Difference between revisions of "Taylor series of cosine"
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| − | + | ==Theorem== | |
| − | + | Let $z_0 \in \mathbb{C}$. The following [[Taylor series]] holds for all $z \in \mathbb{C}$: | |
| − | $$\cos(z)= \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k | + | $$\cos(z)= \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k z^{2k}}{(2k)!},$$ |
where $\cos$ denotes the [[cosine]] function. | where $\cos$ denotes the [[cosine]] function. | ||
| − | + | ||
| − | + | ==Proof== | |
| − | + | Using the [[Taylor series of the exponential function]] and the definition of $\cos$, | |
| − | + | $$\begin{array}{ll} | |
| + | \cos(z) &= \dfrac{e^{iz}+e^{-iz}}{2} \\ | ||
| + | &= \dfrac{1}{2} \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}{2} \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})=2(-1)^k,$$ | ||
| + | and if $n=2k+1$ is a positive odd integer, then | ||
| + | $$i^n(1+(-1)^n)=i^{2k+1}(1+(-1)^{2k+1})=0.$$ | ||
| + | Hence we have derived | ||
| + | $$\begin{array}{ll} | ||
| + | \cos(z)&=\dfrac{1}{2} \displaystyle\sum_{n=0}^{\infty} \dfrac{(z-z_0)^n}{n!}i^n (1+(-1)^n) \\ | ||
| + | &=\dfrac{1}{2} \displaystyle\sum_{n \mathrm{\hspace{2pt} even},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}}{(2k)!}, | ||
| + | \end{array}$$ | ||
| + | as was to be shown. █ | ||
| + | |||
| + | ==References== | ||
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Proven]] | ||
Latest revision as of 03:18, 1 July 2017
Theorem
Let $z_0 \in \mathbb{C}$. The following Taylor series holds for all $z \in \mathbb{C}$: $$\cos(z)= \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k z^{2k}}{(2k)!},$$ where $\cos$ denotes the cosine function.
Proof
Using the Taylor series of the exponential function and the definition of $\cos$, $$\begin{array}{ll} \cos(z) &= \dfrac{e^{iz}+e^{-iz}}{2} \\ &= \dfrac{1}{2} \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}{2} \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})=2(-1)^k,$$ and if $n=2k+1$ is a positive odd integer, then $$i^n(1+(-1)^n)=i^{2k+1}(1+(-1)^{2k+1})=0.$$ Hence we have derived $$\begin{array}{ll} \cos(z)&=\dfrac{1}{2} \displaystyle\sum_{n=0}^{\infty} \dfrac{(z-z_0)^n}{n!}i^n (1+(-1)^n) \\ &=\dfrac{1}{2} \displaystyle\sum_{n \mathrm{\hspace{2pt} even},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}}{(2k)!}, \end{array}$$ as was to be shown. █
