Elementary Functions - 4.19 Maclaurin Series and Laurent Series

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DLMF Formula Constraints Maple Mathematica Symbolic
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4.19.E7 ln ( sin z z ) = n = 1 ( - 1 ) n 2 2 n - 1 B 2 n n ( 2 n ) ! z 2 n 𝑧 𝑧 superscript subscript 𝑛 1 superscript 1 𝑛 superscript 2 2 𝑛 1 Bernoulli-number-B 2 𝑛 𝑛 2 𝑛 superscript 𝑧 2 𝑛 {\displaystyle{\displaystyle\ln\left(\frac{\sin z}{z}\right)=\sum_{n=1}^{% \infty}\frac{(-1)^{n}2^{2n-1}B_{2n}}{n(2n)!}z^{2n}}}
\ln@{\frac{\sin@@{z}}{z}} = \sum_{n=1}^{\infty}\frac{(-1)^{n}2^{2n-1}\BernoullinumberB{2n}}{n(2n)!}z^{2n}
| z | < π 𝑧 𝜋 {\displaystyle{\displaystyle|z|<\pi}}
ln((sin(z))/(z)) = sum(((- 1)^(n)* (2)^(2*n - 1)* bernoulli(2*n))/(n*factorial(2*n))*(z)^(2*n), n = 1..infinity)
Log[Divide[Sin[z],z]] == Sum[Divide[(- 1)^(n)* (2)^(2*n - 1)* BernoulliB[2*n],n*(2*n)!]*(z)^(2*n), {n, 1, Infinity}, GenerateConditions->None]
Failure Failure Successful [Tested: 7] Successful [Tested: 7]
4.19.E8 ln ( cos z ) = n = 1 ( - 1 ) n 2 2 n - 1 ( 2 2 n - 1 ) B 2 n n ( 2 n ) ! z 2 n 𝑧 superscript subscript 𝑛 1 superscript 1 𝑛 superscript 2 2 𝑛 1 superscript 2 2 𝑛 1 Bernoulli-number-B 2 𝑛 𝑛 2 𝑛 superscript 𝑧 2 𝑛 {\displaystyle{\displaystyle\ln\left(\cos z\right)=\sum_{n=1}^{\infty}\frac{(-% 1)^{n}2^{2n-1}(2^{2n}-1)B_{2n}}{n(2n)!}z^{2n}}}
\ln@{\cos@@{z}} = \sum_{n=1}^{\infty}\frac{(-1)^{n}2^{2n-1}(2^{2n}-1)\BernoullinumberB{2n}}{n(2n)!}z^{2n}
| z | < 1 2 π 𝑧 1 2 𝜋 {\displaystyle{\displaystyle|z|<\frac{1}{2}\pi}}
ln(cos(z)) = sum(((- 1)^(n)* (2)^(2*n - 1)*((2)^(2*n)- 1)*bernoulli(2*n))/(n*factorial(2*n))*(z)^(2*n), n = 1..infinity)
Log[Cos[z]] == Sum[Divide[(- 1)^(n)* (2)^(2*n - 1)*((2)^(2*n)- 1)*BernoulliB[2*n],n*(2*n)!]*(z)^(2*n), {n, 1, Infinity}, GenerateConditions->None]
Failure Failure Manual Skip! Successful [Tested: 6]
4.19.E9 ln ( tan z z ) = n = 1 ( - 1 ) n - 1 2 2 n ( 2 2 n - 1 - 1 ) B 2 n n ( 2 n ) ! z 2 n 𝑧 𝑧 superscript subscript 𝑛 1 superscript 1 𝑛 1 superscript 2 2 𝑛 superscript 2 2 𝑛 1 1 Bernoulli-number-B 2 𝑛 𝑛 2 𝑛 superscript 𝑧 2 𝑛 {\displaystyle{\displaystyle\ln\left(\frac{\tan z}{z}\right)=\sum_{n=1}^{% \infty}\frac{(-1)^{n-1}2^{2n}(2^{2n-1}-1)B_{2n}}{n(2n)!}z^{2n}}}
\ln@{\frac{\tan@@{z}}{z}} = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}2^{2n}(2^{2n-1}-1)\BernoullinumberB{2n}}{n(2n)!}z^{2n}
| z | < 1 2 π 𝑧 1 2 𝜋 {\displaystyle{\displaystyle|z|<\frac{1}{2}\pi}}
ln((tan(z))/(z)) = sum(((- 1)^(n - 1)* (2)^(2*n)*((2)^(2*n - 1)- 1)*bernoulli(2*n))/(n*factorial(2*n))*(z)^(2*n), n = 1..infinity)
Log[Divide[Tan[z],z]] == Sum[Divide[(- 1)^(n - 1)* (2)^(2*n)*((2)^(2*n - 1)- 1)*BernoulliB[2*n],n*(2*n)!]*(z)^(2*n), {n, 1, Infinity}, GenerateConditions->None]
Failure Failure Manual Skip! Successful [Tested: 6]