1
$\begingroup$

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?

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

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

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

$\endgroup$
  • $\begingroup$ I was given this hint too but how does that help with anything? That 1 is going to be annoying to deal with. $\endgroup$ – Future Math person Oct 22 '19 at 15:12
  • $\begingroup$ @FutureMathperson You can use the series for $\cosh x$ in that way. $\endgroup$ – user Oct 22 '19 at 15:15
  • $\begingroup$ Just deal with it separately: $$1+\sum_{n=0}^{\infty} a_nx^n=1+a_0+\sum_{n=1}^{\infty} a_nx^n$$ $\endgroup$ – b00n heT Oct 22 '19 at 15:16
  • $\begingroup$ I will try it out later! Thank you! $\endgroup$ – Future Math person Oct 22 '19 at 15:16
3
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.