       The standard deviation of Three fitting parameters is bigger than

• To: mathgroup at smc.vnet.net
• Subject: [mg100355] The standard deviation of Three fitting parameters is bigger than
• From: holybse at gmail.com
• Date: Mon, 1 Jun 2009 07:12:42 -0400 (EDT)

```I try to fit a formula,
0.00138 t + 74.830*(t/a)^3*Integrate[(E^x x^4)/(E^x - 1)^2, {x, 0, a/
t}] + n 3*8.314 (w/ t)^2 E^(w/ t)/(-1 + E^(w/t) )^2
a,w,n are fitting parameters
t is indenpendt veriable

First, I let a=375.362 to fit w and n.
FindFit[dataAl210K, {0.00138 t + 74.830*(t/375.362)^3*Integrate[(E^x
x^4)/(E^x - 1)^2, {x, 0, 375.362/t}] + n 3*8.314 (w/ t)^2 E^(w/ t)/(-1
+ E^(w/t) )^2}, {n, w},t]
I got {n -> 0.152124, w -> 1043.57} and the S.D.=0.150984 (standard
deviation)

Then, I relax these there fitting parameters and give initial
conditions.
FindFit[dataAl210K, {0.00138 t + 74.830*(t/a)^3*NIntegrate[(E^x x^4)/
(E^x - 1)^2, {x, 0, a/t}]
+n 3*8.314 (w/ t)^2 E^(w/ t)/(-1 + E^(w/t) )^2, {n > 0, a > 0, w >
0}}, {{n, 0.15}, {a, 375}, {w, 1043}}, t]
Intuitively, the S.D. of three parameters should be smaller than two.
However, the S.D. of {n -> 0.904794, a -> 1921.19, w -> 259.28} is
0.273892.

I feel the answer is very close to the first one.
I think Mathematica may change the parameters too quickly,so the
caculation escape the local minimun of the first condition by only few
steps.

Q:Can I constraint Mathematica to change the fitting parameters
slower? Don't escape the initial conditions I give too quickly.
(Can Nonlinearmodelfit do this?)

```

• Prev by Date: Re: treatment on C[i]
• Next by Date: Re: two questions
• Previous by thread: Re: problem
• Next by thread: Re: The standard deviation of Three fitting parameters is bigger than