How can I solve an equation using fixed point method?

g(x)=2^-x;
[1/3,1];
tol=10^-4;
max=100

[David Miller]

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]

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

Answer 1

[Radu Trimbitas]

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

Answer 2

[HUY]

f = @(x) 2*x^3 - 11.7*x^2 + 17.7*x - 5;