how to fit a curve in the form of A = (L^x)(D^y)

hi, i have some response data as vector A where the variables are L and D.

I just want to find the coefficients for L and D which will fit my data in the form mentioned in the title.

I want to fit a curved line, and not a surface.

I feel it should be fairly simple, but reading a few old answers also didn't help my case.

Is there some easy way to do this?

In case u want to see the data, here it is:

A = [0 0.06 0.12 0.44 0.56 0.94 1 1 0 0.04 0.58 0.74 0.86 1 1]

L = [100 100 100 100 100 100 100 100 43.7 49.7 56 61.5 65 77 93.8]

D = [11.3 10.1 8.9 8.5 8.1 7.7 6.5 5.3 5 5 5 5 5 5 5]

Thanks a lot.

More info:

I wrote the above equation as logA = xlogL + ylogD, and tried to use

X = [ones(size(logL)) logL logD];

b = regress(logA,X);

but Matlab didn't return any coefficients, it just gave b = NaN NaN NaN

[Matt J]

Your A has 15 data points while your L,D have only 14.

[Roule]

Actually, all of them have 15 data points. Maybe there are too close to each other, that's why u felt so. U can copy paste in excel to verify though.

[Jos (10584)]

log(A(1)) → -Inf, causing the NaNs

[Roule]

ya, instead of 0 i can use a very small number close to zero, or just ignore that particular data point altogether. in cftool, matlab just ignores that log(0).

Answer 1

[Jos (10584)]

nlm = fitnlm([L(:) D(:)], A, 'y~(x1^b1)*(x2^b2)',[0 0])

[Roule]

Hey Jos, thanks so much. Your one line really did the work.

I am using 2013a so I had to just replace 'fitnlm' with 'NonLinearModel.fit'. I guess both will give the same results.

However the result shows very low R-squared: 0.48. Is there any way to improve this. I want to use only L and D may be in some other combination if that gives better R-squared value.

Below is the complete output:

nlm =

Nonlinear regression model: y ~ (x1^b1)*(x2^b2)

Estimated Coefficients:

          Estimate    SE         tStat      pValue  
    b1    0.92059     0.34937      2.635    0.02059
    b2    -2.5503     0.90462    -2.8192    0.014488

Number of observations: 15, Error degrees of freedom: 13

Root Mean Squared Error: 0.31

R-Squared: 0.48, Adjusted R-Squared 0.44

F-statistic vs. zero model: 30.2, p-value = 1.31e-05

[Jos (10584)]

I had the same outcome, so that's good. If you think you can come up with a better model you can fit that as well, of course. By the way, always plot your data, your fit and your residuals (fittedY - Y) to see how your model is doing.