Bisection method relative error

10 次查看（过去 30 天）
Sazcl
Sazcl 2022-3-17
评论： Jan 2023-8-2
Hello everyone, I don't use MATLAB very well. I have a question. If you can help, I'd appreciate.
I have a function below that I have to find its roots using bisection method. I want the for loop to stop on the point where relative error is lower than %0.05. I couldn't understand how I can define n.
f=@(x) log(x)-cos(x)-exp(-x);
x1=1;
x2=2;
xmid=(x1+x2)/2
for i=1:n;
if (f(xmid)*f(x2))<0
x1=xmid;
else
x2=xmid;
end
xmid=(x1+x2)/2;
end
fprintf('The root is: %3.8g\n',xmid)

请先登录，再回答此问题。

采纳的回答

Mohammed Hamaidi
Mohammed Hamaidi 2022-3-17
编辑：Mohammed Hamaidi 2022-3-18
Hi
Just use "while" loop with your condition as follows:
f=@(x) log(x)-cos(x)-exp(-x);
x1=1;
x2=2;
xmid=(x1+x2)/2;
while (x2-x1)>0.0005
if (f(xmid)*f(x2))<0
x1=xmid;
else
x2=xmid;
end
xmid=(x1+x2)/2;
end
fprintf('The root is: %3.8g\n',xmid)
  4 个评论
显示 2更早的评论
Sazcl
Sazcl 2022-3-17
It really helped, I got it done.
Thank you both.
Jan
Jan 2022-3-18
If this answer solves the problem, please accept it.

请先登录，再进行评论。

更多回答（1 个）

John
John 2023-7-31
function [p, pN] = Bisection_371(a,b,N, tol)
if f(a)*f(b) > 0
disp("IVT does not guarantee a root in [a,b]")
elseif f(a)*f(b) == 0
disp("The root is either a or b")
else
for n = 1:N
p = (a+b)/2;
pN(n) = p;
if f(p) == 0 || (b-a)/2 < tol
break
elseif f(p)*f(a) < 0
b = p;
else
a = p;
end
end
end
end
%f = @(x)x^2 - 1;
function y = f(x)
y = x^2 - 1;
end
  1 个评论
Jan
Jan 2023-8-2
For numerical reasons it is rather unlikely that the condiotion f(p) == 0 is met exactly. Use a tolerance instead.

请先登录，再进行评论。

类别

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

标签

产品

Community Treasure Hunt

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

Start Hunting!

Translated by