Taylor series proof

Question

I would like to ask for some tips and where can I find something to prove that:

$f''(x_0)= \frac{-f(x_0-3h)+4f(x_0-2h)-5f(x_0-h)+2f(x_0)}{h^2}+O(h^2)$

Thanks ${}$

Answer 1

Using second order Taylor-Lagrange formula, as $f''(x_0)$ exists, we can write

$$f(x_0-3h)=f(x_0)-3hf'(x_0)+\frac{9h^2}{2}f''(x_0)+O(h^3)$$

$$f(x_0-2h)=f(x_0)-2hf'(x_0)+\frac{4h^2}{2}f''(x_0)+O(h^3)$$

$$f(x_0-h)=f(x_0)-hf'(x_0)+\frac{h^2}{2}f''(x_0)+O(h^3)$$

$\implies$

$$-f(x_0-3h)+4f(x_0-2h)-5f(x0-h)+2f(x_0)=h^2f''(x_0)+O(h^3)$$

qed.