User talk:PaoloMartini

Show that if $p(x)$ is a polynomial of degree $n$, then $p(x - 1)$ is a polynomial of the same degree.

Definition of polynomial.

$p(x) = \sum_{i=0}^n a_i x^i$

Binomial theorem.

$(a + b)^n = \sum_{i=0}^n {n \choose i} a^{n-i} b^i$

Special case.

$(x - 1)^n = \sum_{i=0}^n {n \choose i} x^{n-i} (-1)^i$

Binomial coefficient simmetry.

${n \choose k} = {n \choose n-k}$

Hence:

$(x - 1)^n = \sum_{i=0}^n {n \choose i} x^i (-1)^i$

$p(x-1) = \sum_{i=0}^n a_i (x - 1)^i = \sum_{i=0}^n \left[ a_i \left( \sum_{k=0}^n {n \choose k} x^k (-1)^k \right) \right] = \sum_{i=0}^n \sum_{k=0}^i (-1)^k a_i {n \choose k} x^k = \sum_{i=0}^n \sum_{k=0}^i a_k {n \choose i} x^i (-1)^i.$

QED.

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