18.6: Difference between revisions
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Admin moved page Main Page to Verifying DLMF with Maple and Mathematica |
Admin moved page Main Page to Verifying DLMF with Maple and Mathematica |
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! scope="col" style="position: sticky; top: 0;" | Numeric<br>Mathematica | ! scope="col" style="position: sticky; top: 0;" | Numeric<br>Mathematica | ||
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| [https://dlmf.nist.gov/18.6.E1 18.6.E1] | | | [https://dlmf.nist.gov/18.6.E1 18.6.E1] || <math qid="Q5564">\LaguerrepolyL[\alpha]{n}@{0} = \frac{\Pochhammersym{\alpha+1}{n}}{n!}</math><br><syntaxhighlight lang="tex" style="font-size: 75%;" inline>\LaguerrepolyL[\alpha]{n}@{0} = \frac{\Pochhammersym{\alpha+1}{n}}{n!}</syntaxhighlight> || <math></math> || <syntaxhighlight lang=mathematica>LaguerreL(n, alpha, 0) = (pochhammer(alpha + 1, n))/(factorial(n))</syntaxhighlight> || <syntaxhighlight lang=mathematica>LaguerreL[n, \[Alpha], 0] == Divide[Pochhammer[\[Alpha]+ 1, n],(n)!]</syntaxhighlight> || Missing Macro Error || Successful || - || Successful [Tested: 9] | ||
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| [https://dlmf.nist.gov/18.6.E2 18.6.E2] | | | [https://dlmf.nist.gov/18.6.E2 18.6.E2] || <math qid="Q5565">\lim_{\alpha\to\infty}\frac{\JacobipolyP{\alpha}{\beta}{n}@{x}}{\JacobipolyP{\alpha}{\beta}{n}@{1}} = \left(\frac{1+x}{2}\right)^{n}</math><br><syntaxhighlight lang="tex" style="font-size: 75%;" inline>\lim_{\alpha\to\infty}\frac{\JacobipolyP{\alpha}{\beta}{n}@{x}}{\JacobipolyP{\alpha}{\beta}{n}@{1}} = \left(\frac{1+x}{2}\right)^{n}</syntaxhighlight> || <math></math> || <syntaxhighlight lang=mathematica>limit((JacobiP(n, alpha, beta, x))/(JacobiP(n, alpha, beta, 1)), alpha = infinity) = ((1 + x)/(2))^(n)</syntaxhighlight> || <syntaxhighlight lang=mathematica>Limit[Divide[JacobiP[n, \[Alpha], \[Beta], x],JacobiP[n, \[Alpha], \[Beta], 1]], \[Alpha] -> Infinity, GenerateConditions->None] == (Divide[1 + x,2])^(n)</syntaxhighlight> || Failure || Aborted || Successful [Tested: 27] || Skipped - Because timed out | ||
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| [https://dlmf.nist.gov/18.6.E3 18.6.E3] | | | [https://dlmf.nist.gov/18.6.E3 18.6.E3] || <math qid="Q5566">\lim_{\beta\to\infty}\frac{\JacobipolyP{\alpha}{\beta}{n}@{x}}{\JacobipolyP{\alpha}{\beta}{n}@{-1}} = \left(\frac{1-x}{2}\right)^{n}</math><br><syntaxhighlight lang="tex" style="font-size: 75%;" inline>\lim_{\beta\to\infty}\frac{\JacobipolyP{\alpha}{\beta}{n}@{x}}{\JacobipolyP{\alpha}{\beta}{n}@{-1}} = \left(\frac{1-x}{2}\right)^{n}</syntaxhighlight> || <math></math> || <syntaxhighlight lang=mathematica>limit((JacobiP(n, alpha, beta, x))/(JacobiP(n, alpha, beta, - 1)), beta = infinity) = ((1 - x)/(2))^(n)</syntaxhighlight> || <syntaxhighlight lang=mathematica>Limit[Divide[JacobiP[n, \[Alpha], \[Beta], x],JacobiP[n, \[Alpha], \[Beta], - 1]], \[Beta] -> Infinity, GenerateConditions->None] == (Divide[1 - x,2])^(n)</syntaxhighlight> || Failure || Failure || Error || Successful [Tested: 27] | ||
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| [https://dlmf.nist.gov/18.6.E4 18.6.E4] | | | [https://dlmf.nist.gov/18.6.E4 18.6.E4] || <math qid="Q5567">\lim_{\lambda\to\infty}\frac{\ultrasphpoly{\lambda}{n}@{x}}{\ultrasphpoly{\lambda}{n}@{1}} = x^{n}</math><br><syntaxhighlight lang="tex" style="font-size: 75%;" inline>\lim_{\lambda\to\infty}\frac{\ultrasphpoly{\lambda}{n}@{x}}{\ultrasphpoly{\lambda}{n}@{1}} = x^{n}</syntaxhighlight> || <math></math> || <syntaxhighlight lang=mathematica>limit((GegenbauerC(n, lambda, x))/(GegenbauerC(n, lambda, 1)), lambda = infinity) = (x)^(n)</syntaxhighlight> || <syntaxhighlight lang=mathematica>Limit[Divide[GegenbauerC[n, \[Lambda], x],GegenbauerC[n, \[Lambda], 1]], \[Lambda] -> Infinity, GenerateConditions->None] == (x)^(n)</syntaxhighlight> || Failure || Aborted || Successful [Tested: 9] || Skipped - Because timed out | ||
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| [https://dlmf.nist.gov/18.6.E5 18.6.E5] | | | [https://dlmf.nist.gov/18.6.E5 18.6.E5] || <math qid="Q5568">\lim_{\alpha\to\infty}\frac{\LaguerrepolyL[\alpha]{n}@{\alpha x}}{\LaguerrepolyL[\alpha]{n}@{0}} = (1-x)^{n}</math><br><syntaxhighlight lang="tex" style="font-size: 75%;" inline>\lim_{\alpha\to\infty}\frac{\LaguerrepolyL[\alpha]{n}@{\alpha x}}{\LaguerrepolyL[\alpha]{n}@{0}} = (1-x)^{n}</syntaxhighlight> || <math></math> || <syntaxhighlight lang=mathematica>limit((LaguerreL(n, alpha, alpha*x))/(LaguerreL(n, alpha, 0)), alpha = infinity) = (1 - x)^(n)</syntaxhighlight> || <syntaxhighlight lang=mathematica>Limit[Divide[LaguerreL[n, \[Alpha], \[Alpha]*x],LaguerreL[n, \[Alpha], 0]], \[Alpha] -> Infinity, GenerateConditions->None] == (1 - x)^(n)</syntaxhighlight> || Missing Macro Error || Aborted || - || Skipped - Because timed out | ||
|} | |} | ||
</div> | </div> | ||
Latest revision as of 11:44, 28 June 2021
| DLMF | Formula | Constraints | Maple | Mathematica | Symbolic Maple |
Symbolic Mathematica |
Numeric Maple |
Numeric Mathematica |
|---|---|---|---|---|---|---|---|---|
| 18.6.E1 | Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle \LaguerrepolyL[\alpha]{n}@{0} = \frac{\Pochhammersym{\alpha+1}{n}}{n!}}
\LaguerrepolyL[\alpha]{n}@{0} = \frac{\Pochhammersym{\alpha+1}{n}}{n!} |
Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle } | LaguerreL(n, alpha, 0) = (pochhammer(alpha + 1, n))/(factorial(n))
|
LaguerreL[n, \[Alpha], 0] == Divide[Pochhammer[\[Alpha]+ 1, n],(n)!]
|
Missing Macro Error | Successful | - | Successful [Tested: 9] |
| 18.6.E2 | Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle \lim_{\alpha\to\infty}\frac{\JacobipolyP{\alpha}{\beta}{n}@{x}}{\JacobipolyP{\alpha}{\beta}{n}@{1}} = \left(\frac{1+x}{2}\right)^{n}}
\lim_{\alpha\to\infty}\frac{\JacobipolyP{\alpha}{\beta}{n}@{x}}{\JacobipolyP{\alpha}{\beta}{n}@{1}} = \left(\frac{1+x}{2}\right)^{n} |
Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle } | limit((JacobiP(n, alpha, beta, x))/(JacobiP(n, alpha, beta, 1)), alpha = infinity) = ((1 + x)/(2))^(n)
|
Limit[Divide[JacobiP[n, \[Alpha], \[Beta], x],JacobiP[n, \[Alpha], \[Beta], 1]], \[Alpha] -> Infinity, GenerateConditions->None] == (Divide[1 + x,2])^(n)
|
Failure | Aborted | Successful [Tested: 27] | Skipped - Because timed out |
| 18.6.E3 | Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle \lim_{\beta\to\infty}\frac{\JacobipolyP{\alpha}{\beta}{n}@{x}}{\JacobipolyP{\alpha}{\beta}{n}@{-1}} = \left(\frac{1-x}{2}\right)^{n}}
\lim_{\beta\to\infty}\frac{\JacobipolyP{\alpha}{\beta}{n}@{x}}{\JacobipolyP{\alpha}{\beta}{n}@{-1}} = \left(\frac{1-x}{2}\right)^{n} |
Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle } | limit((JacobiP(n, alpha, beta, x))/(JacobiP(n, alpha, beta, - 1)), beta = infinity) = ((1 - x)/(2))^(n)
|
Limit[Divide[JacobiP[n, \[Alpha], \[Beta], x],JacobiP[n, \[Alpha], \[Beta], - 1]], \[Beta] -> Infinity, GenerateConditions->None] == (Divide[1 - x,2])^(n)
|
Failure | Failure | Error | Successful [Tested: 27] |
| 18.6.E4 | Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle \lim_{\lambda\to\infty}\frac{\ultrasphpoly{\lambda}{n}@{x}}{\ultrasphpoly{\lambda}{n}@{1}} = x^{n}}
\lim_{\lambda\to\infty}\frac{\ultrasphpoly{\lambda}{n}@{x}}{\ultrasphpoly{\lambda}{n}@{1}} = x^{n} |
Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle } | limit((GegenbauerC(n, lambda, x))/(GegenbauerC(n, lambda, 1)), lambda = infinity) = (x)^(n)
|
Limit[Divide[GegenbauerC[n, \[Lambda], x],GegenbauerC[n, \[Lambda], 1]], \[Lambda] -> Infinity, GenerateConditions->None] == (x)^(n)
|
Failure | Aborted | Successful [Tested: 9] | Skipped - Because timed out |
| 18.6.E5 | Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle \lim_{\alpha\to\infty}\frac{\LaguerrepolyL[\alpha]{n}@{\alpha x}}{\LaguerrepolyL[\alpha]{n}@{0}} = (1-x)^{n}}
\lim_{\alpha\to\infty}\frac{\LaguerrepolyL[\alpha]{n}@{\alpha x}}{\LaguerrepolyL[\alpha]{n}@{0}} = (1-x)^{n} |
Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle } | limit((LaguerreL(n, alpha, alpha*x))/(LaguerreL(n, alpha, 0)), alpha = infinity) = (1 - x)^(n)
|
Limit[Divide[LaguerreL[n, \[Alpha], \[Alpha]*x],LaguerreL[n, \[Alpha], 0]], \[Alpha] -> Infinity, GenerateConditions->None] == (1 - x)^(n)
|
Missing Macro Error | Aborted | - | Skipped - Because timed out |