How is the Taylor expansion for $f(x + h)$ derived? According to this
Wikipedia article, the expansion for $f(x\pm h)$ is:
$$f(x \pm h) = f(x) \pm hf'(x) + \frac{h^2}{2}f''(x) \pm \frac{h^3}{6}f^{(3)}(x) + O(h^4)$$
I'm not understanding how you are left with $f(x)$ terms on the right hand side.
I tried working out, for example, the Taylor expansion for $f(x + h)$ (using $(x+h)$ as $x_0$) and got this:
$$ f(x + h) = f(x+h) + f'(x + h)(x-(x+h)) + \frac{f''(x+h)}{2!}(x-(x+h))^2 + \frac{f'''(x+h)}{3!} (x - (x + h))^3 + \cdots $$
$$ = f(x + h) - hf'(x+h) + \frac{h^2}{2!}f''(x + h) - \frac{h^3}{3!} f'''(x+h) + \cdots$$
Am I doing this correctly?
 A: It looks as if the notation you are accustomed to for the Taylor expansion is something like
$$f(x)\approx f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+ \frac{f'''(x_0)}{3!}(x-x_0)^3+\cdots.$$
Now write $x$ instead of $x_0$, and $x\pm h$ instead of $x$. 
A: This has already been answered, but I have seen some of the comments to the answer from Andre Nicolas and I thought I could make it clear with a step by step explanation.
Instead of just doing a substitution, as Andre mentioned, think about the meaning of it.
Start with the standard Taylor series expansion,
$$ f(x) \approx f(x_0)+f′(x_0)(x−x_0)+\frac{f′′(x_0)}{2!}(x−x_0)^2+\frac{f′′′(x_0)}{3!}(x−x_0)^3+⋯. \qquad (*)$$
Now what does $x-x_0$ mean? For convergence, we usually need this to be small, so we can call this $h$. Now substitute $x-x_0=h$ (and obviously $x=x_0+h$) into $(*)$ to get:
$$ f(x_0+h) \approx f(x_0)+f′(x_0)h+\frac{f′′(x_0)}{2!}h^2+\frac{f′′′(x_0)}{3!}h^3+⋯.$$
This is in the required form but with $x_0$ instead of $x$ and since this is just a variable we can just substitute it for $x$.
