# Proving that : $(\forall n: \sum_{i=1}^{m} a_i x_i^{n} =\sum_{i=1}^{m} a_i y_i^{n} )\rightarrow (\sum_{i=1}^{m}x_i =\sum_{i=1}^{m}y_i )$

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 $$n$$: $$$$\sum_{i=1}^{m}a_ix_i^{n} = \sum_{i=1}^{m}a_iy_i^{n} \tag{1}$$$$ Then $$\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: $$$$\label{eq: lemma_1_equivalence} \forall n \in \mathbb{Z^{+}}: \sum_{i=1}^{m'}c_iu_i^{n} = \sum_{i=1}^{m''}d_iv_i^{n} \tag{2}$$$$ As $$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 : $$$$\label{eq: lemma_1_equivalence_1} \forall n \in \mathbb{Z^{+}}: \sum_{i=2}^{m'}c_iu_i^{n} = \sum_{i=2}^{m''}d_iv_i^{n}$$$$ We may now repeat the aforementioned argument and infer that $$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}$$.

• 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... Mar 11, 2022 at 13:47
• @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 !) Mar 11, 2022 at 13:56

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

• Are there any examples without repeating terms? Mar 11, 2022 at 17:12
• @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). Mar 11, 2022 at 17:14
• 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 $a_i.$ Mar 11, 2022 at 17:49
• @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. Mar 11, 2022 at 19:22
• The counterexample works as the measures $2\delta_1+\delta_2+\delta_2$ and $2\delta_2+\delta_1+\delta_1$ are equal Mar 13, 2022 at 19:57

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.