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Let $(x_i)^{m}_{i=1}$,$(y_i)^{m}_{i=1}$ and $(a_i)^{m}_{i=1}$ be tuples of positive non-zero reals. If for all positive integers $n$: \begin{equation} \sum_{i=1}^{m}a_ix_i^{n} = \sum_{i=1}^{m}a_iy_i^{n} \tag{1} \end{equation} Then \begin{equation*} \sum_{i=1}^{m}x_i = \sum_{i=1}^{m}y_i \end{equation*} NOTE: This is not a textbook problem statement. It could potentially be false, so I would accept a counter example as an answer too.

MY WORK

I have been trying to prove this statement. Through help of MSE, I have been able to prove that if $\{u_{i}\}^{m'}_{i=1}$ and $\{v_{i}\}^{m''}_{i=1}$ are the set of unique entries in $(x_i)^{m}_{i=1}$ and $(y_i)^{m}_{i=1}$ respectively. Then $\{u_{i}\}^{m'}_{i=1}$ and $\{v_{i}\}^{m''}_{i=1}$ are the same set, the proof goes as follows:

Let $\{u_{i}\}^{m'}_{i=1}$ and $\{v_{i}\}^{m''}_{i=1}$ be the set of unique entries in $(x_i)^{m}_{i=1}$ and $(y_i)^{m}_{i=1}$ respectively. Also, without loss of generality, we may assume an ordering such that $u_1 > u_2> ... >u_{m'} $ and $v_1 > v_2> ...>v_{m''} $ and also that $m''\geq m'$. We can rewrite $(1)$ as: \begin{equation} \label{eq: lemma_1_equivalence} \forall n \in \mathbb{Z^{+}}: \sum_{i=1}^{m'}c_iu_i^{n} = \sum_{i=1}^{m''}d_iv_i^{n} \tag{2} \end{equation} As $n$ grows the leading term on LHS is $c_1u_{1}^{n}$ and on the RHS is $d_1v_{1}^{n}$. Hence, it must be :

\begin{equation*} \forall n \in \mathbb{Z^{+}}: c_1 u_{1}^{n} = d_1 v_{1}^{n} \end{equation*} Since, $u_1,v_1,c_1$ and $d_1$ are non-zero positive reals, we can conclude that $u_1=v_1$ and $c_1 = d_1$. Hence, we may subtract $c_1 u_{1}^{n}$ from both sides in $(2)$ to get : \begin{equation} \label{eq: lemma_1_equivalence_1} \forall n \in \mathbb{Z^{+}}: \sum_{i=2}^{m'}c_iu_i^{n} = \sum_{i=2}^{m''}d_iv_i^{n} \end{equation} We may now repeat the aforementioned argument and infer that $u_2=v_2$ and $c_2 = d_2$. Furthermore, repeating this argument $m'$ times, we can infer that $\{u_i\}^{m'}_{i=1} = \{v_i\}^{m'}_{i=1}$, leaving us with $0 =\sum_{i=m''-m'+1}^{m''}d_iv_i^{n}$, which is a contradiction, hence, $m'=m''$. Hence, we have that $\{u_{i}\}^{m'}_{i=1}$ = $\{v_{i}\}^{m''}_{i=1}$.

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    $\begingroup$ I think I'm very close to the answer, but I'm just thinking about how to write it. But I started by thinking what would happen if $(x_i)^{m}_{i=1}$ was not a rearrangement of $(y_i)^{m}_{i=1}$. For example if $\max (x_i)^{m}_{i=1}>\max (y_i)^{m}_{i=1}$ then the left-hand side of $(1)$ dominates the right-hand side for $n\geq N$ where $N$ is some integer. Then think about the second largest number in each tuple... $\endgroup$ Mar 11, 2022 at 13:47
  • $\begingroup$ @AdamRubinson Thanks. I tried a similar approach and as you can see in the proof I provided, it allows me to show that the entries in $(x_i)^{m}_{i=1}$ and $(y_i)^{m}_{i=1}$ form the same set. But I am really not able to write this idea to show that they are just a re-arrangment of each other i.e. the multiplicity of each unique element is the same (which is actually a stricter condition than I want, which would be great !) $\endgroup$
    – SagarM
    Mar 11, 2022 at 13:56

2 Answers 2

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The statement is not true. Consider $m=3$, and $$(x_0,x_1,x_2)=(1,2,2), \quad \quad (y_0,y_1,y_2)=(2,1,1), \quad \quad \text{and } \quad (a_0,a_1,a_2)=(2,1,1)$$

Then for every $n \geq 0$, one has $$\sum_{i=1}^m a_ix_i^n = 2 \times 1^n + 1 \times 2^n + 1 \times 2^n = 2 + 2^{n+1}$$ and $$\sum_{i=1}^m a_iy_i^n = 2 \times 2^n + 1 \times 1^n + 1 \times 1^n = 2 + 2^{n+1}$$

but obviously, $$\sum_{i=1}^3 x_i = 5 \neq 4 = \sum_{i=1}^3 y_i$$

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  • $\begingroup$ Are there any examples without repeating terms? $\endgroup$ Mar 11, 2022 at 17:12
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    $\begingroup$ @AdamRubinson No, the statement is correct if there are no repeating terms. (actually, I thought first that the statement was correct in general, and posted an incorrect proof of it, which however works in the case where there is no repetition). $\endgroup$ Mar 11, 2022 at 17:14
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    $\begingroup$ cool - maybe add this proof to your answer then! Also I am interested in if there any examples for $a_i=1\forall i$ in other words if we can ignore the $a_i.$ $\endgroup$ Mar 11, 2022 at 17:49
  • $\begingroup$ @AdamRubinson the summation being the same given that all the elements are distinct follows from the proof in the question, that shows that the {x_i} and {y_i} form the same set. $\endgroup$
    – SagarM
    Mar 11, 2022 at 19:22
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    $\begingroup$ The counterexample works as the measures $2\delta_1+\delta_2+\delta_2 $ and $2\delta_2+\delta_1+\delta_1$ are equal $\endgroup$ Mar 13, 2022 at 19:57
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Though the statement is not true, we can get some simpler conclusions on $a,x,y$: $$ \forall r,\sum_{i=1}^m[x_i=r]a_i=\sum_{i=1}^m[y_i=r]a_i $$

where $[P]$ is $1$ if $P$ if true, and $0$ otherwise.

Considering the generating function using the placeholder $z$. We have

$$ \sum_{i=1}^m\frac{a_i}{1-x_iz}=\sum_{i=1}^m\sum_{n=0}^{+\infty}a_iz^nx_i^n=\sum_{n=0}^{+\infty}z^n\sum_{i=1}^ma_ix_i^n=\sum_{n=0}^{+\infty}z^n\sum_{i=1}^ma_iy_i^n=\sum_{i=1}^m\frac{a_i}{1-y_iz} $$

Lemma: If the g.f. of $z$ satisfies

$$ \sum_{i=1}^t\frac{a_i}{1-x_iz}=0 $$

and $x_i$ has no repeating terms, then $\forall i,a_i=0$.

Proof: For a fixed $i$, multiply both sides with $(1-x_iz)$, and then let $z$ take the value $x_i^{-1}$, we get $a_i=0$. $\blacksquare$

Finally, it is not hard to get the conclusion considering coefficient of $1/(1-rz)$ on both sides.

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