# Does this system of infinite equations has an (almost) unique solution?

Let $$a_1,\dots ,a_n \in \Bbb C$$, consider the following system of equations $$\begin{cases} x_1+ \cdots+ x_n=a_1 \\ {x_1}^2+\cdots+{x_n}^2=a_2 \\ \qquad \qquad \vdots \\ {x_1}^n+\cdots+{x_n}^n=a_n \end{cases}$$ Its "easy" to prove this system has a unique solution up to permutations. The reason is due to Newton identities that allow us to create an equivalent system $$\begin{cases} e_1(x_1,\dots,x_n)=b_1 \\ e_2(x_1,\dots,x_n)=b_2 \\ \qquad \qquad \vdots \\ e_n(x_1,\dots,x_n)=b_n \end{cases}$$ Where $$e_1, \dots ,e_n$$ are the elementary symmetric polynomials and $$b_1, \dots ,b_n \in \Bbb C$$ are numbers which can be calculated in terms of $$a_1, \dots ,a_n$$. If we consider the polynomial $$P(X)=X^n-b_1 \cdot X^{n-1}+b_2 \cdot X^{n-2} + \cdots+(-1)^n \cdot b_0$$ Then, due to Vieta's Formulas, the solution to our system are precisely the roots of $$P$$ (counted with multiplicity) which are unique up to permutations. My question is the following, let $$(a_n)_{n\in \Bbb N}$$ be complex numbers and consider the following system of infinite equations in $$l^1(\Bbb C)$$ $$\begin{cases} \sum_{n \in \Bbb N} x_n = a_1 \\ \sum_{n \in \Bbb N} {x_n}^2 = a_2 \\ \qquad \quad \vdots \\ \sum_{n \in \Bbb N} {x_n}^k = a_k \\ \qquad \quad \vdots \end{cases}$$ Are there any necessary/sufficienct conditions over $$(a_n)_{n\in \Bbb N}$$ for a solution to exist? If $$(x_n)_{n\in \Bbb N}$$ is a solution to our system and $$\sigma : \Bbb N \rightarrow \Bbb N$$ is a bijection then $$(x_{\sigma (n)})_{n\in \Bbb N}$$ is a solution to our system so any permutation of a solution is again a solution. Also, if we take a solution $$(x_n)_{n\in \Bbb N}$$ and we "add" zeros to our sequence then we get another solution, for example, the sequence $$(y_n)_{n\in \Bbb N}$$ defined as $$y_{2n}=0 \qquad ; \qquad y_{2n-1}=x_n \qquad \forall n \in \Bbb N$$ Is another solution to our system of equations. My second question would be the following. If our system of infinite equations has two solutions, $$(x_n)_{n\in \Bbb N}$$ and $$(y_n)_{n\in \Bbb N}$$, is it true that I can get $$(y_n)_{n\in \Bbb N}$$ by taking $$(x_n)_{n\in \Bbb N}$$ and adding zeros and making permutations?

Inspired by the finite case, I did the following approach. Its easy to prove (again, using Newton Identities) that if $$(x_n)_{n\in \Bbb N} \in l^1(\Bbb C)$$ then the following limit exists for every $$k \in \Bbb N_0$$ $$e_k(x):=\lim_{n \to \infty} e_k(x_1,\dots ,x_n) < \infty$$ And it can be calculated in terms of $$\sum_{n \in \Bbb N} x_n , \dots , \sum_{n \in \Bbb N} {x_n}^k$$ so we get the following (equivalent) system of equations $$\begin{cases} e_1(x) = b_1 \\ e_2(x) = b_2 \\ \qquad \vdots \\ e_k(x) = b_k \\ \qquad \vdots \end{cases}$$ Where $$(b_n)_{n\in \Bbb N}$$ are complex numbers which can be calculated in terms of $$(a_n)_{n\in \Bbb N}$$. To reproduce the following step in the finite case, we will make some modifications in the approach. By Vieta's formulas, it's easy to check the following polynomial equality $$\prod_{k=1}^n (1-x_i \cdot X) = \sum_{k=0}^n (-1)^k \cdot e_k(x_1,\dots, x_n) \cdot X^k$$ With $$x_1,\dots ,x_n \in \Bbb C$$. Note that the roots of this polynomial are $${x_i}^{-1}$$ for every $$x_i \not = 0$$. I would like to say (and have no idea how to prove) that if $$(x_n)_{n \in \Bbb N} \in l^1(\Bbb C)$$ then $$\prod_{k=1}^\infty (1-x_i \cdot z) = \sum_{k=0}^\infty (-1)^k \cdot e_k(x) \cdot z^k \qquad \forall \; z \in \Bbb C$$ If this is true, going back to our infinite system of equations, we could consider the series $$f(z)=1+ \sum_{k=1}^\infty (-1)^k \cdot b_k \cdot z^k$$ Wich is a function we can "calculate" since we know $$(b_k)_{k \in \Bbb N}$$. We should ask some conditions over $$(b_k)_{k \in \Bbb N}$$ for this to be well defined over an open set around zero (or all over $$\Bbb C$$). Let $$(r_n)_{n \in \Bbb N}$$ be the roots of $$f$$ counted with multiplicity (Does this have a meaning???), let $$x_n = {r_n}^{-1} \quad \forall n \in \Bbb N$$ (Note $$r_n \not = 0 \quad \forall n \in \Bbb N$$ because $$f(0)=1$$), if $$f$$ has finite roots, we fill the sequence with zeros. I affirm that $$(x_n)_{n \in \Bbb N}$$ is a solution to our system. There are a LOT of gaps in my reasoning and Im not sure how to continue. My Complex analysis knowledge is really weak.

• For me $l^1(\Bbb C)$ is the space of absolute convergent sequences. Although the original question does not involve complex analysis, my approach does. I suggest finding the roots of the (holomorfic?) function I defined and each root multiplicity can be a way to find the solution to our system of equations (for example, we know the roots are countable thanks to complex analysis). – Marcos Martínez Wagner Sep 12 '19 at 7:28

• That $$x_n \in \Bbb{C}^*, \qquad \sum_{n=1}^\infty \frac1{|x_n|} < \infty$$ means $$f_x(z) = \sum_{n=1}^\infty \frac{1}{z-x_n}$$ converges absolutely and locally uniformly away from the $$x_n$$ thus it is meromorphic on $$\Bbb{C}$$ with simple poles at each $$x_n$$ of residue $$\# \{m, x_m=x_n\}$$.
• For $$|z| < \inf_n |x_n|$$, from the absolute convergence of $$\sum_{n=1}^\infty \frac{1}{|x_n|-|z|}=\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{|z|^{k-1}}{|x_n|^k}$$ we obtain $$f_x(z) = -\sum_{k=1}^\infty z^{k-1} \sum_{n=1}^\infty \frac1{x_n^k}$$
• If $$\sum_{n=1}^\infty \frac1{|y_n|} < \infty$$ and $$\forall k, \sum_{n=1}^\infty \frac1{x_n^k}= \sum_{n=1}^\infty \frac1{y_n^k}$$ then $$f_x(z)=f_y(z)$$ for $$|z| < \inf_n \min (|x_n|,|y_n|)$$ thus by the identity theorem $$f_x=f_y$$ and hence $$(y_n)$$ is a permutation of $$(x_n)$$.