Hi fortrans, yes, but, tau = (sxx(3)/d(3))^0.0834

Tau = (sxx(3)/d(3))^0.0834

d(3)

____________ * sxx(3)^(tau-1)

sxx(3)^tau

To make easier to understand, I calculated it in wolfram alpha like this

Renamed:

d(3) = x

sxx(3) = y

Which lead us to:

x

____________ * y^(tau-1)

y^tau

Since, tau = (y/x)^ 0.0834, the equation becomes:

x

____________________ * y^(((y/x)^ 0.0834)-1)

y^((y/x)^ 0.0834)

Putting that on wolfram alpha results in:

x

__

y

https://www.wolframalpha.com/input/?i=x+%2F+(y%5E((y%2Fx)%5E+0.0834))+*+(y%5E(((y%2Fx)%5E+0.0834)-1))

And..after renaming back from x and y result in:

d(3)/sxx(3)

I suceeded to make it work in matlab, but they produces the very same result. So, why make it different ffrom diag and diag_nl if they are the same ?

Either in Wolfram Alpha and MatLab the result is the same, proofing that both diag and diag_nl are the same thing. That´s why i´m confused...Why displaying diag and diag_nl as different equations if they produces the same result (And, in fact, are the same mathematical equation) ?

Since the exponential is used in the diag_nl matrix, it results in:

diag_nl = [d(1)/sxx(1) 0 0;

0 d(2)/sxx(2) 0;

0 0 d(3)/sxx(3)]; % non linear coeff in S

That is the same as the linear formula used on diag

diag = [d(1)/sxx(1) 0 0; % diagonal linear matrix

0 d(2)/sxx(2) 0;

0 0 d(3)/sxx(3)];