False position infinite loop

13 vues (au cours des 30 derniers jours)
Allie
Allie le 18 Nov 2014
Réponse apportée : My le 21 Déc 2025 à 20:01
Hi there.
This function gets stuck in an infinite loop. Do you all have any suggestions for me?
function [R, E] = myFalsePosition(f, xL, xR, tol)
if sign (f(xL)) == sign(f(xR))
error 'you are arrested!!!'
end
yL = f(xL);
yR = f(xR);
new_x = ((xR*yL) - (xL*yR))/(yL - yR);
new_y = f(new_x);
e = abs(new_y);
E = e;
while e > tol
if f(xL)*f(new_x) > 0
xL = new_x;
yL = f(xL);
else
xR = new_x;
yR = f(xR);
end
end
new_x = ((xR*yL) - (xL*yR))/(yL - yR);
new_y = f(new_x);
R = [R new_x];
e = abs(new_y);
E = [E e];
end

Réponse acceptée

Image Analyst
Image Analyst le 18 Nov 2014
Well you could tell us the values you used for f, xL, xR, tol when you called it. And you can use the debugger to figure out why "e" never falls below "tol". Using the debugger yourself will be your fastest course of action , rather than trying to debug it via back-and-forth Answers forum postings, which can take hours.

Plus de réponses (2)

per isakson
per isakson le 18 Nov 2014
Modifié(e) : per isakson le 19 Nov 2014
I reformatted your function.
Neither e nor tol is changed in the while-loop. If &nbsp e > tol &nbsp is true when entering the loop it will remain true.
Possible, the end of the loop is not in the position, which you intended.

My
My le 21 Déc 2025 à 20:01
function p = myfalseposition(f,p0,p1,TOL,N0)
q0 = f(p0);
q1 = f(p1);
for i = 1:N0
% False Position formülü
p = p1 - q1*(p1-p0)/(q1-q0);
if abs(p-p1) < TOL
return
end
q = f(p);
if q*q1 < 0
p0 = p1;
q0 = q1;
end
p1 = p;
q1 = q;
end
disp('Method failed');
end
%Main
clear; clc;
f = @(x) cos(x) - x;
p0 = 0.5;
p1 = pi/2;
TOL = 1e-3;
N0 = 100;
p = myfalseposition(f,p0,p1,TOL,N0)

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