Find function whose power series is $\sum\frac{(-1)^n}{(2n)!}x^n$

Question

Let $(a_n)_{n \ge 0}$ be a real sequence such that $\forall n \in \mathbb N, a_n = \frac{(-1)^n}{(2n)!}$.

Find the radius of convergence $R$ of $\displaystyle \sum a_nx^n$ and the function $f$ such that $ \displaystyle \forall x \in ]-R,R[, f(x) = \sum_{n = 0}^{\infty}a_nx^n$.

The first part of the question is pretty easy. Since $\forall x \in \mathbb R$ the series $\displaystyle \sum \frac{x^n}{(2n)!}$ converges, we find $R = \infty$.

But I can't figure out how to find the function $f$. I think it has something to do with $\cos$, because $\displaystyle \forall x \in \mathbb R, \cos x = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}$.

Any help would be apreciated.

Answer 1

Hint $$(x^{\frac12})^{2n}=x^n$$ Replace $x$ with $x^{1/2}$ in the $\cos x$ series...

As has been noted, this only works for $x\geq0$

Edit

For $x\geq0$, $$f(x)=\cos\sqrt x$$ For $x<0$, $$f(x)=\cosh\sqrt{-x}$$

Answer 2

Hint For $x \geq 0$

$$\sum\frac{(-1)^n}{(2n)!}x^n= \sum\frac{(-1)^n}{(2n)!}(\sqrt{x})^{2n}$$

For $x <0$ you have $$\sum\frac{(-1)^n}{(2n)!}x^n= \sum\frac{1}{(2n)!}(\sqrt{-x})^{2n}$$ which is related to the Taylor series of $\cosh(x)$.