# Bisection method relative error

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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)
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### Risposta accettata

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)
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Sazcl il 17 Mar 2022
It really helped, I got it done.
Thank you both.
Jan il 18 Mar 2022

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### Più risposte (1)

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
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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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