Prove that $f(x,y)=g(x+y,xy)$.

Question

Let $f\in \mathbb R[x,y]$ such that $f(x,y)=f(y,x)$. Prove that $f(x,y)=g(x+y,xy)$.

My try.

$f\in \Bbb R[x,y]$ and $f(x,y)=f(y,x)$ .If I take for example $f(x,y)=a_{00}+a_{01}y+a_{11}xy+a_{10}x+a_{13}xy^3$ then only the coefficients which are symmetric w.r.t $x $ and $y$ will remain in the expression of $f$ ,all other terms will cancel out because $f(x,y)=f(y,x)$.

But How can I write down the equation of $f$ from here?

Please provide some hints.

Answer 1

Set x+y=A and xy=G

we get

The resultant is a function since it is not one-many (f(x,y)=f(y,x)) and not many-many , since we have transformed a set of ordered pairs into another.

Answer 2

Write $f = f_{0} + f_{1} + \dots + f_{n}$, where each $f_{i}$ is homogenous of degree $i$ and symmetric.

You want to prove that each $f_{i}(x, y)$ is a polynomial in $x + y, x y$.

This is obvious for $f_{0}$ and $f_{1}$. We do also the case $i = 2$, as an example of the induction step that follows. We have \begin{align} f_{2} &= a_{2} x^{2} + a_{1} x y + a_{2} y^{2} \\&= a_{2} (x^{2} + 2 x y + y^{2}) + (a_{1} - 2 a_{2}) x y \\&= a_{2} (x + y)^{2} + (a_{1} - 2 a_{2}) x y. \end{align}

In general, proceeding by induction on $i \ge 2$, let $$ f_{i} = a_{i} x^{i} + a_{i-1} x^{i-1} y + \dots a_{i-1} x y^{i-1} + a_{i} y^{i} $$ for some $a_{j}$. Then there are $b_{j} = a_{j} - a_{j} \binom{i}{j}$ such that \begin{align} f_{i} &= a_{i} (x + y)^{i} + b_{i-1} x^{i-1} y + \dots b_{i-1} x y^{i-1} \\&= a_{i} (x + y)^{i} + x y (b_{i-1} x^{i-2} + b_{i-2} x^{i-3} y + \dots + b_{i-2} x y^{i-3} + b_{i-1} y^{i-2}), \end{align} and now by induction $$ b_{i-1} x^{i-2} + b_{i-2} x^{i-3} y + \dots + b_{i-2} x y^{i-3} +b_{i-1} y^{i-2} $$ is a polynomial in $x + y, x y$. The point is that as the $a_{j}$ are symmetric, that is, $a_{i-j} = a_{j}$, so are the $b_{j}$, as $b_{i-j} = a_{i - j} - a_{i-j} \binom{i}{i-j} = a_{j} - a_{j} \binom{i}{j} = b_{j}$.

Answer 3

Let $f\in \mathbb R[x,y]$ such that $f(x,y)=f(y,x)$. To prove that $f(x,y)=g(x+y,xy)$.

Let $$f(x,y)=a_{0,0}x^0y^0+a_{0,1}x^0y^1+a_{0,2}x^0y^2+a_{1,0}x^1y^0+a_{1,1}x^1y^1+a_{1,2}x^1y^2+a_{2,0}x^2y^0+a_{2,1}x^2y^1+a_{2,2}x^2y^2$$ We then have

$$f(y,x)=a_{0,0}y^0x^0+a_{0,1}y^0x^1+a_{0,2}y^0x^2+a_{1,0}y^1x^0+a_{1,1}y^1x^1+a_{1,2}y^1x^2+a_{2,0}y^2x^0+a_{2,1}y^2x^1+a_{2,2}x^2y^2$$

As $f(x,y)=f(y,x)$ we have $a_{0,1}=a_{1,0}, a_{2,0}=a_{0,2},a_{1,2}=a_{2,1}$.. So, we have

$$f(x,y)=a_{0,0}x^0y^0+a_{0,1}x^0y^1+a_{0,2}x^0y^2+a_{0,1}x^1y^0+a_{1,1}x^1y^1+a_{1,2}x^1y^2+a_{0,2}x^2y^0+a_{1,2}x^2y^1+a_{2,2}x^2y^2=a_{0,0}x^0y^0+a_{0,1}(x^0y^1+x^1y^0)+a_{0,2}(x^0y^2+x^2y^0)+a_{1,1}x^1y^1+a_{1,2}(x^1y^2+x^2y^1)+a_{2,2}x^2y^2=a +bxy+c(xy)^2+d(x+y)+e((x+y)^2-2xy)+fxy(x+y)$$