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MathGroup Archive 1996

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Comment: Fit in Excel Mathlink & some

  • To: mathgroup at
  • Subject: [mg5039] Comment: Fit in Excel Mathlink & some
  • From: "Daitaro Hagihara" <daiyanh at>
  • Date: Sat, 19 Oct 1996 16:40:39 -0400
  • Sender: owner-wri-mathgroup at

Mark Dowell wrote:
> Dear All,
> Can anyone hellp I'm trying to fit some data under Excel with the =
> 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 =
> 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.


>From: afritzse at (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)


    y1=3D=3Da0+a1 Cos[x1]+b1 Sin[x1]+a2 Cos[2 x1]+b2 Sin[2 x1]+a3 Cos[3 =
        b3 Sin[3 x1], 
    y2=3D=3Da0+a1 Cos[x2]+b1 Sin[x2]+a2 Cos[2 x2]+b2 Sin[2 x2]+a3 Cos[3 =
        b3 Sin[3 x2],
    y3=3D=3Da0+a1 Cos[x3]+b1 Sin[x3]+a2 Cos[2 x3]+b2 Sin[2 x3]+a3 Cos[3 =
        b3 Sin[3 x3],
    y4=3D=3Da0+a1 Cos[x4]+b1 Sin[x4]+a2 Cos[2 x4]+b2 Sin[2 x4]+a3 Cos[3 =
        b3 Sin[3 x4],
    y5=3D=3Da0+a1 Cos[x5]+b1 Sin[x5]+a2 Cos[2 x5]+b2 Sin[2 x5]+a3 Cos[3 =
        b3 Sin[3 x5],
    y6=3D=3Da0+a1 Cos[x6]+b1 Sin[x6]+a2 Cos[2 x6]+b2 Sin[2 x6]+a3 Cos[3 =
        b3 Sin[3 x6],
    y7=3D=3Da0+a1 Cos[x7]+b1 Sin[x7]+a2 Cos[2 x7]+b2 Sin[2 x7]+a3 Cos[3 =
        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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