Simple fixed-point iteration method

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John Smith
John Smith 2019-9-22
回答： Tsega'ab 2023-12-13
My task is to implement (simple) fixed-point interation.
So far, I've got the following and I keep receiving error Undefined function 'fixedpoint' for input arguments of type 'function_handle'.
(I'm new in Matlab, so there may be both syntactical or semantical errors...)
function [ x ] = fixedpoint(g,I,y,tol,m)
% input: g, I, y, tol, max
% g - function
% I - interval
% y - starting point
% tol - tolerance (error)
% m - maximal number of iterations
% x - approximate solution
a=I(1);b=I(2);
if(y<a | y>b)
error('The starting iteration does not lie in I.')
end
x=y;
gx=g(y);
while(abs(x-gx)>tol & m>0)
if(gx<a | gx>b)
error('The point g(x) does not lie in I.')
end
y=x;
x=g(y);
m=m-1;
end

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采纳的回答

Dimitris Kalogiros
Dimitris Kalogiros 2019-9-22
Dear John
Put your function into the same folder with the program (m-file) that calls it.
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John Smith
John Smith 2019-9-23
I am not sure, what I have done, but it is working fine now. ¯\_(ツ)_/¯
Anyway, thank you for your time.
> I wonder , what is the calling program.
It might have been the problem. I guess the function and the file have to share the name, right?(I mean, if I code a function F, then it has to be saved as file F.m.)
Dimitris Kalogiros
Dimitris Kalogiros 2019-9-24
Hm hm, I don't know if it is necessary , but I always follow this rule.

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更多回答（4 个）

Ishita Sharma
Ishita Sharma 2020-8-18
f(x)=x^2 - x -1 =0
emmanuel john Lavarias
  1. Solve one real root of ex−2x−5=0ex−2x−5=0 with x0=−2x0=−2 using the Fixed-Point Iteration Method accurate to four decimal places.
Ahteshamul Hoque Tareq
a=I(1);b=I(2); if(y<a | y>b) error('The starting iteration does not lie in I.') end x=y; gx=g(y); while(abs(x-gx)>tol & m>0) if(gx<a | gx>b) error('The point g(x) does not lie in I.') end y=x; x=g(y); m=m-1; end
Tsega'ab
Tsega'ab 2023-12-13
function [ x ] = fixedpoint(g,I,y,tol,m)
% input: g, I, y, tol, max
% g - function
% I - interval
% y - starting point
% tol - tolerance (error)
% m - maximal number of iterations
% x - approximate solution
a=I(1);b=I(2);
if(y<a | y>b)
error('The starting iteration does not lie in I.')
end
x=y;
gx=g(y);
while(abs(x-gx)>tol & m>0)
if(gx<a | gx>b)
error('The point g(x) does not lie in I.')
end
y=x;
x=g(y);
m=m-1;
end

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