Curve fitting with a constrained y value to Zero

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

Answer 1

[John D'Errico]

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 =
        -0.126909925659631
        
a2 = a12(2)
a2 =
      0.000863233648946335

The model is simple to evaluate.

ypred = a1*(x - 21.1) + a2*(x-21.1).^2
ypred =
          2.36014299087048
                         0
          -1.8786485261612
         -3.26886936348561

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 =
          2.36014299087048
     -6.10622663543836e-16
          -1.8786485261612
         -3.26886936348562

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.

[Khaled Bahnasy]

Thanks

but how it can be achieved with 3 Coefficients (a,b,c )

C must be equal zero if the equation ax^2 + b*x + c, even if x ( x-21.1)

So, no need for the third coeff or what you mean by 3 Coeff ?

Thanks alot John!

[John D'Errico]

Funny. As you were asking me how to write that in terms of 3 coefficients, I was explaining how to do that by amending my answer.

[John D'Errico]

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.

Answer 2

[Serhii Tetora]

clear;close;clc
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
xlabel('x');
ylabel('y');
grid on

[Khaled Bahnasy]

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

Answer 3

[Matt J]

Using lsqlin,

x0=21.1;
x = [4.4  37.8  54.4  ].';
y = [2.367224698  -1.857318083 -3.276015126].';

p=lsqlin(x.^[2,1,0],y,[],[],x0.^[2,1,0],0)

Check:

>> polyval(p,21.1)

ans =

  -4.4409e-16

[Khaled Bahnasy]

Thanks Matt

Still y-value approx. equal to Zero

I am trying to constrain this value to be sharply zero

[Matt J]

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,

'21.10000000000000142108547152020037174224853515625'