Simple Fixed Point Iteration MATLAB

Question

I have attempted to code fixed point iteration to find the solution to (x+1)^(1/3). I keep getting the following error: error: 'g' undefined near line 17 column 6 error: called from fixedpoint at line 17 column 4

    clear -all;
clc;

function f = f(x)

    f = (x+1)^(1/3)

    f = g(x)
end

# Start out iteration loop
x1 = 0;
x2 = g(x1);

iterations = 0; # iteration counter

while abs(x2-x1 > 1e-5)
    plot([x1 x1], [x1 g(x1)], 'k-')
    plot([x1 x1], [x1 g(x1)], 'k--')
    pause
    x1 = x2;
    x2 = g(x1);
    iterations = iterations + 1;
end

iterations
x1
x2

I have no idea what is wrong. My logic seems to be correct.

Hunter Jiang · Accepted Answer · 2018-02-21 16:03:55Z

I modified your code a little, it could get the solution of f(x)=cos(x)-x, and you could change g(x) to whatever you want.

clear;
clc;

%# You function here
g=@(x) cos(x);

%# Start out iteration loop
x1 = 0;
x2 = g(x1);

iterations = 0;% # iteration counter

ezplot(g,[0,1]);
hold on 
ezplot('x',[[0,1]])

while (abs(x2-x1) > 1e-5 && iterations<100)
    plot([x1 x1], [x1 x2], 'k-')
    plot([x1 x2], [x2 x2], 'k--')
     %pause
    iterations = iterations + 1;
    x1 = x2;
    x2 = g(x1);
end
iterations 
[x1 x2]

And the solution will be:

iterations =

    29

ans =

    0.7391    0.7391

With the output image:

However, are you certain that your function is right? It seems that this function could not use Fixed Point Iteration to solve, since f(x)=0 equals to g(x)=x and g(x)=(x+1)^(1/3)+x here. But if we plot g(x)(blue curve) with h(x)=x(red curve), we have:

So if we start at 0, the iteration can't convergence (x1 will increase dramatically but the root is -1).

Hope it helps!

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