Re: Function Fit to 3D data set

• To: mathgroup at smc.vnet.net
• Subject: [mg57764] Re: Function Fit to 3D data set
• From: dh <dh at metrohm.ch>
• Date: Tue, 7 Jun 2005 05:59:52 -0400 (EDT)
• References: <d83eh8\$o6e\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Hi Adrienne,

sincerely, Daniel

Adrienne R wrote:
> I have been trying to do several analyses with an [x,y,z] data set, where z is dependent on x and y.  First, I tried a 3D plot.  However, the only way I was able to get Mathematica to plot my data was to assume a grid form for x and y, and use ListPlot3D.  First question: can Mathematica make a 3D plot from stated x,y,z points?
ListPlot3D will do the job
>
> Secondly, I am trying to fit a function to the data.  I have determined a function to which to fit parameters.  However, Mathematica does not seem to want to accept y as a variable.  Here is a simple suggestion of what I was trying.  Note, my function is much more complicated.  Please also ignore if the given does not work as stated, as I'm making it up.
>
> points = {{1, 1, 1}, {1, 2, 4}, {1, 4, 16}}
> f1 = FindFit[points, a + b*x^0 * y^1 + c*x^0 *y^2 +k*x^2 *y^0 + l*x^2 * y^1, {a, b, c, k, l}, x, y]
do not dispair, it is simply a syntax error. FindFit wants the variable
names as a list like:
FindFit[...,{x,y}]

>
> The error I receive is \!\(\*
>   RowBox[{\(FindFit::"nonopt
>   "\), \(\(:\)\(\ \)\), "\<\"Options expected (
>     instead of \\!\\(y\\)) beyond
>       position \\!\\(4\\) in \\!\\(\[LeftSkeleton] 1 \[RightSkeleton]\\). An \
> option must be a rule or a list of rules. \\!\\(\\*ButtonBox[\\\"Moreâ?¦\\\", \
> ButtonStyle->\\\"RefGuideLinkText\\\", ButtonFrame->None, \
> ButtonData:>\\\"General::nonopt\\\"]\\)\"\>"}]\)
>
> Second question: Can Mathematica take a set of x,y,z data where z is dependent on x and y and fit a function to it?
>
> Secondary question for both: How?
>
> Thanks for the help!
>

```

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