lsqcurve fit precision problem?

I am trying to fit an Arrhenius equation to a set of data points using lsqcurvefit. I keep getting my initial guesses as the resulting parameters from the fits, but it's obviously wrong. I've used lsqcurvefit a bit, so I do not think the problem is in the syntax. I think it's in my data set. The values of my data are in the magnitude of 10^-10; it's very small. My parameters, however, are supposed to be huge. I think it's a precision thing. If I change the units for the data, the resulting parameters are supposed to increase in magnitude.

Here's what I'm doing:

Arrhenius Equation:

function [d]=Arrhenius(constant,t)

a = constant(1);

e = constant(2);

    d=a.*exp(-e.*t); %e=Activation Energy/R

%t is inverse temp. (Kelvin): 1/T (1/Kelvin) end

Code:

x = [25+273,35+273,45+273,50+273]; %temp data

x1= 1./x; %inverse temp.

y = [0.3551 0.6390 1.4327 1.4692];

y2 = y.*10^-10 %convert to S.I. units (m^2/s)

guess = [0.002 4000]; lower = [0 100]; upper = [1 10^4];

const = lsqcurvefit(@Arrhenius,guess,x1,y2,lower,upper)

When I run this, all I get for my parameters is my initial guess, which is 0.002 and 40000. I got my initial guesses by using Mathematica, another math program, to fit the data. I really think the problem is because my y2 values are too low. Is there a way around this?

Answer 1

[Geoff]

Probably, like you say, related to scale.

You can use optimset to adjust the solver:

> opts = optimset('lsqcurvefit');
> opts.TolX

ans =
  1.0000e-006

> opts.TolX = 1.0e-12;
> const = lsqcurvefit(@Arrhenius,guess,x1,y2,lower,upper,opts);

Notice you pass opts into the function now.

The TolX field is "termination tolerance on X"

There is also TolFun: "termination tolerance on function value"

Play around =) You should be able to get it going.

-g-