Re: FindFit
- To: mathgroup at smc.vnet.net
- Subject: [mg92748] Re: [mg92733] FindFit
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Sun, 12 Oct 2008 04:32:21 -0400 (EDT)
- Reply-to: hanlonr at cox.net
As stated in the documentation: "In the nonlinear case, it finds in general only a locally optimal fit."
data = Table[
Simplify[Expand[(1/2) ((1 + Sqrt[13])^(2^n) + (1 - Sqrt[13])^(2^n))]], {n,
1, 6}];
model = (1/2) ((a + Sqrt[b])^(2^x) + (a - Sqrt[b])^(2^x));
FindFit[data, model, {a, b}, x]
{a->1.79057,b->2.58114}
Add constraints and use NMinimize to look beyond the local minima
FindFit[data, {model, a > 0, b > 10}, {a, b}, x, Method -> NMinimize]
{a->1.05257,b->12.6236}
Fix one parameter and look for best fit to other
FindFit[data, model /. a -> 1, b, x, Method -> NMinimize]
{b->13.}
Bob Hanlon
---- Artur <grafix at csl.pl> wrote:
=============
Dear Mathematica Gurus,
Who have idea which procedure I can use inspite FindFit in following:
In[1]: Table[Simplify[Expand[(1/2) ((1 + Sqrt[13])^(2^n) + (1 -
Sqrt[13])^(2^n))]], {n, 1, 6}]
Out[1]: {14, 248, 102272, 20489142272, 839425017825619607552,
1409268686920894404615927074915795024740352}
In[2]: FindFit[{14, 248, 102272, 20489142272, 839425017825619607552,
1409268686920894404615927074915795024740352}, (1/
2) ((a + Sqrt[b])^(2^x) + (a - Sqrt[b])^(2^x)), {a, b}, x,
WorkingPrecision -> 100]
Out[2]:{a -> 2.43785286368448657933626328600,
b -> 4.95343824602875750402007143975}
Should be: {a -> 1.00000000000000000000,
b -> 13.000000000000000000}
Who have idea how find (true) a,b when I have sequence
Best wishes
Artur
--
Bob Hanlon