Finding $\sum_i x_i$ given $\{\sum_i x_i^{2n}\}_{n\in \mathbb{N}}$

Question

Can we find $\sum_i x_i$ given $\{\sum_i x_i^{2n}\}_{n\in \mathbb{N}}$, assuming $\{x_i\}_{i\in\mathbb{N}}$ is a set of positive real numbers? Perhaps an easier question is, can we find $\sum_i x_i$ given $\{\sum_i x_i^{n}\}_{n\in \mathbb{N}\setminus 1}$? Intuitively it seems impossible without further assumptions on $\{x_i\}_{i\in\mathbb{N}}$. Is that true?

Answer 1

If you plot $\log f_n$ versus $n$, with $f_n=\sum_i x_i^{2n}$, then the asymptotic slope for large $n$ will give you the largest of the $x_i$; subtracting that contribution from $f_n$ and repeating will give you the next-to-largest $x_i$, and so on; there will be issues of numerical accuracy, but as a matter of principle it should work.

Answer 2

If $\sum x_i^2$ is finite, the sum $f(z)=\sum \frac{x_i^2}{1-x_i^2z}$ is a meromorphic function on the complex plane, and we know its Taylor series at 0. Thus we know $f$, hence the poles of $f$, hence $x_i$'s.

Answer 3

In terms of the finite and compactly supported measure $\mu=\sum x_j^2 \delta_{x_j^2}$, we are given the moments $\int t^k\, d\mu(t)$, $k=0,1,2 \ldots$.

Moment problems with compactly supported solutions are determinate, that is, they have unique solutions.

In fact, if this is combined with the Muntz-Szasz theorem, we see that it would be enough to know $\sum x_j^{a+bN_n}$ to recover the $x_j$ as long as $\sum 1/N_n=\infty$. (In this case, define $\mu=\sum x_j^a \delta_{x_j^b}$.)

Answer 4

[edit: completed] Assuming $x_i\ge0$ with $ \sum_i x_i <\infty$, we have that $\phi(t):=\sum_i(e^{x_it}-1)=\sum_{k\ge1} \big(\sum_i x_i^k\big)t^k/k!$ is an entire function (we can expand the exponentials and exchange order of summation, by absolute summability).

Now if for an other non-negative sequence $y_i$ we have $\sum y_i <\infty$ and $\sum_i x_i^k=\sum y_i^k$ for all $k> N$ we consider the corresponding entire function $\psi(t):=\sum_i(e^{y_it}-1)$: then $\phi(t)-\psi(t)=\sum_{k=1}^N(\sum_i x_i^k - \sum_i y_i^k)t^k/k!$ is a polynomial. It is sufficient to prove that $\phi(t) -\psi(t) =o(|t|)$ for $t\to-\infty$, and it follows it is identically zero, so $\sum_i x_i^k=\sum_i y_i^k$ for $1\le k\le N$ too.

To show $\phi(t)-\psi(t)=o(|t|)$ for (real) $t\to -\infty$: each term $e^{x_it}-e^{y_it}$ tends to $0$ as $t\to-\infty$, and $|e^{x_it}-e^{y_it}|\le |t||x_i-y_i|\le |t|(x_i+y_i)$ because the function $\exp$ is $1$-Lipschitz on $\mathbb R_-$. Therefore by dominated convergence of series, $(\phi(t)-\psi(t))/t\to0$ as $t\to-\infty$, that is $\phi(t)-\psi(t)=o(|t|)$.

Answer 5

Here is what happens without assumptions.

Functions $p_n:=\sum_i x_i^n$ represent power-sum symmetric polynomials. By Newton's identities, we have $$E(t):=\sum_{k=0}^\infty e_k \,t^k = \exp\left(\sum_{k=1}^\infty \frac{(-1)^{k+1}p_k}{k} \,t^k \right).$$ Knowing all $p_{2n}$ is equivalent to knowing $$E(t)E(-t) = \exp\left(-\sum_{k=1}^\infty \frac{p_{2k}}{k} \,t^{2k} \right).$$ Hence, the question amounts to computing series $E(t)$ from the known series $E(t)E(-t)$.

Let $z:=t^2$ and $E(t) = f(z) + tg(z)$ be the sum of even and odd parts. Then $$E(t)E(-t) = f(z)^2 - zg(z)^2 = \exp\left(-\sum_{k=1}^\infty \frac{p_{2k}}{k} \,z^k \right).$$ To solve this, we can take an arbitrary series $g(z)$, and then compute $f(z)$ as $$f(z) = \sqrt{\exp\left(-\sum_{k=1}^\infty \frac{p_{2k}}{k} \,z^k \right) + zg(z)^2}.$$ Then we recover values $p_n$ from $$\sum_{k=1}^\infty \frac{(-1)^{k+1}p_k}{k} \,t^k = \log( f(t^2) + tg(t^2)).$$

ADDED. If the set $\{x_i\}$ is finite, then $E(t)$ is a polynomial and so is $E(t)E(-t)$, and the problem can be solved by factoring of the latter polynomial.