# §4.19 Maclaurin Series and Laurent Series

 4.19.1 $\sin z=z-\frac{z^{3}}{3!}+\frac{z^{5}}{5!}-\frac{z^{7}}{7!}+\cdots,$ ⓘ Symbols: $!$: factorial (as in $n!$), $\sin\NVar{z}$: sine function and $z$: complex variable A&S Ref: 4.3.65 Referenced by: §4.45(i) Permalink: http://dlmf.nist.gov/4.19.E1 Encodings: TeX, pMML, png See also: Annotations for §4.19 and Ch.4
 4.19.2 $\cos z=1-\frac{z^{2}}{2!}+\frac{z^{4}}{4!}-\frac{z^{6}}{6!}+\cdots.$ ⓘ Symbols: $\cos\NVar{z}$: cosine function, $!$: factorial (as in $n!$) and $z$: complex variable A&S Ref: 4.3.66 Referenced by: §4.45(i) Permalink: http://dlmf.nist.gov/4.19.E2 Encodings: TeX, pMML, png See also: Annotations for §4.19 and Ch.4

In (4.19.3)–(4.19.9), $B_{n}$ are the Bernoulli numbers and $E_{n}$ are the Euler numbers (§§24.2(i)24.2(ii)).

 4.19.3 $\tan z=z+\frac{z^{3}}{3}+\frac{2}{15}z^{5}+\frac{17}{315}z^{7}+\cdots+\frac{(-% 1)^{n-1}2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}z^{2n-1}+\cdots,$ $|z|<\frac{1}{2}\pi$,
 4.19.4 $\csc z=\frac{1}{z}+\frac{z}{6}+\frac{7}{360}z^{3}+\frac{31}{15120}z^{5}+\cdots% +\frac{(-1)^{n-1}2(2^{2n-1}-1)B_{2n}}{(2n)!}z^{2n-1}+\cdots,$ $0<|z|<\pi$,
 4.19.5 $\sec z=1+\frac{z^{2}}{2}+\frac{5}{24}z^{4}+\frac{61}{720}z^{6}+\cdots+\frac{(-% 1)^{n}E_{2n}}{(2n)!}z^{2n}+\cdots,$ $|z|<\frac{1}{2}\pi$, ⓘ Symbols: $E_{\NVar{n}}$: Euler numbers, $\pi$: the ratio of the circumference of a circle to its diameter, $!$: factorial (as in $n!$), $\sec\NVar{z}$: secant function, $n$: integer and $z$: complex variable A&S Ref: 4.3.69 Referenced by: §24.1, §24.19(i), §24.2(ii) Permalink: http://dlmf.nist.gov/4.19.E5 Encodings: TeX, pMML, png See also: Annotations for §4.19 and Ch.4
 4.19.6 $\cot z=\frac{1}{z}-\frac{z}{3}-\frac{z^{3}}{45}-\frac{2}{945}z^{5}-\cdots-% \frac{(-1)^{n-1}2^{2n}B_{2n}}{(2n)!}z^{2n-1}-\cdots,$ $0<|z|<\pi$,
 4.19.7 $\ln\left(\frac{\sin z}{z}\right)=\sum_{n=1}^{\infty}\frac{(-1)^{n}2^{2n-1}B_{2% n}}{n(2n)!}z^{2n},$ $|z|<\pi$,
 4.19.8 $\ln\left(\cos z\right)=\sum_{n=1}^{\infty}\frac{(-1)^{n}2^{2n-1}(2^{2n}-1)B_{2% n}}{n(2n)!}z^{2n},$ $|z|<\frac{1}{2}\pi$,
 4.19.9 $\ln\left(\frac{\tan z}{z}\right)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}2^{2n}(2^{% 2n-1}-1)B_{2n}}{n(2n)!}z^{2n},$ $|z|<\frac{1}{2}\pi$.