Simple polynomial interpolation in two dimensions is not always possible. For example, suppose that the following data are to be represented by a polynomial of first degree in $x$ and $y$, $p(t)=a+bx+cy$, where $t=(x,y):$
Data: $f(1,1) = 3, f(3,2)=2, f(5,3)=6$
Show that it is not possible.
I've been trying to think of ways to use either Newton's Form or Divided Differences to prove this is not possible, but can't come up with how to work it out. Looking for any help :)
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More generally, suppose $p(t)=a+bx+cy$ fits three points $(x_i, y_i, z_i), i=1,2,3$.
Then $a+bx_1+cy_1=z_1$, $a+bx_2+cy_2=z_2$, $a+bx_3+cy_3=z_3$.
Subtracting the first two, $b(x_2-x_1)+c(y_2-y_1) = z_2-z_1$. Subtracting the first and third, $b(x_3-x_1)+c(y_3-y_1) = z_3-z_1$.
For this system of equations for $b$ and $c$ to have a solution, if $z_2\ne z_1$ or $z_3 \ne z_1$, we must have the determinant $(x_2-x_1)(y_3-y_1)-(x_3-x_1)(y_2-y_1) \ne 0$.
In this case, the determinant is $(3-1)(3-1)-(5-1)(2-1) =2\times 2-4\times 1 =0$, so there is no solution.
Since you want a polynomial of degree $\leqslant 1$, you have the three equations
$$\begin{align} a + b + c &= 3\tag{1}\\ a + 3b + 2c &= 2\tag{2}\\ a + 5b + 3c &= 6\tag{3} \end{align}$$
Subtracting $(1)$ from $(2)$ yields $2b + c = -1$, and subtracting $(2)$ from $(3)$ yields $2b + c = 4$. These are incompatible.