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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