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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?

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    It is $x$ which is the "$x_0$", and $x\pm h$ which is the "$x$".2012-12-09

2 Answers 2

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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 $x$.

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    @rjkaplan I think the resulting series is no longer a Taylor series; it is just a series obtained from Taylor expansion.2018-11-24