Double Summation Over all subset of $\{1,2,...n\}$

98 Views Asked by At

In Benson's Book "Polynomial In variants of Finite Groups" It is claimed that(Without any proof):

$$ j! u_1u_2...u_j =\sum_{I \subseteq \{1,2,...,j\} } (-1)^I (\sum_{i \in I}u_i)^j$$

Where $I$ runs over all subsets of $\{1,2,...,n\}$

I tried to prove this claim by induction but it seems it's not easy by induction.Is this equality really easy because Benson does not even say a single word about proof of this.Any ideas for proving this?

1

There are 1 best solutions below

0
On

It is a homogeneous polynomial of degree $j$. Fix some subset $J\subseteq \{1,\ldots,j\}$ and degrees $(d_k \mid k\in J)$. (So $\sum d_k=j$.) Then the coefficient of the monomial $$ \prod_{k\in J }u_k^{d_k}$$ equals $$\frac{j!}{\prod_{k\in J}d_k!}\sum_{I\supseteq J}(-1)^{|I|} = \frac{j!}{\prod_{k\in J}d_k!} \sum_{m=0}^{j-|J|}{j-|J| \choose m} (-1)^{|J|+m} = \begin{cases} j!\,(-1)^j & |J| =j \\[1ex] 0 & \textrm{otherwise}\end{cases}$$