LaTeX to CAS translator

This mockup demonstrates the concept of TeX to Computer Algebra System (CAS) conversion.

The demo-application converts LaTeX functions which directly translate to CAS counterparts.

Functions without explicit CAS support are available for translation via a DRMF package (under development).

The following LaTeX input ...

{\displaystyle \begin{align} \log G(z+1) = {} & \frac{z^2}{2} \log z - \frac{3z^2}{4} + \frac{z}{2}\log 2\pi -\frac{1}{12} \log z \\ & {} + \left(\frac{1}{12}-\log A \right) +\sum_{k=1}^N \frac{B_{2k + 2}}{4k\left(k + 1\right)z^{2k}}~+~O\left(\frac{1}{z^{2N + 2}}\right). \end{align}}

${\displaystyle {\begin{aligned}\log G(z+1)={}&{\frac {z^{2}}{2}}\log z-{\frac {3z^{2}}{4}}+{\frac {z}{2}}\log 2\pi -{\frac {1}{12}}\log z\\&{}+\left({\frac {1}{12}}-\log A\right)+\sum _{k=1}^{N}{\frac {B_{2k+2}}{4k\left(k+1\right)z^{2k}}}~+~O\left({\frac {1}{z^{2N+2}}}\right).\end{aligned}}}$

... is translated to the CAS output ...

Semantic latex: \begin{align}\log \BarnesG@{z + 1} ={} &\frac{z^2}{2} \log z - \frac{3z^2}{4} + \frac{z}{2} \log 2 \cpi - \frac{1}{12} \log z \\ &{} +(\frac{1}{12} - \log A) + \sum_{k=1}^N \frac{\BernoullinumberB{2k + 2}}{4 k(k + 1) z^{2k}} + O(\frac{1}{z^{2N + 2}}) .\end{align}

Confidence: 0.7041471827212

Mathematica

Translation:

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Symbol info

(LaTeX -> Mathematica) Error while translating DLMF/DRMF Macro: Unable to retrieve correct number of arguments for the macro \BarnesG [\BarnesG]

Tests

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SymPy

Translation:

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(LaTeX -> SymPy) No translation possible for given token: Cannot extract information from feature set: \BarnesG [\BarnesG]

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Maple

Translation:

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(LaTeX -> Maple) No translation possible for given token: Cannot extract information from feature set: \BarnesG [\BarnesG]

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Includes

$B_{k}$

$A$

$z$

$\,2\pi$

$G$

$B_{2k}$

${\begin{aligned}\log G(z+1)={}&{\frac {z^{2}}{2}}\log z-{\frac {3z^{2}}{4}}+{\frac {z}{2}}\log 2\pi -{\frac {1}{12}}\log z\\&{}+\left({\frac {1}{12}}-\log A\right)+\sum _{k=1}^{N}{\frac {B_{2k+2}}{4k\left(k+1\right)z^{2k}}}+O\left({\frac {1}{z^{2N+2}}}\right).\end{aligned}}$

$G(z+1)$

Is part of

${\begin{aligned}\log G(z+1)={}&{\frac {z^{2}}{2}}\log z-{\frac {3z^{2}}{4}}+{\frac {z}{2}}\log 2\pi -{\frac {1}{12}}\log z\\&{}+\left({\frac {1}{12}}-\log A\right)+\sum _{k=1}^{N}{\frac {B_{2k+2}}{4k\left(k+1\right)z^{2k}}}+O\left({\frac {1}{z^{2N+2}}}\right).\end{aligned}}$

Complete translation information:

{
  "id" : "FORMULA_493debcd6fe4e50af1dcd315dd88006a",
  "formula" : "\\begin{align}\n\\log G(z+1) = {} & \\frac{z^2}{2} \\log z - \\frac{3z^2}{4} + \\frac{z}{2}\\log 2\\pi -\\frac{1}{12} \\log z \\\\\n            & {} + \\left(\\frac{1}{12}-\\log A \\right)\n            +\\sum_{k=1}^N \\frac{B_{2k + 2}}{4k\\left(k + 1\\right)z^{2k}}~+~O\\left(\\frac{1}{z^{2N + 2}}\\right).\n\\end{align}",
  "semanticFormula" : "\\begin{align}\\log \\BarnesG@{z + 1} ={} &\\frac{z^2}{2} \\log z - \\frac{3z^2}{4} + \\frac{z}{2} \\log 2 \\cpi - \\frac{1}{12} \\log z \\\\ &{} +(\\frac{1}{12} - \\log A) + \\sum_{k=1}^N \\frac{\\BernoullinumberB{2k + 2}}{4 k(k + 1) z^{2k}} + O(\\frac{1}{z^{2N + 2}}) .\\end{align}",
  "confidence" : 0.7041471827211964,
  "translations" : {
    "Mathematica" : {
      "translation" : "",
      "translationInformation" : {
        "tokenTranslations" : {
          "Error" : "(LaTeX -> Mathematica) Error while translating DLMF/DRMF Macro: Unable to retrieve correct number of arguments for the macro \\BarnesG [\\BarnesG]"
        }
      }
    },
    "SymPy" : {
      "translation" : "",
      "translationInformation" : {
        "tokenTranslations" : {
          "Error" : "(LaTeX -> SymPy) No translation possible for given token: Cannot extract information from feature set: \\BarnesG [\\BarnesG]"
        }
      }
    },
    "Maple" : {
      "translation" : "",
      "translationInformation" : {
        "tokenTranslations" : {
          "Error" : "(LaTeX -> Maple) No translation possible for given token: Cannot extract information from feature set: \\BarnesG [\\BarnesG]"
        }
      }
    }
  },
  "positions" : [ ],
  "includes" : [ "B_k", "A", "z", "\\, 2\\pi", "G", "B_{2k}", "\\begin{align}\\log G(z+1) = {} & \\frac{z^2}{2} \\log z - \\frac{3z^2}{4} + \\frac{z}{2}\\log 2\\pi -\\frac{1}{12} \\log z \\\\            & {} + \\left(\\frac{1}{12}-\\log A \\right)            +\\sum_{k=1}^N \\frac{B_{2k + 2}}{4k\\left(k + 1\\right)z^{2k}}+O\\left(\\frac{1}{z^{2N + 2}}\\right).\\end{align}", "G(z+ 1)" ],
  "isPartOf" : [ "\\begin{align}\\log G(z+1) = {} & \\frac{z^2}{2} \\log z - \\frac{3z^2}{4} + \\frac{z}{2}\\log 2\\pi -\\frac{1}{12} \\log z \\\\            & {} + \\left(\\frac{1}{12}-\\log A \\right)            +\\sum_{k=1}^N \\frac{B_{2k + 2}}{4k\\left(k + 1\\right)z^{2k}}+O\\left(\\frac{1}{z^{2N + 2}}\\right).\\end{align}" ],
  "definiens" : [ ]
}

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Formula input
Wikitext context	[[File:Barnes G Function.png\|thumb\|The Barnes G function along part of the real axis]] In [[mathematics]], the '''Barnes G-function''' ''G''(''z'') is a [[function (mathematics)\|function]] that is an extension of [[superfactorial]]s to the [[complex number]]s. It is related to the [[gamma function]], the [[K-function]] and the [[Glaisher–Kinkelin constant]], and was named after [[mathematician]] [[Ernest William Barnes]].<ref>E. W. Barnes, "The theory of the G-function", ''Quarterly Journ. Pure and Appl. Math.'' '''31''' (1900), 264–314.</ref> It can be written in terms of the [[double gamma function]]. Formally, the Barnes ''G''-function is defined in the following [[Weierstrass product]] form: :<math> G(1+z)=(2\pi)^{z/2} \exp\left(- \frac{z+z^2(1+\gamma)}{2} \right) \, \prod_{k=1}^\infty \left\{ \left(1+\frac{z}{k}\right)^k \exp\left(\frac{z^2}{2k}-z\right) \right\}</math> where <math>\, \gamma </math> is the [[Euler–Mascheroni constant]], [[exponential function\|exp]](''x'') = ''e''<sup>''x''</sup>, and ∏ is [[capital pi notation]]. ==Functional equation and integer arguments== The Barnes ''G''-function satisfies the [[functional equation]] :<math> G(z+1)=\Gamma(z)\, G(z) </math> with normalisation ''G''(1) = 1. Note the similarity between the functional equation of the Barnes G-function and that of the Euler [[gamma function]]: :<math> \Gamma(z+1)=z \, \Gamma(z) .</math> The functional equation implies that ''G'' takes the following values at [[integer]] arguments: :<math>G(n)=\begin{cases} 0&\text{if }n=0,-1,-2,\dots\\ \prod_{i=0}^{n-2} i!&\text{if }n=1,2,\dots\end{cases}</math> (in particular, <math>\,G(0)=0, G(1)=1</math>) and thus :<math>G(n)=\frac{(\Gamma(n))^{n-1}}{K(n)}</math> where <math>\,\Gamma(x)</math> denotes the [[gamma function]] and ''K'' denotes the [[K-function]]. The functional equation uniquely defines the G function if the convexity condition: <math>\, \frac{d^3}{dx^3}G(x)\geq 0</math> is added.<ref>M. F. Vignéras, ''L'équation fonctionelle de la fonction zêta de Selberg du groupe mudulaire SL<math>(2,\mathbb{Z})</math>'', Astérisque '''61''', 235–249 (1979).</ref> ==Value at 1/2== <math>G\left(\frac12\right) = 2^{\frac{1}{24}} e^{\frac32 \zeta'(-1)}\pi^{-\frac14}.</math> ==Reflection formula 1.0== The [[difference equation]] for the G-function, in conjunction with the [[functional equation]] for the [[gamma function]], can be used to obtain the following [[reflection formula]] for the Barnes G-function (originally proved by [[Hermann Kinkelin]]): :<math> \log G(1-z) = \log G(1+z)-z\log 2\pi+ \int_0^z \pi x \cot \pi x \, dx.</math> The logtangent integral on the right-hand side can be evaluated in terms of the [[Clausen function]] (of order 2), as is shown below: :<math>2\pi \log\left( \frac{G(1-z)}{G(1+z)} \right)= 2\pi z\log\left(\frac{\sin\pi z}{\pi} \right) + \operatorname{Cl}_2(2\pi z)</math> The proof of this result hinges on the following evaluation of the cotangent integral: introducing the notation <math>\operatorname{Lc}(z)</math> for the logcotangent integral, and using the fact that <math>\,(d/dx) \log(\sin\pi x)=\pi\cot\pi x</math>, an integration by parts gives :<math>\begin{align} \operatorname{Lc}(z) &= \int_0^z\pi x\cot \pi x\,dx \\ &= z\log(\sin \pi z)-\int_0^z\log(\sin \pi x)\,dx \\ &= z\log(\sin \pi z)-\int_0^z\Bigg[\log(2\sin \pi x)-\log 2\Bigg]\,dx \\ &= z\log(2\sin \pi z)-\int_0^z\log(2\sin \pi x)\,dx . \end{align}</math> Performing the integral substitution <math>\, y=2\pi x \Rightarrow dx=dy/(2\pi)</math> gives :<math>z\log(2\sin \pi z)-\frac{1}{2\pi}\int_0^{2\pi z}\log\left(2\sin \frac{y}{2} \right)\,dy.</math> The [[Clausen function]] – of second order – has the integral representation :<math>\operatorname{Cl}_2(\theta) = -\int_0^{\theta}\log\Bigg\|2\sin \frac{x}{2} \Bigg\|\,dx.</math> However, within the interval <math>\, 0 < \theta < 2\pi </math>, the [[absolute value]] sign within the [[integrand]] can be omitted, since within the range the 'half-sine' function in the integral is strictly positive, and strictly non-zero. Comparing this definition with the result above for the logtangent integral, the following relation clearly holds: :<math>\operatorname{Lc}(z)=z\log(2\sin \pi z)+\frac{1}{2\pi} \operatorname{Cl}_2(2\pi z).</math> Thus, after a slight rearrangement of terms, the proof is complete: :<math>2\pi \log\left( \frac{G(1-z)}{G(1+z)} \right)= 2\pi z\log\left(\frac{\sin\pi z}{\pi} \right)+\operatorname{Cl}_2(2\pi z)\, . \, \Box </math> Using the relation <math>\, G(1+z)=\Gamma(z)\, G(z) </math> and dividing the reflection formula by a factor of <math>\, 2\pi </math> gives the equivalent form: :<math> \log\left( \frac{G(1-z)}{G(z)} \right)= z\log\left(\frac{\sin\pi z}{\pi} \right)+\log\Gamma(z)+\frac{1}{2\pi}\operatorname{Cl}_2(2\pi z) </math> Ref: see '''Adamchik''' below for an equivalent form of the [[reflection formula]], but with a different proof. ==Reflection formula 2.0== Replacing '''''z''''' with '''(1/2) − ''z''''''' in the previous reflection formula gives, after some simplification, the equivalent formula shown below (involving [[Bernoulli polynomials]]): :<math>\log\left( \frac{ G\left(\frac{1}{2}+z\right) }{ G\left(\frac{1}{2}-z\right) } \right) =</math> :<math> \log \Gamma \left(\frac{1}{2}-z \right) + B_1(z) \log 2\pi+\frac{1}{2}\log 2+\pi \int_0^z B_1(x) \tan \pi x \,dx</math> ==Taylor series expansion== By [[Taylor's theorem]], and considering the logarithmic [[derivative]]s of the Barnes function, the following series expansion can be obtained: :<math>\log G(1+z) = \frac{z}{2}\log 2\pi -\left( \frac{z+(1+\gamma)z^2}{2} \right) + \sum_{k=2}^{\infty}(-1)^k\frac{\zeta(k)}{k+1}z^{k+1}.</math> It is valid for <math>\, 0 < z < 1 </math>. Here, <math>\, \zeta(x) </math> is the [[Riemann Zeta function]]: :<math> \zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}. </math> Exponentiating both sides of the Taylor expansion gives: :<math>\begin{align} G(1+z) &= \exp \left[ \frac{z}{2}\log 2\pi -\left( \frac{z+(1+\gamma)z^2}{2} \right) + \sum_{k=2}^{\infty}(-1)^k\frac{\zeta(k)}{k+1}z^{k+1} \right] \\ &=(2\pi)^{z/2}\exp\left[ -\frac{z+(1+\gamma)z^2}{2} \right] \exp \left[\sum_{k=2}^{\infty}(-1)^k\frac{\zeta(k)}{k+1}z^{k+1} \right].\end{align}</math> Comparing this with the [[Weierstrass product]] form of the Barnes function gives the following relation: :<math>\exp \left[\sum_{k=2}^\infty (-1)^k\frac{\zeta(k)}{k+1}z^{k+1} \right] = \prod_{k=1}^{\infty} \left\{ \left(1+\frac{z}{k}\right)^k \exp \left(\frac{z^2}{2k}-z\right) \right\}</math> ==Multiplication formula== Like the gamma function, the G-function also has a multiplication formula:<ref>I. Vardi, ''Determinants of Laplacians and multiple gamma functions'', SIAM J. Math. Anal. '''19''', 493–507 (1988).</ref> :<math> G(nz)= K(n) n^{n^{2}z^{2}/2-nz} (2\pi)^{-\frac{n^2-n}{2}z}\prod_{i=0}^{n-1}\prod_{j=0}^{n-1}G\left(z+\frac{i+j}{n}\right) </math> where <math>K(n)</math> is a constant given by: :<math> K(n)= e^{-(n^2-1)\zeta^\prime(-1)} \cdot n^{\frac{5}{12}}\cdot(2\pi)^{(n-1)/2}\,=\, (Ae^{-\frac{1}{12}})^{n^2-1}\cdot n^{\frac{5}{12}}\cdot (2\pi)^{(n-1)/2}.</math> Here <math>\zeta^\prime</math> is the derivative of the [[Riemann zeta function]] and <math>A</math> is the [[Glaisher–Kinkelin constant]]. ==Asymptotic expansion== The [[logarithm]] of ''G''(''z'' + 1) has the following asymptotic expansion, as established by Barnes: :<math>\begin{align} \log G(z+1) = {} & \frac{z^2}{2} \log z - \frac{3z^2}{4} + \frac{z}{2}\log 2\pi -\frac{1}{12} \log z \\ & {} + \left(\frac{1}{12}-\log A \right) +\sum_{k=1}^N \frac{B_{2k + 2}}{4k\left(k + 1\right)z^{2k}}~+~O\left(\frac{1}{z^{2N + 2}}\right). \end{align}</math> Here the <math>B_k</math> are the [[Bernoulli numbers]] and <math>A</math> is the [[Glaisher–Kinkelin constant]]. (Note that somewhat confusingly at the time of Barnes <ref>[[E. T. Whittaker]] and [[G. N. Watson]], "[[A Course of Modern Analysis]]", CUP.</ref> the [[Bernoulli number]] <math>B_{2k}</math> would have been written as <math>(-1)^{k+1} B_k </math>, but this convention is no longer current.) This expansion is valid for <math>z </math> in any sector not containing the negative real axis with <math>\|z\|</math> large. ==Relation to the Loggamma integral== The parametric Loggamma can be evaluated in terms of the Barnes G-function (Ref: this result is found in '''Adamchik''' below, but stated without proof): :<math> \int_0^z \log \Gamma(x)\,dx=\frac{z(1-z)}{2}+\frac{z}{2}\log 2\pi +z\log\Gamma(z) -\log G(1+z) </math> The proof is somewhat indirect, and involves first considering the logarithmic difference of the [[gamma function]] and Barnes G-function: :<math>z\log \Gamma(z)-\log G(1+z)</math> where :<math>\frac{1}{\Gamma(z)}= z e^{\gamma z} \prod_{k=1}^\infty \left\{ \left(1+\frac{z}{k}\right)e^{-z/k} \right\}</math> and <math>\,\gamma</math> is the [[Euler–Mascheroni constant]]. Taking the logarithm of the [[Weierstrass product]] forms of the Barnes function and gamma function gives: :<math> \begin{align} & z\log \Gamma(z)-\log G(1+z)=-z \log\left(\frac{1}{\Gamma (z)}\right)-\log G(1+z) \\[5pt] = {} & {-z} \left[ \log z+\gamma z +\sum_{k=1}^\infty \Bigg\{ \log\left(1+\frac{z}{k} \right) -\frac{z}{k} \Bigg\} \right] \\[5pt] & {} -\left[ \frac{z}{2}\log 2\pi -\frac{z}{2}-\frac{z^2}{2} -\frac{z^2 \gamma}{2} + \sum_{k=1}^\infty \Bigg\{k\log\left(1+\frac{z}{k}\right) +\frac{z^2}{2k} -z \Bigg\} \right] \end{align} </math> A little simplification and re-ordering of terms gives the series expansion: :<math> \begin{align} & \sum_{k=1}^\infty \Bigg\{ (k+z)\log \left(1+\frac{z}{k}\right)-\frac{z^2}{2k}-z \Bigg\} \\[5pt] = {} & {-z}\log z-\frac{z}{2}\log 2\pi +\frac{z}{2} +\frac{z^2}{2}- \frac{z^2 \gamma}{2}- z\log\Gamma(z) +\log G(1+z) \end{align} </math> Finally, take the logarithm of the [[Weierstrass product]] form of the [[gamma function]], and integrate over the interval <math>\, [0,\,z]</math> to obtain: :<math> \begin{align} & \int_0^z\log\Gamma(x)\,dx=-\int_0^z \log\left(\frac{1}{\Gamma(x)}\right)\,dx \\[5pt] = {} & {-(z\log z-z)}-\frac{z^2 \gamma}{2}- \sum_{k=1}^\infty \Bigg\{ (k+z)\log \left(1+\frac{z}{k}\right)-\frac{z^2}{2k}-z \Bigg\} \end{align} </math> Equating the two evaluations completes the proof: :<math> \int_0^z \log \Gamma(x)\,dx=\frac{z(1-z)}{2}+\frac{z}{2}\log 2\pi +z\log\Gamma(z) -\log G(1+z)</math> And since <math>\, G(1+z)=\Gamma(z)\, G(z) </math> then, :<math> \int_0^z \log \Gamma(x)\,dx=\frac{z(1-z)}{2}+\frac{z}{2}\log 2\pi -(1-z)\log\Gamma(z) -\log G(z)\, .</math> ==References== <references/> {{dlmf\|first=R.A. \|last=Askey\|first2=R.\|last2=Roy\|id=5.17}} {{DEFAULTSORT:Barnes G-Function}} [[Category:Number theory]] [[Category:Special functions]] {{cite arXiv\|last=Adamchik\|first=Viktor S.\|title=Contributions to the Theory of the Barnes function\|year=2003\|eprint=math/0308086}}
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