9
$\begingroup$

We know the followings :

$$\color{red}{{x_1}^2+{x_2}^2-2x_1x_2}=\color{blue}{(x_1-x_2)^2}$$ $$\color{red}{{x_1}^3+{x_2}^3+{x_3}^3-3x_1x_2x_3}$$$$=\frac{1}{2}(x_1+x_2+x_3)\color{blue}{(x_1-x_2)^2}+\frac{1}{2}(x_1+x_2+x_3)\color{blue}{(x_2-x_3)^2}+\frac{1}{2}(x_1+x_2+x_3)\color{blue}{(x_3-x_1)^2}$$

I've been interested in generalizing these identities. Then, I got the followings :

$$\color{red}{{x_1}^4+{x_2}^4+{x_3}^4+{x_4}^4-4x_1x_2x_3x_4}$$$$=\frac{1}{6}\left(2({x_1}^2+x_1x_2+{x_2}^2)+(x_1+x_2)(x_3+x_4)+2x_3x_4\right)\color{blue}{(x_1-x_2)^2}$$$$ +\frac{1}{6}\left(2({x_1}^2+x_1x_3+{x_3}^2)+(x_1+x_3)(x_2+x_4)+2x_2x_4\right)\color{blue}{(x_1-x_3)^2}$$$$+\frac{1}{6}\left(2({x_1}^2+x_1x_4+{x_4}^2)+(x_1+x_4)(x_2+x_3)+2x_2x_3\right)\color{blue}{(x_1-x_4)^2}$$$$+\frac{1}{6}\left(2({x_2}^2+x_2x_3+{x_3}^2)+(x_2+x_3)(x_1+x_4)+2x_1x_4\right)\color{blue}{(x_2-x_3)^2}$$$$+\frac{1}{6}\left(2({x_2}^2+x_2x_4+{x_4}^2)+(x_2+x_4)(x_1+x_3)+2x_1x_3\right)\color{blue}{(x_2-x_4)^2}$$$$+\frac{1}{6}\left(2({x_3}^2+x_3x_4+{x_4}^2)+(x_3+x_4)(x_1+x_2)+2x_1x_2\right)\color{blue}{(x_3-x_4)^2}$$

$$\color{red}{{x_1}^5+{x_2}^5+{x_3}^5+{x_4}^5+{x_5}^5-5x_1x_2x_3x_4x_5}$$$$\small=F_{1,2}\color{blue}{(x_1-x_2)^2}+F_{1,3}\color{blue}{(x_1-x_3)^2}+F_{1,4}\color{blue}{(x_1-x_4)^2}+F_{1,5}\color{blue}{(x_1-x_5)^2}+F_{2,3}\color{blue}{(x_2-x_3)^2}$$$$\small +F_{2,4}\color{blue}{(x_2-x_4)^2}+F_{2,5}\color{blue}{(x_2-x_5)^2}+F_{3,4}\color{blue}{(x_3-x_4)^2}+F_{3,5}\color{blue}{(x_3-x_5)^2}+F_{4,5}\color{blue}{(x_4-x_5)^2}$$ where $$12F_{1,2}=3({x_1}^3+{x_1}^2{x_2}+{x_1}{x_2}^2+{x_2}^3)+({x_1}^2+{x_1}{x_2}+{x_2}^2)(x_3+x_4+x_5)+(x_1+x_2)(x_3x_4+x_4x_5+x_5x_3)+3x_3x_4x_5$$ $$12F_{1,3}=3({x_1}^3+{x_1}^2{x_3}+{x_1}{x_3}^2+{x_3}^3)+({x_1}^2+{x_1}{x_3}+{x_3}^2)(x_2+x_4+x_5)+(x_1+x_3)(x_2x_4+x_4x_5+x_5x_2)+3x_2x_4x_5$$$$12F_{1,4}=3({x_1}^3+{x_1}^2{x_4}+{x_1}{x_4}^2+{x_4}^3)+({x_1}^2+{x_1}{x_4}+{x_4}^2)(x_2+x_3+x_5)+(x_1+x_4)(x_2x_3+x_3x_5+x_5x_2)+3x_2x_3x_5$$ $$12F_{1,5}=3({x_1}^3+{x_1}^2{x_5}+{x_1}{x_5}^2+{x_5}^3)+({x_1}^2+{x_1}{x_5}+{x_5}^2)(x_2+x_3+x_4)+(x_1+x_5)(x_2x_3+x_3x_4+x_4x_2)+3x_2x_3x_4$$$$12F_{2,3}=3({x_2}^3+{x_2}^2{x_3}+{x_2}{x_3}^2+{x_3}^3)+({x_2}^2+{x_2}{x_3}+{x_3}^2)(x_1+x_4+x_5)+(x_2+x_3)(x_1x_4+x_4x_5+x_5x_1)+3x_1x_4x_5$$$$12F_{2,4}=3({x_2}^3+{x_2}^2{x_4}+{x_2}{x_4}^2+{x_4}^3)+({x_2}^2+{x_2}{x_4}+{x_4}^2)(x_1+x_3+x_5)+(x_2+x_4)(x_1x_3+x_3x_5+x_5x_1)+3x_1x_3x_5$$$$12F_{2,5}=3({x_2}^3+{x_2}^2{x_5}+{x_2}{x_5}^2+{x_5}^3)+({x_2}^2+{x_2}{x_5}+{x_5}^2)(x_1+x_3+x_4)+(x_2+x_5)(x_1x_3+x_3x_4+x_4x_1)+3x_1x_3x_4$$$$12F_{3,4}=3({x_3}^3+{x_3}^2{x_4}+{x_3}{x_4}^2+{x_4}^3)+({x_3}^2+{x_3}{x_4}+{x_4}^2)(x_1+x_2+x_5)+(x_3+x_4)(x_1x_2+x_2x_5+x_5x_1)+3x_1x_2x_5$$$$12F_{3,5}=3({x_3}^3+{x_3}^2{x_5}+{x_3}{x_5}^2+{x_5}^3)+({x_3}^2+{x_3}{x_5}+{x_5}^2)(x_1+x_2+x_4)+(x_3+x_5)(x_1x_2+x_2x_4+x_4x_1)+3x_1x_2x_4$$$$12F_{4,5}=3({x_4}^3+{x_4}^2{x_5}+{x_4}{x_5}^2+{x_5}^3)+({x_4}^2+{x_4}{x_5}+{x_5}^2)(x_1+x_2+x_3)+(x_4+x_5)(x_1x_2+x_2x_3+x_3x_1)+3x_1x_2x_3$$ Here, I have a conjecture.

Question : Is the following conjecture true?

Conjecture : For any $n\ (\ge 2\in\mathbb N)$ variables $x_1,\cdots,x_n$, the following holds.

$$\sum_{i=1}^{n}{x_i}^n-n\prod_{i=1}^{n}x_i=\sum_{1\le i\lt j\le n}(x_i-x_j)^2\sum_{k=0}^{n-2}\frac{\left(\sum_{m=0}^{k}{x_i}^m{x_j}^{k-m}\right)\cdot s_{n-2,n-2-k}(\not= x_i,x_j)}{{(n-1)\binom{n-2}{k}}}$$

where $s_{n-2,n-2-k}(\not= x_i,x_j)=s_{n-2,n-2-k}(x_1,\cdots,x_{i-1},x_{i+1},\cdots,x_{j-1},x_{j+1},\cdots,x_n)$ and $s_{n,k}(x_1,\cdots,x_n)$ is the sum of $\binom{n}{k}$ products of every $k$ elements chosen from $\{x_1,\cdots,x_n\}$ with $s_{n,0}(x_1,\cdots,x_n)=1$.

I've already checked that this is true for $n=2,3,4,5,6$ and $7$, but I don't have any good idea to prove that. Can anyone help?

$\endgroup$
  • $\begingroup$ have you tried induction on $n$? $\endgroup$ – danimal Mar 9 '15 at 11:57
  • $\begingroup$ @danimal: I tried, but I could not find any good way using induction on $n$. $\endgroup$ – mathlove Mar 9 '15 at 12:09
  • 4
    $\begingroup$ I have know <inequalitys> by G.Hardy,J,E,Littlewood& G.Polya, page 55,The identity of Hurwitz and Muirhead proved $$a^n_{1}+a^n_{2}+\cdots+a^n_{n}-na_{1}a_{2}\cdots a_{n}$$ can be expressed as a sum in which every term is obviously non-negative,maybe you can find their paper $\endgroup$ – math110 Mar 9 '15 at 12:37
  • $\begingroup$ @math110: Thank you for your comment. So far I can find this paper whose main theorems are very similar to my conjecture. $\endgroup$ – mathlove Mar 11 '15 at 11:02
6
$\begingroup$

It suffices to prove, that after reducing the entire thing, there are no terms left with $x_1^a$ for any exponent $2\leq a\leq n-1$. Then it is easy to check that the coefficients of $x_1^n$ and $x_1x_2\cdots x_n$ are $1$ and $-n$ respectively. The rest follows from the symmetry of the expression.


It is easy to see why there can be no expression left containing $x_1^a x_2^b$ where $a,b\geq 2$ since this can only stem from the $(x_1-x_2)^2$-part from the subexpression $$ (x_1-x_2)^2\sum_{i+j=a+b-2}x_1^ix_2^j $$ in which we can identify $$ x_1^2(x_1^{a-2} x_2^b)+x_2^2(x_1^a x_2^{b-2})-2x_1x_2(x_1^{a-1}x_2^{b-1})=0 $$ as the only terms resulting in $x_1^a x_2^b$ for $a,b\geq 2$. By symmetry, this shows that there will be no terms left of the form $x_i^a x_j^b$ with $a,b\geq 2$.


So it remains to show that for $2\leq a\leq n-1$ the terms of the form $$ x_1^a\prod_{i\in S} x_i $$ for any given subset $S\subseteq\{2,3,...,n\}$ all cancel. For a given $a$, these can only come either from expressions of the form $$ q(x_1^{a-1}+x_1^{a-2}x_2+...)(\text{products of other }x_i\text{'s})(x_1^2+x_2^2-2x_1x_2) $$ from which we get $-qx_1^a x_2(\text{products of other }x_i\text{'s})$ or from expressions of the form $$ z(x_1^{a-2}+x_1^{a-3}x_2+...)(\text{products of other }x_i\text{'s})(x_1^2+x_2^2-2x_1x_2) $$ from which we get $z x_1^a(\text{products of other }x_i\text{'s})$. Now consider a specific product $$ x_1^a \prod_{i\in S}x_i $$ for a fixed set $S\subseteq\{2,3,...,n\}$. This appears once in every $(x_1-x_s)^2$-part as $-qx_1^a\prod_{i\in S}x_i$ for $s\in S$. And it appears once in every $(x_1-x_t)^2$-part as $zx_1^a\prod_{i\in S}x_i$ for $t\in\{2,3,...,n\}\setminus S$.

Note that we have $|S|=n-a$ choices for $s$ and $a-1$ choices for $t$, so if these contributions are to cancel this happens iff $(a-1)z-(n-a)q=0$ so that the terms cancel if and only if $$ \frac{q}{z}=\frac{a-1}{n-a} $$ This uniquely determines the ratio between the coefficients $q$ and $z$ of successive parts $q(x_1^{a-1}+x_1^{a-2}x_2+...)$ and $z(x_1^{a-2}+x_1^{a-3}x_2+...)$ for $2\leq a\leq n-1$ which actually covers all the relevant coefficients.


Finally note that, since in your general formula $q=\frac{1}{(n-1)\binom{n-2}{a-1}}$ and $z=\frac{1}{(n-1)\binom{n-2}{a-2}}$, these actually satisfy $$ \frac{q}{z}=\binom{n-2}{a-2}/\binom{n-2}{a-1}=\frac{a-1}{n-a} $$ as desired.

$\endgroup$
  • 1
    $\begingroup$ Thank you for your answer. I think I can get what you mean. By the way, $q=\frac{1}{(n-1)\binom{n-2}{a-1}},z=\frac{1}{(n-1)\binom{n-2}{a-2}}$, right? $\endgroup$ – mathlove Mar 11 '15 at 10:22
  • $\begingroup$ @mathlove: Yes, those are $q$ and $z$. I am sorry if my rendering of it is not entirely clear. I just wrote down some principles about the terms on my paper that made sense to me, but to communicate those without giving too many or too few details to maintain a clear course, did seem quite challenging! $\endgroup$ – String Mar 11 '15 at 10:28
  • $\begingroup$ BTW, I actually thought of $q,z$ as being scaled by the least common multiply of all those coefficient's denominators and moving $\frac{1}{\operatorname{(n-1)lcm}\left(\binom{n-2}{0},\binom{n-2}{1},...,\binom{n-2}{n-2}\right)}$ out in front of each part - just like you did for $n=3,4$ and $5$. After all, only the proportions are relevant in order to make the terms cancel. $\endgroup$ – String Mar 11 '15 at 10:36
  • $\begingroup$ Yes, that was what I implied in my first comment. Since $q,z$ can be expressed as my first comment, $(a-1)z-(n-a)q$ does equal $0$, so the terms cancel. To be honest, it took a lot of time to understand your answer:) but I think your answer is well elaborated. Thank you! $\endgroup$ – mathlove Mar 11 '15 at 10:45
  • $\begingroup$ Technical details about rather clear ideas tend to be tedious to read through. The main idea I presented here was to divide into cases that were more distinct and thus simpler to manage. You probably would have had a a simpler job simply managing those yourself than to struggle with my write-up of it ;) $\endgroup$ – String Mar 11 '15 at 10:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.