Bernoulli and Euler Polynomials - 24.16 Generalizations

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DLMF Formula Constraints Maple Mathematica Symbolic
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24.16.E5 t ln ( 1 + t ) = n = 0 b n t n 𝑡 1 𝑡 superscript subscript 𝑛 0 subscript 𝑏 𝑛 superscript 𝑡 𝑛 {\displaystyle{\displaystyle\frac{t}{\ln\left(1+t\right)}=\sum_{n=0}^{\infty}b% _{n}t^{n}}}
\frac{t}{\ln@{1+t}} = \sum_{n=0}^{\infty}b_{n}t^{n}
| t | < 1 𝑡 1 {\displaystyle{\displaystyle|t|<1}}
(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 t ( 1 + λ t ) 1 / λ - 1 = n = 0 β n ( λ ) t n n ! 𝑡 superscript 1 𝜆 𝑡 1 𝜆 1 superscript subscript 𝑛 0 subscript 𝛽 𝑛 𝜆 superscript 𝑡 𝑛 𝑛 {\displaystyle{\displaystyle\frac{t}{(1+\lambda t)^{\ifrac{1}{\lambda}}-1}=% \sum_{n=0}^{\infty}\beta_{n}(\lambda)\frac{t^{n}}{n!}}}
\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 β n ( λ ) = n ! b n λ n + k = 1 n / 2 n 2 k B 2 k s ( n - 1 , 2 k - 1 ) λ n - 2 k subscript 𝛽 𝑛 𝜆 𝑛 subscript 𝑏 𝑛 superscript 𝜆 𝑛 superscript subscript 𝑘 1 𝑛 2 𝑛 2 𝑘 Bernoulli-number-B 2 𝑘 Stirling-number-first-kind-S 𝑛 1 2 𝑘 1 superscript 𝜆 𝑛 2 𝑘 {\displaystyle{\displaystyle\beta_{n}(\lambda)=n!b_{n}\lambda^{n}+\sum_{k=1}^{% \left\lfloor\ifrac{n}{2}\right\rfloor}\frac{n}{2k}B_{2k}s\left(n-1,2k-1\right)% \lambda^{n-2k}}}
\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 , χ ( x ) = k = 0 n ( n k ) B k , χ x n - k subscript 𝐵 𝑛 𝜒 𝑥 superscript subscript 𝑘 0 𝑛 binomial 𝑛 𝑘 subscript 𝐵 𝑘 𝜒 superscript 𝑥 𝑛 𝑘 {\displaystyle{\displaystyle B_{n,\chi}(x)=\sum_{k=0}^{n}{n\choose k}B_{k,\chi% }x^{n-k}}}
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 B n ( x ) = B n , χ 0 ( x - 1 ) Bernoulli-polynomial-B 𝑛 𝑥 subscript 𝐵 𝑛 subscript 𝜒 0 𝑥 1 {\displaystyle{\displaystyle B_{n}\left(x\right)=B_{n,\chi_{0}}(x-1)}}
\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 E n ( x ) = - 2 1 - n n + 1 B n + 1 , χ 4 ( 2 x - 1 ) Euler-polynomial-E 𝑛 𝑥 superscript 2 1 𝑛 𝑛 1 subscript 𝐵 𝑛 1 subscript 𝜒 4 2 𝑥 1 {\displaystyle{\displaystyle E_{n}\left(x\right)=-\frac{2^{1-n}}{n+1}B_{n+1,% \chi_{4}}(2x-1)}}
\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