       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)

```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)^(2^n) + (1 - Sqrt)^(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: Table[Simplify[Expand[(1/2) ((1 + Sqrt)^(2^n) + (1 -
Sqrt)^(2^n))]], {n, 1, 6}]
Out: {14, 248, 102272, 20489142272, 839425017825619607552,
1409268686920894404615927074915795024740352}
In: 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:{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

```

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