Difference between revisions of "Taylor series of sine"

Revision as of 00:27, 4 June 2016

Theorem

Let $z_0 \in \mathbb{C}$. The following Taylor series holds: $$\sin(z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k(z-z_0)^{2k+1}}{(2k+1)!},$$ where $\sin$ denotes the sine function.

Proof

Using the Taylor series of the exponential function and the definition of $\sin$, $$\begin{array}{ll} \sin(z) &= \dfrac{e^{iz}-e^{-iz}}{2i} \\ &= \dfrac{1}{2i} \left[ \displaystyle\sum_{n=0}^{\infty} \dfrac{i^n (z-z_0)^n}{n!} - \displaystyle\sum_{n=0}^{\infty} \dfrac{(-1)^n i^n (z-z_0)^n}{n!} \right] \\ &= \dfrac{1}{2i} \displaystyle\sum_{n=0}^{\infty} \dfrac{(z-z_0)^n}{n!}i^n (1-(-1)^n). \end{array}$$ Note that if $n=2k$ is a positive even integer, then $$i^n(1-(-1)^n)=i^{2k}(1-(-1)^{2k})=0,$$ and if $n=2k+1$ is a positive odd integer, then $$i^n(1-(-1)^n)=i^{2k+1}(1-(-1)^{2k+1})=2i(-1)^k.$$ Hence we have derived $$\begin{array}{ll} \sin(z)&=\dfrac{1}{2i} \displaystyle\sum_{n=0}^{\infty} \dfrac{(z-z_0)^n}{n!}i^n (1-(-1)^n) \\ &=\displaystyle\sum_{n \mathrm{\hspace{2pt} odd},n>0}^{\infty} \dfrac{(z-z_0)^n}{n!}i^n (1-(-1)^n) \\ &= \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k (z-z_0)^{2k+1}}{(2k+1)!}, \end{array}$$ as was to be shown. █

References