Interpolating multivariable functions Assume I have two functions, $f_1$ and $f_2$, that both depend on $x$ and $y$, so that $f_1(x,y)$ and $f_2(x,y)$.
I don't know the exact functions, but know values of each function at some points (actually, any points I want).
So, for example, let's say that:
at $x=+1$ and $y=-1$,
        $$f_1=+9,$$
        $$f_2=+7,$$
at $x=-1$ and $y=+1$,
        $$f_1=-2,$$
        $$f_2=-6,$$
at $x=+1$ and $y=+1,$
        $$f_1=+11,$$
        $$f_2=+9.$$
at $x=-1$ and $y=-1,$
        $$f_1=-7,$$
        $$f_2=-8.$$
I need to find the values for $x$ and $y$ where both functions equal $0$.
If each function only depended on one variable a linear interpolation would suffice. But as both functions depend on $2$ variables I'm getting a bit confused.
I've been searching on bilinear and trilinear interpolation, but I can't really pinpoint what I actually need to use.
Thank you all.
 A: How about this way? Suppose the function can be approximated by a polynomial of degree 1, that is, 
$$f_1(x, y) = c_{00} + c_{10}x + c_{01}y$$
for some constants $c_{00}$, $c_{10}$, and $c_{01}$.
From the first three conditions, we have
$$
\begin{bmatrix}
1 & 1 & -1 \\
1 & -1 & 1 \\
1 & 1 & 1
\end{bmatrix}
\begin{bmatrix}
c_{00} \\
c_{10} \\
c_{01}
\end{bmatrix} =
\begin{bmatrix}
f_1(1, -1) \\
f_1(-1, 1) \\
f_1(1, 1)
\end{bmatrix} =
\begin{bmatrix}
9 \\
-2 \\
11
\end{bmatrix}
$$
and its solution is $c_{00} = 7/2$, $c_{10} = 13/2$, and $c_{01} = 1$.
Hence
$$ f_1(x, y) \approx \frac{7}{2} + \frac{13}{2}x + y. $$
Do it again for $f_2$ and we get
$$ f_2(x, y) \approx \frac{1}{2} + \frac{15}{2}x + y. $$
Lastly, solving
$$\begin{cases}
f_1(x, y) = 0 \\
f_2(x, y) = 0
\end{cases}$$
gives you $x = 3$ and $y = -23$.
If you would like to get more accurate approximation, try to do this by polynomial with higher degree.
A: Just another try. Since it seems, the coordinates are of a plane,I interpret them as complex values. Then from the four given arguments and the four resulting values I do an ordinary polynomial interpolation procedure. This gives the complex polynomial:
$$ \operatorname{ cofu } (z)= (-25/16 - 37/16 I) z^3 + 3/8 I z^2 + (33/8 + 23/8 I) z + (11/4 + 1/2 I) $$
This gives then, for instance, $$ \operatorname{cofu}(1-1 I) = 9 + 7I
 $$
One can plot this using mathematica, getting a rough impression - don't find the best command-parameters,though... 
Then one can use a complex root-finder, as for instance in Pari/GP (or again in Wolfram-mathematica). We get using Pari/GP:
$$ \operatorname{polroots}(\operatorname{cofu}(z)) = \\ [1.57353592868 - 0.292970871914 I, \\ -0.931132188421 - 0.0424173327510 I, \\ -0.531069738252 + 0.410613881696 I] $$ 
