MHB Polynomial in n variables: Prove the identity

AI Thread Summary
The discussion focuses on proving the identity involving a polynomial in n variables of degree less than or equal to n-1. The identity states that the alternating sum of the polynomial evaluated at all combinations of binary inputs equals zero. It is emphasized that the proof can be simplified by considering the polynomial in a specific monomial form. The key point is that the linearity of the identity allows for this reduction. Ultimately, the identity holds true for all such polynomials, confirming the initial assertion.
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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 $p_1+p_2+p_3 + ... + p_n \le n-1$.
 
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Solution:
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 $S = 0$.
 
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