I decided to use the mean value theorem.Suppose $\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}$ is twwice differentiable and $\displaystyle a \in \mathbb{R}$. Prove that $\displaystyle \forall \ h>0 \ \exists \ \xi \in(a-h,a+h)$ with

$\displaystyle f(a+h)-2f(a)+f(a-h)=h^2f''(\xi)$

Hint: consider the function $\displaystyle \phi$ given by

$\displaystyle \phi(t)=f(a+t)-2f(a)+f(a-t)-\left( \frac{t}{h} \right)^2(f(a+h)-2f(a)+f(a-h))$.

Let $\displaystyle \xi_1 \in(0,h]$ and $\displaystyle \xi_2 \in [-h,0)$.

Then $\displaystyle \xi=\xi_1 \cup \xi_2$.

(this is done so that $\displaystyle \xi \neq 0$).

MVT gives: $\displaystyle \phi'(t_0)=\frac{f(a+h)+f(a-h)-f(a+h)-f(a-h)-f(a)+2f(a)-f(a)}{h}=0$ using the interval $\displaystyle (0,h]$.

I have no idea why this helps me (if it does!).