Combinatorial Identities for Stirling Numbers: The by Jocelyn Quaintance, Henry W. Gould

By Jocelyn Quaintance, Henry W. Gould

This booklet is a special paintings which gives an in-depth exploration into the mathematical services, philosophy, and data of H W Gould. it really is written in a mode that's obtainable to the reader with simple mathematical wisdom, and but includes fabric that may be of curiosity to the professional in enumerative combinatorics. This e-book starts off with exposition at the combinatorial and algebraic recommendations that Professor Gould makes use of for proving binomial identities. those recommendations are then utilized to improve formulation which relate Stirling numbers of the second one sort to Stirling numbers of the 1st type. Professor Gould's suggestions additionally offer connections among either sorts of Stirling numbers and Bernoulli numbers. Professor Gould believes his learn good fortune comes from his instinct on find out how to detect combinatorial identities.This e-book will attract a large viewers and should be used both as lecture notes for a starting graduate point combinatorics classification, or as a examine complement for the expert in enumerative combinatorics.

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Additional resources for Combinatorial Identities for Stirling Numbers: The Unpublished Notes of H W Gould

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27) k=0 i=⌊ k ⌋(k−n) n+1 where we have suppressed the bars after substitution. 27) is called Gould’s convolution formula for the product of two finite series. Gould’s convolution formula is useful for calculating the product of two polynomials. Let f (i) = Ai xi and ϕ(j) = Bj xj . 27) implies Bj xj = Ai xi j=0 i=0 k+1 k−⌊ n+1 ⌋(k−n) n n n xk k=0 Ai Bk−i . 28) k i=⌊ n+1 ⌋(k−n) Another application of the Gould convolution formula calculates the square of a series. Let ϕ(j) = f (j). 29) k=0 i=⌊ k ⌋(k−n) n+1 In particular, if f (k) = n f (i)f (k − i).

25) page 18 September 15, 2015 12:0 ws-book9x6 Combinatorial Identities for Stirling Nu... 25) shows us that n Sn = k=0 n+k 1 = 2n . 23) and notice that ∞ k=0 n+k 1 = 2n+1 = 2 · 2n = 2 2k k Thus the sum of the first n + 1 terms, namely one half of the total value of the infinite sum. n k=0 n+k 1 . 2k k n n+k 1 , k=0 k 2k is precisely page 19 May 2, 2013 14:6 BC: 8831 - Probability and Statistical Theory This page intentionally left blank PST˙ws September 15, 2015 12:0 ws-book9x6 Combinatorial Identities for Stirling Nu...

27) is zero for all x ∈ {1, 3, 5, 7, . . , 2n − 1} = {2m + 1}m=0 . 26) is zero for these same n values. Let x = 2m + 1 for 0 ≤ m ≤ n − 1. 26) 2n−2m−1 becomes nj=0 (−1)j 2m+1 . If 2m + 1 > n, we can extend the j n−j range of summation to 2m + 1 since 2n−2m−1 vanishes whenever j > n. n−j If 2m + 1 < n, we can truncate the range of summation at 2m + 1 since 2m+1 = 0 whenever j > 2m + 1. If 2m + 1 = n, the range of summation j 2n−2m−1 stays the same. This analysis implies that nj=0 (−1)j 2m+1 = j n−j 2m+1 j 2m+1 j=0 (−1) j 2n−2m−1 n−j .

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