$f'(x) - $ $f(x+h) - f(x-h) \over 2h$ $\in O(h^2)$

Question

Let $f$ be 3 times continuously differentiable. How do I show that $f'(x) - {f(x+h) - f(x-h) \over 2h}\in O(h^2)$?

By the intermediate value theorem, I know that there exists a $\xi \in [x-h,x+h]: f'(\xi) =\frac{f(x+h) - f(x-h)}{2h}$. I think I could use Taylor but I doesn't lead me to any solution.

Answer 1

By Taylor,

$$f(x+h)=f(x)+f'(x)h+f''(x){h^2 \over 2} + O(h^3)$$ $$f(x-h)=f(x)-f'(x)h+f''(x){h^2 \over 2} + O(h^3)$$

So,

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

Answer 2

Apply Taylor's theorem thusly:

$$f(x+h) = f(x) + hf'(x) + h^2f''(x) + O(h^3)$$ and $$f(x-h) = f(x) - h f'(x) + (-h)^2f''(x) + O((-h)^3) = f(x) - h f'(x) + h^2f''(x) + O(h^3)$$ for sufficiently small $|h|$. Now subtract:

$$f(x+h) - f(x-h) = 2h f'(x) + O(h^3).$$

Answer 3

Note that $g(h)=f(x+h)-f(x-h)$ is an odd function of $h$, so any constant terms and quadratic terms in $h$ will cancel. Expanding about $x$ then gives

$$g(h)=2hf'(x)+O(h^3)$$ which can be rearranged to the form you want. The conditions on differentiability ensure that the expansion is valid.

This is perhaps not the indicated proof, but the observation that $\frac {g(h)}2$ is the odd part of $f(x+h)$ considered as a function of $h$ perhaps illuminates what is going on here.