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 A}

${\displaystyle A}$

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

Semantic latex: A

Confidence: 0

Mathematica

Translation: A

Information

Sub Equations

A

Free variables

A

Tests

Symbolic

Numeric

SymPy

Translation: A

Information

Sub Equations

A

Free variables

A

Tests

Symbolic

Numeric

Maple

Translation: A

Information

Sub Equations

A

Free variables

A

Tests

Symbolic

Numeric

Dependency Graph Information

Is part of

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

Description

Glaisher -- Kinkelin

derivative of the Riemann zeta function

Bernoulli number

asymptotic expansion

logarithm

Barnes

Complete translation information:

{
  "id" : "FORMULA_7fc56270e7a70fa81a5935b72eacbe29",
  "formula" : "A",
  "semanticFormula" : "A",
  "confidence" : 0.0,
  "translations" : {
    "Mathematica" : {
      "translation" : "A",
      "translationInformation" : {
        "subEquations" : [ "A" ],
        "freeVariables" : [ "A" ]
      },
      "numericResults" : {
        "overallResult" : "SKIPPED",
        "numberOfTests" : 0,
        "numberOfFailedTests" : 0,
        "numberOfSuccessfulTests" : 0,
        "numberOfSkippedTests" : 0,
        "numberOfErrorTests" : 0,
        "wasAborted" : false,
        "crashed" : false,
        "testCalculationsGroups" : [ ]
      },
      "symbolicResults" : {
        "overallResult" : "SKIPPED",
        "numberOfTests" : 0,
        "numberOfFailedTests" : 0,
        "numberOfSuccessfulTests" : 0,
        "numberOfSkippedTests" : 0,
        "numberOfErrorTests" : 0,
        "crashed" : false,
        "testCalculationsGroup" : [ ]
      }
    },
    "SymPy" : {
      "translation" : "A",
      "translationInformation" : {
        "subEquations" : [ "A" ],
        "freeVariables" : [ "A" ]
      },
      "numericResults" : {
        "overallResult" : "SKIPPED",
        "numberOfTests" : 0,
        "numberOfFailedTests" : 0,
        "numberOfSuccessfulTests" : 0,
        "numberOfSkippedTests" : 0,
        "numberOfErrorTests" : 0,
        "wasAborted" : false,
        "crashed" : false,
        "testCalculationsGroups" : [ ]
      },
      "symbolicResults" : {
        "overallResult" : "SKIPPED",
        "numberOfTests" : 0,
        "numberOfFailedTests" : 0,
        "numberOfSuccessfulTests" : 0,
        "numberOfSkippedTests" : 0,
        "numberOfErrorTests" : 0,
        "crashed" : false,
        "testCalculationsGroup" : [ ]
      }
    },
    "Maple" : {
      "translation" : "A",
      "translationInformation" : {
        "subEquations" : [ "A" ],
        "freeVariables" : [ "A" ]
      },
      "numericResults" : {
        "overallResult" : "SKIPPED",
        "numberOfTests" : 0,
        "numberOfFailedTests" : 0,
        "numberOfSuccessfulTests" : 0,
        "numberOfSkippedTests" : 0,
        "numberOfErrorTests" : 0,
        "wasAborted" : false,
        "crashed" : false,
        "testCalculationsGroups" : [ ]
      },
      "symbolicResults" : {
        "overallResult" : "SKIPPED",
        "numberOfTests" : 0,
        "numberOfFailedTests" : 0,
        "numberOfSuccessfulTests" : 0,
        "numberOfSkippedTests" : 0,
        "numberOfErrorTests" : 0,
        "crashed" : false,
        "testCalculationsGroup" : [ ]
      }
    }
  },
  "positions" : [ {
    "section" : 6,
    "sentence" : 1,
    "word" : 11
  }, {
    "section" : 7,
    "sentence" : 0,
    "word" : 24
  } ],
  "includes" : [ ],
  "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" : [ {
    "definition" : "Glaisher -- Kinkelin",
    "score" : 0.8869384888466118
  }, {
    "definition" : "derivative of the Riemann zeta function",
    "score" : 0.6687181434333315
  }, {
    "definition" : "Bernoulli number",
    "score" : 0.6042528684491243
  }, {
    "definition" : "asymptotic expansion",
    "score" : 0.5377290372506534
  }, {
    "definition" : "logarithm",
    "score" : 0.5377290372506534
  }, {
    "definition" : "Barnes",
    "score" : 0.441749596053091
  } ]
}

Specify your own input

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}}
	Purge cached results.

LaTeX to CAS translator

Mathematica

Information

Tests

Symbolic

Numeric

SymPy

Information

Tests

Symbolic

Numeric

Maple

Information

Tests

Symbolic

Numeric

Dependency Graph Information

Specify your own input

Navigation menu

Search