# How can I solve an equation using fixed point method?

10 views (last 30 days)
Carla Coutinho on 2 Jul 2016
Answered: HUY on 22 May 2024 at 9:53
g(x)=2^-x;
[1/3,1];
tol=10^-4;
max=100
David Miller on 2 Jul 2016
Edited: David Miller on 2 Jul 2016
Please be more specific. What does [1/3,1] and max=100 represent? I assume tol is the tolerance for convergence, sometimes referred to as epsilon.
Radu Trimbitas on 20 Mar 2020
Yes, David, tol is the tolerance, or epsilon; max is the maximum number of iterations; x0 is the initial approximation (starting value)
Code for sapmeth
function [z,ni]=sapmeth(f,x0,tol,maxit)
for k=1:maxit x1=f(x0);
if abs(x1-x0) < tol %success
z=x1; ni=k;
return;
end;
x0=x1;
end error('iteration number exceeded')
Call:
f=@(x) 2^(-x);
x0=0.6;
[z,ni]=sapmeth(f,x0,1e-6,100)
Results
z=0.641185
n=15

Radu Trimbitas on 4 Jul 2016
If you wish to solve x=2^(-x) use successive approximation method. Provide a function, a starting value and a tolerance. function [z,ni]=sapmeth(f,x0,tol,maxit) for k=1:maxit x1=f(x0); if abs(x1-x0) < tol %success z=x1; ni=k; return; end; x0=x1; end error('iteration number exceeded')
if you provide the input parameters f=@(x) 2^(-x), x0=0.6 and call [z,ni]=[z,ni]=sapmeth(f,x0,1e-6,100) after 15 iterations one obtains the fixpoint z=0.641185

HUY on 22 May 2024 at 9:53
f = @(x) 2*x^3 - 11.7*x^2 + 17.7*x - 5;

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