Bisection Method vs Fixed-Point Iteration

How much faster is FPI compared to the Bisection Method? For that we need to think about accuracy. Or the error compared to the true solution. Now I know we don't know the true solution otherwise we would not be needing these numeric methods! But we can enclose the true solution inside an interval where we know that it lies in it (well for a continuous functions in the interval of interest).

Starting from The Bisection Method we know for each step the interval [a,b] is cut in half! After n steps our new interval [an, bn] have the length (b-a)/2^n. But choosing the solution to be the midpoint of the n-interval we know the true solution (if it is not the midpoint) is within an interval of length (b-a)/2^(n+1).

A solution is correct within p decimal places if the error |xc-r|<0.5*10^-p.

How many times do we need to cut the interval [0,1] in half to get an approximation with 6 correct digits for the problem f(x)=x^3+x-1 using the Bisection Method?

Solving (b-a)/2^(n+1) < 0.5*10^(-6) => n = 19.9. Which means The Bisection method will need 20 steps for 6 correct digits. The approximated solution with 6 correct digits is xc=0.682327.

Using FPI with g(x)=(1+2x^3)/(1+3x^2) and an initial guess x0=1. After 4 steps we get the same solution. A major improvement of number of steps by using FPI with a "nice" g(x)!