Essentially what the title says. I'm asked to find the Taylor polynomial of degree $n$ for $f(x)=\sinh^2(x)$ about $a=0$.
This is essentially a Maclaurin series.
I could use the fact that I know what the Maclaurin series of $\sinh(x)$ which is $\sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$ and then I could expand term by term.
Is there a better way of doing this though?
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$\begingroup$Note that $$\sinh ^2 x = (\frac {e^x-e^{-x}}{2})^2=$$
$$(1/4)(e^{2x} +e^{-2x} -2)$$
Now use $$e^{2x} = 1+(2x) + (2x)^2/2 + (2x)^3 / {3!}+.....$$ and $$e^{-2x} = 1+(-2x) + (-2x)^2/2 + (-2x)^3 / {3!}+.....$$ to get your result.
$\endgroup$ $\begingroup$We have that by hyperbolic function identities
$$\sinh^2 x = \frac12\left(\cosh(2x)-1\right)$$
then use that
$$\cosh x = \sum_{n=0}^\infty \frac{x^{2n}}{(2n)!}$$
that is
$$\sinh^2 x=-\frac12+\frac12\sum_{n=0}^\infty \frac{{(2x)}^{2n}}{(2n)!}=-\frac12+\frac12+\frac12\sum_{n=1}^\infty \frac{{(2x)}^{2n}}{(2n)!}=\sum_{n=1}^\infty \frac{2^{2n-1}{x}^{2n}}{(2n)!}$$
$\endgroup$ 4 $\begingroup$Let the subscript $n\ge 1$ denote the nth-derivative, then$$y(x)=y_{0}(x)=\sinh^2 x, y_1(x)=\sinh 2x, y_{n}(x)=2^{n-2}[e^{2x}+(-1)^n e^{-2x}.$$We have $$y_{0}(0)=0, y_{1}(0)=0, y_2(0)= 2, y_3(0)=0,y_(4)(0)=2^3,....y_{2m+1}(0)=0, y_{2m}(0)=2^{2m-1}.$$^h3 McLaurib seies is given by $$y(x)=\sum_{k=0}^{\infty}\frac{y_k(0)~ x^k}{k!}.$$Finally $$\sinh^2 x=\sum_{m=1}^{\infty} \frac{2^{2m-1} x^{2m}}{(2m)!} $$
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