Re: complex function fitting?

• To: mathgroup at smc.vnet.net
• Subject: [mg98331] Re: complex function fitting?
• From: "Ned Lieb" <okty.gy.ned at gmail.com>
• Date: Wed, 8 Apr 2009 02:44:13 -0400 (EDT)
• References: <grcgjg\$pml\$1@smc.vnet.net> <49D9DDF3.30605@metrohm.com>

```Just a clarification of my original question: I'm referring to a specific
issue I'm having using the NonlinearModelFit function, which fits lists of
data-points to functions of a type specified by the user (for example, say I
knew that my data fit an exponential equation (it doesn't, by the way, this
is just an example) with x as an independent variable: NonLinearModelFit
could find the constant A in the equation A*e^x that would give the function
that most closely approximated my data). My list of data is complex-valued
(with non-zero imaginary components). Mathematica returned an error message
saying I could only use real numbers. I was wondering whether there was a
way to get around this.

Thanks

-----Original Message-----
From: dh [mailto:dh at metrohm.com]
Sent: Monday, April 06, 2009 6:48 AM
To: Ned Lieb
Subject: [mg98331] Re: complex function fitting?

Hi Ned,
Mathematica works by default with complex numbres. E.g. using "Fit":
d = Table[{x + I y, Exp[x + y I ]}, {x, 0, 1, .1}, {y, 0, 1, .1}];
d = Flatten[d, 1];
pol = Fit[d, {1, x, x^2}, x]
yc = pol /. x -> d[[All, 1]];
ListPlot[{Re[#], Im[#]} & /@  (yc - d[[All, 2]]), PlotRange -> All]
Daniel

Ned Lieb wrote:
> Does anyone know how I could extend the domain of Mathematica's
> function-fitting functions so I can use complex-valued data?
>
>
>

```

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