# How to prove that $f(x)$ is equal to $\frac{1}{2}(\exp(x) -\exp(-x))$?

$$f(x) = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$$

Since we know that $\frac{1}{2}(\exp(x) - \exp(-x))$ is same as $$\frac{1}{2}\left(\sum_{n=0}^{\infty}\frac{x^{n}}{n!} - \sum_{n=0}^{\infty}\frac{(-x)^{n}}{n!}\right)$$

we can place our $f(x)$ in this form. But first we can do more simplifying, we can write that form as $\displaystyle \frac{1}{2}\sum_{n=0}^{\infty}\frac{x^{n}-(-x)^n}{n!}$. In my book I found next steps, we are placing our $f(x)$ in this last form and we get $$\frac{1}{2}\left(\sum_{n=0}^{\infty}\frac{x^{2n+1}-(-x)^{2n+1}}{(2n+1)!} + \frac{x^{2n}-(-x)^{2n}}{(2n)!}\right)$$

The part which I don't understand is, from where do we get that $\displaystyle\frac{x^{2n}-(-x)^{2n}}{(2n)!}$ ?

• Well, $$\frac{1}{2}\sum_{n=0}^{\infty}\frac{x^{n}-(-x)^n}{n!}=\sum_{n=0}^{\infty}\frac{1-(-1)^n}{2}\frac{x^{n}}{n!}$$ right? So all there is to do is to compute $$\frac{1-(-1)^n}{2}$$ for every $n$. Can you do that? – Did Nov 17 '16 at 10:48

They split the sum in an even and an odd part. In general

$$\sum_{n = 0}^{+\infty} a_n = \sum_{n=0,\;n\; {\rm even}}^{+\infty}a_n + \sum_{n=0,\;n\; {\rm odd}}^{+\infty}a_n = \sum_{n=0}^{+\infty}a_{2n} + \sum_{n=0}^{+\infty}a_{2n+1}$$

This is particularly useful here because $1-(-1)^{2n} = 1-1 = 0$ and $1 -(-1)^{2n+1} = 1 + 1 = 2$.

For $n\in\mathbb N$, you have two cases:
1. $n$ is odd. Then, $n=2k+1$ for some $k=0,1,2,\dots$
2. $n$ is even. Then, $n=2k$ for some $k=0,1,2,\dots$
Hence, $\exp(x)-\exp(-x)$ can be written as (LHS) \begin{align}\sum_{n=0}^{\infty}\frac{x^{n}-(-x)^n}{n!}&=\sum_{n \text{ odd}}^{\infty}\frac{x^{n}-(-x)^n}{n!}+\sum_{n \text{ even}}^{\infty}\frac{x^{n}-(-x)^n}{n!}\\[0.3cm]&=\sum_{k=0}^{\infty}\frac{x^{2k+1}-(-x)^{2k+1}}{(2k+1)!}+\sum_{k=0}^{\infty}\frac{x^{2k}-(-x)^{2k}}{(2k)!}\\[0.3cm]&=\sum_{k=0}^{\infty}\frac{x^{2k+1}+x^{2k+1}}{(2k+1)!}+\sum_{k=0}^{\infty}\frac{x^{2k}-x^{2k}}{(2k)!}=\sum_{k=0}^{\infty}\frac{2x^{2k+1}}{(2k+1)!}\end{align}