# How do I find the Maclaurin series of $\sinh^2(x)$?

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?

• Hint: \begin{eqnarray*} \cosh(2x)=1+2 \sinh^2(x). \end{eqnarray*} – Donald Splutterwit Oct 22 '19 at 15:11
• That 1 is really annoying though. How do I deal with those constants? – Future Math person Oct 22 '19 at 15:13
• That $1$ just cancels out with the first term in the expansion of $\cosh$. – Donald Splutterwit Oct 22 '19 at 15:16
• I will try it out later then! Thank you! – Future Math person Oct 22 '19 at 15:17

## 3 Answers

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

$$\sinh^2 x=-\frac12+\frac12\sum_{n=0}^\infty \frac{{(2x)}^{2n}}{(2n)!}=-\frac12+\frac12+\frac12\sum_{n=1}^\infty \frac{{(2x)}^{2n}}{(2n)!}=\sum_{n=1}^\infty \frac{2^{2n-1}{x}^{2n}}{(2n)!}$$

• I was given this hint too but how does that help with anything? That 1 is going to be annoying to deal with. – Future Math person Oct 22 '19 at 15:12
• @FutureMathperson You can use the series for $\cosh x$ in that way. – user Oct 22 '19 at 15:15
• Just deal with it separately: $$1+\sum_{n=0}^{\infty} a_nx^n=1+a_0+\sum_{n=1}^{\infty} a_nx^n$$ – b00n heT Oct 22 '19 at 15:16
• I will try it out later! Thank you! – Future Math person Oct 22 '19 at 15:16

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.

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 $$\sinh^2 x=\sum_{m=1}^{\infty} \frac{2^{2m-1} x^{2m}}{(2m)!}$$