I am not seeing what is wrong with my code, I've checked doc and help.

function [ root, error_bound ] = bisection(f, a0, b0, ep, max_iterate)
%function bisection(f, a0, b0, ep, max_iterate)
%This is the bisection method for solving an equation f(x)=0.
% Input:
%       f:  function to find root.
%       a0: Left endpoint of initial interval.
%       b0: Right endpoint of initial interval.
%       ep: Error tolerance.
%       max_iterate: Maximum iterates in loop, in case the code has an
%           infinite cycle.
% Output:
%       root
%       error_bound

      % =================Initializing=====================
      root = 0;
      error_bound = 0;
      % ==================================================
      % ============= YOUR CODE BEGINS HERE ==============
     a = a0;
     b = b0;
     for n = 0:max_iterate
      c = (a + b) / 2;
      if b - c <= ep
          disp('c is the root')
          break;
      end
      if sign(f(b)) * sign(f(c)) <= 0  %%this is where I am having an issue but I am sure this is right%%
          a = c;
      else
          b = c;
          return
      end
      root = c;
      error_bound = b - c;
      % ============== YOUR CODE ENDS HERE ===============
  end

[Andrei Bobrov]

Please read here.

[James T]

so, after reading what you showed me, is this the code I am looking for?

if sign(f(a0))*sign(f(b0)) > 0
    return
end
for n = 1:max_iterate

      c = (a0 + b0) / 2;
      if b0 - c <= ep
          break;
      end
      if sign(f(b0)) * sign(f(a0)) <= 0
          a0 =c;
      else
          b0 = c;
      end
  end
  root = c;
  error_bound = b0 - c;

Answer 1

[James Tursa]

You have an inadvertent return statement:

    else
        b = c;
        return  <-- get rid of this line

The way you have it now, the first time you enter this branch the function returns without doing any more bisection iterations.

[James T]

thanks man, is this a little better?

if sign(f(a0))*sign(f(b0)) > 0
      return
  end

    for n = 1:max_iterate
        c = (a0 + b0) / 2;
        if b0 - c <= ep
            break;
        end
        if sign(f(b0)) * sign(f(a0)) <= 0
            a0 =c;
        else
            b0 = c;
        end
    end
    root = c;
    error_bound = b0 - c;

[James Tursa]

I like the fact that you have added a check to see that the inputs have bracketed the root. However, for this case I would advise generating an error instead of returning with bogus 0 values. E.g.,

if sign(f(a0))*sign(f(b0)) > 0
    error('Inputs do not bracket the root');
end

You might also think about other issues to check for. E.g., what happens if a0 > b0 on input? What happens if ep < 0 on input? Etc.

But you have changed your working middle point check for an incorrect test. I.e., you have changed this line, which correctly tests where c is:

        if sign(f(b)) * sign(f(c)) <= 0

for this test, which doesn't test c at all:

        if sign(f(b0)) * sign(f(a0)) <= 0

Why did you make that change? Go back to the form of the original test which was working to test which side of the root c is on.