I cannot the error in the bisection method...

1 次查看（过去 30 天）
llucia
llucia 2023-4-6
评论： llucia 2023-4-6
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

请先登录，再回答此问题。

采纳的回答

John D'Errico
John D'Errico 2023-4-6
编辑：John D'Errico 2023-4-6
Ran in:
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.

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 NaNs 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by