Assuming that 
has a power-series representation in terms of powers of 
, verify that it has the form
![\[ \sinh x = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-248b3fabdb257393ccdede2695610620_l3.png "Rendered by QuickLaTeX.com")
valid for all real .
First, we recall the definition of the hyperblic sine in terms of the exponential,
![\[ \sinh x = \frac{e^x - e^{-x}}{2}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-eb729d2dc37c0f9fe1273bcdc7d295ab_l3.png "Rendered by QuickLaTeX.com")
Further, we know
![\[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-d4fef7f4db4f4c6d1adb3d93afbe4c08_l3.png "Rendered by QuickLaTeX.com")
Putting these together we have
![\begin{align*} \sinh x &= \frac{e^x - e^{-x}}{2} \\[9pt] &= \frac{1}{2} \left( \sum_{n=0}^{\infty} \frac{x^n}{n!} - \sum_{n=0}^{\infty} \frac{(-x)^n}{n!} \right) \\[9pt] &= \frac{1}{2} \left( \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} + \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} - \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} + \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} \right) \\[9pt] &= \frac{1}{2} \left( 2 \cdot \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} \right)\\[9pt] &= \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-89f2269c4e50751229e25c0ebf047743_l3.png "Rendered by QuickLaTeX.com")
This is valid for all 
since the expansions for 
are valid for all .