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

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Pragyan Kumar Sarma 2023-4-7

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https://ww2.mathworks.cn/matlabcentral/answers/1943549-how-do-i-take-natural-logarithm-of-a-set-of-datas-and-do-a-least-square-fitting-and-get-an-expressio

评论： Shree Charan 2023-6-2

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?

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Pragyan Kumar Sarma 2023-4-8

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 2023-6-2

在 MATLAB Online 中打开

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'.

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