24.16: 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/24.16.E5 24.16.E5] | | | [https://dlmf.nist.gov/24.16.E5 24.16.E5] || <math qid="Q7575">\frac{t}{\ln@{1+t}} = \sum_{n=0}^{\infty}b_{n}t^{n}</math><br><syntaxhighlight lang="tex" style="font-size: 75%;" inline>\frac{t}{\ln@{1+t}} = \sum_{n=0}^{\infty}b_{n}t^{n}</syntaxhighlight> || <math>|t| < 1</math> || <syntaxhighlight lang=mathematica>(t)/(ln(1 + t)) = sum(b[n]*(t)^(n), n = 0..infinity)</syntaxhighlight> || <syntaxhighlight lang=mathematica>Divide[t,Log[1 + t]] == Sum[Subscript[b, n]*(t)^(n), {n, 0, Infinity}, GenerateConditions->None]</syntaxhighlight> || Failure || Failure || <div class="toccolours mw-collapsible mw-collapsed">Failed [20 / 20]<div class="mw-collapsible-content"><syntaxhighlight lang=mathematica>Result: .1439972511-.3333333333*I | ||
Test Values: {t = -1/2, b[n] = 1/2*3^(1/2)+1/2*I}</syntaxhighlight><br><syntaxhighlight lang=mathematica>Result: 1.054680854-.5773502693*I | Test Values: {t = -1/2, b[n] = 1/2*3^(1/2)+1/2*I}</syntaxhighlight><br><syntaxhighlight lang=mathematica>Result: 1.054680854-.5773502693*I | ||
Test Values: {t = -1/2, b[n] = -1/2+1/2*I*3^(1/2)}</syntaxhighlight><br>... skip entries to safe data</div></div> || <div class="toccolours mw-collapsible mw-collapsed">Failed [20 / 20]<div class="mw-collapsible-content"><syntaxhighlight lang=mathematica>Result: Complex[0.14399725125485596, -0.33333333333333326] | Test Values: {t = -1/2, b[n] = -1/2+1/2*I*3^(1/2)}</syntaxhighlight><br>... skip entries to safe data</div></div> || <div class="toccolours mw-collapsible mw-collapsed">Failed [20 / 20]<div class="mw-collapsible-content"><syntaxhighlight lang=mathematica>Result: Complex[0.14399725125485596, -0.33333333333333326] | ||
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Test Values: {Rule[t, -0.5], Rule[Subscript[b, n], Power[E, Times[Complex[0, Rational[2, 3]], Pi]]]}</syntaxhighlight><br>... skip entries to safe data</div></div> | Test Values: {Rule[t, -0.5], Rule[Subscript[b, n], Power[E, Times[Complex[0, Rational[2, 3]], Pi]]]}</syntaxhighlight><br>... skip entries to safe data</div></div> | ||
|- style="background: #dfe6e9;" | |- style="background: #dfe6e9;" | ||
| [https://dlmf.nist.gov/24.16.E7 24.16.E7] | | | [https://dlmf.nist.gov/24.16.E7 24.16.E7] || <math qid="Q7577">\frac{t}{(1+\lambda t)^{\ifrac{1}{\lambda}}-1} = \sum_{n=0}^{\infty}\beta_{n}(\lambda)\frac{t^{n}}{n!}</math><br><syntaxhighlight lang="tex" style="font-size: 75%; background: inherit;" inline>\frac{t}{(1+\lambda t)^{\ifrac{1}{\lambda}}-1} = \sum_{n=0}^{\infty}\beta_{n}(\lambda)\frac{t^{n}}{n!}</syntaxhighlight> || <math></math> || <div class="mw-highlight mw-highlight-lang-mathematica mw-content-ltr" dir="ltr"><pre style="background: inherit;">(t)/((1 + lambda*t)^((1)/(lambda))- 1) = sum(beta[n](lambda)*((t)^(n))/(factorial(n)), n = 0..infinity)</pre></div> || <div class="mw-highlight mw-highlight-lang-mathematica mw-content-ltr" dir="ltr"><pre style="background: inherit;">Divide[t,(1 + \[Lambda]*t)^(Divide[1,\[Lambda]])- 1] == Sum[Subscript[\[Beta], n][\[Lambda]]*Divide[(t)^(n),(n)!], {n, 0, Infinity}, GenerateConditions->None]</pre></div> || Skipped - no semantic math || Skipped - no semantic math || - || - | ||
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| [https://dlmf.nist.gov/24.16.E8 24.16.E8] | | | [https://dlmf.nist.gov/24.16.E8 24.16.E8] || <math qid="Q7578">\beta_{n}(\lambda) = n!b_{n}\lambda^{n}+\sum_{k=1}^{\floor{\ifrac{n}{2}}}\frac{n}{2k}\BernoullinumberB{2k}\Stirlingnumbers@{n-1}{2k-1}\lambda^{n-2k}</math><br><syntaxhighlight lang="tex" style="font-size: 75%;" inline>\beta_{n}(\lambda) = n!b_{n}\lambda^{n}+\sum_{k=1}^{\floor{\ifrac{n}{2}}}\frac{n}{2k}\BernoullinumberB{2k}\Stirlingnumbers@{n-1}{2k-1}\lambda^{n-2k}</syntaxhighlight> || <math></math> || <syntaxhighlight lang=mathematica>beta[n](lambda) = factorial(n)*b[n]*(lambda)^(n)+ sum((n)/(2*k)*bernoulli(2*k)*Stirling1(n - 1, 2*k - 1)*(lambda)^(n - 2*k), k = 1..floor((n)/(2)))</syntaxhighlight> || <syntaxhighlight lang=mathematica>Subscript[\[Beta], n][\[Lambda]] == (n)!*Subscript[b, n]*\[Lambda]^(n)+ Sum[Divide[n,2*k]*BernoulliB[2*k]*StirlingS1[n - 1, 2*k - 1]*\[Lambda]^(n - 2*k), {k, 1, Floor[Divide[n,2]]}, GenerateConditions->None]</syntaxhighlight> || Failure || Failure || <div class="toccolours mw-collapsible mw-collapsed">Failed [300 / 300]<div class="mw-collapsible-content"><syntaxhighlight lang=mathematica>Result: .3333333331-1.133974598*I | ||
Test Values: {beta = 3/2, lambda = 1/2*3^(1/2)+1/2*I, b[n] = 1/2*3^(1/2)+1/2*I, beta[n] = 1/2*3^(1/2)+1/2*I, n = 2}</syntaxhighlight><br><syntaxhighlight lang=mathematica>Result: -1.032692071-1.500000002*I | Test Values: {beta = 3/2, lambda = 1/2*3^(1/2)+1/2*I, b[n] = 1/2*3^(1/2)+1/2*I, beta[n] = 1/2*3^(1/2)+1/2*I, n = 2}</syntaxhighlight><br><syntaxhighlight lang=mathematica>Result: -1.032692071-1.500000002*I | ||
Test Values: {beta = 3/2, lambda = 1/2*3^(1/2)+1/2*I, b[n] = 1/2*3^(1/2)+1/2*I, beta[n] = -1/2+1/2*I*3^(1/2), n = 2}</syntaxhighlight><br>... skip entries to safe data</div></div> || Skipped - Because timed out | Test Values: {beta = 3/2, lambda = 1/2*3^(1/2)+1/2*I, b[n] = 1/2*3^(1/2)+1/2*I, beta[n] = -1/2+1/2*I*3^(1/2), n = 2}</syntaxhighlight><br>... skip entries to safe data</div></div> || Skipped - Because timed out | ||
|- style="background: #dfe6e9;" | |- style="background: #dfe6e9;" | ||
| [https://dlmf.nist.gov/24.16.E11 24.16.E11] | | | [https://dlmf.nist.gov/24.16.E11 24.16.E11] || <math qid="Q7581">B_{n,\chi}(x) = \sum_{k=0}^{n}{n\choose k}B_{k,\chi}x^{n-k}</math><br><syntaxhighlight lang="tex" style="font-size: 75%; background: inherit;" inline>B_{n,\chi}(x) = \sum_{k=0}^{n}{n\choose k}B_{k,\chi}x^{n-k}</syntaxhighlight> || <math></math> || <div class="mw-highlight mw-highlight-lang-mathematica mw-content-ltr" dir="ltr"><pre style="background: inherit;">B[n , chi](x) = sum(binomial(n,k)*(B[k , chi](x))^(n - k), k = 0..n)</pre></div> || <div class="mw-highlight mw-highlight-lang-mathematica mw-content-ltr" dir="ltr"><pre style="background: inherit;">Subscript[B, n , \[Chi]][x] == Sum[Binomial[n,k]*(Subscript[B, k , \[Chi]][x])^(n - k), {k, 0, n}, GenerateConditions->None]</pre></div> || Skipped - no semantic math || Skipped - no semantic math || - || - | ||
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| [https://dlmf.nist.gov/24.16.E12 24.16.E12] | | | [https://dlmf.nist.gov/24.16.E12 24.16.E12] || <math qid="Q7582">\BernoullipolyB{n}@{x} = B_{n,\chi_{0}}(x-1)</math><br><syntaxhighlight lang="tex" style="font-size: 75%;" inline>\BernoullipolyB{n}@{x} = B_{n,\chi_{0}}(x-1)</syntaxhighlight> || <math></math> || <syntaxhighlight lang=mathematica>bernoulli(n, x) = B[n , chi[0]]*(x - 1)</syntaxhighlight> || <syntaxhighlight lang=mathematica>BernoulliB[n, x] == Subscript[B, n , Subscript[\[Chi], 0]]*(x - 1)</syntaxhighlight> || Failure || Failure || <div class="toccolours mw-collapsible mw-collapsed">Failed [280 / 300]<div class="mw-collapsible-content"><syntaxhighlight lang=mathematica>Result: .5669872980-.2500000000*I | ||
Test Values: {chi = 1/2*3^(1/2)+1/2*I, x = 3/2, B[n,chi[0]] = 1/2*3^(1/2)+1/2*I, chi[0] = 1/2*3^(1/2)+1/2*I, n = 1}</syntaxhighlight><br><syntaxhighlight lang=mathematica>Result: .4836539647-.2500000000*I | Test Values: {chi = 1/2*3^(1/2)+1/2*I, x = 3/2, B[n,chi[0]] = 1/2*3^(1/2)+1/2*I, chi[0] = 1/2*3^(1/2)+1/2*I, n = 1}</syntaxhighlight><br><syntaxhighlight lang=mathematica>Result: .4836539647-.2500000000*I | ||
Test Values: {chi = 1/2*3^(1/2)+1/2*I, x = 3/2, B[n,chi[0]] = 1/2*3^(1/2)+1/2*I, chi[0] = 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 [280 / 300]<div class="mw-collapsible-content"><syntaxhighlight lang=mathematica>Result: Complex[0.5669872981077806, -0.24999999999999997] | Test Values: {chi = 1/2*3^(1/2)+1/2*I, x = 3/2, B[n,chi[0]] = 1/2*3^(1/2)+1/2*I, chi[0] = 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 [280 / 300]<div class="mw-collapsible-content"><syntaxhighlight lang=mathematica>Result: Complex[0.5669872981077806, -0.24999999999999997] | ||
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Test Values: {Rule[n, 2], Rule[x, 1.5], Rule[χ, Power[E, Times[Complex[0, Rational[1, 6]], Pi]]], Rule[Subscript[χ, 0], Power[E, Times[Complex[0, Rational[1, 6]], Pi]]], Rule[Subscript[B, n, Subscript[χ, 0]], Power[E, Times[Complex[0, Rational[1, 6]], Pi]]]}</syntaxhighlight><br>... skip entries to safe data</div></div> | Test Values: {Rule[n, 2], Rule[x, 1.5], Rule[χ, Power[E, Times[Complex[0, Rational[1, 6]], Pi]]], Rule[Subscript[χ, 0], Power[E, Times[Complex[0, Rational[1, 6]], Pi]]], Rule[Subscript[B, n, Subscript[χ, 0]], 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/24.16.E13 24.16.E13] | | | [https://dlmf.nist.gov/24.16.E13 24.16.E13] || <math qid="Q7583">\EulerpolyE{n}@{x} = -\frac{2^{1-n}}{n+1}B_{n+1,\chi_{4}}(2x-1)</math><br><syntaxhighlight lang="tex" style="font-size: 75%;" inline>\EulerpolyE{n}@{x} = -\frac{2^{1-n}}{n+1}B_{n+1,\chi_{4}}(2x-1)</syntaxhighlight> || <math></math> || <syntaxhighlight lang=mathematica>euler(n, x) = -((2)^(1 - n))/(n + 1)*B[n + 1 , chi[4]]*(2*x - 1)</syntaxhighlight> || <syntaxhighlight lang=mathematica>EulerE[n, x] == -Divide[(2)^(1 - n),n + 1]*Subscript[B, n + 1 , Subscript[\[Chi], 4]]*(2*x - 1)</syntaxhighlight> || Failure || Failure || <div class="toccolours mw-collapsible mw-collapsed">Failed [290 / 300]<div class="mw-collapsible-content"><syntaxhighlight lang=mathematica>Result: 1.866025404+.5000000000*I | ||
Test Values: {chi = 1/2*3^(1/2)+1/2*I, x = 3/2, B[n+1,chi[4]] = 1/2*3^(1/2)+1/2*I, chi[4] = 1/2*3^(1/2)+1/2*I, n = 1}</syntaxhighlight><br><syntaxhighlight lang=mathematica>Result: 1.038675135+.1666666667*I | Test Values: {chi = 1/2*3^(1/2)+1/2*I, x = 3/2, B[n+1,chi[4]] = 1/2*3^(1/2)+1/2*I, chi[4] = 1/2*3^(1/2)+1/2*I, n = 1}</syntaxhighlight><br><syntaxhighlight lang=mathematica>Result: 1.038675135+.1666666667*I | ||
Test Values: {chi = 1/2*3^(1/2)+1/2*I, x = 3/2, B[n+1,chi[4]] = 1/2*3^(1/2)+1/2*I, chi[4] = 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 [290 / 300]<div class="mw-collapsible-content"><syntaxhighlight lang=mathematica>Result: Complex[1.8660254037844388, 0.49999999999999994] | Test Values: {chi = 1/2*3^(1/2)+1/2*I, x = 3/2, B[n+1,chi[4]] = 1/2*3^(1/2)+1/2*I, chi[4] = 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 [290 / 300]<div class="mw-collapsible-content"><syntaxhighlight lang=mathematica>Result: Complex[1.8660254037844388, 0.49999999999999994] |
Latest revision as of 12:03, 28 June 2021
DLMF | Formula | Constraints | Maple | Mathematica | Symbolic Maple |
Symbolic Mathematica |
Numeric Maple |
Numeric Mathematica |
---|---|---|---|---|---|---|---|---|
24.16.E5 | \frac{t}{\ln@{1+t}} = \sum_{n=0}^{\infty}b_{n}t^{n} |
(t)/(ln(1 + t)) = sum(b[n]*(t)^(n), n = 0..infinity)
|
Divide[t,Log[1 + t]] == Sum[Subscript[b, n]*(t)^(n), {n, 0, Infinity}, GenerateConditions->None]
|
Failure | Failure | Failed [20 / 20] Result: .1439972511-.3333333333*I
Test Values: {t = -1/2, b[n] = 1/2*3^(1/2)+1/2*I}
Result: 1.054680854-.5773502693*I
Test Values: {t = -1/2, b[n] = -1/2+1/2*I*3^(1/2)}
... skip entries to safe data |
Failed [20 / 20]
Result: Complex[0.14399725125485596, -0.33333333333333326]
Test Values: {Rule[t, -0.5], Rule[Subscript[b, n], Power[E, Times[Complex[0, Rational[1, 6]], Pi]]]}
Result: Complex[1.054680853777815, -0.5773502691896257]
Test Values: {Rule[t, -0.5], Rule[Subscript[b, n], Power[E, Times[Complex[0, Rational[2, 3]], Pi]]]}
... skip entries to safe data | |
24.16.E7 | \frac{t}{(1+\lambda t)^{\ifrac{1}{\lambda}}-1} = \sum_{n=0}^{\infty}\beta_{n}(\lambda)\frac{t^{n}}{n!} |
|
(t)/((1 + lambda*t)^((1)/(lambda))- 1) = sum(beta[n](lambda)*((t)^(n))/(factorial(n)), n = 0..infinity) |
Divide[t,(1 + \[Lambda]*t)^(Divide[1,\[Lambda]])- 1] == Sum[Subscript[\[Beta], n][\[Lambda]]*Divide[(t)^(n),(n)!], {n, 0, Infinity}, GenerateConditions->None] |
Skipped - no semantic math | Skipped - no semantic math | - | - |
24.16.E8 | \beta_{n}(\lambda) = n!b_{n}\lambda^{n}+\sum_{k=1}^{\floor{\ifrac{n}{2}}}\frac{n}{2k}\BernoullinumberB{2k}\Stirlingnumbers@{n-1}{2k-1}\lambda^{n-2k} |
|
beta[n](lambda) = factorial(n)*b[n]*(lambda)^(n)+ sum((n)/(2*k)*bernoulli(2*k)*Stirling1(n - 1, 2*k - 1)*(lambda)^(n - 2*k), k = 1..floor((n)/(2)))
|
Subscript[\[Beta], n][\[Lambda]] == (n)!*Subscript[b, n]*\[Lambda]^(n)+ Sum[Divide[n,2*k]*BernoulliB[2*k]*StirlingS1[n - 1, 2*k - 1]*\[Lambda]^(n - 2*k), {k, 1, Floor[Divide[n,2]]}, GenerateConditions->None]
|
Failure | Failure | Failed [300 / 300] Result: .3333333331-1.133974598*I
Test Values: {beta = 3/2, lambda = 1/2*3^(1/2)+1/2*I, b[n] = 1/2*3^(1/2)+1/2*I, beta[n] = 1/2*3^(1/2)+1/2*I, n = 2}
Result: -1.032692071-1.500000002*I
Test Values: {beta = 3/2, lambda = 1/2*3^(1/2)+1/2*I, b[n] = 1/2*3^(1/2)+1/2*I, beta[n] = -1/2+1/2*I*3^(1/2), n = 2}
... skip entries to safe data |
Skipped - Because timed out |
24.16.E11 | B_{n,\chi}(x) = \sum_{k=0}^{n}{n\choose k}B_{k,\chi}x^{n-k} |
|
B[n , chi](x) = sum(binomial(n,k)*(B[k , chi](x))^(n - k), k = 0..n) |
Subscript[B, n , \[Chi]][x] == Sum[Binomial[n,k]*(Subscript[B, k , \[Chi]][x])^(n - k), {k, 0, n}, GenerateConditions->None] |
Skipped - no semantic math | Skipped - no semantic math | - | - |
24.16.E12 | \BernoullipolyB{n}@{x} = B_{n,\chi_{0}}(x-1) |
|
bernoulli(n, x) = B[n , chi[0]]*(x - 1)
|
BernoulliB[n, x] == Subscript[B, n , Subscript[\[Chi], 0]]*(x - 1)
|
Failure | Failure | Failed [280 / 300] Result: .5669872980-.2500000000*I
Test Values: {chi = 1/2*3^(1/2)+1/2*I, x = 3/2, B[n,chi[0]] = 1/2*3^(1/2)+1/2*I, chi[0] = 1/2*3^(1/2)+1/2*I, n = 1}
Result: .4836539647-.2500000000*I
Test Values: {chi = 1/2*3^(1/2)+1/2*I, x = 3/2, B[n,chi[0]] = 1/2*3^(1/2)+1/2*I, chi[0] = 1/2*3^(1/2)+1/2*I, n = 2}
... skip entries to safe data |
Failed [280 / 300]
Result: Complex[0.5669872981077806, -0.24999999999999997]
Test Values: {Rule[n, 1], Rule[x, 1.5], Rule[χ, Power[E, Times[Complex[0, Rational[1, 6]], Pi]]], Rule[Subscript[χ, 0], Power[E, Times[Complex[0, Rational[1, 6]], Pi]]], Rule[Subscript[B, n, Subscript[χ, 0]], Power[E, Times[Complex[0, Rational[1, 6]], Pi]]]}
Result: Complex[0.4836539647744474, -0.24999999999999997]
Test Values: {Rule[n, 2], Rule[x, 1.5], Rule[χ, Power[E, Times[Complex[0, Rational[1, 6]], Pi]]], Rule[Subscript[χ, 0], Power[E, Times[Complex[0, Rational[1, 6]], Pi]]], Rule[Subscript[B, n, Subscript[χ, 0]], Power[E, Times[Complex[0, Rational[1, 6]], Pi]]]}
... skip entries to safe data |
24.16.E13 | \EulerpolyE{n}@{x} = -\frac{2^{1-n}}{n+1}B_{n+1,\chi_{4}}(2x-1) |
|
euler(n, x) = -((2)^(1 - n))/(n + 1)*B[n + 1 , chi[4]]*(2*x - 1)
|
EulerE[n, x] == -Divide[(2)^(1 - n),n + 1]*Subscript[B, n + 1 , Subscript[\[Chi], 4]]*(2*x - 1)
|
Failure | Failure | Failed [290 / 300] Result: 1.866025404+.5000000000*I
Test Values: {chi = 1/2*3^(1/2)+1/2*I, x = 3/2, B[n+1,chi[4]] = 1/2*3^(1/2)+1/2*I, chi[4] = 1/2*3^(1/2)+1/2*I, n = 1}
Result: 1.038675135+.1666666667*I
Test Values: {chi = 1/2*3^(1/2)+1/2*I, x = 3/2, B[n+1,chi[4]] = 1/2*3^(1/2)+1/2*I, chi[4] = 1/2*3^(1/2)+1/2*I, n = 2}
... skip entries to safe data |
Failed [290 / 300]
Result: Complex[1.8660254037844388, 0.49999999999999994]
Test Values: {Rule[n, 1], Rule[x, 1.5], Rule[χ, Power[E, Times[Complex[0, Rational[1, 6]], Pi]]], Rule[Subscript[χ, 4], Power[E, Times[Complex[0, Rational[1, 6]], Pi]]], Rule[Subscript[B, Plus[1, n], Subscript[χ, 4]], Power[E, Times[Complex[0, Rational[1, 6]], Pi]]]}
Result: Complex[1.0386751345948129, 0.16666666666666663]
Test Values: {Rule[n, 2], Rule[x, 1.5], Rule[χ, Power[E, Times[Complex[0, Rational[1, 6]], Pi]]], Rule[Subscript[χ, 4], Power[E, Times[Complex[0, Rational[1, 6]], Pi]]], Rule[Subscript[B, Plus[1, n], Subscript[χ, 4]], Power[E, Times[Complex[0, Rational[1, 6]], Pi]]]}
... skip entries to safe data |