simple Fixed Point Iteration

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지민 이
지민 이 on 7 Nov 2021
Commented: 지민 이 on 7 Nov 2021
clear all;
clc;
x(1)=0;
g=@(x) exp(-x);
f=@(x) exp(-x)-x;
true_root=0.56714329;
disp('----------------------------------------------------');
disp('i xi Ea(%) Et(%) ');
disp('----------------------------------------------------');
for i=1:10
x(i+1)=g(x(i));
end
if x~=0
ea=100*abs(x(i+1)-g(x(i))/x(i+1));
end
e_t=100*abs(x-true_root)/true_root;
res=[[0:10]' x' ea' e_t'];
fprintf('%d %1.6f %3.3f %3.3f\n', res');
i want to add the ea. but i can. how can i this??

Accepted Answer

Alan Stevens
Alan Stevens on 7 Nov 2021
Probably more like this (though you don't seem to have used function f anywhere):
n = 11;
x = zeros(1,numel(n));
ea = zeros(1,numel(n));
g=@(x) exp(-x);
f=@(x) exp(-x)-x;
true_root=0.56714329;
disp('----------------------------------------');
----------------------------------------
disp('i xi Ea(%) Et(%) ');
i xi Ea(%) Et(%)
disp('----------------------------------------');
----------------------------------------
for i=1:n-1
x(i+1)=g(x(i));
if x(i+1)~= 0
ea(i+1) = 100*abs(x(i+1)-x(i))/x(i+1);
end
end
e_t = 100*abs(x-true_root)/true_root;
res=[(0:n-1)' x' ea' e_t'];
fprintf('%d %10.6f %10.3f %10.3f\n', res');
0 0.000000 0.000 100.000 1 1.000000 100.000 76.322 2 0.367879 171.828 35.135 3 0.692201 46.854 22.050 4 0.500474 38.309 11.755 5 0.606244 17.447 6.894 6 0.545396 11.157 3.835 7 0.579612 5.903 2.199 8 0.560115 3.481 1.239 9 0.571143 1.931 0.705 10 0.564879 1.109 0.399

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