proof of generalized Ruiz’s identity

Theorem.

Consider the polynomials $c_{i\mathrm{,}j}\mathit{}\mathrm{(}x\mathrm{)}\mathrm{=}{\mathrm{(}x\mathrm{+}i\mathrm{)}}^j\mathrm{-}{\mathrm{(}x\mathrm{+}i\mathrm{-}\mathrm{1}\mathrm{)}}^j$. 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 $M_{i,j}={(-1)}^j\left(\genfrac{}{}{0pt}{}{i-1}{j-1}\right)$ and $C_{i,j}={(x+i)}^j-{(x+i-1)}^j$.

${(MC)}_{i,j}$ $={\displaystyle \sum _{k=1}^n}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{i-1}{k-1}}\right)\left({(x+k)}^j-{(x+k-1)}^j\right)$

$={\displaystyle \sum _{k=0}^n}{(-1)}^k\left(\left({\displaystyle \genfrac{}{}{0pt}{}{i-1}{k-1}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{i-1}{k}}\right)\right){(x+k)}^j$

$={\displaystyle \sum _{k=0}^n}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{i}{k}}\right){(x+k)}^j$

$={(-1)}^j{\displaystyle \sum _{k=0}^i}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{i}{k}}\right){(-x-k)}^j$

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

Title	proof of generalized Ruiz’s identity
Canonical name	ProofOfGeneralizedRuizsIdentity
Date of creation	2013-03-22 14:32:02
Last modified on	2013-03-22 14:32:02
Owner	GeraW (6138)
Last modified by	GeraW (6138)
Numerical id	8
Author	GeraW (6138)
Entry type	Proof
Classification	msc 11B65
Classification	msc 05A10