Taylor's Theorem Expansions to equate centered difference formula with O(h^2) As I am really struggling with my Numerical Method's class, I am trying to prepare for an upcoming exam with some example problems. This being one of them, I am not exactly sure how to even approach this fully. I have looked at this post Using Taylor Polynomial To Derive Centered Difference and think it is fairly similar with what I am trying to do, but there was no finishing responses to this post.
The question is:
Use Taylor's Theorem expansions about both $$f(x + h) $$$$ f(x - h)$$ to show that the centered difference formula $$f^`(x) = \frac{f(x+h)-f(x-h)}{2h}$$ is $$O(h^2)$$

Like the post, I started off with
$$f(x+h) = f(x) + f^`(x)h + O(h^2)$$
$$f(x-h) = f(x) - f^`(x)h + O(h^2)$$
and subtracted them and end up getting the centered difference
$$\frac{f(x+h)-f(x-h)}{2h}$$
Does this just mean that because I was able to get this from Taylor's Theorem expansions, I can just set this value equal to the remaining $$O(h^2)$$ and be done?

I am very confused on how this problem can be finished and hoping to get an idea around it.
 A: Starting from Taylor's series definition for a real value a (arbitrarily selected):
$$f(x)=\frac{f^{(0)}(a)}{0!}(x-a)^0+\frac{f^{(1)}(a)}{1!}(x-a)^1+\frac{f^{(2)}(a)}{2!}(x-a)^2+...$$
which can be written simpler as 
$$f(x)={f(a)}+\frac{f^{(1)}(a)}{1!}(x-a)+\frac{f^{(2)}(a)}{2!}(x-a)^2+...$$
In our case it's important to remember that h is a very small number, so the bigger exponent is over h, the smaller resulting value.
Now let's add some substitutions:


*

*Our function will be f(x+h)

*Expansion at point zero


$$f(x+h)={f(0)}+\frac{f^{(1)}(0)}{1!}(x+h)+\frac{f^{(2)}(0)}{2!}(x+h)^2+...$$
Same thing for f(x-h):
$$f(x-h)={f(0)}+\frac{f^{(1)}(0)}{1!}(x-h)+\frac{f^{(2)}(0)}{2!}(x-h)^2+...$$
Let's make a substitution:
$$\phi_i=\frac{f^{(i)}(0)}{i!}$$
Then
$$f(x+h)=\phi_0+\phi_1(x+h)+\phi_2(x+h)^2+\phi_3(x+h)^3+...$$
And
$$f(x-h)=\phi_0+\phi_1(x-h)+\phi_2(x-h)^2+\phi_3(x-h)^3+...$$
Let's subtract:
$$f(x+h)-f(x-h)=\phi_1(2h)+\phi_2(4hx)+\phi_3(6hx^2+2h^3)+...$$
Let's replace dots with big O:
$$f(x+h)-f(x-h)=\phi_1(2h)+\phi_2(4hx)+\phi_3(6hx^2+2h^3)+O(h^4)$$
It's useful to extract factor 2h since that's the denominator in final equation:
$$f(x+h)-f(x-h)={2h}{(\phi_1+\phi_2(2x)+\phi_3(3x^2+h^2)+O(h^3))}$$
So:
$$\frac{f(x+h)-f(x-h)}{2h}={\phi_1+\phi_2(2x)+\phi_3(3x^2+h^2)+O(h^3)}$$
Since on the right side we still have squared h:
$$\frac{f(x+h)-f(x-h)}{2h}={\phi_1+\phi_2(2x)+O(h^2)}$$
Since h is a very small value, big O for smaller exponent dominates over one with larger exponent.
I guess that's a verbose answer to the question. 
Question's importance may come from the fact that standard form of numerical derivative has error of O(h) order. Discussed under this wiki page.
