positive index error in matlab

clc;
clear all;
R=3;
beta=8;
x1=1:1:10;
for x=1:length(x1)
x=round(x);
f(x)=1-(exp(-((2^(R))-1)/x)^(beta/2));
df(x)=(beta/2)*(((2^R)-1)^beta)*(x^(-(beta+1)))*(exp(-((2^(R))-1)/x)^(beta/2));
format compact
disp(' iterate x f(x) est. error ')
for n = 0:5
s =( f(x)/df(x));
sprintf(' %2d %2.10f %2.10f % 2.10f \n', n,x,f(x),s)
x = x-s;
end
end
error is:
Attempted to access f(-62718.3); index must be a positive integer or logical.
Error in newton (line 15)
s =( f(x)/df(x));
plzz suggest a solution

 Risposta accettata

Amit
Amit il 25 Gen 2014

0 voti

In MATLAB f(x) usage is to (if f is a matrix) access the xth element in the matrix. And Here thats why you're getting this error.

2 Commenti

If I was you, I'd do something like this:
R=3;
beta=8;
x1=1:1:10;
x = 1; % Intial value of x
for n=1:length(x1)
f=1-(exp(-((2^(R))-1)/x)^(beta/2));
df=(beta/2)*(((2^R)-1)^beta)*(x^(-(beta+1)))*(exp(-((2^(R))-1)/x)^(beta/2));
format compact
disp(' iterate x f(x) est. error ')
% for n = 0:5
s =( f/df);
disp(sprintf(' %2d %2.10f %2.10f % 2.10f \n', n,x,f,s));
x = x-s;
%end
end
Thanks alot..dis code is working

Accedi per commentare.

Più risposte (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by