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I just realized that there is a nice pattern in multinomial formulas (which I'd say is related to notions of interference). I am sure this is something well known, but as I haven't seen this before I don't know whats to look for.

So to be more specific, my questions are:

a) is there any name for the following types of equations?

b) what would be a reference for such equations, and a proof of the general case (sadly, I currently don't find the time to do the multinomial-juggling)?

The Equations

This is the general pattern of equations (think of them as over $\mathbb{R}$ or $\mathbb{C}$). They relate the sum of $k$ terms raised to the $k-1$ order with sums of shorter sums each raised to the $k-1$ order.

Starting with $k=2$: $$(u+v)^1=u^1+v^1$$ $k=3$: $$(u+v+w)^2=(u+v)^2 + (v+w)^2 +(w+u)^2 -u^2 -v^2-w^2$$ $k=4$: $$(u+v+w+x)^3 = (u+v+w)^3 + (u+v+x)^3 + (u+x+w)^3 + (x+v+w)^3 - (u+v)^3 - (v+w)^3 - (w+x)^3 - (u+w)^3 - (x+u)^3 -(v+x)^3 + u^3 +v^3+w^3+x^3$$ I expect the general pattern to be along the lines: $$(x_1+\dots+x_{k+1})^k = \sum_i (x_1 +\dots \hat{x_i}\dots+x_{k+1})^k - \sum_{ij} (x_1 +\dots \hat{x_i},\hat{x_j}\dots+x_{k+1})^k + \dots (-1)^{k-1} \sum_i x_i^k$$ where the hatted $x$'s are ommitted in the sums.

uzer.name
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1 Answers1

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A) I'm not aware of any special name given to these equations. I've seen them before.

B) Here's a quick proof of the statement, generalized to the exponent having degree $ k < n$. OP stated the $k = n-1$ case.

Shift all the terms to one side, and we have a polynomial in $ f(x_1, x_2, x_3, \ldots, x_n)$. This is homogenous of degree $k$.

Focus on one term, WLOG $x_{\color{red} 1 } $.
For each subset $ S \subset \{ x_1, x_2 , x_2, \ldots x_n\}$, create a bijection with $ S \triangle \{ x_{\color{red} 1 } \} $ (symmetric difference).
Each pair of terms $ \pm [( \sum_{x_i \in S } x_i )^{k} - ( \sum_{ x_i \in S \triangle \{ x_{\color{red} 1 } \} } x_i ) ^ {k}]$ has a factor $ [( \sum_{x_i \in S } x_i ) - ( \sum_{ x_i \in S \triangle \{ x_{\color{red} 1 }\} } x_i )] = \pm x_{\color{red} 1 }$.
Summing up across all terms, we conclude that $x_{\color{red} 1 } \mid f$.

This is true for each $x_{\color{red} 1 }$, so $ x_1 x_2 \ldots x_n \mid f(x_1, x_2, x_3, \ldots, x_n)$.
Since the polynomial has degree $k < n$, it is the zero polynomial.

Extension: Furthermore, when the exponent is $n$, a similar analysis shows that the expression is equal to $n! \prod x_i$.
Again, $ x_1 x_2 \ldots x_n \mid f(x_1, x_2, x_3, \ldots, x_n)$ and so it has the form $Ax_1 x_2 \ldots x_n$ for some constant $A$. There is only one term that contributes this.

$n=1: a = a $
$n = 2: (a+b)^2 - a^2 - b^2 = 2ab$
$n=3: (a+b+c)^3 - (a+b)^3-(b+c)^3-(c+a)^2 +a^3+b^3+c^3 = 6abc$

Further extension: What happens when the exponent is $n+1$ for $n > 1$?

Calvin Lin
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  • Thanks for the great answer! Since I am not used to the notation you are using, let me double check: What is the set $[x_n]$? And what is the meaning of $x\vert f$? Also, I assume that $\triangle$ denotes the symmetric difference of sets? – uzer.name Jan 05 '25 at 21:32
  • 1/ I've edited it for clarity. $[x_n]$ is my shorthand for $ { x_1, x_2, x_3, \ldots x_n }$, like how $[n]$ is shorthand for $ { 1, 2, 3, \ldots n }$. $\quad$ 2/ That's divisibility notation. It means that $x$ is a factor of the polynomial $ f(x)$. $\quad$ 3/ Ah yes, symmetric difference. Updated. – Calvin Lin Jan 05 '25 at 21:37
  • Cool, that all makes sense! Thanks for the quick reply! – uzer.name Jan 05 '25 at 21:40
  • PS on the sentence about the bijection: If I am right you form a bijection between the subsets containing $x_1$ and those without $x_1$, i.e. a one to one map between sets of sets. That is a nice trick, though took me a bit because I confused the sentence as a bijection between the sets $S$ and $S\triangle{x_1}$ -- which cant work due to the constructino of the sets ($S\triangle {x_1}$ is either 1 element bigger or smaller than $S$). – uzer.name Jan 05 '25 at 21:51
  • Regarding your extension $k=n+1$: your analysis provides that the remainder again is some polynomial that is divisible by $x_1\cdots x_n$. By symmetry arguments it must be symmetric under exchanging the $x_i$'s. With knowledge of the multinomial formula it should be (modulo mistakes) $(x_1+\dots+x_n)\frac{n!}{2} x_1\cdots x_n$. – uzer.name Jan 05 '25 at 22:08
  • 1/ Correct, it's a bijection of subsets. (Not a bijection of their elements) $\quad$ 2/ Correct. Again this only arises from one term, but we can obtain the coefficient through the multinomial theorem. – Calvin Lin Jan 05 '25 at 22:49