Difference between revisions of "Taylor series of sine"
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where $\sin$ denotes the [[sine]] function. | where $\sin$ denotes the [[sine]] function. | ||
<div class="mw-collapsible-content"> | <div class="mw-collapsible-content"> | ||
| − | <strong>Proof:</strong> █ | + | <strong>Proof:</strong> Using the [[Taylor series of the exponential function]], and the definition of $\sin$, |
| + | $$\begin{array}{ll} | ||
| + | \sin(z) &= \dfrac{e^{iz}-e^{-iz}}{2i} \\ | ||
| + | &= \dfrac{1}{2i} \left[ \displaystyle\sum_{k=0}^{\infty} \dfrac{i^k z^k}{k!} - \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k i^k z^k}{k!} \right] \\ | ||
| + | &= \dfrac{1}{2i} \displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{k!}i^k (1-(-1)^k). | ||
| + | \end{array}$$ | ||
| + | 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 | ||
| + | $$\begin{array}{ll} | ||
| + | \sin(z)&=\dfrac{1}{2i} \displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{k!}i^k (1-(-1)^k) \\ | ||
| + | &=\displaystyle\sum_{k \mathrm{\hspace{2pt} odd},k>0}^{\infty} \dfrac{z^k}{k!}i^k (1-(-1)^k) \\ | ||
| + | &= \displaystyle\sum_{n=0}^{\infty} \dfrac{(-1)^n z^{2n+1}}{(2n+1)!}, | ||
| + | \end{array}$$ | ||
| + | as was to be shown. █ | ||
</div> | </div> | ||
</div> | </div> | ||
Revision as of 07:26, 25 March 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 $\sin$, $$\begin{array}{ll} \sin(z) &= \dfrac{e^{iz}-e^{-iz}}{2i} \\ &= \dfrac{1}{2i} \left[ \displaystyle\sum_{k=0}^{\infty} \dfrac{i^k z^k}{k!} - \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k i^k z^k}{k!} \right] \\ &= \dfrac{1}{2i} \displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{k!}i^k (1-(-1)^k). \end{array}$$ 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 $$\begin{array}{ll} \sin(z)&=\dfrac{1}{2i} \displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{k!}i^k (1-(-1)^k) \\ &=\displaystyle\sum_{k \mathrm{\hspace{2pt} odd},k>0}^{\infty} \dfrac{z^k}{k!}i^k (1-(-1)^k) \\ &= \displaystyle\sum_{n=0}^{\infty} \dfrac{(-1)^n z^{2n+1}}{(2n+1)!}, \end{array}$$ as was to be shown. █
