Attempt:
a) $g(x) = x- f(x)/f’(x) = x – [(x-2)^4]/[4(x-2)^3] = (2-x)/4+x = (3x+2)/4$
So, $p_k = [3p_{k-1}+2] /4$
b)
p(1)=2.1; for j=2:5, p(j) = (3*p(j-1)+2)/4 end
p = 2.1000 2.0750 2.0563 2.0422 2.0316
So, p(0)=1
p(1)=2.0750
p(2)=2.0563
p(3)=2.0422
p(4)=2.0316
c) Tried this:
for i=1:4 p(i+1)-p(i) end
ans = -0.0250
ans = -0.0187
ans = -0.0141
ans = -0.0105
So, p(1)-p(0) = -0.0250
p(2)-p(1)= -0.0187
p(3)-p(2)= -0.0141
p(4)-p(3) = -0.0105
What next?
The equation $(x-2)^4=0$ has the obvious root $x=2$.
We have $$p_{n+1}-2=\frac{3p_n+2}{4}-2=\frac{3}{4}(p_n-2).$$ Thus $$|p_{n+1}-2|=\frac{3}{4}|p_n-2|.$$
So the distance from the root is multiplied by the factor $\frac{3}{4}$ for each iteration. This is linear convergence.
Remark: Note that the behaviour of the Newton Method in this case is worse than the behaviour of the Bisection Method.
The reason for the bad behaviour is that the derivative of our function $(x-2)^4$ is $0$ at the root.
Given the recurrence you have derived, it is apparent that
$$p_n = 2 + \left ( \frac{3}{4} \right )^n (p_1 - 2)$$
so that
$$p_{n+1}-p_n = \frac{3}{4} (p_n-p_{n-1})$$
Your recurrence converges to the root $x=2$ linearly.
