Difference between revisions of "Taylor series of sine"
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| − | + | ==Theorem== | |
| − | + | Let $z_0 \in \mathbb{C}$. The following [[Taylor series]] holds: | |
| − | + | $$\sin(z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k z^{2k+1}}{(2k+1)!},$$ | |
| − | + | where $\sin$ denotes the [[sine]] function. | |
| − | + | ||
| − | + | ==Proof== | |
| + | Using the [[Taylor series of the exponential function]] and the definition of $\sin$, | ||
| + | $$\begin{array}{ll} | ||
| + | \sin(z) &= \dfrac{e^{iz}-e^{-iz}}{2i} \\ | ||
| + | &= \dfrac{1}{2i} \left[ \displaystyle\sum_{k=0}^{\infty} \dfrac{i^k (z-z_0)^k}{k!} - \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k i^k (z-z_0)^k}{k!} \right] \\ | ||
| + | &= \dfrac{1}{2i} \displaystyle\sum_{k=0}^{\infty} \dfrac{(z-z_0)^k}{k!}i^k (1-(-1)^k). | ||
| + | \end{array}$$ | ||
| + | Note that if $k=2n$ is a positive even integer, then | ||
| + | $$i^k(1-(-1)^k)=i^{2n}(1-(-1)^{2n})=0,$$ | ||
| + | and if $k=2n+1$ is a positive odd integer, then | ||
| + | $$i^k(1-(-1)^k)=i^{2n+1}(1-(-1)^{2n+1})=2i(-1)^n.$$ | ||
| + | Hence we have derived | ||
| + | $$\begin{array}{ll} | ||
| + | \sin(z)&=\dfrac{1}{2i} \displaystyle\sum_{k=0}^{\infty} \dfrac{(z-z_0)^k}{k!}i^k (1-(-1)^k) \\ | ||
| + | &=\displaystyle\sum_{k \mathrm{\hspace{2pt} odd},k>0}^{\infty} \dfrac{(z-z_0)^k}{k!}i^k (1-(-1)^k) \\ | ||
| + | &= \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k (z-z_0)^{2k+1}}{(2k+1)!}, | ||
| + | \end{array}$$ | ||
| + | as was to be shown. █ | ||
| + | |||
| + | ==References== | ||
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Proven]] | ||
Latest revision as of 03:19, 1 July 2017
Theorem
Let $z_0 \in \mathbb{C}$. The following Taylor series holds: $$\sin(z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k z^{2k+1}}{(2k+1)!},$$ where $\sin$ denotes the sine function.
Proof
Using the Taylor series of the exponential function and the definition of $\sin$, $$\begin{array}{ll} \sin(z) &= \dfrac{e^{iz}-e^{-iz}}{2i} \\ &= \dfrac{1}{2i} \left[ \displaystyle\sum_{k=0}^{\infty} \dfrac{i^k (z-z_0)^k}{k!} - \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k i^k (z-z_0)^k}{k!} \right] \\ &= \dfrac{1}{2i} \displaystyle\sum_{k=0}^{\infty} \dfrac{(z-z_0)^k}{k!}i^k (1-(-1)^k). \end{array}$$ Note that if $k=2n$ is a positive even integer, then $$i^k(1-(-1)^k)=i^{2n}(1-(-1)^{2n})=0,$$ and if $k=2n+1$ is a positive odd integer, then $$i^k(1-(-1)^k)=i^{2n+1}(1-(-1)^{2n+1})=2i(-1)^n.$$ Hence we have derived $$\begin{array}{ll} \sin(z)&=\dfrac{1}{2i} \displaystyle\sum_{k=0}^{\infty} \dfrac{(z-z_0)^k}{k!}i^k (1-(-1)^k) \\ &=\displaystyle\sum_{k \mathrm{\hspace{2pt} odd},k>0}^{\infty} \dfrac{(z-z_0)^k}{k!}i^k (1-(-1)^k) \\ &= \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k (z-z_0)^{2k+1}}{(2k+1)!}, \end{array}$$ as was to be shown. █
