フィルターのクリア

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

1 回表示 (過去 30 日間)
Joe
Joe 2013 年 4 月 11 日
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 件のコメント
Matt J
Matt J 2013 年 4 月 11 日
編集済み: Matt J 2013 年 4 月 11 日
Show us what error you're getting and how to reproduce it.
Joe
Joe 2013 年 4 月 12 日
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.

サインインしてコメントする。

回答 (1 件)

Roger Stafford
Roger Stafford 2013 年 4 月 12 日
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.

カテゴリ

Help Center および File ExchangeShifting and Sorting Matrices についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by