Home » Blog » Verify that sinh x = ∑ x2n+1 / (2n+1)!

Verify that sinh x = ∑ x2n+1 / (2n+1)!

Assuming that \sinh x has a power-series representation in terms of powers of x, verify that it has the form

    \[ \sinh x = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} \]

valid for all real x.

First, we recall the definition of the hyperblic sine in terms of the exponential,

    \[ \sinh x = \frac{e^x - e^{-x}}{2}. \]

Further, we know

    \[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}. \]

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*}

This is valid for all x \in \mathbb{R} since the expansions for e^x are valid for all x \in \mathbb{R}.

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):