Bernoulli and Euler Polynomials - 24.8 Series Expansions

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DLMF Formula Constraints Maple Mathematica Symbolic
Maple 		Symbolic
Mathematica 		Numeric
Maple 		Numeric
Mathematica
24.8.E1 B 2 ⁢ n ⁡ ( x ) = ( - 1 ) n + 1 ⁢ 2 ⁢ ( 2 ⁢ n ) ! ( 2 ⁢ π ) 2 ⁢ n ⁢ ∑ k = 1 ∞ cos ⁡ ( 2 ⁢ π ⁢ k ⁢ x ) k 2 ⁢ n Bernoulli-polynomial-B 2 𝑛 𝑥 superscript 1 𝑛 1 2 2 𝑛 superscript 2 𝜋 2 𝑛 superscript subscript 𝑘 1 2 𝜋 𝑘 𝑥 superscript 𝑘 2 𝑛 {\displaystyle{\displaystyle B_{2n}\left(x\right)=(-1)^{n+1}\frac{2(2n)!}{(2% \pi)^{2n}}\sum_{k=1}^{\infty}\frac{\cos\left(2\pi kx\right)}{k^{2n}}}}
\BernoullipolyB{2n}@{x} = (-1)^{n+1}\frac{2(2n)!}{(2\pi)^{2n}}\sum_{k=1}^{\infty}\frac{\cos@{2\pi kx}}{k^{2n}}		 

bernoulli(2*n, x) = (- 1)^(n + 1)*(2*factorial(2*n))/((2*Pi)^(2*n))*sum((cos(2*Pi*k*x))/((k)^(2*n)), k = 1..infinity)

BernoulliB[2*n, x] == (- 1)^(n + 1)*Divide[2*(2*n)!,(2*Pi)^(2*n)]*Sum[Divide[Cos[2*Pi*k*x],(k)^(2*n)], {k, 1, Infinity}, GenerateConditions->None]
Aborted Failure
Failed [6 / 9]
Result: 1.000000000
Test Values: {x = 3/2, n = 1}


Result: .5000000000
Test Values: {x = 3/2, n = 2}

... skip entries to safe data
Failed [6 / 9]
Result: Complex[1.0, 0.0]
Test Values: {Rule[n, 1], Rule[x, 1.5]}


Result: Complex[0.5000000000000001, 0.0]
Test Values: {Rule[n, 2], Rule[x, 1.5]}

... skip entries to safe data
24.8.E2 B 2 ⁢ n + 1 ⁡ ( x ) = ( - 1 ) n + 1 ⁢ 2 ⁢ ( 2 ⁢ n + 1 ) ! ( 2 ⁢ π ) 2 ⁢ n + 1 ⁢ ∑ k = 1 ∞ sin ⁡ ( 2 ⁢ π ⁢ k ⁢ x ) k 2 ⁢ n + 1 Bernoulli-polynomial-B 2 𝑛 1 𝑥 superscript 1 𝑛 1 2 2 𝑛 1 superscript 2 𝜋 2 𝑛 1 superscript subscript 𝑘 1 2 𝜋 𝑘 𝑥 superscript 𝑘 2 𝑛 1 {\displaystyle{\displaystyle B_{2n+1}\left(x\right)=(-1)^{n+1}\frac{2(2n+1)!}{% (2\pi)^{2n+1}}\sum_{k=1}^{\infty}\frac{\sin\left(2\pi kx\right)}{k^{2n+1}}}}
\BernoullipolyB{2n+1}@{x} = (-1)^{n+1}\frac{2(2n+1)!}{(2\pi)^{2n+1}}\sum_{k=1}^{\infty}\frac{\sin@{2\pi kx}}{k^{2n+1}}		 

bernoulli(2*n + 1, x) = (- 1)^(n + 1)*(2*factorial(2*n + 1))/((2*Pi)^(2*n + 1))*sum((sin(2*Pi*k*x))/((k)^(2*n + 1)), k = 1..infinity)

BernoulliB[2*n + 1, x] == (- 1)^(n + 1)*Divide[2*(2*n + 1)!,(2*Pi)^(2*n + 1)]*Sum[Divide[Sin[2*Pi*k*x],(k)^(2*n + 1)], {k, 1, Infinity}, GenerateConditions->None]
Aborted Failure -
Failed [6 / 9]
Result: Complex[0.75, 0.0]
Test Values: {Rule[n, 1], Rule[x, 1.5]}


Result: Complex[0.3125, 0.0]
Test Values: {Rule[n, 2], Rule[x, 1.5]}

... skip entries to safe data
24.8.E4 E 2 ⁢ n ⁡ ( x ) = ( - 1 ) n ⁢ 4 ⁢ ( 2 ⁢ n ) ! π 2 ⁢ n + 1 ⁢ ∑ k = 0 ∞ sin ⁡ ( ( 2 ⁢ k + 1 ) ⁢ π ⁢ x ) ( 2 ⁢ k + 1 ) 2 ⁢ n + 1 Euler-polynomial-E 2 𝑛 𝑥 superscript 1 𝑛 4 2 𝑛 superscript 𝜋 2 𝑛 1 superscript subscript 𝑘 0 2 𝑘 1 𝜋 𝑥 superscript 2 𝑘 1 2 𝑛 1 {\displaystyle{\displaystyle E_{2n}\left(x\right)=(-1)^{n}\frac{4(2n)!}{\pi^{2% n+1}}\sum_{k=0}^{\infty}\frac{\sin\left((2k+1)\pi x\right)}{(2k+1)^{2n+1}}}}
\EulerpolyE{2n}@{x} = (-1)^{n}\frac{4(2n)!}{\pi^{2n+1}}\sum_{k=0}^{\infty}\frac{\sin@{(2k+1)\pi x}}{(2k+1)^{2n+1}}		 

euler(2*n, x) = (- 1)^(n)*(4*factorial(2*n))/((Pi)^(2*n + 1))*sum((sin((2*k + 1)*Pi*x))/((2*k + 1)^(2*n + 1)), k = 0..infinity)

EulerE[2*n, x] == (- 1)^(n)*Divide[4*(2*n)!,(Pi)^(2*n + 1)]*Sum[Divide[Sin[(2*k + 1)*Pi*x],(2*k + 1)^(2*n + 1)], {k, 0, Infinity}, GenerateConditions->None]
Aborted Failure
Failed [6 / 9]
Result: .5000000000+0.*I
Test Values: {x = 3/2, n = 1}


Result: .1249999998+0.*I
Test Values: {x = 3/2, n = 2}

... skip entries to safe data
Failed [6 / 9]
Result: Complex[0.4999999999999999, -6.717074394942855*^-17]
Test Values: {Rule[n, 1], Rule[x, 1.5]}


Result: Complex[0.1250000000000001, -1.3482715791848248*^-17]
Test Values: {Rule[n, 2], Rule[x, 1.5]}

... skip entries to safe data
24.8.E5 E 2 ⁢ n - 1 ⁡ ( x ) = ( - 1 ) n ⁢ 4 ⁢ ( 2 ⁢ n - 1 ) ! π 2 ⁢ n ⁢ ∑ k = 0 ∞ cos ⁡ ( ( 2 ⁢ k + 1 ) ⁢ π ⁢ x ) ( 2 ⁢ k + 1 ) 2 ⁢ n Euler-polynomial-E 2 𝑛 1 𝑥 superscript 1 𝑛 4 2 𝑛 1 superscript 𝜋 2 𝑛 superscript subscript 𝑘 0 2 𝑘 1 𝜋 𝑥 superscript 2 𝑘 1 2 𝑛 {\displaystyle{\displaystyle E_{2n-1}\left(x\right)=(-1)^{n}\frac{4(2n-1)!}{% \pi^{2n}}\sum_{k=0}^{\infty}\frac{\cos\left((2k+1)\pi x\right)}{(2k+1)^{2n}}}}
\EulerpolyE{2n-1}@{x} = (-1)^{n}\frac{4(2n-1)!}{\pi^{2n}}\sum_{k=0}^{\infty}\frac{\cos@{(2k+1)\pi x}}{(2k+1)^{2n}}		 

euler(2*n - 1, x) = (- 1)^(n)*(4*factorial(2*n - 1))/((Pi)^(2*n))*sum((cos((2*k + 1)*Pi*x))/((2*k + 1)^(2*n)), k = 0..infinity)

EulerE[2*n - 1, x] == (- 1)^(n)*Divide[4*(2*n - 1)!,(Pi)^(2*n)]*Sum[Divide[Cos[(2*k + 1)*Pi*x],(2*k + 1)^(2*n)], {k, 0, Infinity}, GenerateConditions->None]
Aborted Failure
Failed [6 / 9]
Result: 1.
Test Values: {x = 3/2, n = 1}


Result: .2500000000
Test Values: {x = 3/2, n = 2}

... skip entries to safe data
Failed [6 / 9]
Result: Complex[1.0, -2.3810929344395102*^-33]
Test Values: {Rule[n, 1], Rule[x, 1.5]}


Result: Complex[0.25000000000000006, 5.146963577016199*^-33]
Test Values: {Rule[n, 2], Rule[x, 1.5]}

... skip entries to safe data
24.8.E6 B 4 ⁢ n + 2 = ( 8 ⁢ n + 4 ) ⁢ ∑ k = 1 ∞ k 4 ⁢ n + 1 e 2 ⁢ π ⁢ k - 1 Bernoulli-number-B 4 𝑛 2 8 𝑛 4 superscript subscript 𝑘 1 superscript 𝑘 4 𝑛 1 superscript 𝑒 2 𝜋 𝑘 1 {\displaystyle{\displaystyle B_{4n+2}=(8n+4)\sum_{k=1}^{\infty}\frac{k^{4n+1}}% {e^{2\pi k}-1}}}
\BernoullinumberB{4n+2} = (8n+4)\sum_{k=1}^{\infty}\frac{k^{4n+1}}{e^{2\pi k}-1}		 

bernoulli(4*n + 2) = (8*n + 4)*sum(((k)^(4*n + 1))/(exp(2*Pi*k)- 1), k = 1..infinity)

BernoulliB[4*n + 2] == (8*n + 4)*Sum[Divide[(k)^(4*n + 1),Exp[2*Pi*k]- 1], {k, 1, Infinity}, GenerateConditions->None]
Failure Failure Successful [Tested: 1]
Failed [3 / 3]
Result: Plus[0.023809523809523808, Times[-12.0, NSum[Times[Power[Plus[-1, Power[E, Times[2, k, Pi]]], -1], Power[k, 5]]
Test Values: {k, 1, DirectedInfinity[1]}, Rule[GenerateConditions, None]]]], {Rule[n, 1]}


Result: Plus[0.07575757575757576, Times[-20.0, NSum[Times[Power[Plus[-1, Power[E, Times[2, k, Pi]]], -1], Power[k, 9]]
Test Values: {k, 1, DirectedInfinity[1]}, Rule[GenerateConditions, None]]]], {Rule[n, 2]}

... skip entries to safe data
24.8.E7 B 2 ⁢ n = ( - 1 ) n + 1 ⁢ 4 ⁢ n 2 2 ⁢ n - 1 ⁢ ∑ k = 1 ∞ k 2 ⁢ n - 1 e π ⁢ k + ( - 1 ) k + n Bernoulli-number-B 2 𝑛 superscript 1 𝑛 1 4 𝑛 superscript 2 2 𝑛 1 superscript subscript 𝑘 1 superscript 𝑘 2 𝑛 1 superscript 𝑒 𝜋 𝑘 superscript 1 𝑘 𝑛 {\displaystyle{\displaystyle B_{2n}=\frac{(-1)^{n+1}4n}{2^{2n}-1}\sum_{k=1}^{% \infty}\frac{k^{2n-1}}{e^{\pi k}+(-1)^{k+n}}}}
\BernoullinumberB{2n} = \frac{(-1)^{n+1}4n}{2^{2n}-1}\sum_{k=1}^{\infty}\frac{k^{2n-1}}{e^{\pi k}+(-1)^{k+n}}		 

bernoulli(2*n) = ((- 1)^(n + 1)* 4*n)/((2)^(2*n)- 1)*sum(((k)^(2*n - 1))/(exp(Pi*k)+(- 1)^(k + n)), k = 1..infinity)

BernoulliB[2*n] == Divide[(- 1)^(n + 1)* 4*n,(2)^(2*n)- 1]*Sum[Divide[(k)^(2*n - 1),Exp[Pi*k]+(- 1)^(k + n)], {k, 1, Infinity}, GenerateConditions->None]
Failure Failure Successful [Tested: 1]
Failed [3 / 3]
Result: Plus[0.16666666666666666, Times[-1.3333333333333333, NSum[Times[Power[Plus[Power[-1, Plus[1, k]], Power[E, Times[k, Pi]]], -1], k]
Test Values: {k, 1, DirectedInfinity[1]}, Rule[GenerateConditions, None]]]], {Rule[n, 1]}


Result: Plus[-0.03333333333333333, Times[0.5333333333333333, NSum[Times[Power[Plus[Power[-1, Plus[2, k]], Power[E, Times[k, Pi]]], -1], Power[k, 3]]
Test Values: {k, 1, DirectedInfinity[1]}, Rule[GenerateConditions, None]]]], {Rule[n, 2]}

... skip entries to safe data
24.8.E8 B 2 ⁢ n 4 ⁢ n ⁢ ( α n - ( - β ) n ) = α n ⁢ ∑ k = 1 ∞ k 2 ⁢ n - 1 e 2 ⁢ α ⁢ k - 1 - ( - β ) n ⁢ ∑ k = 1 ∞ k 2 ⁢ n - 1 e 2 ⁢ β ⁢ k - 1 Bernoulli-number-B 2 𝑛 4 𝑛 superscript 𝛼 𝑛 superscript 𝛽 𝑛 superscript 𝛼 𝑛 superscript subscript 𝑘 1 superscript 𝑘 2 𝑛 1 superscript 𝑒 2 𝛼 𝑘 1 superscript 𝛽 𝑛 superscript subscript 𝑘 1 superscript 𝑘 2 𝑛 1 superscript 𝑒 2 𝛽 𝑘 1 {\displaystyle{\displaystyle\frac{B_{2n}}{4n}\left(\alpha^{n}-(-\beta)^{n}% \right)=\alpha^{n}\sum_{k=1}^{\infty}\frac{k^{2n-1}}{e^{2\alpha k}-1}-(-\beta)% ^{n}\sum_{k=1}^{\infty}\frac{k^{2n-1}}{e^{2\beta k}-1}}}
\frac{\BernoullinumberB{2n}}{4n}\left(\alpha^{n}-(-\beta)^{n}\right) = \alpha^{n}\sum_{k=1}^{\infty}\frac{k^{2n-1}}{e^{2\alpha k}-1}-(-\beta)^{n}\sum_{k=1}^{\infty}\frac{k^{2n-1}}{e^{2\beta k}-1}		 

(bernoulli(2*n))/(4*n)*((alpha)^(n)-(- beta)^(n)) = (alpha)^(n)* sum(((k)^(2*n - 1))/(exp(2*alpha*k)- 1), k = 1..infinity)-(- beta)^(n)* sum(((k)^(2*n - 1))/(exp(2*beta*k)- 1), k = 1..infinity)

Divide[BernoulliB[2*n],4*n]*(\[Alpha]^(n)-(- \[Beta])^(n)) == \[Alpha]^(n)* Sum[Divide[(k)^(2*n - 1),Exp[2*\[Alpha]*k]- 1], {k, 1, Infinity}, GenerateConditions->None]-(- \[Beta])^(n)* Sum[Divide[(k)^(2*n - 1),Exp[2*\[Beta]*k]- 1], {k, 1, Infinity}, GenerateConditions->None]
Failure Aborted
Failed [6 / 9]
Result: 1.443014212
Test Values: {alpha = 3/2, beta = 1/2, n = 2}


Result: -.7774267192e-1
Test Values: {alpha = 3/2, beta = 2, n = 2}

... skip entries to safe data
 Skipped - Because timed out
24.8.E9 E 2 ⁢ n = ( - 1 ) n ⁢ ∑ k = 1 ∞ k 2 ⁢ n cosh ⁡ ( 1 2 ⁢ π ⁢ k ) - 4 ⁢ ∑ k = 0 ∞ ( - 1 ) k ⁢ ( 2 ⁢ k + 1 ) 2 ⁢ n e 2 ⁢ π ⁢ ( 2 ⁢ k + 1 ) - 1 Euler-number-E 2 𝑛 superscript 1 𝑛 superscript subscript 𝑘 1 superscript 𝑘 2 𝑛 1 2 𝜋 𝑘 4 superscript subscript 𝑘 0 superscript 1 𝑘 superscript 2 𝑘 1 2 𝑛 superscript 𝑒 2 𝜋 2 𝑘 1 1 {\displaystyle{\displaystyle{}E_{2n}=(-1)^{n}\sum_{k=1}^{\infty}\frac{k^{2n}}{% \cosh\left(\tfrac{1}{2}\pi k\right)}-4\sum_{k=0}^{\infty}\frac{(-1)^{k}(2k+1)^% {2n}}{e^{2\pi(2k+1)}-1}}}
{}\EulernumberE{2n} = (-1)^{n}\sum_{k=1}^{\infty}\frac{k^{2n}}{\cosh@{\tfrac{1}{2}\pi k}}-4\sum_{k=0}^{\infty}\frac{(-1)^{k}(2k+1)^{2n}}{e^{2\pi(2k+1)}-1}		 

*euler(2*n) = (- 1)^(n)* sum(((k)^(2*n))/(cosh((1)/(2)*Pi*k)), k = 1..infinity)- 4*sum(((- 1)^(k)*(2*k + 1)^(2*n))/(exp(2*Pi*(2*k + 1))- 1), k = 0..infinity)

*EulerE[2*n] == (- 1)^(n)* Sum[Divide[(k)^(2*n),Cosh[Divide[1,2]*Pi*k]], {k, 1, Infinity}, GenerateConditions->None]- 4*Sum[Divide[(- 1)^(k)*(2*k + 1)^(2*n),Exp[2*Pi*(2*k + 1)]- 1], {k, 0, Infinity}, GenerateConditions->None]
Translation Error Translation Error - -
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