data fitting with custom equation

Hi

I am facing some issue in the data fitting with the equation. I tried but somehow I am not able manage it. I wanted to fit to my data

t = 1.00E-07 2.00E-07 5.00E-07 1.00E-06 2.00E-06 5.00E-06 1.00E-05 2.00E-05 5.00E-05 1.00E-04 2.00E-04 5.00E-04 1.00E-03 2.00E-03 5.00E-03

p(t) = 0.0824 0.1515 0.2339 0.3229 0.4505 0.6173 0.7434 0.822 0.9151 0.9795 0.982 0.9829 0.9861 0.9846 0.9856

with equation: (n =2)

if possible let me know the code also.

Thanks in advance

somnath

Answer 1

[William Rose]

@Somnath Kale

Here is code that fits w and t1 as you requested. For a starting guess [w,t1]=[1,1], the console output and plot are below.

>> SomnathMain2

Best fit: SumSqError=0.377, w=3.791, t1=0.00000217

That does not look like a great fit, so I changed line 4 of SomnathMain2.m to a different initial guess:

a0=[1;1e-6];

The fit was a lot better from this initial guess - see output below, and note the reduction in SumSqError.

>> SomnathMain2

Best fit: SumSqError=0.020, w=1.036, t1=0.00000330

To find a better fit, I suggest A. try different starting points, to make sure you have not gotten stuck in a local minimu that is not a global minimum, or B. modify the code to fit "A" and/or "n" in addition to "w" and "t1".

Option B. should be pretty obvious, based on the code I have provided. You need to make the necessary adjustments in both SomnathsFunc2.m and SomnathMain2.m.

Option A. is a bit trickier, to do in a programmatic way, but worth doing. I do option A. whenever I fit parameters of a non-trivial model. The basic idea for option A is:

Define a generous physically reasonable range for each parameter (err on the side of too big a range). Then make an array of inital guesses (i.e. each row is a possible value to a0, the initial guess). The values might be chosen to be at 20% and 80% of the range, and you try all the combinations. So if you are fitting two parameters, and the reasonable ranges are a1=[-10,+10] and a2=[0, 1] respectively, then you'll have four starting points, which are

[-9, 0.1; +9, 0.1; -9, 0.9; +9, 0.9];

You do the fit four times. Save the residual error and fitted parameters from each initial guess. After doing the four fits, pick the parameters that gave the best fit. Sometimes I try 10%, 50%, and 90% of the range as my initial guesses. In that case, I have 3^N initial guesses to try, where N=number of fitted parameters.

[Somnath Kale]

@William Rose

Once again thanku very much! Initially I tried using function fminsearch but I was unable to manage integration. I modified the code for A and n fitting now it fits well and respective error is also reduced.

Best fit: SumSqError=0.0030, w=0.242, t1=0.00000549, n=0.5910, A=0.9971

Moreover, I was looking to plot firther F(logto) vs log(to). Is it possible to get the values for x (i. e. log(t0)) from fitted data so I can plot them ?

[William Rose]

@Somnath Kale,

log t0 is the variable of integration in the integral. It does not depend on the data values. You can plot F(log(t0)) versus log(t0), as follows:

Flogt0=@(x,A,w,t1) (A*w/pi)/((x-log(t1))^2+w^2);
A=.9971; w=.242; t1=0.00000549;
x=-10:.1:10;
for i=1:201, y(i)=Flogt0(x(i),A,w,t1); end
plot(x,y,'r.-'); xlabel('log(t_0)'); ylabel('F(log(t_0))');

It makes the plot above.

Answer 2

[William Rose]

@Somnath Kale

What is the value of A?

Am I correct to understand that the variable of integration is log t0, and it varies from -Inf to +Inf? Make the substitution

Then , and the integral can be written

which looks challenging but should be doable with integral(). But my initial attempt yields an error.

[Somnath Kale]

@William Rose

A is normalization constant. Lets take A=1. And now we can fit the following equation for n, w and t1