# Proof of Approximation Formula for a Second Derivative

Using the formula [f(a+h)-f(a-h)]/2h = f prime of a, prove that
[f(a+h) - 2f(a) + f(a-h)]/h^2 = f double prime of a.

I get [f(a+h) - 2f(a) + f(a-h)]/2h^2

Could someone show me the proof of this? It's driving me crazy.

My work
[[2f(a+h) - 2f(a)] - [f(a+h) - f(a-h)]]/2h/h = the above (I assume incorrect) answer

#### zzzoak

I think that
$$\displaystyle f'(a+h/2)=[f(a+h)-f(a)]/h$$
$$\displaystyle f'(a-h/2)=[f(a)-f(a-h)]/h$$
$$\displaystyle f''(a)=[f'(a+h/2)-f'(a-h/2)]/h$$

$$\displaystyle f'(a+h/2)=[f(a+h)-f(a)]/h$$
$$\displaystyle f'(a-h/2)=[f(a)-f(a-h)]/h$$
$$\displaystyle f''(a)=[f'(a+h/2)-f'(a-h/2)]/h$$