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)];