I cannot the error in the bisection method...

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llucia
llucia 2023 年 4 月 6 日
コメント済み: llucia 2023 年 4 月 6 日
Hi! I am trying to detect why my code doesn´t work properly.
This is the code.
f = @(x) x^3 + 2*x^2 + 10*x - 20;
a = 0;
b = 1;
eps_f = 10e-10;
eps_x = 10e-10;
maxits = 100;
[x, n, error] = biseccion(f,a,b,eps_x, eps_f, maxits)
function [x,n,error] = biseccion(f,a,b,eps_x, eps_f, maxits);
x0 = a;
x1 = b;
n = 0;
error (1) = abs(x1-x0);
while n < maxits && (error(n+1) > eps_x || abs(f(x0)) > eps_f)
x = 0.5 * (x0 + x1);
if f(x0)*f(x) < 0
x1 = x;
else
x0 = x;
end
n = n+1;
error(n+1) = abs(x1-x0);
end
end

採用された回答

John D'Errico
John D'Errico 2023 年 4 月 6 日
編集済み: John D'Errico 2023 年 4 月 6 日
Get used to using the dotted operators for some things. That allows MATLAB to use vectorization in tools like fplot.
f = @(x) x.^3 + 2*x.^2 + 10*x - 20;
a = 0;
b = 1;
But the big problem is in f, and your limits.
fplot(f,[a,b])
grid on
Do you see a zero crossing in that interval? I don't.
Bisection CANNOT find a root in an interval where no root exists.
If you expand the interval, you might find it works better.
fplot(f,[0,2])
grid on
yline(0,'r')
A basic rule of using MATLAB is to plot EVERYTHING. Then plot something else. Then think about what you see. Do all of this before you just run code blindly.
  1 件のコメント
llucia
llucia 2023 年 4 月 6 日
Thanks a lot! I will do that.

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