Error using nlinfit for logarithmic model

Hello,

I'm trying to fit a logarithmic model to my data but I keep geting the following error:

Error using nlinfit>checkFunVals (line 641)

The function you provided as the MODELFUN input has returned Inf or NaN values.

clear all, clc;

y=[0;0.1244; 0.1569; 0.1016; 0.2784; 0.4066; 0.2746; 0.7044; 0.7061;...
    0.7054; 0.6752; 0.7102; 0.707];

x=[0; 1; 11; 13; 19; 25; 31; 38; 55; 61; 74; 92; 109];

fun = @(a,x)a(1)*log(x);
a0=0.15;
f = nlinfit(x,y,fun,a0);

Answer 1

[Star Strider]

Of course it will return Inf or NaN values!

Note that:

x=[0; 1; 11; 13; 19; 25; 31; 38; 55; 61; 74; 92; 109];

and

log(0) = -Inf

Try this simple transformation:

fun = @(a,x) x.^a;
a0=0.15;
f = nlinfit(x,exp(y),fun,a0)

figure
plot(x, y, '+')
hold on
plot(x, log(fun(f,x)), '-r')
hold off

to get this reasonably decent fit:

[Image Analyst]

Very nice. I question that the data follows a log process though. It looks to me like the "rate equation" might be a better choice since the data seem to level off after a while.

[Star Strider]

Thank you.

I agree that something like this might be more appropriate:

however we do not have that information.

[Oriol Sanchez Rovira]

Thank you for your help and your quick response.

[Star Strider]

My pleasure.

Answer 2

[Image Analyst]

OK, I've done it 4 ways, two models with slightly different rate equations, Star's way, and a log fit with fitnlm(). None of them look like a great fit. Star's model couln't get the fit at (0,0) so I took the residuals at just the second and later points to see which one fit the best. See the plot, created by the attached demo.

Note that the two rate equations are pretty much overlapped so you only see the dark green one.

The residuals are

residuals1 = 1.02835678164379

residuals2 = 1.02842292127149

residualsStarStrider = 1.51350802831932

residualsLog = 1.07621627525508

So it looks like the rate equation is the best. Note that the rate equation will level off at some assymptote (which your data seem to do), while the log fits will head up to y=infinity with increasing x, so that may be another reason to favor the rate equation over the log fit. But, like I said, none of those fits look great and you might want to try a different model, like the s-curve shaped "Boltzman equation".

[Oriol Sanchez Rovira]

Thank you for the advise.

[Steven Manz]

Hi @Image Analyst, I tried your example for nonlinear fitting. I used Rate Model 2. I tried using a while loop and allowing beta0 to change to the new coefficient values on every iteration, hoping the fitter would work like the curve fitting toolbox and simply keep iterating until the residuals were under a value of 2.

However, the coefficients/residuals do not seem to change. No matter what I put for beta0, I get the same exact coefficients/residuals every time. Could I be doing something incorrectly? Would you like me to create a separate question for this with my data?

[Image Analyst]

@Steven Manz start your own question and attach your data and script of mine that you modified.