## Calculus 8th Edition

$f(x)+f''(x)=0$
$f(x)=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{x^{2n}}{2n}$, $f'(x)=\Sigma_{n=1}^{\infty}(-1)^{n}\frac{2nx^{2n-1}}{(2n)!}$ $f''(x)=\Sigma_{n=1}^{\infty}(-1)^{n}\frac{2n(2n-1)x^{2n-2}}{(2n)!}$ Let $f(x)+f''(x)=I$ Then $I=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{x^{2n}}{2n}+\Sigma_{n=1}^{\infty}(-1)^{n}\frac{2n(2n-1)x^{2n-2}}{(2n)!}$ After simplification $f''(x)=-f(x)$ Thus, $f(x)+(-f(x))=0$ or Hence, $f(x)+f''(x)=0$