Re: Help in parameter estimation

[Nicholas Jankowski]

On Tue, Aug 30, 2016 at 6:00 AM, Bharath R <address@hidden> wrote:

 

function m=f(x,p)

  m(1,:)=20;

  for i=1:length(x)

    m(i+1,:)=((p(1)-m(i,:))/(p(2)*p(3))+((p(4)*x(i,2)*(x(i,3)-m(i,:)))+(p(5)*x(i,4))+(p(6)*x(i,5)))/(p(3))+p(7))+m(i,:);

  endfor

  mat=m(2:end);

endfunction

is your intent to return the variable m or the variable mat from function f? you are currently returning m and never using mat. since mat is a different size than m, this could make a significant difference when you use the results of f later.

your declaration for m is odd. it asks to make all of the columns in row1 equal to 20.  since m has not yet been initialized, the number of rows is 1. so the output is a 1x1 array, or a scalar, and m = 20. then in the for loop, if the result of the m(i+1,:) equation was 6, you would be appending it as a new row to m, creating a 2x1 array [20 ; 6] . Since you went to the trouble of using the : operator, but have it doing nothing, I suspect this was not your intent?