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I would like to find taylor expansion of $sh(x)$

My thoughts indeed,

note that : $\sinh(x)=\dfrac{e^{x}-e^{-x}}{2}$ then

\begin{align} \sinh(x)&=\frac{e^x-e^{-x}}{2} \\ &=\frac{1}{2}\left( e^x-e^{-x} \right)\\ &\underset{x\to 0}=\frac{1}{2}\left(\sum_{k=0}^n\frac{x^k}{k!}+o(x^n)-\sum_{k=0}^n\frac{(-1)^k x^{k}}{k!}-o(x^n)\right) \\ &\underset{x\to 0}=\frac{1}{2}\left(\sum_{k=0}^{n}\left( \frac{x^{k}}{k!}-\frac{(-1)^{k}x^{k}}{k!}\right)+o(x^{n})\right) \\ &\underset{x\to 0}=\frac{1}{2}\left(\sum_{k=0}^{n}\left( \frac{(1+(-1)^{k+1})x^{k}}{k!}\right)+o(x^{n})\right) \\ &\underset{x\to 0}=\frac{1}{2}\left(\sum_{k=0}^{n}\left( \frac{(1+(-1)^{k+1})x^{k}}{k!}\right)+o(x^{n})\right)\\ &\underset{x\to 0}=\begin{cases}\dfrac{1}{2}\left(\sum\limits_{k'=0}^{n}\left( \dfrac{(1+(1)^{2(k'+1)})x^{2k'+1}}{k!}\right)+o(x^{n})\right) & \text{ if } k \text{ odd }\, k=2k'+1 \text{ with } k'\in\mathbb{Z}\\\dfrac{1}{2}\left(\sum\limits_{k=0}^{n}\left( \dfrac{(1+(-1)^{2k'+1})x^{k}}{k!}\right)+o(x^{n})\right) & \text{ if } k \text{ even }\, k=2k' \text{ with } k'\in\mathbb{Z}\end{cases}\\ &\underset{x\to 0}=\begin{cases}\dfrac{1}{2}\left(\sum\limits_{k'=0}^{2n'+1}\left( \dfrac{2x^{2k'+1}}{k!}\right)+o(x^{2n'+1})\right) & \text{ if } k \text{ odd }\, k=2k'+1 \text{ with } k'\in\mathbb{Z}\\\dfrac{1}{2}\left(\sum\limits_{k=0}^{2n'}\left( \dfrac{(1-1)x^{2k'}}{2k'!}\right)+o(x^{2n'})\right) & \text{ if } k \text{ even }\, k=2k' \text{ with } k'\in\mathbb{Z}\end{cases}\\ &\underset{x\to 0}=\begin{cases}\left(\sum\limits_{k'=0}^{2n'+1}\left( \dfrac{x^{2k'+1}}{(2k'+1)!}\right)+o(x^{2n'+1})\right) & \text{ if } k \text{ odd }\, k=2k'+1 \text{ with } k'\in\mathbb{Z}\\ 0 & \text{ if } k \text{ even }\, k=2k' \text{ with } k'\in\mathbb{Z}\end{cases} \end{align}

Update

  • but if I want from there

$$\sinh(x)\underset{x\to 0}=\frac{1}{2}\sum_{k'=0}^{E(n/2)} \frac{(1+(-1)^{2k'+1})x^{2k'}}{(2k')!} + \frac{1}{2}\sum_{k'=0}^{E((n-1)/2)} \frac{(1+(-1)^{2k'+2})x^{2k'+1}}{(2k'+1)!} + o(x^{n}) $$

  • to get the desired result:

$$\sinh(x)\underset{x\to 0}=\left(\sum_{k=0}^{n}\left(\dfrac{x^{2k+1}}{(2k+1)!}\right)+o(x^{2n+1})\right)$$

indeed,

\begin{align} sh(x)&\underset{x\to 0}=\frac{1}{2}\sum_{k'=0}^{E(n/2)} \frac{(1+(-1)^{2k'+1})x^{2k'}}{(2k')!} + \frac{1}{2}\sum_{k'=0}^{E((n-1)/2)} \frac{(1+(-1)^{2k'+2})x^{2k'+1}}{(2k'+1)!} + o(x^{n})\\ &\underset{x\to 0}=0 + \frac{1}{2}\sum_{k'=0}^{E((n-1)/2)} \frac{2)x^{2k'+1}}{(2k'+1)!} + o(x^{n})\\ &\underset{x\to 0}= \sum_{k'=0}^{E((n-1)/2)} \frac{x^{2k'+1}}{(2k'+1)!} + o(x^{n})\\ &\underset{x\to 0}=\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\\ &\text{These few passing I would like to know them. }\\ sh(x)&\underset{x\to 0}=\left(\sum_{k=0}^{n}\left(\dfrac{x^{2k+1}}{(2k+1)!}\right)+o(x^{2n+1})\right) \end{align}

  • Is my proof correct
  • I'm interested in more ways of finding taylor expansion of $\sinh(x)$.
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2 Answers 2

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How about a rather simple derivation like the one below:

$$\exp(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots $$

and

$$\exp(-x)= 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$

So when you subtract the two equations:

$$\exp(x) - \exp(-x) = 2x + 2\frac{x^3}{3!} + 2\frac{x^5}{5!}$$

Finally,

$$\frac{\exp(x) - \exp(-x)}{2} = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots = \sum_{n=0}^{\infty} \dfrac{x^{2n+1}}{(2n+1)!}$$

Odd powers remain and sine is an odd function.

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  • $\begingroup$ Thank you but i need to understand other way $\endgroup$
    – Educ
    Oct 29, 2015 at 15:05
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There are a couple things wrong here.

The big sum has not two values according to the even-ness of $k$, as it does not depend on any variable called $k$. And in those sums, the upper bound value is false.


You really wanted to write: \begin{align} S &=\frac{1}{2}\sum_{k=0}^{n} \frac{(1+(-1)^{k+1})x^{k}}{k!}+o(x^{n}) \\ &=\frac{1}{2}\sum_{k'=0}^{E(n/2)} \frac{(1+(-1)^{2k'+1})x^{2k'}}{(2k')!} + \frac{1}{2}\sum_{k'=0}^{E((n-1)/2)} \frac{(1+(-1)^{2k'+2})x^{2k'+1}}{(2k'+1)!} + o(x^{n}) \\ \end{align}

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