# 3D interpolation

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 August 19, 2004, 09:13 3D interpolation #1 Lipo Wang Guest   Posts: n/a Now I want to do parabolic interpolation in 3-D space Cartesian Coordiate. The form of parabolic should be: a0+a1*x+a2*x^2+b0*y+b1*y^2+b2*x*y+c0*z+ c1*z^2+c2*y*z+c3*x*z Totally 10 unknown coefficients. But for a cube, there are totally 8 end points. And if we also consider the information of 1st derivatives, then totally we have 32 known conditions. So the number of known coditions are not equal to that of unknowns. How to to this problem? Thanks

 August 19, 2004, 10:01 Re: 3D interpolation #2 Markus Lummer Guest   Posts: n/a Try a local tensor product of Lagrange polynomials. Regards, Markus

 August 19, 2004, 11:40 Re: 3D interpolation #3 Lipo Wang Guest   Posts: n/a Hello,Markus Could you explain in more details? Lagrange polynomials is parobolic? What is the form of tensor you said? Thanks a lot Lipo

 August 19, 2004, 12:54 Re: 3D interpolation #4 ag Guest   Posts: n/a You may want to consider a least-squares approach.

 August 20, 2004, 01:12 Re: 3D interpolation #5 Markus Lummer Guest   Posts: n/a 1) 1d case. Let the grid points be x_i, i=0,1,...,nx and the function values f_i = f(x_i) The Langrange polynomials L_p(x) depend only on the x_i (The L_p are of order nx. The definition of L_p should be found in any book about basic numerical mathematics.) and have the property L_p(x_i) = \delta_{ip} The interpolating polynomial then reads f(x) = \sum_{p=0}^nx f_p L_p(x) 2) 3d case. Cartesian grid points (x_i, i=0,1,...,nx), (y_j, j=0,1,...,ny), (z_k, k=0,1,...,nz), and function values f_{ijk} = f(x_i,y_j,z_k) You calculate the Lagrange polynomials L_p(x),L_q(y),L_r(z) and write for the interpolating polynomial f(x,y,z) = \sum_{p=0}^nx \sum_{q=0}^ny \sum_{r=0}^nz f_{pqr} L_p(x)L_q(y)L_r(z) You can apply this formula locally. I.e., e.g. choose nx=ny=nz=2 (quadratic polynomials) and consider the grid points (x_{i-1},x_{i},x_{i+1}) (y_{j-1},y_{j},y_{j+1}) (z_{k-1},z_{k},z_{k+1})

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