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