Curve fitting with a constrained y value to Zero

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Khaled Bahnasy
Khaled Bahnasy on 13 Aug 2020
Edited: Matt J on 13 Aug 2020
I need to fit a curve ( Second-order polynomial ) with a constraint that a specific y-value equal to Zero
4.4 2.367224698
21.1 0
37.8 -1.857318083
54.4 -3.276015126
X & Y values as an example attached X = [ 4.4 21.1 37.8 54.4 ]
I want to fit the cuve where the y-value at x= 21.1 equal to zero
I am new to matlab and i have tried Curve fitting toolbox, I think it is not provided as a constraint in the toolbox

Accepted Answer

John D'Errico
John D'Errico on 13 Aug 2020
Edited: John D'Errico on 13 Aug 2020
This is quite easy, actually.
xy = [4.4 2.367224698
21.1 0
37.8 -1.857318083
54.4 -3.276015126];
x = xy(:,1);
y = xy(:,2);
Now, you want to force a quadratic polynomial to go through y==0, at x == 21.1.
The simple solution is to fir a quadratic model of the form
y = a1*(x - 21.1) + a2*(x - 21.1)^2
So with effectively no constant term. What happens when x == 21.1? WE GET ZERO! See how nicely it works? Yes, it is true, that IF you expand it out there would be a constant term. And that is why you use this form for the model.
That quadratic model is not difficult to fit, either.
a12 = [x - 21.1,(x-21.1).^2]\y;
a1 = a12(1)
a1 =
a2 = a12(2)
a2 =
The model is simple to evaluate.
ypred = a1*(x - 21.1) + a2*(x-21.1).^2
ypred =
As you can see, it passes EXACTLY through the point (21.1,0).
Could we have gotten this in the form of a 2 parameter model, so with 3 coefficients? Well yes. That would take slightly more effort, but still quit doable.
Perhaps simplest, if we wanted to find the 3 parameter quadratic that has the indicated behavior from a1 and a2, we could just expand the polynomial. Pencil and paper will suffice. Thus is we have this function:
a1*(x - 21.1) + a2*(x-21.1).^2
then in the form
b0 + b1*x + b2*x.^2
We would have
b2 = a2
b1 = a1 - 42.2*a2
b0 = 445.21*a2 - 21.1*a1
Again, if we use this to predict the value for the vector x, we see:
b0 + b1*x + b2*x.^2
ans =
The second element of that vector is non-zero, but only by an amount that corresponds to floating point crap in the last significant bits of the number. 6e-16 is effectively zero here.
John D'Errico
John D'Errico on 13 Aug 2020
Note that if I am feeling too lazy to do the pencil and paper (not uncommon for me) then I might have let MATLAB do the work.
syms a1 a2 X
vpa(expand(a1*(X - 21.1) + a2*(X-21.1).^2))
ans =
445.21*a2 - 21.1*a1 + X*a1 - 42.2*X*a2 + X^2*a2
Now by collecting the terms, we should get the same conversion I wrote.

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More Answers (2)

Serhii Tetora
Serhii Tetora on 13 Aug 2020
Edited: Serhii Tetora on 13 Aug 2020
x = [4.4 21.1 37.8 54.4 ];
y = [2.367224698 0 -1.857318083 -3.276015126];
w = [1 1000 1 1];
[xData, yData, weights] = prepareCurveData( x, y, w );
% Set up fittype and options.
ft = fittype( 'poly2' );
opts = fitoptions( 'Method', 'LinearLeastSquares' );
opts.Weights = weights;
% Fit model to data.
[fitresult, gof] = fit( xData, yData, ft, opts );
% Plot fit with data.
h = plot( fitresult, xData, yData, 'o' );
legend( h, 'y vs. x', 'fit', 'Location', 'NorthEast');
% Label axes
grid on
  1 Comment
Khaled Bahnasy
Khaled Bahnasy on 13 Aug 2020
Thanks for reply
This approximates the y-value to Zero
I put a constraint in excel solver that this value to be equal to zero
is this applicable in matlab ?
I calculated the y-value at x=21.1, y= 0.000675272
Thanks in advance

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Matt J
Matt J on 13 Aug 2020
Edited: Matt J on 13 Aug 2020
Using lsqlin,
x = [4.4 37.8 54.4 ].';
y = [2.367224698 -1.857318083 -3.276015126].';
>> polyval(p,21.1)
ans =
Matt J
Matt J on 13 Aug 2020
The approximation error given by my approach is at the limit of floating point precision. It's meaningless to aspire beyond that. The value 21.1 doesn't even have an exact representation in a binary floating point computer. To 47 decimal places, the number that the computer is really holding when you enter x0=21.1 is,

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