Interpolation with 2 variables

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The discussion focuses on the interpolation of polynomial functions with two variables, specifically how five points in a plane can uniquely determine a quadratic equation of the form Ax² + Bxy + Cy² + Dx + Ey + F = 0, provided no three points are collinear. The coefficients of this quadratic can be derived by solving a linear system similar to the one-variable case. The mathematical representation of the coefficients is established through a matrix equation involving the coordinates of the points. This method extends the concept of polynomial interpolation to multiple variables.

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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}##

And this ideia can extended for ##P_n## points... so, analogously, given a set of points, exist a relationship between the coefficients of ##Ax^2+Bxy+Cy^2+Dx+Ey+F=0## and the coordinates of the points?
 
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Yes, that is a standard result. Five points in the plane determine a quadratic in two variables, if no three are collinear the quadratic will be unique and non degenerate. The coefficients can be found by solving a linear system like in the one variable case.

http://en.wikipedia.org/wiki/Five_points_determine_a_conic
 
lurflurf said:
Yes, that is a standard result. Five points in the plane determine a quadratic in two variables, if no three are collinear the quadratic will be unique and non degenerate. The coefficients can be found by solving a linear system like in the one variable case.

http://en.wikipedia.org/wiki/Five_points_determine_a_conic

Yeah! But don't exist the equation of the coefficients in function of the points' coordinates in the wikipage...
 
Last edited:

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