Show that the power series is solution to f ''(x) + f(x) = 0 Show that

$$f(x)=\sum_{n=0}^∞ \frac{(-1)^nx^{2n}}{(2n)!}$$

is a solution of:
$$f''(x)+f(x)=0$$



It seems that $f''x$ is just $-f(x)$ $=&gt;$ $-f(x)+f(x)=0$, but when I try to get the solution  I get a weird power series that is not $-f(x)$. Can anyone tell me what went wrong? See below:

$$f'(x)=\sum_{n=1}^∞ \frac{(-1)^nx^{2n-1}*(2n)}{(2n)!}$$
 $$f''(x)=\sum_{n=2}^∞ \frac{(-1)^nx^{2n-2}*(2n)*(2n-1)}{(2n)!}$$

$$f''(x)=\sum_{n=0}^∞ \frac{(-1)^{n+2}x^{2n+2}*(2n+4)*(2n+3)}{(2n+4)!}$$
 $$=\sum_{n=0}^∞ \frac{(-1)^{n+2}x^{2n+2}}{(2n+2)!}$$

Anyone know what mistake I made? I need $f''(x)$ to be $-f(x)$
You have made mistakes in changing $n$ to $n+2$. In $f''(x)$, $x^{2n-2}$ the becomes $x^{2n+2}$ and $(2n)(2n-1)$ becomes $(2n+4)(2n+3)$. Also $(2n)!$ becomes $(2n+4)!$

Answer for the edited version. To compare your series for $f''(x)$ with the series for $f(x)$ you have to change $n$ to $n-1$ (so that $x^{2n+2}$ becomes $x^{2n}$).
‘’ has to star with n=1 not n=2. And be more careful when reindexing to 0, alternating power will switch by 1, x power by two.