Difference between revisions of "Taylor series of cosine"

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

References