False position infinite loop

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Allie
Allie am 18 Nov. 2014
Beantwortet: My am 21 Dez. 2025 um 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

Akzeptierte Antwort

Image Analyst
Image Analyst am 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.

Weitere Antworten (2)

per isakson
per isakson am 18 Nov. 2014
Bearbeitet: per isakson am 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 vor etwa 2 Stunden
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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