Here is my code for false position method and im getting a infinite loop, i want it to stop at tol=10^-4, can someone help?
a=0.5;
b=1;
tol=0.0001;
counter=0;
f =@(x) exp(x)-3*x^2;
while abs(b-a)>tol
c= b - (f(b)* (b-a)/(f(b)-f(a)));
if f(a)*f(c)<0
b=c;
end
if f(a)*f(c)>0
a=c;
end
counter=counter+1;
er=abs(a-b);
errorcounter(counter)=er;
end
figure
hold on
plot(errorcounter)
title('Convergence of False Position')
ylabel('Error')
xlabel('Iterations')
axis ([0 5 0 9e-04])

1 commentaire

Cris LaPierre
Cris LaPierre le 27 Oct 2020
btw, hold on is unnecessary here. When used, it should always be paired with hold off.

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 Réponse acceptée

I don't understand what your approach is, but by converting your while loop to a for loop, I can see that your error stops decreasing at 0.09. Here's a simplified version of your code.
a=0.5;
b=1;
f =@(x) exp(x)-3*x^2;
% while abs(b-a)>tol
for loop = 1:10
c= b - (f(b)* (b-a)/(f(b)-f(a)));
if f(a)*f(c)<0
b=c;
elseif f(a)*f(c)>0
a=c;
end
er=abs(a-b)
end
er = 0.1193
er = 0.0915
er = 0.0901
er = 0.0900
er = 0.0900
er = 0.0900
er = 0.0900
er = 0.0900
er = 0.0900
er = 0.0900

2 commentaires

I want the loop to stop at tol=10^-4, for example here is my code for bisection method which works perfectly but when i tried for false postion it doesn't
f=@(x) exp(x)-3*(x)^2;
a=0.5;
b=1;
tol=10^(-4);
counter = 0;
while abs(b-a) > tol
c = (b+a)/2;
if (f(a)*f(c)) <0
b = c;
end
if (f(a)*f(c)) >0
a = c;
end
counter = counter + 1;
err(counter) = abs(a - b);
end
disp(b)
figure
hold on
plot(err)
title('Convergence of Bisection')
ylabel('Error')
xlabel('Iterations')
axis ([0 13 0 0.0001])
Cris LaPierre
Cris LaPierre le 27 Oct 2020
Modifié(e) : Cris LaPierre le 27 Oct 2020
I understand. Check your algorithm. You have not implemented it correctly.

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