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 \pi^{1/2} H(y) = \operatorname{Im} \int_0^\infty dk \, \exp[-k^2/4+iky] }

${\displaystyle \pi ^{1/2}H(y)=\operatorname {Im} \int _{0}^{\infty }dk\,\exp[-k^{2}/4+iky]}$

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

Semantic latex: \cpi^{1/2} \HeavisideH@{y} = \imagpart \int_0^\infty \diff{k} \exp [- k^2 / 4 + \iunit ky]

Confidence: 0.52589992733497

Mathematica

Translation:

Information

Symbol info

(LaTeX -> Mathematica) The input LaTeX is invalid: Limited expression in the end are illegal!

Tests

Symbolic

Numeric

SymPy

Translation:

Information

Symbol info

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

Tests

Symbolic

Numeric

Maple

Translation:

Information

Symbol info

(LaTeX -> Maple) The input LaTeX is invalid: Limited expression in the end are illegal!

Tests

Symbolic

Numeric

Dependency Graph Information

Includes

$H(y)$

$y$

$k$

$\pi ^{1/2}$

$\pi ^{1/2}H(y)=\operatorname {Im} \int _{0}^{\infty }dk\,\exp[-k^{2}/4+iky]$

Is part of

$\pi ^{1/2}H(y)=\operatorname {Im} \int _{0}^{\infty }dk\,\exp[-k^{2}/4+iky]$

Complete translation information:

{
  "id" : "FORMULA_0d0a4a6e3754c72ce3ae7dd5133d1fb4",
  "formula" : "\\pi^{1/2} H(y) = \\operatorname{Im} \\int_0^\\infty dk  \\exp[-k^2/4+iky]",
  "semanticFormula" : "\\cpi^{1/2} \\HeavisideH@{y} = \\imagpart \\int_0^\\infty \\diff{k} \\exp [- k^2 / 4 + \\iunit ky]",
  "confidence" : 0.5258999273349684,
  "translations" : {
    "Mathematica" : {
      "translation" : "",
      "translationInformation" : {
        "tokenTranslations" : {
          "Error" : "(LaTeX -> Mathematica) The input LaTeX is invalid: Limited expression in the end are illegal!"
        }
      }
    },
    "SymPy" : {
      "translation" : "",
      "translationInformation" : {
        "tokenTranslations" : {
          "Error" : "(LaTeX -> SymPy) No translation possible for given token: Cannot extract information from feature set: \\HeavisideH [\\HeavisideH]"
        }
      }
    },
    "Maple" : {
      "translation" : "",
      "translationInformation" : {
        "tokenTranslations" : {
          "Error" : "(LaTeX -> Maple) The input LaTeX is invalid: Limited expression in the end are illegal!"
        }
      }
    }
  },
  "positions" : [ ],
  "includes" : [ "H(y)", "y", "k", "\\pi^{1/2}", "\\pi^{1/2} H(y) = \\operatorname{Im} \\int_0^\\infty dk \\, \\exp[-k^2/4+iky]" ],
  "isPartOf" : [ "\\pi^{1/2} H(y) = \\operatorname{Im} \\int_0^\\infty dk \\, \\exp[-k^2/4+iky]" ],
  "definiens" : [ ]
}

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Formula input
Wikitext context	[[Image:DawsonDp.svg\|thumb\|300px\|right\|The Dawson function, <math>F(x) = D_+(x)</math>, around the origin]] [[Image:DawsonDm.svg\|thumb\|300px\|right\|The Dawson function, <math>D_-(x)</math>, around the origin]] In [[mathematics]], the '''Dawson function''' or '''Dawson integral'''<ref>{{dlmf\|id=7\|title=Error Functions, Dawson's and Fresnel Integrals\|first=N. M. \|last=Temme}}</ref> (named after [[H. G. Dawson]]<ref>{{cite journal \| author = Dawson, H. G. \| title = On the Numerical Value of <math>\textstyle\int_0^h \exp(x^2) \, dx</math> \| volume = s1-29 \| number = 1 \| pages = 519–522 \| year = 1897 \| doi=10.1112/plms/s1-29.1.519 \| url = http://plms.oxfordjournals.org/content/s1-29/1/519.short \| journal = Proceedings of the London Mathematical Society }}</ref>) is the one-sided Fourier–Laplace [[sine transform]] of the Gaussian function. ==Definition== The Dawson function is defined as either: :<math>D_+(x) = e^{-x^2} \int_0^x e^{t^2}\,dt,</math> also denoted as ''F''(''x'') or ''D''(''x''), or alternatively :<math>D_-(x) = e^{x^2} \int_0^x e^{-t^2}\,dt.\!</math> The Dawson function is the one-sided Fourier–Laplace [[sine transform]] of the [[Gaussian function]], :<math>D_+(x) = \frac12 \int_0^\infty e^{-t^2/4}\,\sin(xt)\,dt.</math> It is closely related to the [[error function]] erf, as :<math id="exp(-x^2) was downstairs, should be upstairs"> D_+(x) = {\sqrt{\pi} \over 2} e^{-x^2} \operatorname{erfi} (x) = - {i \sqrt{\pi} \over 2 }e^{-x^2} \operatorname{erf} (ix) </math> where erfi is the imaginary error function, {{nowrap\|1=erfi(''x'') = −''i'' erf(''ix'').}} Similarly, :<math>D_-(x) = \frac{\sqrt{\pi}}{2} e^{x^2} \operatorname{erf}(x)</math> in terms of the real error function, erf. In terms of either erfi or the [[Faddeeva function]] ''w''(''z''), the Dawson function can be extended to the entire [[complex plane]]:<ref>Mofreh R. Zaghloul and Ahmed N. Ali, "[https://dx.doi.org/10.1145/2049673.2049679 Algorithm 916: Computing the Faddeyeva and Voigt Functions]," ''ACM Trans. Math. Soft.'' '''38''' (2), 15 (2011). Preprint available at [https://arxiv.org/abs/1106.0151 arXiv:1106.0151].</ref> : <math>F(z) = {\sqrt{\pi} \over 2} e^{-z^2} \operatorname{erfi} (z) = \frac{i\sqrt{\pi}}{2} \left[ e^{-z^2} - w(z) \right],</math> which simplifies to :<math>D_+(x) = F(x) = \frac{\sqrt{\pi}}{2} \operatorname{Im}[ w(x) ]</math> :<math>D_-(x) = i F(-ix) = -\frac{\sqrt{\pi}}{2} \left[ e^{x^2} - w(-ix) \right]</math> for real ''x''. For \|''x''\| near zero, {{nowrap\|1=''F''(''x'') ≈ ''x''.}} For \|''x''\| large, {{nowrap\|1=''F''(''x'') ≈ 1/(2''x'').}} More specifically, near the origin it has the series expansion :<math> F(x) = \sum_{k=0}^\infty \frac{(-1)^k \, 2^k}{(2k+1)!!} \, x^{2k+1} = x - \frac{2}{3} x^3 + \frac{4}{15} x^5 - \cdots,</math> while for large ''x'' it has the asymptotic expansion :<math> F(x) = \sum_{k=0}^{\infty} \frac{(2k-1)!!}{2^{k+1} x^{2k+1}} = \frac{1}{2 x} + \frac{1}{4 x^3} + \frac{3}{8 x^5} + \cdots,</math> where ''n''!! is the [[double factorial]]. ''F''(''x'') satisfies the differential equation :<math> \frac{dF}{dx} + 2xF=1\,\!</math> with the initial condition ''F''(0) = 0. Consequently, it has extrema for :<math> F(x) = \frac{1}{2 x},</math> resulting in ''x'' = ±0.92413887... ({{OEIS2C\|id=A133841}}), ''F''(''x'') = ±0.54104422... ({{OEIS2C\|id=A133842}}). Inflection points follow for :<math> F(x) = \frac{x}{2 x^2 - 1},</math> resulting in ''x'' = ±1.50197526... ({{OEIS2C\|id=A133843}}), ''F''(''x'') = ±0.42768661... ({{OEIS2C\|id=A245262}}). (Apart from the trivial inflection point at ''x'' = 0, ''F''(''x'') = 0.) == Relation to Hilbert transform of Gaussian == The [[Hilbert transform]] of the Gaussian is defined as :<math> H(y) = \pi^{-1} \operatorname{P.V.} \int_{-\infty}^\infty \frac{e^{-x^2}}{y-x} \, dx </math> P.V. denotes the [[Cauchy principal value]], and we restrict ourselves to real <math>y</math>. <math>H(y)</math> can be related to the Dawson function as follows. Inside a principal value integral, we can treat <math>1/u</math> as a [[generalized function]] or distribution, and use the Fourier representation :<math> {1 \over u} = \int_0^\infty dk \, \sin ku = \int_0^\infty dk \, \operatorname{Im} e^{iku} </math> With <math>1/u=1/(y-x)</math>, we use the exponential representation of <math>\sin(ku)</math> and complete the square with respect to <math>x</math> to find :<math> \pi H(y) = \operatorname{Im} \int_0^\infty dk \,\exp[-k^2/4+iky] \int_{-\infty}^\infty dx \, \exp[-(x+ik/2)^2] </math> We can shift the integral over <math>x</math> to the real axis, and it gives <math>\pi^{1/2}</math>. Thus :<math> \pi^{1/2} H(y) = \operatorname{Im} \int_0^\infty dk \, \exp[-k^2/4+iky] </math> We complete the square with respect to <math>k</math> and obtain :<math> \pi^{1/2}H(y) = e^{-y^2} \operatorname{Im} \int_0^\infty dk \, \exp[-(k/2-iy)^2] </math> We change variables to <math>u=ik/2+y</math>: :<math> \pi^{1/2}H(y) = -2e^{-y^2} \operatorname{Im} i \int_y^{i\infty+y} du\ e^{u^2} </math> The integral can be performed as a contour integral around a rectangle in the complex plane. Taking the imaginary part of the result gives :<math> H(y) = 2\pi^{-1/2} F(y) </math> where <math>F(y)</math> is the Dawson function as defined above. The Hilbert transform of <math>x^{2n}e^{-x^2}</math> is also related to the Dawson function. We see this with the technique of differentiating inside the integral sign. Let :<math>H_n = \pi^{-1} \operatorname{P.V.} \int_{-\infty}^\infty \frac{x^{2n}e^{-x^2}}{y-x} \, dx </math> Introduce :<math>H_a = \pi^{-1} \operatorname{P.V.} \int_{-\infty}^\infty {e^{-ax^2} \over y-x} \, dx </math> The ''n''th derivative is :<math>{\partial^nH_a \over \partial a^n} = (-1)^n\pi^{-1} \operatorname{P.V.} \int_{-\infty}^\infty \frac{x^{2n}e^{-ax^2}}{y-x} \, dx </math> We thus find :<math> \left . H_n=(-1)^n \frac{\partial^nH_a}{\partial a^n} \right \|_{a=1} </math> The derivatives are performed first, then the result evaluated at <math>a=1</math>. A change of variable also gives <math>H_a=2\pi^{-1/2}F(y\sqrt a)</math>. Since <math>F'(y)=1-2yF(y)</math>, we can write <math>H_n = P_1(y)+P_2(y)F(y)</math> where <math>P_1</math> and <math>P_2</math> are polynomials. For example, <math>H_1=-\pi^{-1/2}y+2\pi^{-1/2}y^2F(y)</math>. Alternatively, <math>H_n</math> can be calculated using the recurrence relation (for <math>n \geq 0</math>) :<math> H_{n+1}(y) = y^2 H_n(y) - \frac{(2n-1)!!}{\sqrt{\pi} 2^n} y. </math> ==References== <references /> == External links == * [https://www.gnu.org/software/gsl/manual/html_node/Dawson-Function.html gsl_sf_dawson] in the [[GNU Scientific Library]] * [http://jugit.fz-juelich.de/mlz/libcerf libcerf], numeric C library for complex error functions, provides a function ''voigt(x, sigma, gamma)'' with approximately 13–14 digits precision. It is based on the [[Faddeeva function]] as implemented in the [http://ab-initio.mit.edu/Faddeeva MIT Faddeeva Package] * [http://mathworld.wolfram.com/DawsonsIntegral.html Dawson's Integral] ''(at Mathworld)'' * [http://nlpc.stanford.edu/nleht/Science/reference/errorfun.pdf Error functions] [[Category:Special functions]] [[Category:Gaussian function]]
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LaTeX to CAS translator

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SymPy

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Maple

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