I do not know what you are going to write for Part 2 and Part 3. For me, the logic to use (1-x)^(1/3) is clear since the original equation x^3+x-1=0 is equivalent to x=(1-x)^(1/3). Therefore, the iteration is naturally derived as:

x_k+1 = (1-x_k)^(1/3)

Certainly, there are many ways to work out an iteration like this, e.g. from x = 1 - x^3, we can have iteration

x_k+1 = 1 - x_k^3

However, this iteration is unstable for 3*x_k^2 > 1, i.e. x_k > 0.5774. Since the solution is x=0.68233 > 0.5774, the second iteration is not suitable for this problem.

The advantage of my solution is that

1. It works for fractional order, e.g. x^3.25 + x - 1 =0, which cannot be solved by roots function.
2. It can be vectorised for different constants other than 1:

x^3.25 + x - a = 0, for a = 1:10.

a=1:10;
x0=0.5;
x1=0;
while norm(x1-x0)>1e-6,
   x1=x0;
   x0=(a-x1).^(1/3.25);
end
disp(x0)
err=x0.^3.25+x0-a

If we use fzero:

fz=zeros(1,10);
for a=1:10
  f=@(x)x.^3.25+x-a;
  fz(a)=fzero(f,0.5);
end
disp(fz)
err=fz.^3.25+fz-(1:10)

Only the first a=1 is solved. Other a’s end up with NaN!

Yi

True, many ways, but I am thinking of how to derive and explain an algorithm to students. How would someone know know to use (1-x)^(1/3)? Wait for my next post to my ideas on how to explore and teach some of these concepts.

–Loren

x0=0.5;
x1=0;
while abs(x1-x0)>1e-6,
   x1=x0;
   x0=(1-x1)^(1/3);
end
disp(x0)