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Re: 3D interpolating

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  • Subject: [mg109119] Re: 3D interpolating
  • From: Bill Rowe <readnews at>
  • Date: Mon, 19 Apr 2010 04:06:31 -0400 (EDT)

On 4/12/10 at 11:01 PM, ibmichuco at (michuco) wrote:

>Hi all,

>I am trying to fit a series of 3D points, say

>data = {{{0, 0}, 0}, {{1, 1}, 1}, {{2, 2}, 2}};


>f = Interpolation[data]

>which gives me the error

>Interpolation::indim: The coordinates do not lie on a structured \
>tensor product grid.

>I think that I got the data format wrong, but I can't tell from the
>Help library of Mathematica 6.

>Ideas? Thanks in advance,

The following which has the same structure but more points works:

In[17]:= data = Flatten[Table[{{x, y}, LCM[x, y]}, {x, 4}, {y,
4}], 1];
f = Interpolation[data]

Out[18]= InterpolatingFunction[{{1,4},{1,4}},<>]

By default Interpolation does a cubic interpolation. Three data
points simply isn't enough to uniquely define the parameters of
a cubic curve.

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