# proof of generalized Ruiz’s identity

###### Theorem.

Consider the polynomials  $c_{i,j}(x)=(x+i)^{j}-(x+i-1)^{j}$. Then, for every positive natural number $n$,

 $\det(c_{i,j})_{i,j=1}^{n}=\prod_{k=1}^{n}k!$
###### Proof.

Consider the matrices $M,C$ defined by $M_{i,j}=(-1)^{j}{i-1\choose j-1}$ and $C_{i,j}=(x+i)^{j}-(x+i-1)^{j}$.

 $\displaystyle(MC)_{i,j}$ $\displaystyle=\sum_{k=1}^{n}(-1)^{k}{i-1\choose k-1}\bigl{(}(x+k)^{j}-(x+k-1)^% {j}\bigr{)}$ $\displaystyle=\sum_{k=0}^{n}(-1)^{k}\left({i-1\choose k-1}+{i-1\choose k}% \right)(x+k)^{j}$ $\displaystyle=\sum_{k=0}^{n}(-1)^{k}{i\choose k}(x+k)^{j}$ $\displaystyle=(-1)^{j}\sum_{k=0}^{i}(-1)^{k}{i\choose k}(-x-k)^{j}$

Therefore, by Ruiz’s identity  , $(MC)_{i,i}=(-1)^{i}i!$ for every $i\in\{1,...,d\}$ and $(MC)_{i,j}=0$ for every $i,j\in\{1,...,n\}$ such that $i>j$. This means that $MC$ is an upper triangular matrix  whose main diagonal is $-1!,2!,-3!,...,(-1)^{n}n!$. Since the determinant  of such a matrix is the product of the elements in the main diagonal, we get that $\det MC=(-1)^{n}\prod_{k=1}^{n}k!$. It is easy to see that $M$ itself is lower triangular with determinant $(-1)^{n}$. Therefore $\det C=\prod_{k=1}^{n}k!$. ∎

Title proof of generalized Ruiz’s identity ProofOfGeneralizedRuizsIdentity 2013-03-22 14:32:02 2013-03-22 14:32:02 GeraW (6138) GeraW (6138) 8 GeraW (6138) Proof msc 11B65 msc 05A10