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Bisection method relative error

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Sazcl
Sazcl il 17 Mar 2022
Commentato: Jan il 2 Ago 2023
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)

Risposta accettata

Mohammed Hamaidi
Mohammed Hamaidi il 17 Mar 2022
Modificato: Mohammed Hamaidi il 18 Mar 2022
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 Commenti
Sazcl
Sazcl il 17 Mar 2022
It really helped, I got it done.
Thank you both.
Jan
Jan il 18 Mar 2022
If this answer solves the problem, please accept it.

Accedi per commentare.

Più risposte (1)

John
John il 31 Lug 2023
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 Commento
Jan
Jan il 2 Ago 2023
For numerical reasons it is rather unlikely that the condiotion f(p) == 0 is met exactly. Use a tolerance instead.

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