find all rational functions such that $f(x,y,z)=f(\frac{y+z}{2},\frac{x+z}{2},\frac{x+y}{2})$

Question

Find all rational functions such that
$f(x,y,z)=f(\frac{y+z}{2},\frac{x+z}{2},\frac{x+y}{2})$

I'm pretty sure it implies it must be a rational function in $x+y+z$ but I've not been able to prove it.

Answer 1

I'm assuming you are working on the real field, but the following ideas should generalize. Forget for a moment that the functions you're looking for should be rational. They are just functions defined on some subset of the vector field $V=\mathbb{R}^3$ to $V$ itself. You're then looking for those functions which are preserved by the transformation $T:V\rightarrow V$ such that $T(x)=\frac{y+z}{2}$ and so on as you defined.

Now, $T$ is a linear map and it's symmetric. You can find its eigenvalues and eigenvectors; they are $\lambda_1=1$, with eigenvectors spanned by $u = (1,1,1)$, and $\lambda_2=-\frac{1}{2}$, with eigenvectors spanned by $w = (-1,1,0)$ and $v = (-1,0,1)$.

Notice now that the condition you want your functions to respect becomes a little easier to understand. Actually, you're looking for those functions $f$ such that $$f(u,w,v)=f(u,-\frac{w}{2},-\frac{v}{2}).\qquad\qquad (\star)$$

Now let's focus on rational functions. Things are going to become quite messy. Let's say $f(u, w, v) = \frac{P(u, w, v)}{Q(u, w, v)}$ is a rational function with coefficients in $\mathbb{R}$ that satisfies $(\star)$. After a few calculations you find that it must satisfy $$P(u, w, v)Q(u, -\frac{w}{2},-\frac{v}{2}) - P(u, -\frac{w}{2},-\frac{v}{2})Q(u, w, v) = 0. \quad (\star\star)$$ You can write: $$P(u,w,v) = \sum_{(i_1,i_2,i_3)\in I}a_{(i_1,i_2,i_3)}u^{i_1}w^{i_2}v^{i_3}$$ and $$Q(u,w,v) = \sum_{(j_1,j_2,j_3)\in J}b_{(j_1,j_2,j_3)}u^{j_1}w^{j_2}v^{j_3},$$ where $I, J$ are finite subsets of $\mathbb{N}^3$ and each $a_{(i_1,i_2,i_3)}$ and $b_{(j_1,j_2,j_3)}$ is in $\mathbb{R}$. You can then put these espressions in $(\star\star)$ and you'll get this scary equation: $$\sum_{(l_1, l_2, l_3)\in L} u^{l_1}w^{l_2}v^{l_3}\sum_{\substack{i_1 + j_1 = l_1 \\i_2 + j_2 = l_2 \\i_3 + j_3 = l_3}} a_{(i_1,i_2,i_3)}b_{(j_1,j_2,j_3)}\bigg(\bigg(-\frac{1}{2}\bigg)^{j_2 + j_3} - \bigg(-\frac{1}{2}\bigg)^{i_2 + i_3}\bigg) = 0$$ where $L\subset\mathbb{N}^3$ is actually $I+J$ (the set of the sums of $I$ and $J$) and is finite.

This is a system with $a_{(i_1,i_2,i_3)}$ and $b_{(j_1,j_2,j_3)}$ as variables. Once you fixed the degrees of $P$ and $Q$, you can in principle solve it (I'll kill myself rather then trying to find a direct expression for the solutions).

Anyway, it suggests that other solutions in addition to those found by you are possible. For example, anytime $P$ and $Q$ are homogeneous in $w$ and $v$ with same degree we get a solution (it's easy to see it using $(\star\star)$). Let's take $P(u, w, v)=w$ and $Q(u, w, v) = v$ and we get $f(u, v, w) = \frac{w}{v}$. Going back to the $\{x, y, z\}$ basis, we get $$ f(x, y, z) = \frac{-x+y}{-x+z}. $$ This can't be expressed as a function of $x+y+z$, otherwise it would be a function of $u$ in the other basis.