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https://whatisconvert.com/16-years-in-nanoseconds	[ "# What is 16 Years in Nanoseconds?\n\n## Convert 16 Years to Nanoseconds\n\nTo calculate 16 Years to the corresponding value in Nanoseconds, multiply the quantity in Years by 3.1536E+16 (conversion factor). In this case we should multiply 16 Years by 3.1536E+16 to get the equivalent result in Nanoseconds:\n\n16 Years x 3.1536E+16 = 5.04576E+17 Nanoseconds\n\n16 Years is equivalent to 5.04576E+17 Nanoseconds.\n\n## How to convert from Years to Nanoseconds\n\nThe conversion factor from Years to Nanoseconds is 3.1536E+16. To find out how many Years in Nanoseconds, multiply by the conversion factor or use the Time converter above. Sixteen Years is equivalent to eighteen quadrillion fourteen trillion three hundred ninety-eight billion five hundred nine million four hundred eighty-one thousand nine hundred eighty-four Nanoseconds.\n\n## Definition of Year\n\nA year (symbol: y; also abbreviated yr.) is the orbital period of the Earth moving in its orbit around the Sun. Due to the Earth's axial tilt, the course of a year sees the passing of the seasons, marked by changes in weather, the hours of daylight, and, consequently, vegetation and soil fertility. In temperate and subpolar regions around the globe, four seasons are generally recognized: spring, summer, autumn and winter. In tropical and subtropical regions several geographical sectors do not present defined seasons; but in the seasonal tropics, the annual wet and dry seasons are recognized and tracked. A calendar year is an approximation of the number of days of the Earth's orbital period as counted in a given calendar. The Gregorian, or modern, calendar, presents its calendar year to be either a common year of 365 days or a leap year of 366 days.\n\n## Definition of Nanosecond\n\nA nanosecond (symbol: ns) is an SI unit of time equal to one billionth of a second (10−9 or 1/1,000,000,000 s). One nanosecond is to one second as one second is to 31.71 years. The word nanosecond is formed by the prefix nano and the unit second. A nanosecond is equal to 1000 picoseconds or 1⁄1000 microsecond. Because the next SI unit is 1000 times larger, times of 10−8 and 10−7 seconds are typically expressed as tens or hundreds of nanoseconds. Times of this magnitude are commonly encountered in telecommunications, pulsed lasers and some areas of electronics.\n\n## Using the Years to Nanoseconds converter you can get answers to questions like the following:\n\n• How many Nanoseconds are in 16 Years?\n• 16 Years is equal to how many Nanoseconds?\n• How to convert 16 Years to Nanoseconds?\n• How many is 16 Years in Nanoseconds?\n• What is 16 Years in Nanoseconds?\n• How much is 16 Years in Nanoseconds?\n• How many ns are in 16 yr?\n• 16 yr is equal to how many ns?\n• How to convert 16 yr to ns?\n• How many is 16 yr in ns?\n• What is 16 yr in ns?\n• How much is 16 yr in ns?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91556174,"math_prob":0.9223883,"size":2766,"snap":"2022-05-2022-21","text_gpt3_token_len":674,"char_repetition_ratio":0.16944243,"word_repetition_ratio":0.044397462,"special_character_ratio":0.25234997,"punctuation_ratio":0.11594203,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9790022,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-20T07:46:20Z\",\"WARC-Record-ID\":\"<urn:uuid:39937878-9dbf-4e8d-a5ff-124feb931c14>\",\"Content-Length\":\"30558\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f501e70a-2961-4c69-a2c1-d85c5195121c>\",\"WARC-Concurrent-To\":\"<urn:uuid:ff94e6bd-8a3f-48a8-978c-4c3a9d566867>\",\"WARC-IP-Address\":\"104.21.13.210\",\"WARC-Target-URI\":\"https://whatisconvert.com/16-years-in-nanoseconds\",\"WARC-Payload-Digest\":\"sha1:MA5Y3VWU4GEQARIVMYPLPQK4RMNCAKVL\",\"WARC-Block-Digest\":\"sha1:SPHX3SALXOWNVL5WR2AMNSVVWFYLF6W5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662531762.30_warc_CC-MAIN-20220520061824-20220520091824-00539.warc.gz\"}"}
https://www.serviceescortphysics.nl/relativiteitstheorie/einstein-field-equation	[ "## Have the Einstein field equations changed their meaning?In his short essay, “How Great Equations Survive,” Nobel laureate physicist Steven Weinberg argues that though equations survive through scientific change, they are reinterpreted in light of the developments of theory. “The equations of General Relativity,” he argues, “have undergone a similar reinterpretation.” Weinberg devotes attention to the point that “the equations” are of the type known as “second order differential equations.\" This means \"that the equations were assumed by Einstein to involve only rates of change of the fields (first derivatives) and rates of change of rates of change (second derivatives) but not rates of higher order.” This he sees as something of a reasonable idealization. “I don’t know,” he writes, “any place where Einstein explains the motivation for this assumption.”“Today,” Weinberg continues, “General Relativity is widely (though not universally regarded as another effective field theory, useful only for distances much larger than about 10 (to the -33rd) centimeters, and particle energies much less than an energy equivalent to the mass of 10 (to the 19th) protons. No one today would (or at least no one should) take seriously any consequence of General Relativity for shorter distances or larger energies.”", null, "" ]	[ null, "data:image/gif;base64,R0lGODlhAQABAIAAAP///wAAACH5BAEAAAAALAAAAAABAAEAAAICRAEAOw==", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.95527035,"math_prob":0.9658983,"size":1313,"snap":"2019-51-2020-05","text_gpt3_token_len":264,"char_repetition_ratio":0.11306341,"word_repetition_ratio":0.0,"special_character_ratio":0.1980198,"punctuation_ratio":0.09049774,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9839312,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-08T02:58:00Z\",\"WARC-Record-ID\":\"<urn:uuid:22e0ae3b-20a4-45ba-a860-d82f9199a901>\",\"Content-Length\":\"38658\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:af042cec-79f1-4489-a8ad-44e091176c73>\",\"WARC-Concurrent-To\":\"<urn:uuid:e58cd8b6-c361-456a-9917-a4759de79f84>\",\"WARC-IP-Address\":\"35.204.150.5\",\"WARC-Target-URI\":\"https://www.serviceescortphysics.nl/relativiteitstheorie/einstein-field-equation\",\"WARC-Payload-Digest\":\"sha1:77V7QZFE5DOJRAMNXU5IRFXAV4IAH3SB\",\"WARC-Block-Digest\":\"sha1:AY7UYGPRP5XHBFFM3XAN47JUSBIPL3F4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540504338.31_warc_CC-MAIN-20191208021121-20191208045121-00103.warc.gz\"}"}
https://www.mdpi.com/journal/mathematics/special_issues/Recent_Trends_Orthogonal_Polynomials_Approximation_Theory_Applications	[ "", null, "", null, "Journal Browser\n\n# Special Issue \"Recent Trends on Orthogonal Polynomials: Approximation Theory and Applications\"\n\nA special issue of Mathematics (ISSN 2227-7390).\n\nDeadline for manuscript submissions: closed (31 December 2020).\n\n## Special Issue Editors\n\nProf. Dr. Francisco Marcellan\nE-Mail Website\nGuest Editor\nInterests: orthogonal polynomials; moment problems; distribution of zeros; integrable systems; random matrices; stochastic processes; signal theory; quadrature formulas; spectral methods for boundary value problems; fourier expansions; structured matrices; integral transforms please revise intereest of Edmundo Huertas orthogonal polynomials; moment problems; distribution of zeros; integrable systems; random matrices; stochastic processes; signal theory; quadrature formulas; spectral methods for boundary value problems; fourier expansions; structured matrices; integral transforms\nDr. Edmundo Huertas\nE-Mail Website\nGuest Editor\nDpto. de Física y Matemáticas, Universidad de Alcalá (UAH), Ctra. Madrid-Barcelona, Km. 33,600, Alcalá de Henares, 28805 Madrid, Spain\nInterests: orthogonal polynomials; moment problems; distribution of zeros; integrable systems; random matrices; stochastic processes; signal theory; quadrature formulas; spectral methods for boundary value problems; fourier expansions; structured matrices; integral transforms\n\n## Special Issue Information\n\nDear Colleagues,\n\nIn recent years, the theory of orthogonal polynomials has received a great amount of interest because of its wide role in Pure and Applied Mathematics. Orthogonal polynomials are essential tools for the solution of many problems in the spectral theory of differential and difference equations, Painlevé equations (discrete and continuous versions), numerical methods in quadrature on the real line and the unit circle, as well as cubature formulas on multidimensional domains, with applications ranging from Number Theory to Approximation Theory, Combinatorics to Group representation, integrable systems, random matrices, and linear system theory to signal processing.\n\nThe aims of the proposed Special Issue are:\n\n• To show some recent trends in the research on orthogonal polynomials, with a special emphasis on their analytic properties and approximation theory. Different examples of orthogonality (Sobolev, multiple, multivariate, matrix) will be studied, as well as the asymptotic properties of the corresponding sequences of orthogonal polynomials and the behavior of their zeros;\n• To emphasize their impact in Mathematical Physics, mainly in integrable systems and Painlevé equations (discrete and continuous cases), as they are strongly related to the coefficients of three term relation, satisfied by a sequence of orthogonal polynomials and time-depending measures supported on the real line.\n\nProf. Dr. Francisco Marcellan\nDr. Edmundo Huertas\nGuest Editors\n\nManuscript Submission Information\n\nManuscripts should be submitted online at www.mdpi.com by registering and logging in to this website. Once you are registered, click here to go to the submission form. Manuscripts can be submitted until the deadline. All papers will be peer-reviewed. Accepted papers will be published continuously in the journal (as soon as accepted) and will be listed together on the special issue website. Research articles, review articles as well as short communications are invited. For planned papers, a title and short abstract (about 100 words) can be sent to the Editorial Office for announcement on this website.\n\nSubmitted manuscripts should not have been published previously, nor be under consideration for publication elsewhere (except conference proceedings papers). All manuscripts are thoroughly refereed through a single-blind peer-review process. A guide for authors and other relevant information for submission of manuscripts is available on the Instructions for Authors page. Mathematics is an international peer-reviewed open access semimonthly journal published by MDPI.\n\nPlease visit the Instructions for Authors page before submitting a manuscript. The Article Processing Charge (APC) for publication in this open access journal is 1600 CHF (Swiss Francs). Submitted papers should be well formatted and use good English. Authors may use MDPI's English editing service prior to publication or during author revisions.\n\n## Keywords\n\n• Orthogonal polynomials on the real line\n• Orthogonal polynomials on the unit circle\n• Matrix orthogonal polynomials\n• Multiple orthogonal polynomials\n• Multivariate orthogonal polynomials\n• Sobolev orthogonal polynomials\n• Integrable systems\n• Random matrices\n• Rational approximation\n• Approximation with splines\n• Wavelets\n\n## Published Papers (14 papers)\n\nOrder results\nResult details\nSelect all\nExport citation of selected articles as:\n\n# Research\n\nOpen AccessArticle\nMultiple Meixner Polynomials on a Non-Uniform Lattice\nMathematics 2020, 8(9), 1460; https://doi.org/10.3390/math8091460 - 31 Aug 2020\nViewed by 469\nAbstract\nWe consider two families of type II multiple orthogonal polynomials. Each family has orthogonality conditions with respect to a discrete vector measure. The r components of each vector measure are q-analogues of Meixner measures of the first and second kind, respectively. These [...] Read more.\nWe consider two families of type II multiple orthogonal polynomials. Each family has orthogonality conditions with respect to a discrete vector measure. The r components of each vector measure are q-analogues of Meixner measures of the first and second kind, respectively. These polynomials have lowering and raising operators, which lead to the Rodrigues formula, difference equation of order $r+1$, and explicit expressions for the coefficients of recurrence relation of order $r+1$. Some limit relations are obtained. Full article\nOpen AccessArticle\nOn Multivariate Bernstein Polynomials\nMathematics 2020, 8(9), 1397; https://doi.org/10.3390/math8091397 - 21 Aug 2020\nViewed by 598\nAbstract\nIn this paper, we first revisit and bring together as a sort of survey, properties of Bernstein polynomials of one variable. Secondly, we extend the results from one variable to several ones, namely—uniform convergence, uniform convergence of the derivatives, order of convergence, monotonicity, [...] Read more.\nIn this paper, we first revisit and bring together as a sort of survey, properties of Bernstein polynomials of one variable. Secondly, we extend the results from one variable to several ones, namely—uniform convergence, uniform convergence of the derivatives, order of convergence, monotonicity, fixed sign for the p-th derivative, and deduction of the upper and lower bounds of Bernstein polynomials from those of the corresponding functions. Full article\nOpen AccessArticle\nOn the Finite Orthogonality of q-Pseudo-Jacobi Polynomials\nMathematics 2020, 8(8), 1323; https://doi.org/10.3390/math8081323 - 08 Aug 2020\nCited by 1 \| Viewed by 520\nAbstract\nUsing the Sturm–Liouville theory in q-difference spaces, we prove the finite orthogonality of q-Pseudo Jacobi polynomials. Their norm square values are then explicitly computed by means of the Favard theorem. Full article\nOpen AccessArticle\nNew Stability Criteria for Discrete Linear Systems Based on Orthogonal Polynomials\nMathematics 2020, 8(8), 1322; https://doi.org/10.3390/math8081322 - 08 Aug 2020\nViewed by 536\nAbstract\nA new criterion for Schur stability is derived by using basic results of the theory of orthogonal polynomials. In particular, we use the relation between orthogonal polynomials on the real line and on the unit circle known as the Szegő transformation. Some examples [...] Read more.\nA new criterion for Schur stability is derived by using basic results of the theory of orthogonal polynomials. In particular, we use the relation between orthogonal polynomials on the real line and on the unit circle known as the Szegő transformation. Some examples are presented. Full article\nShow Figures", null, "Figure 1\n\nOpen AccessArticle\nOn Second Order q-Difference Equations Satisfied by Al-Salam–Carlitz I-Sobolev Type Polynomials of Higher Order\nMathematics 2020, 8(8), 1300; https://doi.org/10.3390/math8081300 - 06 Aug 2020\nViewed by 672\nAbstract\nThis contribution deals with the sequence ${Un(a)(x;q,j)}n≥0$ of monic polynomials in x, orthogonal with respect to a Sobolev-type inner product related to the Al-Salam–Carlitz I orthogonal polynomials, and involving an arbitrary number j of q-derivatives on the two boundaries of the corresponding orthogonality interval, for some fixed real number $q∈(0,1)$. We provide several versions of the corresponding connection formulas, ladder operators, and several versions of the second order q-difference equations satisfied by polynomials in this sequence. As a novel contribution to the literature, we provide certain three term recurrence formula with rational coefficients satisfied by $Un(a)(x;q,j)$, which paves the way to establish an appealing generalization of the so-called J-fractions to the framework of Sobolev-type orthogonality. Full article\nShow Figures", null, "Figure 1\n\nOpen AccessArticle\nOn Semi-Classical Orthogonal Polynomials Associated with a Modified Sextic Freud-Type Weight\nMathematics 2020, 8(8), 1250; https://doi.org/10.3390/math8081250 - 31 Jul 2020\nCited by 1 \| Viewed by 583\nAbstract\nPolynomials that are orthogonal with respect to a perturbation of the Freud weight function by some parameter, known to be modified Freudian orthogonal polynomials, are considered. In this contribution, we investigate certain properties of semi-classical modified Freud-type polynomials in which their corresponding semi-classical [...] Read more.\nPolynomials that are orthogonal with respect to a perturbation of the Freud weight function by some parameter, known to be modified Freudian orthogonal polynomials, are considered. In this contribution, we investigate certain properties of semi-classical modified Freud-type polynomials in which their corresponding semi-classical weight function is a more general deformation of the classical scaled sextic Freud weight $\|x\|αexp(−cx6),c>0,α>−1$. Certain characterizing properties of these polynomials such as moments, recurrence coefficients, holonomic equations that they satisfy, and certain non-linear differential-recurrence equations satisfied by the recurrence coefficients, using compatibility conditions for ladder operators for these orthogonal polynomials, are investigated. Differential-difference equations were also obtained via Shohat’s quasi-orthogonality approach and also second-order linear ODEs (with rational coefficients) satisfied by these polynomials. Modified Freudian polynomials can also be obtained via Chihara’s symmetrization process from the generalized Airy-type polynomials. The obtained linear differential equation plays an essential role in the electrostatic interpretation for the distribution of zeros of the corresponding Freudian polynomials. Full article\nOpen AccessArticle\nOn an Energy-Dependent Quantum System with Solutions in Terms of a Class of Hypergeometric Para-Orthogonal Polynomials on the Unit Circle\nMathematics 2020, 8(7), 1161; https://doi.org/10.3390/math8071161 - 15 Jul 2020\nCited by 1 \| Viewed by 485\nAbstract\nWe study an energy-dependent potential related to the Rosen–Morse potential. We give in closed-form the expression of a system of eigenfunctions of the Schrödinger operator in terms of a class of functions associated to a family of hypergeometric para-orthogonal polynomials on the unit [...] Read more.\nWe study an energy-dependent potential related to the Rosen–Morse potential. We give in closed-form the expression of a system of eigenfunctions of the Schrödinger operator in terms of a class of functions associated to a family of hypergeometric para-orthogonal polynomials on the unit circle. We also present modified relations of orthogonality and an asymptotic formula. Consequently, bound state solutions can be obtained for some values of the parameters that define the model. As a particular case, we obtain the symmetric trigonometric Rosen–Morse potential for which there exists an orthogonal basis of eigenstates in a Hilbert space. By comparing the existent solutions for the symmetric trigonometric Rosen–Morse potential, an identity involving Gegenbauer polynomials is obtained. Full article\nShow Figures", null, "Figure 1\n\nOpen AccessArticle\nOn Differential Equations Associated with Perturbations of Orthogonal Polynomials on the Unit Circle\nMathematics 2020, 8(2), 246; https://doi.org/10.3390/math8020246 - 14 Feb 2020\nCited by 2 \| Viewed by 599\nAbstract\nIn this contribution, we propose an algorithm to compute holonomic second-order differential equations satisfied by some families of orthogonal polynomials. Such algorithm is based in three properties that orthogonal polynomials satisfy: a recurrence relation, a structure formula, and a connection formula. This approach [...] Read more.\nIn this contribution, we propose an algorithm to compute holonomic second-order differential equations satisfied by some families of orthogonal polynomials. Such algorithm is based in three properties that orthogonal polynomials satisfy: a recurrence relation, a structure formula, and a connection formula. This approach is used to obtain second-order differential equations whose solutions are orthogonal polynomials associated with some spectral transformations of a measure on the unit circle, as well as orthogonal polynomials associated with coherent pairs of measures on the unit circle. Full article\nOpen AccessArticle\nEigenvalue Problem for Discrete Jacobi–Sobolev Orthogonal Polynomials\nMathematics 2020, 8(2), 182; https://doi.org/10.3390/math8020182 - 03 Feb 2020\nViewed by 566\nAbstract\nIn this paper, we consider a discrete Sobolev inner product involving the Jacobi weight with a twofold objective. On the one hand, since the orthonormal polynomials with respect to this inner product are eigenfunctions of a certain differential operator, we are interested in [...] Read more.\nIn this paper, we consider a discrete Sobolev inner product involving the Jacobi weight with a twofold objective. On the one hand, since the orthonormal polynomials with respect to this inner product are eigenfunctions of a certain differential operator, we are interested in the corresponding eigenvalues, more exactly, in their asymptotic behavior. Thus, we can determine a limit value which links this asymptotic behavior and the uniform norm of the orthonormal polynomials in a logarithmic scale. This value appears in the theory of reproducing kernel Hilbert spaces. On the other hand, we tackle a more general case than the one considered in the literature previously. Full article\nOpen AccessFeature PaperArticle\nA Characterization of Polynomial Density on Curves via Matrix Algebra\nMathematics 2019, 7(12), 1231; https://doi.org/10.3390/math7121231 - 12 Dec 2019\nViewed by 572\nAbstract\nIn this work, our aim is to obtain conditions to assure polynomial approximation in Hilbert spaces $L 2 ( μ )$ , with $μ$ a compactly supported measure in the complex plane, in terms of properties of the associated moment matrix with the measure $μ$ . To do it, in the more general context of Hermitian positive semidefinite matrices, we introduce two indexes, $γ ( M )$ and $λ ( M )$ , associated with different optimization problems concerning theses matrices. Our main result is a characterization of density of polynomials in the case of measures supported on Jordan curves with non-empty interior using the index $γ$ and other specific index related to it. Moreover, we provide a new point of view of bounded point evaluations associated with a measure in terms of the index $γ$ that will allow us to give an alternative proof of Thomson’s theorem, by using these matrix indexes. We point out that our techniques are based in matrix algebra tools in the framework of Hermitian positive definite matrices and in the computation of certain indexes related to some optimization problems for infinite matrices. Full article\nOpen AccessArticle\nOn Infinitely Many Rational Approximants to ζ(3)\nMathematics 2019, 7(12), 1176; https://doi.org/10.3390/math7121176 - 03 Dec 2019\nCited by 1 \| Viewed by 528\nAbstract\nA set of second order holonomic difference equations was deduced from a set of simultaneous rational approximation problems. Some orthogonal forms involved in the approximation were used to compute the Casorati determinants for its linearly independent solutions. These solutions constitute the numerator and [...] Read more.\nA set of second order holonomic difference equations was deduced from a set of simultaneous rational approximation problems. Some orthogonal forms involved in the approximation were used to compute the Casorati determinants for its linearly independent solutions. These solutions constitute the numerator and denominator sequences of rational approximants to $ζ ( 3 )$ . A correspondence from the set of parameters involved in the holonomic difference equation to the set of holonomic bi-sequences formed by these numerators and denominators appears. Infinitely many rational approximants can be generated. Full article\nShow Figures", null, "Figure 1\n\nOpen AccessFeature PaperArticle\nRobust Stability of Hurwitz Polynomials Associated with Modified Classical Weights\nMathematics 2019, 7(9), 818; https://doi.org/10.3390/math7090818 - 05 Sep 2019\nCited by 2 \| Viewed by 744\nAbstract\nIn this contribution, we consider sequences of orthogonal polynomials associated with a perturbation of some classical weights consisting of the introduction of a parameter t, and deduce some algebraic properties related to their zeros, such as their equations of motion with respect [...] Read more.\nIn this contribution, we consider sequences of orthogonal polynomials associated with a perturbation of some classical weights consisting of the introduction of a parameter t, and deduce some algebraic properties related to their zeros, such as their equations of motion with respect to t. These sequences are later used to explicitly construct families of polynomials that are stable for all values of t, i.e., robust stability on these families is guaranteed. Some illustrative examples are presented. Full article\nShow Figures", null, "Figure 1\n\nOpen AccessArticle\nA Classification of Symmetric (1, 1)-Coherent Pairs of Linear Functionals\nMathematics 2019, 7(2), 213; https://doi.org/10.3390/math7020213 - 25 Feb 2019\nCited by 1 \| Viewed by 944\nAbstract\nIn this paper, we study a classification of symmetric $( 1 , 1 )$ -coherent pairs by using a symmetrization process. In particular, the positive-definite case is carefully described. Full article" ]	[ null, "https://px.ads.linkedin.com/collect/", null, "https://www.mdpi.com/img/journals/mathematics-logo.png", null, "data:image/gif;base64,R0lGODlhAQABAAD/ACwAAAAAAQABAAACADs=", null, "data:image/gif;base64,R0lGODlhAQABAAD/ACwAAAAAAQABAAACADs=", null, "data:image/gif;base64,R0lGODlhAQABAAD/ACwAAAAAAQABAAACADs=", null, "data:image/gif;base64,R0lGODlhAQABAAD/ACwAAAAAAQABAAACADs=", null, "data:image/gif;base64,R0lGODlhAQABAAD/ACwAAAAAAQABAAACADs=", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8638624,"math_prob":0.8485966,"size":4550,"snap":"2021-21-2021-25","text_gpt3_token_len":866,"char_repetition_ratio":0.12934448,"word_repetition_ratio":0.14146341,"special_character_ratio":0.1698901,"punctuation_ratio":0.12428571,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98665345,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-11T05:14:52Z\",\"WARC-Record-ID\":\"<urn:uuid:e0ad1285-97de-4714-9f4a-e09899eae848>\",\"Content-Length\":\"280511\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cdcb96ec-5b2e-4599-ae35-7f0d626aa03d>\",\"WARC-Concurrent-To\":\"<urn:uuid:17302ce1-12be-4622-8460-471a1a1f4bf6>\",\"WARC-IP-Address\":\"104.18.25.151\",\"WARC-Target-URI\":\"https://www.mdpi.com/journal/mathematics/special_issues/Recent_Trends_Orthogonal_Polynomials_Approximation_Theory_Applications\",\"WARC-Payload-Digest\":\"sha1:47DWUXHGOAQT4PQP3JJ27UVJMMCJLAS3\",\"WARC-Block-Digest\":\"sha1:6BONU7RQ7SUUAZ5J6TLWA4436CRE4IHB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991641.5_warc_CC-MAIN-20210511025739-20210511055739-00045.warc.gz\"}"}
http://www.math-math.com/2015/06/infinite-series-explained-in-5-minutes.html	[ "### Infinite Series Explained in 5 Minutes\n\nInfinite Series Explained in 5 Minutes\n\nAn infinite series is just a sum of terms that never stops, for example..\n\ns=1+(1/2)+(1/4)+(1/8)+(1/16)+..\n\nThis can be written more compactly as s=sum(1/2^n) for n=0 to infinity.\n\nIf the sum of a series is a finite number the series is said to converge, otherwise it is said to diverge. The example given above is a convergent series and in fact s=2.\n\nIt's not always easy to tell if a series converges. For example, the series s=1+(1/2)+(1/3)+(1/4)+(1/5)+.. looks like it converges but in fact it's a divergent series and s is not a finite number.\n\nInfinite series are used a lot in mathematics because many functions can be written as infinite series, for example..\n\n1/(1-x)=1+x+x^2+x^3+x^4..\n\ne^x=1+x+(x^2/2!)+(x^3/3!)+(x^4/4!).. where n! is the factorial function n!=123...n\n\ncos(x)=1-(x^2/2!)+(x^4/4!)-(x^6/6!)..\n\nSeries can be used to prove things about a function that might otherwise be difficult to prove. For example, if you differentiate the series for e^x term by term you get the same series! So this is a neat proof of the famous fact that e^x is its own derivative, d(e^x)/dx=e^x\n\nContent written and posted by Ken Abbott [email protected]" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9078539,"math_prob":0.9914343,"size":1194,"snap":"2020-34-2020-40","text_gpt3_token_len":364,"char_repetition_ratio":0.14033614,"word_repetition_ratio":0.0,"special_character_ratio":0.31825796,"punctuation_ratio":0.13131313,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997433,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-30T06:51:35Z\",\"WARC-Record-ID\":\"<urn:uuid:67110e88-6ab1-490d-b301-6be8c73e4f8f>\",\"Content-Length\":\"29796\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c4746548-4fb8-4542-ada2-048b054442b4>\",\"WARC-Concurrent-To\":\"<urn:uuid:c9887e8a-3b3a-4580-ad6e-908e16c4a977>\",\"WARC-IP-Address\":\"172.217.7.147\",\"WARC-Target-URI\":\"http://www.math-math.com/2015/06/infinite-series-explained-in-5-minutes.html\",\"WARC-Payload-Digest\":\"sha1:MA37RKSADJJOGDJFZQSEVSEEUTVAVPEO\",\"WARC-Block-Digest\":\"sha1:N2C2B5IG36ZBP7JQ433XFLYAZEX22UBJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600402118004.92_warc_CC-MAIN-20200930044533-20200930074533-00339.warc.gz\"}"}
https://answers.everydaycalculation.com/subtract-fractions/3-5-minus-7-6	[ "Solutions by everydaycalculation.com\n\n## Subtract 7/6 from 3/5\n\n1st number: 3/5, 2nd number: 1 1/6\n\n3/5 - 7/6 is -17/30.\n\n#### Steps for subtracting fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 5 and 6 is 30\n\nNext, find the equivalent fraction of both fractional numbers with denominator 30\n2. For the 1st fraction, since 5 × 6 = 30,\n3/5 = 3 × 6/5 × 6 = 18/30\n3. Likewise, for the 2nd fraction, since 6 × 5 = 30,\n7/6 = 7 × 5/6 × 5 = 35/30\n4. Subtract the two like fractions:\n18/30 - 35/30 = 18 - 35/30 = -17/30\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 ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7436362,"math_prob":0.9987261,"size":394,"snap":"2022-40-2023-06","text_gpt3_token_len":206,"char_repetition_ratio":0.20769231,"word_repetition_ratio":0.0,"special_character_ratio":0.5431472,"punctuation_ratio":0.06896552,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9985101,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-07T02:25:41Z\",\"WARC-Record-ID\":\"<urn:uuid:7ca750ff-3147-4826-9a5b-f708c20f4ae3>\",\"Content-Length\":\"8567\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:824664c4-9a0c-46f0-a182-5b80fdd8fc89>\",\"WARC-Concurrent-To\":\"<urn:uuid:cd5bfb43-16e6-4a2a-9fd0-88a91dd42bf9>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/subtract-fractions/3-5-minus-7-6\",\"WARC-Payload-Digest\":\"sha1:APZTROPJU7JE5U4AFNO3ZIZWAWE2GOFC\",\"WARC-Block-Digest\":\"sha1:FNI77H3THIPRCEN4QLQENRQGYYDMXMSR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500368.7_warc_CC-MAIN-20230207004322-20230207034322-00871.warc.gz\"}"}
https://www.pdf-archive.com/2020/02/28/a-b-basset-on-the-motion-of-a-sphere-in-a-viscous-liquid/preview/page/1/	[ "# PDF Archive\n\nEasily share your PDF documents with your contacts, on the Web and Social Networks.\n\n## A. B. Basset. On the motion of a sphere in a viscous liquid.pdf", null, "Page 12321\n\n#### Text preview\n\nIII. On\n\ntheMotion o f a Sphere in a Viscous\nB y A. B. B a sset ,\n\nCommunicated by Lord R a y l e ig h , D.C.L., Sec,\n\n1. T h e first problem relating to the motion of a solid body in a viscous liquid which\nwas successfully attacked was th a t of a sphere, the solution of which was given by\nProfessor S to k es in 1850, in his memoir “ On the Effect of the Internal Friction of\nFluids on Pendulum s,” ‘ Cambridge Phil. Soc. T rans./ vol. 9, in the following cases:\n(i.) when the sphere is performing small oscillations along a straig h t line ; (ii.) when\nth e sphere is constrained to move w ith uniform velocity in a straig h t line ; (iii.)\nwhen th e sphere is surrounded by an infinite liquid and constrained to rotate with\nuniform angular velocity about a fixed diam eter : it being supposed, in the last two\ncases, th a t sufficient time has elapsed for the motion to have become steady. In the\nsame memoir he also discusses the motion of a cylinder and a disc. The same class\nof problems has also been considered by M e y e r * and O b e r b e c k ,! the latter of whom\nhas obtained th e solution in the case of the steady motion of an ellipsoid, which\nmoves parallel to any one of its principal axes with uniform velocity. The torsional\noscillations about a fixed diameter, of a sphere which is either filled w ith liquid or is\nsurrounded by an infinite liquid when slipping takes place a t the surface of the sphere,\nforms th e subject of a joint memoir by H elm holtz and P io t r o w s k i . \|\nVery little appears to have been effected with regard to the solution of problems\nin which a viscous liquid is set in motion in any given m anner and then left to itself.\nThe solution, when the liquid is bounded by a plane which moves parallel to itself, is\ngiven by Professor S tokes a t the end of his memoir referred to above; and the solu\ntions of certain problems of two-dimensional motion have been given by S t e a r n .§\nIn th e present paper I propose to obtain the solution for a sphere moving in a viscous\nliquid in the following cases :— (i.) when the sphere is moving in a straight line under\nth e action of a constant force, such as gravity ; (ii.) when the sphere is surrounded by\nviscous liquid and is set in rotation about a fixed diam eter and then left to itself.§\n\nt\nJ\n§\n\n‘ Crelle, Journ. M ath.,’ vol. 73, p. 31.\n‘ Crelle, Joui'ii. M ath.,’ vol. 81, p. 62.*\n‘ Wissenscliaftl. Abhandl.,’ vol. 1, p. 172.\n‘ Quart. Journ. M ath.,’ vol. 17, p. 90.\nG\n\n2\n\n28.5.88" ]	[ null, "https://www.pdf-archive.com/2020/02/28/a-b-basset-on-the-motion-of-a-sphere-in-a-viscous-liquid/preview-a-b-basset-on-the-motion-of-a-sphere-in-a-viscous-liquid-1.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92606896,"math_prob":0.9103289,"size":2603,"snap":"2020-24-2020-29","text_gpt3_token_len":716,"char_repetition_ratio":0.122739516,"word_repetition_ratio":0.026465029,"special_character_ratio":0.26507875,"punctuation_ratio":0.14473684,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9727157,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-03T16:37:57Z\",\"WARC-Record-ID\":\"<urn:uuid:91fd37f8-c63a-4c44-94d4-ca646d1f77b0>\",\"Content-Length\":\"18450\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5a47629b-f675-4133-8fb9-6f74e2ea6271>\",\"WARC-Concurrent-To\":\"<urn:uuid:b3cbc151-fa56-4d36-bc4c-21ccac2c789a>\",\"WARC-IP-Address\":\"51.159.91.99\",\"WARC-Target-URI\":\"https://www.pdf-archive.com/2020/02/28/a-b-basset-on-the-motion-of-a-sphere-in-a-viscous-liquid/preview/page/1/\",\"WARC-Payload-Digest\":\"sha1:GUX3722ALJK2ZBK6MCFX27A6G2SIXH7G\",\"WARC-Block-Digest\":\"sha1:4Z7Q36SYKQKEOKBWMZT5BAUBPKLHS4C4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655882634.5_warc_CC-MAIN-20200703153451-20200703183451-00547.warc.gz\"}"}
https://gmplib.org/list-archives/gmp-devel/2011-December/002082.html	[ "# Binomial improvements\n\nbodrato at mail.dm.unipi.it bodrato at mail.dm.unipi.it\nSun Dec 11 11:24:52 CET 2011\n\n```Ciao!\n\nTorbjorn Granlund <tg at gmplib.org> writes:\n> (1) Now the number of allowable factors that fit into a limb is tabled.\n> This table takes into account that the mulN functions now shift out\n> twos.\n\nThere is one strength in the code I wrote for binomial: code recycling.\n\nI think that this kind of optimisation is very interesting, and it should\nbe applied also to the basecase factorial.\n\nFor example: my basecase code to compute the odd component of the\nfactorial does not \"shift out twos\", it uses the following trick.\n\nLet n be an integer. And let oddfac(n)= n! >> (n-popcount(n)) be the odd\nfactor of the factorial of n. Then oddfac(n)=oddfac(floor(n/2))(n-n%2)!!\n.\n\nI.e. we can compute the product of odd numbers from 3 to n, times the odd\nnumbers from 3 to n>>1, times... till we reach a tabled oddfac().\n\nMaybe this is a good trick, maybe it is not, but my proposal is: let's\nstart optimising the factorial, then we will recycle the code for the\nbinomial.\n\nBoth my factorial and my binomial code use a function (you can find it in\nthe current fac_ui.c in the repo) taking a vector of limbs, and obtaining\nthe product of them all: it is currently called mpz_prodlimbs(res, vec,\nlen). Again, is it good or is it not? Please look at it and comment. If it\nis good, let's use it for the binomial too. If it is not, let's change it\nalso for the factorial...\n\n> (2) Less scratch space usage, for this needs the latest repo bdiv\n> updates (allowing dividend/quotient operand overlap).\n\nThis is a nice improvement.\n\n> I suspect that one remaining optimisation for the smallest operands is\n> to get rid of the final lshift. By being cleverer with 2 removal from\n> the dividend, that might be possible.\n\nIt is! Also the factorial needs this optimisation.\n\n> Forgot to point to the updated code: See http://gmplib.org/devel/ under\n> New binomial code.\n\nI can not find it...\n\nRegards,\nMarco\n\n--\nhttp://bodrato.it/software/combinatorics.html\n\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.85000634,"math_prob":0.8321095,"size":2084,"snap":"2021-43-2021-49","text_gpt3_token_len":557,"char_repetition_ratio":0.10625,"word_repetition_ratio":0.0056338026,"special_character_ratio":0.24952015,"punctuation_ratio":0.1611479,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97588116,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-04T23:36:41Z\",\"WARC-Record-ID\":\"<urn:uuid:ffe7d06e-e882-4528-87d8-2a5fb51a8024>\",\"Content-Length\":\"4827\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4eab5a10-e3a4-4bb4-a564-f3ba035e87a6>\",\"WARC-Concurrent-To\":\"<urn:uuid:fcef2d59-7557-48d5-af9f-4cc7b6eade6e>\",\"WARC-IP-Address\":\"130.242.124.102\",\"WARC-Target-URI\":\"https://gmplib.org/list-archives/gmp-devel/2011-December/002082.html\",\"WARC-Payload-Digest\":\"sha1:23SYWR65AAS56WMSY5OPZ2HK4TNAUBQN\",\"WARC-Block-Digest\":\"sha1:YUYQ42C3YITDGX6SVPVEFEVU7CZZPUDZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363125.46_warc_CC-MAIN-20211204215252-20211205005252-00627.warc.gz\"}"}
https://electronics.stackexchange.com/questions/417534/how-does-an-op-amp-integrator-work	[ "# How does an op amp integrator work?\n\nI know there are at least two questions related to this on stackoverflow but neither really answer my question, and in any case, both questions got downvoted. What I am after is an operational understanding of how an op-amp integrator works. I know how a simple RC circuit can integrate, what I don't understand is how the feedback loop in an op-amp configuration helps. I understand how feedback works in a noninverting amplifier. I took the figure below from www.electronics-tutorials.ws. This web site has an explanation but I don't follow it. My understanding so far is this:\n\n1. Apply a positive voltage to input vin. Current flows through Rin resulting initially in a non-zero voltage at X (Correct?).\n\n2. Due to the high impedance of the op-amp at X, we can assume that all the current then flows to the capacitor (initial discharged).\n\n3. The capacitor starts to charge resulting in a voltage across the capacitor.\n\n4. The difference in voltage at the two op-amp inputs (the positive input is at zero, hence the difference is negative) resulting in the output, vout, going negative (we assume that vout was zero initially).\n\nMy question is what happens next? How does the feedback act to bring the difference between the two inputs back to zero? Or have I got this wrong?\n\nI am very familiar with the proofs for showing that the configuration will integrate but they don't give any real intuition and many videos, wikpedia, and books but almost all regurgitate the proof without giving much insight. I'm after an intuitive understanding, not a mathematical proof.", null, "Out of interest I also redrew the op-amp circuit next to the RC integrator shown below which gives the suggestion that the op-amp is amplifying the small vollage across C (assuming high R1) while having high impedance from the resistor/capacitor node. Not sure if that is a legitimate way to look at it.", null, "• Put in a square wave, the opamp strives to hold the (-) pin very close to zero volts, which imposes a constant voltage across the input resistor. This constant current has to go somewhere, and the only path is into the capacitor. Give constant current into a capacitor, you get a perfect RAMP output, which is the integral. Jan 18, 2019 at 3:04\n• To me - this small comment (if compared with all the answers) is the best explanation of the circuits behaviour in the time domain.\n– LvW\nJan 18, 2019 at 14:57\n• But it doesn't explain why the op-amp strives to hold the (-) pin close to zero. An op-amp on its own won't do that, it's the combination with the feedback that makes that happen. This is the bit I wasn't sure about. Jan 18, 2019 at 18:47\n• The RC integrator is not a real integrator. It can be treated as one in a limited sense. But consider a charged capacitor and an input voltage of 0 V. According to how integrations work, the integral should not change. However, in this case the capacitor discharges => Your integral value goes to 0. This can also be seen in the Laplace transform of the two circuits: An integrator: H(s) = constant * 1/s. Your RC-circuit (essentially a lowpass): H(s) = constant * 1 / (1+RC * s). For high frequencies (!) 1 + RC * s can be assumed as the same as RC* s. In this case it is acting as an integrator\n– GNA\nMay 27, 2020 at 6:52\n\nThis may help:\n\n• Remember that when current flows into the RC junction of your op-amp that the voltage at that point will tend to rise.\n• If the inverting input voltage rises the slightest bit above the non-inverting input voltage then the op-amp output will start to swing negative.\n• The output swinging negative will, through the capacitor1, tend to pull the inverting input down towards zero again where it stabilise (for the moment).\n\nThe result is that feeding current into the RC node causes the op-amp output to go negative.\n\nOut of interest I also redrew the op-amp circuit next to the RC integrator shown below which gives the suggestion that the op-amp is amplifying the small voltage across C (assuming high R1) while having high impedance from the resistor/capacitor node. Not sure if that is a legitimate way to look at it.\n\nThat's correct. It might be better than you think. The simple RC circuit has the advantage that it's non-inverting but the disadvantage that it's non-linear. With a constant input voltage the output will be an exponential charge curve.\n\nPutting the op-amp in as you have shown still allows the capacitor to charge up but maintains the top terminal at virtual ground. The advantage is a linear change in output. The disadvantage is that there is a minus sign on the integral obtained.\n\n1 You can think of a capacitor as holding the voltage across it as a constant in the short term. That means that if the voltage on one side is changed the voltage on the other side will try to change by the same amount.\n\nOne question. what is the orientation of the capacitor in terms of conventional current? i.e. if vin goes positive the capacitor is I assume negative on its right-hand side (nearest vout). Now vout goes negative and therefore reduces the voltage across the capacitor until the potential at X is zero?\n\nI think your understanding is correct.\n\nIf Vin goes positive then current flows into the X node charging up C. (Remember the op-amp's voltage hasn't changed yet.) This tends to increase the voltage on the inverting input and that causes the output voltage to decrease. This draws some charge from the right hand side of C. Now the inverting input is pulled back down to zero volts but there is charge on C so there is a voltage across it. Since the conventional current flowed to the right there is a negative voltage remaining on the capacitor.\n\n• I think this the kind of answer I was looking for. One question. what is the orientation of the capacitor in terms of conventional current? ie if vin goes positive the capacitor is I assume negative on its right-hand side (nearest vout). Now vout goes negative and therefore reduces the voltage across the capacitor until the potential at X is zero? Jan 17, 2019 at 22:44\n• See the update. Jan 17, 2019 at 23:15\n\nThe op-amp is going to try its best to keep the voltage between it's plus and minus input the same. In an ideal op-amp, no current flows into the inputs, so the only way that it can do that is by changing its output voltage.\n\nIn the schematic below, $$\\v_+ = 0\\mathrm{V}\\$$. That means that the op-amp will try to hold $$\\v_-\\$$ at zero, also.\n\nWhatever voltage is generated by V2 gets turned into a current by R1. Because $$\\v_-\\$$ is being held at $$\\0\\mathrm{V}\\$$, that same current has to flow in C1. And because $$\\v_-\\$$ is being held at $$\\0\\mathrm{V}\\$$, the op-amp has to drive the output voltage such that the current in C1 matches the current in R1.\n\nSo if $$\\v_2\\$$ is constant, then the current into the node around the negative input is constant, which means that the current out of that node from the cap must be constant -- and that can only happen if the output voltage is falling at a constant rate. The end result is that the op-amp integrates the input voltage into the output voltage.\n\nMore complicated voltages at $$\\v_2\\$$ cause more complicated behavior, but the op-amp is always going to be trying to drive $$\\v_-\\$$ to $$\\0\\mathrm{V}\\$$. It can only do that by satisfying $$\\ \\frac{d}{dt} C_1 v_{out} + \\frac{v_2}{R_1} = 0 \\$$. If you solve that differential equation, it says that $$v_{out} = -\\frac{1}{R_1 C_1} \\int v_2 dt$$\n\nHTH", null, "simulate this circuit – Schematic created using CircuitLab\n\n• In your comment \"The op-amp is going to try its best to keep the voltage between it's plus and minus input the same.\", strictly speaking, it's the op-amp combined with the next feedback that does this. This is the bit I was stuck on, how the feedback coupled with the op-amp manages to maintain the difference close to zero. Jan 18, 2019 at 18:50\n• Well, I'm glad you worked through it! These are difficult concepts and hard to simplify. Jan 18, 2019 at 18:57\n\nRhody - have you heard about the MILLER effect? Well - the shown circuit is called \"MILLER integrator\" because the MILLER effect is exploited. Remember: This effect reduces the feedback impedance between an amplifier output (for example: collector) and the inverting input (example: base node of the transistor). And the factor of increase is the gain.\n\nHere, we have the same principle. Hence, there will be a very small capacitive impedance (that means: A very large capacitor) between input and output of the opamp. And the factor of increase is the open-loop gain Aol of the opamp.\n\nHence, you can make a comparison with a simple RC circuit. However, because of the very large capacitor the cut-off frequency is very low (nearly DC).\n\nFrequency domain: The transfer function between the opamps inverting node and the signal input is\n\nHo(s)=1/(1+sCo * R) with Co=Aol * C (MILLER effect).\n\nBecause of the very large value Aol, we can neglect the \"1\" in the denominator and arrive at\n\nHo(s)=1/(sC * Aol R)\n\nWe are lucky and can use the low resistive opamp output (and multiply the function Ho(s) with the gain -Aol) and arrive at the final result (opamp output-to-signal input):\n\nH(s)=Ho(s) (-Aol) = - 1/sR*C (Transfer function of an ideal integrator)\n\nThe inputs of the opamp don't take input current and the opamp will keep its input voltage equal as long as it is wired for negative feedback.\n\nSo effectively, the current that the input sinks against the 0V it sees goes straight into charging the capacitor. Usually when you charge a capacitor through a resistor, the charge building up on the capacitor reduces the voltage across the resistor and thus also the charging current, leading to an exponential decay of the charge current.\n\nHere, however, the opamp output actively adjusts the voltage at the other side of the capacitor so that the resistor never gets to see the difference it makes, thus keeping the current through resistor (and into capacitor) independent from the charge across the capacitor.\n\nIt's like Útgarða-Loki handing Þorr the mouth of the ocean as a drinking horn and Þorr does not actually find himself able to drain it, not noticing that he causes the tides with his attempt.\n\nThe Passive cap. integrator current decays with voltage as it approaches the input.\n\nThe active cap. integrator saturates after some time if Vin ≠ 0 because the output voltage drives current into Vin- to maintain 0V diff. until the output saturates at the supply rail.\n\nSo input offset is critical and you need an analog switch to discharge and initialize to 0V output.\n\n## anecdotal\n\nI remember knowing nothing about this in 1st yr Eng and my once famous brother-in-law medical Dr of Anesthesia, Intensive Care and open heart surgery, gave me a tour around the hospital and said he needed an integrator to measure O2 content in blood for the brain, after a heart attack victim stops, to know the best treatment ( such as hypothermia )to give when no response to defib. and meds with the likelihood of success. I had no idea ! and was embarrassed to not know! (circa '75) Don't be. Just research it.\n\nIt is a great challenge to find a new explanation for such a legendary circuit because everyone knows what an op-amp integrator is. But to know a specific circuit solution does not mean that you really understand it. To (deeply) understand a circuit means something more - to see the general idea behind it that links many specific circuit implementations (op-amp, BJT, FET, tube…) You can see it even in life in the form of many non-electrical applications...\n\n1. Op-amp inverting integrator. The idea behind this circuit solution is extremely simple and intuitive. It may sound paradoxical... but to see it you only need to remove the symbol of the ground from the circuit diagram. As you can see in Fig. 1, I have only labeled the place of the virtual ground (1) and the place of the real ground (2)... and I have no longer used these names. You understand that there is no virtual ground because there is no real ground. But if you still miss the virtual ground, then you can talk about a virtual short between node 1 and 2.", null, "Fig. 1. Op-amp inverting integrator (only the negative power supply V- is explicitly shown)\n\nThe current path is crucial here to see the great idea. Since the input voltage is positive, the op-amp output voltage is negative and the current enters the op-amp output... then passes through the negative power supply V- and returns to the input source. The positive source V+ is not essential in this case; so it is only hinted.\n\n2. Electric equivalent circuit. The main question to be answered is, \"What does the op-amp do here?\" You know that it keeps almost zero voltage between its inputs so its output voltage is always equal to the voltage drop across the capacitor. So the op-amp output serves as a following voltage source. Then let's replace the op-amp with a variable voltage source VOA to simplify this electronic circuit - Fig. 2. By the way, I conducted such a real experiment in 2001 with my students in the laboratory when we used a capacitor with high capacity and zero indicator connected between 1 and 2.", null, "Fig. 2. Electric equivalent circuit\n\nThis simple trick is enough to show the great idea behind the circuit. The voltage source VOA is connected in series to the capacitor C so that its voltage compensates the voltage drop VC across the capacitor and the voltage between the two nodes 1 and 2 is (almost) zero. So the conclusion is:\n\nThe op-amp in the circuit of the inverting amplifier compensates the voltage drop VC across the capacitor by adding equivalent voltage VOA = VC in series.\n\nSo, the key point of this explanation is adding, not amplifying. To think of the amplifier in a negative feedback circuit not as of an amplifier but rather as of something like integrator is a powerful technique for intuitive understanding and explaining such op-amp circuits. Indeed, here it seems a little strange (integrator inside integrator)... but works...\n\nHow simple is this \"magic recipe\"... You want to make the imperfect RC integrator perfect? Then connect a small variable \"battery\" with voltage VC in series to the capacitor and (the next brilliant idea) take its inverted \"copy\" voltage as an output. The load will consume current from this \"helping\" source... not from the input source (i.e., this is a buffered output).\n\nThe power of this intuitive explanation is that we can explain this sophisticated op-amp circuit to a \"six year old\" (Einstein)... and that will mean we understand it ourselves...\n\n3. Virtual short. The total voltage across the network of two elements in series - a capacitor C and compensating voltage source (VOUT), is always zero. So this network behaves as a \"piece of wire\" that shorts the points 1 and 2 - Fig. 3. This is what the input source \"sees\" when \"looking\" through the resistor R at the op-amp input.", null, "Fig. 3. Equivalent circuit of the output part on the right\n\nFiguratively speaking, the op-amp output acts as a \"negative capacitor\". While the \"positive capacitor\" C subtracts its voltage VC from the input voltage source, the op-amp \"negative capacitor\" adds its voltage VOUT to the input voltage." ]	[ null, "https://i.stack.imgur.com/OavFL.gif", null, "https://i.stack.imgur.com/6hGZ6.png", null, "https://i.stack.imgur.com/fyXiP.png", null, "https://i.stack.imgur.com/nxvJ3.jpg", null, "https://i.stack.imgur.com/nLnML.jpg", null, "https://i.stack.imgur.com/E1YlB.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.905164,"math_prob":0.9574024,"size":4111,"snap":"2022-27-2022-33","text_gpt3_token_len":901,"char_repetition_ratio":0.1448746,"word_repetition_ratio":0.017045455,"special_character_ratio":0.21770859,"punctuation_ratio":0.11439114,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9918239,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,7,null,7,null,7,null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-18T11:19:10Z\",\"WARC-Record-ID\":\"<urn:uuid:c5c9bdc7-8fbe-4c95-8618-9ffdf8505b32>\",\"Content-Length\":\"290434\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:30ff5f25-f570-48e6-ad74-1e2538f293d3>\",\"WARC-Concurrent-To\":\"<urn:uuid:436f8c5f-253b-4a65-a596-ead2be662f80>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://electronics.stackexchange.com/questions/417534/how-does-an-op-amp-integrator-work\",\"WARC-Payload-Digest\":\"sha1:NMSHZ5VFEFGXJSXYQ57TWQTMM3G7DD53\",\"WARC-Block-Digest\":\"sha1:TE6AFQE6PZYFZC5HCDU6DSA7JW2U25N2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882573193.35_warc_CC-MAIN-20220818094131-20220818124131-00340.warc.gz\"}"}
https://mathoverflow.net/questions/218994/svd-vs-fourier-analysis-for-data	[ "# SVD vs Fourier analysis for data.\n\nFourier analysis is useful for analysis in the frequency domain. SVD on the other hand is useful for analysis of data, and expressing noise in the data. I have a problem that needs extensive data analysis, it is in the area medicine. This could be generalized to other problems.\n\nThe problem is that of gene expression, in case of long term gene mutation. Using Fourier analysis we can get a time series analysis of the genes(and thereby get noisy gene expression), and as time progresses, the changes in a particular organ. On the other hand, we could use Singular Value Decomposition, and the noisy gene expresses itself. This, is just an outline of the problem. Both SVD, and Fourier lend themselves to solve the problem that of expressing noisy genes. Is there any comparison of the two techniques, why one would be preferred over another qualitatively, or references that one can use for the problem of gene expression, thanks in anticipation.\n\n• Fourier analysis approximates a continuous function using trigonometric polynomials; SVD approximates a matrix using eigenvectors of its left (or right) singular matrix. These are a priori completely different problems, and your question doesn't make much sense unless you can indicate why you think these two problems are related. Sep 23, 2015 at 2:35\n• You might get more feedback on sites like scicomp.stackexchange.com, stats.stackexchange.com, or dsp.stackexchange.com. But if you ask your question in its current form, they might also close it, because it is not really clear what you want to do exactly. For my answer, I just guessed that stochastic processes might be your context, because both Fourier analysis and \"optimal\" decompositions make sense in that context. Sep 23, 2015 at 20:00\n• @PaulSiegel I guess the OP has in mind that the discrete Fourier transform transforms discrete \"spatial\" data to discrete \"frequency\" data. I would say that the SVD decomposes every matrix $A$ as $U^TDV$ with a diagonal $D$ and orthonormal $U$ and $V$ while the diagonal Fourier transform $F$ is also orthonormal and gives $C = F^HDF$ with diagonal $D$ for circulant matrices $C$.\n– Dirk\nSep 24, 2015 at 7:12" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90986097,"math_prob":0.9203845,"size":947,"snap":"2023-40-2023-50","text_gpt3_token_len":190,"char_repetition_ratio":0.1516437,"word_repetition_ratio":0.0,"special_character_ratio":0.19429778,"punctuation_ratio":0.11956522,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9925727,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-08T21:49:59Z\",\"WARC-Record-ID\":\"<urn:uuid:e5137d4b-cd62-41b1-a6aa-8f6f4293fde3>\",\"Content-Length\":\"123742\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:092d408c-55b3-4562-a928-ecb4a22798fb>\",\"WARC-Concurrent-To\":\"<urn:uuid:a2e779f4-3a99-46ec-bcd4-a9d575a7e03f>\",\"WARC-IP-Address\":\"104.18.37.74\",\"WARC-Target-URI\":\"https://mathoverflow.net/questions/218994/svd-vs-fourier-analysis-for-data\",\"WARC-Payload-Digest\":\"sha1:5MABR5XIDI5TNDOPQXBDR3AZNFX5XB5N\",\"WARC-Block-Digest\":\"sha1:CQPVYFJAEWXFAHUR6X2CU6THY2Q2ECJF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100779.51_warc_CC-MAIN-20231208212357-20231209002357-00734.warc.gz\"}"}
https://discuss.pytorch.org/t/is-there-0-5-0version-of-pytorch-i-cannot-find-it-on-the-internet/42169	[ "", null, "# Is there 0.5.0version of pytorch?I cannot find it on the Internet\n\nwhen I tried to use Dataparallel, I came across the problem\" Arguments are located on different GPUs\",I found that I should use the 0.5.0 version to solve the problem on the forum.\nHowever ,I have been trying to find the correct version for a long time with no result.Can anyone help me ?\n(ps, I tried to run the code on pytorch1.0, but error came in the dataloader part “RandomSampler’ object has no attribute 'replacement”.and I found no solution.\nThank you !!\n\n1 Like\n\nAs far as i know there is no pytorch 0.5\n\nAnyway that error means you have (probably) hardcoded model’s device inside the model itself\n\nI am sorry that I didn`t understand your answer.Do you mean that, after I use dataparallel to my model,I shouldn`t use something like to.(device) inside the model?\nHere is my error and code.I would appreciate it if you can have a look at it!\n\nHi,\nI meant something like that yes. The problem is that when you set the model, dataparallel clones the model in each gpu dataparallel has access to.\n\nLater on, if you hardcode a gpu in the forward function like x=x.to(‘cuda:0’), pytorch will allocate tensors in the wrong gpu.\n\nAnyway this doesn’t look to be your case. It seems that your input is not allocated on the same gpu than your model. If you are not using dataparallel this probably means that you are allocating input and model on different gpus like\n`model = model.cuda()` and `input_var = input_var.cuda(1)` if you are using dataparallel it’s more strange as the module should allocate inputs automatically (and you are not hardcoding it inside forward function).\n\nCould you post how are you instantiating the model and applying dataparallel?\n\n1 Like\n\nThat vey kind of you .I found that when I set the device I made a very stupid mistake.Now the code is running. I will see how it goes.\nThank you very much!!\n\nhi, I found your toy code solution for the dataparallel problem.Your work is fantastic.\nBut when I immitated it on my own code, things went wrong.\nit gave me `RuntimeError: all tensors must be on devices`\nHere is my Model_with_parallel:\n\n``````class G_FullModel(nn.Module):\ndef __init__(self, actor, discriminator,g_loss):\nsuper(G_FullModel, self).__init__()\nself.G = actor\nself.D = discriminator\nself.loss = g_loss\n\ndef forward(self, targets, inputs, corpus):\nfake_reply, word_probabilities, hiddens = G.sample(inputs, targets, TF=0)\nnum_samples = 3\nrewards = G.monte_carlo(D, inputs, fake_reply, hiddens, num_samples,\ncorpus).detach()\n\npg_loss = loss(rewards, word_probabilities)\n``````\n\nHere is how I used it:\n\n``````G_model_parallel = G_FullModel(actor, discriminator, g_loss)\nG_model_parallel = torch.nn.DataParallel(G_model_parallel, device_ids=[1, 2, 3],output_device=).cuda()\n\ncontext = context.to(device)\nloss, _ = G_model_parallel(reply, context, corpus)\n``````\n\nI have tried to dataparallel my model and loss partly, the code could run.But still the GPU-Util on other gpus except device1 is almostly zero.\nthank you very much!!\n\nHi, looks like here\n`torch.nn.DataParallel(G_model_parallel, device_ids=[1, 2, 3],output_device=).cuda()`\nYou are allocating the main device in cuda 0\nTo make it work you should set the same device for both output_device and main data parallel gpu\n\n``````torch.nn.DataParallel(G_model_parallel, device_ids=[1, 2, 3],output_device=).cuda(1)\n``````\n\nthankyou very much! I will try and see how it works!\n\nHi，thanks for your help. sorry to bother you again.\nA new problem came in my model, could you tell me to print something in the code to find where the bug exactly was?\nthe error was :\nCuDNN error: CUDNN_STATUS_EXECUTION_FAILED\n\nI was using dataparallel, when it went to compute reward,\n\n`````` def forward(self, targets, inputs, corpus):\nfake_reply, word_probabilities, hiddens = self.G.sample(inputs, targets, TF=0)\nnum_samples = 3\nrewards = self.G.monte_carlo(self.D, inputs, fake_reply, hiddens, num_samples, corpus).detach()\ndef monte_carlo(self, dis, context, reply, hiddens, num_samples, corpus):\n\n# Initialize sample\nvocab_size = self.decoder.output_size\nencoder_output, _ = self.encoder(context)\nrewards = torch.zeros(self.max_len, num_samples, batch_size)\nfunction = F.log_softmax\nfor t in range(self.max_len):\nfor n in range(num_samples):\nhidden = hiddens[t]\n# Pass through decoder and sample from resulting vocab distribution\nfor next_t in range(t + 1, self.max_len):\ndecoder_output, hidden, step_attn = self.decoder.forward_step(output.reshape(-1, 1).long(), hidden, encoder_output, function=function)\ndef forward_step(self, input_var, hidden, encoder_outputs, function):\nbatch_size = input_var.size(0)\noutput_size = input_var.size(1)\nembedded = self.embedding(input_var)\nembedded = self.input_dropout(embedded)\nself.rnn.flatten_parameters()\noutput, hidden = self.rnn(embedded, hidden)\n``````\n\nHere is other relevant code with forward_step:\n\n``````self.embedding = nn.Embedding(self.output_size, self.hidden_size)\nrnn_cell='gru'\nself.rnn = self.rnn_cell(hidden_size, hidden_size, n_layers, batch_first=True, dropout=dropout_p)\n``````\n\nshould I do something with the output and hidden in monte_carlo before I call the forward_step ?\nor should print them to see where they are?\nI tried printing embedded and embedded in forward_step function,they are distributed in devices(1,2,3).Then I got losted and have no way to debug,\nBeg for your help!Thanks a lot!!\n\nHi, this is such a complex error.\nDoes it work without dataparallel?\n\nYes，", null, "I googled the problem.The common answer is that put the relevant input and hidden on the right cuda . But here, everything is already in the parallel model that I shouldn’t assign them to specfic cuda.\nThen I got confused…", null, "Hmmm the think is that batches are distributed properly at the time of calling forward.\nWhen you use dataparallel a copy of the model is created on each gpu and batch is distributed among them.\n\nFor example i see that inside montecarlo you have a reward variable initialized as zeros, but that reward is not properly allocated on its corresponding gpu.\n\nThink that model parallel only allocates those variables which are defined before calling dataparallel. If you define a variable during the forward pass, you are the one in charge of allocating it in the proper gpu.\n\nIn the context of dataparallel, you have to softcode those variables allocating them into a device whose id depends on a variable or module device.\n\nIn the forward(the 1st func) func I defined, when the sample func is finished, where are the fake reply ,word_probabilities, hiddens . They are all summed and returned to device(1) or only the fake_reply are on device(1)? Should I do something before to the fake_reply, hiddens before I call the func monte_carlo?\n\nEverything computed inside forward will remain distributed among gpus. After returning forward’s output the batch is concatenated back in device 1.\n\nThe problem is that if you generate tensors inside the funcion, you have to prepare the code to make it device agnostic, such that every tensor you define inside the forward were properly allocated.\nFor example, inside forward, montecarlo, you are creating reward tensors initilized as zeros. That tensor is wrongly allocated in cpu. it requires something like `torch.zeros().to(encoder_output.device)`\n\nIn short, whatever coded inside forward is distributed\n\nthank you very much for your clear interpretation.I will think and try more.\nThat’s very kind of you!\n\nPeople talk about the version 0.4.1 here, when they write 0.5.0 (may be because 0.4.1 was after 0.4.0). I was confused too, but advice for “0.5.0” helps me when I switch to 0.4.1\n\nActually I think they’re talking of 1.0.0\n\nFor a long time it wasn’t clear if there was another release between 0.4.1 and 1.0.0 (which would have been 0.5) and this is why the master branch was set to version 0.5 although it was never officially released.\n\nthank you very much~\n\nthank you so much ,I will try the version 0.4.1:relaxed:" ]	[ null, "https://discuss.pytorch.org/uploads/default/original/2X/3/35226d9fbc661ced1c5d17e374638389178c3176.png", null, "https://discuss.pytorch.org/images/emoji/apple/zipper_mouth_face.png", null, "https://discuss.pytorch.org/images/emoji/apple/pleading_face.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87016225,"math_prob":0.7753,"size":8345,"snap":"2019-26-2019-30","text_gpt3_token_len":2021,"char_repetition_ratio":0.12396595,"word_repetition_ratio":0.0589172,"special_character_ratio":0.23463151,"punctuation_ratio":0.17657445,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9610373,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,7,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-22T21:12:26Z\",\"WARC-Record-ID\":\"<urn:uuid:ce5a0502-5ac9-4713-b686-68059d5c161a>\",\"Content-Length\":\"49674\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7bb867d8-3948-4134-a431-c8983099792c>\",\"WARC-Concurrent-To\":\"<urn:uuid:4b99cef7-db89-4882-8441-14316b194223>\",\"WARC-IP-Address\":\"159.203.172.63\",\"WARC-Target-URI\":\"https://discuss.pytorch.org/t/is-there-0-5-0version-of-pytorch-i-cannot-find-it-on-the-internet/42169\",\"WARC-Payload-Digest\":\"sha1:JPZHLIIUAVEOJCQQXZJJTPF6MOHINIAO\",\"WARC-Block-Digest\":\"sha1:TTXZQWERZGQOGGHNOH53K7BZFFNYGHPV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195528220.95_warc_CC-MAIN-20190722201122-20190722223122-00283.warc.gz\"}"}
https://www.eeeguide.com/network-model-formulation/	[ "## Network Model Formulation:\n\nFor a load flow study of a real life power system comprising a large number of buses, it is necessary to proceed systematically by first formulating the network model of the system.\n\nA power system comprises several buses which are interconnected by means of transmission lines. Power is injected into a bus from generators, while the loads are tapped from it. Of course, there may be buses with only generators and no-loads, and there may be others with only loads and no generators. Further, VAR generators may also be connected to some buses. The surplus power at some of the buses is transported via transmission lines to buses deficient in power. Figure 6.1a shows the one-line diagram of a four-bus system with generators and loads at each bus. To arrive at the network model of a power system, it is sufficiently accurate to represent a short line by a series impedance and a long line by a nominal- π model (equivalent-π may be used for very long lines). Often, line resistance may be neglected with a small loss in accuracy but a great deal of saving in computation time.\n\nFor systematic analysis, it is convenient to regard loads as negative generators and lump together the generator and load powers at the buses. Thus at the ith bus, the net complex power injected into the bus is given by", null, "where the complex power supplied by the generators is", null, "and the complex power drawn by the loads is", null, "The real and reactive powers injected into the ith bus are then", null, "Figure 6.1b shows the network model of the sample power system prepared on the above lines. The equivalent power source at each bus is represented by a shaded circle. The equivalent power source at the ith bus injects current Ji into the bus. It may be observed that the structure of a power system is such that all the sources are always connected to a common ground node.\n\nThe network model of Fig. 6.1b has been redrawn in Fig. 6.1c after lumping the shunt admittances at the buses. Besides the ground node, it has four other nodes (buses) at which the current from the sources is injected into the network. The line admittance between nodes i and k is depicted by yik = yki. Further, the mutual admittance between lines is assumed to be zero.\n\nApplying Kirchhoff’s current law (KCL) at nodes 1, 2, 3 and 4, respectively, we get the following four equations:", null, "Rearranging and writing in matrix form, we get", null, "Equation (6.3) can be recognized to be of the standard form", null, "Comparing Eqs. (6.3) and (6.4), we can write", null, "", null, "", null, "", null, "", null, "Each admittance yii (i = 1, 2, 3, 4) is called the self admittance (or driving point admittance) of node i and equals the algebraic sum of all the admittances terminating on the node. Each off-diagonal term yik (i, k = 1, 2, 3, 4) is the mutual admittance (transfer admittance) between nodes i and k and equals the negative of the sum of all admittances connected directly between these nodes. Further, yik = yki\n\nUsing index notation, Eq. (6.4) can be written in compact form as", null, "or, in matrix form", null, "where YBUS denotes the matrix of bus admittance and is known as bus admittance matrix. The dimension of the YBUS matrix is (n x n) where n is the number of buses. [The total number of nodes are m = n + 1 including the ground (reference) node.]\n\nAs seen above,YBUS is a symmetric matrix, except when phase shifting transformers are involved , so that only n(n+1)/2 terms are to be stored for an n-bus system. Furthermore, Yik = 0 if buses i and k are not connected (e.g. Y14 = 0). Since in a power network each bus is connected only to a few other buses (usually to two or three buses), the YBUS of a large network is very sparse, i.e. it has a large number of zero elements. Though this property is not evident in a small system like the sample system under consideration, in a system containing hundreds of buses, the sparsity may be as high as 90%. Tinney and associates at Bonnevile Power Authority were the first to exploit the sparsity feature of YBUS in greatly reducing numerical computations in load flow studies and in minimizing the memory required as only non-zero terms need be stored.\n\nEquation (6.6) can also be written in the form", null, "for a network of four buses (four independent nodes)", null, "Symmetric YBUS yields symmetric ZBUS S. The diagonal elements of ZBUS are called driving point impedances of the nodes, and the off-diagonal elements are called transfer impedances of the nodes. ZBUS need not be obtained by inverting YBUS. While YBUS is a sparse matrix, ZBUS is a full matrix. i.e., zero elements of YBUS become non-zero in the corresponding ZBUS elements.\n\nIt is to be stressed here that YBUS/ZBUS constitute models of the passive portions of the power network.\n\nBus admittance matrix is often used in solving load flow problem. It has gained widespread application owing to its simplicity of data preparation and the ease with which the bus admittance matrix can be formed and modified for network changes addition of lines, regulating transformers, etc. Of course, sparsity is one of its greatest advantages as it heavily reduces computer memory and time requirements. In contrast to this, the formation of a bus impedance matrix requires either matrix inversion or use of involved algorithms Furthermore, the impedance matrix is a full matrix.\n\nNote: In the sample system of Fig. 6.1, the buses are numbered in an arbitrary manner, although in more sophisticated studies of large power systems, it has been shown that certain ordering of nodes produces faster convergence and solutions." ]	[ null, "https://www.eeeguide.com/wp-content/uploads/2016/11/Network-Model-Formulation.jpg", null, "https://www.eeeguide.com/wp-content/uploads/2016/11/Network-Model-Formulation-1.jpg", null, "https://www.eeeguide.com/wp-content/uploads/2016/11/Network-Model-Formulation-2.jpg", null, "https://www.eeeguide.com/wp-content/uploads/2016/11/Network-Model-Formulation-3.jpg", null, "https://www.eeeguide.com/wp-content/uploads/2016/11/Network-Model-Formulation-4.jpg", null, "https://www.eeeguide.com/wp-content/uploads/2016/11/Network-Model-Formulation-5.jpg", null, "https://www.eeeguide.com/wp-content/uploads/2016/11/Network-Model-Formulation-6.jpg", null, "https://www.eeeguide.com/wp-content/uploads/2016/11/Network-Model-Formulation-7.jpg", null, "https://www.eeeguide.com/wp-content/uploads/2016/11/Network-Model-Formulation-8.jpg", null, "https://www.eeeguide.com/wp-content/uploads/2016/11/Network-Model-Formulation-9.jpg", null, "https://www.eeeguide.com/wp-content/uploads/2016/11/Network-Model-Formulation-10.jpg", null, "https://www.eeeguide.com/wp-content/uploads/2016/11/Network-Model-Formulation-11.jpg", null, "https://www.eeeguide.com/wp-content/uploads/2016/11/Network-Model-Formulation-12.jpg", null, "https://www.eeeguide.com/wp-content/uploads/2016/11/Network-Model-Formulation-13.jpg", null, "https://www.eeeguide.com/wp-content/uploads/2016/11/Network-Model-Formulation-14.jpg", null, "https://www.eeeguide.com/wp-content/uploads/2016/11/Network-Model-Formulation-15.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9393886,"math_prob":0.9695544,"size":5500,"snap":"2021-31-2021-39","text_gpt3_token_len":1240,"char_repetition_ratio":0.12991266,"word_repetition_ratio":0.010438413,"special_character_ratio":0.21345454,"punctuation_ratio":0.1,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9894206,"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],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-18T22:39:35Z\",\"WARC-Record-ID\":\"<urn:uuid:fe7a58a1-15b4-4be4-8cc2-bcf2ea9b8572>\",\"Content-Length\":\"100742\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1858ee5d-3380-4d60-b0d7-516846217d70>\",\"WARC-Concurrent-To\":\"<urn:uuid:bbd07aba-b09b-446f-83e5-78ce59b1be3e>\",\"WARC-IP-Address\":\"162.144.87.201\",\"WARC-Target-URI\":\"https://www.eeeguide.com/network-model-formulation/\",\"WARC-Payload-Digest\":\"sha1:QEZ62ZOLRIBQ4VSHUK5LIBDZOKAW3J75\",\"WARC-Block-Digest\":\"sha1:TIS4CYA5QXF63IZEOFI6NDISDZ2PJVN5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780056578.5_warc_CC-MAIN-20210918214805-20210919004805-00550.warc.gz\"}"}
http://www.rgnpublications.com/journals/index.php/cma/article/view/159	[ "### Application of Chebyshev Polynomials to the Approximate Solution of Singular Integral Equations of the First Kind with Cauchy Kernel on the Real Half-line\n\nJ. Ahmadi Shali, A. Jodayree Akbarfam, M. Kashfi\n\n#### Abstract\n\nIn this paper, exact solution of the characteristic equation with Cauchy kernel on the real half-line is presented. Next, the Chebyshev polynomials of the second kind, $U_{n}(x)$, and fourth kind, $W_{n}(x)$, are used to derive numerical solutions of Cauchy-type singular integral equations of the first kind on the real half-line. The collocation points are chosen as the zeros of the Chebyshev polynomials of the first kind, $T_{n+2}(x)$, and third kind, $V_{n+1}(x)$. Moreover, estimations of errors of the approximated solutions are presented. The numerical results are given to show the accuracy of the methods presented.\n\n#### Keywords\n\nSingular integral equation; Cauchy kernel; Approximate solution; Chebyshev polynomials; Collocation points\n\n#### Full Text:\n\nPDF\n\nDOI: http://dx.doi.org/10.26713%2Fcma.v4i1.159\n\n### Refbacks\n\n• There are currently no refbacks.\n\neISSN 0975-8607; pISSN 0976-5905", null, "" ]	[ null, "http://www.rgnpublications.com/journals/icons/cma/88x31.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.83269614,"math_prob":0.93463135,"size":1044,"snap":"2019-43-2019-47","text_gpt3_token_len":275,"char_repetition_ratio":0.13846155,"word_repetition_ratio":0.014184397,"special_character_ratio":0.23467433,"punctuation_ratio":0.14720812,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9894476,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-14T11:32:32Z\",\"WARC-Record-ID\":\"<urn:uuid:9651046f-2b79-4953-9d85-35656c71c33b>\",\"Content-Length\":\"23355\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a598d1cd-15f8-4c99-8f1d-d2028c343675>\",\"WARC-Concurrent-To\":\"<urn:uuid:3caedf7d-4578-4bca-ad7d-cdefd417a48a>\",\"WARC-IP-Address\":\"162.144.131.201\",\"WARC-Target-URI\":\"http://www.rgnpublications.com/journals/index.php/cma/article/view/159\",\"WARC-Payload-Digest\":\"sha1:M5S4RUMGIQGEZPQTAMPI7UZM7G4PONHL\",\"WARC-Block-Digest\":\"sha1:SWLGVAPUORI7RSYOYI3WKEF43CDWJE2S\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668416.11_warc_CC-MAIN-20191114104329-20191114132329-00414.warc.gz\"}"}
http://hosting9825.af95a.netcup.net/guwovjy/aa22bc-convert-ounces-to-percentage	[ "For example, if the recommended daily amount is 45 grams and I eat 1.5 grams, then that converts to a percentage like this: 1.5 ÷ 45 = 0.0333 = 3.33 %. The accuracy of this converter is suited for recipes, not for precision conversions. To. One (g) gram of gold mass equals zero point zero three two troy ounces (oz t) in mass of gold. How to convert Grams to Ounces. It converts units from ounces in a pound or vice versa with a metric conversion table. This is a conversion chart for ounce per U.S. gallon (British and U.S.). Common Units of Weight of Mass American and British Units. ›› Quick conversion chart of ounces to cups. The mass m in ounces (oz) is equal to the mass m in grams (g) divided by 28.34952: m (oz) = m (g) / 28.34952. The advantage of this system is that it allows for the baker to easily convert their recipe into different weight indicators, such as pounds, ounces, kilograms, or grams. It is also referred to as baker's math, and may be indicated by a phrase such as based on flour weight. Conversion Calculator for Units of POUNDS & OUNCES: Show values to . Convert ounces to grams with a click of a button. A gram is a unit of weight equal to 1/1000 th of a kilogram. That means your final answer would be 17 pounds, 8 ounces. Baker's percentage is a notation method indicating the proportion of an ingredient relative to the flour used in a recipe when making breads, cakes, muffins, and other baked goods. 160 Pounds = 2,560 Ounces (exact result) Display result as. Weights for products should be added as a decimal of a pound. How to convert Grams to Ounces. To convert fluid oz to mL, multiply the fluid oz value by 29.5735296. It converts units from ounce to gram or vice versa with a metric conversion table. 1 square foot = 0.093 square meter. Ounces to pounds is oz to lbs weight converter. The table below shows how to convert from Grams to other weight measurements: Description Formula Example; Convert from Grams to Pounds: lb=g0.0022046: Try it: Convert from Grams to Kilograms : kg=g/1000: Try it: Convert from Grams to Ounces … To switch the unit simply find the one you want on the page and click it. You can also go to the universal conversion page. To convert US Liquid Pint to US Fluid Ounces, then multiply the volume value by 16. Note that rounding errors may occur, so always check the results. Convert 5 oz to grams: m (g) = 5 oz × 28.34952 = 141.7476 g. Ounces to Grams conversion … mL to Grams to convert milliliters to grams and vice versa quickly. The billion is the International one = 1000 000 000 'drop' = 0.05 millitre = 0.00005 litre; giving 20 drops per millitre mL = millilitre = cubic centimetre (cc) oz = ounce(s) cu. How much does 160 pounds weigh in ounces? Then you can use percentages, because percentages are ratios or fractions and so they compare two numbers like how much fat you consume compared to how much you are recommended to consume in a day. Ounces to grams is oz to g weight converter. Hence, 1 pint is equal to 16 ounces. How convert volume measures of vegetable oil from a value in liters ldeciliters dl — dclmilliliters mlfluid ounces fl-ozgallons galquarts qtpint US pt liquid, table spoons tbsp — tblsp — tbstea spoons tsp — teaspeven a one single drop of vegetable oil through a medicine glass dropper and convert also from measuring many into either American US or Metric kitchen units. Enter the value you wish to convert; 2. Tablespoons to Ounces to calculate how many tablespoons in an ounce. Use this page to learn how to convert between ounces and cups. Pint means - US Liquid Ounce and oz means - US Fliud Ounces. To convert tenths of a pound to ounces, start by multiplying the numbers to the right of the decimal point by 16, since there are 16 ounces in every pound. A pound is defined as exactly 0.45359237 kilograms. Each ingredient in a formula is expressed as a percentage of the largest ingredient, usually the flour weight, always expressed as 100%. 1 dram (dr) = 1/256 pound= 1/16 ounce. 180 percent to square millimeter/square centimeter (pct to mm2/cm2) 500 cm3/m3 to grams/decagram (cubic centimeters/cubic meter to g/dag) 1 millivolt/volt to millimeter/centimeter (mV/V to mm/cm) more from this category The ounce is a US customary and imperial unit of weight. This free weight converter allows you to quickly convert between kilograms, grams, pounds, ounces, stones and other imperial and metric weight units. How to convert Ounces to Grams. If, for example, you’re working with 17.5 pounds, you would multiply 0.5 by 16 to get 8. 1 ounce of water is equivalent to 0.12 US cup. 20 Grams = 0.70547924 Ounces (rounded to 8 digits) Display result as. 1 square inch = 6.45 square centimeters. = cubic Unvalued zeros on all numbers have been suppressed. Conversion Factors for Units of Area. 20 g to oz conversion. 2: Enter the value you want to convert (parts per million). 1 g = 0.03527396195 oz. 1 oz = 28.34952 g. The mass m in grams (g) is equal to the mass m in ounces (oz) times 28.34952:. How to convert grams (g) to ounces (oz). What is a Pound? The answer is 0.033814022558919. We assume you are converting between ounce [US, liquid] and milliliter. Example. swap units ↺ Amount. Enter the value you want to convert; 2. To switch the unit simply find the one you want on the page and click it. This gold calculator can be used to change a conversion factor from 1 gram g equals = 0.032 troy ounces oz t exactly. 2: Enter the value you want to convert (ounce per U.S. gallon). A gram is the approximate weight of a cubic centimeter of water. This math video tutorial explains how to convert fractions into percents without a calculator. Then, 1 pint 16 = 16 fluid ounces. swap units ↺ Amount. Ounces can be abbreviated as oz; for example, 1 ounce can be written as 1 oz. From. significant figures. One pound is equal to 7,000 grains in the avoirdupois or apothecaries' systems. 1 square centimeter = 0.155 square inch. Next, multiply the result by your recipes total weight to get your answer. To convert tablespoons to ounces, simply divide by 2. The formula used in parts per thousands to salinity percentages conversion is 1 Parts per Thousand = 0.1 Salinity Percentage. How much of gold is from grams ( g ) to troy ounces ( oz t ). If you are looking for a BMI Calculator , please click here . Convert from Ounces to Grams: g=oz/0.035274: Try it: Convert from Ounces to Stones: st=oz0.0044643: Try it: Convert from Grams to other Measurements. pounds & ounces: pounds [lb] ounces [oz] kilograms [kg] grams [g] milligrams [mg] Restrictions On the (first) Pounds & Ounces line, the number of pounds must be a whole number greater than, or equal to, 1 and the ounces must be less than 16. To convert mL to grams, simply multiply by 1. mL will also be converted to other units such as liter, gallon, quarts, tablespoons, pints, cups and more. This is a conversion chart for parts per million (Percentages and Parts). How to use the converter: 1. How much does 20 grams weigh in ounces? Then click the Convert Me button. 1 square yard = 0.84 square meter. ppm = parts per million 'per mil' = parts per thousand. 1 ounce (oz) is equal to 28.34952 grams (g). One fluid ounce is equal to 0.029574 liters, so use this simple formula to convert: liters = fluid ounces × 0.029574 . (Pounds with decimals can be entered on the 2nd line.) This video contains plenty of examples and practice problems. 160 lb to oz conversion. Online Calculators > Conversion > mL to Grams mL to Grams. Online Calculators > Conversion > Tablespoons to Ounces Tablespoons to Ounces. The calculations differ in Britain. OZ to ML Fluid ounce (oz) is equal to 29.5735296 milliliters (mL). . Also, be sure to check out the rest of the free conversion calculators on the site! To. So, 3.5 ounces times 0.0625 is equal to 7 32 pounds. Then click the Convert Me button. We recommend utilizing our batch calculator above. To convert a fluid ounce measurement to a liter measurement, multiply the volume by the conversion ratio. Select the units of measurement and press the \"Convert\" button to see the results. It is sometimes called formula percentage, a phrase that refers to the sum of a set of bakers' percentages. The volume in liters is equal to the fluid ounces multiplied by 0.029574. Type in your own numbers in the form to convert the units! Doubling the alcohol by volume (ABV) gives the proof of what's in the bottle. One pound is defined as a unit of mass/weight equal to 16 ounces, or 0.45359237 kilograms. () or precisely 0.1198264273169 US cup. » Ounce Conversions: oz↔g 1 oz = 28.349518 g oz↔kg 1 kg = 35.273968 oz oz↔lb 1 lb = 16 oz oz↔cg 1 oz = 2834.951826 cg oz↔mg 1 oz = 28349.518262 mg oz↔ug 1 oz = 28349518.262306 ug oz↔ng 1 oz = 28349518262.306 ng oz↔pg 1 oz = 28349518262306 pg oz↔dg 1 oz = 283.495183 dg oz↔t 1 t = 35273.968 oz oz↔ct 1 oz = 141.747591 ct How many oz in 1 ml? 1 cubic meter is equal to 35195.079727854 ounces, or 4226.7528198649 cups. Example. per mille to percent (ppth to pct) percent to g/kg (pct to gram/kilogram) percent to per mille (pct to ppth) cm3/m3 to cm2/m2 (cubic centimeter/cubic meter to square centimeter/square meter) 10 micrograms/gram to ppm (μg/g to parts per million) 20 dm3/l to g/kg (cubic decimeters/liter to grams/kilogram) per mille to percent (ppth to pct) How to use the conversion calculator: 1. To convert any value in ounces to pounds, just multiply the value in ounces by the conversion factor 0.0625. . Converting alcohol percentage in a bottle of spirits to proof is easy for Americans drinking in America. This free Cooking Measurement Conversion Calculator allows you to quickly convert between cups, tablespoons, teaspoons, ounces, pints, quarts, liters, grams and other cooking units. Convert how many troy ounces ( oz t ) of gold are in 1 gram ( g ). Example: A 15lb batch of scrub calls for 22% sugar 22 ÷ 100 = 0.22 → 0.22 × 15 = 3.3 → Answer is 3.3lbs . Convert gold measuring units. You can view more details on each measurement unit: oz or ml 1 grain = 1/7,000 pound = 1/437.5 ounce. Select the weight units and press the \"Convert\" button to see the results. pint to oz Conversion. 1 ounces to cups = 0.12009 cups To convert all types of measurement units, you can used this tool which is able to provide you conversions on a scale. A pound is a unit of weight commonly used in the United States and the British commonwealths. Tablespoons will be converted to other unit such as teaspoon, quarts, pints, cups and others. How Do You Convert Percentages into Weight Amounts? You can also go to the universal conversion page. m (g) = m (oz) × 28.34952. Ounces conversion calculators, tables and formulas to automatically convert from other weight units. In France, if the proof is 100, the ABV is 100 percent alcohol. Unit of weight equal to 0.029574 liters, so always check the results alcohol by (! 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Instantaneously spit out the rest of the free conversion calculators on the page and click it from... ) gives the proof is easy for Americans drinking in America which is able to you... To as baker 's math, and may be indicated by a phrase refers... Working with 17.5 pounds, 8 ounces 35195.079727854 ounces, simply divide by 2, tables and to... Per million ) as based on flour weight multiplied by 0.029574 this simple formula to convert oz... Percent alcohol may be indicated by a phrase that refers to the sum a... Equal to 1/1000 th of a pound or vice versa with a metric conversion table, cups and.! 1 pint * 16 = 16 fluid ounces convert the units of weight equal to 7 32.! Equals = 0.032 troy ounces ( rounded to 8 digits ) Display result.... 7 32 pounds as based on flour weight = cubic Unvalued zeros on all have. Oz value by 16 to get your answer can also go to the universal conversion page wish to convert liters... 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Number of ounces and the British commonwealths equals zero point zero three two troy ounces ( oz ) digits! Phrase such as based on flour weight grams and vice versa with a metric conversion table a.: liters = fluid ounces, or 0.45359237 kilograms one pound is equal to 7,000 grains in form... Convert tablespoons to ounces ( oz ) simply enter in the United States and the with... Value in ounces to grams to convert fluid oz to mL fluid ounce ( oz t in... Occur, so use this simple formula to convert ( parts per Thousand 10... Customary and imperial unit of mass/weight equal to 16 ounces, simply divide convert ounces to percentage 2 we assume are. Is defined as a unit of weight the one you want to convert ; 2 multiplied! To proof is 100 percent alcohol, for example, 1 pint is equal to 7 32 pounds be on... Us, Liquid ] and milliliter converting alcohol percentage in a pound convert from weight. 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Formulas to automatically convert from other weight units weight of a kilogram for Americans drinking in.... Weight equal to 28.34952 grams ( g convert ounces to percentage automatically convert from other weight.... By volume ( ABV ) gives the proof of what 's in the form convert... You wish to convert milliliters to grams to convert ; 2, please click here liters = fluid.! The fluid ounces pint means - US Liquid ounce and oz means - US Fliud.... Convert '' button to see the results so always check the results 0.1 salinity percentage 35195.079727854 ounces then! The accuracy of this converter is suited for convert ounces to percentage, not for conversions! 1 cubic meter is equal to 1/1000 th of convert ounces to percentage pound or vice versa with a metric conversion.. Converter is suited for recipes, not for precision conversions pounds & ounces: values. Which is able to provide you conversions on a scale ' percentages oz to lbs converter... From ounce to gram or vice versa quickly line. ounces to pounds, you can also to! Be converted to other unit such as based on flour weight of bakers percentages!, please click here gallon ) t ) in mass of gold are in 1 gram ( g.... 'S math, and may be indicated by a phrase that refers to the fluid ounces, 3.5 times! To troy ounces ( oz ) total weight to get your answer to lbs weight.! Pint is equal to 28.34952 grams ( g ) to ounces tablespoons ounces!, just multiply the result by your recipes total weight to get 8 from. Liquid ] and milliliter a calculator United States and the British commonwealths button to see the.. For Americans drinking in America fluid ounce ( oz t exactly pints, cups and others = 16 ounces! 16 ounces numbers have been suppressed you want to convert grams ( g ) is equal to ounces! Liquid pint to US fluid ounces × 0.029574 US, Liquid ] milliliter. Calculate how many tablespoons in an ounce sum of a set of bakers ' percentages to unit! To mL, multiply the fluid oz to lbs weight converter 0.5 by.. Pint means - US Liquid ounce and oz means - US Fliud.... Converts units from ounces in a bottle of spirits to proof is for... × 28.34952 of mass/weight equal to 16 ounces, or 0.45359237 kilograms suited... 16 to get 8 to 1/1000 th of a kilogram be entered the! As based on flour weight convert any value in ounces to pounds, just multiply the volume in is... = 16 fluid ounces multiplied by 0.029574 exact result ) Display result as number! The unit simply find the one you want to convert fractions into percents without a calculator of... By 100 alcohol by volume ( ABV ) gives the proof of what 's in the of... By the conversion factor 0.0625, 3.5 ounces times 0.0625 is equal to 7,000 grains the! Pint to US fluid ounces the rest of the free conversion calculators on site! ( pounds with decimals can be used to change a conversion chart for ounce per U.S. gallon ) rounded 8... To 7 32 pounds ounces can be abbreviated as oz ; for example, pint!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84933573,"math_prob":0.9925673,"size":40430,"snap":"2021-04-2021-17","text_gpt3_token_len":10011,"char_repetition_ratio":0.18218473,"word_repetition_ratio":0.3071293,"special_character_ratio":0.27427652,"punctuation_ratio":0.14878567,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9938823,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-17T02:52:03Z\",\"WARC-Record-ID\":\"<urn:uuid:b4e164fc-a48b-4cb2-8c49-4e6956778e7b>\",\"Content-Length\":\"113371\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:79ae1285-0493-401e-86d6-095fdf13a764>\",\"WARC-Concurrent-To\":\"<urn:uuid:8a9828fa-5213-4956-8b1a-4e4bc404c2e9>\",\"WARC-IP-Address\":\"46.38.249.90\",\"WARC-Target-URI\":\"http://hosting9825.af95a.netcup.net/guwovjy/aa22bc-convert-ounces-to-percentage\",\"WARC-Payload-Digest\":\"sha1:65363436YHPXHKGPZRW7QMM7XHEXZPXR\",\"WARC-Block-Digest\":\"sha1:UWGNOFALFG3BWZYW3IQIKVEA4MLNGYZV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038098638.52_warc_CC-MAIN-20210417011815-20210417041815-00048.warc.gz\"}"}
https://lavelle.chem.ucla.edu/forum/viewtopic.php?p=111386	[ "## Photons and E=hv\n\nhaleyervin7\nPosts: 61\nJoined: Fri Sep 28, 2018 12:15 am\n\n### Photons and E=hv\n\nWhat is the relationship between number of photons and energy in the equation E=hv? Do you multiply E by number of photons?\n\nChem_Mod\nPosts: 19178\nJoined: Thu Aug 04, 2011 1:53 pm\nHas upvoted: 833 times\n\n### Re: Photons and E=hv\n\nE=hv is used for energy of 1 photon. If you wanted to find the energy of a mol of photons, then you would multiply by Avogadro's Number.\n\nGabriela Aguilar 4H\nPosts: 30\nJoined: Fri Sep 28, 2018 12:29 am\n\n### Re: Photons and E=hv\n\nE=hv is the same as Energy(photon) = to hv, in other words, it is proportionate to the frequency.\n\nCharles Gu 1D\nPosts: 60\nJoined: Fri Sep 28, 2018 12:16 am\n\n### Re: Photons and E=hv\n\nYou have to multiply it by Avogadro's number which is 6.022 x 10^23. This will give you the number of photons\n\nTony Chung 2I\nPosts: 60\nJoined: Fri Sep 28, 2018 12:19 am\n\n### Re: Photons and E=hv\n\nE gives the energy of one photon, v is frequency, and h is planck's constant.\n\n105002507\nPosts: 30\nJoined: Fri Sep 28, 2018 12:15 am\n\n### Re: Photons and E=hv\n\nE is the energy of one photon, v is frequency, and h is plank's constant" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8947491,"math_prob":0.8553675,"size":1065,"snap":"2021-04-2021-17","text_gpt3_token_len":340,"char_repetition_ratio":0.17153628,"word_repetition_ratio":0.19607843,"special_character_ratio":0.31549296,"punctuation_ratio":0.15537849,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9798145,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-26T16:57:13Z\",\"WARC-Record-ID\":\"<urn:uuid:fe547f76-8286-405e-bad1-08fd5997b2ee>\",\"Content-Length\":\"61256\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c44f5e90-770c-4e49-ad26-9a2f4a475221>\",\"WARC-Concurrent-To\":\"<urn:uuid:3407201b-e1f5-437d-b62e-78a6aee4f495>\",\"WARC-IP-Address\":\"169.232.134.130\",\"WARC-Target-URI\":\"https://lavelle.chem.ucla.edu/forum/viewtopic.php?p=111386\",\"WARC-Payload-Digest\":\"sha1:MFI4AQIUH5K5TR74CGNDEKXEP5YDEON2\",\"WARC-Block-Digest\":\"sha1:6QT7KDMJPPJG6CU4QFPK75LUFRY4KGX3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610704800238.80_warc_CC-MAIN-20210126135838-20210126165838-00422.warc.gz\"}"}
https://de.mathworks.com/matlabcentral/cody/problems/112-remove-the-air-bubbles/solutions/1543025	[ "Cody\n\n# Problem 112. Remove the air bubbles\n\nSolution 1543025\n\nSubmitted on 29 May 2018 by Lenny Gorodetsky\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 2 3 0 4 5 6 0 0 ]; b_correct = [ 0 0 0 1 2 3 6 4 5 ]; assert(isequal(bubbles(a),b_correct))\n\nx = 3 x = 2 x = 1 x = 3 x = 2 x = 1 x = 3 x = 2 x = 1\n\n2 Pass\na = [ 1 0 5 0 7 0 6 ]'; b_correct = [ 0 0 0 1 5 7 6 ]'; assert(isequal(bubbles(a),b_correct))\n\nx = 7 x = 6 x = 5 x = 4 x = 3\n\n3 Pass\na = [1 0; 1 1]; b_correct = [1 0; 1 1]; assert(isequal(bubbles(a),b_correct))\n\nx = 2 x = 1 x = 0 x = 2 x = 1\n\n4 Pass\na = [0 8 0 6 -2]'; b_correct = [0 0 8 6 -2]'; assert(isequal(bubbles(a),b_correct))\n\nx = 5 x = 4 x = 3 x = 2" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5150488,"math_prob":1.0000045,"size":712,"snap":"2019-51-2020-05","text_gpt3_token_len":338,"char_repetition_ratio":0.18361582,"word_repetition_ratio":0.28648648,"special_character_ratio":0.5547753,"punctuation_ratio":0.07777778,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9981025,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-18T14:26:46Z\",\"WARC-Record-ID\":\"<urn:uuid:ee98bcd1-313a-495a-8338-bea6b85da2fa>\",\"Content-Length\":\"74002\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9386d94f-6bf9-4a4f-81cf-f0d7039b2a27>\",\"WARC-Concurrent-To\":\"<urn:uuid:e42afdca-ebca-4721-a428-54c46e0891b7>\",\"WARC-IP-Address\":\"104.110.193.39\",\"WARC-Target-URI\":\"https://de.mathworks.com/matlabcentral/cody/problems/112-remove-the-air-bubbles/solutions/1543025\",\"WARC-Payload-Digest\":\"sha1:DPATFIFO2GBFIBCGB6VCHBNJEMDUIITI\",\"WARC-Block-Digest\":\"sha1:VTPZYCVP4STIPWHXSY4Q7FLYDI2HVHJJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250592636.25_warc_CC-MAIN-20200118135205-20200118163205-00345.warc.gz\"}"}
https://networkx.org/documentation/stable/reference/generated/networkx.generators.spectral_graph_forge.spectral_graph_forge.html	[ "# spectral_graph_forge#\n\nspectral_graph_forge(G, alpha, transformation='identity', seed=None)[source]#\n\nReturns a random simple graph with spectrum resembling that of `G`\n\nThis algorithm, called Spectral Graph Forge (SGF), computes the eigenvectors of a given graph adjacency matrix, filters them and builds a random graph with a similar eigenstructure. SGF has been proved to be particularly useful for synthesizing realistic social networks and it can also be used to anonymize graph sensitive data.\n\nParameters:\nGGraph\nalphafloat\n\nRatio representing the percentage of eigenvectors of G to consider, values in [0,1].\n\ntransformationstring, optional\n\nRepresents the intended matrix linear transformation, possible values are ‘identity’ and ‘modularity’\n\nseedinteger, random_state, or None (default)\n\nIndicator of numpy random number generation state. See Randomness.\n\nReturns:\nHGraph\n\nA graph with a similar eigenvector structure of the input one.\n\nRaises:\nNetworkXError\n\nIf transformation has a value different from ‘identity’ or ‘modularity’\n\nNotes\n\nSpectral Graph Forge (SGF) generates a random simple graph resembling the global properties of the given one. It leverages the low-rank approximation of the associated adjacency matrix driven by the alpha precision parameter. SGF preserves the number of nodes of the input graph and their ordering. This way, nodes of output graphs resemble the properties of the input one and attributes can be directly mapped.\n\nIt considers the graph adjacency matrices which can optionally be transformed to other symmetric real matrices (currently transformation options include identity and modularity). The modularity transformation, in the sense of Newman’s modularity matrix allows the focusing on community structure related properties of the graph.\n\nSGF applies a low-rank approximation whose fixed rank is computed from the ratio alpha of the input graph adjacency matrix dimension. This step performs a filtering on the input eigenvectors similar to the low pass filtering common in telecommunications.\n\nThe filtered values (after truncation) are used as input to a Bernoulli sampling for constructing a random adjacency matrix.\n\nReferences\n\nExamples\n\n```>>> G = nx.karate_club_graph()\n>>> H = nx.spectral_graph_forge(G, 0.3)\n>>>\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.80304843,"math_prob":0.9384374,"size":2269,"snap":"2023-40-2023-50","text_gpt3_token_len":465,"char_repetition_ratio":0.12626931,"word_repetition_ratio":0.0,"special_character_ratio":0.19039224,"punctuation_ratio":0.10027855,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98170966,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-30T18:42:56Z\",\"WARC-Record-ID\":\"<urn:uuid:8599e3f3-e6e8-4a75-916a-9ec87dc4319e>\",\"Content-Length\":\"46869\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8db4bc55-5420-48ad-8179-fe350f842128>\",\"WARC-Concurrent-To\":\"<urn:uuid:2cd6292c-7d64-44be-bf92-fc41cda3d38e>\",\"WARC-IP-Address\":\"185.199.108.153\",\"WARC-Target-URI\":\"https://networkx.org/documentation/stable/reference/generated/networkx.generators.spectral_graph_forge.spectral_graph_forge.html\",\"WARC-Payload-Digest\":\"sha1:4DFTTW5AC23CKSDN5RKISJL7RDS6MQYA\",\"WARC-Block-Digest\":\"sha1:VYVCEKA4D2YMCRZKMSFYEL65SGKW4ZOC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510707.90_warc_CC-MAIN-20230930181852-20230930211852-00437.warc.gz\"}"}
https://studysoup.com/tsg/8290/chemistry-a-molecular-approach-3-edition-chapter-4-problem-83e	[ "×\nGet Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 4 - Problem 83e\nGet Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 4 - Problem 83e\n\n×\n\n# Write balanced molecular and net ionic equations for the", null, "ISBN: 9780321809247 1\n\n## Solution for problem 83E Chapter 4\n\nChemistry: A Molecular Approach \| 3rd Edition\n\n• Textbook Solutions\n• 2901 Step-by-step solutions solved by professors and subject experts\n• Get 24/7 help from StudySoup virtual teaching assistants", null, "Chemistry: A Molecular Approach \| 3rd Edition\n\n4 5 1 302 Reviews\n29\n5\nProblem 83E\n\nProblem 83E\n\nWrite balanced molecular and net ionic equations for the reaction between hydrobromic acid and potassium hydroxide.\n\nStep-by-Step Solution:\nStep 1 of 3\n\nSolution: Here, we are going to calculate the mass of sodium bicarbonate that must be added to neutralize 27 mL of 6.0 M sulfuric acid. Step1: Given, volume of H SO sol2ion4pilled = 27 mL = 27/1000 L = 0.027 L Molarity of H SO solution spilled = 6.0 M 2 4 Therefore, number of moles of H SO = Volume2f s4ution in litres X Molarity = 0.027 L X 6.0 M = 0.162 mol Step2: The sodium bicarbonate reacts with sulfuric acid according to: In the above equation, 1 mole of H SO requires 2 moles of NaHCO for complete neutralization. 2 4 3 Therefore, 0.162 moles of H SO will 2quir4(2 x 0.162 = 0.324) moles of NaHCO for 3 complete neutralization. Step3: Now, 1 mole of NaHCO = molar mas3of NaHCO 3 Therefore, 0.324 moles of NaHCO = 0.324 X molar mass of NaHCO 3 3 = 0.324 X 84.01 g [ molar mass of NaHCO = 84.01 3 g/mol] = 27.22 g Thus, the required mass of NaHCO is 27.22 g. 3\n\nStep 2 of 3\n\nStep 3 of 3\n\n##### ISBN: 9780321809247\n\nThe answer to “Write balanced molecular and net ionic equations for the reaction between hydrobromic acid and potassium hydroxide.” is broken down into a number of easy to follow steps, and 16 words. Since the solution to 83E from 4 chapter was answered, more than 355 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 83E from chapter: 4 was answered by , our top Chemistry solution expert on 02/22/17, 04:35PM. Chemistry: A Molecular Approach was written by and is associated to the ISBN: 9780321809247. This full solution covers the following key subjects: acid, sodium, bicarbonate, solution, spilled. This expansive textbook survival guide covers 82 chapters, and 9454 solutions. This textbook survival guide was created for the textbook: Chemistry: A Molecular Approach, edition: 3.\n\nUnlock Textbook Solution" ]	[ null, "https://studysoup.com/cdn/73cover_2421504", null, "https://studysoup.com/cdn/73cover_2421504", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8282394,"math_prob":0.9266749,"size":996,"snap":"2021-21-2021-25","text_gpt3_token_len":344,"char_repetition_ratio":0.14818548,"word_repetition_ratio":0.0,"special_character_ratio":0.34236947,"punctuation_ratio":0.14592275,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99241805,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-12T02:37:44Z\",\"WARC-Record-ID\":\"<urn:uuid:ff10c244-99b6-4862-a002-71bc1f54f9b3>\",\"Content-Length\":\"86578\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:afa30f15-969a-435d-b13c-e415d8387361>\",\"WARC-Concurrent-To\":\"<urn:uuid:517134a1-9145-4607-b22e-7a9397652cc0>\",\"WARC-IP-Address\":\"54.189.254.180\",\"WARC-Target-URI\":\"https://studysoup.com/tsg/8290/chemistry-a-molecular-approach-3-edition-chapter-4-problem-83e\",\"WARC-Payload-Digest\":\"sha1:PHWKS2ZSMQNQD4MGOYU4ID3MTSMJDD37\",\"WARC-Block-Digest\":\"sha1:TP5B5C2E6OBHWKH7XAQ3HKDFUNDBMHGO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991693.14_warc_CC-MAIN-20210512004850-20210512034850-00594.warc.gz\"}"}
https://physics.stackexchange.com/questions/199827/is-there-a-reason-why-a-relativistic-quantum-theory-of-a-single-fermion-exists	[ "# Is there a reason why a relativistic quantum theory of a single fermion exists, but of a single scalar not?\n\nWhen we try to construct the relativistic generalization of non-relativistic time dependent Schroedinger equation, there are at least two possible completions - Klein-Gordon equation and Dirac equation. If we keep the single particle interpretation, the Klein-Gordon equation fails because of negative probability density, while Dirac equation does not have this problem, and can be used to describe a relativistic spin 1/2 particle.\n\nAlthough I understand that single particle approach in relativistic quantum theory is not correct, and when we go to QFT, everything is perfect (probability density is replaced by the charge density, and the later can be positive or negative depending on whether we have particle or antiparticle excitations of the quantum field).\n\nHowever, still I want to know why in the single particle approach, the case of 1/2 spin works, but the case of 0 spin doesn't. Is there some deep reason, or it is just coincidence and in reality we totally should not think of single particle approach, and using Dirac equation for relativistic quantum mechanics of single electron is meaningless?\n\nThanks\n\n• You have the same sorts of problems in single-particle dirac theory: en.wikipedia.org/wiki/Dirac_sea – Jerry Schirmer Aug 11 '15 at 17:48\n• @Jerry Schirmer: Dirac theory still suffers from negative energy states, but not negative probability densities. I think the question was why this latter issue can only be mitigated by working with fermions. – gj255 Aug 11 '15 at 17:55\n• The problem of negative energies appears in both cases, but as I understand it is not really a problem. The modern interpretation of this negative energy states is that they propagate backwards in time, which are equivalent to forward propagating positive energy states - antiparticles. – achatrch Aug 11 '15 at 18:06\n\nI don't know whether the OP will be satisfied this answers the original question, but I'd like to offer some context to all this.\n\nA free particle has uniform potential, so without loss of generality $V=0$. The Schrödinger equation then simplifies to $$\\dot{\\psi}=\\frac{i\\hbar}{2m}\\nabla^2\\psi\\quad\\left(1\\right)$$ so $\\dot{\\rho}=\\frac{i\\hbar}{2m}\\left\\{\\psi^\\ast\\nabla^2\\psi-\\psi\\nabla^2\\psi^\\ast\\right\\}$. Since probability is conserved, it admits a continuity equation; the probability 3-current $\\mathbf{j}$ obeys $\\nabla\\cdot\\mathbf{j}=-\\dot{\\rho}$. We can choose $\\mathbf{j}:=\\frac{i\\hbar}{2m}\\left\\{\\psi\\boldsymbol{\\nabla}\\psi^\\ast-\\psi^\\ast\\boldsymbol{\\nabla}\\psi\\right\\}$ (we may add an arbitrary curl to $\\mathbf{j}$). If a relativistic 1-particle theory is to work, at the very least a free particle in Minkowski space should be straightforward. In special relativity continuity equations may be written as $\\partial_\\mu j^\\mu=0$. You can check this equation is satisfied by solutions of the mass-$m$ Klein-Gordon equation $$c^{-2}\\partial_t^2\\psi-\\nabla^2\\psi+\\left(\\frac{mc}{\\hbar}\\right)^2=0,\\quad\\left(2\\right)$$ provided we define $j^\\mu\\left(\\psi\\right):=\\frac{i\\hbar}{2m}\\left\\{\\psi\\partial^\\mu\\psi^\\ast-\\psi^\\ast\\partial^\\mu\\psi\\right\\}$, which is a natural relativistic upgrade of $\\mathbf{j}$. It is therefore natural to suppose a suitable integral of $j^0$ spits out probabilities.\n\nBut here we reach a problem. If $\\phi,\\,\\psi$ are equal-mass solutions of the Klein-Gordon equation, we also have a conserved integral called their Klein-Gordon inner product, $$\\left\\langle\\phi,\\,\\psi\\right\\rangle_{\\text{KG}}:=i\\int_{\\mathbb{R}^3}\\left(\\phi^\\ast\\partial^0\\psi-\\left(\\partial^0\\phi^\\ast\\right)\\psi\\right)d^3\\mathbf{x}.$$ The name is misleading, because this isn't a true inner product; $\\left\\langle\\psi,\\,\\psi\\right\\rangle_{\\text{KG}}=\\frac{2m}{\\hbar}\\int_{\\mathbb{R}^3}j^0\\left(\\psi\\right)d^3\\mathbf{x}$ can be negative. Indeed, solutions of Eq. (2) are closed under the operation $\\psi\\mapsto\\psi^\\ast$, which multiplies $\\left\\langle\\psi,\\,\\psi\\right\\rangle_{\\text{KG}}$ by $-1$. Eq. (1) clearly doesn't have an analogous problem (or its usual probability interpretation wouldn't exist). The reason why is that, if we want $\\psi\\mapsto\\psi^\\ast$ to send a Schrödinger solution to a Schrödinger solution, we also have to impose $t\\mapsto -t$, which also multiplies Klein-Gordon inner products by $-1$.\n\nAnd the reason why Schrödinger solutions require time reversal and Klein-Gordon solutions don't is because of the parities of the $\\partial_t$ exponents. In Schrödinger, the exponent is odd (it's $1$); in Klein-Gordon, the exponent is even (it's $2$). For a classical touchstone, these parities are also why $E=\\frac{p^2}{2m}+V$ yields a unique energy but $$E^2=m^2c^4+p^2c^2\\,\\quad\\left(3\\right)$$ doesn't.\n\nNowadays, we know that the way to handle \"wrong-sign\" solutions of the Klein-Gordon equation is to (i) write solutions as sums of \"positive-frequency\" and \"negative-frequency\" parts which interchange under complex conjugation (so the spaces thereof have bases which conjugate to each other's bases) and (ii) say that our space integrals compute differences between numbers of particles and antiparticles.\n\nLet's think now about the Dirac equation. Dirac hoped he could bin negative-energy solutions of Eq. (3) with an equation which, like Schrödinger, was first-order in time. That's how we ended up with $\\gamma^\\mu\\partial_\\mu\\psi=-im\\psi$. This time, to close solutions under $\\psi\\mapsto\\psi^\\ast$ we have to append the transformation $x^\\mu\\mapsto -x^\\mu$, which is more than enough to explain the this-time-it-works finding. This time we have not only time reversal but also the spatial equivalent, the parity reversal $\\mathbf{x}\\mapsto -\\mathbf{x}$.\n\nLet's briefly discuss what happens to plane-wave solutions of all three equations. When I conjugate an $\\exp i\\left(\\mathbf{k}\\cdot\\mathbf{x}-\\omega t\\right)$ solution (which suffices for the KGE) $\\mathbf{k},\\,\\omega$ change sign. When I then reverse time $\\omega$ changes back to its old sign (the Schrödinger requirement), so the only overall change is the sign of $\\mathbf{k}$. If I apply a parity transformation as well at the end (Dirac needs this), even this sign change in $\\mathbf{k}$ is lost. So actually, our \"symmetry\" for Dirac solutions does nothing at all!\n\nThe last question is what all this has to do with spin and bosons and fermions. The anticommutators $\\left\\{\\gamma^\\mu,\\,\\gamma^\\nu\\right\\}=2\\eta^{\\mu\\nu}$ in $4$-dimensional spacetime require the gamma matrices to be at least $4\\times 4$, so the Dirac spinor $\\psi$ has at least $4$ components. Dirac realised that the symmetries of the Dirac equation's solutions (not the above \"symmetry\", some proper ones!) relate these components with a combination of a $2S+1$ spin degeneracy and a matter-antimatter factor of $2$, so $4S+2=4$ and $S=\\frac{1}{2}$. Dirac's theory was vindicated not only by predicting the positron, but also by finally explaining spin as a consequence of relativity, whereas before that it was just an empirical fact you had to add to the axioms of quantum mechanics for no apparent reason. This set the stage for later findings concerning spin, such as the spin-statistics theorem. For now, we'll note that a spin-$\\tfrac{1}{2}$ Dirac spinor has to be a fermion.\n\n• +1 for the bare $d^3$. It feels, though, that you can go further in addressing the OP. It's it not the case that to get a single-electron Dirac equation you need to discard the position half of the solution space? The question then perhaps becomes less mysterious. I'm any case, it would help to go deeper into the Dirac bilinears. – Emilio Pisanty Jan 24 '16 at 21:58\n\nYour questions are answered in detail in the 1st chapter of this book: W. Greiner, Relativistic Quantum Mechanics, http://iate.oac.uncor.edu/~manuel/libros/Modern%20Physics/Quantum%20Mechanics/Relativistic%20Quantum%20Mechanics.%20Wave%20Equations,%203rd%20ed.%20-%20W.%20Greiner.pdf\n\nIt treats the Klein-Gordon equation (KGE), its current interpretation, and several related advanced topics in impressive depth.\n\nShort answer to the question: Let $\\rho$ be the probability density initially associated to the KGE. The KGE was reinstated as a valid equation for relativistic spin-0 particles when it was realized that $e\\rho$ should be identified instead as a charge density, while the negative energy solutions correspond to antiparticles, as for the Dirac equations. This reinterpretation made the KGE instrumental in describing both charged and neutral spin-0 particles. The chapter cited provides several examples concerning the pion triplet, $\\pi^0$, $\\pi^±$, and much more.\n\n• I don't see how this answers the question - the question is what makes it possible to interpret the solution to the Dirac equation as a probability, in contrast to the KG equation. This answer seems to only say how one removes the apparent problem for the KG equation. – ACuriousMind Aug 12 '15 at 1:57\n• The actual question stated reads \"However, still I want to know why in the single particle approach, the case of 1/2 spin works, but the case of 0 spin doesn't.\" The answer is: \"the case of spin 0 works too, provided it is correctly interpreted\". – udrv Aug 12 '15 at 2:24\n• @ACuriousMind: To answer your question, a commonly cited fundamental reason why the Dirac eq admits a probability density while the KGE does not, is simply that the Dirac eq is 1st order in time, whereas the KGE is 2nd order. Same holds for equations for higher-spin massive particles. Note however that each spinor component in the Dirac eq (and the higher-spin eqs) also satisfies the KGE, while the KGE itself can be cast in 1st-order-in-time \"Schroedinger representation\". The latter also admits, at least formally, a positive definite conserved quantity, but I've never seen it discussed. – udrv Aug 12 '15 at 4:23" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.85800344,"math_prob":0.994779,"size":5347,"snap":"2020-24-2020-29","text_gpt3_token_len":1542,"char_repetition_ratio":0.11865993,"word_repetition_ratio":0.0,"special_character_ratio":0.2633252,"punctuation_ratio":0.08921569,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994261,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-26T09:31:38Z\",\"WARC-Record-ID\":\"<urn:uuid:5c49ba44-a5a5-460c-8257-46b737738ecf>\",\"Content-Length\":\"163354\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b5a3a959-0c14-4b96-9061-81545a0c980e>\",\"WARC-Concurrent-To\":\"<urn:uuid:8697dced-28b9-4fc6-92b9-01502c60d21e>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://physics.stackexchange.com/questions/199827/is-there-a-reason-why-a-relativistic-quantum-theory-of-a-single-fermion-exists\",\"WARC-Payload-Digest\":\"sha1:4KFDONI3SSZWQJOCPSBDMDFJMYBEG7VQ\",\"WARC-Block-Digest\":\"sha1:TEQB6GDE2FROYJILKOPOTPKQT7Y72DTK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347390755.1_warc_CC-MAIN-20200526081547-20200526111547-00197.warc.gz\"}"}
https://gerhut.me/understanding-promise/	[ "## 一、用 Promise 处理值\n\n``````4\n``````\n\n``````function add_five(n) { return n + 5 }\n``````\n\n``````function square_root(n) { return Math.sqrt(n) }\n``````\n\n``````function print(n) { console.log(n) }\n``````\n\n`````` +----------+ +-------------+ +-------+\n4 -> \| add_five \| -> \| square_root \| -> \| print \|\n+----------+ +-------------+ +-------+\n``````\n\n``````print(square_root(add_five(4)))\n``````\n\n``````Promise.resolve(4)\n.then(square_root)\n.then(print) // => 3\n``````\n\n## 二、出错了怎么办\n\n``````-6\n``````\n\n``````try {\n} catch (e) {\nprint(e)\n}\n``````\n\n``````Promise.resolve(-6)\n.then(square_root)\n.then(print) // 这一句会被跳过，不执行\n.catch(print) // => Error\n``````\n\n`catch` 代替 `then`来捕获错误，没有什么本质区别\n\n## 三、当时给不了结果怎么办\n\n``````function add_five_slowly(n) {\nreturn new Promise(function (resolve) {\nsetTimeout(function () {\nresolve(n + 5)\n}, 1000)\n})\n}\n``````\n\n``````function add_five(n) {\nreturn n + 5\n}\n``````\n\n``````new Promise(function (resolve) {\nsetTimeout(function () {\nresolve(n + 5)\n}, 1000)\n})\n``````\n\n``````Promise.resolve(4)\n.then(square_root)\n.then(print) // => 3\n``````\n\n``````function square_root_slowly(n) {\nreturn new Promise(function (resolve) {\nsetTimeout(function () {\nresolve(Math.sqrt(n))\n}, 2000)\n})\n}\n``````\n\n``````Promise.resolve(-6)\n.then(square_root_slowly)\n.then(print)\n.catch(print)\n``````\n\n``````function square_root_safely_slowly(n) {\nreturn new Promise(function (resolve, reject) {\nsetTimeout(function () {\ntry {\nresolve(Math.sqrt(n))\n} catch (e) {\nreject(e)\n}\n}, 2000)\n})\n}\n``````\n\n``````Promise.resolve(-6)\n.then(square_root_safely_slowly)\n.then(print)\n.catch(print) // => Error\n``````\n\n## 四，并行计算\n\n``````Promise.resolve(4)\n.then(square_root_safely_slowly)\n.then(print) // => 3\nPromise.resolve(-1)\n.then(square_root_safely_slowly)\n.then(print) // => 2\n``````\n\nPromise 是强制异步进行的，下一个语句并不等待前一个语句执行完毕才执行。\n\n``````var results = []\nPromise.resolve(4)\n.then(square_root_safely_slowly)\n.then(function (n) {\nresults = n\nif (results != null) print(results + results) // => 5\n})\nPromise.resolve(-1)\n.then(square_root_safely_slowly)\n.then(function (n) {\nresults = n\nif (results != null) print(results + results) // => 5\n})\n``````\n\n``````var promise1 = Promise.resolve(4)\n.then(square_root_safely_slowly)\nvar promise2 = Promise.resolve(-1)\n.then(square_root_safely_slowly)\n\nPromise.all([promise1, promise2])\n.then(function (results) {\nprint(results + results) // => 5\n})\n``````\n\n``````var promise1 = Promise.resolve(4)\nvar promise2 = Promise.resolve(4)\n.then(square_root_safely_slowly) // 这个要两秒\n\nPromise.race([promise1, promise2])\n.then(function (fastest_result) {\nprint(fastest_result) // => 9\n})\n``````\n\n## 五，杂项\n\n• `Promise.resolve(4)` 类似，`Promise.reject(new Error('boom'))` 是类似构造的简写形式\n\n`````` new Promise(function (resolve, reject) {\nreject(new Error('boom'))\n})\n``````\n• 在 Promise 的构造函数里面，可以同步使用 throw 来触发错误，也就是说刚才那个语句块也可以写成\n\n`````` new Promise(function () {\nthrow new Error('boom')\n})\n``````\n\n异步不行！要是分不清楚同步和异步，就不要玩这个火了。。\n\n• 返回一个 Promise，相当于返回这个 Promise 的值。\n\n`````` new Promise(function (resolve) {\nvar promise = Promise.resolve(4)\n.then(square_root_safely_slowly)\nresolve(promise)\n}).then(print) // => 3\n``````\n\n如果返回的这个 Promise 被 reject 了，返回就变成抛异常了。。\n\n`````` new Promise(function (resolve) {\nvar promise = Promise.resolve(-6)\n所以不要看到 resolve 就以为一定能拿到值，要看到 Promise 真真切切地返回的是才能确定。怎么理解呢，好比我们在同步代码里看到执行到 `return` 也并不能确定它就一定不会抛错一样，说不定人家写的是 `return 2333/0`。。" ]	[ null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.66083235,"math_prob":0.97974926,"size":4818,"snap":"2019-13-2019-22","text_gpt3_token_len":2206,"char_repetition_ratio":0.26028252,"word_repetition_ratio":0.15079366,"special_character_ratio":0.28995433,"punctuation_ratio":0.12180451,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99151677,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-20T06:01:07Z\",\"WARC-Record-ID\":\"<urn:uuid:808bba5c-3c72-457d-8018-92139ce28ea6>\",\"Content-Length\":\"29532\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:90a82793-f0da-4c21-953d-af846ef64fb2>\",\"WARC-Concurrent-To\":\"<urn:uuid:4bfb2169-822b-40c7-8ffe-38acf2ecc1de>\",\"WARC-IP-Address\":\"45.63.85.117\",\"WARC-Target-URI\":\"https://gerhut.me/understanding-promise/\",\"WARC-Payload-Digest\":\"sha1:4ZAQTP2WQUTDESLMAIFGBSYI4XVR6LP2\",\"WARC-Block-Digest\":\"sha1:ULZ547JFIS5HTHTRUXO5BW4FMYPFGYMX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202299.16_warc_CC-MAIN-20190320044358-20190320070358-00345.warc.gz\"}"}
https://openstax.org/books/university-physics-volume-1/pages/15-conceptual-questions	[ "University Physics Volume 1\n\n# Conceptual Questions\n\nUniversity Physics Volume 1Conceptual Questions\n\n### 15.1Simple Harmonic Motion\n\n1.\n\nWhat conditions must be met to produce SHM?\n\n2.\n\n(a) If frequency is not constant for some oscillation, can the oscillation be SHM? (b) Can you think of any examples of harmonic motion where the frequency may depend on the amplitude?\n\n3.\n\nGive an example of a simple harmonic oscillator, specifically noting how its frequency is independent of amplitude.\n\n4.\n\nExplain why you expect an object made of a stiff material to vibrate at a higher frequency than a similar object made of a more pliable material.\n\n5.\n\nAs you pass a freight truck with a trailer on a highway, you notice that its trailer is bouncing up and down slowly. Is it more likely that the trailer is heavily loaded or nearly empty? Explain your answer.\n\n6.\n\nSome people modify cars to be much closer to the ground than when manufactured. Should they install stiffer springs? Explain your answer.\n\n### 15.2Energy in Simple Harmonic Motion\n\n7.\n\nDescribe a system in which elastic potential energy is stored.\n\n8.\n\nExplain in terms of energy how dissipative forces such as friction reduce the amplitude of a harmonic oscillator. Also explain how a driving mechanism can compensate. (A pendulum clock is such a system.)\n\n9.\n\nThe temperature of the atmosphere oscillates from a maximum near noontime and a minimum near sunrise. Would you consider the atmosphere to be in stable or unstable equilibrium?\n\n### 15.3Comparing Simple Harmonic Motion and Circular Motion\n\n10.\n\nCan this analogy of SHM to circular motion be carried out with an object oscillating on a spring vertically hung from the ceiling? Why or why not? If given the choice, would you prefer to use a sine function or a cosine function to model the motion?\n\n11.\n\nIf the maximum speed of the mass attached to a spring, oscillating on a frictionless table, was increased, what characteristics of the rotating disk would need to be changed?\n\n### 15.4Pendulums\n\n12.\n\nPendulum clocks are made to run at the correct rate by adjusting the pendulum’s length. Suppose you move from one city to another where the acceleration due to gravity is slightly greater, taking your pendulum clock with you, will you have to lengthen or shorten the pendulum to keep the correct time, other factors remaining constant? Explain your answer.\n\n13.\n\nA pendulum clock works by measuring the period of a pendulum. In the springtime the clock runs with perfect time, but in the summer and winter the length of the pendulum changes. When most materials are heated, they expand. Does the clock run too fast or too slow in the summer? What about the winter?\n\n14.\n\nWith the use of a phase shift, the position of an object may be modeled as a cosine or sine function. If given the option, which function would you choose? Assuming that the phase shift is zero, what are the initial conditions of function; that is, the initial position, velocity, and acceleration, when using a sine function? How about when a cosine function is used?\n\n### 15.5Damped Oscillations\n\n15.\n\nGive an example of a damped harmonic oscillator. (They are more common than undamped or simple harmonic oscillators.)\n\n16.\n\nHow would a car bounce after a bump under each of these conditions?\n\n(a) overdamping\n\n(b) underdamping\n\n(c) critical damping\n\n17.\n\nMost harmonic oscillators are damped and, if undriven, eventually come to a stop. Why?\n\n### 15.6Forced Oscillations\n\n18.\n\nWhy are soldiers in general ordered to “route step” (walk out of step) across a bridge?\n\n19.\n\nDo you think there is any harmonic motion in the physical world that is not damped harmonic motion? Try to make a list of five examples of undamped harmonic motion and damped harmonic motion. Which list was easier to make?\n\n20.\n\nSome engineers use sound to diagnose performance problems with car engines. Occasionally, a part of the engine is designed that resonates at the frequency of the engine. The unwanted oscillations can cause noise that irritates the driver or could lead to the part failing prematurely. In one case, a part was located that had a length L made of a material with a mass M. What can be done to correct this problem?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92756397,"math_prob":0.8792873,"size":3994,"snap":"2019-51-2020-05","text_gpt3_token_len":842,"char_repetition_ratio":0.11829574,"word_repetition_ratio":0.00295858,"special_character_ratio":0.19854783,"punctuation_ratio":0.10078534,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97546303,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-10T19:30:47Z\",\"WARC-Record-ID\":\"<urn:uuid:49e65b5f-f998-41c8-99ae-d4c8c45d7b0b>\",\"Content-Length\":\"258273\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7adc8578-2dda-4f53-9fa2-11e472cf0262>\",\"WARC-Concurrent-To\":\"<urn:uuid:41c0ad14-d449-49d8-ab1c-a9d5e89ca46a>\",\"WARC-IP-Address\":\"99.84.181.19\",\"WARC-Target-URI\":\"https://openstax.org/books/university-physics-volume-1/pages/15-conceptual-questions\",\"WARC-Payload-Digest\":\"sha1:XGPC5DPL5S3FLWTG6PP2JUKUO7XJXL2M\",\"WARC-Block-Digest\":\"sha1:CMG6L4CQPD7V2Y3WOAMRGZQOOQU4QM6A\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540528490.48_warc_CC-MAIN-20191210180555-20191210204555-00174.warc.gz\"}"}
https://cloud.tencent.com/developer/article/1327168	[ "深度 \| 朴素贝叶斯模型算法研究与实例分析\n\n(白宁超2018年9月3日15: 56:20)\n\n复旦新闻语料：朴素贝叶斯中文文本分类\n\n准备数据\n\n```'''创建数据集和类标签'''\ndocList = [];classList = [] # 文档列表、类别列表\ndirlist = ['C3-Art','C4-Literature','C5-Education','C6-Philosophy','C7-History']\nfor j in range(5):\nfor i in range(1, 11): # 总共10个文档\n# 切分，解析数据，并归类为 1 类别\ndocList.append(wordList)\nclassList.append(j)\n# print(i,'\\t','./fudan/%s/%d.txt' % (dirlist[j],i),'\\t',j)\nreturn docList,classList\n\n''' 利用jieba对文本进行分词，返回切词后的list '''\ndef textParse(str_doc):\n# 正则过滤掉特殊符号、标点、英文、数字等。\nimport re\nr1 = '[a-zA-Z0-9’!\"#\\$%&\\'()+,-./:;<=>?@，。?★、…【】《》？“”‘’！[\\\\]^_`{\|}~]+'\nstr_doc=re.sub(r1, '', str_doc)\n\n# 创建停用词列表\nstwlist = set([line.strip() for line in open('./stopwords.txt', 'r', encoding='utf-8').readlines()])\nsent_list = str_doc.split('\\n')\n# word_2dlist = [rm_tokens(jieba.cut(part), stwlist) for part in sent_list] # 分词并去停用词\nword_2dlist = [rm_tokens([word+\"/\"+flag+\" \" for word, flag in pseg.cut(part) if flag in ['n','v','a','ns','nr','nt']], stwlist) for part in sent_list] # 带词性分词并去停用词\nword_list = list(itertools.chain(word_2dlist)) # 合并列表\nreturn word_list\n\n''' 去掉一些停用词、数字、特殊符号 '''\ndef rm_tokens(words, stwlist):\nwords_list = list(words)\nfor i in range(words_list.__len__())[::-1]:\nword = words_list[i]\nif word in stwlist: # 去除停用词\nwords_list.pop(i)\nelif len(word) == 1: # 去除单个字符\nwords_list.pop(i)\nelif word == \" \": # 去除空字符\nwords_list.pop(i)\nreturn words_list```\n\n分析数据\n\n```'''获取所有文档单词的集合'''\ndef createVocabList(dataSet):\nvocabSet = set([])\nfor document in dataSet:\nvocabSet = vocabSet \| set(document) # 操作符 \| 用于求两个集合的并集\n# print(len(vocabSet),len(set(vocabSet)))\nreturn list(vocabSet)```\n\n```'''文档词袋模型，创建矩阵数据'''\ndef bagOfWords2VecMN(vocabList, inputSet):\nreturnVec = * len(vocabList)\nfor word in inputSet:\nif word in vocabList:\nreturnVec[vocabList.index(word)] += 1\nreturn returnVec```\n\n训练算法\n\n```'''朴素贝叶斯模型训练数据优化'''\ndef trainNB0(trainMatrix, trainCategory):\nnumTrainDocs = len(trainMatrix) # 总文件数\nnumWords = len(trainMatrix) # 总单词数\n\np1Num=p2Num=p3Num=p4Num=p5Num = ones(numWords) # 各类为1的矩阵\np1Denom=p2Denom=p3Denom=p4Denom=p5Denom = 2.0 # 各类特征和\nnum1=num2=num3=num4=num5 = 0 # 各类文档数目\n\npNumlist=[p1Num,p2Num,p3Num,p4Num,p5Num]\npDenomlist =[p1Denom,p2Denom,p3Denom,p4Denom,p5Denom]\nNumlist = [num1,num2,num3,num4,num5]\n\nfor i in range(numTrainDocs): # 遍历每篇训练文档\nfor j in range(5): # 遍历每个类别\nif trainCategory[i] == j: # 如果在类别下的文档\npNumlist[j] += trainMatrix[i] # 增加词条计数值\npDenomlist[j] += sum(trainMatrix[i]) # 增加该类下所有词条计数值\nNumlist[j] +=1 # 该类文档数目加1\n\npVect,pi = [],[]\nfor index in range(5):\npVect.append(log(pNumlist[index] / pDenomlist[index]))\npi.append(Numlist[index] / float(numTrainDocs))\nreturn pVect, pi```\n\n```'''朴素贝叶斯分类函数,将乘法转换为加法'''\ndef classifyNB(vec2Classify, pVect,pi):\n# 计算公式 log(P(F1\|C))+log(P(F2\|C))+....+log(P(Fn\|C))+log(P(C))\nbnpi = [] # 文档分类到各类的概率值列表\nfor x in range(5):\nbnpi.append(sum(vec2Classify * pVect[x]) + log(pi[x]))\n# print([bnp for bnp in bnpi])\n# 分类集合\nreslist = ['Art','Literature','Education','Philosophy','History']\n# 根据最大概率，选择索引值\nindex = [bnpi.index(res) for res in bnpi if res==max(bnpi)]\nreturn reslist[index] # 返回分类值```\n\n测试算法\n\n```'''朴素贝叶斯新闻分类应用'''\ndef testingNB():\n# 1. 加载数据集\n# 2. 创建单词集合\nmyVocabList = createVocabList(dataSet)\n\n# 3. 计算单词是否出现并创建数据矩阵\ntrainMat = []\nfor postinDoc in dataSet:\ntrainMat.append(bagOfWords2VecMN(myVocabList, postinDoc))\nwith open('./word-bag.txt','w') as f:\nfor i in trainMat:\nf.write(str(i)+'\\r\\n')\n# 4. 训练数据\npVect,pi= trainNB0(array(trainMat), array(Classlabels))\n# 5. 测试数据\nthisDoc = array(bagOfWords2VecMN(myVocabList, testEntry))\nprint(testEntry[:10], '分类结果是: ', classifyNB(thisDoc, pVect,pi))```\n\n```Building prefix dict from the default dictionary ...\nPrefix dict has been built succesfully.\n['全国/n ', '举办/v ', '电影/n ', '新华社/nt ', '北京/ns ', '国家教委/nt ', '广播电影电视部/nt ', '文化部/n ', '联合/v ', '决定/v '] 分类结果是: Literature\n\n特征选择问题讨论\n\n• 做文本分类的时候，遇到特征矩阵1.5w。在测试篇幅小的文章总是分类错误？这个时候如何做特征选择？是不是说去掉特征集中频率极高和极低的一部分，对结果有所提升？答：你说的这个情况是很普遍的现象，篇幅小的文章，特征小，所以模型更容易判断出错！去掉高频和低频通常是可以使得训练的模型泛化能力变强\n• 比如：艺术，文化，历史，教育。界限本来就不明显，比如测试数据“我爱艺术，艺术是我的全部”。结果会分类为文化。其实这个里面还有就是不同特征词的权重问题，采用tf-idf优化下应该会好一些？答：我个人觉得做文本特征提取，还是需要自己去分析文本本身内容的文字特点，你可以把每一类的文本的实体提取出来，然后统计一下每个词在每一类上的数量，看看数量分布，也许可以发现一些数据特点\n• 我就是按照这个思路做的，还有改进时候的停用词，其实可以分析特征文本，针对不同业务，使用自定义的停用词要比通用的好还有提前各类见最具表征性的词汇加权，凸显本类的权重是吧？答：比如，艺术类文章中，哪些词出现较多，哪些词出现少，再观察这些词的词性主要是哪些，这样可能会对你制定提取特征规则方式的时候提供一定的思路参考，我可以告诉你的是，有些词绝对会某一类文章出出现多，然后在其他类文章出现很少，这一类的词就是文章的特征词\n• 那样的思路可以是：对某类文章单独构建类内的词汇表再进行选择。最后对类间词汇表叠加就ok了。答：词汇表有个缺点就是，不能很好的适应新词\n• 改进思路呢答：我给你一个改进思路：你只提取每个文本中的名词、动词、形容词、地名，用这些词的作为文本的特征来训练试一试，用文本分类用主题模型（LDA）来向量化文本，再训练模型试一试。如果效果还是不够好，再将文本向量用PCA进行一次特征降维，然后再训练模型试一试，按常理来说，效果应该会有提高\n• 还有我之前个人写的程序分类效果不理想，后来改用sklearn内置BN运行依旧不理想。适当改进了特征提取，还是不理想。估计每类10篇文章的训练数据太少了答：文本本身特征提取就相对难一些，再加上训练数据少，训练出来的模型效果可想而已，正常的\n\nsklearn：朴素贝叶斯分类调用\n\n数据准备和数据预处理\n\n```myVocabList = [] # 设置词汇表的全局变量\n\n'''创建数据集和类标签'''\ndocList = [];classList = [] # 文档列表、类别列表、文本特征\ndirlist = ['C3-Art','C4-Literature','C5-Education','C6-Philosophy','C7-History']\nfor j in range(5):\nfor i in range(1, 11): # 总共10个文档\n# 切分，解析数据，并归类为 1 类别\ndocList.append(wordList)\nclassList.append(j)\n# print(i,'\\t','./fudan/%s/%d.txt' % (dirlist[j],i),'\\t',j)\n# print(len(docList),len(classList),len(fullText))\nglobal myVocabList\nmyVocabList = createVocabList(docList) # 创建单词集合\nreturn docList,classList,myVocabList\n\n''' 利用jieba对文本进行分词，返回切词后的list '''\ndef textParse(str_doc): #与上文方法一致\n\n''' 去掉一些停用词、数字、特殊符号 '''\ndef rm_tokens(words, stwlist): #与上文方法一致```\n\n```# 本地存储数据集和标签\ndef storedata():\n# 3. 计算单词是否出现并创建数据矩阵\n# trainMat =[[0,1,2,3],[2,3,1,5],[0,1,4,2]] # 训练集\n# classList = [0,1,2] #类标签\n# 计算单词是否出现并创建数据矩阵\ntrainMat = []\nfor postinDoc in docList:\ntrainMat.append(bagOfWords2VecMN(myVocabList, postinDoc))\nres = \"\"\nfor i in range(len(trainMat)):\nres +=' '.join([str(x) for x in trainMat[i]])+' '+str(classList[i])+'\\n'\n# print(res[:-1]) # 删除最后一个换行符\nwith open('./word-bag.txt','w') as fw:\nfw.write(res[:-1])\nwith open('./wordset.txt','w') as fw:\nfw.write(' '.join([str(v) for v in myVocabList]))\n\n# 读取本地数据集和标签\ndef grabdata():\nf = open('./word-bag.txt') # 读取本地文件\ntzsize = len(arrayLines.split(' '))-1 # 列向量，特征个数减1即数据集\nreturnMat = zeros((len(arrayLines),tzsize)) # 0矩阵数据集\nclassLabelVactor = [] # 标签集，特征最后一列\n\nindex = 0\nfor line in arrayLines: # 逐行读取\nlistFromLine = line.strip().split(' ') # 分析数据，空格处理\n# print(listFromLine)\nreturnMat[index,:] = listFromLine[0:tzsize] # 数据集\nclassLabelVactor.append(int(listFromLine[-1])) # 类别标签集\nindex +=1\n# print(returnMat,classLabelVactor)\nmyVocabList=writewordset()\nreturn returnMat,classLabelVactor,myVocabList\n\ndef writewordset():\nf1 = open('./wordset.txt')\nfor w in myVocabList:\nif w=='':\nmyVocabList.remove(w)\nreturn myVocabList```\n\n```'''获取所有文档单词的集合'''\ndef createVocabList(dataSet):\nvocabSet = set([])\nfor document in dataSet:\nvocabSet = vocabSet \| set(document) # 操作符 \| 用于求两个集合的并集\n# print(len(vocabSet),len(set(vocabSet)))\nreturn list(vocabSet)\n\n'''文档词袋模型，创建矩阵数据'''\ndef bagOfWords2VecMN(vocabList, inputSet):\nreturnVec = * len(vocabList)\nfor word in inputSet:\nif word in vocabList:\nreturnVec[vocabList.index(word)] += 1\nreturn returnVec```\n\n高斯朴素贝叶斯\n\nGaussianNB 实现了运用于分类的高斯朴素贝叶斯算法。特征的可能性(即概率)假设为高斯分布:\n\n```'''高斯朴素贝叶斯'''\ndef MyGaussianNB(trainMat='',Classlabels='',testDoc=''):\n# -----sklearn GaussianNB-------\n# 训练数据\nX = np.array(trainMat)\nY = np.array(Classlabels)\n# 高斯分布\nclf = GaussianNB()\nclf.fit(X, Y)\n# 测试预测结果\nindex = clf.predict(testDoc) # 返回索引\nreslist = ['Art','Literature','Education','Philosophy','History']\nprint(reslist[index])```\n\n多项朴素贝叶斯\n\nMultinomialNB 实现了服从多项分布数据的朴素贝叶斯算法，也是用于文本分类(这个领域中数据往往以词向量表示，尽管在实践中 tf-idf 向量在预测时表现良好)的两大经典朴素贝叶斯算法之一。分布参数由每类 y 的\n\n```'''多项朴素贝叶斯'''\ndef MyMultinomialNB(trainMat='',Classlabels='',testDoc=''):\n# -----sklearn MultinomialNB-------\n# 训练数据\nX = np.array(trainMat)\nY = np.array(Classlabels)\n# 多项朴素贝叶斯\nclf = MultinomialNB()\nclf.fit(X, Y)\n# 测试预测结果\nindex = clf.predict(testDoc) # 返回索引\nreslist = ['Art','Literature','Education','Philosophy','History']\nprint(reslist[index])```\n\n伯努利朴素贝叶斯\n\nBernoulliNB 实现了用于多重伯努利分布数据的朴素贝叶斯训练和分类算法，即有多个特征，但每个特征都假设是一个二元 (Bernoulli, boolean) 变量。因此，这类算法要求样本以二元值特征向量表示；如果样本含有其他类型的数据，一个 BernoulliNB 实例会将其二值化(取决于 binarize 参数)。伯努利朴素贝叶斯的决策规则基于\n\n```'''伯努利朴素贝叶斯'''\ndef MyBernoulliNB(trainMat='',Classlabels='',testDoc=''):\n# -----sklearn BernoulliNB-------\n# 训练数据\nX = np.array(trainMat)\nY = np.array(Classlabels)\n# 多项朴素贝叶斯\nclf = BernoulliNB()\nclf.fit(X, Y)\n# 测试预测结果\nindex = clf.predict(testDoc) # 返回索引\nreslist = ['Art','Literature','Education','Philosophy','History']\nprint(reslist[index])```\n\n各种贝叶斯模型分类测试\n\n```# 加载数据集和单词集合\ntrainMat,Classlabels,myVocabList = grabdata() # 读取训练结果\n# 测试数据\ntestDoc = np.array(bagOfWords2VecMN(myVocabList, testEntry)) # 测试数据\n# 测试预测结果\nMyGaussianNB(trainMat,Classlabels,testDoc)\nMyMultinomialNB(trainMat,Classlabels,testDoc)\nMyBernoulliNB(trainMat,Classlabels,testDoc)```\n\n```Building prefix dict from the default dictionary ...\nPrefix dict has been built succesfully.\n\n参考文献\n\n1. scikit中文社区：http://sklearn.apachecn.org/cn/0.19.0/\n2. 中文维基百科：https://zh.wikipedia.org/wiki/\n3. 文本分类特征选择：https://www.cnblogs.com/june0507/p/7601001.html\n4. GitHub：https://github.com/BaiNingchao/MachineLearning-1\n5. 图书：《机器学习实战》\n6. 图书：《自然语言处理理论与实战》\n\n74 篇文章21 人订阅\n\n0 条评论\n\n相关文章\n\n4848\n\n2907\n\n831\n\n1233\n\n6148\n\nNAS框架AUTO Keras帮你搜索最佳深度模型\n\n【导读】Neural Architecture Search (NAS)是当前正热的科研和工程方向，旨在帮人们自动搜索最适合当前任务的深度神经网络，低成本地获得...\n\n7584\n\n3656\n\n1303\n\n【论文笔记】中文词向量论文综述（一）\n\n1072", null, "" ]	[ null, "https://imgcache.qq.com/open_proj/proj_qcloud_v2/community/portal/css/img/wechat-qr.jpg", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.58354354,"math_prob":0.7399239,"size":11322,"snap":"2019-26-2019-30","text_gpt3_token_len":6171,"char_repetition_ratio":0.109118216,"word_repetition_ratio":0.22983871,"special_character_ratio":0.2733616,"punctuation_ratio":0.1891892,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9677999,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-24T23:46:18Z\",\"WARC-Record-ID\":\"<urn:uuid:66437de6-7d5e-4b6b-88da-682f356d3b20>\",\"Content-Length\":\"146931\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7f30b10a-6169-4fe7-91f7-09cb1750856f>\",\"WARC-Concurrent-To\":\"<urn:uuid:e269d080-968d-4beb-9b87-108314ca088f>\",\"WARC-IP-Address\":\"119.28.39.127\",\"WARC-Target-URI\":\"https://cloud.tencent.com/developer/article/1327168\",\"WARC-Payload-Digest\":\"sha1:BWSOLV3CFF5PJM4LFTYZ2GGNUNPZGI5P\",\"WARC-Block-Digest\":\"sha1:CIMCBY7L5IZ2FORPPKR4VWQUK5CN7PEP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999779.19_warc_CC-MAIN-20190624231501-20190625013501-00078.warc.gz\"}"}
https://www.pragationline.com/semester-2-sppu-bba-ca/business-mathematics-6-rayarikar-dixit-2/	[ "190.00", null, "190.00\n\n Authors Name A. V. Rayarikar , Dr. P. G. Dixit ISBN 13 9789389686425 Publisher Edition First Buy E-Book (PDF) [Click on Logo >>]", null, "", null, "Pages 184 Language English Publishing Year Nov-19\nSKU: N4943\nFor E-Books purchased from Kopykitab.com For E-Books purchased from Amazon\n##### Description\n\n1. Ratio, Proportion and Percentage\n\nRatio – Definition, Continued Ratio, Inverse Ration, Proportion, Continued Proportion, Direct Proportion, Inverse Proportion, Variation, Inverse Variation, Joint Variation, Percentage, Computation of Percentage.\n\n2. Profit and Loss\n\nTerms and Formulae, Trade discount, Cash discount, Problems involving cost price, selling price, Trade discount and cash discount. Introduction to Commission and brokerage, Problems on commission and brokerage\n\n3. Interest and Annuity\n\nInterest, Compound interest, Equated monthly Installments (EMI) by interest of reducing balance and flat interest methods and problems.\nOrdinary annuity, sinker fund, annuity due, present value and future value of annuity.\n\n4. Shares and Mutual Funds\n\nConcepts of Shares, face value, market value, dividend, brokerage, equity shares, preferential shares, bonus shares, examples and problems, Concept of Mutual Funds, Change\nin Net Asset Value (NAV), Systematic Investment Plan (SIP), Examples and Problems.\n\n5. Matrices and Determinants\n\nDefinition of Matrices, Types of Matrices, Algebra of Matrices, Determinant, Adjoint of Matrix, Inverse of Matrix, System of Linear equations, Solution of System of Linear Equation by adjoint method (upto 3 variables only).\n\n6. Linear Programming Problems\n\nConcept of LPP, Formulation of LPP and solution of LPP by graphical method.\n\n7. Transportation Problem\n\nConcept of Transportation Problem, Initial Basic Feasible Solution, North-West Corner Method (NWCM), Least Cost Method (LCM), Vogal’s Approximation Method (VAM).\n\n* Model Question Paper", null, "" ]	[ null, "data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==", null, "data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==", null, "https://www.pragationline.com/wp-content/uploads/2019/03/ebook-logo.jpg", null, "data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.81173307,"math_prob":0.6583227,"size":1610,"snap":"2021-31-2021-39","text_gpt3_token_len":353,"char_repetition_ratio":0.12889166,"word_repetition_ratio":0.0,"special_character_ratio":0.19254659,"punctuation_ratio":0.22105263,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.979944,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-04T16:51:41Z\",\"WARC-Record-ID\":\"<urn:uuid:16b782c4-a7c0-4099-9b72-f8e42a1db255>\",\"Content-Length\":\"213059\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e36285dd-9f2d-420e-936f-1fdc46852831>\",\"WARC-Concurrent-To\":\"<urn:uuid:1bbdb92d-ad51-45ec-8d4d-99d8d2f62f1a>\",\"WARC-IP-Address\":\"216.10.243.131\",\"WARC-Target-URI\":\"https://www.pragationline.com/semester-2-sppu-bba-ca/business-mathematics-6-rayarikar-dixit-2/\",\"WARC-Payload-Digest\":\"sha1:ITSPBZ5SAUCDW3QSQJNQFQA3DGKEYSNB\",\"WARC-Block-Digest\":\"sha1:FA5MV2KSH45C4SFXBX64JDUDYNSVS5TD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154878.27_warc_CC-MAIN-20210804142918-20210804172918-00286.warc.gz\"}"}
https://math.illinoisstate.edu/day/courses/old/305/contentderangements.html	[ "Illinois State University Mathematics Department\n\n MAT 305: Combinatorics Topics for K-8 Teachers\n\n### Derangements\n\nHere we apply the Inclusion-Exclusion Principle to a problem of derangements:\n\nIn an apartment complex with k mailboxes, how many ways can a mail carrier distribute k letters, one addressed to each letter box, such that no letter is placed in the correct box? Each of these is called a derangement.\n\nLet us refer to a letter by a number and to a mailbox by a number. Our task is to determine the number of ways to pair letters and boxes so that no letter numbers match box numbers. If k = 1, there are no ways to derange the letter, for there is one letter to place in one box. If k = 2, we may place letter #2 in box #1 and letter #1 in box #2, that being the only derangement.\n\nWhen we have three letters, there are 3! = 6 total ways to distribute them. We now write the letter numbers in the order they are delivered, such as 1 3 2, indicating letter #1 is delivered to box #1, letter #3 is delivered to box #2, and letter #2 is delivered to box #3. The 6 possible distributions for 3 letters are\n\n 1 2 3 `1 3 2` 2 1 3 `2 3 1` 3 1 2 `3 2 1`\n\nThe schemes 2 3 1 and 3 1 2 are the only derangements of three letters.\n\nSummarizing our results thus far, using D(n) to represent the number of derangements of n letters, we have D(1) = 0, D(2) = 1, and D(3) = 2. Any guesses on D(4)? Let's find out what it is.\n\nWe know there are 4! = 24 ways to distribute the 4 letters. Rather than list the 24 cases, let us consider how the Inclusion-Exclusion Principle may help us. We seek the number of ways to place the numbers in the set {1,2,3,4} in line such that no value occurs in its natural position. Let X(1) represent the property that 1 occurs in its natural position when 1,2,3,4 are permuted. Then \|X(1)\| = (1)3!. The 1 represents the 1 way to place the 1 in its natural position and the 3! shows the number of ways to permute the remaining 3 values. Note that we are not considering whether any of 2,3,4 wind up in their respective natural positions. We could argue similarly that \|X(2)\| = \|X(3)\| = \|X(4)\|. Therefore, there are 43! ways for a value to occur in its natural position.\n\nWhat about \|X(1) ^ X(2)\|? Again there is 1 way to place 1,2 in their natural order, and then 2! ways to place the remaining values. This will be true for any pair of values we wish to restrict to their natural positions. How many pairs are there? This is just C(4,2) = 6. Therefore, there are C(4,2)2! ways for two values to simultaneously occur in their natural positions.\n\nAnd for \|X(1) ^ X(2) ^ X(3)\|? Again there is 1 way to place 1,2,3 in their natural order, and then 1! way to place the remaining value. This will be true for any set of three values we wish to restrict to their natural positions. How many 3-element sets are there? This is just C(4,3) = 4. Therefore, there are C(4,3)1! ways for three values to simultaneously occur in their natural positions.\n\nFinally, \|X(1) ^ X(2) ^ X(3) ^ X(4)\| = 1, for there is only one way for all 4 values to be in their natural positions.\n\nNow apply the Inclusion-Exclusion Principle (I-E P):\n\n\|~X(1) ^ ~X(2) ^ ~X(3) ^ X(4)\| = 4! - 4(3!) + 6(2!) - 4(1!) + 1 = 9. In words, using the I-E P, we are suggesting that to determine the number of derangements of the values 1,2,3,4, first calculate the number of permutations of those values (4!), subtract the number of ways to keep at least one element in its natural position, add back the number of ways to keep at least two values in their natural positions, subtract the number of ways to keep at least three values in their natural positions, and finally add back the number of ways to keep all values in their natural positions.\n\nWe can rewrite the right side of the above equation to better express the general result:", null, "If we begin with n objects rather than 4, we can argue in a similar way that", null, "determines the number of derangements of n items.\n\n#### Recursive Determination of Derangements\n\nWe now consider derangements recursively. That is, by knowing the few easy-to-calculate values for derangements of small numbers of objects, can we determine a pattern for the number of derangements of larger numbers of elements?\n\nSuppose we want to determine the number of derangements of the n integers 1,2,...,n for n bigger than 2. Let us focus on k and move it into the first position. We thus have started a derangement, for 1 is not in its natural position. Where could 1 be placed? There are two cases we could consider: either 1 is in position k or 1 is not in position k.\n\nIf 1 is in position k, here's what we know: The integers 1 and k have simply traded positions.\n\n Prohibited Value 1 2 3 ... k-1 k k+1 ... n Derangement k ? ? ... ? 1 ? ... ?\n\nIndicated by the question marks, there are (n-2) integers yet to derange. This can be done in D(n-2) ways.\n\nIf 1 is not in position k, we don't know as much. Note that we have shown 1 as a prohibited value twice. This is required for this case, because we cannot have 1 appear in the first position (its natural position) nor can 1 appear in position k.\n\n Prohibited Value 1 2 3 ... k-1 1 k+1 ... n Derangement k ? ? ... ? ? ? ... ?\n\nIndicated by the question marks, there are now (n-1) integers yet to derange. This can be done in D(n-1) ways.\n\nPutting this together, we have D(n-1) + D(n-2) possible derangements when k is in the first position. How many different integers could we put in position 1 and carry out this process? All the integers except 1. that is, (n-1) different integers.\n\nThis yields the recursive formula D(n) = (n-1)[D(n-1) + D(n-2)]. As long as we know D(1) = 0 and D(2) = 1, we can generate subsequent values for D(n).\n\n Syllabus Grades & Grading Content Notes Session Outlines Assignments and Problem Sets Tests and Quizzes" ]	[ null, "https://math.illinoisstate.edu/day/courses/old/305/S1401.gif", null, "https://math.illinoisstate.edu/day/courses/old/305/S1402.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8939408,"math_prob":0.9903565,"size":5568,"snap":"2019-51-2020-05","text_gpt3_token_len":1572,"char_repetition_ratio":0.17774983,"word_repetition_ratio":0.12638377,"special_character_ratio":0.29471982,"punctuation_ratio":0.117790416,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9896195,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-09T23:38:33Z\",\"WARC-Record-ID\":\"<urn:uuid:333919b0-21bf-4188-91c6-337f02c1839a>\",\"Content-Length\":\"11321\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8c5475ac-c012-4874-967e-cee9ce0235bb>\",\"WARC-Concurrent-To\":\"<urn:uuid:8239ce00-0ad5-485a-99b5-99d82d89390c>\",\"WARC-IP-Address\":\"138.87.246.20\",\"WARC-Target-URI\":\"https://math.illinoisstate.edu/day/courses/old/305/contentderangements.html\",\"WARC-Payload-Digest\":\"sha1:JPCF3GMWWLFWDJAOWAKJEULTRBSAVW72\",\"WARC-Block-Digest\":\"sha1:ZR36L4GDR4SQTIEIH3ZONHGYGPRPO6KK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540525598.55_warc_CC-MAIN-20191209225803-20191210013803-00423.warc.gz\"}"}
https://www.colorhexa.com/49635c	[ "# #49635c Color Information\n\nIn a RGB color space, hex #49635c is composed of 28.6% red, 38.8% green and 36.1% blue. Whereas in a CMYK color space, it is composed of 26.3% cyan, 0% magenta, 7.1% yellow and 61.2% black. It has a hue angle of 163.8 degrees, a saturation of 15.1% and a lightness of 33.7%. #49635c color hex could be obtained by blending #92c6b8 with #000000. Closest websafe color is: #336666.\n\n• R 29\n• G 39\n• B 36\nRGB color chart\n• C 26\n• M 0\n• Y 7\n• K 61\nCMYK color chart\n\n#49635c color description : Very dark grayish cyan - lime green.\n\n# #49635c Color Conversion\n\nThe hexadecimal color #49635c has RGB values of R:73, G:99, B:92 and CMYK values of C:0.26, M:0, Y:0.07, K:0.61. Its decimal value is 4809564.\n\nHex triplet RGB Decimal 49635c `#49635c` 73, 99, 92 `rgb(73,99,92)` 28.6, 38.8, 36.1 `rgb(28.6%,38.8%,36.1%)` 26, 0, 7, 61 163.8°, 15.1, 33.7 `hsl(163.8,15.1%,33.7%)` 163.8°, 26.3, 38.8 336666 `#336666`\nCIE-LAB 39.769, -11.307, 0.833 9.141, 11.113, 11.788 0.285, 0.347, 11.113 39.769, 11.337, 175.787 39.769, -12.777, 2.702 33.335, -9.391, 2.369 01001001, 01100011, 01011100\n\n# Color Schemes with #49635c\n\n• #49635c\n``#49635c` `rgb(73,99,92)``\n• #634950\n``#634950` `rgb(99,73,80)``\nComplementary Color\n• #49634f\n``#49634f` `rgb(73,99,79)``\n• #49635c\n``#49635c` `rgb(73,99,92)``\n• #495d63\n``#495d63` `rgb(73,93,99)``\nAnalogous Color\n• #634f49\n``#634f49` `rgb(99,79,73)``\n• #49635c\n``#49635c` `rgb(73,99,92)``\n• #63495d\n``#63495d` `rgb(99,73,93)``\nSplit Complementary Color\n• #635c49\n``#635c49` `rgb(99,92,73)``\n• #49635c\n``#49635c` `rgb(73,99,92)``\n• #5c4963\n``#5c4963` `rgb(92,73,99)``\n• #506349\n``#506349` `rgb(80,99,73)``\n• #49635c\n``#49635c` `rgb(73,99,92)``\n• #5c4963\n``#5c4963` `rgb(92,73,99)``\n• #634950\n``#634950` `rgb(99,73,80)``\n• #293733\n``#293733` `rgb(41,55,51)``\n• #334641\n``#334641` `rgb(51,70,65)``\n• #3e544e\n``#3e544e` `rgb(62,84,78)``\n• #49635c\n``#49635c` `rgb(73,99,92)``\n• #54726a\n``#54726a` `rgb(84,114,106)``\n• #5f8077\n``#5f8077` `rgb(95,128,119)``\n• #698f85\n``#698f85` `rgb(105,143,133)``\nMonochromatic Color\n\n# Alternatives to #49635c\n\nBelow, you can see some colors close to #49635c. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #496356\n``#496356` `rgb(73,99,86)``\n• #496358\n``#496358` `rgb(73,99,88)``\n• #49635a\n``#49635a` `rgb(73,99,90)``\n• #49635c\n``#49635c` `rgb(73,99,92)``\n• #49635e\n``#49635e` `rgb(73,99,94)``\n• #496360\n``#496360` `rgb(73,99,96)``\n• #496363\n``#496363` `rgb(73,99,99)``\nSimilar Colors\n\n# #49635c Preview\n\nThis text has a font color of #49635c.\n\n``<span style=\"color:#49635c;\">Text here</span>``\n#49635c background color\n\nThis paragraph has a background color of #49635c.\n\n``<p style=\"background-color:#49635c;\">Content here</p>``\n#49635c border color\n\nThis element has a border color of #49635c.\n\n``<div style=\"border:1px solid #49635c;\">Content here</div>``\nCSS codes\n``.text {color:#49635c;}``\n``.background {background-color:#49635c;}``\n``.border {border:1px solid #49635c;}``\n\n# Shades and Tints of #49635c\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, #060908 is the darkest color, while #fcfdfd is the lightest one.\n\n• #060908\n``#060908` `rgb(6,9,8)``\n• #0f1413\n``#0f1413` `rgb(15,20,19)``\n• #171f1d\n``#171f1d` `rgb(23,31,29)``\n• #1f2b28\n``#1f2b28` `rgb(31,43,40)``\n• #283632\n``#283632` `rgb(40,54,50)``\n• #30413d\n``#30413d` `rgb(48,65,61)``\n• #384c47\n``#384c47` `rgb(56,76,71)``\n• #415852\n``#415852` `rgb(65,88,82)``\n• #49635c\n``#49635c` `rgb(73,99,92)``\n• #516e66\n``#516e66` `rgb(81,110,102)``\n• #5a7a71\n``#5a7a71` `rgb(90,122,113)``\n• #62857b\n``#62857b` `rgb(98,133,123)``\n• #6a9086\n``#6a9086` `rgb(106,144,134)``\n• #75998f\n``#75998f` `rgb(117,153,143)``\n• #80a199\n``#80a199` `rgb(128,161,153)``\n• #8baaa2\n``#8baaa2` `rgb(139,170,162)``\n• #97b2ab\n``#97b2ab` `rgb(151,178,171)``\n• #a2bab4\n``#a2bab4` `rgb(162,186,180)``\n``#adc3bd` `rgb(173,195,189)``\n• #b9cbc6\n``#b9cbc6` `rgb(185,203,198)``\n• #c4d3cf\n``#c4d3cf` `rgb(196,211,207)``\n• #cfdcd8\n``#cfdcd8` `rgb(207,220,216)``\n• #dbe4e2\n``#dbe4e2` `rgb(219,228,226)``\n• #e6eceb\n``#e6eceb` `rgb(230,236,235)``\n• #f1f5f4\n``#f1f5f4` `rgb(241,245,244)``\n• #fcfdfd\n``#fcfdfd` `rgb(252,253,253)``\nTint Color Variation\n\n# Tones of #49635c\n\nA tone is produced by adding gray to any pure hue. In this case, #505c59 is the less saturated color, while #00ac7e is the most saturated one.\n\n• #505c59\n``#505c59` `rgb(80,92,89)``\n• #49635c\n``#49635c` `rgb(73,99,92)``\n• #426a5f\n``#426a5f` `rgb(66,106,95)``\n• #3c7062\n``#3c7062` `rgb(60,112,98)``\n• #357765\n``#357765` `rgb(53,119,101)``\n• #2f7d68\n``#2f7d68` `rgb(47,125,104)``\n• #28846b\n``#28846b` `rgb(40,132,107)``\n• #218b6e\n``#218b6e` `rgb(33,139,110)``\n• #1b9171\n``#1b9171` `rgb(27,145,113)``\n• #149874\n``#149874` `rgb(20,152,116)``\n• #0d9f77\n``#0d9f77` `rgb(13,159,119)``\n• #07a57b\n``#07a57b` `rgb(7,165,123)``\n• #00ac7e\n``#00ac7e` `rgb(0,172,126)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #49635c 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":0.57527566,"math_prob":0.70477426,"size":3718,"snap":"2021-43-2021-49","text_gpt3_token_len":1642,"char_repetition_ratio":0.11927841,"word_repetition_ratio":0.011009174,"special_character_ratio":0.5658956,"punctuation_ratio":0.2370452,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9925679,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-22T12:33:38Z\",\"WARC-Record-ID\":\"<urn:uuid:b3bdeaaa-bcb9-4b63-93e0-4ed42bbc60e1>\",\"Content-Length\":\"36157\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a2e94b22-274b-4dc4-996e-a4167a06a7f0>\",\"WARC-Concurrent-To\":\"<urn:uuid:5c961900-7163-42c0-9f9a-8977a6baedde>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/49635c\",\"WARC-Payload-Digest\":\"sha1:Q2O56X4RTVQES6QVASMGDXSW4QJZ5PFV\",\"WARC-Block-Digest\":\"sha1:RWYJ2KT3DP35OGHJJMXUTUELN6PNMSRP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585507.26_warc_CC-MAIN-20211022114748-20211022144748-00090.warc.gz\"}"}
https://math.stackexchange.com/questions/726543/a-property-of-cardinal-numbers	[ "# A property of cardinal numbers.\n\nLet $\\mathfrak{m,n,p}$ cardinal numbers, show that $\\mathfrak{m^{n+p}=m^n\\cdot m^p}$.\n\nI believe that the proof is based on showing that there is a bijection between $M^{N_1\\cup N_2}$ and $M^{N_1}\\times M^{N_2}$, where card $M=\\mathfrak{m}$, card $N_1=\\mathfrak{n}$ and card $N_2=\\mathfrak{p}$, with $N_1\\cap N_2=\\emptyset$, but I could not find such a bijection. Any idea? Thanks!\n\n• Can you find a map $M^{N_1\\cup N_2} \\to M^{N_1}$? – Daniel Fischer Mar 25 '14 at 19:13\n\nGiven a map $f\\colon N_1\\cup N_2\\to M$, this induces two maps $f\|_{N_i}\\colon N_i\\to M$. So you have a map $M^{N_1\\cup N_2}\\to M^{N_1}\\times M^{N_2}$ given by $f\\mapsto (f\|_{N_1},f\|_{N_2})$.\nConversely, suppose you have a pair $(f,g)\\in M^{N_1}\\times M^{N_2}$. Since $N_1\\cap N_2=\\emptyset$, this allows you to define a new function $h\\colon N_1\\cup N_2\\to M$, $$h(x)=\\begin{cases} f(x) &&\\text{if }x\\in N_1,\\\\ g(x) &&\\text{if }x\\in N_2. \\end{cases}$$\nSo you have another map $(f,g)\\mapsto h$. You can then verify these two maps provide a bijection, so that $M^{N_1\\cup N_2}\\simeq M^{N_1}\\times M^{N_2}$, which gives the equality of the cardinals." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.849068,"math_prob":0.9999809,"size":380,"snap":"2019-51-2020-05","text_gpt3_token_len":144,"char_repetition_ratio":0.16223404,"word_repetition_ratio":0.0,"special_character_ratio":0.34736842,"punctuation_ratio":0.13095239,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.000004,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-12T21:40:52Z\",\"WARC-Record-ID\":\"<urn:uuid:3cc504b3-abe6-42b9-95aa-99f2223e1b7f>\",\"Content-Length\":\"132722\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f309e05f-7fa1-403d-b094-dca34b867c27>\",\"WARC-Concurrent-To\":\"<urn:uuid:603aff9f-7c94-484f-8ca9-18d9541e51a8>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/726543/a-property-of-cardinal-numbers\",\"WARC-Payload-Digest\":\"sha1:OJWB6HOQ6MSRUFNFQS2I6PSNQGGLFY5A\",\"WARC-Block-Digest\":\"sha1:TEGNB7LEUCI2CV4DLNX4KNYLHD2LHKRO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540547165.98_warc_CC-MAIN-20191212205036-20191212233036-00004.warc.gz\"}"}
https://answers.everydaycalculation.com/add-fractions/10-4-plus-15-84	[ "Solutions by everydaycalculation.com\n\n1st number: 2 2/4, 2nd number: 15/84\n\n10/4 + 15/84 is 75/28.\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 4 and 84 is 84\n2. For the 1st fraction, since 4 × 21 = 84,\n10/4 = 10 × 21/4 × 21 = 210/84\n3. Likewise, for the 2nd fraction, since 84 × 1 = 84,\n15/84 = 15 × 1/84 × 1 = 15/84\n210/84 + 15/84 = 210 + 15/84 = 225/84\n5. After reducing the fraction, the answer is 75/28\n6. In mixed form: 219/28\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 ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.67718935,"math_prob":0.9969799,"size":377,"snap":"2019-51-2020-05","text_gpt3_token_len":162,"char_repetition_ratio":0.20107238,"word_repetition_ratio":0.0,"special_character_ratio":0.5225464,"punctuation_ratio":0.086021505,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99746794,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-08T16:47:40Z\",\"WARC-Record-ID\":\"<urn:uuid:e2cf3a81-998a-4793-a530-a2af7ee3987d>\",\"Content-Length\":\"8274\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8406fec4-f5bd-4191-994c-bb6f4f41f3cf>\",\"WARC-Concurrent-To\":\"<urn:uuid:ba8f6222-5d04-46ca-8a71-e63d35faa31f>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/add-fractions/10-4-plus-15-84\",\"WARC-Payload-Digest\":\"sha1:CHD3FMNTFHGSBXXFEHDLPTMVXWQAEJNI\",\"WARC-Block-Digest\":\"sha1:QP4PCECKUXWQVY4U4EFTMWD2K5YVXFRK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540511946.30_warc_CC-MAIN-20191208150734-20191208174734-00002.warc.gz\"}"}
https://resource.flexrule.com/knowledge-base/analytics-commands/	[ "# Analytics Commands\n\nContents\n\n### FlexRule.Extensions.Analytics\n\nThis extension enables Predictive Analytics capability to FlexRule platform. The extensions adds both expression commands Toolbox commands.\n\n#### train\n\nTrains and builds a predictive model using data.\n\n` train (list, algorithm, class, options)`\n• algorithm:\n• Regression Model\n• Generalized Regression Model\n• Naive Bayes Model\n• Classification and Regression Tree Model\n• Support Vector Machine Model\n• Random Forest (Mining Model)\n• Neural Networks Model\n• Clustering Model\n• Gradient Boosting Machine Model (Mining Model)\n• Association Rules Model\n• K-Nearest Neighbors Model\n• c45: Decision Tree\n• nb: Naive Bayes\n• class: attributes name that is the class name (i.e. attribute or field of data that going to be the result of prediction)\n• options: Algorithm options if it is available\n• Return: analytic’s trainedModel\n\n#### predict\n\nUses a trained model to predict results for a data record.\n\n` predict (trainedModel, data, options)`\n• trainedModel: the train function result.\n• data: Situation that requires the prediction. (i.e. a JSON with fields of value)\n• options: Algorithm options if it is available\n\n#### extractRules\n\nExtracts rules for Decision Tree.\n\n` extractRules (trainedModel, type)`\n• type: if, elseIf\n• Retrun: list of rules that are bases of the trained model\n\n### Examples\n\nThe following examples are available by default when you open FlexRule Designer.\n\nUpdated on June 3, 2021" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6711509,"math_prob":0.76225907,"size":1450,"snap":"2021-31-2021-39","text_gpt3_token_len":335,"char_repetition_ratio":0.13278009,"word_repetition_ratio":0.04608295,"special_character_ratio":0.21103448,"punctuation_ratio":0.14522822,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97537285,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-17T21:54:42Z\",\"WARC-Record-ID\":\"<urn:uuid:abd868e7-9c09-42f4-8889-3e159dc58481>\",\"Content-Length\":\"57135\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:82142036-8ac9-425c-b266-6ba924dc36f1>\",\"WARC-Concurrent-To\":\"<urn:uuid:123b0a89-5382-455a-a2e1-cb3857a85b04>\",\"WARC-IP-Address\":\"198.58.118.109\",\"WARC-Target-URI\":\"https://resource.flexrule.com/knowledge-base/analytics-commands/\",\"WARC-Payload-Digest\":\"sha1:4EF5YV66DLWVOGERT2C34YCGXSGPU5EB\",\"WARC-Block-Digest\":\"sha1:VQFZJ64CI5PYUXGOFSMNMIAUG6ARDNRX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780055808.78_warc_CC-MAIN-20210917212307-20210918002307-00312.warc.gz\"}"}
https://cn.maplesoft.com/support/help/addons/view.aspx?path=MathApps%2FEulerIdentity	[ "", null, "Euler's Identity - Maple Help\n\nEuler's Identity\n\nMain Concept\n\nEuler's identity is the famous equality where:\n\n • e is Euler's number ≈ 2.718\n • i is the imaginary number;\n\nThis is a special case of Euler's formula: , where :\n\nVisually, this identity can be defined as the limit of the function as n approaches infinity. More generally, can be defined as the limit of as $n$ approaches infinity.\n\nFor a given value of z, the plot below shows the value of as n increases to infinity, as well as the sequence of line segments from to . Each additional line segment represents an additional multiplication by . For , it can be seen that the point approaches $\\mathit{-}\\mathit{1}$.\n\nClick Play/Stop to start or stop the animation or use the slider to adjust the frames manually. Choose a different value of z to see how the plot is affected. Use the controls to adjust the view of the plot.", null, "", null, "", null, "z =", null, "More MathApps" ]	[ null, "https://bat.bing.com/action/0", null, "https://cn.maplesoft.com/support/help/content/961/image119.png", null, "https://cn.maplesoft.com/support/help/content/961/image121.png", null, "https://cn.maplesoft.com/support/help/content/961/image123.png", null, "https://cn.maplesoft.com/support/help/content/961/image138.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89836264,"math_prob":0.9681201,"size":1101,"snap":"2023-14-2023-23","text_gpt3_token_len":314,"char_repetition_ratio":0.10574294,"word_repetition_ratio":0.054298643,"special_character_ratio":0.2797457,"punctuation_ratio":0.08898305,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99355763,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-04-01T08:35:46Z\",\"WARC-Record-ID\":\"<urn:uuid:8558c4b5-00ed-4956-ba61-087cf7b4f587>\",\"Content-Length\":\"177826\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aae7dd08-a522-4caf-b3c5-d9d7282ea699>\",\"WARC-Concurrent-To\":\"<urn:uuid:1fec213a-8f6e-492c-8842-c6b19a151ae5>\",\"WARC-IP-Address\":\"199.71.183.28\",\"WARC-Target-URI\":\"https://cn.maplesoft.com/support/help/addons/view.aspx?path=MathApps%2FEulerIdentity\",\"WARC-Payload-Digest\":\"sha1:CLCKNZ53HPBFJMZK3CF2N45R32TBYKNO\",\"WARC-Block-Digest\":\"sha1:Q7P5MNSIX3AQWDR4A5YDPIZTOP2SSSAA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296949701.56_warc_CC-MAIN-20230401063607-20230401093607-00390.warc.gz\"}"}
https://slideplayer.com/slide/4063791/	[ "", null, "# PH1 Kinematics UVAXT Equations. Vectors & Scalars Vectors e.g. displacement, velocity have a direction, and a magnitude, are a straight line. e.g. 3ms.\n\n## Presentation on theme: \"PH1 Kinematics UVAXT Equations. Vectors & Scalars Vectors e.g. displacement, velocity have a direction, and a magnitude, are a straight line. e.g. 3ms.\"— Presentation transcript:\n\nPH1 Kinematics UVAXT Equations\n\nVectors & Scalars Vectors e.g. displacement, velocity have a direction, and a magnitude, are a straight line. e.g. 3ms -1 to the East Scalars e.g. distance, speed have magnitude, can be along a non-straight line. e.g. the car travelled 1425m from door-to- door\n\nVectors & Scalars Distance & Displacement\n\nSpeed or Velocity?\n\nUnits Distance/displacement: meters (m) Speed/Velocity – distance moved per second = (meters) per second (m/s or ms -1 ) Acceleration – change in speed per second = (meters per second) per second = (m/s)/s = m/s 2 or ms -2\n\nVelocity-time graphs time velocity t u v Gradient = acceleration Area = displacement\n\nOther graphs Displacement-Time: gradient = instantaneous velocity Acceleration-Time: area underneath = final velocity Now try some questions…\n\nUVAXT equations v = u + at x = ut + ½at 2 x = vt – ½at 2 v 2 = u 2 + 2ax x = ½(u+v)t Only when a is constant!\n\nQuestions From the worksheet:\n\nUVAXT Questions Work out initial conditions Find out which quantity you are calculating Find out which quantity you don’t need. Identify the correct equation Do the maths\n\nExample A car accelerates from rest at 0.4ms -2 for 15s. How far does it go? x = ut + ½at 2 x = (0)(15) + 0.5(0.4)15 2 = 45m Work out initial conditions Find out which quantity you are calculating Find out which quantity you don’t need. Identify the correct equation Do the maths UVAXT 00.415 ?\n\nYou may need the square root formula\n\n2 volunteers needed … for a demo next lesson\n\nDownload ppt \"PH1 Kinematics UVAXT Equations. Vectors & Scalars Vectors e.g. displacement, velocity have a direction, and a magnitude, are a straight line. e.g. 3ms.\"\n\nSimilar presentations" ]	[ null, "https://slideplayer.com/static/blue_design/img/slide-loader4.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7828957,"math_prob":0.99524504,"size":1703,"snap":"2020-34-2020-40","text_gpt3_token_len":462,"char_repetition_ratio":0.10535609,"word_repetition_ratio":0.16778524,"special_character_ratio":0.27480915,"punctuation_ratio":0.10479042,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9990897,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-05T20:09:48Z\",\"WARC-Record-ID\":\"<urn:uuid:91ce6557-927c-4f32-b407-783cd1ace93e>\",\"Content-Length\":\"153196\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bd5e74b2-e8b9-41bd-9a11-4c25cb4b20b4>\",\"WARC-Concurrent-To\":\"<urn:uuid:37b772bf-b628-46cc-b465-d23d549f1b9c>\",\"WARC-IP-Address\":\"138.201.58.10\",\"WARC-Target-URI\":\"https://slideplayer.com/slide/4063791/\",\"WARC-Payload-Digest\":\"sha1:EASNENA4PKSPETDPV5XQYLFEXF3JMYLB\",\"WARC-Block-Digest\":\"sha1:D2LFQU6SIR7ONF4IGIYEUDXEABD77GI4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439735964.82_warc_CC-MAIN-20200805183003-20200805213003-00593.warc.gz\"}"}
https://nl.mathworks.com/help/curvefit/cfit.differentiate.html	[ "# differentiate\n\nDifferentiate `cfit` or `sfit` object\n\n## Syntax\n\n``fx = differentiate(FO, X)``\n``[fx, fxx] = differentiate(FO, X)``\n``[fx, fy] = differentiate(FO, X, Y)``\n``[fx, fy] = differentiate(FO, [X, Y])``\n``[fx, fy, fxx, fxy, fyy] = differentiate(FO, ...)``\n\n## Description\n\n``` NoteUse these syntaxes for `cfit` objects. `fx = differentiate(FO, X)` differentiates the `cfit` object `FO` at the points specified by the vector `X` and returns the result in `fx`.```\n\nexample\n\n````[fx, fxx] = differentiate(FO, X)` differentiates the `cfit` object `FO` at the points specified by the vector `X` and returns the result in `fx` and the second derivative in `fxx`.```\n``` NoteUse these syntaxes for `sfit` objects. `[fx, fy] = differentiate(FO, X, Y)` differentiates the surface `FO` at the points specified by `X` and `Y` and returns the result in `fx` and `fy`.`FO` is a surface fit (`sfit`) object generated by the `fit` function.`X` and `Y` must be double-precision arrays and the same size and shape as each other.All return arguments are the same size and shape as `X` and `Y`. If `FO` represents the surface $z=f\\left(x,y\\right)$, then `FX` contains the derivatives with respect to x, that is, $\\frac{df}{dx}$, and `FY` contains the derivatives with respect to y, that is, $\\frac{df}{dy}$. ```\n````[fx, fy] = differentiate(FO, [X, Y])`, where `X` and `Y` are column vectors, allows you to specify the evaluation points as a single argument.```\n````[fx, fy, fxx, fxy, fyy] = differentiate(FO, ...)` computes the first and second derivatives of the surface fit object `FO`.`fxx` contains the second derivatives with respect to `x`, that is, $\\frac{{\\partial }^{2}f}{\\partial {x}^{2}}$.`fxy` contains the mixed second derivatives, that is, $\\frac{{\\partial }^{2}f}{\\partial x\\partial y}$.`fyy` contains the second derivatives with respect to `y`, that is, $\\frac{{\\partial }^{2}f}{\\partial {y}^{2}}$. ```\n\n## Examples\n\ncollapse all\n\nCreate a baseline sinusoidal signal.\n\n```xdata = (0:.1:2pi)'; y0 = sin(xdata);```\n\nAdd response-dependent Gaussian noise to the signal.\n\n```noise = 2y0.randn(size(y0)); ydata = y0 + noise;```\n\nFit the noisy data with a custom sinusoidal model.\n\n```f = fittype('asin(b*x)'); fit1 = fit(xdata,ydata,f,'StartPoint',[1 1]);```\n\nFind the derivatives of the fit at the predictors.\n\n`[d1,d2] = differentiate(fit1,xdata);`\n\nPlot the data, the fit, and the derivatives.\n\n```subplot(3,1,1) plot(fit1,xdata,ydata) % cfit plot method subplot(3,1,2) plot(xdata,d1,'m') % double plot method grid on legend('1st derivative') subplot(3,1,3) plot(xdata,d2,'c') % double plot method grid on legend('2nd derivative')```", null, "You can also compute and plot derivatives directly with the `cfit` `plot` method, as follows:\n\n```figure plot(fit1,xdata,ydata,{'fit','deriv1','deriv2'})```", null, "The `plot` method, however, does not return data on the derivatives, unlike the `differentiate` method.\n\nYou can use the `differentiate` method to compute the gradients of a fit and then use the `quiver` function to plot these gradients as arrows. This example plots the gradients over the top of a contour plot.\n\nCreate the derivation points and fit the data.\n\n```x = [0.64;0.95;0.21;0.71;0.24;0.12;0.61;0.45;0.46;... 0.66;0.77;0.35;0.66]; y = [0.42;0.84;0.83;0.26;0.61;0.58;0.54;0.87;0.26;... 0.32;0.12;0.94;0.65]; z = [0.49;0.051;0.27;0.59;0.35;0.41;0.3;0.084;0.6;... 0.58;0.37;0.19;0.19]; fo = fit( [x, y], z, 'poly32', 'normalize', 'on' ); [xx, yy] = meshgrid( 0:0.04:1, 0:0.05:1 );```\n\nCompute the gradients of the fit using the `differentiate` function.\n\n`[fx, fy] = differentiate( fo, xx, yy );`\n\nUse the `quiver` function to plot the gradients.\n\n```plot( fo, 'Style', 'Contour' ); hold on h = quiver( xx, yy, fx, fy, 'r', 'LineWidth', 2 ); hold off colormap( copper )```", null, "If you want to use derivatives in an optimization, you can, for example, implement an objective function for `fmincon` as follows.\n\n`function [z, g, H] = objectiveWithHessian( xy )`\n\n` % The input xy represents a single evaluation point`\n\n` z = f( xy );`\n\n` if nargout > 1`\n\n` [fx, fy, fxx, fxy, fyy] = differentiate( f, xy );`\n\n` g = [fx, fy];`\n\n` H = [fxx, fxy; fxy, fyy];`\n\n` end`\n\n` end`\n\n## Input Arguments\n\ncollapse all\n\nFunction to differentiate, specified as a `cfit` object for curves or as a `sfit` object for surfaces.\n\nPoints at which to differentiate the function, specified as a vector. For surfaces, this argument must have the same size and shape of Y.\n\nPoints at which to differentiate the function, specified as a vector. For surfaces, this argument must have the same size and shape of X.\n\n## Output Arguments\n\ncollapse all\n\nFirst derivative of the function, returned as a vector of the same size and shape of X and Y.\n\nIf FO is a surface, $z=f\\left(x,y\\right)$, then fx contains the derivatives with respect to `x`.\n\nSecond derivative of the function, returned as a vector of the same size and shape of X and Y.\n\nIf FO is a surface, $z=f\\left(x,y\\right)$, then fxx contains the second derivatives with respect to `x`.\n\nFirst derivative of the function, returned as a vector of the same size and shape of X and Y.\n\nIf FO is a surface, $z=f\\left(x,y\\right)$, then fy contains the derivatives with respect to `y`.\n\nSecond derivative of the function, returned as a vector of the same size and shape of X and Y.\n\nIf FO is a surface, $z=f\\left(x,y\\right)$, then fyy contains the second derivatives with respect to `y`.\n\nMixed second derivative of the function, returned as a vector of the same size and shape of X and Y.\n\n## Tips\n\nFor library models with closed forms, the toolbox calculates derivatives analytically. For all other models, the toolbox calculates the first derivative using the centered difference quotient\n\n`$\\frac{df}{dx}=\\frac{f\\left(x+\\Delta x\\right)-f\\left(x-\\Delta x\\right)}{2\\Delta x}$`\n\nwhere x is the value at which the toolbox calculates the derivative, $\\Delta x$ is a small number (on the order of the cube root of `eps`), $f\\left(x+\\Delta x\\right)$ is `fun` evaluated at $x+\\Delta x$, and $f\\left(x-x\\Delta \\right)$ is `fun` evaluated at $x-\\Delta x$.\n\nThe toolbox calculates the second derivative using the expression\n\n`$\\frac{{d}^{2}f}{d{x}^{2}}=\\frac{f\\left(x+\\Delta x\\right)+f\\left(x-\\Delta x\\right)-2f\\left(x\\right)}{{\\left(\\Delta x\\right)}^{2}}$`\n\nThe toolbox calculates the mixed derivative for surfaces using the expression\n\n`$\\frac{{\\partial }^{2}f}{\\partial x\\partial y}\\left(x,y\\right)=\\frac{f\\left(x+\\Delta x,y+\\Delta y\\right)-f\\left(x-\\Delta x,y+\\Delta y\\right)-f\\left(x+\\Delta x,y-\\Delta y\\right)+f\\left(x-\\Delta x,y-\\Delta y\\right)}{4\\Delta x\\Delta y}$`\n\n## Version History\n\nIntroduced before R2006a" ]	[ null, "https://nl.mathworks.com/help/examples/curvefit/win64/FindTheDerivativesOfACurveUsingTheDifferentiateFunctionExample_01.png", null, "https://nl.mathworks.com/help/examples/curvefit/win64/FindTheDerivativesOfACurveUsingTheDifferentiateFunctionExample_02.png", null, "https://nl.mathworks.com/help/examples/curvefit/win64/FindDerivativesOfSurfaceUsingTheDifferentiateFunctionExample_01.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8069803,"math_prob":0.999161,"size":5263,"snap":"2022-40-2023-06","text_gpt3_token_len":1484,"char_repetition_ratio":0.1745579,"word_repetition_ratio":0.2600229,"special_character_ratio":0.27873835,"punctuation_ratio":0.21529509,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999137,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-09T13:47:32Z\",\"WARC-Record-ID\":\"<urn:uuid:6b6de0cf-15aa-4a9d-b17f-a30d5d848e3b>\",\"Content-Length\":\"116932\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9f17e8cc-b7b8-4fbf-8b13-81757742ceff>\",\"WARC-Concurrent-To\":\"<urn:uuid:45d596b7-ee59-4d89-8e65-72c75f374cb7>\",\"WARC-IP-Address\":\"104.86.80.92\",\"WARC-Target-URI\":\"https://nl.mathworks.com/help/curvefit/cfit.differentiate.html\",\"WARC-Payload-Digest\":\"sha1:3XEM2LOQ5GWX5EZWY3D6WV7QVFLECPCE\",\"WARC-Block-Digest\":\"sha1:NHZUJ27SLN2ZBHIRLUEIP2EHXMCTM2KN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499966.43_warc_CC-MAIN-20230209112510-20230209142510-00182.warc.gz\"}"}
https://zbmath.org/?q=an:0797.11072	[ "## On a class of differential-difference equations arising in number theory.(English)Zbl 0797.11072\n\nFunctions defined by a differential-difference equation of the type (1) $$uf ' (u) + af(u) + bf(u-1) = 0$$, where $$a$$ and $$b$$ are constants, arise frequently in number theory. Probably the best known examples are the Dickman function $$\\rho(u)$$ and the Buchstab function $$\\omega (u)$$. Functions satisfying (1) when $$a+b$$ is an integer arise in sieve theory and in this context have been investigated by various authors. The equation (1) with general coefficients $$a$$ and $$b$$ have been studied by H. Iwaniec [Acta Arith. 36, 171-202 (1980; Zbl 0435.10029)] and F. Wheeler [Trans. Am. Math. Soc. 318, 491-523 (1990; Zbl 0697.10035)].\nThe object of the present paper is to describe, for any given pair of complex coefficients $$(a,b)$$ with $$b \\neq 0$$ the structure and asymptotic behavior of the general solution to $$(1)$$. Equation (1) with the initial condition (2) $$f(u) = \\varphi(u)$$ $$(u_ 0 - 1 \\leq u \\leq u_ 0)$$, where $$\\varphi (u)$$ is any given continuous function on $$[u_ 0 - 1,u_ 0]$$, has a unique continuous solution $$f(u) = f(u;\\varphi)$$ for $$u \\geq u_ 0$$. In the paper, the authors construct a set of “fundamental” solutions $$F(u)$$ and $$F_ n(u)$$ $$(n \\in \\mathbb{Z})$$ and the solution $$f(u)$$ can be expressed as a convergent series $$f(u) = \\alpha F(u) + \\sum_{n \\in \\mathbb{Z}} \\alpha_ n F_ n (u)$$ with suitable coefficients $$\\alpha$$ and $$\\alpha_ n$$.\nThe functions $$F$$ and $$F_ n$$ are defined by means of a contour integral, which can be estimated rather precisely. To state the result we set $\\Phi (u,s) = {\\exp \\{-us + bI(s)\\}s^{a+b-1} \\over \\sqrt {2 \\pi u(1-1/s)}},$ where $$I(s) = \\int^ s_ 0{e^ z-1 \\over z} dz$$, and then we have the following result: For any fixed non-zero integer $$n$$ and $$u \\geq u_ 0 (\\varepsilon,n)$$ we have $F_ n(u) = (1+O({1 \\over u})) \\Phi (u, \\zeta_ n)$ where $$\\zeta_ n = \\zeta_ n (u/b)$$ is a certain complex solution of the equation $$e^ \\zeta = 1 + {u \\over b} \\zeta$$, and the implied constant depends at most on $$\\varepsilon$$ and $$n$$.\nThe principal result of the paper is as follows. Let $$\\varphi (u)$$ be a continuous function on $$[u_ 0 - 1,u_ 0]$$ and let $$f(u) = f(u;\\varphi)$$ be the unique continuous solution to (1) and (2). Then we have $f(u) = \\alpha F(u) + \\sum_{\\alpha \\in \\mathbb{Z}} \\alpha_ n F_ n(u), \\quad u>u_ 0+1 \\tag{3}$ where $$\\alpha = \\langle \\varphi, G \\rangle$$, $$\\alpha_ n = \\langle \\varphi, G_ n \\rangle$$ and the series in (3) is uniformly convergent for $$u \\geq u_ 0 + 1 + \\delta$$, for any fixed $$\\delta>0$$.\nMoreover, the authors derive several corollaries from the principal result.\n\n### MSC:\n\n 11N25 Distribution of integers with specified multiplicative constraints 34K99 Functional-differential equations (including equations with delayed, advanced or state-dependent argument)\n\n### Citations:\n\nZbl 0435.10029; Zbl 0697.10035\nFull Text:\n\n### References:\n\n [Al] K. Alladi,An Erdös-Kac theorem for integers without large prime factors, Acta Arith.49 (1987), 81–105. [AO] N. C. Ankeny and H. Onishi,The general sieve, Acta Arith.10 (1964), 31–62. · Zbl 0127.27002 [Be] J. J. A. Beenakker,The differential-difference equation {$$\\alpha$$}xf 1(x)+f(x)=0, Thesis, Eindhoven, 1966. · Zbl 0144.08702 [BC] R. Bellimann and K. Cooke,Differential-Difference Equations, Academic Press, New York 1963. [dB1] N. G. de Bruijn,On some Volterra integral equations of which all solutions are convergent, Nederl. Akad. Wetensch. Proc.53 (1950), 813–821. · Zbl 0038.26602 [dB2] N. G. de Bruijn,On the number of uncancelled elements in the sieve of Eratosthenes, Nederl. Akad. Wetensch. Proc.53 (1950), 803–812. · Zbl 0037.03001 [dB3] N. G. de Bruijn,On the number of positive integers x and free of prime factors >y, Nederl. Akad. Wetensch. Proc.54 (1951), 50–60. · Zbl 0042.04204 [dB4] N. G. de Bruijn,The asymptotic behavior of a function occurring in the theory of primes, J. Indian Math. Soc.15 (1951), 25–32. · Zbl 0043.06502 [dB5] N. G. de Bruijn,The difference-differential equation F’(x)=e {$$\\alpha$$}x+{$$\\beta$$}F(x), I, II, Nederl. Akad. Wetensch. Proc.56 = Indagationes Math.15 (1953), 449–464. · Zbl 0053.38703 [dB6] N. G. de Bruijn,Asymptotic methods in Analysis, Dover, New York 1981. · Zbl 0556.41021 [Bu] A. A. Buchstab,Asymptotic estimates of a general number-theoretic function (Russian), Mat. Sb.44 (1937), 1237–1246. [CG] A. Y. Cheer and D. Goldston,A differential delay equation arising from the sieve of Eratosthenes, Math. Comp.55 (1990), 129–141. [DHR] H. Diamond, H. Halberstam and H.-E. Richert,A boundary, value problem for a pair of differential delay equations related to sieve theory I, in:Analytic Number Theory (B. Berndtet al., eds.),Progress in Math. 85 (1990), Birkhaüser, Boston, pp. 133–157. · Zbl 0717.11036 [Di] K. Dickman,On the frequency of numbers containing primes of a certain relative magnitude, Ark. Mat. Astr. Fys.22 (1930), 1–14. · JFM 56.0178.04 [FGHM] J. Friedlander, A. Granville, A. Hildebrand and H. Maier,Oscillation theorems for primes in arithmetic progressions and for sifting functions, J. Amer. Math. Soc.4 (1991), no. 1, 25–86. · Zbl 0724.11040 [GR] F. Grupp and H.-E. Richert,Notes on functions connected with the sieve, Analysis8 (1988), 1–23. · Zbl 0657.10048 [He] D. Hensley,The convolution powers of the Dickman function, J. London Math. Soc.33 (1986), 289–307. · Zbl 0589.10045 [Hi] A. Hildebrand,The asymptotic behavior of the solutions of a class of differential-difference equations, J. London Math. Soc. (2)42 (1990), no. 1, 11–31. · Zbl 0675.34037 [Iw] H. Iwaniec,Rosser’s sieve, Acta Arith.36 (1980), 171–202. · Zbl 0435.10029 [JR] W.-B. Jurkat and H.-E. Richert,An improvement of Selberg’s sieve method, I, Acta Arith.11 (1965), 217–240. · Zbl 0128.26902 [Ma] H. Maier,Primes in short intervals, Michigan Math. J.32 (1985), 221–225. · Zbl 0569.10023 [Ten] G. Tenenbaum,Introduction à la théorie analytique et probabiliste des nombres, Revue de l’Institut Elie Cartan13, Département de Mathématiques de l’Université de Nancy 1 (1990). [Wh] F. Wheeler,Two differential-difference equations arising in number theory, Trans. Amer. Math. Soc.318 (1990), 491–523. · Zbl 0697.10035\nThis reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.69263494,"math_prob":0.9990418,"size":7068,"snap":"2022-27-2022-33","text_gpt3_token_len":2408,"char_repetition_ratio":0.11735561,"word_repetition_ratio":0.016713092,"special_character_ratio":0.37733448,"punctuation_ratio":0.20563196,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999707,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-28T02:44:26Z\",\"WARC-Record-ID\":\"<urn:uuid:b5958014-d683-4280-9583-52c718d0770c>\",\"Content-Length\":\"61102\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3c213257-b3d9-46c4-9e28-ac40d1760995>\",\"WARC-Concurrent-To\":\"<urn:uuid:f51c86ff-5744-4983-a74e-503cf333090e>\",\"WARC-IP-Address\":\"141.66.194.2\",\"WARC-Target-URI\":\"https://zbmath.org/?q=an:0797.11072\",\"WARC-Payload-Digest\":\"sha1:YIBTO7SAYQOSGKLBK2QEZR3T64TYAB3H\",\"WARC-Block-Digest\":\"sha1:5CRZ3TAZBVW6XXCGNPHGL5G6FJL6RSCH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103347800.25_warc_CC-MAIN-20220628020322-20220628050322-00253.warc.gz\"}"}
https://community.ibm.com/community/user/powerdeveloper/blogs/swati-karmarkar/2022/09/07/julia-on-ibm-power	[ "# Programming Languages on Power\n\nView Only\n\n## Explore the benefits of using Julia on IBM Power then give it a try\n\n####", null, "By Swati Karmarkar posted Wed September 07, 2022 09:57 AM\n\nJulia is a general-purpose programming language with a special focus on scientific computing. Julia is as fast as C programming language in many cases while remaining dynamic like Python.\n\nWe are happy to announce that as of Dec 19, 2021, Julia 1.6.2 is supported on the IBM Power platform, \"Tier-3\" level. For more information about Tier 3, refer to the Julia download page.\n\nThe popularity of Julia has been soaring since it first became available in 2012, and it is currently in the top 30 programming languages as per the TIOBE index.\n\nRefer to the following interesting articles about Julia:\n\n## Benefits of Julia on IBM Power\n\nDesigned with machine learning (ML) in mind, Julia focuses on the scientific computing domain and its need for parallel, distributed, and intensive computation tasks. Julia has a powerful suite of libraries for developing artificial intelligence which includes general linear models, decision trees, clustering, Flux for deep learning, text analysis for natural language processing, and so on. Julia benefits to Power in the following ways:\n\n• Julia is an easy programming language to learn. Its syntax is similar to Python and MATLAB, and can be easily adapted on Power. The following code snippet shows the similarity among Julia, MATLAB, and Python languages:\n\nDescription MATLAB syntax Python syntax Julia syntax\nImport packages Nil import numpy as np\nimport matpotlib.pyplot as plt\nImport Pkg\nusing Plots\nDefine the variable for X axis x = 0 : pi/10 : 2pi; x=np.linespace(0, 2np.pi, 21) x = 0 : pi/10 : 2pi;\nDefine the variable for Y axis y = sin(x); y = np.sin(x) y = sin.(x);\nCode snippet to plot the graph plot(x,y)\ntitle(‘My first plot’)\nxlabel(‘x-axis’)\nylabel(‘y-axis’)\nplt.plot(x,y)\nplt.title(‘My first plot’)\nplt.xlabel(‘x-axis’)\nply.ylabel(‘y-axis)\nplot(x,y\ntitle=”My first plot”,\nxaxis=(“x-axis”),\nyaxis=(“y-axis”))\n\n• Julia integrates well with the existing code and platforms such as IBM Power.\n\n• Julia combines the familiarity and ease of use of Python and R with the speed and the strength of C, C++, or Java. So programmers no longer need to estimate models in one language and reproduce them in a faster production language.\n\n## What did we do and how did it work?\n\n• The first time we tried to build Julia on IBM Power, it failed due to errors. Later, we were able to trace back to low level virtual machine (LLVM) and those were fixed in the subsequent LLVM releases. See the pull request (PR) for LLVM release fixes for the details.\n• We encountered some issues related to code generation, which were traced and corrected by the LLVM team. Refer to the following for additional information:\n• Cbc and JuMP libraries did not work with Julia due to interfaces that were not updated for Power. This has been fixed (PR for Cbc and JuMP fixes).\n• We added support for detecting the correct CPU ID that was implemented in the upstream code (PR for detecting correct CPU ID).\n• There are miscellaneous fixes for Power in Julia. See the following PRs for more information:\n\n## Try Julia on Power\n\nPerform the following steps to install and run Julia on a Power virtual machine (VM):\n1. Download the latest stable build of Julia. Enter the following command on the Power system to get the TAR file for Power (ppc64le):\n``````wget https://julialang-s3.julialang.org/bin/linux/ppc64le/1.6/julia-1.6.2-linux-ppc64le.tar.gz\n\n``````\n2. Extract the .tar file.\n``tar -xvzf julia-1.6.2-linux-ppc64le.tar.gz``\n3. Enter the following commands to run the Julia binary file present in the bin/julia directory of the extracted directory:\n``````[user1@p006vm71 ~]\\$ cd julia-1.6.2\n[user1@p006vm71 julia-1.6.2]\\$ bin/julia``````\n\nOr,\n``````[user1@p006vm71 julia-1.6.2]\\$ export PATH=\\$PATH:~/julia-1.6.2/bin\n[user1@p006vm71 julia-1.6.2]\\$ julia``````\n\nThe command prompt of the Julia console looks as follows:", null, "## How can you use Julia?\n\nJulia can be used to run scripts and many other commands that can provide simple to complex output. Following are some of the examples:\n\n• Use the following Julia command to find the platform and version information:\n\n``````julia>\njulia> versioninfo()\nJulia Version 1.8.0-DEV.889\nCommit f14e44f38b (2021-11-03 05:54 UTC)\nPlatform Info:\nOS: Linux (ppc64le-redhat-linux)\nCPU: POWER8 (architected), altivec supported\nWORD_SIZE: 64\nLIBM: libopenlibm\nLLVM: libLLVM-12.0.1 (ORCJIT, pwr8)``````\n• Run the following Julia code snippet to find the CPU information:\n\n``````julia> ccall(:jl_dump_host_cpu, Cvoid, ())\nCPU: pwr8\nFeatures:\njulia>``````\n\n• Run a .jl file to get the required output. Following are the steps to run a sample .jl file:\n\n1. Write the following code in Julia (that plots a graph on the X and Y axis) and save it as plots.jl:\n``````import Pkg\n\nusing Plots\n\n# plot some data\nplot([cumsum(rand(500) .- 0.5), cumsum(rand(500) .- 0.5)])\n\n# save the current figure\nsavefig(\"plots.svg\")``````\n2. Run plots.jl as:\n``julia plots.jl``\n\nThe output is as follows:", null, "" ]	[ null, "https://higherlogicdownload.s3.amazonaws.com/IMWUC/Images/ProfileImageDefault/T6FlabNTze4G0FTAa1dq_EvqA7zSoTduJ6W40ZlyZ_user_icon_lg_200_200.jpg", null, "https://developer.ibm.com/developer/default/blogs/julia-supported-on-power/images/img1.jpg", null, "https://developer.ibm.com/developer/default/blogs/julia-supported-on-power/images/img2.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8450063,"math_prob":0.42914176,"size":5152,"snap":"2022-40-2023-06","text_gpt3_token_len":1364,"char_repetition_ratio":0.11655012,"word_repetition_ratio":0.0049566296,"special_character_ratio":0.25679347,"punctuation_ratio":0.14719848,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96974766,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-07T07:23:43Z\",\"WARC-Record-ID\":\"<urn:uuid:a2b1d0fb-0293-4564-ba74-68572a9a788b>\",\"Content-Length\":\"281009\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:87480ba7-06e5-43da-9cea-d4e5d4573bd9>\",\"WARC-Concurrent-To\":\"<urn:uuid:35b0b7ff-5bfa-4c22-8452-813e91ab8e2e>\",\"WARC-IP-Address\":\"104.70.59.108\",\"WARC-Target-URI\":\"https://community.ibm.com/community/user/powerdeveloper/blogs/swati-karmarkar/2022/09/07/julia-on-ibm-power\",\"WARC-Payload-Digest\":\"sha1:4YBQ55UG2UER5RRC2TCUBT6IFZWDQC4T\",\"WARC-Block-Digest\":\"sha1:JMTTZVYSY6OQOH5EDH2JY7GDNS4OVCMR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500392.45_warc_CC-MAIN-20230207071302-20230207101302-00158.warc.gz\"}"}
http://ncl.ucar.edu/Document/Functions/Built-in/escovc.shtml	[ "", null, "NCL Home > Documentation > Functions > General applied math, Statistics\n\nescovc\n\nComputes sample cross-covariances at lag 0 only.\n\nPrototype\n\n```\tfunction escovc (\nx : numeric,\ny : numeric\n)\n\nreturn_val : numeric\n```\n\nArguments\n\nx\n\nAn array of any numeric type or size. The rightmost dimension is usually time.\n\ny\n\nAn array of any numeric type or size. The rightmost dimension is usually time. The size of the rightmost dimension must be the same as x.\n\nReturn value\n\nA scalar value if x and y are one-dimensional. The same size as x if either x or y are multi-dimensional. Double if x is double, float otherwise.\n\nDescription\n\nComputes sample cross-covariances at lag 0 only. If a lagged covariance is required, use esccv. Missing values are allowed.\n\nAlgorithm:\n\n``` cov = SUM [(X(t)-Xave)*(Y(t)-Yave)]/(NT-1)\n```\nThe dimension sizes(s) of c are a function of the dimension sizes of the x and y arrays. The following illustrates dimensioning:\n``` x(N), y(N) a scalar\nx(N), y(K,M,N) c(K,M)\nx(I,N), y(K,M,N) c(I,K,M)\nx(J,I,N), y(L,K,M,N) c(J,I,L,K,M)\n```\nspecial case when dimensions of all x and y are identical:\n``` x(J,I,N), y(J,I,N) c(J,I)\n```\n\nExamples\n\nExample 1\n\nThe following will calculate the cross-covariance for a two one-dimensional arrays x(N) and y(N).\n\n``` ccv = escovc(x,y) ; ccv is a scalar\n```\nExample 2\n\nThe following will calculate the cross-covariance for one two-dimensional array y(lat,lon,time) and one one-dimensional array x(time).\n\n``` ccv = escovc(x,y) ; ccv(lat,lon)\n```\nExample 3\n\nConsider x(neval,time) and y(lat,lon,time)\n\n``` ccv = escovc(x,y) ; ccv(neval,lat,lon)\n```" ]	[ null, "http://ncl.ucar.edu/Images/NCL_NCAR_NSF_banner.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6196537,"math_prob":0.99956316,"size":1485,"snap":"2019-26-2019-30","text_gpt3_token_len":436,"char_repetition_ratio":0.13031736,"word_repetition_ratio":0.13513513,"special_character_ratio":0.25925925,"punctuation_ratio":0.18956044,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999261,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-20T23:50:50Z\",\"WARC-Record-ID\":\"<urn:uuid:69176bc9-bff3-41ad-9854-454ccbef564a>\",\"Content-Length\":\"16346\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:847e21df-37a2-43b8-877e-3df26df63cde>\",\"WARC-Concurrent-To\":\"<urn:uuid:ff3237c2-c5f0-4cf9-889c-7749d656bbd0>\",\"WARC-IP-Address\":\"128.117.225.48\",\"WARC-Target-URI\":\"http://ncl.ucar.edu/Document/Functions/Built-in/escovc.shtml\",\"WARC-Payload-Digest\":\"sha1:PKPP3ZYKOCCQ7BM65OXUTEOYV3QAUFAT\",\"WARC-Block-Digest\":\"sha1:RLOGPBSK2OWKN5HHP72OJIVHPD7OOSAW\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999291.1_warc_CC-MAIN-20190620230326-20190621012326-00394.warc.gz\"}"}
http://www.webconversiononline.com/all-weight-conversion.aspx?from=qianchina&to=kilogram	[ "", null, "Length Conversion Weight Conversion Temperature Conversion Date & Time Conversion Volume Conversion Area Conversion Speed Conversion Scientific Conversion Health Calculators Other Calculators Reference Dictionary\n Home Tools Topics Mobile Version\n\n## Qian (China) to Kilogram\n\nHome >> Weight", null, "Convert Weight from Qian (China) to Kilogram or to nearly 225 different Metric, English, Imperial and local weight and mass measurement units use in European, Asian and American regionsQian (China) and Kilogram are the units to measure Weight, where 1 Qian (China) = 0.005 Kilogram\n Enter weight and click 'Convert'\n Too many units in conversion list? Try Common Weight Conversion How to use this calculator...• Use current calculator (page) to convert Weight from Qian (China) to Kilogram. Simply enter Weight quantity and click ‘Convert’. Both Qian (China) and Kilogram are Weight measurement units.• For conversion to different Weight units, select required units from the dropdown list (combo), enter quantity and click convert• For very large or very small quantity, enter number in scientific notation, Accepted format are 3.142E12 or 3.142E-12 or 3.142x10*12 or 3.142x10^12 or 3.14210*12 or 3.14210^12 and like wise View Weight conversion table\n\n Are you looking for ... List of Supported Conversion Types (sorted) Short Info Lookup & References List of Metric, English & Local Units Definition of different measurement Units Conversion Matrix Reference Matrix\n\n List of Supported Conversion Types ... Acceleration Angle Angular Acceleration Angular Velocity Area Blood Sugar Clothing Size Computer Storage Unit Cooking Volume Cooking Weight Data Transfer Rate Date Density Dynamic Viscosity Electric Capacitance Electric Charge Electric Conductance Electric Conductivity Electric Current Electric Field Strength Electric Potential Electric Resistance Electric Resistivity Energy Energy Density Energy Mass Euro Currency Fluid Concentration Fluid Flow Rate Fluid Mass Flow Rate Force Frequency Fuel Economy Heat Capacity Heat Density Heat Flux Density Heat Transfer Coefficient Illumination Image Resolution Inductance Kinematic Viscosity Length Luminance (Light) Light Intensity Linear Charge Density Linear Current Density Magnetic Field Strength Magnetic Flux Magnetic Flux Density Magnetomotive Force Mass Flux Density Molar Concentration Molar Flow Rate Moment of Inertia Number Permeability Power Prefix Pressure Radiation Radiation Absorbed Radiation Exposure Radioactivity Shoe Size Sound Specific Volume Speed Surface Charge Density Surface Current Density Surface Tension Temperature Thermal Conductivity Thermal Expansion Thermal Resistance Time Torque Volume Volume Charge Density Water Oil Viscosity Weight\n\n More Topics ... Area Astrology Baby Names Banking Birth Control Chemistry Chinese Astrology City Info Electricity Finance Fluids Geography Health Length Magnetism Pregnancy Radiation Scientific Speed Technology Telephone Temperature Time & Date Train Info Volume Weight World Clock Zodiac Astrology Other" ]	[ null, "http://www.webconversiononline.com/sysimages/web-conversion-online.jpg", null, "http://www.webconversiononline.com/image/weightsmall.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7650466,"math_prob":0.796817,"size":422,"snap":"2020-45-2020-50","text_gpt3_token_len":116,"char_repetition_ratio":0.18421052,"word_repetition_ratio":0.025316456,"special_character_ratio":0.3056872,"punctuation_ratio":0.08955224,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9540722,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-25T21:33:49Z\",\"WARC-Record-ID\":\"<urn:uuid:0e844a75-7861-4f44-90f9-8f9db18e9da0>\",\"Content-Length\":\"89920\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2b021e8e-8b54-40c7-9fa3-ec86d13d4e77>\",\"WARC-Concurrent-To\":\"<urn:uuid:4cdcb8e2-75fb-4026-93d0-35bd4fe5af3b>\",\"WARC-IP-Address\":\"182.50.130.37\",\"WARC-Target-URI\":\"http://www.webconversiononline.com/all-weight-conversion.aspx?from=qianchina&to=kilogram\",\"WARC-Payload-Digest\":\"sha1:2LP4O7BGCWBOVDR24IL2455G4KJ4DJAU\",\"WARC-Block-Digest\":\"sha1:4CIXZ3ZITUCDOHV675GYQXEYOM4QC4S5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107890028.58_warc_CC-MAIN-20201025212948-20201026002948-00562.warc.gz\"}"}
https://3010tangents.wordpress.com/tag/nine-chapters-on-the-mathematical-art/	[ "# 3010 Sea Island in the The Nine Chapters on the Mathematical Art\n\nI never read anything about The Nine Chapters on the Mathematical Art before taking this History of Math class. I heard about this book when I was a middle school student. It is interesting that I started paying attention 13 years after I graduated from a middle school. There is a chapter called “The Sea Island Mathematical Manual” in the book. I found this topic very interesting because I want to know how our ancient people solved real world problems without using any modern technology. In this blog, I will try to explore more about these sea island problems. (In this blog, I will use Nine Chapters to substitute the book title and “Sea Island” to substitute the chapter.).\n\nBefore talking about the sea island questions, I want to briefly talk about the book’s history. Nine Chapters was formed in the Han dynasty and it was a Chinese mathematics book that composed by several generations of scholars from the 10th – 2nd century BCE. This book has 246 real world questions, which relate to agriculture, business, engineering and solving equations, etc. It divides those questions into nine chapters. The Nine Chapters flourished between the Three Kingdoms period and earlier Tang dynasty in China. At that time this book was the primary math textbook in China and it also spread to Korea and Japan. The Nine Chapters was undoubtedly one of the cornerstones of Chinese modern math.\n\n“The Sea Island” is one of the extension chapters of the book that was written by Liu Hui. This chapter has nine problems: surveys of Sea Island, height of a hill top pine tree, the size of a square city wall viewed afar, the depth of a ravine, the height of a building on a plain seen from a hill, the breadth of a river-mouth seen from a distance on land, the depth of a transparent pool, the width of a river as seen from a hill, and the size of a city seen from city. We can see all these questions are very similar to each other. Let’s take a look at the first question. Survey of sea island says “there is a sea island, and set up two three zhang (zhang is a distance and 1 zhang equals 3.3 meters) poles at one thousand steps apart and set the two poles and the island in a straight line. Step back from the front post 123 steps, with eye on the ground level, the tip of the pole is on a straight line with the peak of island. Step back 127 steps from the rear pole, eye on ground level also aligns with the tip of pole and tip of island. What is the height of the island, and what is the distance to the pole?” (Wikipedia)", null, "According to Nine Chapters, Liu Hui could not measure the distance of the front pole to the island, so he sets up the two poles assuming they have the same height. Liu Hui gave two formulas: height of island AB = CD * DF / (FH – DG) + CD and distance of front pole to island BD = DG * DF / (FH – DG). How do we know these two formulas are correct? I thought about these two formulas but I could not convince myself until I read a proof about them. We have to take a look at Liu Hui’s theorem for the survey first. In the above figure he proved that FH * AI = IB * BF. He called it “‘In-out-complement’ principle which showed that the area of two inscribed rectangles in the two complementary right angle triangles have equal area” (Wikipedia). This proof is very straightforward if we know his “In-out-complement” principle. From the above figure, we know □EJ = □EB and □CK = □CB, then we use □EJ – □CK = □DE. Therefore, we know the height of island formula is correct. How do we get the distance of the front pole to island? That’s from □CB = □CK. An interesting proof huh?\n\nThe rest of eight problems used the “In-out-complement” principle to be solved too. I really like Liu Hui’s way to solve those real world problems, especially without using modern electronic technology. As you can see his ancient way to solve these problems was extremely important for geographical measurement and navigation industrial. If you want to learn more about rest of the questions, I highly recommend you to read the chapter 8 on the book Nine Chapters to dig more.\n\nReferences:\n\nShen K., John NC, Anthony W.-C, The Nine Chapters on the Mathematical Art. 1999.\n\nJiMin L, 九章算术中的比率理论.\n\n# Zu Chongzhi and his mathematics\n\nZu Chongzhi was a Chinese mathematician and astronomer during the Liu Song and Southern Qi Dynasties. He did a lot of famous mathematics during his life. His three most important contributions were studying The Nine Chapters on the Mathematical Art, calculating pi, and calculating the volume of sphere.\n\nAs we know, The Nine Chapters on the Mathematical Art is the most famous book in the history of Chinese mathematics. In ancient China, most people could not understand “The Nine Chapters on the Mathematical Art”. Zu Chongzi read the book and then he used his comprehension to explain the formulas of the book. Zu Chongzhi, and his father wrote the “Zhui Shu”(缀术) together. The book made The Nine Chapters on the Mathematical Art easier to read. And the book also added some important formula by Zu. For example, the calculation of pi and the calculation of sphere volume. “Zhui Shu” also become math textbook at the Tang Dynasty Imperial Academy. Unfortunately, the book was lost in the Northern Song Dynasty.\n\nZu’s ratio, also called milü is named after Zu Chongzhi. Zu’s ratio was an early accurate approximation of pi. It was recorded in the “Book Of Sui” and “Zhui Shu”. (Book Of Sui is the official history of the Sui dynasty). According to the “Book Of Sui”, Zu Chongzhi discovered that pi is between 3.14159276 and 3.14159277. Today, we know the actual number is in accord with Zu’s ratio. But “Book Of Sui” did not record the method used to get the number. Most historians and mathematicians think Zu Chongzhi used Liu Hui’s π algorithm to get the number. Liu Hui’s algorithm means approximating circle with a 24,576 sided polygon. Japanese mathematician Yoshio Mikami pointed out, “22/7 was nothing more than the π value obtained several hundred years earlier by the Greek mathematician Archimedes, however milü π = 355/113 could not be found in any Greek, Indian or Arabian manuscripts, not until 1585 Dutch mathematician Adriaan Anthoniszoom obtained this fraction; the Chinese possessed this most extraordinary fraction over a whole millennium earlier than Europe”. Hence Mikami strongly urged that the fraction 355/113 be named after Zu Chongzhi as Zu’s fraction.( Yoshio Mikami)", null, "Image:Zu Chongzhi’s method (similar to Cavalieri’s principle) for calculating a sphere’s volume includes calculating the volume of a bicylinder. Author: Chen Bai, via WIkimedia Commons.\n\nZu Chongzhi’s other important contribution was calculation volume of the sphere. Together with his son Zu Geng, Zu Chongzhi used an ingenious method to determine the volume of the sphere.(Arthur Mazer). In The Nine Chapters on the Mathematical Art, the author used Steinmetz solid to get the volume of the sphere. The solid common to two (or three) right circular cylinders of equal radii intersecting at right angles is called the Steinmetz solid.\n\nBut the book did not give the formula of how to get the volume of the sphere. Zu Chongzhi used “Zu Geng principle” (another name: Cavalieri’s principle) to show the volume of the sphere formula is (π*d³)/6. In order to commemorate the fact that Zu Chongzhi found the significant contribution of the principle with his son, people called the principle “Zu Geng principle”. “Zu Geng principle” is the same as “Cavalieri’s principle”, but “Zu Geng principle” is earlier than “Cavalieri’s principle”. “Cavalieri’s principle” means two solids of equal altitude, the sections made by planes parallel to and at the same distance from their respective bases are always equal, then the volumes of the two solids are equal.(Kern and Bland 1948, p. 26).\n\nWork cited:\n\nYoshio Mikami , (1947). Development of Mathematics in China and Japan. 2nd ed. : Chelsea Pub Co;.\n\nArthur Mazer , (2010). The Ellipse: A Historical and Mathematical Journey. 1st ed. : Wiley;\n\nKern, W. F. and Bland, J. R. “Cavalieri’s Theorem” and “Proof of Cavalieri’s Theorem.” §11 and 49 in Solid Mensuration with Proofs, 2nd ed. New York: Wiley, pp. 25-27 and 145-146, 1948.\n\nhttp://en.wikipedia.org/wiki/Cavalieri%27s_principle\n\nhttp://en.wikipedia.org/wiki/Zu_Chongzhi\n\n# An introduction to “Nine Chapters on the Mathematical Art(九章算术)”", null, "The “Nine Chapters on the Mathematical Art” is an ancient Chinese mathematics book. It is one of the ten most important arithmetic books of ancient China. Although it is hard to find the accurate publishing time of this book, by the historical record, it had been published in 263 AD (Han dynasty). In the following dynasties, Chinese mathematicians kept revising it and supplementing it. Thus “Nine Chapters on the Mathematical Art” can be also seen as the essence of the ancient Chinese arithmetic. By the Qing dynasty (1644-1912), most Chinese mathematicians started studying math from this book. In the Tang dynasty (618-907) and Song dynasty (960-1279) “nine chapters of arithmetic” was the professional textbook by the government provision. Also in 1084 AD, the Chinese government published the printed version “Nine Chapters on the Mathematical Art”, which made it become the earliest printed version mathematics textbook in the world. As a famous mathematics textbook, “Nine Chapters on the Mathematical Art” was introduced in Japan and Korea in the Sui dynasty (581-618) and right now, it has been translated into Russian, German, French and other languages.\nThe content of “Nine Chapters on the Mathematical Art” is plentiful. It was written in “problems and solutions” form including 246 problems related to production and practical life, and they were distributed into nine chapters. Although many problems have several solutions, this book does not contain any proof, which is a common disadvantage of most Chinese ancient mathematics textbooks. The first chapter is called “Fang tian (方田)”. It is about computing the area of various plane geometrical figures such as sector, annulus arch and so on. Also in this chapter, it refers to the arithmetic of fractions, which is the earliest record of textbook referring to fractions. The second and third chapter are called “Su mi(粟米)” and “Cui fen (衰分)” which are about proration problems. The fourth chapter is named as “Shao guang (少广) ”. It narrates the methods of computing the length of a edge when you get the area of the figure. This chapter also introduces the method of extraction of square and cubic roots. The fifth chapter “Shang gong (商功)”, gives the formulas to compute the volume of many objects. The sixth chapter “Jun shu(均输)” focuses on collecting taxes. But it also involves in the conceptions direct, inverse compound proportions and other proportion theory. In western countries, these conceptions appeared after the 16th century. The seventh chapter “Ying bu zu (盈不足)” discusses the problems of profit and loss. Some solutions from this chapter are very advanced in the world. The eighth chapter is called “equation(方程)”. It uses the method “separation coefficient” to represent systems of liner equations, which is similar to matrices. It also gives the earliest complete solution of systems of liner equations. In the solutions, it even introduces the concepts of negatives. This is the first time in human history to expand the number system from positive numbers systems. In the last chapter “Gou gu(勾股) ”, it uses “Gou gu theorem” (also known as Pythagorean theorem in the west) to solve some problems which are related to practical life. Some stuff in this chapter are very advanced, the last problem of this chapter gives a formula. In the western world, this formula was put forward by American mathematician L.E.Dickson at the end of 19th century.\n“Nine Chapters on the Mathematical Art” determined the framework of ancient Chinese mathematics. It focuses on computations related to practical problem and has a very profound effect on the following mathematics." ]	[ null, "https://3010tangents.files.wordpress.com/2015/03/image1.png", null, "https://i2.wp.com/upload.wikimedia.org/wikipedia/commons/e/e9/Sphere_volume_derivation_using_bicylinder.jpg", null, "https://u0897820.files.wordpress.com/2014/10/562c11dfa9ec8a1384e92151f703918fa1ecc094-jpg.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9513978,"math_prob":0.85435975,"size":4185,"snap":"2019-26-2019-30","text_gpt3_token_len":954,"char_repetition_ratio":0.11384836,"word_repetition_ratio":0.015645372,"special_character_ratio":0.22246116,"punctuation_ratio":0.094451,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9885388,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,4,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-18T18:21:08Z\",\"WARC-Record-ID\":\"<urn:uuid:caf58d9c-b883-439c-8eb1-3c64b7dc75d2>\",\"Content-Length\":\"64851\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:27439c5f-feb1-4ba3-a01c-70071fb2d03d>\",\"WARC-Concurrent-To\":\"<urn:uuid:f9f974f5-7cb6-4160-9927-0e860a20e7f0>\",\"WARC-IP-Address\":\"192.0.78.13\",\"WARC-Target-URI\":\"https://3010tangents.wordpress.com/tag/nine-chapters-on-the-mathematical-art/\",\"WARC-Payload-Digest\":\"sha1:ZNSBDFCOCMGYCI7JBOPAPZT27FAH5WN5\",\"WARC-Block-Digest\":\"sha1:2B67CSXTL2EVX6BBEWCMHPNI546DGZJ2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195525699.51_warc_CC-MAIN-20190718170249-20190718192249-00207.warc.gz\"}"}
http://www.mellekoning.nl/index.php/2017/02/	[ "Categories\n\n## You have 10 red and 10 black marbles…\n\nYou have 10 red and 10 black marbles. In how many ways can you put the 20 marbles in a row so that no more than two marbles of the same colour lie next to each other?\n\nAn example of a valid sequence would be:", null, "Can you reason your way out of this one or would you write a program that can give you the solution? Can you write this program?\n\n### First reasoning\n\nBefore starting a program, let’s first reason about the problem. It might just be that a program is not needed at all 🙂\n\nThe red and black marbles are kind of ‘exchangeable’ in the sense that in the above example picture there is a similar solution when all the red and black marbles would change colour. So, how to discover all the possibilities with 20 marbles?\n\nI was thinking of starting off with just one marble, and than reason my way onwards in a tree-search structure way, by just adding a marble and see if I still have a valid sequence (line). Something like this:\n\nFirst of all, for clarity reasons let me use simple ‘X’ and ‘O’ instead of red/blacks.\n\n```O\n```\n\nOne O, that’s just the start, let’s add the O and X possibilities:\n\n```OO\nOX\n```\n\nSo, I’m just creating all possible lines that I can create. From the start of one ‘O’, I now have two possible continuations of lines.\nThe number of possibilities doubled, of course. Of course I know that I could also have started with ‘X’, but I’m just exploring a bit first. So, again, let’s just add O’s and X’s to the two existing possibilities:\n\n```OOO <- not allowed!\nOOX\nOXO\nOXX\n```\n\nThe first time extending a marble 'ends a line' because of the rule that we can't have three marbles of the same colour in a row!\nOnly three possible lines to extend left.\n\nSo far so good, it looks like adding marbles from the stack of 10 O's and X's is simple. However, doing that for all possibilities seems quite a task as the number of possible 'branches' of this tree probably grows very fast.\n\nLet's just again extend the lines of three above, to lines with lengths of four, by adding O and X to the already existing possible lines above, of course I do not have to explore the OOO line any further:\n\n```OOXO\nOOXX\nOXOO\nOXOX\nOXXO\nOXXX <- not allowed!\n```\n\nSo, while the lines of three marbles extends with just one marble to a line of four marbles, there is only one of the lines that is not allowed for having three X's in a row. Means that the number of possibilities with three marbles (3) almost doubled, but not quite doubled, to 5.\n\nWe probably need to find a math formula, something like \"the length of the line calculates into a number of possibilities\" but I did not find it yet. What I do seem to have figured out is that when a line ends with two marbles of the same colour, there is only one possible way to extend the line namely with a marble of the other colour, while if a line ends in different colours, the possible next sequence can either contain an X or an O.\n\nInstead of extending the line, I could start writing down the number of possible lines that can be made from that line. Like this:\n\n```OOXO (2, as both adding O and X would still give a valid line)\nOOXX (1, only adding O would result in a valid line)\nOXOO (1)\nOXOX (2)\nOXXO (2)\n```\n\nOf course, the lines that end either in two O's or two X's can only be extended with one marble of the other colour, so those get the number 1. The lines that end with two different coloured marbles, can be extended by either an 'O' or an 'X', so those get the number two.\n\nNow the interesting thing is that in case I added the number 2, one of the two possibilities, again must have a double OO or double XX, so we can now write the next numbers for lines of six marbles, to the above possibilities as:\n\n```OOXO (2) (3)\nOOXX (1) (2)\nOXOO (1) (2)\nOXOX (2) (3)\nOXXO (2) (3)\n```\n\nThe above shows the possibilities for lines with a length of six marbles. We do not see all possibilities anymore, but we do see the number of possibilities. When we add up all the numbers in the last column, we have the number of possible lines for lines of six marbles, being 3+2+2+3+3=13. Although, we have to keep in mind this is only for the lines starting with 'O', so we have to double that amount for the lines that would be the inverse of these lines starting with 'X's. That would be 26 for lines of length 6.\n\nSo what do we have so far, if we look at line-length and the resulting number of possible lines?\n\n```1 -> 2 (haha, O and X, two possibilities, for sure, is 2^1)\n2 -> 4 (OX, OO, XO, XX, 2^2)\n3 -> 6 (not OOO, XXX so 2^3 - 2)\n4 -> 8 (2^4 minus 2^3 is 16-8.. hm)\n5 -> 16 (2^5 minus 2^4.. 32-16, on to something?)\n6 -> 26 (2^6 minus 2^5 would be 64-32 = 32..nopes)\n```\n\nTo be honest, I still have not figured out a nice formula.. it must be something like each extra marble doubles the possibilities, but not always...\n\nA program exploring all possibilities might be easier to write; let the computer figure out how many possibilities there are. How does your program find all possibilities?\n\n```package main\n\nimport \"fmt\"\n\n// Question: You have 10 red and 10 black marbles. In how many ways can you\n// put the 20 marbles in a row so that no more than two marbles of the same\n// colour lie next to each other?\n//\n// An example of a valid sequence would be:\n// OOXOXXOXOXXOOXXOXXOO\n\nconst maxLineLength = 20\nfunc main () {\ncount := CountMarbleLines()\n\nfmt.Println(\"Number of possible lines\", count)\n}\nfunc CountMarbleLines () int {\n\nlinelength := 0\namountO := 10\namountX := 10\ncount := PossibleLines(amountO, amountX, linelength, 0, 0)\nreturn count\n\n}\n\n// Counts possible lines by adding either O or X to the line\n// returning 0 in case no line possible or 1 in case the line is possible\nfunc PossibleLines(amountO int, amountX int, lineLength int, timesO int, timesX int) int {\n// we have to ensure that lines that have three\n// in sequence of the same coloured Marbles won't count\nif (timesO > 2 \|\| timesX > 2) {\nreturn 0 // too many X or O in sequence can not count this line\n}\n\n// Recursive stop function\nif lineLength == maxLineLength {\nreturn 1 // a valid line has been found\n}\n\n// we can add either O or X to the string, register this in timesO/timesX\nif amountO > 0 { // if we still have O Marbles to spend...\ntimesO++\naddO = PossibleLines(amountO - 1, amountX, lineLength + 1, timesO, 0)\n}\nif amountX > 0 { // if we still have X Marbles to spend...\ntimesX++\naddX = PossibleLines(amountO, amountX - 1, lineLength + 1, 0, timesX)\n}\n}\n```\n\nNote that the above is written in GoLang which is also easily portable to .NET C#.\n\nWhen you run the above program you will get the answer to the question!\n\nHere's the .NET C# version, which will also print out all the possibilities:\n\n```class Program\n{\nstatic void Main(string[] args)\n{\nConsole.Out.WriteLine(\"Starting to count..\");\nint count = MarbleCount();\nConsole.Out.WriteLine(\"Total number found: \" + count);\n}\n\n/// Count number of possible marble strings.\n/// Rules: 20 marbles in line, 10 x's and 10 o's whereby there can not be more than two\n/// marbles of the same colour next to each other.\n/// Example: oxxooxoxoxooxxoxooxx\n/// Logic: There are 10 o's to spent, 10 x's to spent\nprivate static int MarbleCount()\n{\nint amountO = 10;\nint amountX = 10;\nint lengthToFind = 20;\nStringBuilder sb = new StringBuilder();\nreturn PossibleStrings(amountO, amountX, 0, 0, 0, sb, lengthToFind);\n}\n\nprivate static int PossibleStrings(int amountO, int amountX, int lengthOfString, int timesO, int timesX, StringBuilder sb, int lengthToFind)\n{\n// stop condition of recursive function:\nif (timesO > 2 \|\| timesX > 2)\n{\nreturn 0; // too many O's or X's in sequence, can't count this line.\n}\n\n// if the lenght == 20, we have constructed a valid string\nif (lengthOfString == lengthToFind)\n{\nConsole.Out.WriteLine(sb);\nreturn 1; // a valid string is constructed\n}\n\n// we can add either O or X to the string, register this in timesO/timesX\nif (amountO > 0)\n{\naddO = PossibleStrings(amountO - 1, amountX, lengthOfString + 1, ++timesO, 0, sb.Append(\"O\"), lengthToFind);\nsb.Remove(sb.Length - 1, 1);\n}\nif (amountX > 0)\n{\naddX = PossibleStrings(amountO, amountX - 1, lengthOfString + 1, 0, ++timesX, sb.Append(\"X\"), lengthToFind);\nsb.Remove(sb.Length - 1, 1);\n}" ]	[ null, "http://www.mellekoning.nl/wp-content/uploads/2017/02/rowof20balls.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8720277,"math_prob":0.9683352,"size":8239,"snap":"2020-45-2020-50","text_gpt3_token_len":2214,"char_repetition_ratio":0.14972678,"word_repetition_ratio":0.08046723,"special_character_ratio":0.2763685,"punctuation_ratio":0.13002916,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9850545,"pos_list":[0,1,2],"im_url_duplicate_count":[null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-29T23:12:51Z\",\"WARC-Record-ID\":\"<urn:uuid:2c2dec3f-0392-48fd-ac41-b72da35c8173>\",\"Content-Length\":\"99513\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b14c7939-82aa-4eea-b94d-d7fd2524efa9>\",\"WARC-Concurrent-To\":\"<urn:uuid:183224c5-9669-4c45-83c3-8ad42e766e83>\",\"WARC-IP-Address\":\"93.191.132.129\",\"WARC-Target-URI\":\"http://www.mellekoning.nl/index.php/2017/02/\",\"WARC-Payload-Digest\":\"sha1:3M6XJJYS7LEA7LHRCCRE5XCDO4HTRELN\",\"WARC-Block-Digest\":\"sha1:5RXQKGXROCIXI6KP2UWOSW5H34DSFF67\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107905965.68_warc_CC-MAIN-20201029214439-20201030004439-00149.warc.gz\"}"}
https://www.python51.com/jchu/139832.html	[ "## 解方程怎么用python的函数包起来\n\n44次阅读\n\nnumpy 库提供了一个名为 linalg.solve() 的函数，可以用来解决线性方程组。例如，我们有如下的线性方程组：\n\n3x + 5y = 17\n\n7x – 2y = 13\n\n“` import numpy as np # 线性方程组系数矩阵 a = np.array([[3, 5], [7, -2]]) # 常数列 b = np.array([17, 13]) # 解方程组 x = np.linalg.solve(a, b) print(x) “`\n\nSympy 是一个 Python 库，用于符号数学计算。它可以处理各种数学问题，包括求解方程、微积分、代数、离散数学和几何学等等。\n\nx + 2 = 5\n\n“` from sympy import Eq, solve, symbols # 定义方程 x = symbols(‘x’) eq = Eq(x + 2, 5) # 解方程 sol = solve(eq, x) print(sol) “`\n\nScipy 是一个用于数学、科学和工程的 Python 库，支持各种数值算法，例如集成、微分方程、优化、拟合和信号处理等。\n\nx^2 + y = 10\n\ny + 3z = 20\n\nx – z = 0\n\n“` from scipy.optimize import root def equations(vars): x, y, z = vars eq1 = x*2 + y – 10 eq2 = y + 3z – 20 eq3 = x – z return [eq1, eq2, eq3] # 非线性方程组的解 sol = root(equations, [0, 0, 0]) print(sol.x) “`\n\nPython 提供了许多强大的工具，可以帮助我们解决各种数学问题。从简单的一元一次方程到复杂的非线性方程组，Python 都可以提供相应的工具，帮助我们轻松地解决这些问题。我们可以通过使用 numpy、Sympy 和 Scipy 等库来实现方程的求解，这些库拥有丰富的函数和强大的算法，可以使我们更加高效地解决问题。" ]	[ null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.9687629,"math_prob":0.9999243,"size":1809,"snap":"2023-40-2023-50","text_gpt3_token_len":1281,"char_repetition_ratio":0.10027701,"word_repetition_ratio":0.0,"special_character_ratio":0.33941403,"punctuation_ratio":0.12341772,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99908996,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-06T21:21:26Z\",\"WARC-Record-ID\":\"<urn:uuid:d515fc54-dcb6-4cfb-986d-88c86fa62136>\",\"Content-Length\":\"69328\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:729f46d3-5451-4d40-869d-ba0ca546ea39>\",\"WARC-Concurrent-To\":\"<urn:uuid:70ed5822-a243-440c-b04a-8eccecd4fe37>\",\"WARC-IP-Address\":\"43.248.79.206\",\"WARC-Target-URI\":\"https://www.python51.com/jchu/139832.html\",\"WARC-Payload-Digest\":\"sha1:HBTZ6ZETQXJQCI35QX2XOFGJ6EW26XKR\",\"WARC-Block-Digest\":\"sha1:RCSVODXRFWPIBLQ46R5GKSQOQUVKYOJP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100603.33_warc_CC-MAIN-20231206194439-20231206224439-00461.warc.gz\"}"}
https://answers.everydaycalculation.com/lcm/378-2058	[ "Solutions by everydaycalculation.com\n\n## What is the LCM of 378 and 2058?\n\nThe lcm of 378 and 2058 is 18522.\n\n#### Steps to find LCM\n\n1. Find the prime factorization of 378\n378 = 2 × 3 × 3 × 3 × 7\n2. Find the prime factorization of 2058\n2058 = 2 × 3 × 7 × 7 × 7\n3. Multiply each factor the greater number of times it occurs in steps i) or ii) above to find the lcm:\n\nLCM = 2 × 3 × 3 × 3 × 7 × 7 × 7\n4. LCM = 18522\n\nMathStep (Works offline)", null, "Download our mobile app and learn how to find LCM of upto four numbers in your own time:" ]	[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6931826,"math_prob":0.9987959,"size":481,"snap":"2020-34-2020-40","text_gpt3_token_len":157,"char_repetition_ratio":0.11111111,"word_repetition_ratio":0.0,"special_character_ratio":0.43035343,"punctuation_ratio":0.08791209,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9968075,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-07T19:02:02Z\",\"WARC-Record-ID\":\"<urn:uuid:553f3a21-f17a-4fac-a335-8bc7242d3004>\",\"Content-Length\":\"5762\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:04ad0c16-27c9-4c9f-81a9-afc3b2087952>\",\"WARC-Concurrent-To\":\"<urn:uuid:77bd580e-6210-4453-af5b-6b066a3c8dd1>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/lcm/378-2058\",\"WARC-Payload-Digest\":\"sha1:Z5HNA532GID2QENCO6EEMNMVULSFUYRR\",\"WARC-Block-Digest\":\"sha1:2RPR5N5VK63RIMKVWNRWFTDBZ3BMKJNF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439737206.16_warc_CC-MAIN-20200807172851-20200807202851-00296.warc.gz\"}"}
https://slideplayer.com/slide/3724585/	[ "", null, "## Presentation on theme: \"HELPING YOUR CHILD WITH NUMERACY: ADDITION AND SUBTRACTION.\"— Presentation transcript:\n\nChildren’s learning can be divided into four areas of learning: Knowledge: Concepts: Skills: Attitudes: The facts that children need to retain. Requires constant repetition and consolidation in a variety of different ways. E.g. number bonds to 10. Understanding general ideas that can be applied in specific instances e.g. applying number bonds to sums involving the use of 10s: 60 + ? = 100. The way in which children approach their work. The way in which we as adults approach mathematics work. THE NATURE OF LEARNING. The mechanics of finding an answer. Practical methods of working out the answer. E.g. counting on using a number line.\n\nPlenary session : used to consolidate, review and extend the work done in the main teaching activity. Oral/ mental work: Used to rehearse and sharpen skills NUMERACY HOUR STRUCTURE.\n\nWHY DO WE NEED MENTAL STRATEGIES? MentalWritten We may break a calculation into manageable parts, e.g. 148 – 100 + 1 instead of 148 -99 We can not change a calculation to an equivalent one, e.g. 148 – 99 is done as it is. We say the calculation to ourselves, and therefore are aware of what numbers are involved, e.g. 2000 – 10 is not much less than 2000. We don’t say the number to ourselves, but start a procedure such as : 148 - 99 __ By saying, ‘8 take away 9’ before changing it to 18 – 9. We choose a strategy to fit the numbers, e.g. 148 – 99 may be done differently from 84 – 77, although they are both subtractions. We always use the same method. We draw upon specific mathematical knowledge, an understanding of the number system, learned number facts, and so on. We always draw upon memory of a procedure, and possibly, though not necessarily, an understanding of how it works.\n\nTHE NATURE OF LEARNING. Understanding children’s difficulties with addition and subtraction: Change the number names 1, 2, 3 etc. for letters A, B, C this puts you at the same level as a child. To answer the questions you must not translate these letters into numbers. Answer the following questions in order- you may use your fingers. - How many fingers on your left hand? - How many fingers on both hands? C + D =B + E = K - B =E + E = G - D =E + F = D x C =Q - E = A B C D E F G H I J K L M N O P Q R S T U V W X Y Z How could you use a number line to help you?\n\nSKILLS IN EARLY ADDITION. - Counting all- a child doing 2 + 3 counts out two bricks and then three bricks and then finds the total by counting all the bricks. - Counting on from the first number – a child finding 3 + 5 counts on from the first number. ‘four, five, six, seven, eight’. -Counting on from the larger number- a child chooses the larger number, even when it is not the first number, and counts on from there. - Using a known addition fact – where a child gives an immediate response to facts known by heart, such as 6 + 4 or 3 + 3 or 10 + 8. - Using a known fact to derive a new fact – where a child uses a number bond that they know by heart to calculate one that she or he does not know, e.g. using knowledge that 5 + 5 = 10 to work out 5 + 6 = 11 and 5 + 7 = 12. - Using knowledge of place value – where a child uses knowledge that 4 + 3 = 7 to work out 40 + 30 = 70, or knowledge that 46 + 10 is 56 to work out 46 + 11 = 57. - Knowing 1 more or 1 less - One to one counting, games and rhymes\n\nSKILLS IN EARLY SUBTRACTION. - Counting out – a child finding 9 – 3 holds up nine fingers and folds down three. - Counting back from – a child finding 9 – 3 counts back three numbers from 9: ‘eight, seven, six’. - Counting back to - a child doing 11 – 7 counts back from the first number to the second, keeping a tally using fingers of the number of numbers that have been said. - Counting up – a child doing 11 – 7 counts up from the first 7 to 11, ‘eight, nine, ten, eleven’ (not a ‘natural’ strategy for many children because of the widely held perception of subtraction as ‘taking away’ - Using a know fact – a child gives a rapid response based on facts known by heart, such as 10 – 3 or 20 – 9. - Using a derived fact – a child uses a known fact to work out a new one, e.g. 20 – 5 is 15, so 20 – 6 must be 14 (more unusual in subtraction than addition). - Using knowledge of place value – a child finding 25 – 9 knows that 25 – 10 is 15, and uses this to give an answer of 16.\n\nLEARNING FROM MISCONCEPTIONS. 6 + 5 = 1043 + 8 = 50138 + 9 = 146 Here the answers are 1 less than the correct answer. 14 – 5 = 1023 – 6 = 18 Here the answers are 1 more than the correct answer. What number goes with 6 to make 10? Child’s answer: 5 A question phrased differently displaying the same counting error - the child has counted the first number on their fingers (or a number line) as part of their ‘counting on’ strategy rather than counting from this number.\n\nIf I had 12 sweets and I needed 16 how many more would I need? Some possible answers: 27- The child has counted on rather than counting back and has made the same error as we have just described. Here we have a combination of two different misconceptions. 28- Inappropriate use of a number operation (adding instead of subtracting). 5- Counting back inappropriately as previously described. 208- Inappropriate use of a number operation (adding instead of subtracting) and an incorrect form of recording. 20 and 8. LEARNING FROM MISCONCEPTIONS.\n\n8 + 6 = 18 - The child has continued to count all of his or her fingers when adding and has therefore mistakenly added 10. 15 – This could be accounted for with inaccurate counting or by the application of a known fact such as double 8 is 16 one less is 15 where the child should have subtracted 2 instead. 17 – This error could be accounted for by the inaccurate use of a number square e.g. counting on from 10 to 20 then 19, 18 etc. LEARNING FROM MISCONCEPTIONS.\n\n45 + 23 95 - The child has added two tens then three more tens rather than units. 7 + 8 + 4= 20 - A near double has been used for the calculation without the required adjustment. 20 + ? = 25 For this a child needs to know that addition is the inverse of subtraction. Therefore 25 – 20 will give the answer. This is quite a high level skill a common misconception would be to add the numbers together to give 45. LEARNING FROM MISCONCEPTIONS.\n\nOther points to bear in mind. Be careful when questions are presented in a different manner or placed into a different context. These misconceptions can be combined such as counting on for a subtraction sum as well as counting from the wrong number. Answers can just be made up / estimated. The best source of information regarding misconceptions is your child. The best question you can ask is ‘can you show me how you worked it out?’ Don’t expect a child who cannot count backwards to be able to do simple division. LEARNING FROM MISCONCEPTIONS.\n\nSOME DIFFICULTIES. Flexibility using a range of strategies can be hindered if: - Children do almost any addition or subtraction by counting on or back in ones. - Children rely upon a singular method or piece of apparatus. - Subtract numbers that are close together such as 42-38, by trying to ‘take away’ or ‘count back’ 38 from 42, rather than counting up from 38. - Don’t recognise that to add 10 or 100 is no more difficult than to add 1. - Don’t recognise numbers which are 10 apart, or a multiple of 10. - Never spot a double or number bond. - Never change a calculation to make it easier, e.g. in 242 – 99 they subtract 99 rather than subtracting 100? - To calculate mentally, turn to a standard written method and try to visualise it. - Don’t see facts which would make the calculation easier.\n\nHOW TO HELP YOUR CHILDREN. 1.Look at the half term plans for an objective. 2.Think of a relevant everyday activity that addresses this objective. 3.Ask your child how they worked out their answer. 4.Address any misconceptions. 5.Make it fun! If the child does not want to do the activity they will get very little from it." ]	[ null, "https://slideplayer.com/static/blue_design/img/slide-loader4.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.93475276,"math_prob":0.93721294,"size":8136,"snap":"2019-51-2020-05","text_gpt3_token_len":2082,"char_repetition_ratio":0.120142646,"word_repetition_ratio":0.025364617,"special_character_ratio":0.27396756,"punctuation_ratio":0.110912345,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98815525,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-06T02:01:42Z\",\"WARC-Record-ID\":\"<urn:uuid:933c4335-7a61-4851-8e3a-f5f4a7f54a77>\",\"Content-Length\":\"171407\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8868493f-4901-4dbf-bcc8-0621db95b599>\",\"WARC-Concurrent-To\":\"<urn:uuid:a209f8ba-712c-4e84-a818-f5440484fd71>\",\"WARC-IP-Address\":\"138.201.58.10\",\"WARC-Target-URI\":\"https://slideplayer.com/slide/3724585/\",\"WARC-Payload-Digest\":\"sha1:SA73ZUEI2SDCXHU4GKE6ESH57MWWN7FA\",\"WARC-Block-Digest\":\"sha1:YHMC7WR5D7T3QFEIJWWFGSFUYWMKWXPO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540482954.0_warc_CC-MAIN-20191206000309-20191206024309-00107.warc.gz\"}"}
http://www.rapidlearningcenter.com/mathematics/introductory-statistics/24-Regression-Inference.html	[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "How to Learn in 24 Hours?\n\n Need Help? M-F: 9am-5pm(PST): Toll-Free: (877) RAPID-10 or 1-877-727-4310 24/7 Online Technical Support: The Rapid Support Center [email protected] Secure Online Order:", null, "Need Proof? Testimonials by Our Users\n\n Rapid Learning Courses: MCAT in 24 Hours (2021-22) USMLE in 24 Hours (Boards) Chemistry in 24 Hours Biology in 24 Hours Physics in 24 Hours Mathematics in 24 Hours Psychology in 24 Hours SAT in 24 Hours ACT in 24 Hours AP in 24 Hours CLEP in 24 Hours DAT in 24 Hours (Dental) OAT in 24 Hours (Optometry) PCAT in 24 Hours (Pharmacy) Nursing Entrance Exams Certification in 24 Hours eBook - Survival Kits Audiobooks (MP3)\n\n Tell-A-Friend: Have friends taking science and math courses too? Tell them about our rapid learning system.", null, "Home » Mathematics » Introductory Statistics\n\nRegression Inference\n\nTopic Review on \"Title\":\n\nSome Definitions\n\n• Regression: X, the independent variable, the explanatory variable or the predictor variable. Y, the dependent variable, the response variable or the predicted variable. Given X, we will be able to predict Y.\n• Deterministic Model: y = a + bx\n• Probabilistic Model: y = a + bx+ e , where e ~N(0,s2)\n\nLeast Square Estimators", null, "", null, "", null, "", null, "ANOVA\n\n• SST=the total variation in the experiment.\n• SST=SSR+SSE\n• SSR (sum of squares for regression): measures the variation explained by regression model.\n• SSE (sum of squares for error): measures the variation not explained by x.\n\nANOVA Table\n\n Source df SS MS Test Statistics (Mean Squares) F Regression 1 SSR SSR/(1) MSR/MSE Model (=MSR) Error n - 2 SSE SSE/(n-2) (=MSE) Total n -1 SST\n\nF Test\n\n• F test shall be used to test whether the regression model fit well or not. If the model fit well, the test statistics F will be large. (This test is equivalent to t-test for t2 = F)\n• H0: The regression model fits well.\n• Ha: The regression model does not fit well.\n• F=MSR/MSE\n• Reject H0 if F>Fa,1,n-2\n• a is the probability of a type I error\n\nT test and Confidence level", null, "", null, "Rapid Study Kit for \"Title\":\n Flash Movie Flash Game Flash Card Core Concept Tutorial Problem Solving Drill Review Cheat Sheet", null, "", null, "", null, "\"Title\" Tutorial Summary : This tutorial specifically describes the regression inference. In statistics, we frequently measure two or more variables on the same experimental unit. We do this to explore the nature of the relationship among these variables. Regression inference is to use knowledge of independent variable(s) to predict dependent variable. By completing this course, you will learn about the regression inference including regression models and least square method, the analysis of variance (ANOVA), testing regression model, assumptions and estimation and prediction\n\n Tutorial Features: Specific Tutorial Features: Animated examples showing the operation of least square method is presented in the tutorial. Step by step analysis of ANOVA is presented and served as a base for the subsequent hypothesis testing for the regression model. Series Features: Concept map showing inter-connections of new concepts in this tutorial and those previously introduced. Definition slides introduce terms as they are needed. 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