I would like to know if it is possible to tweak the coefficients of an $n$-th order bivariate polynomial such that the function passes through a certain set of points on the boundary?
So if I have a function of the form:
\begin{equation} f(x,y) = \sum^{n}_{i,j} a_{ij}x^{i}y^{j} \end{equation}
that when plotted looks something like this:
is it possible to choose coefficients so that along the lines $x=1$ and $x=-1$ the function is equal to some constant value?
We have \begin{align} f(1,y)&=\sum_{j=0}^n\Bigl(\sum_{i=0}^na_{ij}\Bigr)y^j\\ f(-1,y)&=\sum_{j=0}^n\Bigl(\sum_{i=0}^n(-1)^i\,a_{ij}\Bigr)y^j. \end{align} Since we want the right hand side to be constants, we are lead to a system of $2(n+1)$ equations with $(n+1)^2$ unknowns: \begin{align} \sum_{i=0}^na_{i0}&=C_1\\ \sum_{i=0}^n(-1)^i\,a_{i0}&=C_2\\ \sum_{i=0}^na_{ij}&=0,\quad 1\le j\le n\\ \sum_{i=0}^n(-1)^ia_{ij}&=0,\quad 1\le j\le n \end{align} For $n=1$, the only solution is a polynomial independent of $y$. For $n=2$ the solutions are $$ f(x,y)=a_{00}+a_{10}\,x+a_{20}\,x^2+\alpha(1-x^2)y+\beta(1-x^2)y^2. $$ It is clear now that a solution for arbitrary $n>1$ is $$ P(x)+(1-x^{2})Q(x)\sum_{i=0}^nb_{i}\,y^i. $$ where $P$ is a polynomial of degree $n$ and $Q$ is a polynomial of degree $n-2$.