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.


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

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

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