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Prove the following claim: Let $[a, b] ⊂ R $ and $x ∈ (a, b)$ be given. And E is the error of numerical differentation. If $f ∈ C^3 [a, b]$, then it holds that:

$| f'(x) - \frac{f(x+h)-f(x-h)}{2h}| \leq max_{E \in [a,b]} \frac{|f'''(E)|}{3!} h^2$

As I see it: I have to show the left side is less or equal to the right side. LHS is differentiated one time, RHS differentiated 3 times. My guess is to use Taylor polynomial, is that right? And if, then how to apply as a proof?

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  • $\begingroup$ "$E$ is the error of numerical differentation": nope. $\endgroup$
    – user65203
    Mar 21, 2021 at 19:14
  • $\begingroup$ No? Please elaborate? $\endgroup$
    – bestmate21
    Mar 21, 2021 at 19:17
  • $\begingroup$ Who told you so ? $\endgroup$
    – user65203
    Mar 21, 2021 at 19:17

1 Answer 1

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hint

$f $ is $ C^3([a,b],\Bbb R) $ . By Taylor-Lagrange formula, and for very small $ h $,

$$f(x+h)=f(x)+hf'(x)+\frac{h^2}{2}f''(x)+\frac{h^3}{6}f'''(c_1)$$ and $$f(x\color{red}{-h})=f(x)\color{red}{-h}f'(x)+....$$

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