Question

Explain why fitting a cubic through the mpoints ${\mathbf{P}}_{\mathbf{1}}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{y}}_{\mathbf{1}}\mathbf{)}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathbf{P}}_{\mathbf{m}}\mathbf{(}{\mathbf{x}}_{\mathbf{m}}\mathbf{,}{\mathbf{y}}_{\mathbf{m}}\mathbf{)}$amounts to finding the kernel of an mx10matrix A. Give the entries of theof row A.

Short answer

Thus, m points is same as finding the kernel of a m x 10 matrix and the ithrow entries are $\left[1x_i{y}_j{x}_i^2{x}_i{y}_i{y}_i^2{x}_i^3{x}_i^2{y}_i{x}_i{y}_i^2{y}_i^3\right]$.

Step-by-step solution

Explanation for points of the cubic.

If nine points to fit to the cubic as follows:

$\mathbf{f}\left(x,y\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{xc}}_{\mathbf{2}}\mathbf{+}{\mathbf{yc}}_{\mathbf{3}}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}{\mathbf{c}}_{\mathbf{4}}\mathbf{+}{\mathbf{xyc}}_{\mathbf{5}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}{\mathbf{c}}_{\mathbf{6}}\mathbf{+}{\mathbf{x}}^{\mathbf{3}}{\mathbf{c}}_{\mathbf{7}}\mathbf{+}{\mathbf{x}}^{\mathbf{3}}{\mathbf{yc}}_{\mathbf{8}}\mathbf{+}{\mathbf{xy}}^{\mathbf{2}}{\mathbf{c}}_{\mathbf{9}}\mathbf{+}{\mathbf{y}}^{\mathbf{3}}{\mathbf{c}}_{\mathbf{10}}\mathbf{=}\mathbf{0}$

A system of m distinct points is same as obtaining the kernel of a m x 10 matrix.

The first ith row of A is:

$\left[1x_i{y}_j{x}_i^2{x}_i{y}_i{y}_i^2{x}_i^3{x}_i^2{y}_i{x}_i{y}_i^2{y}_i^2\right]$

With columns represent as $c_1,...,c_{10}$.

Determine the solutions of all cubics.

Since the form of cubic always equals to zero, to obtain all solutions$c_1,...,c_{10}$such the cubic equal 0 for the given points. By definition, this is the equivalent to obtain the kernel solutions to$A\stackrel{\rightharpoonup }{x}=0$of the matrix m x 10

Hence, m points is same as finding the kernel of a m x 10 matrix and the ith row entries are

$\left[1x_i{y}_j{x}_i^2{x}_i{y}_i{y}_i^2{x}_i^3{x}_i^2{y}_i{x}_i{y}_i^2{y}_i^3\right]$.