What's wrong with my bisection method function???

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Joe
Joe el 11 de Abr. de 2013
I'm not quite sure what's exactly wrong with my bisection method function that I have written. Can someone please help me figure out what the error is?
function root = bisectIter(f,a,b,tol)
if sign(f(a))==sign(f(b))
error('a and b do not bracket the root');
end
k = 1;
x(k) = (a+b)/2;
while ((k<=tol)&&((b-a)/2)>= tol)
if f(x(k)) == 0
error('bisection condition didnt apply')
end
if (f(x(k))*f(a))<0
b = x(k);
else
a = x(k);
end
k = k + 1;
x(k) = (a+b)/2;
root = x(k);
end
end
  2 comentarios
Matt J
Matt J el 11 de Abr. de 2013
Editada: Matt J el 11 de Abr. de 2013
Show us what error you're getting and how to reproduce it.
Joe
Joe el 12 de Abr. de 2013
The error "Output argument 'root' not assigned during call..." shows up. I've already checked for the possible causes of this error and none seem to fit. When I alter my code so that it should produce an error when I run it, it does in fact produce that error instead of the output argument error.

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Respuestas (1)

Roger Stafford
Roger Stafford el 12 de Abr. de 2013
In the 'while' condition "((k<=tol)&&((b-a)/2)>= tol)" presumably you have set 'tol' to some very small number to allow a and b to approach each other closely. That means the "k<=tol" part will fail at the very beginning and you will never enter the while-loop. That is why the 'root' argument is never assigned.
In my opinion your 'while' criterion should be based on how close to zero the f function gets instead of the width b-a or the number of trips through the loop.
Also the "f(x(k)) == 0" condition that produces an error message is rather self-defeating. This is the very condition you are striving for, namely to find a root.

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