maclaurin series: sinh^2 (x)

Question

anonymous · Answer

Using a power reduction formula, you have
$$\sinh^2x=\frac{1}{2}\cosh2x-\frac{1}{2}$$
Next, since $\cosh x=\dfrac{e^{x}+e^{-x}}{2}$, you get
$$\begin{align*}\sinh^2x&=\frac{1}{2}\left(\frac{e^{2x}+e^{-2x}}{2}\right)-\frac{1}{2}\
&=\frac{1}{4}e^{2x}+\frac{1}{4}e^{-2x}-\frac{1}{2}
\end{align*}$$
Recall the power series for $e^x$:
$$e^{x}=\sum_{n=0}^\infty \frac{x^n}{n!}$$
Substituting $2x$ for $x$, you have
$$\begin{align*}\sinh^2x&=\frac{1}{4}\sum_{n=0}^\infty \frac{(2x)^n}{n!}+\frac{1}{4}\sum_{n=0}^\infty \frac{(-2x)^n}{n!}-\frac{1}{2}\
&=\small \frac{1}{4}\left(1+2x+\frac{(2x)^2}{2!}+\frac{(2x)^3}{3!}+\cdots\right)+\frac{1}{4}\left(1-2x+\frac{(2x)^2}{2!}-\frac{(2x)^3}{3!}+\cdots\right)-\frac{1}{2}\
&=\frac{1}{4}\left(\frac{(2x)^2}{2!}+\frac{(2x)^4}{4!}+\cdots\right)+\frac{1}{4}\left(\frac{(2x)^2}{2!}+\frac{(2x)^4}{4!}+\cdots\right)\
&=\frac{1}{2}\left(\frac{(2x)^2}{2!}+\frac{(2x)^4}{4!}+\cdots\right)\
&=\frac{2^1x^2}{2!}+\frac{2^3x^4}{4!}+\cdots\
&=\large\sum_{n=1}^\infty\frac{2^{2n-1}x^{2n}}{(2n)!}
\end{align*}$$