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|$\frac{f(x+h)+f(x-h)-2f(x)}{h^{2}}-f''(x)$| is less than or equal to $h^{3}$ times some constant times the supremum of $|f''(y)|$ over y in R

Answer

$f'''(x)$ = $lim_{h\rightarrow 0} \frac{f(x+3h)-f(x+2h)}{h^{3}}-\frac{2f(x+2h)-f(x+h)}{h^{3}}+\frac{f(x+h)-f(x)}{h^{3}}$

$f'''(x)$= $lim_{h\rightarrow 0} \frac{f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^{3}}$

not sure how to proceed

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1 Answer 1

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Hint: Taylor formulas yield that $f(x \pm h)=f(x) \pm hf’(x)+\frac{h^2}{2}\int_0^1{(1-u)f’’(x \pm hu)du}$.

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  • $\begingroup$ A bit vague , but thats for trying $\endgroup$ Dec 18, 2018 at 21:55
  • $\begingroup$ That is why I wrote it was a hint (you even asked how to proceed, not the full solution). You can figure out the solution from this, especially if you realize, say, that $f’’$ is Lipschitz continuous with constant $\sup_y |f’’’|$. $\endgroup$
    – Aphelli
    Dec 18, 2018 at 23:47
  • $\begingroup$ thanks a lot for the hint $\endgroup$ Dec 20, 2018 at 9:19

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