Finding a function from the derivatives at 0

Question

I'm looking for a function like that:$$\frac{d^nf(x)}{dx^n}\mid_{x=0}=\frac{n!}{(2n)!}$$I know, there are infinitely many of those, I just need one of them

Answer 1

If $f^{(n)}(0)=\frac{n!}{(2n)!}$, then $\frac{f^{(n)}(0)}{n!}=\frac1{(2n)!}$. So, you can take$$f(x)=\sum_{n=0}^\infty\frac{x^n}{(2n)!}.$$You can deduce from the ratio test that this definition makes sense, since this series converges everywhere. You will get a function $f$ such that $(\forall x\in\Bbb R):f(x^2)=\cosh(x)$.