Modify the code to verify numerically that the approximation of $f^{\text{'}}(x_0)$ given by the formula

$\frac{f(x_0-2h)-8f(x_0-h)+8f(x_0+h)-f(x_0+2h)}{12h}$

is of order $4,$ namely the truncation error goes to zero as $h^4.$

from numpy import sin$,$ cos

from matplotlib import pyplot as plt

$x0=1$

$N=8$

$h=[0]**N$

err $=[0]**N$

for $i$ in range $(N)$ :

$h[i]=\frac{1}{10}****i$

err$[i]\frac{?}{a}=bs(\frac{sin(x+h[i])-sin(x-h[i])}{h}\frac{i}{2}-cos(x))$

plt$.$clf$()$

plt$.$ for $i$ in $h$

plt$.$ legend$([\text{'}$error$\text{'},$ 'slope $1\text{'},$ 'slope $2\text{'}],$ loc$=$'upper right'$)$

plt$.$show$()$