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I try to solve the following task. It is given a finite difference: $$g(x)=f'(x)\approx\frac{f(x+h)-f(x)}{h}$$ $$g'(x)\approx\frac{g(x)-g(x-h)}{h}\approx\frac{\frac{f(x+h)-f(x)}{h}\;-\;\frac{f(x)-f(x-h)}{h}}{h}$$ $$\iff g'(x)\approx\frac{f(x+h)-2f(x)+f(x-h)}{h^2}\approx f''(x)$$ I shall determine the order of approximation of $f''$ depending on Big-O.

The order of approximation should say something about the quality of the approximation (size of the possible error?). There wasn't much explanation in the lecture so I am not sure. But they want us to give the order of approximation depending on $\mathcal(O)$ (Landau symbol). (Definition of Big-O is clear)

On our frames is only a barely explained example in which something with Taylor Expansions is done. Still without further explanations it is hard for me to understand what I have to do...

I really hope for your help - please explain very detailed because I am a computer scientist and not a mathematician and could probably missing some more in detail information about certain subjects.

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Using the Taylor expansion around $h=0$ $$f(x+n h)=f(x)+h n f'(x)+\frac{1}{2} h^2 n^2 f''(x)+\frac{1}{6} h^3 n^3 f'''(x)+O\left(h^4\right)$$ then $$\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=\frac{h^2 f''(x)+O\left(h^4\right)}{h^2}=f''(x)+O\left(h^2\right)$$

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  • $\begingroup$ Okay, now I feel dumb... I completly missed the fact that it is just insert and calculate... Thank you very much - in the end pretty simple :D $\endgroup$
    – baxbear
    Apr 29, 2018 at 14:22
  • $\begingroup$ I just have on open question in mind: a more precise Taylor expansion should lead to a smaller error I suppose. But If the error is in $\mathcal{O}(h^2)$ instead of in $\mathcal{O}(h)$ for example shouldn't this implicate a growing error instead of a decreasing one? Where is my mistake? $\endgroup$
    – baxbear
    Apr 29, 2018 at 14:34
  • $\begingroup$ @baxbear. You are very welcome ! Higher $n$ in $\mathcal{O}(h^n)$, better ! Have a look at math.stackexchange.com/questions/2461506/… $\endgroup$
    – Claude Leibovici
    Apr 29, 2018 at 14:53

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