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I've got a question that I've become curious about as a result of supervising some undergraduate research. Let's suppose we have some sequence of polynomials $f_0, f_1, f_2, \cdots \in \mathbb{Z}[X]$, where $f_0=1$. Now, define the following sequence of symmetric polynomials on variables $x_1, \ldots x_n$:

$$P_m(x_1, \ldots, x_n)=\sum\limits_{m_1+\cdots+m_n=m}f_{m_1}(x_1) \cdots f_{m_n}(x_n)$$

If you want, you can think of this as the coefficient of $y^m$ in the two-variable generating function $\prod\limits_{i=1}^{n}\left(\sum\limits_{j \ge 0}f_j(x_i)y^j\right)$. Now my question is, if you know $x_1, \ldots, x_n$ are positive integers, and you know the values of $P_m(x_1, \ldots, x_n)$ for $m=0, 1, \ldots$, what conditions need to be true on $f_1, f_2, \ldots$ in order for the values of of $x_1, \ldots, x_n$ to be completely determined (up to ordering)? I think it's key that we need to work with inputs in positive integers - if we work over $\mathbb{C}$, the answer may be different (see at the bottom).

For example, if $f_1(x)=x$, and $f_2=f_3=\cdots=0$, then $P_m$ is just the $m$-th elementary symmetric polynomial $\sigma_m$, and then obviously, $x_1, \ldots, x_n$ are determined by $\sigma_1, \sigma_2, \ldots$, being the roots of the polynomial with coefficients $(-1)^i\sigma_i$. If $f_i(x)=x^i$ for each $i$, then it's a little less straightforward, but still not hard: $P_m=\sum\limits_{j=1}^{m}(-1)^{j-1}\sigma_j P_{m-j}$, and so we can show by induction on $m$ that $P_1, \ldots, P_m$ together determine $\sigma_1, \ldots, \sigma_m$, and hence, $x_1, \ldots, x_n$ are again determined. For the general case, we can equivalently ask whether the values of $P_1, P_2, \ldots$ uniquely determine the values of $\sigma_1, \sigma_2, \ldots$ (and this is perhaps a more natural question).

As long as the polynomials $f_1$ aren't all constants, I don't, off the top of my head, know any sequences of polynomials $f_1, f_2, \ldots$ for which the values of $P_1, P_2, \ldots$ don't determine $x_1, \ldots, x_n$. So I'm wondering if it's true given that at least one of the $f_i$'s is nontrivial.

I feel it's worth pointing out that what I'm asking isn't the same thing as asking that the $P_i$ generate the ring of symmetric polynomials. For example, if the values of the $P_i$ were to determine the values of $\sigma_1^2, \sigma_2^2, \ldots$, that would determine the values of $\sigma_1, \sigma_2, \ldots$, even if the $P_i$ don't generate the ring. This is why the condition that the $x_i$'s are positive integers is important. That said, I don't know the answer over $\mathbb{C}$ either.

Anyone have any ideas? If it's of any relevance, in the student's research, the polynomial $f_i$ has degree $2i$ for each $i$.

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Taking the $\log$ of the two-variable generating function might help. It converts it to $$ \sum_{i=1}^n \log(1 + F(x_i)) = \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k} \sum_{i=1}^n F(x_i)^k,$$ where $F(x)= f_1(x)y + f_2(x)y^2 + \dots $. So for instance, the coefficient of $y$ tells you $f_1(x_1) + \dots + f_1(x_n)$, the coefficient of $y^2$ tells you $f_2(x_1) + \dots + f_2(x_n)- (f_1(x_1)^2 + \dots + f_1(x_n)^2)/2$, etc.

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  • $\begingroup$ Thanks for the comment! It turns out there's a counterexample to the statement I gave (courtesy of Bjorn Poonen), but this statement is stronger than what I needed. I started fiddling around with something else involving symmetric polynomials, and I started getting a calculation quite similar to what you've got here, which I think may help out the student. $\endgroup$ – Krishanu Sankar Jul 31 '13 at 5:10

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