Difference between revisions of "Taylor series of sine"

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==Theorem==
 
==Theorem==
 
Let $z_0 \in \mathbb{C}$. The following [[Taylor series]] holds:
 
Let $z_0 \in \mathbb{C}$. The following [[Taylor series]] holds:
$$\sin(z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k(z-z_0)^{2k+1}}{(2k+1)!},$$
+
$$\sin(z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k z^{2k+1}}{(2k+1)!},$$
 
where $\sin$ denotes the [[sine]] function.
 
where $\sin$ denotes the [[sine]] function.
  
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$$\begin{array}{ll}
 
$$\begin{array}{ll}
 
\sin(z) &= \dfrac{e^{iz}-e^{-iz}}{2i} \\
 
\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} \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_{n=0}^{\infty} \dfrac{(z-z_0)^n}{n!}i^n (1-(-1)^n).
+
&= \dfrac{1}{2i} \displaystyle\sum_{k=0}^{\infty} \dfrac{(z-z_0)^k}{k!}i^k (1-(-1)^k).
 
\end{array}$$
 
\end{array}$$
Note that if $n=2k$ is a positive even integer, then
+
Note that if $k=2n$ is a positive even integer, then
$$i^n(1-(-1)^n)=i^{2k}(1-(-1)^{2k})=0,$$
+
$$i^k(1-(-1)^k)=i^{2n}(1-(-1)^{2n})=0,$$
and if $n=2k+1$ is a positive odd integer, then
+
and if $k=2n+1$ is a positive odd integer, then
$$i^n(1-(-1)^n)=i^{2k+1}(1-(-1)^{2k+1})=2i(-1)^k.$$
+
$$i^k(1-(-1)^k)=i^{2n+1}(1-(-1)^{2n+1})=2i(-1)^n.$$
 
Hence we have derived
 
Hence we have derived
 
$$\begin{array}{ll}
 
$$\begin{array}{ll}
\sin(z)&=\dfrac{1}{2i} \displaystyle\sum_{n=0}^{\infty} \dfrac{(z-z_0)^n}{n!}i^n (1-(-1)^n) \\
+
\sin(z)&=\dfrac{1}{2i} \displaystyle\sum_{k=0}^{\infty} \dfrac{(z-z_0)^k}{k!}i^k (1-(-1)^k) \\
&=\displaystyle\sum_{n \mathrm{\hspace{2pt} odd},n>0}^{\infty} \dfrac{(z-z_0)^n}{n!}i^n (1-(-1)^n) \\
+
&=\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)!},
 
&= \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k (z-z_0)^{2k+1}}{(2k+1)!},
 
\end{array}$$  
 
\end{array}$$  
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==References==
 
==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. █

References