Proof of Approximation Formula for a Second Derivative

May 2010
2
0
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
 
Mar 2010
280
122
I think that
\(\displaystyle f'(a+h/2)=[f(a+h)-f(a)]/h\)
\(\displaystyle f'(a-h/2)=[f(a)-f(a-h)]/h\)
\(\displaystyle f''(a)=[f'(a+h/2)-f'(a-h/2)]/h\)