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单词 ProofOfGeneralizedRuizsIdentity
释义

proof of generalized Ruiz’s identity


Theorem.

Consider the polynomialsMathworldPlanetmath ci,j(x)=(x+i)j-(x+i-1)j. Then, for every positive natural number n,

det(ci,j)i,j=1n=k=1nk!
Proof.

Consider the matrices M,C defined by Mi,j=(-1)j(i-1j-1) and Ci,j=(x+i)j-(x+i-1)j.

(MC)i,j=k=1n(-1)k(i-1k-1)((x+k)j-(x+k-1)j)
=k=0n(-1)k((i-1k-1)+(i-1k))(x+k)j
=k=0n(-1)k(ik)(x+k)j
=(-1)jk=0i(-1)k(ik)(-x-k)j

Therefore, by Ruiz’s identityPlanetmathPlanetmath, (MC)i,i=(-1)ii! for every i{1,,d} and (MC)i,j=0 for every i,j{1,,n} such that i>j. Thismeans that MC is an upper triangular matrixMathworldPlanetmath whose main diagonal is -1!,2!,-3!,,(-1)nn!. Since the determinantMathworldPlanetmath of such a matrix isthe product of the elements in the main diagonal, we get that detMC=(-1)nk=1nk!. It is easy to see that M itself is lowertriangular with determinant (-1)n. Therefore detC=k=1nk!.∎

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