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Lets say that I have a function $F(x,y)$ that is defined on nonnegative integers (or at least those are the values I care about) and is symmetric, so that $F(x,y)=F(y,x)$. Moreover, I know that for any fixed value of $y$ I have that $F(x,y)$ is a polynomial in $x$ of degree $y$. What can I say about the form of the function $F(x,y)$?

Does it change what we would know if I just know that for any fixed $y$ we have that $F(x,y) = O(x^y)$ rather than a polynomial?

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  • $\begingroup$ The first guess would of course be to guess that $F(x,y)=x^y$ but then you aren't symmetric. And if you say $F(x,y)=x^y+y^x$ then you are no longer $O(x^y)$... $\endgroup$ – user61388 Jan 11 at 20:06
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    $\begingroup$ $F(x,y)$ is a polynomial of degree exactly $y$ or at most $y$? $\endgroup$ – Fedor Petrov Jan 11 at 20:09
  • $\begingroup$ An example would be ${x+y\choose y}$ $\endgroup$ – juan Jan 11 at 20:20
  • $\begingroup$ @juan, $\binom{x + y}y$ is not symmetric in $x$ and $y$. $\endgroup$ – LSpice Jan 11 at 22:53
  • $\begingroup$ @LSpice is not it, really? $\endgroup$ – Fedor Petrov Jan 12 at 0:32
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With the polynomiality assumption a characterization is possible. I doubt that much can be said without it.

Theorem: Assuming that $F(x,y)$ is a polynomial in $x$ of degree $y$ for any $y\in \mathbb Z_{\geq 0}$ we have that $$F(x,y)=\sum_{k\geq 0}\alpha_k (x+y)^k\binom{x+y-2k}{y-k}$$ for some arbitrary sequence $\alpha_0,\alpha_1,\dots.$

Proof: There exist coefficients $\alpha_{k,y}$ such that for each $y\in \mathbb Z_{\geq 0}$ we can write $$F(x,y)=\sum_{0\le k\le y} \alpha_{k,y}(x+y)^k \binom{x+y-2k}{y-k}.$$ Notice that the terms in this expression are nonzero only if $k\le \min{x,y}$. We can prove by induction on $k$ that $\alpha_{k,y}$ doesn't depend on $y$. For the base case $k=0$ we have $F(x,0)=F(0,x)$ which gives $\alpha_{0,0}=\alpha_{0,x}$. For the general case, assume we have shown our claim for $k\le m-1$. Looking at $F(x,m)=F(m,x)$ for arbitrary $x\geq m$, we cancel out the equal terms that we have from the induction hypothesis and we are left with $$\alpha_{m,m}(x+m)^m\binom{x-m}{0}=\alpha_{m,x}(m+x)^m\binom{x-m}{x-m}\implies \alpha_{m,m}=\alpha_{m,x}$$ finishing our proof.

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  • $\begingroup$ If you're stating the theorem in generality, you should mention that $F$ is symmetric. $\endgroup$ – mathworker21 Jan 18 at 3:48
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The general solution for the polynomial case is $$F(x,y) := \sum_{k=0}^{\textrm{min}(x,y)} a_k {x \choose k}{y \choose k}. \tag1$$ where $a_k$ are any coefficients. Clearly it is symmetric in $x$ and $y$. For a fixed $y$ the function is a polynomial in $x$ with maximum degree $y$, and iff all of the $a_k$ are non-zero, then the degree is exactly $y$.

For the $O(x^y)$ case, $F(x,y)$ as defined in equation $(1)$ is a solution, but otherwise the solutions are very general. In brief, they should not grow too fast, but there does not seem to be a simple way to describe the general solution except to state the obvious. That is, $$F(x,0) = O(1),\, F(x,1) = O(x),\, F(x,2) = O(x^2),\, \dots. \tag2$$

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