Re: 3D interpolating

*To*: mathgroup at smc.vnet.net*Subject*: [mg109119] Re: 3D interpolating*From*: Bill Rowe <readnews at sbcglobal.net>*Date*: Mon, 19 Apr 2010 04:06:31 -0400 (EDT)

On 4/12/10 at 11:01 PM, ibmichuco at hotmail.com (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}}; >with >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.