Surface interpolation in 3D

1. Jun 6, 2014

Jhenrique

If given three points $P_0 = (x_0, y_0)$, $P_1 = (x_1, y_1)$ and $P_2 = (x_2, y_2)$, the polynomial function $f(x)$ that intersect those points is $f(x) = a_2 x^2 + a_1 x^1 + a_0 x^0$.

where:
$\begin{bmatrix} a_0\\ a_1\\ a_2\\ \end{bmatrix} = \begin{bmatrix} x_0^0 & x_1^0 & x_2^0 \\ x_0^1 & x_1^1 & x_2^1 \\ x_0^2 & x_1^2 & x_2^2 \\ \end{bmatrix}^{-T} \begin{bmatrix} y_0\\ y_1\\ y_2\\ \end{bmatrix}$

Ok, but if is given points in the space with coordinates $P_1 = (x_1 ,y_1 ,z_1)$, $P_2 = (x_2 ,y_2 ,z_2)$, $P_3 = (x_3 ,y_3 ,z_3)$, ..., $P_n = (x_n ,y_n ,z_n)$... is possible to determine the coefficients of the polynomial function of 2 variables, $f(x,y) = Ax^2+Bxy+Cy^2+Dx+Ey+F$, in function of the point's coordinates (like above in the 2D case)?

2. Jun 27, 2014

JJacquelin

Hi !
The method to solve this problem is given in section 4, pages 8-9 of the paper : http://fr.scribd.com/doc/14819165/Regressions-coniques-quadriques-circulaire-spherique
The correspondance of notations with your notations is as follows :
The polynomial function of 2 variables is: $f(x,y) = a_1x^2+a_2xy+a_3y^2+a_4x+a_5y+a_6$
The given points are $P_1 = (x_1 ,y_1 ,z_1)$, $P_2 = (x_2 ,y_2 ,z_2)$, $P_3 = (x_3 ,y_3 ,z_3)$, ..., $P_n = (x_n ,y_n ,z_n)$...
$F_1(x,y)=x^2$
$F_2(x,y)=xy$
$F_3(x,y)=y^2$
$F_4(x,y)=x$
$F_5(x,y)=y$
$F_6(x,y)=1$
The number of coefficients is $p=6$
Then with the formulas given in the paper, compute the sums $B_{i,j}$ and $C_i$ where $0<i<7$ and $0<j<7$
In the formula of $C_i$ written in the paper, the symbol $y_k$ must be replaced by your symbol $z_k$
Then, solve the matrix system as shown in the paper page 9, leading to the values of the coefficients.