Matlab while loop issues bisection

Question

I am new to MATLAB and have encountered a problem on an assignment. The assignment is to use the bisection method. My original code was:

a = -0.5;
b = 1.1;
f=@(x) x^5-2*x^4+3*x^3-2*x^2+x-1-cos(30*x);
val1 = f(a);
val2 = f(b);
p1a1 = val1*val2;
save('A1.dat','p1a1','-ascii')

%bisection

for i = 0:100
    c = (a+b)/2;
    if f(c) >= -0.000000001 && f(c) <= 0.000000001
        answer = c;
    elseif f(c) > 0
        b = c;
    else a = c;
    end
end
answer = c;
save('A2.dat','answer','-ascii')

However, I needed to count the number of iterations it took, so I changed the second part of the code (after "bisection") to:

tolerance = 0.000000001;
count = 0;
c =(a+b)/2;
while abs(f(c))>tolerance
    count=count+1;
    if f(c) > 0
        b = c;
    else
        a = c;
    end
end
answer = c;

Where c would show the zero of the function, within the prescribed tolerance. The new code won't run however, and I cannot seem to figure out where i have gone wrong. Apologies if it is something simple

to avoid potential infinite loop, stopping criterion must be |b-a|<eps. Reason: for numerically inaccurate f(.) function, function without zero or for very small eps you are not sure that |f(c)|<eps is ever fulfilled. On the other side |b-a|<eps will happen for sure. — Picaud Vincent
– Picaud Vincent, Commented Oct 19, 2018 at 19:21
@PicaudVincent Note that <= would probably be better than <, given that (by definition) eps is the smallest representable value. Also eps = 2.2e-16 on my machine, the tolerance used in the question should be of no concern for numerical accuracy. — Wolfie
– Wolfie, Commented Oct 19, 2018 at 19:58

Wolfie · Accepted Answer · 2018-10-19 19:56:24Z

1

You're not updating the value of c inside the loop, it never changes which means you're stuck in the while loop. Copy the c = (a+b)/2 definition to inside the loop, as you had it before.

c = (a+b)/2;
while abs(f(c))>tolerance
    count=count+1;        
    if f(c) > 0
        b = c;
    else
        a = c;
    end
    c = (a+b)/2; % <<<<<< This is needed to avoid an infinite loop!
end
answer = c;

The bisection method is guaranteed to converge to a root (if it exists), but you should be careful with while loops for numerical methods. In particular, add a maximum iteration counter

maxIters = 1000;
c = (a+b)/2;
while abs(f(c)) > tolerance && count < maxIters
    % Code as before ...
end

answered Oct 19, 2018 at 19:56

Wolfie

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2 Comments

Pat LaChapelle Over a year ago

Awesome, thank you very much. I tried to put c in the while loop, but I put it at the top, before the "count=count+1. Now it makes sense. Also, thank you for the helpful tip about setting the max number of iterations. This will undoubtedly help with future coding!

Wolfie Over a year ago

@Pat No problem, consider marking this answer as accepted (the green tick mark under the voting buttons) if it answers your question!

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