Bisection method relative error

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)

Answer 1

[Mohammed Hamaidi]

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)

[Sazcl]

Thanks for your time.

But as far as I know, error tolerance is calculated by (upper limit-lower limit)/(2^iteration number) when the exact root is not given in the question.

Actually your code gives the right answer but I don't think it's what the question asks.

[Jan]

@Sazcl: If you do have the mathematical definition of "relative error", it should be easy to insert it in the posted code. Simply use it as condition in the WHILE command.

To avoid an infinite loop, add a counter, which stops the loop after a certain limit, e.g.:

kMax = 1e6;
k    = 0;
while <InsertYourCondition> && k < kMax
    ...
    k = k + 1;
end
if k == kMax
    error('No convergence')
end

But think twice: Under which circumstances is this possible?

By the way, f(xmid)*f(x2) < 0 does not catch the cases in which xmid or x2 is exactly the root. What a pity, if the root way found and the iteration goes on.

[Sazcl]

It really helped, I got it done.

Thank you both.

[Jan]

If this answer solves the problem, please accept it.

Answer 2

[John]

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

[Jan]

For numerical reasons it is rather unlikely that the condiotion f(p) == 0 is met exactly. Use a tolerance instead.