Find the Taylor series for $\sinh(x)$ and indicate why it converges to $\sinh(x)$.
By John Parsons •
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 $t$.
$\endgroup$More in general
"Zoraya ter Beek, age 29, just died by assisted suicide in the Netherlands. She was physically healthy, but psychologically depressed. It's an abomination that an entire society would actively facilitate, even encourage, someone ending their own life because they had no hope. Th…"
