Find the Taylor series for $\sinh(x)$ and indicate why it converges to $\sinh(x)$.
Find the Taylor series for $\sinh(x)=\frac 1 2(e^x-e^{-x})$ and indicate why it converges to $\sinh(x)$.
$\endgroup$ 42 Answers
$\begingroup$$$\sinh x = \frac{1}{2} ( e^{x} - e^ {-x}) = \frac{1}{2}(\sum_{n=0}^\infty \frac{x^n}{n!} - \sum_{n=0}^\infty \frac{(-x)^n}{n!}) = \frac{1}{2}\sum_{n=0}^\infty \frac{(1-(-1)^n ) x^n}{n!} = \sum_{n=0}^\infty \frac{ x^{2n+1}}{(2n+1)!}$$
$\endgroup$ 2 $\begingroup$To show that the series does indeed converge to $\sinh x$, for all $x$, you may want to use Taylor's Theorem. The Lagrange form of the remainder will do the job. You will need to prove, or recall, that $\displaystyle\lim_{n\to\infty} \frac{|x|^n}{n!}=0$.
Alternately, you can recall (if that has already been proved in your course) that the usual power series for $e^t$ converges to $e^t$ for all $t$.
$\endgroup$
