# Thread: Proof of Approximation Formula for a Second Derivative

1. ## 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

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

3. Originally Posted by zzzoak
I think that
$\displaystyle f'(a+h/2)=[f(a+h)-f(a)]/h$
$\displaystyle f'(a-h/2)=[f(a)-f(a-h)]/h$
$\displaystyle f''(a)=[f'(a+h/2)-f'(a-h/2)]/h$
Genius!