How do I take natural logarithm of a set of datas and do a least square fitting and get an expression?

I have a set of 3 datas. i.e. one set of dependent variables and 2 set of independent variables. I want to use natural logarithms on these datas and use least square fitting to obtain a relationship between these three set of datas

% Dependent varaibales:
psi_o= [1.0690 2.02 1.3474 1.3580 1.3966 1.6701 1.4901 1.669 1.476 2.848 1.3871 2.8008 3.0724 2.3614 2.7726];
psi_1= [1.0431 1.1251 0.9519 0.9691 0.9380 1.5003 1.2390 1.3260 1.2584 2.848 1.2260 1.9754 2.125 1.6842 2.098];
psi_2= [0.9819 0.9916 1.0444 0.9938 0.9807 1.3555 1.2052 1.227 1.1849 1.747 1.2106 1.5462 1.6876 1.5860 1.726];
psi_3= [1.015 0.9875 1.0682 1.0763 1.0021 1.295 1.119 1.1112 1.125 1.681 1.1667 1.6932 1.862 1.677 1.6443 ];

% Independent variables : 
We= [8.518 9.903 16.473 22.68 26.214 58.84 78.77 117.76 138.37 184.842 193.155 272.054 327.49 443.05 615.34];
Lambda= [1 0.701 0.501 0.397];

Here, for the dependent variable data set 'psi_o ' the independent variables are the whole data set of 'We' and only for Lambda =1, likewise 'psi_1' corresponds to We and Lambda = 0.701, similarly for psi_2 ansd psi_3 it the whole data set We and only Lambda = 0.501 and 0.397 respectively.

So how do I take the logarithms of all the values and obtain a relationship between psi_1,2,3 ; We; Lambda through least square fitting?

[Walter Roberson]

What is your model function?

In the case that you have a single model function with unknown parameters then you can fit the parameters given datasets.

In the case where you have multiple specific potential model functions and you want to see which one is "best", then you can fit each of them and decide which is best based on the fit. However the more functions you add, the greater the risk that the "wrong" function will turn out to have the best fit, especially in any situation in which there is noise in the data. And having only three digits for your We or Lambda is effectively having "noise" for those.

If you do not have a list of specific models to test and are hoping that somehow MATLAB can tell you what the relationship is, then, Sorry, mathematically it can be shown that it is not possible to deduce the correct model given only a finite set of data.

[the cyclist]

Do you mean that Lambda is the intercept for each fit? If so, I can make a couple guesses as to what model you mean, depending on whether you want the same coefficient of We for each fit:

Different beta for each equation

log(psi_0) = 1.000 + beta_0 * log(We);
log(psi_1) = 0.701 + beta_1 * log(We);
log(psi_2) = 0.501 + beta_2 * log(We);
log(psi_3) = 0.397 + beta_3 * log(We);

or

Same beta for each equation (and only the intercepts are different)

log(psi_0) = 1.000 + beta * log(We);
log(psi_1) = 0.701 + beta * log(We);
log(psi_2) = 0.501 + beta * log(We);
log(psi_3) = 0.397 + beta * log(We);

Are either of these correct for the fit(s) you are trying to estimate? If not, can you write out what you want in the same style that I just did?

[Pragyan Kumar Sarma]

Yes,

for a particular Lambda..say Lambda= 1, I have the same data set We and dependent variable data set is psi_o (for every psi_o, Lamda value is the same)

For, Lamdba= 0.701, We is the same and the dependent variable data set is psi_2

Lambda = 0.501, We is the same data set and the dependent variable data set is psi_3

Lambda= 0.397, same data set We and the dependent variable data set is psi_4

I have attached two images, where a very similar type of problem is solved, the authors took the natural logarithms and then did a least square fitting and obtained an expression.

Here in this above figure for a particular Lambda value which is an independent variable and We, the dependent variable data set is Psi_max.

I also have the same dataset for independent variable We, (1+L/D) which is my Lambda and independent variable data set Psi_max which is my psi_0,2,3,4

[Shree Charan]

Hi @Pragyan Kumar Sarma,

% Dependent varaibales:
psi_o= [1.0690 2.02 1.3474 1.3580 1.3966 1.6701 1.4901 1.669 1.476 2.848 1.3871 2.8008 3.0724 2.3614 2.7726];
psi_1= [1.0431 1.1251 0.9519 0.9691 0.9380 1.5003 1.2390 1.3260 1.2584 2.848 1.2260 1.9754 2.125 1.6842 2.098];
psi_2= [0.9819 0.9916 1.0444 0.9938 0.9807 1.3555 1.2052 1.227 1.1849 1.747 1.2106 1.5462 1.6876 1.5860 1.726];
psi_3= [1.015 0.9875 1.0682 1.0763 1.0021 1.295 1.119 1.1112 1.125 1.681 1.1667 1.6932 1.862 1.677 1.6443 ];
% Independent variables : 
We= [8.518 9.903 16.473 22.68 26.214 58.84 78.77 117.76 138.37 184.842 193.155 272.054 327.49 443.05 615.34];
lambda= [1 0.701 0.501 0.397];

log_psi_0 = log(psi_o);
log_psi_1 = log(psi_1);
log_psi_2 = log(psi_2);
log_psi_3 = log(psi_3);
log_We = log(We);

X(:,1) = lambda(1)*log_We;
X(:,2) = lambda(2)*log_We;
X(:,3) = lambda(3)*log_We;
X(:,4) = lambda(4)*log_We;

Y = [log_psi_0', log_psi_1', log_psi_2', log_psi_3'];
B = X\Y;

B should have the coefficients for the relationship between 'psi_o', 'psi_1', 'psi_2' and 'psi_3', however the rank of X =1 indicating some sort of dependancy among 'lambda' and 'We'.