24.19: Difference between revisions

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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/24.19#Ex2 24.19#Ex2] || [[Item:Q7594|<math>\BernoullinumberB{2n} = \dfrac{N_{2n}}{D_{2n}}</math>]]<br><syntaxhighlight lang="tex" style="font-size: 75%;" inline>\BernoullinumberB{2n} = \dfrac{N_{2n}}{D_{2n}}</syntaxhighlight> || <math></math> || <syntaxhighlight lang=mathematica>bernoulli(2*n) = (N[2*n])/(D[2*n])</syntaxhighlight> || <syntaxhighlight lang=mathematica>BernoulliB[2*n] == Divide[Subscript[N, 2*n],Subscript[D, 2*n]]</syntaxhighlight> || Failure || Failure || <div class="toccolours mw-collapsible mw-collapsed">Failed [300 / 300]<div class="mw-collapsible-content"><syntaxhighlight lang=mathematica>Result: -.8333333333
| [https://dlmf.nist.gov/24.19#Ex2 24.19#Ex2] || <math qid="Q7594">\BernoullinumberB{2n} = \dfrac{N_{2n}}{D_{2n}}</math><br><syntaxhighlight lang="tex" style="font-size: 75%;" inline>\BernoullinumberB{2n} = \dfrac{N_{2n}}{D_{2n}}</syntaxhighlight> || <math></math> || <syntaxhighlight lang=mathematica>bernoulli(2*n) = (N[2*n])/(D[2*n])</syntaxhighlight> || <syntaxhighlight lang=mathematica>BernoulliB[2*n] == Divide[Subscript[N, 2*n],Subscript[D, 2*n]]</syntaxhighlight> || Failure || Failure || <div class="toccolours mw-collapsible mw-collapsed">Failed [300 / 300]<div class="mw-collapsible-content"><syntaxhighlight lang=mathematica>Result: -.8333333333
Test Values: {D[2*n] = 1/2*3^(1/2)+1/2*I, N[2*n] = 1/2*3^(1/2)+1/2*I, n = 1}</syntaxhighlight><br><syntaxhighlight lang=mathematica>Result: -1.033333333
Test Values: {D[2*n] = 1/2*3^(1/2)+1/2*I, N[2*n] = 1/2*3^(1/2)+1/2*I, n = 1}</syntaxhighlight><br><syntaxhighlight lang=mathematica>Result: -1.033333333
Test Values: {D[2*n] = 1/2*3^(1/2)+1/2*I, N[2*n] = 1/2*3^(1/2)+1/2*I, n = 2}</syntaxhighlight><br>... skip entries to safe data</div></div> || <div class="toccolours mw-collapsible mw-collapsed">Failed [300 / 300]<div class="mw-collapsible-content"><syntaxhighlight lang=mathematica>Result: -0.8333333333333334
Test Values: {D[2*n] = 1/2*3^(1/2)+1/2*I, N[2*n] = 1/2*3^(1/2)+1/2*I, n = 2}</syntaxhighlight><br>... skip entries to safe data</div></div> || <div class="toccolours mw-collapsible mw-collapsed">Failed [300 / 300]<div class="mw-collapsible-content"><syntaxhighlight lang=mathematica>Result: -0.8333333333333334
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Test Values: {Rule[n, 2], Rule[Subscript[D, Times[2, n]], Power[E, Times[Complex[0, Rational[1, 6]], Pi]]], Rule[Subscript[N, Times[2, n]], Power[E, Times[Complex[0, Rational[1, 6]], Pi]]]}</syntaxhighlight><br>... skip entries to safe data</div></div>
Test Values: {Rule[n, 2], Rule[Subscript[D, Times[2, n]], Power[E, Times[Complex[0, Rational[1, 6]], Pi]]], Rule[Subscript[N, Times[2, n]], Power[E, Times[Complex[0, Rational[1, 6]], Pi]]]}</syntaxhighlight><br>... skip entries to safe data</div></div>
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| [https://dlmf.nist.gov/24.19.E3 24.19.E3] || [[Item:Q7595|<math>\frac{t^{2}}{\cosh@@{t}-1} = -2\sum_{n=0}^{\infty}(2n-1)\BernoullinumberB{2n}\frac{t^{2n}}{(2n)!}</math>]]<br><syntaxhighlight lang="tex" style="font-size: 75%;" inline>\frac{t^{2}}{\cosh@@{t}-1} = -2\sum_{n=0}^{\infty}(2n-1)\BernoullinumberB{2n}\frac{t^{2n}}{(2n)!}</syntaxhighlight> || <math></math> || <syntaxhighlight lang=mathematica>((t)^(2))/(cosh(t)- 1) = - 2*sum((2*n - 1)*bernoulli(2*n)*((t)^(2*n))/(factorial(2*n)), n = 0..infinity)</syntaxhighlight> || <syntaxhighlight lang=mathematica>Divide[(t)^(2),Cosh[t]- 1] == - 2*Sum[(2*n - 1)*BernoulliB[2*n]*Divide[(t)^(2*n),(2*n)!], {n, 0, Infinity}, GenerateConditions->None]</syntaxhighlight> || Failure || Aborted || Successful [Tested: 6] || Skipped - Because timed out
| [https://dlmf.nist.gov/24.19.E3 24.19.E3] || <math qid="Q7595">\frac{t^{2}}{\cosh@@{t}-1} = -2\sum_{n=0}^{\infty}(2n-1)\BernoullinumberB{2n}\frac{t^{2n}}{(2n)!}</math><br><syntaxhighlight lang="tex" style="font-size: 75%;" inline>\frac{t^{2}}{\cosh@@{t}-1} = -2\sum_{n=0}^{\infty}(2n-1)\BernoullinumberB{2n}\frac{t^{2n}}{(2n)!}</syntaxhighlight> || <math></math> || <syntaxhighlight lang=mathematica>((t)^(2))/(cosh(t)- 1) = - 2*sum((2*n - 1)*bernoulli(2*n)*((t)^(2*n))/(factorial(2*n)), n = 0..infinity)</syntaxhighlight> || <syntaxhighlight lang=mathematica>Divide[(t)^(2),Cosh[t]- 1] == - 2*Sum[(2*n - 1)*BernoulliB[2*n]*Divide[(t)^(2*n),(2*n)!], {n, 0, Infinity}, GenerateConditions->None]</syntaxhighlight> || Failure || Aborted || Successful [Tested: 6] || Skipped - Because timed out
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Latest revision as of 12:03, 28 June 2021


DLMF Formula Constraints Maple Mathematica Symbolic
Maple
Symbolic
Mathematica
Numeric
Maple
Numeric
Mathematica
24.19#Ex2 B 2 n = N 2 n D 2 n Bernoulli-number-B 2 𝑛 subscript 𝑁 2 𝑛 subscript 𝐷 2 𝑛 {\displaystyle{\displaystyle B_{2n}=\dfrac{N_{2n}}{D_{2n}}}}
\BernoullinumberB{2n} = \dfrac{N_{2n}}{D_{2n}}

bernoulli(2*n) = (N[2*n])/(D[2*n])
BernoulliB[2*n] == Divide[Subscript[N, 2*n],Subscript[D, 2*n]]
Failure Failure
Failed [300 / 300]
Result: -.8333333333
Test Values: {D[2*n] = 1/2*3^(1/2)+1/2*I, N[2*n] = 1/2*3^(1/2)+1/2*I, n = 1}

Result: -1.033333333
Test Values: {D[2*n] = 1/2*3^(1/2)+1/2*I, N[2*n] = 1/2*3^(1/2)+1/2*I, n = 2}

... skip entries to safe data
Failed [300 / 300]
Result: -0.8333333333333334
Test Values: {Rule[n, 1], Rule[Subscript[D, Times[2, n]], Power[E, Times[Complex[0, Rational[1, 6]], Pi]]], Rule[Subscript[N, Times[2, n]], Power[E, Times[Complex[0, Rational[1, 6]], Pi]]]}

Result: -1.0333333333333334
Test Values: {Rule[n, 2], Rule[Subscript[D, Times[2, n]], Power[E, Times[Complex[0, Rational[1, 6]], Pi]]], Rule[Subscript[N, Times[2, n]], Power[E, Times[Complex[0, Rational[1, 6]], Pi]]]}

... skip entries to safe data
24.19.E3 t 2 cosh t - 1 = - 2 n = 0 ( 2 n - 1 ) B 2 n t 2 n ( 2 n ) ! superscript 𝑡 2 𝑡 1 2 superscript subscript 𝑛 0 2 𝑛 1 Bernoulli-number-B 2 𝑛 superscript 𝑡 2 𝑛 2 𝑛 {\displaystyle{\displaystyle\frac{t^{2}}{\cosh t-1}=-2\sum_{n=0}^{\infty}(2n-1% )B_{2n}\frac{t^{2n}}{(2n)!}}}
\frac{t^{2}}{\cosh@@{t}-1} = -2\sum_{n=0}^{\infty}(2n-1)\BernoullinumberB{2n}\frac{t^{2n}}{(2n)!}

((t)^(2))/(cosh(t)- 1) = - 2*sum((2*n - 1)*bernoulli(2*n)*((t)^(2*n))/(factorial(2*n)), n = 0..infinity)
Divide[(t)^(2),Cosh[t]- 1] == - 2*Sum[(2*n - 1)*BernoulliB[2*n]*Divide[(t)^(2*n),(2*n)!], {n, 0, Infinity}, GenerateConditions->None]
Failure Aborted Successful [Tested: 6] Skipped - Because timed out