find which function corresponds to $\displaystyle{\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} x^{2k}}$ I want to find which function corresponds to $\displaystyle{\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} x^{2k}}$.
That's what I have tried:
$$a_{2k-1}=0 \ \ \ \ \ \ \ \ \ \ \ \  \ \  \ \ \  \ a_{2k}=\frac{(-1)^k}{(2k)!}, \ \ \ \ \ \ k\in \mathbb{N}^{*}$$
$$\sqrt[2k-1]{|a_{2k-1}|}=0, \sqrt[2k]{|a_{2k}|}=\sqrt[2k]{|\frac{(-1)^k}{(2k)!}|}=\frac{1}{\sqrt[2k]{(2k)!}} \to 0$$
Therefore, $\sqrt[k]{|a_k|} \to 0$,so $\lim \sup \sqrt[k]{|a_k|}=0 \Rightarrow R=+\infty$,so the power series converges at $(-\infty,+\infty)$.
Let $\displaystyle{s(x)=\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} x^{2k}}, x \in \mathbb{R}$
$$s'(x)=\sum_{k=1}^{\infty} \frac{(-1)^k}{(2k)!}2k x^{2k-1}= \dots=- \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{(2k-1)!} x^{2k-1}$$
We consider the power series:
$$\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{(2k-1)!} x^{2k-1}$$
$$a_{2k}=0 \ \ \ \ \ \ \ \ \ \ \ \  \ \  \ \ \  \ a_{2k-1}=\frac{(-1)^{k-1}}{(2k-1)!}, \ \ \ \ \ \ k\in \mathbb{N}^{*}$$
$$\sqrt[2k]{|a_{2k}|}=0, \sqrt[2k-1]{|a_{2k-1}|}=\sqrt[2k-1]{|\frac{(-1)^{k-1}}{(2k-1)!}|}=\frac{1}{\sqrt[2k-1]{(2k-1)!}} \to 0$$
So,$R=+\infty $,therefore the powerseries converges at  $(-\infty,+\infty)$.
$\exists c(x):\mathbb{R} \to \mathbb{R} \text{ such that } c(x)=\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{(2k-1)!} x^{2k-1}$
We see that $ \displaystyle{ s'(x)=-c(x), x \in \mathbb{R}}$
$$c'(x)= \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{(2k-1)!} (2k-1) x^{2k-2}= \dots= \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k}}{(2k)!}=s(x)$$
We see that $\displaystyle{ c'(x)=s(x) \Rightarrow -s''(x)=s(x) \Rightarrow s''(x)+s(x)=0}$
$$\text{ The characteristic equation is } m^2+1=0 \Rightarrow m=\pm i$$
So,we have $\displaystyle{ s(x)=c_1 \cos{x}+ c_2 \sin{x}}$
$$s(0)=1 \Rightarrow c_1=1 , s'(0)=0 \Rightarrow c_2=0 $$
So,we have $s(x)=\cos{x} \text{ and } c(x)= \sin{x}$.
Could you tell me if it is right???

Hint: Note that $e^{z}=\sum_{k=0}^{\infty}\frac{z^{k}}{k!}$ so
$$e^{z}+e^{-z}=2\sum_{k=0}^{\infty}\frac{z^{2k}}{(2k)!}$$
