Bisection method not returning any values.

Question

David 2013-10-16

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https://ww2.mathworks.cn/matlabcentral/answers/90441-bisection-method-not-returning-any-values

评论： Matt J 2013-10-16

So, I've written the following code for the bisection method, however it doesn't seem to be returning any values.

function [r, n] = bisection(a,b,f,tol)
if (f(a)*f(b) > 0)
    error('Invalid choice of interval');
end
r = 0;
n = 0;
while ((b - a)/2 > tol)
    n = n + 1;
    r = (a + b)/2;
    if(f(r) == 0)
        break;
    elseif (f(a)*f(b) < 0)
        b = r;
    else
        a = r;
    end
end

I'm using it with inputs of (0,1,q,0.5) I also tried tol as 0.1 too. q is some anonymous function that I defined, in this case it's q(x) = 3x - 1. Like, I type the command Bisection(0,1,q,0.5) and it executes without returning an error but doesn't return any numerical values whatsoever. Any ideas why this is?

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Answer 1

Matt J 2013-10-16

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https://ww2.mathworks.cn/matlabcentral/answers/90441-bisection-method-not-returning-any-values#answer_99915

编辑：Matt J 2013-10-16

Works fine for me. It returns the wrong value, but you can fix that by changing f(a)*f(b) < 0 to f(a)*f(r) < 0.

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David 2013-10-16

Actually, one further question. If I enter a quadratic equation, which may have 2 distinct roots,is there anyway to get it to output both roots? Or, would this be a much more complicated program.

Matt J 2013-10-16

编辑：Matt J 2013-10-16

No, there is no general algorithm for finding "all" roots of a general function without at least knowing in advance a lower bound on the spacing between the roots.

I assume, by the way, that this is just a homework exercise. In reality, you would never use your own home-brewed bisection algorithm. You would just use FZERO. So, there is no obvious rationale for giving this code more capabilities then the homework assignment requests.

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