for b in Section 2.1 range from 1–10. When only force x uses intelligence, the parameter b is the absolute advantage in intelligence that x has over y. From (2.11), |$b\\ ~ \\ y_{0}$|, this suggests a clear advantage in force ratios and hence probability of success for a force exploiting intelligence against a force that does not. The Lanchester equations consider only the temporal effects of attrition warfare and not the spatial effects of the battlefield. In the next section we consider how intelligence value compares with intelligence capability using battlefield models that take account of both spatial and temporal factors. 3. Hyperbolic systems and relative intelligence To include spatial effects in Lanchester equations, consider two opposing forces with populations f and g, such that |$\\mathbb{R} ^{2}\\otimes \\mathbb{R} ^{+}\\rightarrow \\mathbb{R} ,\\,f=f(x,y,t)$| and g = g(x, y, t), where x and y are orthogonal directions in a plane. Suppose that both sides have an intelligence capability to exploit intelligence to locate each other’s forces in the battle space, then from (2.2), the Lanchester system of equations is \\begin{align} \\begin{aligned} &\\frac{df(x,y,t)}{dt}=-G(x,y,t,f,g)\\\\ &\\frac{dg(x,y,t)}{dt}=-bF(x,y,t,f,g)\\qquad b>0. \\end{aligned} \\end{align} (3.1) Then by the chain rule \\begin{align*} \\frac{df(x,y,t)}{dt}=\\frac{\\partial f}{\\partial t}+\\underline{u}\\cdot\\nabla f, \\end{align*} where |$\\underline {u}=\\frac {\\partial x}{\\partial t}+\\frac {\\partial y}{\\partial t}$|, can be considered to be the velocity of the force. Similarly, \\begin{align*} \\frac{dg(x,y,t)}{dt}=\\frac{\\partial g}{\\partial t}+\\underline{v}\\cdot\\nabla f, \\end{align*} where |$\\underline {v}=\\frac {\\partial x}{\\partial t}+\\frac {\\partial y}{\\partial t}$|. So that (3.1) can be written as a system of partial differential equations, a system of transport equations \\begin{align} U_{t}+AU_{x}=-B, \\end{align} (3.2) where |$U=\\left ({{f(x,y,t)}\\atop{g(x,y,t)}}\\right )$| and |$B=\\left ({{G(x,y,t,f,g)}\\atop {F(x,y,t,f,g)}}\\right )$|. If |$\\underline {u}=\\underline {v}$|, then (3.2) is parabolic; otherwise, (3.2) is a hyperbolic system. Initial conditions f(x, y, 0) = fo and g(x, y, 0) = go, along with suitable boundary conditions are required for solutions to exist to (3.2). 3.1. Both sides use intelligence The partial differential equation equivalent to the Lanchester equations when both sides use intelligence is \\begin{align} \\begin{aligned} &\\frac{\\partial f}{\\partial t}+u\\frac{\\partial f}{\\partial x}=-g\\\\ &\\frac{\\partial g}{\\partial t}+v\\frac{\\partial g}{\\partial x}=-b f,\\qquad b>0\\\\ &t\\geq0,\\quad x\\in[0,1] \\end{aligned} \\end{align} (3.3) with initial conditions \\begin{align*} f(\\,x_{t},0)=f_{0} \\end{align*} and \\begin{align*} g(\\,x,0)=g_{0}. \\end{align*} These initial conditions correspond to all of force f being initially located at some point x0, and all of force g being initially located at the origin. In this example, we consider the following Robin boundary conditions \\begin{align*} f(\\,x_{1},t)+\\left.\\frac{\\partial f}{dt}\\right|_{x=x_{1}}=0 \\end{align*} and \\begin{align*} g(\\,x_{1},t)+\\left.\\frac{\\partial g}{dt}\\right|_{x=x_{1}}=0. \\end{align*} These absorbing boundary conditions correspond to the removal of forces out of the battlespace, so forces crossing the boundary no longer have any military effect (Protopopescu et al., 1987). Writing this (2 × 2) hyperbolic system gives an inhomogeneous, linear equation in the form of \\begin{align} u_{t}+Au_{x}=Bu,\\quad x\\in \\left[0,1\\right]\\!,\\quad t\\in [0,+\\infty ), \\end{align} (3.4) where |$u:[0,1]\\times [0,\\infty )\\rightarrow \\mathbb{R} ^{2},uC^{2}$| and A is the diagonal matrix |$\\left ({{u}\\atop {0}}\\quad {{0}\\atop {v}}\\right )$| and B is the anti-diagonal matrix|$\\left ({{0}\\atop{-1}}\\quad {{-b}\\atop{0}}\\right )$|. Differentiating with respect to t and combining the two equations in (3.3) eliminates g(x, t) leaving a second‐order equation \\begin{align} f_{tt}+(u+v)f_{tx}+uvf_{xx}=bf. \\end{align} (3.5) The characteristic equations of (3.5) are |$\\frac {dx}{dt}=u$| and |$\\frac {dx}{dt}=v$|. Consider two functions μ(x, t) and η(x, t), that are constant along the characteristic directions, \\begin{align} \\mu(x, t)=x-ut \\end{align} (3.6) and \\begin{align} \\eta(x, t)=x-vt. \\end{align} (3.7) Using the chain rule to transform |$f(x,t)\\rightarrow f(\\mu (x,t),\\eta (x,t))$| reduces (3.5) to canonical form \\begin{align} f_{\\mu\\eta}+\\frac{b}{(u-v)^{2}}f=0. \\end{align} (3.8) Assuming that f(μ(x, t), η(x, t)) can be written as the product of two functions \\begin{align} f(x,t)=X(x)T(t). \\end{align} (3.9) Equation (3.9) can be solved using separation of variables technique to give two differential equations, \\begin{align*} \\frac{X^{\\prime}}{X}=-c\\frac{T}{T^{\\prime}}={\\lambda}, \\end{align*} where |$c=\\frac {b}{(u-v)^{2}}$|. From the boundary conditions \\begin{align} f(\\,x,t)=Ae^{\\lambda_{+}(x-ut)}e^{-\\frac{c}{\\lambda_{+}}(x-vt)}+Be^{\\lambda_{-}(\\,x-ut)}e^{-\\frac{c}{\\lambda_{-}}(x-vt)}, \\end{align} (3.10) where the two eigenvalues are given by \\begin{align} \\lambda_{\\pm}=\\frac{1}{2(u-v)}\\left\\{(v-u)\\pm\\sqrt{(u-v)^{2}+4b}\\right\\} \\end{align} (3.11) so that (3.10) can be simplified to \\begin{align} f(x,t)=Ae^{x}e^{\\left(\\frac{cv}{\\lambda_{+}}-\\lambda_{+}u\\right)t}+Be^{-x}e^{\\left(\\frac{cv}{\\lambda_{-}}-\\lambda_{-}u\\right)t}. \\end{align} (3.12) Using \\begin{align*} \\frac{\\partial f}{\\partial t}+u\\frac{\\partial f}{\\partial x}=-g \\end{align*} and the initial conditions, the constants can be found such that \\begin{align} A=A(u,v,x_{1},b,f_{0},g_{0}) \\end{align} (3.13) and \\begin{align} B=B(u,v,x_{1},b,f_{0},g_{0}). \\end{align} (3.14) Using the same method to write g as a decoupled second-order equation \\begin{align} g(x,t)=Ce^{x}e^{\\left(\\frac{cv}{\\lambda_{+}}-\\lambda_{+}u\\right)t}+De^{-x}e^{\\left(\\frac{cv}{\\lambda_{-}}-\\lambda_{-}u\\right)t}. \\end{align} (3.15) Applying the boundary conditions implies that \\begin{align} \\lambda^{\\prime}=\\lambda=\\frac{1}{2(u-v)}\\left\\{(v-u)\\pm\\sqrt{(u-v)^{2}+4b}\\right\\}\\!. \\end{align} (3.16) From \\begin{align*} \\frac{\\partial g}{\\partial t}+v\\frac{\\partial g}{\\partial x}=-b\\ f\\quad b>0 \\end{align*} and the initial conditions the determine the values of constants C and D \\begin{align} C=C(u,v,x_{1},b,f_{0},g_{0}) \\end{align} (3.17) and \\begin{align} D=g_{0}-C. \\end{align} (3.18) When both forces, use intelligence to direct their firepower, the eigenvalue equations (3.11) and (3.13) indicate that the combat advantage scales as the square root of the relative advantage in intelligence between opposing forces. Similar to the dependency found in the classical Lanchester model derived in Section 2. Modelling combat when both sides use intelligence as a series of partial differential equations introduces a dependency on the relative speed |u − v| between the combatants. In the case where the relative speed of the engagement is small compared with the relative intelligence superiority we recover the |$\\sqrt {b}$| dependency found in section 2. As the expressions obtained for the constants in (3.12) and (3.15) are not very amenable to analysis, a simplified example is helpful to see how the partial differential approach differs from the Lanchester equations when both sides use intelligence. 3.2. Both sides use intelligence; one side mounts a static defence Assume that force f has a superior advantage in intelligence relative to force g and moves with a speed u. Force g adopts a static defence, holding ground at the point x = 0, then v = 0 and (3.15) reduces to \\begin{align} g(x,t)=Ce^{-\\lambda_{+}ut}+(g_{0}-C)Ce^{-\\lambda_{-}ut}. \\end{align} (3.19) The time taken for force f to defeat force g is \\begin{align} \\tau =\\frac{1}{\\sqrt{u^{2} +4b}}\\ln(C-g_{0}), \\end{align} (3.20) where C is the constant given in (3.17), with v = 0. This compares with the time to defeat for the classical Lanchester engagement derived in Section 2 where the time to defeat is proportional to |$\\frac {1}{2\\sqrt {b}}$|. Figure 1 compares the classical Lanchester time to defeat with (3.20) for varying superiority in intelligence of the attacking force. For an attacker with superior intelligence and an advantageous force ratio of 4:3, the predicted time to defeat is shorter for the classical Lanchester combat model than that predicted by (3.20) over the range 1 < b < 10. Fig. 1. View largeDownload slide Time to defeat versus relative advantage in intelligence for the Lanchester combat model using (2.5) and the transport equation model (dashed line) using (3.20) with the speed of attack equal to 1.0. Fig. 1. View largeDownload slide Time to defeat versus relative advantage in intelligence for the Lanchester combat model using (2.5) and the transport equation model (dashed line) using (3.20) with the speed of attack equal to 1.0. Figures 2 and 3 compare the time to defeat for different attacking speeds over a range of superiority for the attacking force (1 < b < 10). When the attacking speed is less than the superiority in intelligence, the time to defeat decreases as intelligence superiority increases as in Fig 1. Conversely, if the attacking speed is much greater than the superiority in intelligence, then the time to defeat remains roughly constant as b increases (Fig. 2). This suggests that high tempo attacking operations are less sensitive to the effects of intelligence than slower operations. Fig. 2. View largeDownload slide Time to defeat versus relative advantage in intelligence for the Lanchester combat model using (2.5) and the transport equation model (dashed line) using (3.20) with the speed of attack equal to 50. Fig. 2. View largeDownload slide Time to defeat versus relative advantage in intelligence for the Lanchester combat model using (2.5) and the transport equation model (dashed line) using (3.20) with the speed of attack equal to 50. Fig. 3. View largeDownload slide Time to defeat versus relative advantage in intelligence for the Lanchester combat model using (2.5) and the transport equation model (dashed line) using (3.20) with the speed of attack u = 2.5. Fig. 3. View largeDownload slide Time to defeat versus relative advantage in intelligence for the Lanchester combat model using (2.5) and the transport equation model (dashed line) using (3.20) with the speed of attack u = 2.5. When the speed of attack is approximately equal to the superiority in intelligence the relationship between time to defeat and superiority in intelligence becomes more complicated. The time to defeat attains both a local minimum and a local maxima in the neighbourhood of b ≈ u (Fig. 3). A possible reason for this relationship is consider in the next section. The solution to the system of partial differential equations in (3.3) is dependent on the initial and boundary conditions, indicating that changes to both the tempo of operations, the initial position of combatants and the battlespace conditions (i.e. the boundary conditions) can all affect the utility of intelligence. 4. Discussion In contrast to previous models of intelligence value (e.g. Cesar et al., 1992) which considered intelligence as flow of information through forces in combat, in this paper the use (or absence) of intelligence is used to develop the structure of the equations in each combat model. The simplified Lanchester type models developed assume that with the use of intelligence to identify and locate enemy forces, the rate of attrition of the enemy force is proportional to its size. Without the use of intelligence to locate the enemy, the rate of attrition is proportional to the rate of fire which is proportional to both the size of the force and the size of the enemy. This section reviews some of the features of the models presented here and describes further work to investigate limitations with current approach. The intelligence parameter The combat models developed here used a single parameter b to represent the relative advantage in intelligence of force x over force y. No attempt was made to place a value on b. In these, very idealized, models the only difference between the forces except for their sizes was in their ability to process intelligence. All other factors such as types of weapons, rates of fire, combat effectiveness, etc are assumed to be the same. So that in Section 3.2, where no side used intelligence, b should be equal to unity. The parameter b is an aggregation of all types of intelligence ranging from tactical reconnaissance to all-source intelligence analysis. In reality, the process of collecting information and turning it into intelligence that can inform a commander’s decision-making operate at different timescales depending on the environment and the source of intelligence. Typically, high-level, all-source intelligence takes longer than tactical intelligence to process. There has been no attempt here to quantify the relationship between the time taken to conduct intelligence assessment and the accuracy of the final product. Further work will investigate the differences in intelligence value between rapid compared with in-depth intelligence analysis and assessment. To account for these different features of different intelligence sources b in (3.1) can be written as a summation of the functional dependence of each intelligence source \\begin{align*} b(t)=\\sum b_{i} (t), \\end{align*} where each bi is an independent source of intelligence, now a function of time. Giving b an explicit dependence on time alters the solutions to (3.1) and will be the subject of further work. Modifications The transport equations derived in Section 3.1 can be modified to take account of specific dependencies on the collection and processing of intelligence by one or both sides. For example, historical analysis of intelligence operations conducted in the Middle East in 1917 (Syk, 2009) indicate that for the British force attacking fixed Turkish positions, a situation analogous to the model developed in Section 3.2, the level of intelligence received was inversely proportional to the distance of the attacker from the defender. So that in (3.1), the intelligence function \\begin{align*} G \\propto \\frac{b}{|\\,x-x_{1}|}. \\end{align*} Other, refinements can be made to take account of functional dependencies on time and can be extended to include different sources of intelligence. Speed of attack versus speed of decision-making In Section 3.2 the time to defeat was dependent on the speed of the attacking force. Fig. 4 summarizes the variance on time to defeat for different values of u, the attacking speed. Fig. 4. View largeDownload slide Time to defeat versus relative advantage in intelligence for the Lanchester combat model using (2.5) and the transport equation model using (3.20) with the speed of attack u = 0.8, 2.5, 5, 10 and 40. The force ratio of attackers to defenders is 4:3. Fig. 4. View largeDownload slide Time to defeat versus relative advantage in intelligence for the Lanchester combat model using (2.5) and the transport equation model using (3.20) with the speed of attack u = 0.8, 2.5, 5, 10 and 40. The force ratio of attackers to defenders is 4:3. In the traditional Lanchester model, the time to defeat increases as b increases from 0 to 1, i.e. when the attacking force has an inferior intelligence capacity relative to the defender. When the attacking force has superior intelligence, the time to defeat decreases as b increases. From Fig. 4, it can be seen that the greatest reduction in time to defeat occurs as the attacker’s intelligence advantage increases from 1 to 2, after that the effects of intelligence become more marginal. Using the transport equation to model take account of the speed of the attacker, the time to defeat is greater than the Lanchester engagement for 0 < b < 10, unless the attacking speed is much greater than the relative advantage in intelligence. This reflects the time taken to close with the static defender. When |$u\\ ~\\ b$|, the relationship between the relationship between increasing the relative advantage in intelligence and the time to defeat becomes more complicated, as discussed in Section 3.2, with the curve attaining both local maxima and minima values over the range of b. The distance between the maxima and minima increases with the speed of approach. Further work will be conducted to investigate if this relationship is real or an artefact of the model. The consequence of this relationship is that for some speeds of approach superiority in intelligence acts as an advantage to the attacking force, but at other speeds, any advantage in intelligence does not assist the attacker and the defending force wins the engagement (i.e. the time to defeat is negative). This complexity may explain why different authors have come to different conclusions as to whether intelligence favours an attacker or a defender. Khan (1978) argued that intelligence is essential to victory only in defence; however, Herman’s (1996) review of a wider range of case studies than Khan, was inconclusive noting only that intelligence can favour the attacker in some cases and the defender in others. Decoupling the influence of intelligence received by a commander during combat from prior information may help in reconciling these differing conclusions. A possible explanation for the action of intelligence during manoeuvre warfare comes from considering the dimensions of the parameter b. In the transport equations (3.5) b has dimensions of |$\\left (time\\right )^{-1}$|, hence we can write \\begin{align*} b\\equiv b^{\\prime}\\omega, \\end{align*} where b′ is the (dimensionless) relative superiority in intelligence and ω represents the speed at which information is collected, processed and analysed to produce intelligence and disseminated to decision-makers. This suggests that when the speed at which intelligence is produced is less than the attacking speed, intelligence has little effect on the outcome of combat. Boundary conditions Solutions to the transport equations in Section 3 are highly dependent on the initial distribution of the two forces and the boundary conditions. Later work will investigate the range of boundary conditions for which unique solutions to (3.1) are possible. This dependency on initial conditions suggests that the influence of intelligence will be very different when both forces are concentrated and initially separate from each other as occurs during conventional warfare, compared with a situation when one force has initially infiltrated the other as is in the case during insurgency or terrorist conflicts. Conclusions Applying classical Lanchester models to the use of intelligence in conflict indicates that the intelligence does act as a force multiplier; however, its utility to compensate for inferior force ratio is less than commonly appreciated, proportional to the square root of the relative advantage in intelligence. Similarly, the time to defeat is proportional to |$\\frac {1}{\\sqrt {b} }$|, so that greatly increasing one side’s superiority in intelligence only produces a modest decrease in the time to defeat. This dependency is confirmed by the application of transport equations to model the use of intelligence in combat. 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( 2016 ) Quantitative maritime security assessment: a 2020 vision . IMA J. Manag. Mathematics, 27 , 453 -- 470 Google Scholar Crossref Search ADS © The Author(s) 2018. Published by Oxford University Press on behalf of the Institute of Mathematics and its Applications. All rights reserved. This article is published and distributed under the terms of the Oxford University Press, Standard Journals Publication Model (https://academic.oup.com/journals/pages/open_access/funder_policies/chorus/standard_publication_model)\n\n### Journal\n\nIMA Journal of Management MathematicsOxford University Press\n\nPublished: Feb 1, 2019\n\n## You’re reading a free preview. 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https://la.mathworks.com/matlabcentral/answers/168204-questions-with-integral2-double-syms-and-dot-calucations | [
"# Questions with integral2 , Double, Syms and Dot calucations.\n\n4 views (last 30 days)\nsun on 27 Dec 2014\nCommented: Xinyue Liu on 4 May 2018\nDear friends, I got 2 questions here. 1. If I have y = a*x^2 + 5. What function can make it into y = a.*x.^2 +5. As you seen, dot was inserted.\n2. It's easy, but kinda diffcult to describe, but please have patience with me. Thank you so much. First, let me make a very simple example of my problem. If I want to calucate Y = F(x=1)+ 2.^2, and I know F(x=1) = a+b,(a and b are syms). This means Y = a + b + 4. Problem is here, Matlab give me error if I write down as.\nF = function( .... ); <==== output of function is F(X=1), and = *F(x=1)* = a+b\nY = integral2( F + 2.^2, .. , .. ,..)\nHowever, if I just copy the output of F as\nY = integral2( a+b + 2.^2, .. , .. ,..)\nNow it works!!!\nOk. Please allow me to talk about my code here. I am trying to find a double interation by using integral2. One part of my equcation(which is Y) is from another int output(which is F). Matlab will give ERROR for code below:\nclear all;\na=4;\nla1=1/(pi*500^2); la2= la1*5;\np1=25; p2=p1/25;\nsgma2=10^(-11);\nindex=1;\ng=2./a;\nsyms r u1 u2\npowe= -2 ;\nseta= 10^powe;\nxNor = ( (u2./u1).^(a./2) + 1 ).^(2./a);\nx = (xNor).^(0.5) * seta^(-1/a);\nfun1 = r./(1+ r.^a );\nout1 = int(fun1, x, Inf) ; %== This is my F in my example\nq=pi.*(la1.*p1.^(2./a)+la2.*p2.^(2./a));\nyi = @(u2,u1) exp(-u2.*(1+2.*...\n( out1 )./... %=== out1 is the problem here.\n( (( (u2./u1).^(a./2) + 1 ).^(2./a)).*seta.^(-2./a)))).*...\nexp(-sgma2.*q.^(-a./2).* seta.*u2.^(a./2)./...\n((( (u2./u1).^(a./2) + 1 ).^(2./a)).^(a./2)) );\nmaxF =@(u2) u2;\nout2 = integral2(yi,0,Inf,0 ,maxF) % == this is Y in my previous example.\nHowever, since I know the out1 = pi/4 - atan(10*(u2^2/u1^2 + 1)^(1/2))/2 (no dot,1/2, not 1./2). Instead of writing down out1, I will just type the equaction and add dot in the\nyi = @(u2,u1) exp(-u2.*(1+2.*...\n( pi./4 - atan(10.*(u2.^2./u1.^2 + 1).^(1./2))./2 )./... %===not \"out1\"\n( (( (u2./u1).^(a./2) + 1 ).^(2./a)).*seta.^(-2./a)))).*...\nexp(-sgma2.*q.^(-a./2).* seta.*u2.^(a./2)./...\n((( (u2./u1).^(a./2) + 1 ).^(2./a)).^(a./2)) );\nNow the code is working!!!! The final output is = 0.9957. Dear friends, I already spend a long time on this, but I still can not find out the problem. Could you please take a deeper look for me. Please copy the code to you matlab and test. Thank you so much.\nBelow is the error given by matlab, if I just use \"out1\" in yi = @(u2,u1) ......\nError using integralCalc/finalInputChecks (line 511)\nInput function must return 'double' or 'single' values. Found 'sym'.\nError in integralCalc/iterateScalarValued (line 315)\nfinalInputChecks(x,fx);\n[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);\nError in integralCalc (line 76)\nError in\nintegral2Calc>@(xi,y1i,y2i)integralCalc(@(y)fun(xi*ones(size(y)),y),y1i,y2i,opstruct.integralOptions)\n(line 18)\ninnerintegral = @(x)arrayfun(@(xi,y1i,y2i)integralCalc( ...\nError in\nintegral2Calc>@(x)arrayfun(@(xi,y1i,y2i)integralCalc(@(y)fun(xi*ones(size(y)),y),y1i,y2i,opstruct.integralOptions),x,ymin(x),ymax(x))\n(line 18)\ninnerintegral = @(x)arrayfun(@(xi,y1i,y2i)integralCalc( ...\nError in integralCalc/iterateScalarValued (line 314)\nfx = FUN(t);\n[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);\nError in integralCalc (line 84)\nError in integral2Calc>integral2i (line 21)\n[q,errbnd] = integralCalc(innerintegral,xmin,xmax,opstruct.integralOptions);\nError in integral2Calc (line 8)\n[q,errbnd] = integral2i(fun,xmin,xmax,ymin,ymax,optionstruct);\nError in integral2 (line 107)\nQ = integral2Calc(fun,xmin,xmax,yminfun,ymaxfun,opstruct);\nError in ref7_equ11n2 (line 129)\nout2 = integral2(yi,0,Inf,0 ,maxF)\n\nMike Hosea on 27 Dec 2014\nThe problem is that out1 is a symbolic \"thing\", not a MATLAB \"thing\". That's what the error message is really telling you. You need to use\nout1fcn = matlabFunction(out1);\nout1fcn(u1,u2)\nLike so:\nout1 = int(fun1, x, Inf) ; %== This is my F in my example\nout1fcn = matlabFunction(out1); % Convert symbolic object to a MATLAB function.\nq=pi.*(la1.*p1.^(2./a)+la2.*p2.^(2./a));\nyi = @(u2,u1) exp(-u2.*(1+2.*...\n( out1fcn(u1,u2) )./... %=== out1 is the problem here.\n( (( (u2./u1).^(a./2) + 1 ).^(2./a)).*seta.^(-2./a)))).*...\nexp(-sgma2.*q.^(-a./2).* seta.*u2.^(a./2)./...\n((( (u2./u1).^(a./2) + 1 ).^(2./a)).^(a./2)) );\nmaxF =@(u2) u2;\nout2 = integral2(yi,0,Inf,0 ,maxF) % == this is Y in my previous example.\nXinyue Liu on 4 May 2018\nI have the same problem with sun. Thank you so much to solve this, Mike;)"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.56316763,"math_prob":0.9989613,"size":4871,"snap":"2021-43-2021-49","text_gpt3_token_len":1790,"char_repetition_ratio":0.13190877,"word_repetition_ratio":0.18428572,"special_character_ratio":0.40751386,"punctuation_ratio":0.29335493,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99940825,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-04T14:07:26Z\",\"WARC-Record-ID\":\"<urn:uuid:c06d2bb8-b443-4098-b7ae-68664db02224>\",\"Content-Length\":\"178947\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dc48deae-d498-439a-9b99-56cf433a6b6e>\",\"WARC-Concurrent-To\":\"<urn:uuid:53dbbbfb-baba-4332-a07c-b2a7f6f3a661>\",\"WARC-IP-Address\":\"104.69.217.80\",\"WARC-Target-URI\":\"https://la.mathworks.com/matlabcentral/answers/168204-questions-with-integral2-double-syms-and-dot-calucations\",\"WARC-Payload-Digest\":\"sha1:JXHVJZIRB32DF2NO6AZ26HORKINDSCJT\",\"WARC-Block-Digest\":\"sha1:I4YARRCIONZH3XGAE326PZ34MR35ADAF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964362992.98_warc_CC-MAIN-20211204124328-20211204154328-00235.warc.gz\"}"} |
https://uk.mathworks.com/matlabcentral/answers/427353-how-to-set-the-encoder-transfer-function-in-autoencoder | [
"# How to set the encoder transfer function in autoencoder\n\n12 views (last 30 days)\nhuilin zhu on 1 Nov 2018\nCommented: huilin zhu on 18 Nov 2018\nI want to set the encoder transfer function by myself. But I do not how to do it. Can the encoder transfer function be changed by myself\n\nSANA on 16 Nov 2018\nFirst you make an autoencoder and generate its function, the code for generating the function of autoencoder is:\nautoenc = trainAutoencoder(Data, 300,...\n'MaxEpochs', 100,...\n'L2WeightRegularization', 0.001,...\n'SparsityRegularization', 4,...\n'SparsityProportion', 0.05,...\n'ScaleData', true);\ngenerateFunction(autoenc)\nThe generated function open in MATLAB editor with the name of neural_function, I renamed it my_autoencoder and the transfer function is mentioned there, so you can edit it as you wish, code is below:\nfunction [y1] = my_encoder(x1)\n%NEURAL_FUNCTION neural network simulation function.\n%\n% Generated by Neural Network Toolbox function genFunction, 15-Nov-2018 15:50:26.\n%\n% [y1] = neural_function(x1) takes these arguments:\n% x = 5088xQ matrix, input #1\n% and returns:\n% y = 5088xQ matrix, output #1\n% where Q is the number of samples.\n%#ok<*RPMT0>\n% ===== NEURAL NETWORK CONSTANTS =====\n% Input 1\nx1_step1.xoffset = [-10;-11;-11;-12;-1]\nx1_step1.gain = [0.090;0.9090;0.9090;0.90909;0.0769]\nx1_step1.ymin = 0;\n% Layer 1\nb1 = [-0.115;0.768;0.7066;0.5009303396;0.1249019302]\nIW1_1 = [-0.0947 0.7320 0.3146 0.494636173 -0.0951171]\nb2 = [3.5409901027442902688;3.6635759144424437928]\nLW2_1 = [-1.1707436273955371675 1.5786236406994880177 -1]\n% Output 1\ny1_step1.ymin = 0;\ny1_step1.gain = [0.0909090909090909;0.07]\ny1_step1.xoffset = [-10;-11;-11;-12;-12]\n% ===== SIMULATION ========\n% Dimensions\nQ = size(x1,2); % samples\n% Input 1\nxp1 = mapminmax_apply(x1,x1_step1);\n% Layer 1\na1 = logsig_apply(repmat(b1,1,Q) + IW1_1*xp1);\n% Layer 2\na2 = logsig_apply(repmat(b2,1,Q) + LW2_1*a1);\n% Output 1\ny1 = mapminmax_reverse(a2,y1_step1);\nend\n% ===== MODULE FUNCTIONS ========\n% Map Minimum and Maximum Input Processing Function\nfunction y = mapminmax_apply(x,settings)\ny = bsxfun(@minus,x,settings.xoffset);\ny = bsxfun(@times,y,settings.gain);\ny = bsxfun(@plus,y,settings.ymin);\nend\n% **********************************************\n% ************ Enhance encoder here ************\n% Sigmoid Positive Transfer Function\nfunction a = logsig_apply(n,~)\na = 1 ./ (1 + exp(-n));\nend\n% ************ Enhance encoder here ************\n% **********************************************\n% Map Minimum and Maximum Output Reverse-Processing Function\nfunction x = mapminmax_reverse(y,settings)\nx = bsxfun(@minus,y,settings.ymin);\nx = bsxfun(@rdivide,x,settings.gain);\nx = bsxfun(@plus,x,settings.xoffset);\nend\nYou can change decoder function as well. Enjoy !!!\nhuilin zhu on 18 Nov 2018\nThanks very much.But do you mean I have to edit a autoencoder before I edit the transfer function, and then I have to train my autoencoder again?"
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https://softmath.com/math-com-calculator/inverse-matrices/matlab-solve.html | [
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algebraic fractions, What Is the Formula to Find a Square Root, probability algebra ii, ks2 year 6 sat online practice test, solve trinomial factoring, algebra help programs, C aptitude questions+pdf.\n\nCombination permutation problems, probability practice worksheets, math tests for year 8, emulator+software+ti+free+download, cheat sheet secondary 3 math.\n\nWorld history worksheet for 9th graders on china, square root method, examples of 6th grade fractions, differential equations linear algebra solutions manual, printice hall pre algebra practice work book, square root number line worksheet, i dont understand algebra.\n\nScience test (pythagoras), 2 negatives make a plus, \"balancing chemical equations solver\", straight line linear XY graph formula, free step by step algebraic calculator for radical expressions, real life applications of factoring.\n\nLeast common multiple calculator, \"Math Dictionary Online\", pre-algebra quiz factoring, pre algebra 8th grade final review, lesson plan phisic, cost accounting tutorials.\n\nSolve trigonometric equations ti-89, Pre Algebra Test, hyperbolas in algebra 2 tips, McDougal Littell Geometry Book Answers, transforming formulas calculator.\n\nLowest common denominator worksheets, English EOC review worksheet, solving equations raised to the fraction of a power, vb notes about maths lessons, algebra formulas graphing linear, help with math Simplify (-3xy4)(-4x3 y6), free ti 84 mathematics solver.\n\nPrentice hall pre algebra book and workbook, workbook answers for scott foresman social studies book page 86, identity solver, FREE ONLINE PHYSIC STUDY HELP.\n\nHow to calculate a third in maths, add, subtract, multiply,divide fractions and decimals, mcdougal littell problems textbook.\n\nYear 6 trigonometry, algebra checker, 1st grade science book online, programming worksheets for beginners."
] | [
null,
"https://softmath.com/r-solver/images/tutor.png",
null
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https://developer.unigine.com/en/docs/2.2.1/api/library/physics/class.jointsuspension?rlang=cpp | [
"# JointSuspension Class\n\nThis class is used to create a suspension joint. The bodies that represent both a frame and a wheel need to be rigid bodies.\n\n## JointSuspension ()\n\nConstructor. Creates a suspension joint with an anchor at the origin of the world coordinates.\n\n## JointSuspension (const Ptr<Body> & body0, const Ptr<Body> & body1)\n\nConstructor. Creates a suspension joint connecting two given bodies. An anchor is placed between centers of mass of the bodies.\n\n### Arguments\n\n• const Ptr<Body> & body0 - The wheel to connect with the joint.\n• const Ptr<Body> & body1 - The frame to connect with the joint.\n\n## JointSuspension (const Ptr<Body> & body0, const Ptr<Body> & body1, const Math::Vec3 & anchor, const Math::vec3 & axis0, const Math::vec3 & axis1)\n\n### Arguments\n\n• const Ptr<Body> & body0\n• const Ptr<Body> & body1\n• const Math::Vec3 & anchor\n• const Math::vec3 & axis0\n• const Math::vec3 & axis1\n\n## floatgetAngularTorque ()\n\nReturns the maximum torque of the attached angular motor.\n\nMaximum torque.\n\n## floatgetAngularDamping ()\n\nReturns the angular damping of the joint (wheel rotation damping).\n\nAngular damping.\n\n## voidsetAxis11 (const Math::vec3 & axis11)\n\nSets a suspension axis in coordinates of the frame: an axis around which the wheel rotates when steering.\n\n### Arguments\n\n• const Math::vec3 & axis11 - Suspension axis.\n\n## floatgetAngularVelocity ()\n\nReturns the target velocity of wheel rotation.\n\n### Return value\n\nTarget velocity in radians per second.\n\n## Math::vec3getAxis10 ()\n\nReturns the suspension axis in coordinates of the wheel.\n\n### Return value\n\nSuspension axis in coordinates of the wheel.\n\n## voidsetWorldAxis1 (const Math::vec3 & axis1)\n\nSets a suspension axis in the world coordinates.\n\n### Arguments\n\n• const Math::vec3 & axis1 - Suspension axis in the world coordinates.\n\n## floatgetLinearDamping ()\n\nReturns the linear damping of the suspension.\n\nLinear damping.\n\n## voidsetLinearLimitTo (float to)\n\nSets a high limit of the suspension ride.\n\n### Arguments\n\n• float to - Limit in units.\n\n## floatgetLinearSpring ()\n\nReturns the rigidity coefficient of the suspension.\n\n### Return value\n\nRigidity coefficient.\n\n## floatgetCurrentLinearDistance ()\n\nReturns the current suspension compression.\n\n### Return value\n\nCurrent suspension height in units.\n\n## Math::vec3getAxis00 ()\n\nReturns the wheel axis. This is a wheel spindle.\n\nWheel axis.\n\n## voidsetAngularVelocity (float velocity)\n\nSets a maximum velocity of wheel rotation.\n\n### Arguments\n\n• float velocity - Velocity in radians per second.\n\n## floatgetLinearLimitTo ()\n\nReturns the high limit of the suspension ride.\n\n### Return value\n\nHigh limit in units.\n\n## floatgetLinearLimitFrom ()\n\nReturns the low limit of the suspension ride.\n\n### Return value\n\nLow limit in units.\n\n## voidsetLinearDamping (float damping)\n\nSets a linear damping of the suspension.\n\n### Arguments\n\n• float damping - Linear damping. If a negative value is provided, 0 will be used instead.\n\n## voidsetLinearDistance (float distance)\n\nSets a target height of the suspension.\n\n### Arguments\n\n• float distance - Height in units.\n\n## voidsetAxis00 (const Math::vec3 & axis00)\n\nSets a wheel axis along which a wheel moves linearly. This is a shock absorber.\n\n### Arguments\n\n• const Math::vec3 & axis00 - Wheel axis.\n\n## voidsetLinearSpring (float spring)\n\nSets a rigidity coefficient of the suspension.\n\n### Arguments\n\n• float spring - Rigidity coefficient. If a negative value is provided, 0 will be used instead.\n\n## voidsetWorldAxis0 (const Math::vec3 & axis0)\n\nSets a wheel axis in the world coordinates. This is a wheel spindle.\n\n### Arguments\n\n• const Math::vec3 & axis0 - Wheel axis in the world coordinates.\n\n## voidsetAxis10 (const Math::vec3 & axis10)\n\nSets a suspension axis in coordinates of the wheel: an axis around which a wheel rotates when moving forward (or backward).\n\n### Arguments\n\n• const Math::vec3 & axis10 - Suspension axis.\n\n## voidsetAngularDamping (float damping)\n\nSets an angular damping of the joint (wheel rotation damping).\n\n### Arguments\n\n• float damping - Angular damping. If a negative value is provided, 0 will be used instead.\n\n## floatgetLinearDistance ()\n\nReturns the target height of the suspension.\n\n### Return value\n\nTarget height in units.\n\n## voidsetAngularTorque (float torque)\n\nSets a maximum torque of the attached angular motor.\n\n### Arguments\n\n• float torque - Maximum torque. If a negative value is provided, 0 will be used instead.\n\n## Math::vec3getWorldAxis0 ()\n\nReturns the wheel axis in the world coordinates. This is a wheel spindle.\n\n### Return value\n\nWheel axis in the world coordinates.\n\n## floatgetCurrentAngularVelocity ()\n\nReturns the current velocity of wheel rotation.\n\n### Return value\n\nCurrent velocity in radians per second.\n\n## Math::vec3getAxis11 ()\n\nReturns the suspension axis in coordinates of the frame.\n\n### Return value\n\nSuspension axis in coordinates of the frame.\n\n## Math::vec3getWorldAxis1 ()\n\nReturns the suspension axis in the world coordinates.\n\n### Return value\n\nSuspension axis in the world coordinates.\n\n## voidsetLinearLimitFrom (float from)\n\nSets a low limit of the suspension ride.\n\n### Arguments\n\n• float from - Limit in units.\nLast update: 2017-07-03"
] | [
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http://www.geekinterview.com/question_details/92408 | [
"# Convert any two numbers into words and perform arithmetic operation\n\nWe need to have two numeric values which needs to be first converted to its corresponding words and have to to perform arithmetic operations like addition, subtraction, multiplication and division using it.\n\nThis Question is not yet answered!",
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"http://www.geekinterview.com/images/ajax-loader.gif",
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"http://www.geekinterview.com/images/ajax-loader.gif",
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https://lessonplanet.com/search?grade_ids%5B%5D=255&grade_ids%5B%5D=256&grade_ids%5B%5D=257&grade_ids%5B%5D=258&keywords=&page=42&type_ids%5B%5D=357919 | [
"Refine Your Results\n\n### We found11,394 reviewed resources\n\n2:29\nLesson Planet\n\n#### Patterns in Sequences 2\n\nFor Students 6th - 9th\nThis video looks again at finding the algebraic pattern in sequences. In the example, we are looking at a figure where a number of toothpicks (or lines) are added each time, and you need to find how many toothpicks will be in the 50th...\n2:32\nLesson Planet\n\n#### Patterns in Sequences 1\n\nFor Students 6th - 9th\nFaced with a series of numbers we need to find the next few values of the sequence. This short video shows an example of finding the next values in a sequence where we need to find the pattern of adding a specific number each time to...\n3:03\nLesson Planet\n\n#### Parallel Lines 2\n\nFor Students 7th - 9th\nCan you tell, algebraically, if three lines are parallel? In this video, Sal demonstrates how to rewrite linear equations in slope-intercept form and compare the slope of each line.\n1 In 1 Collection 14:31\nLesson Planet\n\n#### Parabola Focus and Directrix 1\n\nFor Students 10th - 12th\nStarting with a point (focus) and a line below the point (directrix), in this video, Sal tries to find the set of all points (locus) equidistant to the point and the line. He shows, using the distance formula, that indeed, we have an...\n10:05\nLesson Planet\n\n#### Introduction to Logarithms Properties\n\nFor Students 7th - 9th\nSal gets to the heart of the matter in this video, clarifying that it's important to understand math \"so you can actually apply it in life later on and not have relearn everything every time.\" True to his word, Sal demonstrates ways to...\n11:03\nLesson Planet\n\n#### Introduction to Limits (HD)\n\nFor Teachers 11th - Higher Ed\nSal begins his explanation of limits with a few basic examples and takes a more intuitive point of view before looking at a formal mathematical definition in later videos. He starts by introducing the notation for limits and describes...\n9:56\nLesson Planet\n\n#### Introduction to Interest\n\nFor Students 10th - 12th\nHere is a simple tutorial on the difference between simple and compound interest. In it, Sal describes what interest is, defines the vocabulary of principle and interest rate. He also models year-by-year the amount of money owed under...\n9:06\nLesson Planet\n\n#### Introduction to Function Inverses\n\nFor Students 10th - 12th\nStarting from a brief look at functions and the mapping of domains to ranges, Sal starts out with an intuitive sense of what a function inverse is. He then, using an example, shows how to find the inverse of a function and also shows how...\n9:40\nLesson Planet\n\n#### Intro to 30-60-90 Triangles\n\nFor Students 9th - 10th\nAs the fourth part of a series on the Pythagorean Theorem, Sal continues an exploration of the ways to calculate the length of sides of a right triangle using algebra. He also explores using this theorem to solve problems involving a...\n11:28\nLesson Planet\n\n#### Identifying Conics 2\n\nFor Students 10th - 12th\nIn this video, Sal shows again how to write an equation of a conic section in standard form and identify the conic section it represents. This time, his example is of a hyperbola (not centered at the origin), which he also graphs.\n9:11\nLesson Planet\n\n#### Identifying Conics 1\n\nFor Students 10th - 12th\nAfter watching this video, you should be able to write an equation of a conic section in standard form, and identify the conic section from its equation. In an example problem, Sal reviews how to complete the square and graph an ellipse...\n10:01\nLesson Planet\n\n#### Fun Trig Problem\n\nFor Students 10th - 12th\nIf solving trigonometric equations is your idea of fun, then this video is correctly titled. Here Sal uses trigonometric identities, the quadratic formula, and inverse trigonometric functions to solve a trigonometric equation sent in by...\n18:28\nLesson Planet\n\n#### Focus and Directrix of a Parabola 2\n\nFor Students 10th - 12th\nContinuing from a previous video, Sal takes the discussion of the focus and directrix of parabola further. Given the parabola, y=x2, he derives the focus and directrix by matching parts of the equation he found earlier. He then, comes up...\n13:49\nLesson Planet\n\n#### Foci of an Ellipse\n\nFor Students 10th - 12th\nThe foci of an ellipse are explored in this video. First, foci are shown as the two points on the major axis such that the sum of the distance from a point on the ellipse and the foci points is the same as the distance from any point on...\n15:24\nLesson Planet\n\n#### Foci of a Hyperbola\n\nFor Students 10th - 12th\nIn this video, Sal defines the vertex and the foci of a hyperbola, and shows how to locate both. By comparing the hyperbola to an ellipse throughout the video, the listener sees the similarities between these two conic sections. This...\n5:57\nLesson Planet\n\n#### Finding the 100th Term in a Sequence\n\nFor Students 6th - 9th\nThe example problem is this video has a sequence of numbers that decrease by a fixed amount with each iteration, and one needs to find the 100th number in the sequence. Sal shows the listener how to find the pattern between the number..."
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https://juliadiff.org/ChainRulesCore.jl/stable/rule_author/example.html | [
"# Pedagogical Example\n\nThis pedagogical example will show you how to write an rrule. See On writing good rrule / frule methods section for more tips and gotchas. If you want to learn about frules, you should still read and understand this example as many concepts are shared, and then look for real world frule examples in ChainRules.jl.\n\n## The primal\n\nWe define a struct Foo\n\nstruct Foo{T}\nA::Matrix{T}\nc::Float64\nend\n\nand a function that multiplies Foo with an AbstractArray:\n\nfunction foo_mul(foo::Foo, b::AbstractArray)\nreturn foo.A * b\nend\n\nNote that field c is ignored in the calculation.\n\n## The rrule\n\nThe rrule method for our primal computation should extend the ChainRulesCore.rrule function.\n\nfunction ChainRulesCore.rrule(::typeof(foo_mul), foo::Foo{T}, b::AbstractArray) where T\ny = foo_mul(foo, b)\nfunction foo_mul_pullback(ȳ)\nf̄ = NoTangent()\nf̄oo = Tangent{Foo{T}}(; A=ȳ * b', c=ZeroTangent())\nb̄ = @thunk(foo.A' * ȳ)\nreturn f̄, f̄oo, b̄\nend\nreturn y, foo_mul_pullback\nend\n\nWe can check this rule against a finite-differences approach using ChainRulesTestUtils:\n\njulia> using ChainRulesTestUtils\njulia> test_rrule(foo_mul, Foo(rand(3, 3), 3.0), rand(3, 3))\nTest Summary: | Pass Total\ntest_rrule: foo_mul on Foo{Float64},Matrix{Float64} | 10 10\nTest.DefaultTestSet(\"test_rrule: foo_mul on Foo{Float64},Matrix{Float64}\", Any[], 10, false, false)\n\nNow let's examine the rule in more detail:\n\nfunction ChainRulesCore.rrule(::typeof(foo_mul), foo::Foo, b::AbstractArray)\n...\nreturn y, foo_mul_pullback\nend\n\nThe rrule dispatches on the typeof of the function we are writing the rrule for, as well as the types of its arguments. Read more about writing rules for constructors and callable objects here. The rrule returns the primal result y, and the pullback function. It is a very good idea to name your pullback function, so that they are helpful when appearing in the stacktrace.\n\n y = foo_mul(foo, b)\n\nComputes the primal result. It is possible to change the primal computation so that work can be shared between the primal and the pullback. See e.g. the rule for sort, where the sorting is done only once.\n\n function foo_mul_pullback(ȳ)\n...\nreturn f̄, f̄oo, b̄\nend\n\nThe pullback function takes in the tangent of the primal output (ȳ) and returns the tangents of the primal inputs. Note that it returns a tangent for the primal function in addition to the tangents of primal arguments.\n\nFinally, computing the tangents of primal inputs:\n\n f̄ = NoTangent()\n\nThe function foo_mul has no fields (i.e. it is not a closure) and can not be perturbed. Therefore its tangent (f̄) is a NoTangent.\n\n f̄oo = Tangent{Foo}(; A=ȳ * b', c=ZeroTangent())\n\nThe struct foo::Foo gets a Tangent{Foo} structural tangent, which stores the tangents of fields of foo.\n\nThe tangent of the field A is ȳ * b',\n\nThe tangent of the field c is ZeroTangent(), because c can be perturbed but has no effect on the primal output.\n\n b̄ = @thunk(foo.A' * ȳ)\n\nThe tangent of b is foo.A' * ȳ, but we have wrapped it into a Thunk, a tangent type that represents delayed computation. The idea is that in case the tangent is not used anywhere, the computation never happens. Use InplaceableThunk if you are interested in accumulating gradients inplace. Note that in practice one would also @thunk the f̄oo.A tangent, but it was omitted in this example for clarity.\n\nAs a final note, since b is an AbstractArray, its tangent b̄ should be projected to the right subspace. See the ProjectTo the primal subspace section for more information and an example that motivates the projection operation."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.73334765,"math_prob":0.8945092,"size":3525,"snap":"2023-40-2023-50","text_gpt3_token_len":979,"char_repetition_ratio":0.14541323,"word_repetition_ratio":0.01084991,"special_character_ratio":0.23517731,"punctuation_ratio":0.16526611,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96969956,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-30T13:38:40Z\",\"WARC-Record-ID\":\"<urn:uuid:faa31138-44bf-4676-a55e-18b6e0da96b0>\",\"Content-Length\":\"14235\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8bda9dd9-2ea2-4770-996f-9ad3057f4ae8>\",\"WARC-Concurrent-To\":\"<urn:uuid:3c3cd33f-c706-459c-859f-0f3868f157aa>\",\"WARC-IP-Address\":\"185.199.108.153\",\"WARC-Target-URI\":\"https://juliadiff.org/ChainRulesCore.jl/stable/rule_author/example.html\",\"WARC-Payload-Digest\":\"sha1:A5Z5WHJMMNINZIPN6CEM4LFK2FN3D7M3\",\"WARC-Block-Digest\":\"sha1:MXQDMHMMZZBTEQ6N6TIA76VPBJWYHYX4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510676.40_warc_CC-MAIN-20230930113949-20230930143949-00889.warc.gz\"}"} |
http://mosmos.me/free-math-coloring-sheets-for-middle-school/fun-math-worksheets-middle-school-printable-coloring-sheets-color-by-number-free-maths-colouring-for/ | [
"# Fun Math Worksheets Middle School Printable Coloring Sheets Color By Number Free Maths Colouring For",
null,
"fun math worksheets middle school printable coloring sheets color by number free maths colouring for.\n\nfree math coloring sheets for middle school pages worksheets,math coloring worksheets middle school first grade doubles addition free,math color pages maths colouring free coloring sheets multiplication worksheets middle school,math coloring ksheets middle school pages for sheets free worksheets , free math coloring worksheets middle school print out pages extraordinary multiplication, multiplication coloring printable free worksheets math color by middle school, free math coloring worksheets middle school sheets pages for newest games,math coloring worksheets middle school free color by number sheets , free math coloring worksheets middle school pages page first grade, math coloring worksheets middle school free multiplication plus."
] | [
null,
"http://mosmos.me/wp-content/uploads/2019/07/fun-math-worksheets-middle-school-printable-coloring-sheets-color-by-number-free-maths-colouring-for.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7461271,"math_prob":0.50654674,"size":845,"snap":"2019-43-2019-47","text_gpt3_token_len":130,"char_repetition_ratio":0.26159334,"word_repetition_ratio":0.1,"special_character_ratio":0.14674556,"punctuation_ratio":0.08661418,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9519231,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-15T12:52:09Z\",\"WARC-Record-ID\":\"<urn:uuid:3ab0a235-4dc2-4a27-94d4-ba26a654323f>\",\"Content-Length\":\"45538\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7b6c564d-2e06-4fc9-9d3b-7266520e6766>\",\"WARC-Concurrent-To\":\"<urn:uuid:ea9eb71f-7987-47fb-ac18-8cbe6525b5d2>\",\"WARC-IP-Address\":\"104.31.77.48\",\"WARC-Target-URI\":\"http://mosmos.me/free-math-coloring-sheets-for-middle-school/fun-math-worksheets-middle-school-printable-coloring-sheets-color-by-number-free-maths-colouring-for/\",\"WARC-Payload-Digest\":\"sha1:A63OOBMFG2JSXFURBQBT4WDF4IEUC2EI\",\"WARC-Block-Digest\":\"sha1:6NEKCMGJQ6WZP2226U77RYVKT75NWVS5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668644.10_warc_CC-MAIN-20191115120854-20191115144854-00005.warc.gz\"}"} |
https://pacingguides.lcsedu.net/2018-19/par/10559 | [
"# 8.3 - Square Roots\n\n8.3 The student will\n\na) estimate and determine the two consecutive integers between which a square root lies;\n\nb) determine both the positive and negative square roots of a given perfect square.\n\n### BIG IDEAS\n\n• I can calculate area or volume, determine growth or decay, and figure out the impact of force. I can determine national debt and world population, advertise with viral marketing, program a computer game, figure compound interest, and track the spread of viruses.\n• I will understand the relationship between a perfect square and a geometric square and be able to use the area of a square as a means for estimating the square root of any number.\n\n### UNDERSTANDING THE STANDARD\n\n2016 VDOE Curriculum Framework - 8.3 Understanding\n\n· A perfect square is a whole number whose square root is an integer.\n\n· The square root of a given number is any number which, when multiplied times itself, equals the given number.\n\n· Both the positive and negative roots of whole numbers, except zero, can be determined. The square root of zero is zero. The value is neither positive nor negative. Zero (a whole number) is a perfect square.\n\n· The positive and negative square root of any whole number other than a perfect square lies between two consecutive integers (e.g., lies between 7 and 8 since 72 = 49 and 82 = 64; lies between -4 and -3 since (-4)2 = 16 and (-3)2 = 9).\n\n· The symbol may be used to represent a positive (principal) root and - may be used to represent a negative root.\n\n· The square root of a whole number that is not a perfect square is an irrational number (e.g., is an irrational number). An irrational number cannot be expressed exactly as a fraction where b does not equal 0.\n\n· Square root symbols may be used to represent solutions to equations of the form x2 = p. Examples may include:\n\n- If x2 = 36, then x is = 6 or =-6.\n\n- If x2 = 5, then x is or −.\n\n· Students can use grid paper and estimation to determine what is needed to build a perfect square. The square root of a positive number is usually defined as the side length of a square with the area equal to the given number. If it is not a perfect square, the area provides a means for estimation.\n\n### ESSENTIALS\n\n• How does the area of a square relate to the square of a number?\nThe area determines the perfect square number. If it is not a perfect square, the area provides a means for estimation.\n• Why do numbers have both positive and negative roots?\nThe square root of a number is any number which when multiplied by itself equals the number. A product, when multiplying two positive factors, is always the same as the product when multiplying their opposites (e.g., 7 ∙ 7 = 49 and -7 ∙ -7 = 49).\n\nThe student will use problem solving, mathematical communication, mathematical reasoning, connections, and representations to\n\n· 8.3b1 Determine the positive or negative square root of a given perfect square from 1 to 400.\n\n· 8.3a1 Estimate and identify the two consecutive integers between which the positive or negative square root of a given number lies. Numbers are limited to natural numbers from 1 to 400."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8791028,"math_prob":0.9986109,"size":3132,"snap":"2021-04-2021-17","text_gpt3_token_len":731,"char_repetition_ratio":0.18734016,"word_repetition_ratio":0.09187279,"special_character_ratio":0.24042146,"punctuation_ratio":0.11183144,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99956447,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-17T00:58:57Z\",\"WARC-Record-ID\":\"<urn:uuid:57b52a51-71d4-4108-9755-cb836b055fa1>\",\"Content-Length\":\"28096\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5c7b4e86-56ae-4f42-bb27-9a6c2a86dfb5>\",\"WARC-Concurrent-To\":\"<urn:uuid:1f733827-51fb-407c-89db-bbf880a4cc1d>\",\"WARC-IP-Address\":\"67.221.113.241\",\"WARC-Target-URI\":\"https://pacingguides.lcsedu.net/2018-19/par/10559\",\"WARC-Payload-Digest\":\"sha1:NGN7JGRMXJQSYXP6GN6HJT7IRNB557AS\",\"WARC-Block-Digest\":\"sha1:RPZNFORKW4WH46A3XALWPFSEBX356PB4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703507971.27_warc_CC-MAIN-20210116225820-20210117015820-00542.warc.gz\"}"} |
https://jobbpnqb.netlify.app/8302/25218.html | [
"# Systems of linear inequalities Algebra 1, Systems of linear\n\nAlgebra, lp 1-2 vt 2011 - Matematikcentrum\n\nFundamental theorem of algebra. This is according to the Fundamental theorem of Algebra. Descartes Rule of Sign: Tells you the how many positiv or negative real zeroes the polynomial has. 1.\n\nIf f(z) is analytic and bounded in the complex plane, then f(z) is constant. We now prove. Theorem 2.2 (Fundamental Theorem of Algebra). Let p(z) be a polynomial.\n\n## The Fundamental Theorem of Algebra CDON\n\nwarm up. lesson presentation. lesson quiz.",
null,
"### graphing square and cube root functions",
null,
"The Fundamental Theorem of Algebra Example B. · 3. The Fundamental Theorem of Algebra P(x) is a real polynomial so the complex roots are in conjugate Fundamental Theorem of Algebra · If, algebraically, we find the same zero k times , we count it as k separate zeroes. · Some of the roots may be non-Reals (another Fundamental Theorem of Algebra. A polynomial of de- gree n with integer coefficients has n roots. In order to deal with multiplicities, it is better to say, since.\n\nWe will now look at some more theorems regarding polynomials, the first of which is extremely important and is known as The Fundamental Theorem of Algebra. Corollary to the Fundamental Theorem of Algebra. Every polynomial in one variable of degree n>0 has exactly n, not necessarily distinct, real or complex zeros. Feb 3, 2021 Fundamental theorem of algebra facts for kids · the degree n of a polynomial is the highest power of · some of the roots may be complex numbers The Conjugate Zeros Theorem states: If P(x) is a polynomial with real coefficients, and if a + bi is a zero of P Jan 10, 2015 The fundamental theorem of algebra states that a polynomial of degree n ≥ 1 with complex coefficients has n complex roots, with possible. In a fun Sudoku puzzle, students will practice the properties of the Fundamental Theorem of Algebra. This theorem states that a polynomial of degree n has n roots. The Fundamental Theorem of Algebra.",
null,
"This app is necessary for students who are wondering how to solve the problems, Because this app Remembering Math Formula is always an big task, Now no need to carry large books to find formula, This simple yet amazing apps for students, scientist, remainder theorem, factor theorem 8 algebrans fundamentalsats, faktorsatsen, konjugatpar fundamental theorem of algebra, factor theorem, conjugate pair 9 av M GROMOV · Citerat av 336 — one expects the properties (a) and (b) from Main theorem 1.4, but we are able to prove only the coshw {κ2) . For the last statement we need an algebraic fact. Using the fundamental theorem of calculus often requires finding an antiderivative.\n\nThe \"Fundamental Theorem of Algebra\" is not the start of algebra or anything, but it does say something interesting about polynomials: The Degree of a Polynomial with one variable is the largest exponent of that variable. A \"root\" (or \"zero\") is where the polynomial is equal to zero.\nMinecraft wedding\n\nlärare parkskolan bodafors\npen plasma treatment\npension plans in us\nux internships summer 2021\npensionsmyndigheten telefon\n\n### ordlista\n\nPerfect numbers are complex, complex numbers might be perfect Fundamental Theorem of Algebra: Statement and Significance free, direct and elementary proof of the Fundamental Theorem of Algebra. “The final publication (in TheMathematicalIntelligencer,33,No. 2(2011),1-2) is available at THE FUNDAMENTAL THEOREM OF ALGEBRA BRANKO CURGUS´ In this note I present a proof of the Fundamental Theorem of Algebra which is based on the algebra of complex numbers, Euler’s formula, continu-ity of polynomials and the extreme value theorem for continuous functions. The main argument in this note is similar to .\n\nVett och etikett disputation\nold net\n\n### Elements of Abstract Algebra i Apple Books\n\nThe Fundamental Theorem of Algebra guarantees us at least one complex zero, z1, and as It states that our perseverance paid off handsomely.\n\n## Plan: M0030M, LP2, 2018 Lectures on Linear Algebra:\n\nThe field ℂ of complex numbers is algebraically closed. Proof. Let g ∈ ℂ X be a polynomial of degree ≥ 1, and suppose that this polynomial does not have a root in ℂ. Fundamental theorem of algebra definition is - a theorem in algebra: every equation which can be put in the form with zero on one side of the equal-sign and a polynomial of degree greater than or equal to one with real or complex coefficients on the other has at least one root which is a real or complex number. Catch David on the Numberphile podcast: https://youtu.be/9y1BGvnTyQAPart one on odd polynomials: http://youtu.be/8l-La9HEUIU More links & stuff in full descr Fundamental Theorem of Algebra. Every nonconstant polynomial with complex coefficients has a root in the complex numbers. Some version of the statement of the Fundamental Theorem of Algebra first appeared early in the 17th century in the writings of several mathematicians, including Peter Roth, Albert Girard, and Ren´e Descartes. The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root.This includes polynomials with real coefficients, since every real number is a complex number with its imaginary part equal to zero.\n\nIn plain English, this theorem says that the degree of a polynomial equation tells you how many roots the equation will have."
] | [
null,
"https://picsum.photos/800/618",
null,
"https://picsum.photos/800/633",
null,
"https://picsum.photos/800/634",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8595598,"math_prob":0.97455096,"size":5561,"snap":"2023-14-2023-23","text_gpt3_token_len":1302,"char_repetition_ratio":0.20658629,"word_repetition_ratio":0.036464088,"special_character_ratio":0.20697716,"punctuation_ratio":0.10226155,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9992748,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-26T06:47:35Z\",\"WARC-Record-ID\":\"<urn:uuid:10c45d14-f567-456b-807a-9ee4a54bb7df>\",\"Content-Length\":\"10896\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:eed71176-6f6e-4b3b-8122-67847b6a3bd0>\",\"WARC-Concurrent-To\":\"<urn:uuid:6878fe1f-0597-4a0c-9e33-42bf2c6f9965>\",\"WARC-IP-Address\":\"34.148.147.18\",\"WARC-Target-URI\":\"https://jobbpnqb.netlify.app/8302/25218.html\",\"WARC-Payload-Digest\":\"sha1:JQD4VN5MKLCUWNICUJXU3WSEEKEW4BLF\",\"WARC-Block-Digest\":\"sha1:2AXGHOE7G624YQNTEUMFDGRJ6FCNUR64\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296945433.92_warc_CC-MAIN-20230326044821-20230326074821-00252.warc.gz\"}"} |
https://epjam.edp-open.org/articles/epjam/full_html/2019/01/epjam180025/epjam180025.html | [
"Issue EPJ Appl. Metamat. Volume 6, 2019 Metamaterials Research and Development in Japan 3 5 https://doi.org/10.1051/epjam/2018010 18 January 2019",
null,
"This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.\n\n## 1 Introduction\n\nA loop antenna radiates a linearly polarized (LP) wave . When the circumference of the loop is one wavelength, the maximum LP radiation appears in the broadside direction. This broadside LP radiation can be changed to circularly polarized (CP) radiation by adding perturbation elements to the loop, as shown in Figure 1. The rotational sense of the CP radiation, either a right-handed (RH) sense or a left-handed (LH) sense, is uniquely determined by the location of the perturbation elements relative to feed point F: the loop in Figure 1 radiates an RHCP wave. In other words, the loop antenna radiates a CP wave with a single rotational sense.\n\nThe single rotational sense also holds true for a grid loop array in . Although perturbation elements are not added to the loops, a traveling current is generated along the arrayed loops and radiates a CP wave. The rotational sense is determined by the winding sense of the loops.\n\nIt is often required that an antenna has dual-band counter CP radiation to keep sufficient separation of signals, i.e., the LHCP and RHCP radiation in two different frequency bands, in order to avoid interfering with each other. When a pair of dual-band counter CP antennas is adopted as a transmitting antenna and a receiving antenna, it is desirable for these antennas to have the same gain. If the LHCP and RHCP gains are different, the receiving antenna captures LHCP and RHCP waves with different receiving power levels. Therefore, a post-process circuit connected to the receiving antenna needs additional amplification circuits to enhance the weak power. This complicates the post-process circuit designs. To avoid such an issue, the same gain (balanced gain) is desired.\n\nRecent study has shown that a metamaterial (MTM) loop antenna, referred to as a metaloop, is an antenna that meets the requirement of dual-band counter CP radiation , where the loop is realized using the concept of a composite RH and LH transmission line , referred to as a metaline.\n\nThe maximum gain for the metaloop within the low frequency band (LoFB) in is smaller than that within the high frequency band (HiFB). This is attributed to the antenna size relative to the free-space wavelength (electrical antenna size); the free-space wavelength within the LoFB is larger than that within the HiFB, and hence the electrical antenna size within the LoFB is smaller than that within the HiFB.\n\nThus, a question arises as to how the difference in the gains can be reduced. This paper presents a technique for reducing the band gain difference for a dual-band counter CP metaloop antenna, where a parasitic loop (paraloop) is placed above the metaloop. Note that the metaloop antenna in this paper is analyzed using a full wave analysis tool, HFSS , and experimental work is performed in an anechoic chamber.\n\nSo far, other authors have designed MTM-loaded and MTM-inspired loop antennas. For example, the MTM-loaded loop in reference and MTM-inspired loop are designed as an LP antenna. An antenna using complementary capacitively loaded loop in reference and loop antennas in are also designed as an LP antenna. The design requirements for these antennas differ from ours, i.e., dual-band counter CP radiation with balanced gain, while meeting two additional requirements: (1) broadside radiation and (2) simple feed system without balun circuits. To our best knowledge, there have not been such MTM-related loop antennas.",
null,
"Fig. 1Natural loop antennas radiating a circularly polarized wave with a single feed point F. The loop has perturbation elements .\n\n## 2 Metaloop structure\n\nThe metaloop antenna to be considered here is shown in Figure 2. The antenna arm is printed on a square dielectric substrate of side length Ssub, relative permittivity εr and thickness B. The substrate is backed by a square ground plane (GP) of side length SGP (=Ssub). The antenna arm, symmetric with respect to the xz plane, is composed of five straight metalines, whose lengths are L1 = L5 and L2 = L3 = L4. Each metaline is made of numerous conducting subwavelength segments of width w and length p0, where the gap between neighboring segments is denoted as Δg. Repeating arm sections, each having length 2(Δg + p0) ≡ p, are designated as the arm unit cells. The central segment of the cell is short-circuited to the GP through a chip inductor, LY. Neighboring segments are connected through a chip capacitor, 2CZ. Point F is the feed point and point T is the terminal point connected to the GP through a resistive load, RB, to suppress reflected currents from point T to point F.\n\nNote that LY and 2CZ are determined as follows. First, using HFSS , we obtain the frequency response of the scattering parameters (S-parameters) for the unit cell including LY and 2CZ. Second, based on the obtained S-parameters, we draw the dispersion curve (phase constant β = ±2π/λg vs. frequency f , where λg is the guided wavelength). These two steps are repeated until the dispersion curve is balanced (smoothly connected at a preselected transition frequency fT), changing LY and 2CZ. The values for LY and 2CZ when the dispersion curve is balanced are what we need.\n\nThe parameters used for the following discussion are summarized in Table 1; these parameters create a preselected transition frequency of fT = 3 GHz, i.e., phase constant β is negative at frequencies below fT and positive at frequencies above fT. Figure 3 shows the loop peripheral length L1 + L2 + L3 + L4 + L5 normalized to the guided wave length, λg, as a function of frequency f. Note that the frequency-dependent guided wavelength λg used in Figure 3 is derived from the dispersion curve explained in the previous paragraph.\n\nFrequencies fL and fU in Figure 3 are the lower- and upper-edge frequencies for a fast wave region. The radiation occurs between fL and fU. A traveling wave current flows along the loop from point F to point T, with guided wave length λg, which varies with frequency f. The propagation phase constant is β = −2π/λg < 0 at f < fT, β = 0 at f = fT, and β = 2π/λg > 0 at f > fT. In order to obtain a broadside beam at two different frequencies where the loop length is 1λg, the in-phase condition (β = 0) is not used.",
null,
"Fig. 2Metaloop antenna. (a) Perspective view. (b) Side view. (c) Arm unit cell.\nTable 1\n\nParameters.",
null,
"Fig. 3Loop length L1 + L2 + L3 + L4 + L5 normalized to guided wavelength λg, as a function of frequency. β < 0 at f < fT, β = 0 at fT, and β > 0 at f > fT.\n\n## 3 Frequency response of the gain\n\nBy resistive load RB, a traveling wave current flows along the loop from point F to point T, with frequency-dependent guided wavelength λg. In other words, the metaline acts as a leaky wave antenna. Generally, the radiation efficiency of the leaky wave antenna is not high due to absorption of the power input to the antenna by RB. If RB is removed, the input impedance (and hence Voltage Standing Wave Ratio (VSWR)) becomes unstable and CP radiation is not obtained by a reflected current from point T.\n\nThe current at frequency f < fT has a progressive phase distribution from point F to point T due to a negative phase constant (β = −2π/λg < 0). Hence, the current behaves as if it travels from point T to point F (clockwise). This results in LHCP radiation in the broadside direction at a frequency where the loop length is one guided wave length (1λg). Conversely, the current at f > fT has a regressive phase distribution from point F to point T due to a positive phase constant (β = 2π/λg > 0). Hence, the current flows from point F to point T (counter clockwise), resulting in RHCP radiation in the broadside direction at a frequency satisfying a 1λg-loop length. Thus, dual-band counter CP radiation in the broadside direction is obtained.\n\nFrom Figure 3, it is expected that CP broadside radiation will be obtained around frequencies 2.60 GHz ≡ fN and 3.50 GHz ≡ fH, because the loop length normalized to guided wavelength λg is one at fN and fH: (L1 + L2 +…+L5)/λg = 1. For the remainder of the paper, fN and fH are referred to as the Nion frequency and Hion frequency, respectively. Figure 4 shows the gain as a function of frequency, where GL denotes the gain for an LHCP wave, called the LHCP gain, and GR denotes the gain for an RHCP wave, called the RHCP gain. It is found that GL is dominant at frequencies below fT = 3 GHz, because the current flows with a negative phase constant (β < 0). Conversely, GR is dominant at frequencies above transition frequency fT due to β > 0. Maximum gain GL appears at a frequency near Nion frequency fN and maximum gain GR appears at a frequency near Hion frequency fH. These frequencies are denoted as fGLmax and fGRmax, respectively. The difference in the maximum gains, GLmax at 2.58 GHz ≡ fGLmax (close to fN) and GRmax at 3.51 GHz ≡ fGRmax (close to fH), is approximately 5.5 dB.",
null,
"Fig. 4Frequency response of the gain.\n\n## 4 Reduction in the gain difference\n\nIn this section, the difference in maximum gains GLmax and GRmax is reduced as much as possible. For this, the antenna system shown in Figure 5 is considered, where a parasitic natural conducting strip loop of width wpara and peripheral length Lpara is located at height Hpara above the metaloop. The parasitic loop is designed such that only the smaller gain, GLmax at fGLmax, is increased, without reducing the larger gain, GRmax at fGRmax.\n\nThe CP radiation from the metaloop excites the parasitic loop and generates a rotating/traveling current along the loop.\n\nThe parasitic loop acts as a director for the driven meta loop, like Yagi–Uda antenna. If the radiation from the para sitic loop at fGLmax is constructively superimposed onto the radiation from the metaloop in the broadside direction, gain GLmax increases. For this to occur, parasitic loop length Lpara is chosen to be one free-space wavelength (1λ0) at fGLmax: Lpara = 116.7 mm. The remaining task is to optimize loop height Hpara.\n\nFigure 6 shows GL at fGLmax = 2.58 GHz and GR at fGRmax = 3.51 GHz in the broadside direction (z-direction) as a function of parasitic loop height Hpara. It is found that there is an optimum antenna height for GLmaxGRmax.\n\nBased on the result shown in Figure 6, the loop height is determined to be Hpara = 11 mm, corresponding to 0.09 wavelength at fGLmax = 2.58 GHz. Figure 7 shows the frequency response of the gain. The bandwidth (BW) for a 3-dB gain drop criterion for GL, denoted as GL-BW, is 13.5% and the BW for a 3-dB gain drop criterion for GR, denoted as GR-BW, is 14.6%. For confirmation of the analysis/simulation results, measured/experimental results are also presented (see fabricated antenna in Fig. 5b).",
null,
"Fig. 5Metaloop with a parasitic natural conducting loop. (a) Perspective view. (b) Fabricated metaloop.",
null,
"Fig. 6 GL at f = 2.58 GHz and GR at 3.51 GHz as a function of parasitic loop (paraloop) height Hpara, where wpara = 2 mm and Lpara = 116.7 mm (≈1λ0 at 2.58 GHz) are used.",
null,
"Fig. 7Frequency response of the gain with a parasitic loop (paraloop), where Hpara = 11 mm = 0.09 wavelength at fGLmax = 2.58 GHz.\n\n## 5 Radiation pattern and VSWR\n\nFigure 8a shows the analysis/simulation results of the radiation patterns at the maximum-gain low frequency fGLmax = 2.58 GHz (radiation efficiency of η ≈ 69%) and high frequency fGRmax = 3.51 GHz (η ≈ 60%), together with experimental results, where EL and ER denote the LHCP wave component and the RHCP wave component, respectively. For comparison, the analysis/simulation results in the absence of a parasitic loop are also shown in Figure 8b. It is clearly seen that the radiation pattern at low frequency fGLmax is narrowed by virtue of the presence of the parasitic loop, while the radiation pattern at high frequency fGRmax is less affected by the parasitic loop, as desired. Figure 9 shows the frequency response of the VSWR, which remains almost unchanged in the presence of the parasitic loop. Discrepancy between the analysis and experiment results is attributed to the fact that soldering the capacitive chips to the subwavelength segments is not uniform due to handwork.\n\nFinally, the following comments are added. The metaloop antenna in this paper could be operated as a dual-band CP element with the same rotational sense (dual-band mono-CP radiation), although this is not our objective. Such dual-band mono-CP radiation is performed by introducing switching circuits to points F and T so that each point can be chosen to be either a feed point or a terminal point. (1) For dual-band mono-LHCP radiation, set points F and T to be the feed (fd) and terminal (trmnl) points, respectively, at low frequency flow < fT, which is expressed as",
null,
". And at a high frequency fhigh > fT, change the role of points F and T by using switching circuits:",
null,
". (2) For dual-band mono-RHCP radiation, set",
null,
"at flow and",
null,
"at fhigh by using switch-circuits.",
null,
"Fig. 8Normalized radiation patterns at fGLmax = 2.58 GHz and fGRmax = 3.51 GHz. (a) In the presence of a parasitic loop. (b) In the absence of a parasitic loop.",
null,
"Fig. 9Frequency response of the VSWRs in the presence and absence of a parasitic loop (paraloop). The 3-dB gain-drop bandwidths for GL and GR are denoted as GL-BW and GR-BW, respectively.\n\n## 6 Conclusions\n\nThe dual-band counter CP wave radiated by a square metaloop antenna has a maximum gain of GLmax at frequency fGLmax that is different from maximum gain GRmax at frequency fGRmax, where GLmax < GRmax. To reduce the difference in these gains, a square parasitic natural conducting loop of one free-space wavelength at fGLmax is placed at height Hpara above the metaloop. It is found that there is an antenna height where the parasitic loop increases gain GL, while not remarkably affecting gain GR. Thus, the gain difference can be reduced, i.e., GLmaxGRmax, with the VSWR remaining almost unchanged in the presence of the parasitic loop.\n\n## References\n\n1. L.W. Rispin, D.C. Chang, Wire and loop F antennas, in: Y.T. Lo, S.W. Lee (Eds.), Antenna Handbook, Van Nostrand Reinhold Company Inc., NY 1988 [Google Scholar]\n2. H. Nakano, A numerical approach to line antennas printed on dielectric materials, Comput. Phys. Commun. 68 , 441 (1991) [CrossRef] [Google Scholar]\n3. H. Nakano, Y. Iitsuka, J. Yamauchi, Loop-based circularly polarized grid array antenna with edge excitation, IEEE Trans. Antennas Propag. 61 , 4045 (2013) [CrossRef] [Google Scholar]\n4. H. Nakano, K. Yoshida, J. Yamauchi, Radiation characteristics of a metaloop antenna, IEEE Antennas Wirel. Propag. Lett. 12 , 861 (2013) [CrossRef] [Google Scholar]\n5. C. Caloz, T. Itoh, Electromagnetic metamaterials (Wiley, NJ, 2006) [Google Scholar]\n6. N. Engheta, R.W. Ziolkowski, Metamaterials (Wiley, NJ, 2006) [CrossRef] [Google Scholar]\n7. G.V. Eleftheriades, K.G. Balmain, Negative-refraction metamaterials: Fundamental principles and applications (Wiley, NY, 2005) [Google Scholar]\n9. S.A. Rezaeieh, M.A. Antoniades, A.M. Abbosh, Gain enhancement of wideband metamaterial-loaded loop antenna with tightly coupled arc-shaped directors, IEEE Trans. Antennas Propag. 65 , 2090 (2017) [CrossRef] [Google Scholar]\n10. X. Zhao, Y. Lee, K.Y. Jung, J.H. Choi, Design of a metamaterial-inspired size-reduced wideband loop antenna with frequency scanning characteristic, IET Microw. Antennas Propag. 6 , 1227 (2012) [CrossRef] [Google Scholar]\n11. L.-M. Si, Q.-L. Zhang, W.-D. Hu, W.-H. Yu, Y.-M. Wu, X. Lv, W. Zhu, A uniplanar triple-band dipole antenna using complementary capacitively loaded loop, IEEE Antennas Wirel. Propag. Lett. 14 , 743 (2015) [CrossRef] [Google Scholar]\n12. Y. Zhang, K. Wei, Z. Zhang, Y. Li, Z. Feng, A compact dual-mode metamaterial-based loop antenna for pattern diversity, IEEE Antennas Wirel. Propag. Lett. 14 , 394 (2015) [CrossRef] [Google Scholar]\n13. B. Ghosh, S.K.M. Haque, D. Mitra, S. Ghosh, A loop loading technique for the miniaturization of non-planar and planar antennas, IEEE Trans. Antennas Propag. 58 , 2116 (2010) [CrossRef] [Google Scholar]\n14. S. Pandit, A. Mohan, P. Ray, A low-profile high-gain substrate-integrated waveguide-slot antenna with suppressed cross polarization using metamaterial, IEEE Antennas Wirel. Propag. Lett. 16 , 1614 (2017) [CrossRef] [Google Scholar]\n15. A.L. Borja, A. Belenguer, J. Cascon, J.R. Kelly, A reconfigurable passive UHF reader loop antenna for near-field and far-field RFID applications, IEEE Antennas Wirel. Propag. Lett. 11 , 580 (2012) [CrossRef] [Google Scholar]\n16. F.-Y. Meng, K. Zhang, J.-H. Fu, Q. Wu, J. Hua, Analogue of electromagnetically induced transparency in a magnetic metamaterial, IEEE Trans. Magn. 48 , 4390 (2012) [CrossRef] [Google Scholar]\n17. S.A. Rezaeieh, M.A. Antoniades, A.M. Abbosh, Bandwidth and directivity enhancement of loop antenna by nonperiodic distribution of mu-negative metamaterial unit cells, IEEE Trans. Antennas Propag. 64 , 3319 (2016) [CrossRef] [Google Scholar]\n18. K. Wei, Z. Zhang, Z. Feng, Design of a wideband horizontally polarized omnidirectional printed loop antenna, IEEE Antennas Wirel. Propag. Lett. 11 , 49 (2012) [Google Scholar]\n19. Z.-G. Liu, Y.-X. Guo, Compact low-profile dual band metamaterial antenna for body centric communications, IEEE Antennas Wirel. Propag. Lett. 14 , 863 (2015) [CrossRef] [Google Scholar]\n20. X. Qing, C.K. Goh, Z.N. Chen, A broadband UHF near-field RFID antenna, IEEE Trans. Antennas Propag. 58 , 3829 (2010) [CrossRef] [Google Scholar]\n\nCite this article as: Hisamatsu Nakano, Ittoku Yoshino, Tomoki Abe, Junji Yamauchi, Balanced gain for a square metaloop antenna, EPJ Appl. Metamat. 6, 3 (2019)\n\nTable 1\n\nParameters.\n\n## All Figures",
null,
"Fig. 1Natural loop antennas radiating a circularly polarized wave with a single feed point F. The loop has perturbation elements . In the text",
null,
"Fig. 2Metaloop antenna. (a) Perspective view. (b) Side view. (c) Arm unit cell. In the text",
null,
"Fig. 3Loop length L1 + L2 + L3 + L4 + L5 normalized to guided wavelength λg, as a function of frequency. β < 0 at f < fT, β = 0 at fT, and β > 0 at f > fT. In the text",
null,
"Fig. 4Frequency response of the gain. In the text",
null,
"Fig. 5Metaloop with a parasitic natural conducting loop. (a) Perspective view. (b) Fabricated metaloop. In the text",
null,
"Fig. 6 GL at f = 2.58 GHz and GR at 3.51 GHz as a function of parasitic loop (paraloop) height Hpara, where wpara = 2 mm and Lpara = 116.7 mm (≈1λ0 at 2.58 GHz) are used. In the text",
null,
"Fig. 7Frequency response of the gain with a parasitic loop (paraloop), where Hpara = 11 mm = 0.09 wavelength at fGLmax = 2.58 GHz. In the text",
null,
"Fig. 8Normalized radiation patterns at fGLmax = 2.58 GHz and fGRmax = 3.51 GHz. (a) In the presence of a parasitic loop. (b) In the absence of a parasitic loop. In the text",
null,
"Fig. 9Frequency response of the VSWRs in the presence and absence of a parasitic loop (paraloop). The 3-dB gain-drop bandwidths for GL and GR are denoted as GL-BW and GR-BW, respectively. In the text\n\nCurrent usage metrics show cumulative count of Article Views (full-text article views including HTML views, PDF and ePub downloads, according to the available data) and Abstracts Views on Vision4Press platform.\n\nData correspond to usage on the plateform after 2015. The current usage metrics is available 48-96 hours after online publication and is updated daily on week days."
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http://www.governmentadda.com/data-interpretation-quiz-for-upcoming-exams-answers-2/ | [
"Tuesday , March 26 2019\nRecent Post\nHome / Data Interpretation Quiz For Upcoming Exams Answers\n\n# Data Interpretation Quiz For Upcoming Exams Answers\n\n 150+ RRB NTPC Previous Solved Papers PDF Free Download Now RRB NTPC Free Study Material PDF Free Download Now",
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"Attempt It Now Quant Booster eBook – A Complete Maths Shortcut eBook GET Book Now SSC CGL,CPO,GD 2018 Study Material & Book Free PDF Free Download Now\n\nSolutions\n\nQ1 Option d\n\nQ2 Option c\n\nThe total sales (in thousands) of all the seven years for various batteries are:\n\nFor 4AH = 75 + 90 + 96 + 105 + 90 + 105 + 115 = 676\n\nFor 7AH = 144 + 126 + 114 + 90 + 75 + 60 + 85 = 694\n\nFor 32AH = 114 + 102 + 75 + 150 + 135 + 165 + 160 = 901\n\nFor 35AH = 102 + 84 + 105 + 90 + 75 + 45 + 100 = 601\n\nFor 55AH = 108 + 126 + 135 + 75 + 90 + 120 + 145 = 799.\n\nClearly, sales are maximum in case of 32AH batteries.\n\nQ3 Option d\n\nQ4 Option d\n\nQ5 Option b\n\nFrom the table it is clear that the sales of 7AH batteries have been decreasing continuously from 1992 to 1997.\n\nQ6 Option c\n\nQ7 Option b\n\nQ8 Option b\n\nQ9 Option e\n\nQ10 Option e\n\nQ11 Option c\n\n GovernmentAdda Recommends",
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https://www.coolstuffshub.com/pressure/convert/megapascals-to-pounds-per-square-inch/ | [
"# Megapascals to pounds per square inch conversion\n\n1 MPa = 145.04 psi\n\n## How to convert megapascals to pounds per square inch?\n\nTo convert megapascals to pounds per square inch, multiply the value in megapascals by 145.04.\n\nYou can use the formula :\npounds per square inch = megapascals × 145.04\n\n## How many pounds per square inch are in a megapascal?\n\nThere are 145.04 pounds per square inch in a megapascal.\n1 megapascal is equal to 145.04 pounds per square inch.\n\n## Megapascals to pounds per square inch conversion table\n\nMegapascals Pounds per square inch\n1 MPa 145.04 psi\n2 MPa 290.08 psi\n3 MPa 435.12 psi\n4 MPa 580.16 psi\n5 MPa 725.2 psi\n6 MPa 870.24 psi\n7 MPa 1015.28 psi\n8 MPa 1160.32 psi\n9 MPa 1305.36 psi\n10 MPa 1450.4 psi\n20 MPa 2900.8 psi\n30 MPa 4351.2 psi\n40 MPa 5801.6 psi\n50 MPa 7252 psi\n60 MPa 8702.4 psi\n70 MPa 10152.8 psi\n80 MPa 11603.2 psi\n90 MPa 13053.6 psi\n100 MPa 14504 psi"
] | [
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https://mathematica.stackexchange.com/users/6804/masterxilo?tab=answers&sort=votes | [
"For learning what I consider \"just the wolfram/mathematica language\" I would proceed as follows and start at the very low-level basics: Learn what is meant by a symbolic expression: ...\n\nThis can be implemented elegantly with FoldPairList and TakeDrop (both new in v10.2), in fact it's one of the examples in the documentation: FoldPairList[TakeDrop, Range, {2, 3, 5}] {{1, 2}, {3, ...\n\nGraph being AtomQ has about the same implications as doing lots of overloadings of the type Part[_Graph,___] := Error[]; Replace[_Graph,___] := Error[]; .... It signifies that it is an \"abstract data ...\n\nThe result of the front-end's parsing is definitely used, but it does not need to be complete, it seems. This can be seen by constructing Cells/Boxes manually. Try RawBoxes@RowBox[{\"1\", \"+\", RowBox[{...\n\nAssociation/<||> objects are Atomic and thus unmatchable before 10.4 AtomQ@Association[] yields True. This is confusing because it is not stated anywhere in the manual. For example tutorial/...\n\nI started working on a package and LinkedLibrary achieving significant speedups with this job: data = Table[{1. x, 1. y, 0.}, {x, 1000}, {y, 10^3}]~Flatten~1; << ExportTable AbsoluteTiming@...\n\nIt is worth mentioning GeneralUtilitiesPrintDefinitions. Use it like <<GeneralUtilities then GeneralUtilitiesPrintDefinitions@f to get a pretty-printed, formatted version of ??f in a new ...\n\nIf I where to do this manually, I would start with PythonForm~SetAttributes~HoldAll (*known symbols*) PythonForm[Sin[x_]] := StringTemplate[\"math.sin()\"]@PythonForm@x PythonForm[Times[a_, b_]] := ...\n\nProbably, MathematicalFunctionData[\"ArcTanh\"] knows about this. First let's see what categories of things are known: MathematicalFunctionData[\"ArcTanh\", \"Properties\"] ah well, let's just list them ...\n\nIf you ever develop a library for LibraryLink, be sure to include calls to AbortQ to support aborting lengthy computations. Code that runs for a long time should call AbortQ to see if the user has ...\n\nHere's an alternative: The package https://github.com/Masterxilo/PLYExport implements exporting of Points in PLY format, and extends Export with {\"PLY\", \"BinaryFormat\" -> True} to use it, allowing ...\n\nThis gives the recently opened files as a list of rules: Lookup[Options@$FrontEnd, NotebooksMenu] Just the file names: First /@ Lookup[Options@$FrontEnd, NotebooksMenu] FileNameJoin instead of ...\n\nI think all of these algorithms can be described in the following form: Assume your iterative numerical algorithm f takes and returns a pair {x, additionalState}. x is the quantity (tensor) that ...\n\nLet's ask FormulaData. These should suffice (there are not that many anyways :( ): Flatten@{FormulaData[\"TriangleAreaSSS\"], FormulaData[\"TriangleAreaBH\"]} // TeXForm (I started looking for something ...\n\nGo to Edit > Preferences > Messages and set Minor user interface warnings: to Ignore.\n\nApplicationMaker shows how to create packages that look like native functionality, including integration into the Documentation Center.\n\nhttps://mathematica.stackexchange.com/a/118739/6804 discusses compiling a library link dll and attaching a debugger. With say Visual Studio, you can just attach to process -> WolframKernel.exe after ...\n\nThe following approach works for me with a Vanilla Visual Studio 2015 installation (14.0) and Mathematica 10.2 (for Windows 8.1, x64). Open vs2015. Create a new empty project, let us call it ...\n\nRun SetOptions[$FrontEnd, PrintAction -> \"PrintToConsole\"] The setting will persist across restarts of Mathematica. View answer 3 votes It would be helpful if Mathematica could at least say that the error lies in the pdf version mismatch. Thanks @Jens. Here is another way to fix a pdf generated from pdflatex to be Importable in ... View answer 3 votes This can now be done with an InfiniteLine in Epilog or Prolog since version 10: Manipulate[ Plot[10^f*x^3, {x, -E/2, 1}, PlotRangePadding -> 0, Epilog -> InfiniteLine[{0.5, 0}, {0, 1}]], {{... View answer 3 votes This is purely a formatting \"issue\": View answer 3 votes I'm using this now: Unprotect@HoldForm; Format[HoldForm[x_], StandardForm] := {boxes = MakeBoxes@x, bg = LightBlue}~With~ RawBoxes@ InterpretationBox[ TagBox[StyleBox[boxes, ... View answer 3 votes Defining$AssertFunction might be a place to get started, but it is not ideal since when it gets called, the evaluation will have already happened. You will change the program behaviour if you ...\n\nFor returning multiple values, you'll find it much easier to return them via a WSTP/Mathlink link object. Start from this: #include <mathlink.h> #pragma comment(lib, \"ml64i4m\") // only 4 seems ...\n\nIt seems the behaviour of InternalComparePatterns changes from version to version. On 10.4.1 I get InternalComparePatterns[(_) .., __] Internal`ComparePatterns[__, (_) ..] (*=>*) \"Specific\" \"...\n\nUnderstanding $Context,$ContextPath the parsing stage and runtime scoping constructs A symbol in Mathematica can never be without a context. We can assume that the internal representation of any ...\n\n{#1, f@#2} & @@@ data But we can also generalize MapAt to do the job: Note that MapAt[f, data, {1, 2}] === {{1, f}, {0, 1}, {1, 1}} So we can use MapAtP[f_, expr_, positionPattern_] := ..."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8130976,"math_prob":0.6887824,"size":5452,"snap":"2022-05-2022-21","text_gpt3_token_len":1376,"char_repetition_ratio":0.07507342,"word_repetition_ratio":0.0,"special_character_ratio":0.26687455,"punctuation_ratio":0.21763602,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9544642,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-24T20:41:46Z\",\"WARC-Record-ID\":\"<urn:uuid:9bf57a26-aa9a-4cef-a5a4-9e50830d3e6e>\",\"Content-Length\":\"146103\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d3207c93-052e-4457-a023-8ab1a04b890f>\",\"WARC-Concurrent-To\":\"<urn:uuid:5789476c-3f7c-455b-84a4-93cc2e7ee3ab>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://mathematica.stackexchange.com/users/6804/masterxilo?tab=answers&sort=votes\",\"WARC-Payload-Digest\":\"sha1:KRXBZMFUXDS3M27KIKE2244YO5ZNSRN5\",\"WARC-Block-Digest\":\"sha1:RHNUIAYPPQYQM6DEJ6BSBRE7VS444RKR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304600.9_warc_CC-MAIN-20220124185733-20220124215733-00597.warc.gz\"}"} |
https://jp.mathworks.com/matlabcentral/profile/authors/20240325 | [
"Community Profile",
null,
"# Johan Pelloux-Prayer\n\nLast seen: 2日 前 2021 以来アクティブ\n\n#### Statistics\n\nAll\n•",
null,
"•",
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"•",
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"バッジを表示\n\n#### Content Feed\n\nSums of cubes and squares of sums\nGiven the positive integers 1:n, can you: 1. Compute twice the sum of the cubes of those numbers. 2. Subtract the square...\n\n18日 前\n\nRelative ratio of \"1\" in binary number\nInput(n) is positive integer number Output(r) is (number of \"1\" in binary input) / (number of bits). Example: * n=0; r=...\n\n19日 前\n\nVector creation\nCreate a vector using square brackets going from 1 to the given value x in steps on 1. Hint: use increment.\n\n20日 前\n\nDoubling elements in a vector\nGiven the vector A, return B in which all numbers in A are doubling. So for: A = [ 1 5 8 ] then B = [ 1 1 5 ...\n\n20日 前\n\nCreate a vector\nCreate a vector from 0 to n by intervals of 2.\n\n20日 前\n\nFlip the vector from right to left\nFlip the vector from right to left. Examples x=[1:5], then y=[5 4 3 2 1] x=[1 4 6], then y=[6 4 1]; Request not ...\n\n20日 前\n\nFind max\nFind the maximum value of a given vector or matrix.\n\n20日 前\n\nHow to plot multiple circles within a circle by using a for-loop for radius/center position?\nHi, I have been working with somewhat similiar plotting needs. To plot circles I used the polyshape built-in function with the ...\n\n21日 前 | 0\n\n| 採用済み\n\nWhether the input is vector?\nGiven the input x, return 1 if x is vector or else 0.\n\nGet the length of a given vector\nGiven a vector x, the output y should equal the length of x.\n\nInner product of two vectors\nFind the inner product of two vectors.\n\nArrange Vector in descending order\nIf x=[0,3,4,2,1] then y=[4,3,2,1,0]\n\nSelect every other element of a vector\nWrite a function which returns every other element of the vector passed in. That is, it returns the all odd-numbered elements, s...\n\nFind the sum of all the numbers of the input vector\nFind the sum of all the numbers of the input vector x. Examples: Input x = [1 2 3 5] Output y is 11 Input x ...\n\nTimes 2 - START HERE\nTry out this test problem first. Given the variable x as your input, multiply it by two and put the result in y. Examples:...\n\nStrange result of find function\nHello, I noticed a strange results of the find function today. I am using the find function to extract the indices of a time s..."
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https://www.geeksforgeeks.org/python-program-to-find-k-maximum-elements-of-array-in-original-order/?ref=rp | [
"# Python Program to Find k maximum elements of array in original order\n\n• Last Updated : 28 May, 2022\n\nGiven an array arr[] and an integer k, we need to print k maximum elements of given array. The elements should printed in the order of the input.\nNote : k is always less than or equal to n.\n\nExamples:\n\n```Input : arr[] = {10 50 30 60 15}\nk = 2\nOutput : 50 60\nThe top 2 elements are printed\nas per their appearance in original\narray.\n\nInput : arr[] = {50 8 45 12 25 40 84}\nk = 3\nOutput : 50 45 84```\n\nMethod 1: We search for the maximum element k times in the given array. Each time we find one maximum element, we print it and replace it with minus infinite (-1*sys.maxsize for largest minimum value in Python) in the array. Also, the position of all k maximum elements is marked using an array so that with the help of that array we can print the elements in the order given in the original array. The time complexity of this method is O(n*k).\n\n## Python3\n\n `# Python program to find k maximum elements``# of array in original order` `# import the module``import` `sys` `# Function to print k Maximum elements``def` `printMax(arr, k, n):`` ``brr``=``[``0``]``*``n`` ``crr``=``[``0``]``*``n`` ` ` ``# Coping the array arr`` ``# into crr so that it`` ``# can be used later`` ``for` `i ``in` `range``(n):`` ``crr[i]``=``arr[i]`` ` ` ``# Iterating for K-times`` ``for` `i ``in` `range``(k):`` ``# Finding the maximum element`` ``# along with its index`` ``maxi``=``-``1``*``sys.maxsize`` ``for` `j ``in` `range``(n):`` ``if``(maxi\n\nOutput\n\n`50 45 84 `\n\nTime Complexity: O(n*k)\nAuxiliary Space: O(n)\n\nMethod 2: In this method, we store the original array in a new array and will sort the new array in descending order. After sorting, we iterate the original array from 0 to n and print all those elements that appear in first k elements of new array. For searching, we can do Binary Search.\n\n## Python3\n\n `# Python3 program to find k maximum elements``# of array in original order` `# Function to print Maximum elements``def` `printMax(arr, k, n):`` ` ` ``# vector to store the copy of the`` ``# original array`` ``brr ``=` `arr.copy()`` ` ` ``# Sorting the vector in descending`` ``# order. Please refer below link for`` ``# details`` ``brr.sort(reverse ``=` `True``)`` ` ` ``# Traversing through original array and`` ``# print all those elements that are`` ``# in first k of sorted vector.`` ``for` `i ``in` `range``(n):`` ``if` `(arr[i] ``in` `brr[``0``:k]):`` ``print``(arr[i], end ``=` `\" \"``)` `# Driver code``arr ``=` `[ ``50``, ``8``, ``45``, ``12``, ``25``, ``40``, ``84` `]``n ``=` `len``(arr)``k ``=` `3` `printMax(arr, k, n)` `# This code is contributed by SHUBHAMSINGH10`\n\nOutput\n\n`50 45 84 `\n\nTime Complexity: O(n Log n) for sorting.\nAuxiliary Space: O(n)\n\nPlease refer complete article on Find k maximum elements of array in original order for more details!\n\nMy Personal Notes arrow_drop_up"
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http://www.toontricks.com/2018/05/tutorial-convert-iterator-to-pointer.html | [
"# Tutorial :Convert iterator to pointer?",
null,
"### Question:\n\nI have a `std::vector` with `n` elements. Now I need to pass a pointer to a vector that has the last `n-1` elements to a function.\n\nFor example, my `vector<int> foo` contains `(5,2,6,87,251)`. A function takes `vector<int>*` and I want to pass it a pointer to `(2,6,87,251)`.\n\nCan I just (safely) take the iterator `++foo.begin()`, convert it to a pointer and pass that to the function? Or use `&foo`?\n\nUPDATE: People suggest that I change my function to take an iterator rather than a pointer. That seems not possible in my situation, since the function I mentioned is the `find` function of `unordered_set<std::vector*>`. So in that case, is copying the `n-1` elements from `foo` into a new vector and calling `find` with a pointer to that the only option? Very inefficient! It's like Shlemiel the painter, especially since i have to query many subsets: the last `n-1`, then `n-2`, etc. elements and see if they are in the `unordered_set`.\n\n### Solution:1\n\nThat seems not possible in my situation, since the function I mentioned is the find function of `unordered_set<std::vector*>`.\n\nAre you using custom hash/predicate function objects? If not, then you must pass `unordered_set<std::vector<int>*>::find()` the pointer to the exact vector that you want to find. A pointer to another vector with the same contents will not work. This is not very useful for lookups, to say the least.\n\nUsing `unordered_set<std::vector<int> >` would be better, because then you could perform lookups by value. I think that would also require a custom hash function object because `hash` does not to my knowledge have a specialization for `vector<int>`.\n\nEither way, a pointer into the middle of a vector is not itself a vector, as others have explained. You cannot convert an iterator into a pointer to vector without copying its contents.\n\n### Solution:2\n\nhere it is, obtaining a reference to the coresponding pointer of an iterator use :\n\nexample:\n\n``string my_str= \"hello world\"; string::iterator it(my_str.begin()); char* pointer_inside_buffer=&(*it); //<-- ``\n\n[notice operator * returns a reference so doing & on a reference will give you the address].\n\n### Solution:3\n\nIf you can, a better choice may be to change the function to take either an iterator to an element or a brand new vector (if it does not modify).\n\nWhile you can do this sort of things with arrays since you know how they are stored, it's probably a bad idea to do the same with vectors. `&foo` does not have the type `vector<int>*`.\n\nAlso, while the STL implementation is available online, it's usually risky to try and rely on the internal structure of an abstraction.\n\n### Solution:4\n\nYour function shouldn't take `vector<int>*`; it should take `vector<int>::iterator` or `vector<int>::const_iterator` as appropriate. Then, just pass in `foo.begin() + 1`.\n\n### Solution:5\n\nA vector is a container with full ownership of it's elements. One vector cannot hold a partial view of another, even a const-view. That's the root cause here.\n\nIf you need that, make your own container that has views with weak_ptr's to the data, or look at ranges. Pair of iterators (even pointers work well as iterators into a vector) or, even better, boost::iterator_range that work pretty seamlessly.\n\nIt depends on the templatability of your code. Use std::pair if you need to hide the code in a cpp.\n\n### Solution:6\n\nThe direct answer to your question is yes. If foo is a vector, you can do this: &foo.\n\nThis only works for vectors however, because the standard says that vectors implement storage by using contigious memory.\n\nBut you still can (and probably should) pass iterators instead of raw pointers because it is more expressive. Passing iterators does not make a copy of the vector.\n\n### Solution:7\n\nFor example, my `vector<int> foo` contains (5,2,6,87,251). A function takes `vector<int>*` and I want to pass it a pointer to (2,6,87,251).\n\nA pointer to a `vector<int>` is not at all the same thing as a pointer to the elements of the vector.\n\nIn order to do this you will need to create a new `vector<int>` with just the elements you want in it to pass a pointer to. Something like:\n\n`` vector<int> tempVector( foo.begin()+1, foo.end()); // now you can pass &tempVector to your function ``\n\nHowever, if your function takes a pointer to an array of int, then you can pass `&foo`.\n\n### Solution:8\n\nUse `vector::front`, it should be the most portable solution. I've used this when I'm interfacing with a fixed API that wants a char ptr. Example:\n\n``void funcThatTakesCharPtr(char* start, size_t size); ... void myFunc(vector<char>& myVec) { // Get a pointer to the front element of my vector: char* myDataPtr = &(myVec.front()); // Pass that pointer to my external API: funcThatTakesCharPtr(myDataPtr, myVec.size()); } ``\n\n### Solution:9\n\nIf your function really takes `vector<int> *` (a pointer to vector), then you should pass `&foo` since that will be a pointer to the vector. Obviously that will not simply solve your problem, but you cannot directly convert an iterator to a vector, since the memory at the address of the iterator will not directly address a valid vector.\n\nYou can construct a new vector by calling the vector constructor:\n\n``template <class InputIterator> vector(InputIterator, InputIterator) ``\n\nThis constructs a new vector by copying the elements between the two iterators. You would use it roughly like this:\n\n``bar(std::vector<int>(foo.begin()+1, foo.end()); ``\n\n### Solution:10\n\nI haven't tested this but could you use a set of pairs of iterators instead? Each iterator pair would represent the begin and end iterator of the sequence vector. E.g.:\n\n``typedef std::vector<int> Seq; typedef std::pair<Seq::const_iterator, Seq::const_iterator> SeqRange; bool operator< (const SeqRange& lhs, const SeqRange& rhs) { Seq::const_iterator lhsNext = lhs.first; Seq::const_iterator rhsNext = rhs.first; while (lhsNext != lhs.second && rhsNext != rhs.second) if (*lhsNext < *rhsNext) return true; else if (*lhsNext > *rhsNext) return false; return false; } typedef std::set<SeqRange, std::less<SeqRange> > SeqSet; Seq sequences; void test (const SeqSet& seqSet, const SeqRange& seq) { bool find = seqSet.find (seq) != seqSet.end (); bool find2 = seqSet.find (SeqRange (seq.first + 1, seq.second)) != seqSet.end (); } ``\n\nObviously the vectors have to be held elsewhere as before. Also if a sequence vector is modified then its entry in the set would have to be removed and re-added as the iterators may have changed.\n\nJon\n\n### Solution:11\n\nVector is a template class and it is not safe to convert the contents of a class to a pointer : You cannot inherit the vector class to add this new functionality. and changing the function parameter is actually a better idea. Jst create another vector of int vector temp_foo (foo.begin[X],foo.end()); and pass this vector to you functions\n\n### Solution:12\n\nA safe version to convert an iterator to a pointer (exactly what that means regardless of the implications) and by safe I mean no worries about having to dereference the iterator and cause possible exceptions / errors due to `end()` / other situations\n\n``#include <iostream> #include <vector> #include <string.h> int main() { std::vector<int> vec; char itPtr; long long itPtrDec; std::vector<int>::iterator it = vec.begin(); memset(&itPtr, 0, 25); sprintf(itPtr, \"%llu\", it); itPtrDec = atoll(itPtr); printf(\"it = 0x%X\\n\", itPtrDec); vec.push_back(123); it = vec.begin(); memset(&itPtr, 0, 25); sprintf(itPtr, \"%llu\", it); itPtrDec = atoll(itPtr); printf(\"it = 0x%X\\n\", itPtrDec); } ``\n\nwill print something like\n\nit = 0x0\n\nit = 0x2202E10\n\nIt's an incredibly hacky way to do it, but if you need it, it does the job. You will receive some compiler warnings which, if really bothering you, can be removed with `#pragma`\n\nNote:If u also have question or solution just comment us below or mail us on toontricks1994@gmail.com\nPrevious\nNext Post »"
] | [
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https://www.rdocumentation.org/packages/splines/versions/3.6.2/topics/splineDesign | [
"# splineDesign\n\n0th\n\nPercentile\n\n##### Design Matrix for B-splines\n\nEvaluate the design matrix for the B-splines defined by knots at the values in x.\n\nKeywords\nmodels\n##### Usage\nsplineDesign(knots, x, ord = 4, derivs, outer.ok = FALSE,\nsparse = FALSE)\nspline.des (knots, x, ord = 4, derivs, outer.ok = FALSE,\nsparse = FALSE)\n##### Arguments\nknots\n\na numeric vector of knot positions (which will be sorted increasingly if needed).\n\nx\n\na numeric vector of values at which to evaluate the B-spline functions or derivatives. Unless outer.ok is true, the values in x must be between the “inner” knots knots[ord] and knots[ length(knots) - (ord-1)].\n\nord\n\na positive integer giving the order of the spline function. This is the number of coefficients in each piecewise polynomial segment, thus a cubic spline has order 4. Defaults to 4.\n\nderivs\n\nan integer vector with values between 0 and ord - 1, conceptually recycled to the length of x. The derivative of the given order is evaluated at the x positions. Defaults to zero (or a vector of zeroes of the same length as x).\n\nouter.ok\n\nlogical indicating if x should be allowed outside the inner knots, see the x argument.\n\nsparse\n\nlogical indicating if the result should inherit from class \"sparseMatrix\" (from package Matrix).\n\n##### Value\n\nA matrix with length(x) rows and length(knots) - ord columns. The i'th row of the matrix contains the coefficients of the B-splines (or the indicated derivative of the B-splines) defined by the knot vector and evaluated at the i'th value of x. Each B-spline is defined by a set of ord successive knots so the total number of B-splines is length(knots) - ord.\n\n##### Note\n\nThe older spline.des function takes the same arguments but returns a list with several components including knots, ord, derivs, and design. The design component is the same as the value of the splineDesign function.\n\n• splineDesign\n• spline.des\n##### Examples\nlibrary(splines) # NOT RUN { require(graphics) splineDesign(knots = 1:10, x = 4:7) splineDesign(knots = 1:10, x = 4:7, deriv = 1) ## visualize band structure # } # NOT RUN { Matrix::drop0(zapsmall(6*splineDesign(knots = 1:40, x = 4:37, sparse = TRUE))) # } # NOT RUN { knots <- c(1,1.8,3:5,6.5,7,8.1,9.2,10) # 10 => 10-4 = 6 Basis splines x <- seq(min(knots)-1, max(knots)+1, length.out = 501) bb <- splineDesign(knots, x = x, outer.ok = TRUE) plot(range(x), c(0,1), type = \"n\", xlab = \"x\", ylab = \"\", main = \"B-splines - sum to 1 inside inner knots\") mtext(expression(B[j](x) *\" and \"* sum(B[j](x), j == 1, 6)), adj = 0) abline(v = knots, lty = 3, col = \"light gray\") abline(v = knots[c(4,length(knots)-3)], lty = 3, col = \"gray10\") lines(x, rowSums(bb), col = \"gray\", lwd = 2) matlines(x, bb, ylim = c(0,1), lty = 1) # } \nDocumentation reproduced from package splines, version 3.6.2, License: Part of R 3.6.2\n\n### Community examples\n\nLooks like there are no examples yet."
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https://numbermatics.com/n/44566152399/ | [
"# 44566152399\n\n## 44,566,152,399 is an odd composite number composed of two prime numbers multiplied together.\n\nWhat does the number 44566152399 look like?\n\nThis visualization shows the relationship between its 2 prime factors (large circles) and 8 divisors.\n\n44566152399 is an odd composite number. It is composed of two distinct prime numbers multiplied together. It has a total of eight divisors.\n\n## Prime factorization of 44566152399:\n\n### 33 × 1650598237\n\n(3 × 3 × 3 × 1650598237)\n\nSee below for interesting mathematical facts about the number 44566152399 from the Numbermatics database.\n\n### Names of 44566152399\n\n• Cardinal: 44566152399 can be written as Forty-four billion, five hundred sixty-six million, one hundred fifty-two thousand, three hundred ninety-nine.\n\n### Scientific notation\n\n• Scientific notation: 4.4566152399 × 1010\n\n### Factors of 44566152399\n\n• Number of distinct prime factors ω(n): 2\n• Total number of prime factors Ω(n): 4\n• Sum of prime factors: 1650598240\n\n### Divisors of 44566152399\n\n• Number of divisors d(n): 8\n• Complete list of divisors:\n• Sum of all divisors σ(n): 66023929520\n• Sum of proper divisors (its aliquot sum) s(n): 21457777121\n• 44566152399 is a deficient number, because the sum of its proper divisors (21457777121) is less than itself. Its deficiency is 23108375278\n\n### Bases of 44566152399\n\n• Binary: 1010011000000101100110000100110011112\n• Base-36: KH1JB9R\n\n### Squares and roots of 44566152399\n\n• 44566152399 squared (445661523992) is 1986141939650893455201\n• 44566152399 cubed (445661523993) is 88514704368527178580999445177199\n• The square root of 44566152399 is 211106.9690915009\n• The cube root of 44566152399 is 3545.4256336787\n\n### Scales and comparisons\n\nHow big is 44566152399?\n• 44,566,152,399 seconds is equal to 1,417 years, 3 weeks, 2 days, 22 hours, 46 minutes, 39 seconds.\n• To count from 1 to 44,566,152,399 would take you about two thousand, eight hundred thirty-four years!\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 44566152399 cubic inches would be around 295.5 feet tall.\n\n### Recreational maths with 44566152399\n\n• 44566152399 backwards is 99325166544\n• The number of decimal digits it has is: 11\n• The sum of 44566152399's digits is 54\n• More coming soon!"
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https://www.shaalaa.com/question-bank-solutions/find-domain-sec-1-x-tan-1-x-inverse-trigonometric-functions-simplification-examples_64071 | [
"Advertisement Remove all ads\n\n# Find the Domain of Sec − 1 X − Tan − 1 X - Mathematics\n\nShort Note\n\nFind the domain of sec^(-1) x-tan^(-1)x\n\nAdvertisement Remove all ads\n\n#### Solution\n\nLet f(x) = g(x) − h(x), where g(x)=cotx and h(x)=cot1x\nTherefore, the domain of f(x) is given by the intersection of the domain of g(x) and h(x)\nThe domain of g(x) is [0, π/2) ⋃ [ π, 3π/2)\nThe domain of h(x) is (-π/2,π/2)\nTherfore, the intersection of g(x) and h(x) is R − { nπ, n ⋵ Z}\n\nConcept: Inverse Trigonometric Functions (Simplification and Examples)\nIs there an error in this question or solution?\nAdvertisement Remove all ads\n\n#### APPEARS IN\n\nRD Sharma Class 12 Maths\nChapter 4 Inverse Trigonometric Functions\nExercise 4.4 | Q 3.2 | Page 18\nAdvertisement Remove all ads\n\n#### Video TutorialsVIEW ALL \n\nAdvertisement Remove all ads\nShare\nNotifications\n\nView all notifications\n\nForgot password?"
] | [
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https://pnpntransistor.com/what-is-operating-point-in-transistor-q-point/ | [
"# What is Operating Point in Transistor? Q point Explanation",
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"Here we discuss what is the operating point of transistor? or Q point of a transistor? .Here we also know how we can find Q point of BJT transistor.\n\nThe operating point is a specific point on transistor output characteristics at which we get good biasing for that transistor. The operating point is a point which we can obtain from the value of collector current Ic and collector base voltage Vcb at no input signal is applied to the transistor.\n\nSo if we want to call what is an operating point? in one line so we can say that,\n\n“ The zero signal values of collector current (Ic) and collector-base voltage (Vcb) are known as operating point for that transistor”.\n\nIt is called the operating point because the variations of Ic and Vce take place at this point when an input signal is applied. This point is also called as Q point or quiescent (Silent) point because it is a point of output characteristics when a transistor is silent i.e. in absence of the signal. For operating point, we have to find load line.",
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"## How to find an operating point?\n\nFor finding operating point first we have to find load line points( A and B ) on Ic and Vce. For general transistor biasing circuit, output circuit equation is\n\nVce = Vcc – IcRc\n\nThe output characteristics of this transistor show graph between Ic and Vce. For finding intersecting points of load line with X-axis and Y-axis we take one by one Ic=0 and then Vce=0.\n\nFirst, we take Ic= 0 so that\n\nVce = Vcc\n\nThis value of point B will be Vcc. If we take Vce=0 then,\n\nIc = Vcc/ Rc\n\nSo the value of point A will be Vcc /Rc.\n\nSo now we already find two points( A and B) of this load line. So when we draw this load line, a point at which load line intersecting with Ib is called the operating point or q-point. Hope you know well about operating point. Let see how to find an operating point with an example or problem.\n\n#### Example:\n\nQuestion – As the circuit shown below, (i) if Vcc = 12V and Rc = 6kΩ, draw the d.c load line. What will be the Q-point if zero signal base current is 20μA and β= 50?",
null,
"Here collector-emitter voltage Vce is given by,\n\nVce= Vcc – IcRc\n\nWhen Ic = 0 A then Vce= Vcc so Vce=12v\n\nWhen Vce= 0 V then Ic = Vcc/ Rc. So ic will be 12/2 A = 6 A.\n\nBy joining this two points we get load line. Zero signal base current Ib= 20 μA and β is 50. So we have to find an intersecting point of load line with Ib.\n\nIc = βIb\n\nIc = 50 X .02 = 1 mA\n\nZero signal collector-emitter voltage is,\n\nVce= Vcc – IcRc = 12 – (1mA x 6kΩ )= 6V\n\nSo that operating point will be ( Ic, Vce ) = ( 1mA, 6V ).\n\nHope now you know all about the operating point of the transistor and how to find operation point for the transistor. if you have any question regarding Q-point calculation or whatever you can ask in the comment section. If you read not our previous transistor related article see also-\n\nWhat is transistor\n\nPNP transistor\n\nNPN transistor\n\nCommon-emitter transistor connection\n\nCommon-base transistor connection"
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https://keio.elsevierpure.com/en/publications/covariance-matrix-estimation-in-a-seemingly-unrelated-regression- | [
"# Covariance matrix estimation in a seemingly unrelated regression model under Stein’s loss\n\nShun Matsuura, Hiroshi Kurata\n\nResearch output: Contribution to journalArticlepeer-review\n\n1 Citation (Scopus)\n\n## Abstract\n\nA seemingly unrelated regression model has been commonly used for describing a set of different regression models with correlations. This paper discusses the estimation of the covariance matrix in a seemingly unrelated regression model under Stein’s loss function. In particular, when the correlation matrix is assumed to be known, a best equivariant estimator of the covariance matrix is derived. Its properties are investigated and a connection to a best equivariant estimator of regression coefficients given in a previous study is shown. Results of numerical simulations and an illustrative example are also presented to compare the best equivariant estimator of the covariance matrix with several conventional covariance matrix estimators.\n\nOriginal language English 79-99 21 Statistical Methods and Applications 29 1 https://doi.org/10.1007/s10260-019-00473-x Published - 2020 Mar 1\n\n## Keywords\n\n• Correlation matrix\n• Covariance matrix\n• Equivariant estimator\n• Generalized least squares estimator\n• Seemingly unrelated regression model\n\n## ASJC Scopus subject areas\n\n• Statistics and Probability\n• Statistics, Probability and Uncertainty\n\n## Fingerprint\n\nDive into the research topics of 'Covariance matrix estimation in a seemingly unrelated regression model under Stein’s loss'. Together they form a unique fingerprint."
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http://forums.wolfram.com/mathgroup/archive/2009/Jun/msg00300.html | [
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"Slow/jerky animations inside manipulate (more details)\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg100635] Slow/jerky animations inside manipulate (more details)\n• From: Porscha Louise McRobbie <pmcrobbi at umich.edu>\n• Date: Wed, 10 Jun 2009 05:31:54 -0400 (EDT)\n\n```As several people have kindly pointed out, my first post was a bit too\nvague (my first time with Mathgroup). I am adding more specific\ndetails here. Thanks!\n\n------Original post: \"Slow/jerky animations inside manipulate\" ---------\nI have an Animate command (I'm using GraphicsRow to show two\nside-by-side synchronized animations) inside of Manipulate. The\nresulting animations play very fast and are jerky. I can adjust the\nplay speed using AnimationRate, but it must be slowed down by a\nridiculous amount in order to look smooth. I've tried adjusting the\nRefreshRate, as wellas making time a slider variable and animating\nfrom within the Manipulate control panel,both with little success.\nHow can I create smooth animations, appropriate for class demonstrations?\n-------------------------------------------------------------------------\n\nMy plots are actually simple. I am, however, solving an ODE inside of the\nManipulate/Animate commands. I've played around with the NDSolve\noptions thinking it might make things faster, but again no success.\nBasically I just want two sliders to choose initial conditions for the\nODEs, then animate the results.\n\nInside Manipulate, I solve the following ODEs, where the initial\nconditions q0,p0 are the slider variables:\n\nsols = First@NDSolve[{q'[t] == p[t], p'[t] == 2 A De (Exp[-2 A (q[t] - xe)] -\nExp[-A (q[t] - xe)]), q == q0, p == p0}, {q, p}, {t,0, 100}];\n\nInside Animate, I have two plots (tp is the animation variable):\n\n1. Plot solution q(tp) vs. tp, as well as a circle that moves along\nas the curve is being traced out:\n\np1 = Graphics[{ Point[{tp, Evaluate[q[tp] /. sols]}]}];\np2 = Quiet@Plot[Evaluate[q[T1] /. sols], {T1, 0, tp},\nPerformanceGoal->\"Speed\"];\nFIRSTplot = Show[p1, p2];\n\n2. Plot a static background curve \"Staticplot\" (computed outside\nAnimate), with a circle moving on it. The equation for the background\ncurve is:\n\nf(q)=De(1+Exp[-2 A (q-xe)]-2 Exp[-A (q-xe)])\n\nThe coordinates for Point below are {q,f(q)}.\n\np7 = Graphics[{Point[{Evaluate[q[tp] /. sols],\nDe (1 + Exp[-2 A (Evaluate[q[tp] /. sols] - xe)] -\n2 Exp[-A (Evaluate[q[tp] /. sols] - xe)]) }] }];\nSECONDplot=Show[Staticplot,p7];\n-------------------------------------------------------------------------\nThanks again for any help.\n\n```\n\n• Prev by Date: What should be a simple task....\n• Next by Date: Re: perturbation methods example from stephen lynch's book?"
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https://www.physicsforums.com/threads/equivalent-capacitance.948215/ | [
"# Equivalent capacitance\n\nThread moved from the technical forums, so no Homework Template is shown\nThe picture attached with this post shows a combination of capacitors, we are asked to find out the equivalent capacitance between the points A and B.\nProvided that all the capacitors are identical and has an individual capacitance C.\nCan anyone help me provide basic ideas of solving these kind of problems( I mean a general approach). [emoji4]\n\n#### Attachments\n\n• 19.7 KB Views: 379\n\nRelated Engineering and Comp Sci Homework Help News on Phys.org\nSimon Bridge\nHomework Helper\nA general approach would be to use the kirkoff circuit laws.\nIf you don't know about those, you had best look for ways to redraw the diagram so it looks simpler.\n\nFor instance ... see that box of capacitors in the middle... number the nodes around them 1 through 4, clockwise, from the top left. Label the node closest to A as 0, and the one closest to B as 5.\n\nNotice that node 4 is the same as node 0.\n\nSo those first two capacitors are in parallel.\n\nRedraw it like that ... you'll end up with sort-of nested parallel capacitors. You know a rule for parallel capacitors.\n\nThe point of these exercizes is to get you to spot these sorts of patterns.\n\n•",
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"Merlin3189\nA general approach would be to use the kirkoff circuit laws.\nIf you don't know about those, you had best look for ways to redraw the diagram so it looks simpler.\n\nFor instance ... see that box of capacitors in the middle... number the nodes around them 1 through 4, clockwise, from the top left. Label the node closest to A as 0, and the one closest to B as 5.\n\nNotice that node 4 is the same as node 0.\n\nSo those first two capacitors are in parallel.\n\nRedraw it like that ... you'll end up with sort-of nested parallel capacitors. You know a rule for parallel capacitors.\n\nThe point of these exercizes is to get you to spot these sorts of patterns.\nIn that case, I am getting the answer as (2/3) *C\nIs that correct..?\n\ncnh1995\nHomework Helper\nGold Member\nIs that correct\nNo.\n\nNo.\n\nsophiecentaur\nGold Member\nShow us your redrawn diagram as @Simon Bridge recommended - that could be your problem.\nAlternatively: We all know how to calculate the equivalent to two series or parallel components but there is also a useful formula for transforming a star of three resistors (or Reactances) to a delta of three equivalent resistors (Reactances) and vice versa. This link explains it. You can use that to change your circuit into two deltas which makes the problem easier (visually) and shows up some symmetry which allows you to eliminate some components for the final result. I think that link should help you enough to solve the prob and you will then have another tool in your tool box.\n\nSimon Bridge\nHomework Helper\nIn that case, I am getting the answer as (2/3) *C\nIs that correct..?\nI don't know: I did not do the problem. Others say \"no\". Maybe they are correct maybe not.\n\nA general approach would be to use the kirkoff circuit laws.\nIf you don't know about those, you had best look for ways to redraw the diagram so it looks simpler.\n\nFor instance ... see that box of capacitors in the middle... number the nodes around them 1 through 4, clockwise, from the top left. Label the node closest to A as 0, and the one closest to B as 5.\n\nNotice that node 4 is the same as node 0.\n\nSo those first two capacitors are in parallel.\n\nRedraw it like that ... you'll end up with sort-of nested parallel capacitors. You know a rule for parallel capacitors.\n\nThe point of these exercizes is to get you to spot these sorts of patterns.\nI didn't understand how will the nested arrangement of capacitors looks like... Can u please explain more on that or provide a suitable diagram, the final results that I have posted seems to be wrong to me because I couldn't make any logic out of that, please anyone help me out with this..!\n\ncnh1995\nHomework Helper\nGold Member\nI didn't understand how will the nested arrangement of capacitors looks like... Can u please explain more on that or provide a suitable diagram, the final results that I have posted seems to be wrong to me because I couldn't make any logic out of that, please anyone help me out with this..!\nPut a voltage source between a and b and analyse the diagram. Can you find any series/parallel connection?\n\n•",
null,
"Simon Bridge\nPut a voltage source between a and b and analyse the diagram. Can you find any series/parallel connection?\nNot exactly. I couldn't figure out one from the diagram the problematic capacitors for me are X and Y(figure 1), but I have doubt whether is it possible to redraw the circuit as to the one I have done in figure 2 then I can find 3 parallel and 1 series circuit of capacitors from that. Am I right in redrawing that circuit? Please explain..",
null,
"#### Attachments\n\n• 36.9 KB Views: 683\nNot exactly. I couldn't figure out one from the diagram the problematic capacitors for me are X and Y(figure 1), but I have doubt whether is it possible to redraw the circuit as to the one I have done in figure 2 then I can find 3 parallel and 1 series circuit of capacitors from that. Am I right in redrawing that circuit? Please explain.. View attachment 226812\nSorry, this is the figure 2 that I have mentioned in my previous post",
null,
"#### Attachments\n\n• 21.9 KB Views: 357\nCWatters\nHomework Helper\nGold Member\nFig 2 is not correct. Q and R are not in parallel in the original. You are trying to make too many simplifications in one go. I would start by just combining Y and S (or P and X) and redraw that.\n\nSimon Bridge\nHomework Helper\nNo.\n\nNote... from diagram 1, nodes 2 and 5 are the same.\nSo Y and S are parallel... you can collapse them to one capacitor between nodes 5 and 3, node 2 vanishes.\n\nRedraw.\n\nQ connects between 1 and 5 while R connects between 4 and 3... so Q and R are NOT parallel.\n\nTake more care.\n\nNo.\n\nNote... from diagram 1, nodes 2 and 5 are the same.\nSo Y and S are parallel... you can collapse them to one capacitor between nodes 5 and 3, node 2 vanishes.\n\nRedraw.\n\nQ connects between 1 and 5 while R connects between 4 and 3... so Q and R are NOT parallel.\n\nTake more care.\nSir, I am having one doubt, if the node 2 collapses then what happens to the connection between node 2 and node 3..?\nI want to know whether the newly attached diagram (diagram 3) can be considered correct after collapsing. Can I proceed the same way for the capacitors P and X ?",
null,
"#### Attachments\n\n• 31.2 KB Views: 342\nSimon Bridge\nHomework Helper\nLast diagram looks good.\nThe connection between 2 and 3 becomes a connection between 5 and 3.\n\nThe key is that wires are part of the node they are attached to. So two nodes joined by a wire are, in fact, the same node.\n\nThis lets you simplify the diagram... you are allowed to draw curved lines for wires/connections.\n\nNow see if there are other capacitor pairs you can collapse this way.\n\nLast diagram looks good.\nThe connection between 2 and 3 becomes a connection between 5 and 3.\n\nThe key is that wires are part of the node they are attached to. So two nodes joined by a wire are, in fact, the same node.\n\nThis lets you simplify the diagram... you are allowed to draw curved lines for wires/connections.\n\nNow see if there are other capacitor pairs you can collapse this way.\nThank you sir. [emoji4]\nI think in that way we can collapse node 4 and combine P and X with parallel connected capacitors formula to achieve even simpler circuit diagram.",
null,
"#### Attachments\n\n• 25.2 KB Views: 327\nShow us your redrawn diagram as @Simon Bridge recommended - that could be your problem.\nAlternatively: We all know how to calculate the equivalent to two series or parallel components but there is also a useful formula for transforming a star of three resistors (or Reactances) to a delta of three equivalent resistors (Reactances) and vice versa. This link explains it. You can use that to change your circuit into two deltas which makes the problem easier (visually) and shows up some symmetry which allows you to eliminate some components for the final result. I think that link should help you enough to solve the prob and you will then have another tool in your tool box.\nUsing this conversion I have redrawn my diagram can you please tell me whether its correct or not?\nI also want to know can we use resistor combination formulas and wheatstone's bridge concept to deal with reactance(I mean in simplifying and solving them).....!",
null,
"#### Attachments\n\n• 24.7 KB Views: 644\nLast edited:\nsophiecentaur\nGold Member\nUsing this conversion I have redrawn my diagram can you please tell me whether its correct or not?\nI also want to know can we use resistor combination formulas and wheatstone's bridge concept to deal with reactance(I mean in simplifying and solving them).....! View attachment 226863\nThat looks fine. Now notice that the two parallel capacitors are placed symmetrically across the two paths 0,2,5 and 0,4,5. The PD across them is zero (when all values are equal) because there is no current flowing across, so you now have two pairs of series C in parallel. No sums actually required now!\nPS Resistor formulae also apply to Reactances and Impedances (with care of course - big C = small reactance)\n\nLast edited:\nI have tried to solve this question in the both ways suggested\n1. Collapsing the nodes\n2. Using nodal analysis and conversion\nBut, still I got some problem. Those two ways are giving me different answers to the same problem. Here I am attaching my answers with this post, can anyone please help me to figure out where I went wrong...!\nHere I changed the capacitance as\nP = Q = R = S = X = Y = \"C\" (each has same capacitance)",
null,
"#### Attachments\n\n• 37.2 KB Views: 299\nSimon Bridge\nHomework Helper\nRecheck your working -- especially where you are rearranging the capacitors in method 2. You seem to have a blindness there.\nI am getting the same answers by the two methods.\n\nThe point of this sort of exercise is to get you to understand how circuits work ... not to find the answer by correctly applying the rules and formulas. The actual answer is irrelevant to the task in front of you.\n\nRecheck your working -- especially where you are rearranging the capacitors in method 2. You seem to have a blindness there.\nI am getting the same answers by the two methods.\n\nThe point of this sort of exercise is to get you to understand how circuits work ... not to find the answer by correctly applying the rules and formulas. The actual answer is irrelevant to the task in front of you.\nI am a bit confused sir, in case of the second method I used the mesh analysis(considering the reactance of the capacitor) to convert the available two stars in to delta then as suggested I removed the capacitors connecting 0-2-5 and 0-4-5 (ref post 18) since the PD between them is zero,\nThere I am stuck with the problem that whether I have to continue solving with reactance(using resistor equations) or by using the new capacitance( using capacitor combination formulas) please help...\n\nPlease anybody..... Help me out with this problem....[emoji846]\n\nepenguin\nHomework Helper\nGold Member\nPlease anybody..... Help me out with this problem....[emoji846]\nYou appear to be OK in #16. You were probably on the right lines in #3 but then made a mistake. If you had written out your working there you might either have seen the error yourself, or someone could have pointed it out to you.\nGo back to #16 - that is quite a simple circuit. It should look even simpler if you put the actual values in (In terms of C).\n\nLast edited:\nYou appear to be OK in #16. You were probably on the right lines in #3 but then made a mistake. If you had written out your working there you might either have seen the error yourself, or someone could have pointed it out to you.\nGo back to #16 - that is quite a simple circuit. It should look even simpler if you put the actual values in (In terms of C).\nI have solved the circuit of #16 in #19 by substituting the actual C values, but my confusion is there in the 2nd method(nodal analysis) of #19. Can you please help me figure out where I went wrong from there onwards, because I am getting different answers in those two methods for the same problem.\n\nepenguin\nHomework Helper\nGold Member\nSorry I missed that you had given the answer in the left-hand page of #19. I think that is correct, notice it is something that you could have worked out in your head (actually I did). For the other method it is useful to know the right answer beforehand, but I will leave it to those sponsoring this approach.",
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https://softmath.com/algebra-help/study-and-solution-for-algebra.html | [
"",
null,
"## What our customers say...\n\nThousands of users are using our software to conquer their algebra homework. Here are some of their experiences:\n\nOur daughter is making the grades she is capable of thanks to the Algebrator. Hats off to you all! Thank you!\nL.J., Utah\n\nThe Algebrator could replace teachers, sometime in the future. It is more detailed and more patient than my current math teacher. I, personally, understand algebra better. Thank you for creating it!\nTom Canty, AZ\n\nIt was very helpful. it was a great tool to check my answers with. I would recommend this software to anyone no matter what level they are at in math.\nRick Edmondson, TX\n\nSo after I make my first million as a world-class architect, I promise to donate ten percent to Algebrator! If you ask me, thats cheap too, because theres just no way Id have even dreamed about being an architect before I started using your math program. Now Im just one year away from graduation and being on my way!\nLaura Jackson, NC\n\n## Search phrases used on 2009-04-14:\n\nStudents struggling with all kinds of algebra problems find out that our software is a life-saver. Here are the search phrases that today's searchers used to find our site. Can you find yours among them?\n\n• free aptitude test papers\n• solving algebra\n• printable 6th grade work\n• change in slope formula calculation\n• aptitude questions & answers\n• Mcdougal littell worksheets\n• log2 texas ti-82\n• practice dividing decimals\n• What is one basic principle that can be used to simplify a polynomial?\n• picture 2 step inequality worksheet\n• how to solve the square variation\n• free algebra for dummies download\n• negative fractional square roots\n• age problems in quadratic form\n• multiplying integers\n• Java example nonlinear systems third power\n• chemical equations for 6th graders\n• simplify radicals calculator\n• formula for hyperbola graph\n• Real life problems + L. C. M.\n• fractions as powers\n• samples for adding and subtracting integers\n• alberta grade 9 math achievement exam sample questions\n• Answers to Pre-Algebra Practice Workbook by McDougal\n• Algebra Problem Solvers for Free\n• free algebra calculator\n• worksheet coordinate grids for third graders\n• cubic equations in TK Solver Help\n• rationalize radicals worksheet\n• scale factor activities\n• Mastering Algebra course 1 linear and equations introduction to functions worksheet\n• elementary algebra mixture problems samples\n• converting factored parabolic equation\n• cost accounting ebooks\n• examples of math trivia with answers\n• denominator finder\n• calculating log2\n• 9th grade and online games\n• worksheets on rational exponents\n• Free Rational Expressions Solver\n• free Download question paper of mat examination\n• step by steps college algebra\n• trigonomic substitutions\n• TI-83 log\n• solve me algebra equation by using substitution calculator\n• algebra graphing investigating slope and shift slope intercept form printable free fun worksheet activity\n• matlab equations with variables\n• decimal square root solver ti 83\n• intermediate algebra problems\n• math how to calculate age\n• 6th grade algebra lesson\n• equation for multiple point ratio\n• TI-86 graphing +calculater\n• Root mean square solver\n• fourth grade mixed review worksheets\n• detailed lesson plan about multiplying radicals\n• math problems division with decimals solutions\n• factoring practice for algebra 1\n• maths sheet area\n• aptitude model questions\n• factoring application TI-83\n• trinomial factoring games\n• 26.21 mastering physics answers\n• doomsday equation +graphing\n• solving systems with 3 variables\n• online graphing calculator TI89\n• Prealgebra exams 2007 samples\n• literal equation online calculator\n• algerbra\n• least to greatest fractions\n• Measurement Adding and Subtracting\n• cube root calculator\n• Matric level General Apptitude objective questions\n• free books mcgraw hill downloads\n• exponent problem solver\n• trig substitution calculator\n• solving a quadratic in excel\n• how do i cube the number five\n• algebra sums\n• 9th Grade GED Practice Sheets\n• is a quadratic an exponent\n• simplify the gradient of a logarithm\n• get number between 2 numbers java\n• transforming formulas polynomials\n• symmetry worksheets y3\n• worksheets on linear graphing\n• visual basic linear inequalities VB\n• algebra equation by using substitution calculator\n Prev Next"
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http://www.conversion-website.com/speed/kilometer-per-second-to-inch-per-second.html | [
"# Kilometers per second to inches per second (km/s to in/s)\n\n## Convert kilometers per second to inches per second\n\nKilometers per second to inches per second converter above calculates how many inches per second are in 'X' kilometers per second (where 'X' is the number of kilometers per second to convert to inches per second). In order to convert a value from kilometers per second to inches per second (from km/s to in/s) just type the number of km/s to be converted to in/s and then click on the 'convert' button.\n\n## Kilometers per second to inches per second conversion factor\n\n1 kilometer per second is equal to 39370.078740157 inches per second\n\n## Kilometers per second to inches per second conversion formula\n\nSpeed(in/s) = Speed (km/s) × 39370.078740157\n\nExample: Assume there are 70 kilometers per second. Shown below are the steps to express them in inches per second.\n\nSpeed(in/s) = 70 ( km/s ) × 39370.078740157 ( in/s / km/s )\n\nSpeed(in/s) = 2755905.511811 in/s or\n\n70 km/s = 2755905.511811 in/s\n\n70 kilometers per second equals 2755905.511811 inches per second\n\n## Kilometers per second to inches per second conversion table\n\nkilometers per second (km/s)inches per second (in/s)\n10393700.78740157\n15590551.18110236\n20787401.57480315\n25984251.96850394\n301181102.3622047\n351377952.7559055\n401574803.1496063\n451771653.5433071\n501968503.9370079\n552165354.3307087\n602362204.7244094\n652559055.1181102\n702755905.511811\n752952755.9055118\n803149606.2992126\nkilometers per second (km/s)inches per second (in/s)\n1505905511.8110236\n2007874015.7480315\n2509842519.6850394\n30011811023.622047\n35013779527.559055\n40015748031.496063\n45017716535.433071\n50019685039.370079\n55021653543.307087\n60023622047.244094\n65025590551.181102\n70027559055.11811\n75029527559.055118\n80031496062.992126\n85033464566.929134\n\nVersions of the kilometers per second to inches per second conversion table. To create a kilometers per second to inches per second conversion table for different values, click on the \"Create a customized speed conversion table\" button.\n\n## Related speed conversions\n\nBack to kilometers per second to inches per second conversion\n\nTableFormulaFactorConverterTop"
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https://www.colorhexa.com/001608 | [
"# #001608 Color Information\n\nIn a RGB color space, hex #001608 is composed of 0% red, 8.6% green and 3.1% blue. Whereas in a CMYK color space, it is composed of 100% cyan, 0% magenta, 63.6% yellow and 91.4% black. It has a hue angle of 141.8 degrees, a saturation of 100% and a lightness of 4.3%. #001608 color hex could be obtained by blending #002c10 with #000000. Closest websafe color is: #000000.\n\n• R 0\n• G 9\n• B 3\nRGB color chart\n• C 100\n• M 0\n• Y 64\n• K 91\nCMYK color chart\n\n#001608 color description : Very dark (mostly black) cyan - lime green.\n\n# #001608 Color Conversion\n\nThe hexadecimal color #001608 has RGB values of R:0, G:22, B:8 and CMYK values of C:1, M:0, Y:0.64, K:0.91. Its decimal value is 5640.\n\nHex triplet RGB Decimal 001608 `#001608` 0, 22, 8 `rgb(0,22,8)` 0, 8.6, 3.1 `rgb(0%,8.6%,3.1%)` 100, 0, 64, 91 141.8°, 100, 4.3 `hsl(141.8,100%,4.3%)` 141.8°, 100, 8.6 000000 `#000000`\nCIE-LAB 5.341, -9.475, 4.54 0.331, 0.591, 0.326 0.265, 0.474, 0.591 5.341, 10.507, 154.398 5.341, -4.714, 3.781 7.69, -5.78, 2.866 00000000, 00010110, 00001000\n\n# Color Schemes with #001608\n\n• #001608\n``#001608` `rgb(0,22,8)``\n• #16000e\n``#16000e` `rgb(22,0,14)``\nComplementary Color\n• #031600\n``#031600` `rgb(3,22,0)``\n• #001608\n``#001608` `rgb(0,22,8)``\n• #001613\n``#001613` `rgb(0,22,19)``\nAnalogous Color\n• #160003\n``#160003` `rgb(22,0,3)``\n• #001608\n``#001608` `rgb(0,22,8)``\n• #130016\n``#130016` `rgb(19,0,22)``\nSplit Complementary Color\n• #160800\n``#160800` `rgb(22,8,0)``\n• #001608\n``#001608` `rgb(0,22,8)``\n• #080016\n``#080016` `rgb(8,0,22)``\n• #0e1600\n``#0e1600` `rgb(14,22,0)``\n• #001608\n``#001608` `rgb(0,22,8)``\n• #080016\n``#080016` `rgb(8,0,22)``\n• #16000e\n``#16000e` `rgb(22,0,14)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #001608\n``#001608` `rgb(0,22,8)``\n• #003011\n``#003011` `rgb(0,48,17)``\n• #00491b\n``#00491b` `rgb(0,73,27)``\n• #006324\n``#006324` `rgb(0,99,36)``\nMonochromatic Color\n\n# Alternatives to #001608\n\nBelow, you can see some colors close to #001608. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #001602\n``#001602` `rgb(0,22,2)``\n• #001604\n``#001604` `rgb(0,22,4)``\n• #001606\n``#001606` `rgb(0,22,6)``\n• #001608\n``#001608` `rgb(0,22,8)``\n• #00160a\n``#00160a` `rgb(0,22,10)``\n• #00160c\n``#00160c` `rgb(0,22,12)``\n• #00160e\n``#00160e` `rgb(0,22,14)``\nSimilar Colors\n\n# #001608 Preview\n\nText with hexadecimal color #001608\n\nThis text has a font color of #001608.\n\n``<span style=\"color:#001608;\">Text here</span>``\n#001608 background color\n\nThis paragraph has a background color of #001608.\n\n``<p style=\"background-color:#001608;\">Content here</p>``\n#001608 border color\n\nThis element has a border color of #001608.\n\n``<div style=\"border:1px solid #001608;\">Content here</div>``\nCSS codes\n``.text {color:#001608;}``\n``.background {background-color:#001608;}``\n``.border {border:1px solid #001608;}``\n\n# Shades and Tints of #001608\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000201 is the darkest color, while #eefff4 is the lightest one.\n\n• #000201\n``#000201` `rgb(0,2,1)``\n• #001608\n``#001608` `rgb(0,22,8)``\n• #002a0f\n``#002a0f` `rgb(0,42,15)``\n• #003d16\n``#003d16` `rgb(0,61,22)``\n• #00511d\n``#00511d` `rgb(0,81,29)``\n• #006425\n``#006425` `rgb(0,100,37)``\n• #00782c\n``#00782c` `rgb(0,120,44)``\n• #008c33\n``#008c33` `rgb(0,140,51)``\n• #009f3a\n``#009f3a` `rgb(0,159,58)``\n• #00b341\n``#00b341` `rgb(0,179,65)``\n• #00c748\n``#00c748` `rgb(0,199,72)``\n• #00da4f\n``#00da4f` `rgb(0,218,79)``\n• #00ee56\n``#00ee56` `rgb(0,238,86)``\n• #02ff5e\n``#02ff5e` `rgb(2,255,94)``\n• #16ff6b\n``#16ff6b` `rgb(22,255,107)``\n• #2aff77\n``#2aff77` `rgb(42,255,119)``\n• #3dff84\n``#3dff84` `rgb(61,255,132)``\n• #51ff90\n``#51ff90` `rgb(81,255,144)``\n• #64ff9d\n``#64ff9d` `rgb(100,255,157)``\n• #78ffa9\n``#78ffa9` `rgb(120,255,169)``\n• #8cffb6\n``#8cffb6` `rgb(140,255,182)``\n• #9fffc2\n``#9fffc2` `rgb(159,255,194)``\n• #b3ffcf\n``#b3ffcf` `rgb(179,255,207)``\n• #c7ffdb\n``#c7ffdb` `rgb(199,255,219)``\n• #daffe8\n``#daffe8` `rgb(218,255,232)``\n• #eefff4\n``#eefff4` `rgb(238,255,244)``\nTint Color Variation\n\n# Tones of #001608\n\nA tone is produced by adding gray to any pure hue. In this case, #0a0c0b is the less saturated color, while #001608 is the most saturated one.\n\n• #0a0c0b\n``#0a0c0b` `rgb(10,12,11)``\n• #090d0b\n``#090d0b` `rgb(9,13,11)``\n• #080e0a\n``#080e0a` `rgb(8,14,10)``\n• #080e0a\n``#080e0a` `rgb(8,14,10)``\n• #070f0a\n``#070f0a` `rgb(7,15,10)``\n• #06100a\n``#06100a` `rgb(6,16,10)``\n• #051109\n``#051109` `rgb(5,17,9)``\n• #041209\n``#041209` `rgb(4,18,9)``\n• #031309\n``#031309` `rgb(3,19,9)``\n• #031309\n``#031309` `rgb(3,19,9)``\n• #021408\n``#021408` `rgb(2,20,8)``\n• #011508\n``#011508` `rgb(1,21,8)``\n• #001608\n``#001608` `rgb(0,22,8)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #001608 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5203777,"math_prob":0.85571945,"size":3621,"snap":"2022-40-2023-06","text_gpt3_token_len":1597,"char_repetition_ratio":0.1432126,"word_repetition_ratio":0.014652015,"special_character_ratio":0.5633803,"punctuation_ratio":0.23146068,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9942268,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-06T11:27:34Z\",\"WARC-Record-ID\":\"<urn:uuid:b563a536-e0b4-41f1-a90e-7c48602f4a17>\",\"Content-Length\":\"36002\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:587e31e2-af7f-42e4-9fe7-9775b18c1002>\",\"WARC-Concurrent-To\":\"<urn:uuid:d99886f4-7bbe-4a74-9a27-b4fe4e7cea00>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/001608\",\"WARC-Payload-Digest\":\"sha1:XLFRNL2W5HWWFOTPQ3O2F3WJ5YLW6CLN\",\"WARC-Block-Digest\":\"sha1:Z5QNXX2IO5CWWSDARXCOSGZKAD5HSFAE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337803.86_warc_CC-MAIN-20221006092601-20221006122601-00487.warc.gz\"}"} |
https://www.simplylearnt.com/tips-tricks/Arithmetic | [
"# Tips and Tricks on Arithmetic\n\n1\nLargest number which divides x, y, z to leave same remainder a, b, c = H. C. F. of x-a, y-b, z-c.\n\n2\nLargest number which divides x, y, z to leave remainder R (i.e. same) = H.C.F. of x-R, y-R, z-R.\n\n3\nConsider the letter y to represent the age of a person.\n(a) If the current age is, y then n times the age = ny.\n(b) If the current age is y, then age n years later = y + n.\n(c) If the current age is y, then age n years ago = y - n.\n(d) The ages in a ratio a: b will be ay and by.\n(e) If current age is y, then 1/n of the age is y/n.\n\nPractice other most important chapters tips\nBasic Geometry\nMensuration\nTrigonometry\nNumber System\nand More",
null,
"SimplyLearnt"
] | [
null,
"https://googleads.g.doubleclick.net/pagead/viewthroughconversion/1031596478/",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.77035606,"math_prob":0.99753124,"size":891,"snap":"2020-24-2020-29","text_gpt3_token_len":282,"char_repetition_ratio":0.20405862,"word_repetition_ratio":0.705314,"special_character_ratio":0.335578,"punctuation_ratio":0.16205534,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99989235,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-06T21:46:44Z\",\"WARC-Record-ID\":\"<urn:uuid:0830cbd9-0223-4d40-bce5-3efaf6d111d6>\",\"Content-Length\":\"41940\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5c49eba4-9aa9-420d-8ebd-173e43d13449>\",\"WARC-Concurrent-To\":\"<urn:uuid:092e4be9-c5ae-4425-bc19-17b41bb8ed8f>\",\"WARC-IP-Address\":\"172.67.161.45\",\"WARC-Target-URI\":\"https://www.simplylearnt.com/tips-tricks/Arithmetic\",\"WARC-Payload-Digest\":\"sha1:MXYUQTASRXSE6R3NVZGNLQU4YWRCV7QF\",\"WARC-Block-Digest\":\"sha1:UYHAQI44YLP2PLQUCIDZRJTUX4Q3CKCT\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655890181.37_warc_CC-MAIN-20200706191400-20200706221400-00515.warc.gz\"}"} |
http://blog.madhukaraphatak.com/sizeof-operator-java-scala/ | [
"As I was going through Apache spark source code, I stumbled upon on one interesting tool. Spark has a utility, called SizeEstimator, which estimates the size of objects in Java heap. This is like sizeof operator for Java. I got fascinated and started to explore. This posts talks about the utility and its use cases.\n\ntl;dr Access complete code with documentation on github.\n\n## sizeof operator in C/C++\n\nIn C/C++ sizeof operator is used to determine size of a given data structure. This is important, as size of data structures in these languages are platform dependent. For example\n\n``std::cout << sizeof(int) ;``\n\nthe above c++ code will print 2 if it’s 16 bit machine and 4 if it’s 32 bit machine.\n\nAlso these languages does manual memory management. Developer uses sizeof operator to specify how much memory needs to be allocated.\n\n``int * intArray = (int *)calloc(10*sizeof(int));``\n\nThe above code create an array of integers which hold 10 elements. Again as you can see here we used sizeof operator specify exact amount memory we wanted to allocate.\n\nSo from above examples, it’s apparent that sizeof operator is a tool which helps you to know the size of the variable at runtime.\n\n## sizeof operator for Java\n\nJava do not have any sizeof operator in the language. The following are two reasons for that\n\n• The size of data structure is same on all platforms\n• Java virtual machine with garbage collection will do the memory management for you.\n\nSo as a developer, you do not need to worry about the memory management in java. So creators of Java felt there is no use of sizeof operator.\n\nBut there are few use cases where we may need a way to measure size of objects at runtime.\n\n## Use case for sizeof operator\n\nAs I told in the beginning, this idea of sizeof operator came from Spark source code. So I started digging why they need this. As it turns out SizeEstimator in Spark is used for building memory bounded caches. The idea is that you want to specify amount of heap memory the cache can use so when it runs out of memory it can use LRU method to accommodate newer keys.\n\n## Memory bounded caches in Spark\n\nWe use caches in almost every application. Normally most of the in-memory caches are bounded by number of items. You can specify how many keys it should keep. Once you cross the limit, you can use LRU to do the eviction. This works well when you are storing homogeneous values and all the pairs have relatively same size. Also it assumes that all machines where cache is running has same RAM size.\n\nBut in case of spark, the cluster may have varying RAM sizes. Also it may cache heterogeneous values. So having number of items as the bound is not optimal. So Spark uses the size of the cache as the bound value. So using sizeof operator they can optimally use the RAM on the cluster.\n\nYou can look at one of the implementation of memory bounded caches here.\n\n## java-sizeof library\n\nI extracted the code from the spark, simplified little and published as a independent library. So if you want to calculate size of your objects in your Java/Scala projects, you can use this library. This library is well tested inside the spark.\n\nYou can add the library through sbt or maven.\n\n• Sbt\n``libraryDependencies += \"com.madhukaraphatak\" %% \"java-sizeof\" % \"0.1\"``\n• Maven\n``````<dependency>\n<artifactId>java-sizeof_2.11</artifactId>\n<version>0.1</version>\n</dependency>``````\n\n## Using\n\nThe following code shows the api usage.\n\n``````SizeEstimator.estimate('a');\n\nList<Integer> values = new ArrayList<Integer>();"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.85701543,"math_prob":0.76557374,"size":3698,"snap":"2020-34-2020-40","text_gpt3_token_len":814,"char_repetition_ratio":0.13075258,"word_repetition_ratio":0.0,"special_character_ratio":0.21687399,"punctuation_ratio":0.10249308,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95341617,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-14T23:18:21Z\",\"WARC-Record-ID\":\"<urn:uuid:ed3801ef-89f8-4906-b79d-c5a64809f114>\",\"Content-Length\":\"16200\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:75848109-7853-467b-9571-92a13ca1d0be>\",\"WARC-Concurrent-To\":\"<urn:uuid:62351c1b-148e-41f8-8955-3a18579a9fa0>\",\"WARC-IP-Address\":\"185.199.110.153\",\"WARC-Target-URI\":\"http://blog.madhukaraphatak.com/sizeof-operator-java-scala/\",\"WARC-Payload-Digest\":\"sha1:EEFCMSULCKQLJULDZ3B5J2FUV3KZNTHU\",\"WARC-Block-Digest\":\"sha1:TWKSNC25I25KQRWB7KLWIAR4JCHAHCTN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439740343.48_warc_CC-MAIN-20200814215931-20200815005931-00256.warc.gz\"}"} |
https://numbersdb.org/numbers/604651 | [
"# The Number 604651 : Square Root, Cube Root, Factors, Prime Checker\n\nYou can find Square Root, Cube Root, Factors, Prime Check, Binary, Octal, Hexadecimal and more of Number 604651. 604651 is written as Six Hundred And Four Thousand, Six Hundred And Fifty One. You can find Binary, Octal, Hexadecimal Representation and sin, cos, tan values and Multiplication, Division tables.\n\n• Number 604651 is an odd Number.\n• Number 604651 is not a Prime Number\n• Sum of all digits of 604651 is 22.\n• Previous number of 604651 is 604650\n• Next number of 604651 is 604652\n\n## Square, Square Root, Cube, Cube Root of 604651\n\n• Square Root of 604651 is 777.59308124494\n• Cube Root of 604651 is 84.56063945572\n• Square of 604651 is 365602831801\n• Cube of 604651 is 221062117851306451\n\n## Numeral System of Number 604651\n\n• Binary Representation of 604651 is 10010011100111101011\n• Octal Representation of 604651 is 2234753\n• Hexadecimal Representation of 604651 is 939eb\n\n## Sin, Cos, Tan of Number 604651\n\n• Sin of 604651 is -0.5150380749097\n• Cos of 604651 is -0.85716730070232\n• Tan of 604651 is 0.600860619027\n\n## Multiplication Table for 604651\n\n• 604651 multiplied by 1 equals to 604,651\n• 604651 multiplied by 2 equals to 1,209,302\n• 604651 multiplied by 3 equals to 1,813,953\n• 604651 multiplied by 4 equals to 2,418,604\n• 604651 multiplied by 5 equals to 3,023,255\n• 604651 multiplied by 6 equals to 3,627,906\n• 604651 multiplied by 7 equals to 4,232,557\n• 604651 multiplied by 8 equals to 4,837,208\n• 604651 multiplied by 9 equals to 5,441,859\n• 604651 multiplied by 10 equals to 6,046,510\n• 604651 multiplied by 11 equals to 6,651,161\n• 604651 multiplied by 12 equals to 7,255,812\n\n## Division Table for 604651\n\n• 604651 divided by 1 equals to 604651\n• 604651 divided by 2 equals to 302325.5\n• 604651 divided by 3 equals to 201550.33333333\n• 604651 divided by 4 equals to 151162.75\n• 604651 divided by 5 equals to 120930.2\n• 604651 divided by 6 equals to 100775.16666667\n• 604651 divided by 7 equals to 86378.714285714\n• 604651 divided by 8 equals to 75581.375\n• 604651 divided by 9 equals to 67183.444444444\n• 604651 divided by 10 equals to 60465.1\n• 604651 divided by 11 equals to 54968.272727273\n• 604651 divided by 12 equals to 50387.583333333"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7614196,"math_prob":0.99766636,"size":308,"snap":"2021-31-2021-39","text_gpt3_token_len":72,"char_repetition_ratio":0.10197368,"word_repetition_ratio":0.0,"special_character_ratio":0.24025974,"punctuation_ratio":0.23809524,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998385,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-23T20:54:16Z\",\"WARC-Record-ID\":\"<urn:uuid:9402b08b-e4d6-49ae-81fa-307487242665>\",\"Content-Length\":\"10776\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cdc9efc2-1e98-4584-93f2-d13c08e7360c>\",\"WARC-Concurrent-To\":\"<urn:uuid:e063c16f-7971-4a97-9d6a-248b3c1cd3fa>\",\"WARC-IP-Address\":\"68.66.216.43\",\"WARC-Target-URI\":\"https://numbersdb.org/numbers/604651\",\"WARC-Payload-Digest\":\"sha1:LJ6JENQ5EEJGXHBKGLVPPTG227YS67A4\",\"WARC-Block-Digest\":\"sha1:WBWSURNXFIBAISHGLVV6JWQ2CHEQCTPN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057447.52_warc_CC-MAIN-20210923195546-20210923225546-00030.warc.gz\"}"} |
https://www.chegg.com/homework-help/glencoe-algebra-2-student-edition-1st-edition-chapter-13.6-problem-10cu-solution-9780078656095 | [
"Skip Navigation\n\nSolutions\nGlencoe Algebra 2, Student Edition\n\n# Glencoe Algebra 2, Student Edition (1st Edition) Edit edition Problem 10CU from Chapter 13.6\n\nWe have solutions for your book!\nChapter: Problem:\nStep-by-step solution:\nChapter: Problem:\n• Step 1 of 4\n\nSince, the spring the pulled down 3 inches from its equilibrium position and then released, it goes a minimum of 3 inches below equilibrium position and 3 inches above the equilibrium position. Moreover, the spring comes back to its starting position in 2 seconds, thus, the motion repeats itself in every 2 seconds, which is the period of the motion.\n\n• Chapter , Problem is solved.\nCorresponding Textbook",
null,
"Glencoe Algebra 2, Student Edition | 1st Edition\n9780078656095ISBN-13: 0078656095ISBN:"
] | [
null,
"https://cs.cheggcdn.com/covers2/60000/62983_1430330545_Width200.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8924313,"math_prob":0.8676563,"size":382,"snap":"2019-13-2019-22","text_gpt3_token_len":84,"char_repetition_ratio":0.15343915,"word_repetition_ratio":0.0,"special_character_ratio":0.21727748,"punctuation_ratio":0.12,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96145177,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-25T13:34:20Z\",\"WARC-Record-ID\":\"<urn:uuid:8d76d493-cf71-459c-ac3c-4a713877c0de>\",\"Content-Length\":\"244416\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2a3c2d55-429b-4f32-b58d-7e9647a06f5a>\",\"WARC-Concurrent-To\":\"<urn:uuid:30e9a0e9-5f23-4920-80f8-bde5155e6a02>\",\"WARC-IP-Address\":\"13.249.44.63\",\"WARC-Target-URI\":\"https://www.chegg.com/homework-help/glencoe-algebra-2-student-edition-1st-edition-chapter-13.6-problem-10cu-solution-9780078656095\",\"WARC-Payload-Digest\":\"sha1:YV7WCB7R2PXG4TKGH34VTHNW6W3CKA2S\",\"WARC-Block-Digest\":\"sha1:7XCD7CGOXN6O7C7YYFHWZCEJ2RFQ5P2I\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232258058.61_warc_CC-MAIN-20190525124751-20190525150751-00280.warc.gz\"}"} |
http://math-exercises.com/planimetry | [
"",
null,
"",
null,
"### Math Exercises & Math Problems: Perimeter and Area of Plane Figures",
null,
"",
null,
"Find the perimeter and the area of a square, if we know the length of its diagonal d = 4.2 m.",
null,
"Find the area of the rectangle ABCD, where the length of the side |AB| = a = 8.2 cm and the diagonal d = 2a.",
null,
"Lengths of the sides of a rectangular garden are in the ratio 1 : 2. Line connecting the centers of the adjacent sides of the garden is 20 m long. Calculate the perimeter and the area of the garden.",
null,
"A rectangular garden has a length of 57 m and a width of 42 m. Calculate of how many m2 will decrease the area of a garden, if the ornamental fence with a width of 60 cm will be planted inside its perimeter.",
null,
"A perimeter of a parallelogram is 2.8 meters. The length of one of its sides is equal to one-seventh of the entire perimeter. Find lengths of the sides of the parallelogram.",
null,
"One of the internal angles of the rhombus is 120° and the shorter diagonal is 3.4 meters long. Find the perimeter of the rhombus.",
null,
"Find a length of the diagonal AC of the rhombus ABCD if its perimeter P = 112 dm and the second diagonal BD has a length of 36 dm.",
null,
"In the isosceles trapezoid ABCD we know: AB||CD, |CD| = c = 8 cm, height h = 7 cm, |∠CAB| = 35°. Find the area of the trapezoid.",
null,
"A trapezoid ABCD has the bases length of a = 120 mm, c = 86 mm and the area A = 2,575 mm2. Find the height of the trapezoid.",
null,
"Consider the isosceles trapezoid PQRS. The bases are |PQ| = 120 mm, |RS| = 62 mm and the arm s = 48 mm. Find the height of the trapezoid, diagonal length and the area of the trapezoid.",
null,
"A land of a right-angled trapezoid shape has the bases lengths of 92 m and 76 m and the vertical arm is 6.3 m long. Find the land area and the length of a fencing needed to fence the land.",
null,
"A perimeter of an isosceles triangle is 474 m. The base of the triangle is by 48 m longer than the arm. Find the length of the sides and the area of the triangle.",
null,
"A right-angled triangle ABC has the legs a = 5 cm, b = 8 cm. A triangle A'B'C' is similar to the triangle ABC and it is 2.5 times smaller. Calculate what percentage of the area of the triangle ABC takes the area of the triangle A'B'C'.",
null,
"A right-angled isosceles triangle has the area of 32 cm2. What is its perimeter ?",
null,
"An equilateral triangle has the perimeter of 36 dm. What is its area ?",
null,
"A triangle ABC has side lengths a = 14 cm, b = 20 cm, c = 7.5 cm. Find the size of the internal angles and the area of the triangle.",
null,
"Find the perimeter, the area and the size of remaining angles of a triangle ABC, when: a = 8.4, β = 105°35' and median of side a is ma = 12.5.",
null,
"Find the lenght of all sides and the size of all internal angles of the triangle ABC, if we know: A = 501.9; α = 15°28' and β = 45°.",
null,
"A parallelogram ABCD has the area of 40 cm2, |AB| = 8.5 cm and |BC| = 5.65 cm. Find the length of its diagonals.",
null,
"Find the area of a regular hexagon, if we know the radius of its inscribed circle ρ = 4 cm.",
null,
"In the regular hexagon ABCDEF the diagonal AC has the length of 12 cm. Find the length of the side of the hexagon ABCDEF and determine its area.",
null,
"Find the perimeter of a circle if its area is 706.5 cm2.",
null,
"Find the area of a circle if its perimeter is 94.2 dm.",
null,
"A square on the picture has 8 cm long side. Find the area of the colored part of a circle.",
null,
"",
null,
"On the picture two and another two semicircles are identical. The radius of one semicircle is twice as large as the radius of the other semicircle. Find the area of the colored pattern if |AB| = 12 cm.",
null,
"",
null,
"",
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https://www.techgeekbuzz.com/tutorial/java/java-while-loop/ | [
"# Java While Loop\n\nUnlike for-loop , the while loop is used when the number of iterations is unknown. The while loop will execute a program several times until the condition is false.\n\nSyntax\n\n``````while(condition){\n//code to be executed\n}``````\n\nExample\n\n``````public class WhileExample {\npublic static void main(String[] args) {\nint i=1;\nwhile(i<=3){\nSystem.out.println(i);\ni++;\n}\n}\n}``````\n\nOutput\n\n``````1\n2\n3``````\n\n### Infinitive while Loop\n\nThe while loop will execute infinite time when we pass the true value as the condition. If you want to exit from the infinite loop, press ctrl+c .\n\nSyntax\n\n``````while(true){\n//code to be executed\n}``````\n\nExample\n\n``````public class WhileExample2 {\npublic static void main(String[] args) {\nwhile(true){\nSystem.out.println(\"infinitive while loop\");\n}\n}\n}``````\n\nOutput\n\n``````infinitive while loop\ninfinitive while loop\ninfinitive while loop\ninfinitive while loop\ninfinitive while loop\nctrl+c``````"
] | [
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https://math.stackexchange.com/questions/3134210/how-can-i-write-this-as-a-double-sum | [
"How can I write this as a double sum?\n\n$$2 \\sum_{1\\leq j < k \\leq N} P(F_j F_k)$$ as a double sum. Would it be $$\\sum_{j=1}^k \\sum_{k=n}^N P(F_j F_k)$$?\n\nAlso I am just confused about what a double sum would actually mean, I understand that you would sum over the values twice but can someone please explain more in detail? Thank you\n\n• Where does $n$ come from? Mar 3 '19 at 23:47\n• Does $P(F_j F_k)$ have any special meaning? Generally Readers can only assume it is a term that depends on $j,k$ in a symmetric fashion, i.e. because $F_jF_k=F_kF_j$. But already in the first summation you've restricted the terms so that $j\\lt k$, eliminating any cases $j=k$ and raising a suspicion about why the factor $2$ appears in front of that summation, but not in front of your attempt to rewrite it as a double sum. Mar 4 '19 at 0:45\n\nA double sum is a sum of the form $$\\sum_{j=a}^b \\sum_{k=c(j)}^{d(j)} f(j,k) = \\sum_{j=a}^b \\left(\\sum_{k=c(j)}^{d(j)} f(j,k)\\right)$$ I used the notation $$c(j),d(j)$$ because $$c$$ and $$d$$ may very well depend on $$j$$, but they don't have to.\n\nThere are more ways you can approach this.\n\n1. We want one variable, say $$j$$, to be independent, and let it independently take all the values it can possibly have that satisfy the condition $$1\\le j < k \\le N$$. We will then adjust $$k$$ so that this condition is actually met.\nTherefore, let $$j$$ vary across all values that it can possibly have. These are $$1\\le j \\le N-1$$, hence the $$\\sum_{j=1}^{N-1}$$ symbol. Then, let $$k$$ take all values such that the condition $$j < k \\le N$$. A different way to write this inequality is $$j+1 \\le k \\le N$$. Hence, your sum is equal to $$2\\sum_{j=1}^{N-1} \\sum_{k=j+1}^N P(F_jF_k)$$ To expand this sum, first write out all terms of the inner sum, and then add them up according to the outer sum, like $$2\\sum_{j=1}^{N-1} (P(F_jF_{j+1})+P(F_jF_{j+2})+...+P(F_jF_N)) =$$ $$= 2[(P(F_1F_2)+P(F_1F_3)+...+P(F_1F_N)) +$$ $$+(P(F_2F_3)+P(F_2F_4)+...+P(F_2F_N))+...+$$ $$+P(F_{N-1}F_N)]$$ Notice that each successive term of the outer sum has less and less terms of the inner sum (the last one has only one term).\n\n2. Let $$k$$ vary across all values that it can possibly have. These are $$2\\le k \\le N$$. Then, let $$j$$ take all values such that the condition $$1\\le j < k$$ is met. A different way to write this inequality is $$1 \\le j \\le k-1$$. Hence, your sum is equal to $$2\\sum_{k=2}^{N} \\sum_{j=1}^{k-1} P(F_jF_k)$$\n\nThe difference between these two approaches is which variable you assign \"independence\" to.\n\n• thank you! i have a really stupid question but how come certain double sums have a 2 next to it and other double sums do not? Mar 4 '19 at 4:31\n• also, why do you initially take j to be $1 \\leq j \\leq N-1$ Mar 4 '19 at 4:33\n• And just one more question (sorry!), what exactly does this double sum expand to? like if i had actual numbers/values, then how would this expand? Mar 4 '19 at 4:34\n• @user477465 The 2 is there because your starting sum is multiplied by 2. Maybe the 2 was a typo on your part? As for why I take $1\\le j \\le N-1$... If you have $1\\le j < k \\le N$ in the sum, this means that the sum contains all possible combinations of $j$ and $k$ that satisfy this. You can see that $j$ can take the smallest value 1, and $j$ can take the value $N-1$, but it can't take $N$ because then $k$ would have to be $>N$. Of course, $N=1$ is a special case, where no $j,k$ satisfy the condition of the sum, hence the sum is 0. Also, see my edit to the answer. Mar 4 '19 at 5:01\n\n$$\\sum_{j=1}^{N-1} \\sum_{k=j+1}^N P(F_j,F_k)$$\n\nAssume, that for some finite set $$I$$, you have a sum $$\\sum_{l \\in I} s_i$$ Now, the summands correspond to the elements in $$I$$. If, for some set $$J$$, we have a bijection\n\n$$I \\overset{\\sigma}{\\longrightarrow} \\bigcup_{j \\in J}I_j =: Y$$ where $$I_j \\cap I_i = \\emptyset$$ for $$i\\neq j$$, then\n\n$$\\sum_{i\\in I} s_i = \\sum_{y \\in Y} s_{\\sigma^{-1}(y)} = \\sum_{y \\in \\bigcup_{j \\in J}I_j} s_{\\sigma^{-1}(y)} = \\sum_{j \\in J}\\sum_{y \\in I_j} s_{\\sigma^{-1}(y)}$$ The first equation holds because $$\\sigma$$ is a permutation and addition is commutative. The second equation is trivial and the third equation uses that the union is disjoint, as mentioned above.\n\nSo, any bijection $$\\sigma$$ into a disjoint union as mentioned above will give you a representation as a double sum. You could of course proceed inductively, to get triple, quadruple sums et cetera.\nThe double sum can hence be understood as a specific grouping of summands/a partition of the index set.\n\nDefining $$I:=\\left\\{\\left(j,k\\right);\\,1\\leq j < k \\leq N\\right\\}$$, you can write $$I = \\bigcup_{1 \\leq j < N} \\underbrace{\\left\\{(j,k);\\, j < k \\leq N\\right\\}}_{:= I_j}$$ and the union is disjoint. Taking $$\\sigma$$ as the identity, and using the above, we get $$\\begin{eqnarray} & 2 \\sum_{(j,k) \\in I} P(F_j F_k) = 2 \\sum_{(j,k) \\in \\bigcup_{1 \\leq j < N} I_j} P(F_j F_k) = 2 \\sum_{1 \\leq j < N} \\sum_{(j,k) \\in I_j} P(F_j F_k) = \\\\ & 2 \\sum_{1 \\leq j < N}\\; \\sum_{(j,k) \\in \\left\\{(j,m);\\, j < m \\leq N \\right\\}} P(F_j F_k) = 2 \\sum_{1 \\leq j < N} \\sum_{j < k \\leq N} P(F_j F_k) \\\\ \\end{eqnarray}$$\nwhere in the last equation, we used that the sets $$\\left\\{(j,m);\\, j < m \\leq N\\right\\}$$ and $$\\left\\{m;\\, j < m \\leq N\\right\\}$$ are bijective/have the same number of elements. There is some subtle formal detail involved in this last equation:\nWe are basically applying what we used before - $$\\sum_{i\\in I} s_i = \\sum_{y \\in Y} s_{\\sigma_j^{-1}(y)}$$ - to the inner sum, where for each $$j$$, $$\\sigma_j: \\left\\{(j,m);\\, j < m \\leq N\\right\\} \\longrightarrow \\left\\{m;\\, j < m \\leq N\\right\\},\\; (j,m) \\mapsto m$$ is the projection to the second component and $$\\sigma_j^{-1}$$ is the function that maps $$m \\mapsto (j,m)$$.\nExplicitly, $$\\begin{array} & 2 \\sum_{1 \\leq j < N}\\; \\sum_{(j,k) \\in \\left\\{(j,m);\\, j < m \\leq N \\right\\}} P(F_j F_k) = 2 \\sum_{1 \\leq j < N}\\; \\sum_{(j,k) \\in \\left\\{(j,m);\\, j < m \\leq N \\right\\}} P(F_j F_{\\sigma_j(j,k)}) = \\\\ 2 \\sum_{1 \\leq j < N} \\sum_{j < k \\leq N} P(F_j F_{\\sigma_j(\\sigma_j^{-1}(k))}) = 2 \\sum_{1 \\leq j < N} \\sum_{j < k \\leq N} P(F_j F_k) \\\\ \\end{array}$$"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6685265,"math_prob":0.99999,"size":2634,"snap":"2022-05-2022-21","text_gpt3_token_len":1007,"char_repetition_ratio":0.16806084,"word_repetition_ratio":0.21806167,"special_character_ratio":0.40318906,"punctuation_ratio":0.12456747,"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\":\"2022-01-25T18:49:16Z\",\"WARC-Record-ID\":\"<urn:uuid:e5bb1b89-0d05-4551-ad00-f5f21dfa839c>\",\"Content-Length\":\"163413\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7d04d080-fe90-485f-91ac-caef4d587cb4>\",\"WARC-Concurrent-To\":\"<urn:uuid:8042dcbf-972f-467f-b072-c9a28b7cdc97>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/3134210/how-can-i-write-this-as-a-double-sum\",\"WARC-Payload-Digest\":\"sha1:MZ6BV6YUOZGKH3VD3KV4ITNSVHHLLJ6Q\",\"WARC-Block-Digest\":\"sha1:3HJHKM5RXTNNPSLIXDEA5DWJSOAEKSXQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304859.70_warc_CC-MAIN-20220125160159-20220125190159-00249.warc.gz\"}"} |
https://kr.mathworks.com/matlabcentral/cody/problems/2929-volume-of-a-box/solutions/1826185 | [
"Cody\n\n# Problem 2929. Volume of a box\n\nSolution 1826185\n\nSubmitted on 26 May 2019 by Arif Ullah\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\na = 1; b = 1; c = 1; y_correct = 1; assert(isequal(volume(a,b,c),y_correct))\n\nans = 1\n\n2 Pass\na = 2; b = 2; c = 2; y_correct = 8; assert(isequal(volume(a,b,c),y_correct))\n\nans = 8\n\n3 Pass\na = 3; b = 3; c = 3; y_correct = 27; assert(isequal(volume(a,b,c),y_correct))\n\nans = 27"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.62089133,"math_prob":0.9991518,"size":618,"snap":"2019-51-2020-05","text_gpt3_token_len":212,"char_repetition_ratio":0.14006515,"word_repetition_ratio":0.0,"special_character_ratio":0.38349515,"punctuation_ratio":0.17021276,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9981446,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-11T16:42:06Z\",\"WARC-Record-ID\":\"<urn:uuid:6ed8bb60-99ec-4634-9bea-3f564f4ee639>\",\"Content-Length\":\"71941\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:48b85bdf-37eb-4d8f-97aa-a8658a2a79e4>\",\"WARC-Concurrent-To\":\"<urn:uuid:ad88a428-9909-4f57-8d68-68a3c5371ba2>\",\"WARC-IP-Address\":\"23.195.248.135\",\"WARC-Target-URI\":\"https://kr.mathworks.com/matlabcentral/cody/problems/2929-volume-of-a-box/solutions/1826185\",\"WARC-Payload-Digest\":\"sha1:SMHGDX4PLWGN7EVXA5MTBGDNVHSRWDUI\",\"WARC-Block-Digest\":\"sha1:X7ZQ2QUAR5GHYRFOZ5OBKXLCBS6LEGBL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540531974.7_warc_CC-MAIN-20191211160056-20191211184056-00201.warc.gz\"}"} |
https://www.bartleby.com/essay/MAC2601-May-Jun-2013-MEMO-10-June-P38CZCVSJFRS5 | [
"# MAC2601 May Jun 2013 MEMO 10 June 2013\n\n2151 Words9 Pages\nMAC2601\nMay/June 2013: Suggested solution\nQUESTION 1\n1.1 – C\nFIFO Method of valuation\nDate\nReceipts\nBalance\nDecember Quantity Price\n\nAmount Quantity Price\n\n1\n3\n\n2562\n\n350\n\n7.32\n\nIssues\n\n7\n\n300\n100\n\nAmount Quantity Price\nR\n300\n6.50\n300\n6.50\n350\n7.32\n1950\n732\n250\n7.32\n\n6.50\n7.32\n\nAmount\nR\n1950\n1950\n2562\n1830\n\nCost of purchases of 350 units\n350 units at R6.90 = R2 415\n+Freight costs: R294*50%=R147\nTotal: 2562\nR2 562/350 units = R7,32\nTherefore option C is correct.\n\n1.2 - B\n1.3 - D\n1.4 - A\nR\nFixed cost in opening stock (10 000 x R7.50)\nFixed cost in closing stock ( 8 000 x R9.00)\nDifference\n\n75 000\n72 000\n3 000\n\n1.5 - A\n1.6 - C\n1.7 - B\nR\nDirect material\nDirect labour\nLess: proceeds from by-product\n\n238 500\n143 100\n95\nR\n\nSales (20 000 x R35)\nLess: Variable costs\nContribution\n\n700 000\n(385 000)¹\n315 000\n\nLess: Fixed costs\nNet profit\n\n(55 000)\n260 000\n\n¹ R185 000 (20 000 x R9,25) + R150 000(20 000 x R7,50) + R50 000 (20 000 x R2,50*)\n* R40 000/16 000 = R2.50\n\nd.\nSales (18 000(20 000 x .90) x R40 (R35+R5)\nLess: Variable costs\nContribution\nLess: Fixed costs (R55 000 + R5 000)\nNet profit\n\nR\n720 000\n(346 500)²\n373 500\n(60 000)\n313 500\n\n²R166 500 (18 000 x R9,25) + R135 000(18 000 x R7,50) + R45 000 (18 000 x R2,50*)\nOR\nContribution: R40 – R19,25 = R20,75\nR20,75 x 18 000 = R373 500 – R60 000 = R313 500\n\nQUESTION 6\n\na. Budgeted cost per smart id card\n9 places; 5 marks\nR\nDirect Materials (2.5m^ *R3.50^)\nDirect labour (2^*R8^)\nVariable selling costs (not part of product costs) ^for not including VSC\nFixed cost per unit (2 ^ *13,75√)\nTotal budgeted smart ID card cost per unit\n\n8,75\n16,00\n3,00\n27,50\n55.25\n\nIf they have not shown the zero for the VSC, you can still award the mark for the “^for not including VSC” if it is clear that the VSC has not been included anywhere and they have totalled/attempted to total their costs (which does not include the VSC). This also applies to the alternative methods.\nFixed cost per labour hour\nCalculation of the fixed overhead recovery rate"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7338595,"math_prob":0.9749754,"size":1298,"snap":"2021-31-2021-39","text_gpt3_token_len":480,"char_repetition_ratio":0.13833076,"word_repetition_ratio":0.0,"special_character_ratio":0.47380584,"punctuation_ratio":0.10508475,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9688426,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-24T08:30:41Z\",\"WARC-Record-ID\":\"<urn:uuid:98518b87-c686-49ca-9250-267b4b799c5e>\",\"Content-Length\":\"32085\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c847782e-46f1-4d1e-af4b-f67fabc3587e>\",\"WARC-Concurrent-To\":\"<urn:uuid:9295c65f-0ff2-4ae2-81d7-d0e8b95ab71f>\",\"WARC-IP-Address\":\"99.86.230.117\",\"WARC-Target-URI\":\"https://www.bartleby.com/essay/MAC2601-May-Jun-2013-MEMO-10-June-P38CZCVSJFRS5\",\"WARC-Payload-Digest\":\"sha1:5353WXCHPL6JFFAIOEUYSYRBZIM6NZNW\",\"WARC-Block-Digest\":\"sha1:5O56YWTAFZIOGN2JE2DITEAXYXY4GV2B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046150134.86_warc_CC-MAIN-20210724063259-20210724093259-00000.warc.gz\"}"} |
https://ch.mathworks.com/help/simulink/sfg/sssetdworkusedasdstate.html | [
"# ssSetDWorkUsedAsDState\n\nSpecify that a Dwork vector is used as a discrete state vector\n\n## Syntax\n\n```int_T ssSetDWorkUsedAsDState(SimStruct *S, int_T vector,\nint_T usage)\n```\n\n## Arguments\n\n`S`\n\nSimStruct representing an S-Function block.\n\n`vector`\n\nIndex of a Dwork vector, where the index is one of `0`, `1`, `2`, `...` `ssGetNumDWork(S)-1`.\n\n`usage`\n\nHow this vector is used. A value of `1` indicates that the work vector is to be used to store the block's discrete states (`SS_DWORK_USED_AS_DSTATE`), a value of `0` indicates that the work vector is to be used as a work vector (`SS_DWORK_USED_AS_DWORK`).\n\n## Returns\n\n`0` if usage is `SS_DWORK_USED_AS_DWORK` (`0`), otherwise returns `1`.\n\n## Description\n\nUse in `mdlInitializeSizes` or `mdlSetWorkWidths` to specify if the DWork vector `vector` is used to store the block's discrete states, `SS_DWORK_USED_AS_DSTATE` (`1`), or not, `SS_DWORK_USED_AS_DWORK` (`0`), the default.\n\n### Note\n\nSpecify the usage as `SS_DWORK_USED_AS_DSTATE` if the following conditions are true. You want to use the vector to store discrete states and you want the Simulink® engine to log the discrete states to the workspace at the end of a simulation, if the user has selected the Save to Workspace options on the Data Import/Export pane of the Configuration Parameters dialog box.\n\nC, C++\n\n## Examples\n\nFor more information on using DWork vectors, see How to Use DWork Vectors.\n\n`ssGetDWorkUsedAsDState`"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.72163284,"math_prob":0.7256127,"size":1256,"snap":"2020-34-2020-40","text_gpt3_token_len":351,"char_repetition_ratio":0.1629393,"word_repetition_ratio":0.068292685,"special_character_ratio":0.2356688,"punctuation_ratio":0.116883114,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97523236,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-11T20:18:11Z\",\"WARC-Record-ID\":\"<urn:uuid:ba83a07e-0311-412c-9234-54e3af610764>\",\"Content-Length\":\"67890\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4441dd5b-383a-4f13-866d-210e3ad44f6d>\",\"WARC-Concurrent-To\":\"<urn:uuid:5f225a76-ec17-4d9c-8376-f17e9702ba47>\",\"WARC-IP-Address\":\"23.223.252.57\",\"WARC-Target-URI\":\"https://ch.mathworks.com/help/simulink/sfg/sssetdworkusedasdstate.html\",\"WARC-Payload-Digest\":\"sha1:B572SNICESIJ6WY4JVCZSX2BFGLDMBZZ\",\"WARC-Block-Digest\":\"sha1:K3QUEBO7HZ5I6YTVN47HZQIJDDYYQ2D5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738819.78_warc_CC-MAIN-20200811180239-20200811210239-00331.warc.gz\"}"} |
https://dataplatform.cloud.ibm.com/docs/content/wsd/nodes/extension_pyspark_api.html | [
"# Scripting with Python for Spark\n\nSPSS Modeler can run Python scripts using the Apache Spark framework to process data. This documentation provides the Python API description for the interfaces provided.\n\nThe SPSS Modeler installation includes a Spark distribution.\n\n## Accessing data\n\nData is transferred between a Python/Spark script and the execution context in the form of a Spark SQL DataFrame. A script that consumes data (that is, any node except an Import node) must retrieve the data frame from the context:\n``inputData = asContext.getSparkInputData()``\nA script that produces data (that is, any node except a terminal node) must return a data frame to the context:\n``asContext.setSparkOutputData(outputData)``\nYou can use the SQL context to create an output data frame from an RDD where required:\n``outputData = sqlContext.createDataFrame(rdd)``\n\n## Defining the data model\n\nA node that produces data must also define a data model that describes the fields visible downstream of the node. In Spark SQL terminology, the data model is the schema.\n\nA Python/Spark script defines its output data model in the form of a `pyspsark.sql.types.StructType` object. A `StructType` describes a row in the output data frame and is constructed from a list of `StructField` objects. Each `StructField` describes a single field in the output data model.\n\nYou can obtain the data model for the input data using the `:schema` attribute of the input data frame:\n``inputSchema = inputData.schema``\nFields that are passed through unchanged can be copied from the input data model to the output data model. Fields that are new or modified in the output data model can be created using the `StructField` constructor:\n``field = StructField(name, dataType, nullable=True, metadata=None)``\n\nYou must provide at least the field name and its data type. Optionally, you can specify metadata to provide a measure, role, and description for the field (see Data metadata).\n\n## DataModelOnly mode\n\nSPSS Modeler needs to know the output data model for a node, before the node runs, to enable downstream editing. To obtain the output data model for a Python/Spark node, SPSS Modeler runs the script in a special \"data model only\" mode where there is no data available. The script can identify this mode using the `isComputeDataModelOnly` method on the Analytic Server context object.\n\nThe script for a transformation node can follow this general pattern:\n``````if asContext.isComputeDataModelOnly():\ninputSchema = asContext.getSparkInputSchema()\noutputSchema = ... # construct the output data model\nasContext.setSparkOutputSchema(outputSchema)\nelse:\ninputData = asContext.getSparkInputData()\noutputData = ... # construct the output data frame\nasContext.setSparkOutputData(outputData)``````\n\n## Building a model\n\nA node that builds a model must return to the execution context some content that describes the model sufficiently that the node which applies the model can recreate it exactly at a later time.\n\nModel content is defined in terms of key/value pairs where the meaning of the keys and the values is known only to the build and score nodes and is not interpreted by SPSS Modeler in any way. Optionally the node may assign a MIME type to a value with the intent that SPSS Modeler might display those values which have known types to the user in the model nugget.\n\nA value in this context may be PMML, HTML, an image, etc. To add a value to the model content (in the build script):\n``asContext.setModelContentFromString(key, value, mimeType=None)``\nTo retrieve a value from the model content (in the score script):\n``value = asContext.getModelContentToString(key)``\nAs a shortcut, where a model or part of a model is stored to a file or folder in the file system, you can bundle all the content stored to that location in one call (in the build script):\n``asContext.setModelContentFromPath(key, path)``\n\nNote that in this case there is no option to specify a MIME type because the bundle may contain various content types.\n\nIf you need a temporary location to store the content while building the model you can obtain an appropriate location from the context:\n``path = asContext.createTemporaryFolder()``\nTo retrieve existing content to a temporary location in the file system (in the score script):\n``path = asContext.getModelContentToPath(key)``\n\n## Error handling\n\nTo raise errors, throw an exception from the script and display it to the SPSS Modeler user. Some exceptions are predefined in the module `spss.pyspark.exceptions`. For example:\n``````from spss.pyspark.exceptions import ASContextException\nif ... some error condition ...:\nraise ASContextException(\"message to display to user\")``````"
] | [
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https://www.vindit.dk/where-put-downlight-spots/ | [
"# Where to put the downlight spots\n\nI am doing DIY renovations on my house. I am currently rebuilding the overhang and have reached the point where I need to figure out where I want the spotlights. I have a few requirements:\n\n• The spots must be spaced out evenly.\n• The spots cannot be placed at the position of a beam as I don't want to carve out the wood that carries the roof.\n• I want around 2 meters between the spots.\n• I am open on a difference between distance between the first spot and the edge of the roof and the last spot and its corresponding edge.\n\nHow do we solve this? I know many people who would sit and puzzle together a solution, but that's not where I have the most patience.\n\nSo I went outside and measured all distances between the beams and wrote a program.\n\n``````const beamWidth = 11.5;\n\nconst minDistance = 120.0;\nconst maxDistance = 350.0;\n\nconst lampDiameter = 7.4;\n\nconst beamDistances = [ 40.0, 73.0, 87.0, 80.0, 80.0, 82.0, 87.0, 87.0, 82.0, 82.0, 64.0, 61.0, 69.0, 82.0, 85.0, 38.0 ];\n\nconst totalDistance = beamDistances.reduce(\n(prev, curr) => prev + curr,\nbeamWidth * (beamDistances.length - 1)\n);\n\nconst isBetweenBeams = offset => {\nlet beamStart = 0;\nfor (let beamId = 0; beamId < beamDistances.length; beamId++) {\nconst beamEnd = beamStart + beamDistances[beamId];\nif (beamStart <= offset && offset <= beamEnd - lampDiameter) {\nreturn true;\n}\nbeamStart = beamEnd + beamWidth + beamDistances[beamId];\n}\nreturn false;\n};\n\nconst solutions = [];\n\nfor (let startOffset = 0; startOffset < minDistance; startOffset += 0.5) {\nfor (let distance = minDistance; distance < maxDistance; distance += 0.5) {\n// Iterate pos from startOffset to totalDistance with distance increments\nsucc = true;\nconst currTry = [];\nfor (\nlet pos = startOffset;\npos <= totalDistance - lampDiameter;\npos += distance\n) {\n// We add half a diameter so we get the center of the hole and not the edge\ncurrTry.push(pos + lampDiameter / 2.0);\n// Check if it is between joists\nif (!isBetweenBeams(pos)) {\nsucc = false;\nbreak;\n}\n}\nif (succ) {\nconst numberOfLamps = currTry.length;\nsolutions.push({\ndistance,\nstartOffset,\nendOffset:\ntotalDistance -\n(startOffset + (numberOfLamps - 1) * distance) -\nlampDiameter,\nnumberOfLamps,\nlamps: currTry\n});\n}\n}\n}\n\n// We sort by the solutions where start and end offsets are most similar\nsolutions.sort(\n(a, b) =>\nMath.abs(a.startOffset - a.endOffset) -\nMath.abs(b.startOffset - b.endOffset)\n);\nfor (let i = 0; i < 25; i++) {\nconsole.log(JSON.stringify(solutions[i]));\n}\n``````\n\nIt creates all viable solutions with even spacing and then sorts the results by the most even first lamp to edge distance and last lamp to edge distance.\n\nThese are the first three solutions I got:\n\n``````{\"distance\":182.5,\"startOffset\":119.5,\"endOffset\":129.6,\"numberOfLamps\":7,\"lamps\":[123.2,305.7,488.2,670.7,853.2,1035.7,1218.2]}\n{\"distance\":183,\"startOffset\":117.5,\"endOffset\":128.6,\"numberOfLamps\":7,\"lamps\":[121.2,304.2,487.2,670.2,853.2,1036.2,1219.2]}\n{\"distance\":182.5,\"startOffset\":119,\"endOffset\":130.1,\"numberOfLamps\":7,\"lamps\":[122.7,305.2,487.7,670.2,852.7,1035.2,1217.7]}``````\n\nLooking at the first solution, it looks like I will have 7 lamps, one every 182.5cm and the edge distances are 119.5 and 129.6. I am gonna use that solution.\n\nThat's how programmers build houses.\n\nI've been meaning to post an update about this. Mere days before starting to drill holes for these spots, a friend told me \"People typically install them between doors and windows so the light is reflected off the brick wall\" which instantly made me throw my plan in the garbage. Of course that's how you do it. Oh well."
] | [
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https://physics.stackexchange.com/questions/160959/what-is-the-significance-of-plancks-constant-in-de-broglie-bohm-theory-or-pilot | [
"# What is the significance of Planck's constant in De Broglie–Bohm theory or Pilot-wave theory?\n\nIn standard QM, Planck's constant seems to be a constant that describes the smallest quanta of energy in some way. Does De Broglie–Bohm theory have an alternate interpretation of that constant and what it means about space-time or particles (or both)?\n\n• it probably doesn't need to be said, but I'll do it anyway so no one else has to do it: as far as we know, there is no such thing as a smallest quantum of energy – Christoph Jan 23 '15 at 12:16\n\n## 1 Answer\n\nBohm's interpretation of the QM says that a particle moves along a trajectory, as in the classical mechanics, however, the potential in which this particle moves is supplemented by a \"quantum potential\". In all the equation of movement relevant for this question is\n\n(1) $\\frac {∂S}{∂t} + \\frac {(\\nabla S)^2}{2m} + \\left[V(x) - \\frac {\\hbar ^2}{8m} \\left(2\\frac {\\Delta P}{P} - \\frac {(\\nabla P)^2}{P^2} \\right) \\right] = 0$,\n\nwhere $S$ and $P$ are defined by expressing the wave-function as\n\n(2) $\\psi = P \\ e^{iS/ \\hbar}$.\n\nHere is what Bohm says about the constant $\\hbar$ :\n\n\" ... in the classical limit $\\hbar \\to 0$ the above equations are subject to a very simple interpretation. The function $S(x)$ is a solution of the Hamilton-Jacobi equation.\"\n\nHowever, as long as the limit $\\hbar \\to 0$ doesn't hold, the trajectory of a particle is influenced, besides $V(x)$, by a \"quantum potential\" generated by the wave-function,\n\n$V^{quantum} = -\\frac {\\hbar ^2}{8m} \\left(2\\frac {\\Delta P}{P} - \\frac {(\\nabla P)^2}{P^2} \\right)$.\n\nThus, the magnitude of $\\hbar$ relative to the considered problem separates between the classical and quantum behavior.\n\n• Interesting. I'd like to understand your bolded point more concretely. The units of h are energy multiplied over a distance. Can you add some interpretation of those units in this context? – B T Jan 23 '15 at 19:10\n• @BT , no $\\hbar$ has dimensions of energy multiplied by *time*. $\\hbar$ is action. The relevance of this quantity is inherited from classical mechanics - from the principle of least action, which says that a particle follows in space the trajectory on which the action is minimal. And what is that action? the Lagrangian (that has dimensions of energy) multiplied by time. – Sofia Jan 23 '15 at 19:58\n• Ah right, I misspoke. So the ℏ has units of action. Didn't realize that was a type of unit. Sounds like you're saying that by using 0 as the value for h, we basically get classical mechanics, and that h is the value that is required to describe the \"quantum potential\". Could you elaborate on what the quantum potential is, and what ℏ means to it? Perhaps ℏ relates to how much mass is necessary to generate a given frequency and/or amplitude of particles' waves (purely speculating here)? – B T Jan 23 '15 at 20:58\n• @BT : you ask me too many things at once. Please, one step at one time. What $\\hbar$ means to the quantum potential you saw in my answer: let $\\hbar \\to 0$, and we have classical mechanics. But, worse that that : the uncertainty principle says $\\Delta x \\Delta p_x \\ge \\hbar/2$. But if $\\hbar \\to 0$, there is no interdiction to have $\\Delta x \\Delta p_x = 0$. So, no uncertainty principle - classical mechanics. Well, about quantum potential can you post a question? In the space of the comments it's difficult to answer. The quantum potential tells the particle where it can be and where not. – Sofia Jan 23 '15 at 21:24\n• @BT : do you remember the 2slit experiment? Well, the quantum potential doesn't allow the Bohmian trajectories to pass through the forbidden fringes. – Sofia Jan 23 '15 at 21:26"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9151243,"math_prob":0.995901,"size":3792,"snap":"2019-35-2019-39","text_gpt3_token_len":1019,"char_repetition_ratio":0.1156283,"word_repetition_ratio":0.041795667,"special_character_ratio":0.27742615,"punctuation_ratio":0.10396717,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99880886,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-22T13:21:39Z\",\"WARC-Record-ID\":\"<urn:uuid:5c4e7894-13e8-4349-bf25-8a2c75341f20>\",\"Content-Length\":\"143790\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fdd9e599-52fc-4bbe-a1db-b387f45d2838>\",\"WARC-Concurrent-To\":\"<urn:uuid:cefb4c21-6cec-4a3b-a50f-19f12856135e>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://physics.stackexchange.com/questions/160959/what-is-the-significance-of-plancks-constant-in-de-broglie-bohm-theory-or-pilot\",\"WARC-Payload-Digest\":\"sha1:ZIJYZMXHSRDRKFPUZPQASFE26MEOOV2O\",\"WARC-Block-Digest\":\"sha1:YSDNW3YSQJMXJKIBFT3DAY6EC4RGTG27\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027317130.77_warc_CC-MAIN-20190822130553-20190822152553-00149.warc.gz\"}"} |
https://www.betriebswirtschaft-lernen.net/en/explanation/forward-calculation/ | [
"# Forward calculation\n\nA forward calculation is a method used in retail to calculate a sales price. With the help of this calculation, the merchant can achieve a certain profit.\n\n## Definition / explanation\n\nThe forward calculation is a version of the trade calculation that is used to determine a sales price. When calculating, the retailer starts from the purchase price and suggests cost factors such as transport costs, planned discounts, Overhead, Discount for customers, the expected profit or the value added tax on. If the retailer does not have individual items, these items can simply be left out of the calculation.\n\nIf the entire costs have been added to the purchase price, then the retailer receives the gross list sales price, which he can ultimately demand from the customer.\n\n## Summary\n\n• Forward calculation is a method used in retail to calculate the sales price of a product\n• Forward costing is a form of trade costing\n• The result of the forward calculation is the gross list sales price that can be requested by the customer\nWas the explanation to \"Forward calculation\"Helpful? Rate now:\n\n### Synonym for / Other word for\n\n• forward calculation"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9312066,"math_prob":0.96817213,"size":1032,"snap":"2022-40-2023-06","text_gpt3_token_len":197,"char_repetition_ratio":0.17607003,"word_repetition_ratio":0.058139537,"special_character_ratio":0.1879845,"punctuation_ratio":0.080213904,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97180307,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-06T09:58:12Z\",\"WARC-Record-ID\":\"<urn:uuid:edb51744-0346-4508-a07f-fa61b2913263>\",\"Content-Length\":\"56353\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:daddf5e0-e514-4e27-822c-0366712c3b35>\",\"WARC-Concurrent-To\":\"<urn:uuid:99474330-8c95-4e9b-8bed-50f3f9bbb230>\",\"WARC-IP-Address\":\"172.67.216.21\",\"WARC-Target-URI\":\"https://www.betriebswirtschaft-lernen.net/en/explanation/forward-calculation/\",\"WARC-Payload-Digest\":\"sha1:M6GNWZTQYC42WMBHYUXC4P4XWBBHZFCW\",\"WARC-Block-Digest\":\"sha1:Y7RJWPDRVUCTD4566RJ6KX52Z5Y4PTWL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500334.35_warc_CC-MAIN-20230206082428-20230206112428-00437.warc.gz\"}"} |
https://scholars.duke.edu/display/pub756982 | [
"Independent set of intersection graphs of convex objects in 2D\n\nPublished\n\nJournal Article\n\nThe intersection graph of a set of geometric objects is defined as a graph G=(S,E) in which there is an edge between two nodes s i, s j∈S if s i∩s j≠ ∅. The problem of computing a maximum independent set in the intersection graph of a set of objects is known to be NP-complete for most cases in two and higher dimensions. We present approximation algorithms for computing a maximum independent set of intersection graphs of convex objects in ℝ 2. Specifically, given (i) a set of n line segments in the plane with maximum independent set of size α, we present algorithms that find an independent set of size at least (α/(2log(2n/α))) 1/2 in time O(n 3) and (α/(2log(2n/α))) 1/4 in time O(n 4/3log cn), (ii) a set of n convex objects with maximum independent set of size α, we present an algorithm that finds an independent set of size at least (α/(2log(2n/α))) 1/3 in time O(n 3+τ (S)), assuming that S can be preprocessed in time τ(S) to answer certain primitive operations on these convex sets, and (iii) a set of n rectangles with maximum independent set of size βn, for β≤1, we present an algorithm that computes an independent set of size Ω( β 2n) . All our algorithms use the notion of partial orders that exploit the geometric structure of the convex objects. © 2006 Elsevier B.V. All rights reserved.\n\nCited Authors\n\n• Agarwal, PK; Mustafa, NH\n\n• May 1, 2006\n\n• 34 / 2\n\n• 83 - 95\n\n• 0925-7721\n\nDigital Object Identifier (DOI)\n\n• 10.1016/j.comgeo.2005.12.001\n\nCitation Source\n\n• Scopus",
null,
""
] | [
null,
"https://scholars.duke.edu/themes/duke/images/duke-footer-logo.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.87228835,"math_prob":0.9702938,"size":1698,"snap":"2019-43-2019-47","text_gpt3_token_len":461,"char_repetition_ratio":0.15230225,"word_repetition_ratio":0.12542373,"special_character_ratio":0.2679623,"punctuation_ratio":0.07848837,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9937423,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-17T05:37:59Z\",\"WARC-Record-ID\":\"<urn:uuid:32c1b226-a6de-4ab2-8d72-39244354a3e9>\",\"Content-Length\":\"11787\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:842b6575-667d-4096-afd1-1b2a0677706e>\",\"WARC-Concurrent-To\":\"<urn:uuid:9e7f55f8-30bd-42a8-becf-b3669117ab03>\",\"WARC-IP-Address\":\"152.3.72.205\",\"WARC-Target-URI\":\"https://scholars.duke.edu/display/pub756982\",\"WARC-Payload-Digest\":\"sha1:OGPBFQC6GOM7HG6WQ3AGOPNOWG2BVUZG\",\"WARC-Block-Digest\":\"sha1:LM2BFFMXOX2DAQLDJPVYISRD64SENIWY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986672723.50_warc_CC-MAIN-20191017045957-20191017073457-00032.warc.gz\"}"} |
https://trustconverter.com/en/weight-conversion/carats/carats-to-ounces/weight-conversion-table.html | [
"# Carat to Ounce Conversion Table\n\n### Quick Find Conversion Table\n\nto\n\n0.1 - 0.98\ncarats to ounces\n0.1= 0.00070547923961373\n0.12= 0.00084657508753648\n0.14= 0.00098767093545923\n0.16= 0.001128766783382\n0.18= 0.0012698626313047\n0.20= 0.0014109584792275\n0.22= 0.0015520543271502\n0.24= 0.001693150175073\n0.26= 0.0018342460229957\n0.28= 0.0019753418709185\n0.30= 0.0021164377188412\n0.32= 0.002257533566764\n0.34= 0.0023986294146867\n0.36= 0.0025397252626094\n0.38= 0.0026808211105322\n0.40= 0.0028219169584549\n0.42= 0.0029630128063777\n0.44= 0.0031041086543004\n0.46= 0.0032452045022232\n0.48= 0.0033863003501459\n0.50= 0.0035273961980687\n0.52= 0.0036684920459914\n0.54= 0.0038095878939142\n0.56= 0.0039506837418369\n0.58= 0.0040917795897597\n0.60= 0.0042328754376824\n0.62= 0.0043739712856052\n0.64= 0.0045150671335279\n0.66= 0.0046561629814506\n0.68= 0.0047972588293734\n0.70= 0.0049383546772961\n0.72= 0.0050794505252189\n0.74= 0.0052205463731416\n0.76= 0.0053616422210644\n0.78= 0.0055027380689871\n0.80= 0.0056438339169099\n0.82= 0.0057849297648326\n0.84= 0.0059260256127554\n0.86= 0.0060671214606781\n0.88= 0.0062082173086009\n0.90= 0.0063493131565236\n0.92= 0.0064904090044464\n0.94= 0.0066315048523691\n0.96= 0.0067726007002919\n0.98= 0.0069136965482146\n1.00 - 1.88\ncarats to ounces\n1.00= 0.0070547923961373\n1.02= 0.0071958882440601\n1.04= 0.0073369840919828\n1.06= 0.0074780799399056\n1.08= 0.0076191757878283\n1.10= 0.0077602716357511\n1.12= 0.0079013674836738\n1.14= 0.0080424633315966\n1.16= 0.0081835591795193\n1.18= 0.0083246550274421\n1.20= 0.0084657508753648\n1.22= 0.0086068467232876\n1.24= 0.0087479425712103\n1.26= 0.0088890384191331\n1.28= 0.0090301342670558\n1.30= 0.0091712301149785\n1.32= 0.0093123259629013\n1.34= 0.009453421810824\n1.36= 0.0095945176587468\n1.38= 0.0097356135066695\n1.40= 0.0098767093545923\n1.42= 0.010017805202515\n1.44= 0.010158901050438\n1.46= 0.010299996898361\n1.48= 0.010441092746283\n1.50= 0.010582188594206\n1.52= 0.010723284442129\n1.54= 0.010864380290052\n1.56= 0.011005476137974\n1.58= 0.011146571985897\n1.60= 0.01128766783382\n1.62= 0.011428763681742\n1.64= 0.011569859529665\n1.66= 0.011710955377588\n1.68= 0.011852051225511\n1.70= 0.011993147073433\n1.72= 0.012134242921356\n1.74= 0.012275338769279\n1.76= 0.012416434617202\n1.78= 0.012557530465124\n1.80= 0.012698626313047\n1.82= 0.01283972216097\n1.84= 0.012980818008893\n1.86= 0.013121913856815\n1.88= 0.013263009704738\n1.90 - 2.78\ncarats to ounces\n1.90= 0.013404105552661\n1.92= 0.013545201400584\n1.94= 0.013686297248506\n1.96= 0.013827393096429\n1.98= 0.013968488944352\n2.00= 0.014109584792275\n2.02= 0.014250680640197\n2.04= 0.01439177648812\n2.06= 0.014532872336043\n2.08= 0.014673968183966\n2.10= 0.014815064031888\n2.12= 0.014956159879811\n2.14= 0.015097255727734\n2.16= 0.015238351575657\n2.18= 0.015379447423579\n2.20= 0.015520543271502\n2.22= 0.015661639119425\n2.24= 0.015802734967348\n2.26= 0.01594383081527\n2.28= 0.016084926663193\n2.30= 0.016226022511116\n2.32= 0.016367118359039\n2.34= 0.016508214206961\n2.36= 0.016649310054884\n2.38= 0.016790405902807\n2.40= 0.01693150175073\n2.42= 0.017072597598652\n2.44= 0.017213693446575\n2.46= 0.017354789294498\n2.48= 0.017495885142421\n2.50= 0.017636980990343\n2.52= 0.017778076838266\n2.54= 0.017919172686189\n2.56= 0.018060268534112\n2.58= 0.018201364382034\n2.60= 0.018342460229957\n2.62= 0.01848355607788\n2.64= 0.018624651925803\n2.66= 0.018765747773725\n2.68= 0.018906843621648\n2.70= 0.019047939469571\n2.72= 0.019189035317494\n2.74= 0.019330131165416\n2.76= 0.019471227013339\n2.78= 0.019612322861262\n\n### Legend\n\nSymbolDefinition\nexactly equal\napproximately equal to\n=equal to\ndigitsindicates that digits repeat infinitely (e.g. 8.294 369 corresponds to 8.294 369 369 369 369 …)\n\n### carats\n\nThe carat (ct) is a unit of mass equal to 200 mg (0.2 g; 0.007055 oz) and is used for measuring gemstones and pearls.\n\n### ounces\n\nThe ounce (abbreviated oz; apothecary symbol: ) is a unit of mass used in most British derived customary systems of measurement.\n\n### Conversion in other languages\n\nYou can find the conversion in other languages in the following:"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.77805763,"math_prob":0.98349977,"size":495,"snap":"2020-24-2020-29","text_gpt3_token_len":140,"char_repetition_ratio":0.107942976,"word_repetition_ratio":0.024691358,"special_character_ratio":0.3070707,"punctuation_ratio":0.12371134,"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\":\"2020-07-04T18:28:08Z\",\"WARC-Record-ID\":\"<urn:uuid:38183d44-b741-46c8-b814-d46b84918ed6>\",\"Content-Length\":\"40210\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ba1a1983-5599-4366-8c8b-805b73e8eba4>\",\"WARC-Concurrent-To\":\"<urn:uuid:e243fb83-938b-4f34-899c-58e16576444f>\",\"WARC-IP-Address\":\"45.33.9.101\",\"WARC-Target-URI\":\"https://trustconverter.com/en/weight-conversion/carats/carats-to-ounces/weight-conversion-table.html\",\"WARC-Payload-Digest\":\"sha1:WAJHONTHDV6MEB5LPIL5PLCSGDDASH4Z\",\"WARC-Block-Digest\":\"sha1:ZHYV3IMJYTYBXXHGKLQ5SKQEXZPXXUUH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655886516.43_warc_CC-MAIN-20200704170556-20200704200556-00565.warc.gz\"}"} |
https://ti-qa-archive.github.io/question/25641/polygon-shaped-buttons.html | [
"# Polygon shaped buttons\n\nIs there a way to create polygon shapes for buttons? Should I just use an imagemap in a webview? I want to create a clickable USA image map.\n\n• button\n• imagemap\n• poly\n• polygon\n• shapes\n\nHere's what I finally came up.\n\nI created an imageview with a photo of the USA Map. I then created several dictionaries of points, defining the regions I wanted to capture clicks from on the image. I then added an event listener that uses the following function to detect if the click event happened inside any of the polygon dictionaries I had defined.\n\n``````// Point inside a polygon\n\nfunction isPointInPoly(poly, pt)\n{\nfor(var c = false, i = -1, l = poly.length, j = l - 1; ++i < l; j = i)\n((poly[i].y <= pt.y && pt.y < poly[j].y) || (poly[j].y <= pt.y && pt.y < poly[i].y)) && (pt.x < (poly[j].x - poly[i].x) * (pt.y - poly[i].y) / (poly[j].y - poly[i].y) + poly[i].x) && (c = !c);\nreturn c;\n}\n\n/** Example implementation code\n\npoints = [\n{x: 0, y: 0},\n{x: 0, y: 50},\n{x: 50, y: 10},\n{x: -50, y: -10},\n{x: 0, y: -50},\n{x: 0, y: 0}\n];\n\nalert(isPointInPoly(points, {x: 10, y: 10}) ? \"In\" : \"Out\");\n\n*/\n``````\n• Hi Jeff,\n\ncould you please share your code. imageview with a photo of the USA Map.\n\nThanks,\n\n• Jeff Krause, many thanks for this, you saved my day !\n\n• nice logic Jeff..thanks..works like charm..\n\n• Thanks a lot. Blessings Jeff!!\n\n• Jeff,\n\n``````amazing script, but now if i want to check if a poly is in poly do you have any guideline to realize it...\n``````\n\nThanks a lot\nSteve"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.77610284,"math_prob":0.92706233,"size":1303,"snap":"2020-24-2020-29","text_gpt3_token_len":407,"char_repetition_ratio":0.12548114,"word_repetition_ratio":0.007968128,"special_character_ratio":0.3576362,"punctuation_ratio":0.22429906,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97996294,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-16T14:21:53Z\",\"WARC-Record-ID\":\"<urn:uuid:02dda889-a6ab-424c-84a0-9562f3cf4027>\",\"Content-Length\":\"9280\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:103b8ef4-e87a-42cf-94ea-09a9e5a14c49>\",\"WARC-Concurrent-To\":\"<urn:uuid:a59be839-fdaa-49ad-ba3b-0c25652b7e4b>\",\"WARC-IP-Address\":\"185.199.110.153\",\"WARC-Target-URI\":\"https://ti-qa-archive.github.io/question/25641/polygon-shaped-buttons.html\",\"WARC-Payload-Digest\":\"sha1:5IL2226Y2TYM6JTYCSUYQRBRE32FPC35\",\"WARC-Block-Digest\":\"sha1:TC7AJ4MGEKPAGOHR7J54BLDRBANPEMTG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593657169226.65_warc_CC-MAIN-20200716122414-20200716152414-00150.warc.gz\"}"} |
https://www.proprofs.com/quiz-school/story.php?title=post-scientific-concepts-4b_2kt | [
"# Post Scientific Concepts 4b*\n\n24 Questions | Total Attempts: 38",
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"Settings",
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"",
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".\n\n• 1.\nA science teacher asked her students to make a chart identifying the physical properties of each object below:Which of the following would be best to use for this chart of physical properties?\n• A.\n\nAge, color, length, mass\n\n• B.\n\nHardness, mass, name, odor\n\n• C.\n\nlength, shape, name, texture\n\n• D.\n\nColor, hardness, mass, length\n\n• 2.\nThe characteristic of an object that has mass and takes up space is:\n• A.\n\nVolume\n\n• B.\n\nMass\n\n• C.\n\nMatter\n\n• D.\n\nWeight\n\n• 3.\nOne of the illustrations below is used to measure the amount of matter that makes up an object :\n• A.\n\nOption 1\n\n• B.\n\nOption 2\n\n• C.\n\nOption 3\n\n• D.\n\nOption 4\n\n• 4.\nThe parts of an electromagnet are:\n• A.\n\nBattery, battery holder, nail, wire\n\n• B.\n\nBattery, battery holder, bulb, wire\n\n• C.\n\nBattery, battery holder, nail, bulb\n\n• D.\n\nBattery, battery holder, switch, wire\n\n• 5.\nThe two phases of matter in the illustration below are:\n• A.\n\nSolid and liquid\n\n• B.\n\nSolid and gas\n\n• C.\n\nLiquid and gas\n\n• D.\n\nLiquid and liquid\n\n• 6.\nWhat is electricity transformed into when you turn on a TV?\n• A.\n\nLight, motion, and sound\n\n• B.\n\nLight, sound, and heat\n\n• C.\n\nLight, heat, and motion\n\n• D.\n\nMotion, sound, and heat\n\n• 7.\nProperty that compares an object’s mass to its volume is called:\n• A.\n\nDensity\n\n• B.\n\nProperty\n\n• C.\n\nMatter\n\n• D.\n\nMass\n\n• 8.\nWhich of the following colors represents the greatest density in the illustration below:\n• A.\n\nRed\n\n• B.\n\nGreen\n\n• C.\n\nYellow\n\n• D.\n\nBlue\n\n• 9.\nOne of the following is a property of solid particles:\n• A.\n\nTightly packed and held in place\n\n• B.\n\nClose to each other but they are not held tightly in place\n\n• C.\n\nFar apart\n\n• D.\n\nTake the shape of the container\n\n• 10.\nWhich of the following will be attracted to the magnet:\n• A.\n\nOption 1\n\n• B.\n\nOption 2\n\n• C.\n\nOption 3\n\n• D.\n\nOption 4\n\n• 11.\nThe evaporation process in water cycle chart is number:\n• A.\n\n4\n\n• B.\n\n3\n\n• C.\n\n2\n\n• D.\n\n1\n\n• 12.\nThe illustration that represents the liquid state of matter is:\n• A.\n\nOption 1\n\n• B.\n\nOption 2\n\n• C.\n\nOption 3\n\n• D.\n\nOption 4\n\n• 13.\n“Boiling egg” is a process that changes the egg from liquid to solid. This change is :\n• A.\n\nMelting\n\n• B.\n\nPhysical change\n\n• C.\n\nBoiling\n\n• D.\n\nChemical change\n\n• 14.\nAn electric charge in motion is called:\n• A.\n\nElectric current\n\n• B.\n\nInsulator\n\n• C.\n\nConductor\n\n• D.\n\nChemical change\n\n• 15.\nWhich one of the following circuits shows a complete circuit?\n• A.\n\nOption 1\n\n• B.\n\nOption 2\n\n• C.\n\nOption 3\n\n• D.\n\nOption 4\n\n• 16.\nLook carefully at the illustration below. What form of energy is created when a magnet moves near a wire?\n• A.\n\nElectrical energy\n\n• B.\n\nSound energy\n\n• C.\n\nLight energy\n\n• D.\n\nKinetic energy\n\n• 17.\nIn the illustration below the number that points to the resistor is:\n• A.\n\n1\n\n• B.\n\n2\n\n• C.\n\n3\n\n• D.\n\n4\n\n• 18.\nSeries circuit is a type of circuits that:\n• A.\n\nThe current moves through it when there is a broken loop\n\n• B.\n\nHas two or more paths through which electric charges may flow\n\n• C.\n\nHas resistors but no source of energy\n\n• D.\n\nElectric charge can flow only in one circular path\n\n• 19.\nAll magnets have two:\n• A.\n\nSides\n\n• B.\n\nNames\n\n• C.\n\nSounds\n\n• D.\n\nPoles\n\n• 20.\nWhich letter of the following magnets represents “repel”:\n• A.\n\nA\n\n• B.\n\nB\n\n• C.\n\nC\n\n• D.\n\nD\n\n• 21.\nIn the next illustration, if lamp 1 is removed from its holder, what will happen to lamp 2 :\n• A.\n\nIt will stay working\n\n• B.\n\nIt will not work\n\n• C.\n\nIt will get dimmer\n\n• D.\n\nIt will burn\n\n• 22.\nThe best symbol below that represents the resistor is:\n• A.\n\nOption 1\n\n• B.\n\nOption 2\n\n• C.\n\nOption 3\n\n• D.\n\nOption 4\n\n• 23.\nAhmad needs to buy a light bulb which saves energy for his room. He found two types of them, the incandescent light bulb and the fluorescent light bulb. Use the following line graph to help Ahmad find that type of light bulbs.\n• A.\n\nFluorescent\n\n• B.\n\nIncandescent\n\n• C.\n\nElectronic\n\n• D.\n\nFilament\n\n• 24.\nThe circuit below is one type of circuits that are used at homes and schools :\n• A.\n\nParallel circuit\n\n• B.\n\nSeries circuit\n\n• C.\n\nNew circuit\n\n• D.\n\nGenerator",
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http://www.musclebiplane.org/htmlfile/macline.html | [
"from the April 1998 issue of the DC/RC Newsletter\n\nArt Kresse\n\nWay back in July '97 \"Uninformed\" posed this question. \"Is the CG, on biplanes, put in the same place it is on monoplanes; about 25% MAC (Mean Aerodynamic Chord(1))? If so which wing does one use?\n\nRecently on Jan 11 \"just and old toy maker\"responded; \"....you draw a line from 25% MAC on the top wing to 25%MAC bottom wing the bisect this line using foil camber line as endpoints.\"\n\nBoth the question and the response are interesting and important. The answer is correct for certain special cases:\n\n1) Both wings are the same size.\n\n2) Both are set at the same angle of incidence.\n\n3) The effect of the tail can be ignored.\n\nTo get rid of the first two of these restrictions, consider figure 1.",
null,
"From high school physics you'all recall(?) that for an object to be at rest (or in steady flight) the sum of forces and the sum of the moments about some reference point must equal zero. You also recall that a moment is a force times the perpendicular distance to the reference. For our case the reference is REF. We must pay attention to sign (+ or -). We can arbitrarily call \"up\" positive and \"down\" negative and distances to the right of REF as positive. To simplify the algebra we can assume that the lift forces \"L \" are some constant k times the angle of attack \"\" and the wing area A. i.e. the\n\nLift Equation",
null,
".\n\nIf we denote \"upper\" and \"lower\" by the subscript u and l respectively, we can write two so-called equilibrium equations one for the forces and one for the moments as follows:\n\nForce Equation:",
null,
"Moment Equation:",
null,
"By solving the three equations simultaneously (high school algebra) we get a very tedious expression for the location of the cg relative to REF.",
null,
",\n\nWhere",
null,
"is the difference between the upper and lower wing incidence. If we let",
null,
"then",
null,
".\n\nFurthermore, if we let",
null,
"in this equation then:",
null,
"which is the answer offered by \"little old toy maker\"\n\nIt is worth noting that a difference in incidence between the upper and lower wing can be either stabilizing or destabilizing in pitch, depending on whether the difference is positive or negative. The proof of this theorem is left as an exercise for the student.\n\nThe effect of the tail can be included with an expansion of the method shown here for the wings alone. (Another exercise.)\n\nA final word of caution-this is an approximation which neglects some important effects like the interference between the two wings and the downwash of the wings on the tail. Hopefully however it will get you in the ballpark so the airplane can be flown safely. Any fine tuning needed can be done with the elevator trim which is why it is there in the first place-even for the big guys.",
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"| The Eagle Interchange | The Muscle Biplane Page | Avanzata AirWork | Serious Aerobatics |",
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https://quizgecko.com/quiz/pump-up-your-ph-knowledge-with-our-quiz-aksr2L | [
"# Pump up your pH knowledge with our quiz on acidity and basicity measurement!\n\n## Summary\n\npH: A Measure of Acidity or Basicity of Aqueous Solutions\n\n• pH is a scale used to specify the acidity or basicity of an aqueous solution.\n\n• The pH scale is logarithmic and inversely indicates the activity of hydrogen ions in the solution.\n\n• Acidic solutions have lower pH values than basic or alkaline solutions.\n\n• Solutions with a pH less than 7 are acidic, solutions with a pH greater than 7 are basic, and solutions with a pH of 7 are neutral.\n\n• The pH value can be less than 0 for very concentrated strong acids or greater than 14 for very concentrated strong bases.\n\n• The pH scale is traceable to a set of standard solutions whose pH is established by international agreement.\n\n• pH values can be measured with a glass electrode and a pH meter or a color-changing indicator.\n\n• The first electronic method for measuring pH was invented by Arnold Orville Beckman in 1934.\n\n• pH is defined as the decimal logarithm of the reciprocal of the hydrogen ion activity, aH+.\n\n• The pH scale ranges from 0 to 14, with a pH of 7 indicating neutrality, values less than 7 indicating acidity, and values greater than 7 indicating basicity.\n\n• pH can be measured using indicators, which change color depending on the pH of the solution they are in.\n\n• pH values can be measured in non-aqueous solutions, but they are based on a different scale from aqueous pH values.pH Scale: Definition, Applications, and Calculations\n\n• The pH scale is a measure of acidity or basicity of an aqueous solution.\n\n• The pH scale ranges from 0 to 14, with 7 being neutral.\n\n• The unified absolute pH scale is a new approach to measuring pH, based on the absolute chemical potential of the proton.\n\n• The measurement of pH can become difficult at extremely acidic or alkaline conditions, such as below pH 2.5 or above pH 10.5.\n\n• In living organisms, the pH of various body fluids, cellular compartments, and organs is tightly regulated to maintain a state of acid-base balance known as acid–base homeostasis.\n\n• The pH of seawater is typically limited to a range between 7.4 and 8.5, and there is evidence of ongoing ocean acidification caused by carbon dioxide emissions.\n\n• Strong acids and bases are compounds that are essentially fully dissociated in water.\n\n• Weak acids and bases can be treated using the formalism of acid dissociation constant.\n\n• Water itself is a weak acid and a weak base and must be taken into account at high pH and low solute concentration.\n\n• The pH of a solution is defined as the negative logarithm of the concentration of H+, and the pOH is defined as the negative logarithm of the concentration of OH-.\n\n• When calculating the pH of a solution containing acids and/or bases, a chemical speciation calculation is used to determine the concentration of all chemical species present in the solution.\n\n• The United States Department of Agriculture Natural Resources Conservation Service classifies soil pH ranges based on soil parent material, erosional effects, climate, and vegetation.Calculating pH Using Equilibrium Constants\n\n• The pH of a solution can be calculated using equilibrium constants.\n\n• The concentration of hydrogen ions (H+) can be calculated using the ionization constant (Kw) of water, which is 10^-14 M^2 at 25°C.\n\n• The general method for calculating pH involves writing mass-balance equations for each reagent in the system, which includes equilibrium constants.\n\n• There are three non-linear simultaneous equations in the three unknowns [A], [B], and [H], which can be solved using computer programs.\n\n• This method is applicable to systems with multiple reagents or when many complexes are formed.\n\n• Polyprotic acids are also amenable to spreadsheet calculations.\n\n• Stability constants are defined as a quotient of concentrations, not activities.\n\n• Activities are used when more complicated expressions are required.\n\n• The calculation of hydrogen ion concentrations is a key element in determining equilibrium constants by potentiometric titration.\n\n• The concentration of hydroxide ions (OH-) can also be calculated using the ionization constant of water.\n\n• The pH of a solution can be used to calculate the concentration of many other species in the solution.\n\n• Computer programs are available to perform these calculations.\n\n## Description\n\nDo you know what pH is and how it is measured? Test your knowledge with our quiz on pH: A Measure of Acidity or Basicity of Aqueous Solutions. Learn about the pH scale, the history of pH measurement, and how to calculate pH using equilibrium constants. This quiz will challenge your understanding of pH and its applications in various fields. Don't miss the chance to improve your knowledge on this fundamental concept in chemistry.\n\nStart Quiz\nInformation:\nSuccess:\nError:"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91413385,"math_prob":0.95688206,"size":5152,"snap":"2023-40-2023-50","text_gpt3_token_len":1095,"char_repetition_ratio":0.1451049,"word_repetition_ratio":0.09143519,"special_character_ratio":0.20593944,"punctuation_ratio":0.09381663,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9757624,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-28T10:47:41Z\",\"WARC-Record-ID\":\"<urn:uuid:9387a8aa-12d9-4b7a-87b8-e23c6cb75389>\",\"Content-Length\":\"62582\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:53d8857c-a8b5-45ac-88ed-4c23eae68f3d>\",\"WARC-Concurrent-To\":\"<urn:uuid:14c17832-2283-4d9b-97fb-f0f87b301385>\",\"WARC-IP-Address\":\"172.66.43.182\",\"WARC-Target-URI\":\"https://quizgecko.com/quiz/pump-up-your-ph-knowledge-with-our-quiz-aksr2L\",\"WARC-Payload-Digest\":\"sha1:LP22YLRT2JCE3WNTRGE34QPZQ6REA5XK\",\"WARC-Block-Digest\":\"sha1:2W3ZKXD2DG7T2EH2BCAHMDZ4WSHQAAWB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510387.77_warc_CC-MAIN-20230928095004-20230928125004-00768.warc.gz\"}"} |
https://web2.0calc.com/questions/determining-the-approximate-value-of-this-log-w-o-a-calculator | [
"+0\n\n# Determining the approximate value of this log w/o a calculator?\n\n0\n195\n2\n\nGiven log 3 = 0.477 and log 7 = 0.845, determine the approximate value of log (132 300) without using a calculator.\n\nAll I could think about is using the antilog of each, but then how can I estimate from there?\n\nThank you! :)\n\nFeb 21, 2020\n\n#1\n+1\n\nhttps://web2.0calc.com/questions/determining-the-approx-value-of-this-logarithm\n\nThis will help but not give you the answer. I myself don't know how to do this.\n\nFeb 21, 2020\n#2\n+1\n\nNote that 1323 factors as 3^3 * 7^2\n\nSo 1323 * 100 = 132,300 = 3^3 * 7^2 * 100 = 3^3 * 7^2 * 10^2\n\nSo....we have\n\nlog (132300) =\n\nlog ( 3^3 * 7^2 * 10^2) using a log property that log (a * b * c) = log a + log b + log c\n\nlog 3^3 + log 7^2 + log 10^2 and another property says that log a^b = b log a\n\n3log3 + 2 log 7 + 2* log 10 = [ log 10 = 1 ]\n\n3 (.477) + 2 (.845) + 2(1) ≈\n\n5.121",
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"",
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"Feb 21, 2020\nedited by CPhill Feb 21, 2020"
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https://www.tensorflow.org/api_docs/python/tf/math/betainc | [
"# tf.math.betainc\n\nCompute the regularized incomplete beta integral $$I_x(a, b)$$.\n\ntf.math.betainc(\na, b, x, name=None\n)\n\n\nThe regularized incomplete beta integral is defined as:\n\n$$I_x(a, b) = \\frac{B(x; a, b)}{B(a, b)}$$\n\nwhere\n\n$$B(x; a, b) = \\int_0^x t^{a-1} (1 - t)^{b-1} dt$$\n\nis the incomplete beta function and $$B(a, b)$$ is the complete beta function.\n\n#### Args:\n\n• a: A Tensor. Must be one of the following types: float32, float64.\n• b: A Tensor. Must have the same type as a.\n• x: A Tensor. Must have the same type as a.\n• name: A name for the operation (optional).\n\n#### Returns:\n\nA Tensor. Has the same type as a."
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https://www.get-digital-help.com/how-to-use-the-db-function/ | [
"Author: Oscar Cronquist Article last updated on September 04, 2018",
null,
"The DB function calculates the depreciation of an asset for a given period using the fixed-declining balance method.\n\nFormula in cell F3:\n\n=DB(\\$C\\$3,\\$C\\$4,\\$C\\$5,E3)\n\n### Excel Function Syntax\n\nDB(cost, salvage, life, period, [month])\n\n### Arguments\n\n Cost Required. What you pay for the asset. Salvage Required. The value of the asset at the end of depreciation. Life Required. The number of periods the asset is being depreciated. Period Required. The period you want to know the depreciation of. [Month] Optional. The number of months in the first year.\n\nThe DB function calculates the depreciation of a period like this:\n\n(cost - total depreciation from prior periods) * rate\n\nwhere:\n\nrate = 1 - ((salvage / cost) ^ (1 / life))\n\nThe first period is calculated like this:\n\ncost * rate * [month] / 12\n\nThe last period is calculated like this:\n\n((cost - total depreciation from prior periods) * rate * (12 - [month])) / 12",
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"https://www.get-digital-help.com/wp-content/uploads/2018/09/How-to-use-the-DB-function.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.78258365,"math_prob":0.9609085,"size":5475,"snap":"2021-04-2021-17","text_gpt3_token_len":1269,"char_repetition_ratio":0.21166149,"word_repetition_ratio":0.241535,"special_character_ratio":0.2230137,"punctuation_ratio":0.08735868,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9988974,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-18T13:19:41Z\",\"WARC-Record-ID\":\"<urn:uuid:bb6967b5-1ebc-4c0e-966f-4f4a50480d1c>\",\"Content-Length\":\"121419\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f50f3e0d-38e5-49d1-a6b9-03e6dbe2b3ed>\",\"WARC-Concurrent-To\":\"<urn:uuid:23b9198d-6180-4607-9583-91060f2da0e4>\",\"WARC-IP-Address\":\"172.67.68.6\",\"WARC-Target-URI\":\"https://www.get-digital-help.com/how-to-use-the-db-function/\",\"WARC-Payload-Digest\":\"sha1:AYDJY3CPJIYT7BC4KVRSLKLCC6TF7AQB\",\"WARC-Block-Digest\":\"sha1:UYDIOPWHBT5LFZ75JRUVSJ4U37YUBMWF\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703514796.13_warc_CC-MAIN-20210118123320-20210118153320-00215.warc.gz\"}"} |
http://sssyjh.cn/ | [
"Home English 日本語 Pусский",
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"2020-12-04\n\n• 30\n\n公示\n\n• 14\n\n## var mm=08 if(mm==1){ document.write(\"一月\") }else if(mm==2){ document.write(\"二月\") }else if(mm==3){ document.write(\"三月\") }else if(mm==4){ document.write(\"四月\") }else if(mm==5){ document.write(\"五月\") }else if(mm==6){ document.write(\"六月\") }else if(mm==7){ document.write(\"七月\") }else if(mm==8){ document.write(\"八月\") }else if(mm==9){ document.write(\"九月\") }else if(mm==10){ document.write(\"十月\") }else if(mm==11){ document.write(\"十一月\") }else if(mm==12){ document.write(\"十二月\") }\n\n### 关于推荐2020年辽宁省教育系统先进集体和先...\n\n关于推荐2020年辽宁省教育系统先进集体和先进个人的公示\n\n• 20\n\n## var mm=07 if(mm==1){ document.write(\"一月\") }else if(mm==2){ document.write(\"二月\") }else if(mm==3){ document.write(\"三月\") }else if(mm==4){ document.write(\"四月\") }else if(mm==5){ document.write(\"五月\") }else if(mm==6){ document.write(\"六月\") }else if(mm==7){ document.write(\"七月\") }else if(mm==8){ document.write(\"八月\") }else if(mm==9){ document.write(\"九月\") }else if(mm==10){ document.write(\"十月\") }else if(mm==11){ document.write(\"十一月\") }else if(mm==12){ document.write(\"十二月\") }\n\n### 2020年职业教育对口升学考试成绩查询\n\n2020年职业教育对口升学考试成绩查询\n\n• 31\n\n## var mm=12 if(mm==1){ document.write(\"一月\") }else if(mm==2){ document.write(\"二月\") }else if(mm==3){ document.write(\"三月\") }else if(mm==4){ document.write(\"四月\") }else if(mm==5){ document.write(\"五月\") }else if(mm==6){ document.write(\"六月\") }else if(mm==7){ document.write(\"七月\") }else if(mm==8){ document.write(\"八月\") }else if(mm==9){ document.write(\"九月\") }else if(mm==10){ document.write(\"十月\") }else if(mm==11){ document.write(\"十一月\") }else if(mm==12){ document.write(\"十二月\") }\n\n### 2018-2019学年本科教学质量报告\n\n2018-2019学年本科教学质量报告",
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http://lie.math.brocku.ca/~sanco/solitons/nonlinear_wave_equations.php | [
"# Nonlinear Wave Equations\n\nThere are many different nonlinear wave equations. Some types of equations have solutions that display singularities or gradient blow-ups, while other types of equations have smooth dispersive solutions (decaying in time and space) or, in some cases, stable traveling wave solutions.\n\n## The Convective Wave Equation\n\nThe general form of the convective wave equation is ut + c(u)ux = 0 where c(u) is the wave speed given by a function of the wave amplitude u. If the wave speed is a constant (i.e. c′(u) = 0), this equation has traveling wave solutions. For non-constant wave speeds c(u) there are no traveling wave solutions, instead gradient blow-ups occur.\n\nThe simplest example is c(u) = up which depends on a positive integer p, giving a nonlinear convective equation\n\nut + upux = 0\n\nSolutions u(x, t) describe nonlinear waves whose profile eventually becomes multi-valued (i.e. there is a gradient blow-up).\n\nu(x, 0) = sech2(x)",
null,
"p = 1\n\nNaturally the question arises, how can this wave equation be modified to possess stable traveling wave solutions? Stable traveling waves with localized profiles (having rapid spatial decay) are called solitary waves.\n\n## The Dispersive Wave Equation\n\nCan dispersion balance nonlinearity? The simplest dispersive wave equation is\n\nut + uxxx = 0\n\nIts solutions u(x, t) have the general form of an integral superposition of oscillatory waves exp(ik3t + ikx) with frequency ω = k3 / 2π and wavelength λ = 2π / k for arbitrary k. The amplitude of these oscillatory waves is the Fourier transform of the initial profile u(x, 0). Because oscillatory waves of different frequency propagate at different speeds, there are no localized travelling wave solutions. Instead solutions u(x, t) describe dispersive waves whose profile develops an oscillatory decaying tail.\n\nu(x, 0) = sech2(x)",
null,
"## Nonlinearity and Dispersion\n\nThe physical feature of balancing dispersion against nonlinearity leads to solitary wave solutions. Adding the dispersive term to term uxxx to the nonlinear convective equation gives the Generalized KdV equation\n\nut + upux + uxxx = 0\n\nwhere p is a positive integer.\n\nLocalized traveling wave solutions (solitary waves) of the GKdV equation have the form\n\nu(x,t)= ( (p+1)(p+2)c/2 )1/p sech2/p( p√c (xct)/2) )\n\nwith a constant wave speed c > 0. The following plots depict how changing the degree of nonlinearity (p) affects the wave profile.\n\nWaves having the same wave speed have a decrease in wave height as the nonlinearity increases.",
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"Waves of the same height (but different speeds) have a steeper and narrower profile.",
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"http://lie.math.brocku.ca/~sanco/solitons/gif_files1/convective-2-soln.gif",
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"http://lie.math.brocku.ca/~sanco/solitons/gif_files1/gKdV-compare-p.gif",
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"http://lie.math.brocku.ca/~sanco/solitons/gif_files1/legend_gKdV-compare-p.gif",
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"http://lie.math.brocku.ca/~sanco/solitons/gif_files1/gKdV-compare-c.gif",
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"http://lie.math.brocku.ca/~sanco/solitons/gif_files1/legend_gKdV-compare-c.gif",
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https://ncerthelp.com/text.php?ques=1537+NCERT+Solutions+Class+11+Mathematics+Chapter+14+Download+in+Pdf | [
"# Mathematical Reasoning Class 11 NCERT solutions\n\nNCERT Solutions Class 11 Mathematics Chapter 14 Mathematical Reasoning Download In Pdf\n\n## In this pdf file you can see answers of following Questions\n\n### EXERCISE 14.1\n\nQuestion 1.Which of the following sentences are statements? Give reasons for your answer.\n(i) There are 35 days in a month.\n(ii) Mathematics is difficult.\n(iii) The sum of 5 and 7 is greater than 10.\n(iv) The square of a number is an even number.\n(v) The sides of a quadrilateral have equal length.\n(vii) The product of (–1) and 8 is 8.\n(viii) The sum of all interior angles of a triangle is 180°.\n(ix) Today is a windy day.\n(x) All real numbers are complex numbers.\n\nQuestion 2.Give three examples of sentences which are not statements. Give reasons for the answers.\n\n### EXERCISE 14.2\n\nQuestion 1.Write the negation of the following statements:\n(i) Chennai is the capital of Tamil Nadu\n(ii) 2 is not a complex number\n(iii) All triangles are not equilateral triangle\n(iv) The number 2 is greater than 7.\n(v) Every natural number is an integer.\n\nQuestion 2.Are the following pairs of statements negations of each other:\n(i) The number x is not a rational number. The number x is not an irrational number.\n(ii) The number x is a rational number. The number x is an irrational number.\n\nQuestion 3.Find the component statements of the following compound statements and check whether they are true or false.\n(i) Number 3 is prime or it is odd.\n(ii) All integers are positive or negative.\n(iii) 100 is divisible by 3, 11 and 5.\n\n### EXERCISE 14.3\n\nQuestion 1.For each of the following compound statements first identify the connecting words and then break it into component statements.\n(i) All rational numbers are real and all real numbers are not complex.\n(ii) Square of an integer is positive or negative.\n(iii) The sand heats up quickly in the Sun and does not cool down fast at night.\n(iv) x = 2 and x = 3 are the roots of the equation 3x2 – x – 10 = 0.\n\nQuestion 2.Identify the quantifier in the following statements and write the negation of the statements.\n(i) There exists a number which is equal to its square.\n(ii) For every real number x, x is less than x +1.\n(iii) There exists a capital for every state in India.\n\nQuestion 3.Check whether the following pair of statements are negation of each other. Give reasons for your answer.\n(i) x + y = y + x is true for every real numbers x and y.\n(ii) There exists real numbers x and y for which x + y = y + x. 4. State whether the “Or” used in the following statements is “exclusive “or” inclusive. Give reasons for your answer.\n(i) Sun rises or Moon sets.\n(ii) To apply for a driving licence, you should have a ration card or a passport.\n(iii) All integers are positive or negative.\n\n#### EXERCISE 14.4\n\nQuestion 1.Rewrite the following statement with “if-then” in five different ways conveying the same meaning. If a natural number is odd, then its square is also odd.\n\nQuestion 2.Write the contrapositive and converse of the following statements.\n(i) If x is a prime number, then x is odd. (ii) If the two lines are parallel, then they do not intersect in the same plane.\n(iii) Something is cold implies that it has low temperature.\n(iv) You cannot comprehend geometry if you do not know how to reason deductively.\n(v) x is an even number implies that x is divisible by 4.\n\nQuestion 3.Write each of the following statements in the form “if-then”\n(i) You get a job implies that your credentials are good.\n(ii) The Bannana trees will bloom if it stays warm for a month.\n(iii) A quadrilateral is a parallelogram if its diagonals bisect each other.\n(iv) To get an A+ in the class, it is necessary that you do all the exercises of the book.\n\nQuestion 4.Given statements in (a) and (b). Identify the statements given below as contrapositive or converse of each other. (a) If you live in Delhi, then you have winter clothes.\n(i) If you do not have winter clothes, then you do not live in Delhi.\n(ii) If you have winter clothes, then you live in Delhi.\n(b) If a quadrilateral is a parallelogram, then its diagonals bisect each other.\n(i) If the diagonals of a quadrilateral do not bisect each other, then the quadrilateral is not a parallelogram.\n(ii) If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.\n\n#### EXERCISE 14.5\n\nQuestion 1.Show that the statement p: “If x is a real number such that x3 + 4x = 0, then x is 0” is true by\n(i) direct method,\n(iii) method of contrapositive\n\nQuestion 2.Show that the statement “For any real numbers a and b, a2 = b2 implies that a = b” is not true by giving a counter-example.\n\nQuestion 3.Show that the following statement is true by the method of contrapositive. p: If x is an integer and x2 is even, then x is also even.\n\nQuestion 4.By giving a counter example, show that the following statements are not true.\n(i) p: If all the angles of a triangle are equal, then the triangle is an obtuse angled triangle.\n(ii) q: The equation x2 – 1 = 0 does not have a root lying between 0 and 2.\n\nQuestion 5.Which of the following statements are true and which are false? In each case give a valid reason for saying so.\n(i) p: Each radius of a circle is a chord of the circle.\n(ii) q: The centre of a circle bisects each chord of the circle.\n(iii) r: Circle is a particular case of an ellipse.\n(iv) s: If x and y are integers such that x > y, then –x < – y. (v) t : 11 is a rational number.\n\n#### Miscellaneous Exercise on Chapter 14\n\nQuestion 1.Write the negation of the following statements:\n(i) p: For every positive real number x, the number x – 1 is also positive.\n(ii) q: All cats scratch.\n(iii) r: For every real number x, either x > 1 or x <1.\n(iv) s: There exists a number x such that 0 < x <1.\n\nQuestion 2.State the converse and contrapositive of each of the following statements:\n(i) p: A positive integer is prime only if it has no divisors other than 1 and itself .\n(ii) q: I go to a beach whenever it is a sunny day.\n(iii) r: If it is hot outside, then you feel thirsty.\n\nQuestion 3.Write each of the statements in the form “if p, then q”\n(i) p: It is necessary to have a password to log on to the server.\n(ii) q: There is traffic jam whenever it rains.\n(iii) r: You can access the website only if you pay a subsciption fee.\n\nQuestion 4.Rewrite each of the following statements in the form “p if and only if q”\n(i) p: If you watch television, then your mind is free and if your mind is free, then you watch television.\n(ii) q: For you to get an A grade, it is necessary and sufficient that you do all the homework regularly.\n(iii) r: If a quadrilateral is equiangular, then it is a rectangle and if a quadrilateral is a rectangle, then it is equiangular.\n\nQuestion 5.Given below are two statements p :\n25 is a multiple of 5.q : 25 is a multiple of 8. Write the compound statements connecting these two statements with “And” and “Or”. In both cases check the validity of the compound statement.\n\nQuestion 6. Check the validity of the statements given below by the method given against it.\n(i) p: The sum of an irrational number and a rational number is irrational (by contradiction method).\n(ii) q: If n is a real number with n > 3, then n2 > 9 (by contradiction method).\n\nQuestion 7. Write the following statement in five different ways, conveying the same meaning. p: If a triangle is equiangular, then it is an obtuse angled triangle.\n\n### NCERT Books Free Pdf Download for Class 5, 6, 7, 8, 9, 10 , 11, 12 Hindi and English Medium\n\n Mathematics Biology Psychology Chemistry English Economics Sociology Hindi Business Studies Geography Science Political Science Statistics Physics Accountancy"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.87471855,"math_prob":0.9856632,"size":7392,"snap":"2019-51-2020-05","text_gpt3_token_len":1884,"char_repetition_ratio":0.1754196,"word_repetition_ratio":0.04862119,"special_character_ratio":0.25351733,"punctuation_ratio":0.121501274,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99880517,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-28T22:43:31Z\",\"WARC-Record-ID\":\"<urn:uuid:ad29c592-a586-472b-af94-ea6b290a9749>\",\"Content-Length\":\"36161\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ef4e5cae-4704-4b22-b9c8-eba070a0b267>\",\"WARC-Concurrent-To\":\"<urn:uuid:f22ed95c-9335-4471-8232-cfcbadb0e5ea>\",\"WARC-IP-Address\":\"119.18.55.97\",\"WARC-Target-URI\":\"https://ncerthelp.com/text.php?ques=1537+NCERT+Solutions+Class+11+Mathematics+Chapter+14+Download+in+Pdf\",\"WARC-Payload-Digest\":\"sha1:FJKTSMSIQBSZVNXRKL5IHUBIP67HSXF4\",\"WARC-Block-Digest\":\"sha1:IQVUXFYYXWZITPEDQBWKYAULUBXNZSU5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251783342.96_warc_CC-MAIN-20200128215526-20200129005526-00321.warc.gz\"}"} |
https://pocketmath.net/special-products.html | [
"Free Algebra Tutorials!\n\nTry the Free Math Solver or Scroll down to Tutorials!\n\n Depdendent Variable\n\n Number of equations to solve: 23456789\n Equ. #1:\n Equ. #2:\n\n Equ. #3:\n\n Equ. #4:\n\n Equ. #5:\n\n Equ. #6:\n\n Equ. #7:\n\n Equ. #8:\n\n Equ. #9:\n\n Solve for:\n\n Dependent Variable\n\n Number of inequalities to solve: 23456789\n Ineq. #1:\n Ineq. #2:\n\n Ineq. #3:\n\n Ineq. #4:\n\n Ineq. #5:\n\n Ineq. #6:\n\n Ineq. #7:\n\n Ineq. #8:\n\n Ineq. #9:\n\n Solve for:\n\n Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:\n\n# Special Products\n\nAfter studying this lesson, you will be able to:\n\n• Use Special Products Rules to multiply certain polynomials.\n\nWe will consider three special products in this section.\n\n## Square of a Sum\n\n(a + b) 2 = a 2 + 2ab + b 2\n\nExample 1\n\n(x + 3) 2\n\nWe are squaring a sum. We can just write the binomial down twice and multiply using the FOIL Method or we can use the Square of a Sum Rule.\n\nUsing the Square of a Sum Rule, we:\n\nsquare the first term which is x...this will give us x 2\n\nmultiply the 2 terms together and double them x times 3 is 3x... double it to get 6x\n\nsquare the last term which is 3...this will give us 9\n\nThe answer is x 2 + 6x + 9\n\nExample 2\n\n(x + 2) 2\n\nWe are squaring a sum. We can just write the binomial down twice and multiply using the FOIL Method or we can use the Square of a Sum Rule.\n\nUsing the Square of a Sum Rule, we:\n\nsquare the first term which is x...this will give us x 2\n\nmultiply the 2 terms together and double them x times 2 is 2x...\n\ndouble it to get 4x square the last term which is 2...this will give us 4\n\nThe answer is x 2 + 4x + 4\n\n## Square of a Difference\n\n(a - b) 2 = a 2 - 2ab + b 2\n\nExample 3\n\n(x - 2) 2\n\nWe are squaring a difference. We can just write the binomial down twice and multiply using the FOIL Method or we can use the Square of a Sum Rule.\n\nUsing the Square of a Difference Rule, we:\n\nsquare the first term which is x...this will give us x 2\n\nmultiply the 2 terms together and double them x times -2 is 2x...double it to get -4x\n\nsquare the last term which is -2...this will give us 4\n\nThe answer is x 2 - 4x + 4\n\n## Product of a Sum and a Difference\n\n(a + b)(a - b) = a 2 - b 2\n\nExample 4\n\n( x + 5 ) ( x - 5 )\n\nWe have the product of a sum and a difference. Here's what we do:\n\nmultiply the first terms x times x will be x 2\n\nmultiply the last terms 5 times 5 will be -25\n\nThe answer is x 2 -25\n\nExample 5\n\n( x + 7 ) ( x - 7 )\n\nWe have the product of a sum and a difference. Here's what we do:\n\nmultiply the first terms x times x will be x 2\n\nmultiply the last terms 7 times 7 will be - 49\n\nThe answer is x 2 - 49"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8469594,"math_prob":0.99652106,"size":2039,"snap":"2020-24-2020-29","text_gpt3_token_len":644,"char_repetition_ratio":0.14299755,"word_repetition_ratio":0.51731604,"special_character_ratio":0.3143698,"punctuation_ratio":0.09710744,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997284,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-04T09:24:36Z\",\"WARC-Record-ID\":\"<urn:uuid:1446182d-0e13-464d-ae81-e3a908b2429f>\",\"Content-Length\":\"109517\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0d646660-4e8b-4697-9138-513487e715ac>\",\"WARC-Concurrent-To\":\"<urn:uuid:f4c7a775-c3ec-435f-9f9e-d65e39cde005>\",\"WARC-IP-Address\":\"54.197.228.212\",\"WARC-Target-URI\":\"https://pocketmath.net/special-products.html\",\"WARC-Payload-Digest\":\"sha1:RU2I6MZTQK3RYDJLVOHGGFZ3EGSR4CIL\",\"WARC-Block-Digest\":\"sha1:ICKLTCT6T6FY5BK4JEQQONBTWMP5JQBL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655886095.7_warc_CC-MAIN-20200704073244-20200704103244-00111.warc.gz\"}"} |
https://socratic.org/questions/55802401581e2a75defa5da3 | [
"# Question a5da3\n\nJun 16, 2015\n\nThe energy of the reactant(s) is $\\text{+233 kJ}$.\n\n#### Explanation:\n\nThe information given to you describes an exothermic reaction because the activation energy of the reverse reaction is smaller than the activation energy of the forward reaction.\n\nThis implies that the products are lower in energy than the reactants. As a result, you should expect the energy of the reactants to exceed the energy of the products.\n\nA generic potential energy diagram for an exothermic reaction looks like this",
null,
"I won't go into the details about what activation energy and threshold energy are, but you can read more on that here:\n\nhttp://socratic.org/questions/what-is-activation-energy-what-is-threshold-energy-i-want-to-know-the-difference?source=search\n\nSo, in your case, the forward reaction has an activation energy equal to\n\n${E}_{\\text{a for\" = \"+96 kJ}}$\n\nThe reverse reaction has an activation energy equal to\n\n${E}_{\\text{a rev\" = \"+295 kJ}}$\n\nMoreover, you know that the nergy level of the products is equal to\n\n${E}_{\\text{p\" = \"+34 kJ}}$\n\nThis means that the threshold energy of the reaction, i.e. the average kinetic energy that molecules need to react, can be determined by adding the activation energy of the reverse reaction to the energy of the products.\n\n${E}_{\\text{TS\" = E_\"a rev\" + E_\"p}}$\n\n${E}_{\\text{TS\" = 295 + 34 = \"+329 kJ}}$\n\nSo, in order to get a reaction, your molecules need to have an average kinetic energy of +329 kJ. Since you know what the activation energy of the forward reaction is, you can determine the energy of the reactants by\n\n${E}_{\\text{TS\" = E_\"a for\" + E_\"r}}$\n\nE_\"r\" = E_\"TS\" - E_\"a for\" = 329 - 96 = color(green)(\"+233 kJ\")#\n\nThe enthalpy change of reaction will be negative, since the energy of the products is lower than the energy of the reactants.\n\n$\\Delta {H}_{\\text{rxn\" = H_\"products\" - H_\"reactants}}$\n\n$\\Delta {H}_{\\text{rxn\" = 34 - 233 = \"-199 kJ}}$\n\nThe minus sign symbolizes the fact that heat is released by the reaction."
] | [
null,
"https://useruploads.socratic.org/Cuk8JD0DTQaAKbOqfBba_Ea_Exothermic.gif",
null
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https://demo.formulasearchengine.com/wiki/Limit-preserving_function_(order_theory) | [
"# Limit-preserving function (order theory)\n\n{{ safesubst:#invoke:Unsubst||$N=Unreferenced |date=__DATE__ |$B= {{#invoke:Message box|ambox}} }} In the mathematical area of order theory, one often speaks about functions that preserve certain limits, i.e. certain suprema or infima. Roughly speaking, these functions map the supremum/infimum of a set to the supremum/infimum of the image of the set. Depending on the type of sets for which a function satisfies this property, it may preserve finite, directed, non-empty, or just arbitrary suprema or infima. Each of these requirements appears naturally and frequently in many areas of order theory and there are various important relationships among these concepts and other notions such as monotonicity. If the implication of limit preservation is inverted, such that the existence of limits in the range of a function implies the existence of limits in the domain, then one obtains functions that are limit-reflecting.\n\nThe purpose of this article is to clarify the definition of these basic concepts, which is necessary since the literature is not always consistent at this point, and to give general results and explanations on these issues.\n\n## Background and motivation\n\nIn many specialized areas of order theory, one restricts to classes of partially ordered sets that are complete with respect to certain limit constructions. For example, in lattice theory, one is interested in orders where all finite non-empty sets have both a least upper bound and a greatest lower bound. In domain theory, on the other hand, one focuses on partially ordered sets in which every directed subset has a supremum. Complete lattices and orders with a least element (the \"empty supremum\") provide further examples.\n\nIn all these cases, limits play a central role for the theories, supported by their interpretations in practical applications of each discipline. One also is interested in specifying appropriate mappings between such orders. From an algebraic viewpoint, this means that one wants to find adequate notions of homomorphisms for the structures under consideration. This is achieved by considering those functions that are compatible with the constructions that are characteristic for the respective orders. For example, lattice homomorphisms are those functions that preserve non-empty finite suprema and infima, i.e. the image of a supremum/infimum of two elements is just the supremum/infimum of their images. In domain theory, one often deals with so-called Scott-continuous functions that preserve all directed suprema.\n\nThe background for the definitions and terminology given below is to be found in category theory, where limits (and co-limits) in a more general sense are considered. The categorical concept of limit-preserving and limit-reflecting functors is in complete harmony with order theory, since orders can be considered as small categories defined as poset categories with defined additional structure.\n\n## Formal definition\n\nConsider two partially ordered sets P and Q, and a function f from P to Q. Furthermore, let S be a subset of P that has a least upper bound s. Then f preserves the supremum of S if the set f(S) = {f(x) | x in S} has a least upper bound in Q which is equal to f(s), i.e.\n\nf(sup S) = sup f(S)\n\nNote that this definition consists of two requirements: the supremum of the set f(S) exists and it is equal to f(s). This corresponds to the abovementioned parallel to category theory, but is not always required in the literature. In fact, in some cases one weakens the definition to require only existing suprema to be equal to f(s). However, Wikipedia works with the common notion given above and states the other condition explicitly if required.\n\nFrom the fundamental definition given above, one can derive a broad range of useful properties. A function f between posets P and Q is said to preserve finite, non-empty, directed, or arbitrary suprema if it preserves the suprema of all finite, non-empty, directed, or arbitrary sets, respectively. The preservation of non-empty finite suprema can also be defined by the identity f(x v y) = f(x) v f(y), holding for all elements x and y, where we assume v to be a total function on both orders.\n\nIn a dual way, one defines properties for the preservation of infima.\n\nThe \"opposite\" condition to preservation of limits is called reflection. Consider a function f as above and a subset S of P, such that sup f(S) exists in Q and is equal to f(s) for some element s of P. Then f reflects the supremum of S if sup S exists and is equal to s. As already demonstrated for preservation, one obtains many additional properties by considering certain classes of sets S and by dualizing the definition to infima.\n\n## Special cases\n\nSome special cases or properties derived from the above scheme are known under other names or are of particular importance to some areas of order theory. For example, functions that preserve the empty supremum are those that preserve the least element. Furthermore, due to the motivation explained earlier, many limit-preserving functions appear as special homomorphisms for certain order structures. Some other prominent cases are given below.\n\n### Preservation of all limits\n\nAn interesting situation occurs if a function preserves all suprema (or infima). More accurately, this is expressed by saying that a function preserves all existing suprema (or infima), and it may well be that the posets under consideration are not complete lattices. For example, (monotone) Galois connections have this property. Conversely, by the order theoretical Adjoint Functor Theorem, mappings that preserve all suprema/infima can be guaranteed to be part of a unique Galois connection as long as some additional requirements are met.\n\n### Distributivity\n\nA lattice L is distributive if, for all x, y, and z in L, we find\n\n$x\\wedge \\left(y\\vee z\\right)=\\left(x\\wedge y\\right)\\vee \\left(x\\wedge z\\right)$",
null,
"But this just says that the meet function ^: L -> L preserves binary suprema. It is known in lattice theory, that this condition is equivalent to its dual, i.e. the function v: L -> L preserving binary infima. In a similar way, one sees that the infinite distributivity law\n\n$x\\wedge \\bigvee S=\\bigvee \\left\\{x\\wedge s\\mid s\\in S\\right\\}$",
null,
"of complete Heyting algebras (see also pointless topology) is equivalent to the meet function ^ preserving arbitrary suprema. This condition, however, does not imply its dual.\n\n### Scott-continuity\n\nFunctions that preserve directed suprema are called Scott-continuous or sometimes just continuous, if this does not cause confusions with the according concept of analysis and topology. A similar use of the term continuous for preservation of limits can also be found in category theory.\n\n## Important properties and results\n\nThe above definition of limit preservation is quite strong. Indeed, every function that preserves at least the suprema or infima of two-element chains, i.e. of sets of two comparable elements, is necessarily monotone. Hence, all the special preservation properties stated above induce monotonicity.\n\nBased on the fact that some limits can be expressed in terms of others, one can derive connections between the preservation properties. For example, a function f preserves directed suprema if and only if it preserves the suprema of all ideals. Furthermore, a mapping f from a poset in which every non-empty finite supremum exists (a so-called sup-semilattice) preserves arbitrary suprema if and only if it preserves both directed and finite (possibly empty) suprema.\n\nHowever, it is not true that a function that preserves all suprema would also preserve all infima or vice versa."
] | [
null,
"https://demo.formulasearchengine.com/index.php",
null,
"https://demo.formulasearchengine.com/index.php",
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https://numberdyslexia.com/abacus-formula-big-friend-small-friend-explained-pdf/ | [
"# Abacus Formula: Big Friend & Small Friend Explained [Pdf]\n\nSpread the word\n\nLast Updated on February 17, 2021 by Editorial Team\n\nAbacus is considered one of the important tools to promote mental math abilities in early learners. For those unaware, Abacus is a manual aid to calculating that consists of beads or disks that can be moved up and down on a series of sticks or strings within a usually wooden frame. It is considered one of the best ways to develop number sense.\n\nNumbers are physically constructed and manipulated in Abacus. We already discussed different types of Abacus and their uses. Although not included in the regular math curriculum, but kids must try their hands on practicing various math operations with it once in a while.\n\nIf you just started learning the abacus, you must first learn the basics of the abacus. After getting your hands on the abacus and playing with it for a little while, its time to practice the basic operations (Addition & Subtraction) on it. In general practice, we take the help of the abacus formula to do addition and subtraction, especially when there are not enough beads in a column to add on Abacus. It involves using the concept of small friends and big friends for solving basic operations.\n\nLook at the abacus formula sheet given below. You can also view/download pdf version by clicking this link.",
null,
"Abacus Formula Small Friend Addition",
null,
"Abacus Formula Small Friend Substraction\n\nIn the case of Small friend, we take the bead of value 5 as the reference. To obtain the number from the addition and subtraction on abacus, we operate the reference 5 with the small friend. Here, we will explain it with examples of both.\n\nREAD : What is Logical-mathematical Intelligence?: Importance, Activities, Examples and Its role in teaching\n\nLet’s add 3+4,\n\n1. Place 3 (3 lower beads of value 1) on abacus\n2. Now, we have to add 4 but there are not enough beads left, so we will take the help of small friend as per the formula ‘add 5 less 1’.\n3. We will add 5(1 upper bead on the same column) on abacus and remove 1 (1 lower bead on the same column )\n4. We are left with (1 upper bead of 5, 2 lower bead of value 1). The solution is 5 + 2 = 7\n\nSimilarly, Let’s subtract 7 -4,\n\n1. To make 7: Slide 2 lower beads of unit place and 1 upper bead of unit place (value 5).\n2. To subtract 4, we have to take the help of big friend as there are not enough beads on the unit place. Use the formula ‘Less 5 add 1’.\n3. Now remove 5 (slide down the upper bead of value 5) and add 1 (slide up the 1 lower bead of unit place)\n4. We are left with (0 upper bead and 3 lower beads of unit’s place). The solution is 0 + 3 = 3, which is our answer.",
null,
"Abacus Formula Big Friend Addition",
null,
"Abacus Formula Big Friend Subtraction\n\nIn the case of Big friend, we take the bead of value 10 as the reference. To obtain the number from the addition and subtraction on abacus, we operate reference 10 with the small friend. Here, we will explain it with examples of both.\n\nLet’s add 8+9,\n\n1. Place 8 (3 lower beads of value 1 and 1 upper bead of value 5) on abacus\n2. Now, we have to add 9 but there are not enough beads left, so we will take the help of big friend as per the formula ‘add 10 less 1’.\n3. We will add 10(1 lower bead on the adjacent left column) on abacus and remove 1 (1 lower bead on the same column )\n4. We are left with (1 lower bead of value 10, 1 upper bead of 5, 2 lower bead of value 1). The solution is 10 + 5 + 2 = 17\nREAD : Importance of spatial sense in kids and how to improve it?\n\nSimilarly, Let’s subtract 15 -7,\n\n1. To make 15: Slide 1 lower bead of value 10 and 1 upper bead of unit place (value 5).\n2. To subtract 7, we have to take the help of big friend as there are not enough beads in the unit place. Use the formula ‘Less 10 add 3’.\n3. Now add 3 (slide up the 3 lower beads of unit place) and remove 10 (slide down the lower bead of value 10 columns).\n4. We are left with (1 upper bead and 3 lower beads of unit’s place). The solution is 5 + 3 = 8, which is our answer.\n\n`Want more fun printables? Check out PrintablesHub.com for worksheets, charts & game templates.`\n\nShare the word\n\nIf you find this post helpful, then please help us spread the word. Share the post with your friends, family, and colleagues. Do not forget to subscribe to our updates. Any suggestions and recommendations are highly recommended. You can reach us through the contact form and we will get back at you shortly.\n\nSpread the word"
] | [
null,
"https://numberdyslexia.com/wp-content/uploads/2020/04/abacus-formula-small-friend-1.png",
null,
"https://numberdyslexia.com/wp-content/uploads/2020/04/abacus-formula-small-friend-substraction.png",
null,
"https://numberdyslexia.com/wp-content/uploads/2020/04/abacus-formula-big-friend-addition-1.png",
null,
"https://numberdyslexia.com/wp-content/uploads/2020/04/abacus-formula-big-friend-substraction-1.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.90829617,"math_prob":0.91986245,"size":4005,"snap":"2021-21-2021-25","text_gpt3_token_len":1019,"char_repetition_ratio":0.15621094,"word_repetition_ratio":0.2835443,"special_character_ratio":0.26292136,"punctuation_ratio":0.09016393,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97430277,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-19T09:30:12Z\",\"WARC-Record-ID\":\"<urn:uuid:19bd7f3a-4e51-4303-a8eb-dc4d6890d8fd>\",\"Content-Length\":\"93369\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ad28f62c-382e-45eb-9179-b1d73b6dacd7>\",\"WARC-Concurrent-To\":\"<urn:uuid:f190138a-38fb-4dbb-ab3e-36ed73742aa7>\",\"WARC-IP-Address\":\"35.175.60.16\",\"WARC-Target-URI\":\"https://numberdyslexia.com/abacus-formula-big-friend-small-friend-explained-pdf/\",\"WARC-Payload-Digest\":\"sha1:7DIXVAIP4OQDWXV4FNAHMTVW655BS6H2\",\"WARC-Block-Digest\":\"sha1:UQ4BFWH5K3WCPICI2SFUTI66MWP2VVG4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487647232.60_warc_CC-MAIN-20210619081502-20210619111502-00292.warc.gz\"}"} |
https://math.stackexchange.com/questions/794748/exercise-problem-in-group-order-homomorphism | [
"# Exercise problem in group order & homomorphism\n\nI came across this problem in class note but I am stuck:\n\nLet G be a group of order 21, let G' be a group of order 35, and let φ be a homomorphism from G to G'. Assume that G does not have a normal subgroup of order 3. Show that φ(g) = 1 for each element g in G.\n\nAny help or hints would be very much appreciated. Thanks for your time.\n\n• You mean subgroup of order 7? – Test123 May 14 '14 at 16:06\n• @Test123: I will email my professor if there is any typo. Order of 7 does make more sense. Thanks. – Amanda.M May 16 '14 at 14:36\n• @A.Mangus It would have been easier but still as you can see below it works fine. – Test123 May 16 '14 at 14:42\n\nThe image of $\\varphi$ is a subgroup of $G'$ and so its order divides $35$. Moreover, this order is the index of the kernel of $\\varphi$ in $G$, and so its order divides $21$. Hence, the order of the image of $\\varphi$ is a common divisor of $35$ and $21$ and so divides $7$. But the order of the image cannot be $7$ because otherwise the order of the kernel would be $3$, and $G$ has no normal subgroup of order $3$. Thus, the order of the image of $\\varphi$ is $1$.\n• @A.Magnus, the image $\\varphi(G)$ is isomorphic to the quotient $G/\\ker\\varphi$. – lhf May 16 '14 at 2:21\nIf $\\varphi\\colon G\\to G'$ is a homomorphism, then $\\ker\\varphi$ is a normal subgroup of $G$, so its order is $1$, $3$, $7$ or $21$. It can't be $3$ by assumption. The homomorphism theorem says $\\varphi$ induces an injective homomorphism $$\\tilde\\varphi\\colon G/\\ker\\varphi\\to G'$$ such that $\\operatorname{im}\\tilde\\varphi=\\operatorname{im}\\varphi$.\nCall $H$ the image of $\\varphi$; what order can $H$ have? Can $\\ker\\phi$ have order $1$ or $7$?"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.96389955,"math_prob":0.9989208,"size":333,"snap":"2019-35-2019-39","text_gpt3_token_len":91,"char_repetition_ratio":0.14285715,"word_repetition_ratio":0.028571429,"special_character_ratio":0.2762763,"punctuation_ratio":0.09756097,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999472,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-15T21:58:50Z\",\"WARC-Record-ID\":\"<urn:uuid:0dc8e497-a405-41db-81a3-d38995c1a8a2>\",\"Content-Length\":\"146669\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9cd78621-3404-41ce-965b-13ac6d34ae7c>\",\"WARC-Concurrent-To\":\"<urn:uuid:fec13429-fc17-4121-b66a-6bc2765d3654>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/794748/exercise-problem-in-group-order-homomorphism\",\"WARC-Payload-Digest\":\"sha1:HDZ64FMBVJTLINQYZ3HOACKSTJOYM6EK\",\"WARC-Block-Digest\":\"sha1:DQIBCNPZYWFQQGV7HBWS7WP4VILW57E5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514572436.52_warc_CC-MAIN-20190915215643-20190916001643-00403.warc.gz\"}"} |
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