0
$\begingroup$

I have a question which I'm troubling to solve.

I've been given the following, a Taylor expansion of $\cosh(x)$ around $x=0$; $$\cosh(x) = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}.$$

Now, using that information I'm now supposed to prove that the Taylor expansion of $\cosh^2(x)$ is $$\cosh^2(x) = \frac{1}{2} + \frac{1}{2} \sum_{n=0}^{\infty}\frac{(2x)^{2n}}{(2n)!}$$

I realise that if you square the sum it will give you the Taylor series but I'm really struggling to prove this.

$\endgroup$
1
$\begingroup$

Use $$ \cosh^2(x)=\frac{(e^x+e^{-x})^2}4=\frac12(1+\cosh(2x)) $$ to directly arrive at the given answer.

$\endgroup$
  • $\begingroup$ Wow yes, thank you. $\endgroup$ – the man Jan 17 '16 at 20:22

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.