To be specific, focus on the following cyclic tridiagonal matrix:

$(A|b)=([4,-1,0,0,$dots,$0,0,0,-1,1],[-1,4,-1,0,$dots,$0,0,0,0,1],[0,-1,4,-1,$dots,$0,0,0,0,1],[vdots,$vdots,vdots,vdots,ddots,vdots,vdots,vdots,vdots,vdots$],[0,0,0,0,$dots,$-1,4,-1,0,1],[0,0,0,0,$dots,$0,-1,4,-1,1],[-1,0,0,0,$dots,$0,0,-1,4,1])$

Something similar appears when you study differential equations with periodic bound$-$

ary conditions, as we will see in chapter $8.($Note that $A$ is "almost" a tridiagonal ma$-$

trix.$)$ For this problem, the Jacobi iterative method of Eq$.(4\mathrm{.}169)$ turns into:

$x_0^{(k)}=(1+x_1^{(k-1)}+x_{n-1}^{(k-1)})\frac{1}{4}$

$x_i^{(k)}=(1+x_{i-1}^{(k-1)}+x_{i+1}^{(k-1)})\frac{1}{4},i=1,$dots,$n-2$

$x_{n-1}^{(k)}=(1+x_0^{(k-1)}+x_{n-2}^{(k-1)})\frac{1}{4}$

Implement the Jacobi iterative method for this set of equations. Your input parameters

should be only the total number of iterations, the tolerance, as well as $n($which deter$-$

mines the size of the problem$),$ i$.$e$.,$ there are no longer any A and $b$ to be passed in$.$

Choose a tolerance of ${10}^{-5}$ and print out the solution vectors for $n=10$ and for $n=20,$

each time also comparing with the output of numpy.linalg.solve$().($To produce

the latter, you will have to form A and $b$ explicitly.$)$ In python pls