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Comment: Fit in Excel Mathlink & some
*To*: mathgroup at smc.vnet.net
*Subject*: [mg5039] Comment: Fit in Excel Mathlink & some
*From*: "Daitaro Hagihara" <daiyanh at mindspring.com>
*Date*: Sat, 19 Oct 1996 16:40:39 -0400
*Sender*: owner-wri-mathgroup at wolfram.com
Mark Dowell wrote:
>
> Dear All,
>
> Can anyone hellp I'm trying to fit some data under Excel with the =
Mathlink
> addon the fitting equation is
>
> y=3Da(0)*Exp(-a(1)*(x-400))+a(2)
>
> I just need value for best fit in a(0),a(1),a(2). for a given range in =
Excel
> cells can anyone help with the syntax
Well, the Standard Package "NonlinearFit" is supposed to do it, but I
had very little luck with problems similar to yours. Maybe things will
improve in Version 3.0
Good luck,
-----
The equation is very typical. I also worked on a similar equation once =
in the past, in Mma v. 1.2-2.0 days. The solution was to write external =
Pascal (C these days) program to do variable matrix method for =
optimization, as described in The Book 'Numerical Recipes'. VM method =
solves almost any optimization problem. At the end of VM run, do =
Newton-Raphson to fine-tune the solution as VM method often gets trapped =
when equations are stiff. Mma comes into the picture for finding the =
derivatives of the cost function, not a trivial task for a large number =
of parameter estimates as was the case in mine. It also comes in handy =
to check for parameter independence, esp. when you are engaging in a =
modeling work, again not so trivial. At any rate, there are a lot more =
work for you to do... Just don't throw away The Book yet.
Also:
>From: afritzse at limmat.ch (Arthur Fritzsche)
>Subject: [mg4979] ts ts ts
Hi there.
Does anybody know why Mathematica gets choked trying to solve
this simple equation? (seems like an easy way to calculate Fourier series =
up to n=3D3)
---------------------------------
x1=3D3
y1=3D3
x2=3D4
y2=3D4
x3=3D5
y3=3D5
x4=3D10
y4=3D-20
x5=3D11
y5=3D30
x6=3D12
y6=3D-30
x7=3D18
y7=3D0
Solve[{
y1=3D=3Da0+a1 Cos[x1]+b1 Sin[x1]+a2 Cos[2 x1]+b2 Sin[2 x1]+a3 Cos[3 =
x1]+
b3 Sin[3 x1],
y2=3D=3Da0+a1 Cos[x2]+b1 Sin[x2]+a2 Cos[2 x2]+b2 Sin[2 x2]+a3 Cos[3 =
x2]+
b3 Sin[3 x2],
y3=3D=3Da0+a1 Cos[x3]+b1 Sin[x3]+a2 Cos[2 x3]+b2 Sin[2 x3]+a3 Cos[3 =
x3]+
b3 Sin[3 x3],
y4=3D=3Da0+a1 Cos[x4]+b1 Sin[x4]+a2 Cos[2 x4]+b2 Sin[2 x4]+a3 Cos[3 =
x4]+
b3 Sin[3 x4],
y5=3D=3Da0+a1 Cos[x5]+b1 Sin[x5]+a2 Cos[2 x5]+b2 Sin[2 x5]+a3 Cos[3 =
x5]+
b3 Sin[3 x5],
y6=3D=3Da0+a1 Cos[x6]+b1 Sin[x6]+a2 Cos[2 x6]+b2 Sin[2 x6]+a3 Cos[3 =
x6]+
b3 Sin[3 x6],
y7=3D=3Da0+a1 Cos[x7]+b1 Sin[x7]+a2 Cos[2 x7]+b2 Sin[2 x7]+a3 Cos[3 =
x7]+
b3 Sin[3 x7]}, {a0, a1, b1, a2, b2, a3, b3}]
---------------------------------
Might there be a more straightforward solution? The sinus and cosinus =
terms evidently are fixed, so....
Gratitude beyond expression,
-----
I could not confirm this claim. The solution was obtainable. If you =
take a look at the solution of a0, for 3 var 3 eq case, there were 13 =
occurences of the form (-SinCos+SinCos). For 4 var 4 eq case, this was =
135. I stopped counting after that. But I'm sure that for 7 var 7 eq =
case, there must be an order of 10^3~4 occurences! Better solve it =
numerically to begin with.
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