<s>[INST] Show that $|f'(x)-\frac{\overline{f(x+h)}-\overline{f(x-h)}}{2h}|\le \frac{M_1h^2}{6}+\frac{\epsilon M_2}{h}$ Suppose $f$ is a $C^3$ function on $[x-2h,x+2h]$. Suppose that the computed values $\overline{f(x+h)}$ and $\overline{f(x-h)}$ satisfy $\overline{f(x+h)}=f(x+h)(1+e_1)$ and $\overline{f(x-h)}=f(x-h)(1+e_2)$, where $e_i$ is error coming
from floating-point arithmetic that satisfy $|e_i|\le \epsilon$. 
Show that $|f'(x)-\frac{\overline{f(x+h)}-\overline{f(x-h)}}{2h}|\le \frac{M_1h^2}{6}+\frac{\epsilon M_2}{h}$, where $M_1 = \max_x |f'''(x)|, M_2 = \max_x |f(x)|$. 
I tried taylor expansion for $f(x+h)$ and $f(x-h)$, but I did not get what we want. 
 [/INST] \begin{align*}
\left|f'(x) - \frac{\hat{f}(x+h) - \hat{f}(x-h)}{2h}  \right|
&=
\left|f'(x) - \frac{f(x+h)(1+e_1) - f(x-h)(1+e_2)}{2h}  \right| \\
&= \left|f'(x) - \frac{f(x+h) - f(x-h)}{2h}  -\frac{f(x+h)e_1 - f(x-h)e_2}{2h}  \right| \\
&\le
\left|f'(x) - \frac{f(x+h) - f(x-h)}{2h}\right| + \left|\frac{f(x+h)e_1 - f(x-h)e_2}{2h}  \right| \\
&\le
\left|f'(x) - \frac{f(x+h) - f(x-h)}{2h}\right| + \frac{\epsilon M_2}{h} \\
\end{align*}
We also have
\begin{align*}
\left|f'(x) - \frac{f(x+h) - f(x-h)}{2h}\right| 
&=
\left|f'(x) - \frac{f(x)+ hf'(x) + h^2f''(x)/2 + h^3f'''(\xi)/6 - (f(x) - hf'(x) +h^2f''(x)/2 - h^3f'''(\xi)/6}{2h}\right| \\
&=
\left|f'(x) - \frac{hf'(x) + h^3f'''(\xi)/6 + hf'(x) + h^3f'''(\xi)/6}{2h}\right| \\
&=
h^2|f'''(\xi)|/6 \\
&\le h^2 M_{1}/6 \qquad \blacksquare
\end{align*}
</s>