Bisection method while loop not iterating, only gives the first answer.

Hello,

To start, I would like to say that I'm pretty new to Matlab and coding in general. I've tried to solve my issue for a while now and I guess it might be better to ask somebody that already knows :)

So, I wrote a matlab function of the bisection method and I have to use a while loop that iterates with new values of a and b until the length (|a-b|) < tolerance but my loop is not iterating and I don't understand why. Please help me!

This is what I did:

function [y,iterations] = bisection(f,a,b,tolerance);
f = @(x) 3*cos(4*x);
a = 0;
b = 1;


 FA = f(a);
 FB = f(b);
    
 if FA*FB > 0 %FA and FB must have opposite signs.
     fprintf('The opposite sign requirement is not met, we need new values of a and b') 
 end
    
 if FA == 0 | FB == 0 %Neither FA or FB can be 0 because that would be a root.
     fprintf('the function evaluated at a or b is a root')  
 end

tolerance = 1^(-6); 
length = abs(a-b);
iterations = 0;

while length > tolerance
                  
    c = FA+FB/2; %Find the midsection
    FC = f(c); 
    
    if FC == 0 %Same as above, FC can't be 0, because that means we found a root.
        fprintf('The midsection is a root')
        break
    end
        
    %Updating values of a and b accordingly.
    
    if FC*FA>0 
        c=a;
    else
        c=b;
    end
    
    length; %calculates the new length
    iterations = iterations + 1;
    
end

y = (a+b)/2
iterations

end
    

[Walter Roberson]

    length; %calculates the new length

No it doesn't. That just brings the value of length to attention, then discards it without doing anything with it because there is a semi-colon at the end that tells it not to output anything.

[Sylvina Barreto]

I see, could you provide some more insight on how to solve it then? If I remove the semicolon and add the formula to that line nothing different happens :/

[Walter Roberson]

You should be recalculating length at that point.

Answer 1

[Walter Roberson]

    if FC*FA>0 
        c=a;
    else
        c=b;
    end

Why are you assigning to c there? You never use c afterwards. If you made it to another iteration of the loop, the next loop would do

c = FA+FB/2; %Find the midsection

which would overwrite c based upon FA and FB, neither of which are changed in your loop.

    c = FA+FB/2; %Find the midsection

That line has multiple errors. You need to think a lot more about that line.