How to fit my data with a custom equation meant to come out as a parabola?

I have some data that I'm trying to fit with an equation.

That equation is:

Where y is the variance of my data and x is the average of my data. I'm trying to get out what i and N are. The values should fit to a parabola (as seen in previous literature), but some of my data is not working with what I currently have.

I've been trying to fit this with the lsqcurvefit function, by doing this: (bg is the background variance which I subtract from y before trying to fit to the equation.)

constant = lsqcurvefit(@f3,[min(average);min(variance);max(bg)/2],x,y);


function  y = f3(constant,x)
y = (constant(1)* x)- ((x.^2)/constant(2));
end

I've picked the least squares method since I've seen lots of research in my field fit it this way, but the way this code runs I get numbers for i and N that look very off and the constant is giving me back three values, not two.

Any thoughts?

[Torsten]

Maybe you could explain the part

That equation is:

y = i*x - x^2/N

where y is the variance of my data and x is the average of my data.

It sounds as if you want to fit two parameters to one value pair (mean(data),variance(data)). But I think this is not what you mean, is it ?

Answer 1

[Star Strider]

The ‘constant’ (parameter vector) is giving back three values because you are giving it three values to begin with:

[min(average); min(variance); max(bg)/2]

It has no idea what to do with the third one (and I have no idea what it does with three values when it only uses two). See if just giving it the first two (or the two that are most appropriate) produces a better result.

[Rebecca Somach]

This hasn't solved my problems, but the code doesn't fall apart when I only give it two values, so that's at least a start. Thank you for answering!

[Star Strider]

My pleasure!

I have no idea what you’re doing.

You may not need lsqcurvefit anyway, since your function is essentially linear, and is equivalent to:

B = [x(:) x(:).^2] \ y(:)

constant(1) = B(1)

constant(2) = -1/B(2)

x = 1:10;
y = 2*x(:) - x(:).^2/5;

B = [x(:) x(:).^2] \ y(:)

B = 2×1

2.0000 -0.2000

const(1) = B(1)

const = 2.0000

const(2) = -1/B(2)

const = 1×2

2.0000 5.0000

I’m not certain what you’re doing, so this is simply a guess.

Answer 2

[Matt J]

If there are 3 coefficients, you should just use polyfit(x,y,2) and forget about lsqcurvefit. No reason an iterative solution is required.

If you are deliberately modeling with only 2 unknown coefficients, you should still use polyfit(x,y,2) or maybe polyfit(x,y./x, 1) to derive the initial guess.

Other than that, there's not much to go on, without runnable code.