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Re: Function Fit to 3D data set

  • To: mathgroup at
  • Subject: [mg57767] Re: [mg57748] Function Fit to 3D data set
  • From: Bob Hanlon <hanlonr at>
  • Date: Wed, 8 Jun 2005 03:21:16 -0400 (EDT)
  • Reply-to: hanlonr at
  • Sender: owner-wri-mathgroup at






-3.6191002429477375*y*x^2 + 2.369175079975754*x^2 - 
1.949339720581435*y^2 + 
  4.0513659678579135*y - 1.1985125893996162


Bob Hanlon

> From: Adrienne R <roehrich at>
To: mathgroup at
> Date: 2005/06/07 Tue AM 02:03:52 EDT
> Subject: [mg57767] [mg57748] Function Fit to 3D data set
> 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?
> 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]
> 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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