Polynomial in n variables: Prove the identity

[lfdahl]

Suppose $f$ is a polynomial in $n$ variables, of degree $ \le n − 1$, ($n = 2, 3, 4 ...$ ).Prove the identity:

\[\sum (-1)^{\epsilon_1+\epsilon_2+\epsilon_3+ ...+\epsilon_n}f(\epsilon_1,\epsilon_2,\epsilon_3,...,\epsilon_n) = 0\;\;\;\;\; (1)\]

where $\epsilon_i$ is either $0$ or $1$, and the sum is over all $2^n$ combinations.

Hint: The identity $(1)$ is linear in $f$, so it suffices to prove it for $f$ of the form

$f(x_1, x_2, x_3,..., x_n) = x_1^{p_1}x_2^{p_2}x_3^{p_3}...x_n^{p_n}$, where $p_1+p_2+p_3 + ... + p_n \le n-1$.

 

[lfdahl]

Solution:

Spoiler

Because of the last restriction, at least one of the $p_i$ is $0$, say $p_n = 0$. Then writing the whole sum as the sum of the terms with $\epsilon_n = 0$ and those with $\epsilon_n = 1$, we have:
\[S = (-1)^0 \sum_{\epsilon_1,\epsilon_2,...,\epsilon_{n-1}} (-1)^{\epsilon_1+\epsilon_2+\epsilon_3+ ...+\epsilon_{n-1}}\epsilon_1^{p_1}\epsilon_2^{p_2}...\epsilon_{n-1}^{p_{n-1}} + (-1)^1 \sum_{\epsilon_1,\epsilon_2,...,\epsilon_{n-1}} (-1)^{\epsilon_1+\epsilon_2+\epsilon_3+ ...+\epsilon_{n-1}}\epsilon_1^{p_1}\epsilon_2^{p_2}...\epsilon_{n-1}^{p_{n-1}}\]
- which is the difference of two identical terms, hence $S = 0$.