Help fitting data to an implicit equation

Hello:

I need to fit some data to the following implicit equation:

((1-y)^(1-b)/y)=exp(-kt)

t is a vector containing time values and y is a vector containing current values. for each series of data y vs t, I need to determine b and k

b has to be between 0 and 1, and k needs to be greater than 0.

I have both the optimization and the curve fitting toolboxes.

Any suggestions on what tools to use (lsqcurvefit? something else? would be very appreciated)

Thanks!

Answer 1

[Jeff Miller]

I would suggest using fminsearch. The error function to be minimized would be something like:

function thiserr = err(x,y,t)
   b = x(1);
   k = x(2);
   thiserr = sum( (((1-y).^(1-b)./y) - exp(-kt))^2 );
end

You should be able to find examples of how to use fminsearch if you need more detail on how to call it. In your case y and t are "extra parameters". Look here for information on how to handle that.

[mura0087]

Thank you!

Here is the error function I am using:

function thiserr = err(x,y,t)
   b = x(1);
   k = x(2);
   thiserr = sum( (((1-y).^(1-b)./y) - exp(-k*t)).^2 );
end

And here is how I am calling the function:

fun = @(x)err(x,yvalues,tvalues); %define the function
x0=rand(2,1);%provide a random starting point for your parameters. 
options = optimset('MaxFunEvals',1e1000000000000);
bestx=fminsearch(fun,x0);

I set options "MaxFunEvals" to a very high number because I keep getting the following error:

Exiting: Maximum number of function evaluations has been exceeded
         - increase MaxFunEvals option.
         Current function value: NaN 
         
         

UPDATED- I believe the I am getting MaxFunEvals because the function does not fit my data. Fminsearch is doing its job just fine, this is a bad model. Thanks for your help

[Jeff Miller]

You are welcome. That function value NaN is a bad sign. It means thiserr is NaN for all values of b and k that fminsearch has checked. You don't have any y=0 values, do you? Dividing by 0 would cause nans for all b and k.

[mura0087]

No y=0 values. I even scaled the y values so that everything was nonzero and less than 1, and it still did not fit. I think I need a new function!

I did get this fminsearch method to a fit a different function to different data (that I generated, so I knew the function to fit it to) as a sanity check. I had not fit using the optimization toolbox before, but I now see this is a pretty elegant method for fitting, thanks again for suggesting it!

Answer 2

[John D'Errico]

My thought would be the lazy solution. If your model is:

((1-y)^(1-b)/y)=exp(-kt)

then log the model. That is, we know that

(1-b)*log(1-y) + k*t = log(y)

With one more step, this reduces to

-b*log(1-y) + k*t = log(y) - log(1-y)

You can compute the parameters k and b using a simple linear regression now. Thus, if y and t are column vectors, we have:

bk = [-log(1-y),t] \ (log(y) - log(1-y));

so bk is a vector of length 2, contining the estimates for b and k respectively. If you find that b or k are estimated to be something outside of the valid region, then I would first consider if this is a reasonable model, but then you could just use lsqlin to estimate them, since lsqlin does provide bound constraints.

[mura0087]

This also works! Thank you!