# Fitting a model to my data using non linear least square fit method

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Kashif Naukhez on 21 Jul 2023
Commented: Kashif Naukhez on 27 Sep 2023
I have x values (first column in the excel file) and P values (second column in the excel file). I want to obtain the parameters 'q' and 'm' using the expression: P = [ 1 - (1-q) * m * x ] ^ ( 1 / (1 - q) ) in a log log plot and plot the same using non linear least square fit method.
Diwakar Diwakar on 21 Jul 2023
% Step 1: Load data from the Excel file
x = data(:, 1);
P = data(:, 2);
% Step 2: Define the equation and the error function for nonlinear fit
eqn = @(params, x) (1 - (1 - params(1)) * params(2) * x) .^ (1 / (1 - params(1)));
errorFunc = @(params) eqn(params, x) - P;
% Step 3: Use lsqcurvefit to obtain parameter estimates
initialGuess = [0.5, 1]; % Initial guess for q and m
paramsFit = lsqcurvefit(eqn, initialGuess, x, P);
% Extract the fitted parameters
q = paramsFit(1);
m = paramsFit(2);
% Step 4: Plot the data and the fitted curve on a log-log plot
x_vals_for_plot = logspace(log10(min(x)), log10(max(x)), 100); % Generating log-spaced x values for the plot
P_fit = eqn(paramsFit, x_vals_for_plot);
loglog(x, P, 'o', 'MarkerFaceColor', 'b'); % Plot the original data points with markers
hold on;
loglog(x_vals_for_plot, P_fit, 'r', 'LineWidth', 2); % Plot the fitted curve in red
hold off;
xlabel('x');
ylabel('P');
legend('Data', 'Fitted Curve');
title('Nonlinear Least Squares Fit');
grid on;

Mathieu NOE on 21 Jul 2023
hello
try this , hope it helps
results :
m = 24.5179
q = 1.2078
x = data(:,1);
y = data(:,2);
% sort / unique x
[x,ia,ic] = unique(x);
y = y(ia);
% remove x = 0 data (just in case)
ind = x>eps;
x = x(ind);
y = y(ind);
% remove y = 0 data (just in case)
ind = y>eps;
x = x(ind);
y = y(ind);
% fit using log spaced values
xx = logspace(log10(min(x)),log10(max(x)),100);
yy = interp1(x,y,xx,'linear');
loglog(x,y,'*-',xx,yy,'*-');
% curve fit using fminsearch
% We would like to fit the function :
% P = [ 1 - (1-q) * m * x ] ^ ( 1 / (1 - q) )
f = @(m,q,x) (1-(1-q)*m*x).^(1/(1-q));
obj_fun = @(params) norm(f(params(1), params(2), x)-yy);
C1_guess = [10 1 ];
sol = fminsearch(obj_fun, C1_guess); %
m_sol = sol(1)
q_sol = sol(2)
yfit = f(m_sol, q_sol, xx);
Rsquared = my_Rsquared_coeff(yy,yfit); % correlation coefficient
loglog(xx, yfit, '-',x,y, 'r .', 'MarkerSize', 25)
title(['Fit equation to data : R² = ' num2str(Rsquared) ], 'FontSize', 20)
xlabel('x data', 'FontSize', 20)
ylabel('y data', 'FontSize', 20)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function Rsquared = my_Rsquared_coeff(data,data_fit)
% R2 correlation coefficient computation
% The total sum of squares
sum_of_squares = sum((data-mean(data)).^2);
% The sum of squares of residuals, also called the residual sum of squares:
sum_of_squares_of_residuals = sum((data-data_fit).^2);
% definition of the coefficient of correlation is
Rsquared = 1 - sum_of_squares_of_residuals/sum_of_squares;
end

Alex Sha on 27 Sep 2023
The best solution:
Root of Mean Square Error (RMSE): 0.0143800358786989
Correlation Coef. (R): 0.998885623657614
R-Square: 0.997772489149861
Parameter Best Estimate
--------- -------------
q 1.42974692753178
m 46.7110560131104
Kashif Naukhez on 27 Sep 2023

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