# Why is it that if I adjust a curve with the curve fitting tool, it does not give me the same results as if I fit it without the tool?

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Angelavtc on 11 Dec 2021
Commented: Matt J on 11 Dec 2021
Hello, community,
I am trying to fit a series of points with a power function with and without the tool, using the same starting points and having different results. Maybe it is an obvious question but I would like to know why. Here are the two codes and the data attached:
function y = myfunction(x,a1,b1)
y=a1*x.^b1;
end
ft5 = fittype( 'myfunction(x,a1,c1)' )
f5 = fit( x_inter',y_inter', ft5,'StartPoint', [8.64593401687141e-71 16.7561500073752] )
plot( f5, x_inter, y_inter )
With the following results:
General model:
f5(x) = myfunction(x,a1,c1)
Coefficients (with 95% confidence bounds):
a1 = 3.368e-71 (-1.784e-70, 2.458e-70)
c1 = 16.76 (16.13, 17.38)
An this is the code generated by the tool:
%% Fit: 'untitled fit 1'.
[xData, yData] = prepareCurveData( x_inter, y_inter );
% Set up fittype and options.
ft = fittype( 'power1' );
opts = fitoptions( 'Method', 'NonlinearLeastSquares' );
opts.Display = 'Off';
opts.StartPoint = [8.64593401687141e-71 16.7561500073752];
% Fit model to data.
[fitresult, gof] = fit( xData, yData, ft, opts );
% Plot fit with data.
figure( 'Name', 'untitled fit 1' );
h = plot( fitresult, xData, yData, 'predobs', 0.99 );
legend( h, 'y_inter vs. x_inter', 'untitled fit 1', 'Lower bounds (untitled fit 1)', 'Upper bounds (untitled fit 1)', 'Location', 'NorthEast', 'Interpreter', 'none' );
% Label axes
xlabel( 'x_inter', 'Interpreter', 'none' );
ylabel( 'y_inter', 'Interpreter', 'none' );
grid on
Getting the results:
General model Power1:
f(x) = a*x^b
Coefficients (with 95% confidence bounds):
a = 9.531e-154 (-3.249e-153, 5.155e-153)
b = 35.54 (35.1, 35.97)
Goodness of fit:
SSE: 1.796e+07
R-square: 0.9771
RMSE: 94.82
Thank you!

Matt J on 11 Dec 2021
Edited: Matt J on 11 Dec 2021
In the first case, fit() doesn't know anything about the structure of your curve model. The details are hidden inside myfunction().
In the second case, you are specifying one of the Toolbox's stock library of curve models. It therefore knows more about the curve and uses a different method to estimate the parameters, one that is specifically tailored to power1.
Because a different solution method is used, a different path is taken by the iterative solver, and a different solution can be reached, when the solution is non-unique. Here, when a=0 (as was the case in both versions), the solution for the b parameter is indeed non-unique.
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Matt J on 11 Dec 2021
If you can extract a subset of data in that region, you can just fit it in the normal way.