By Jocelyn Quaintance, Henry W. Gould
This ebook is a different paintings which supplies an in-depth exploration into the mathematical services, philosophy, and data of H W Gould. it truly is written in a mode that's available to the reader with uncomplicated mathematical wisdom, and but includes fabric that might be of curiosity to the professional in enumerative combinatorics. This ebook starts with exposition at the combinatorial and algebraic thoughts that Professor Gould makes use of for proving binomial identities. those strategies are then utilized to boost formulation which relate Stirling numbers of the second one sort to Stirling numbers of the 1st sort. Professor Gould's innovations additionally offer connections among either kinds of Stirling numbers and Bernoulli numbers. Professor Gould believes his examine luck comes from his instinct on how one can detect combinatorial identities.This publication will attract a large viewers and will be used both as lecture notes for a starting graduate point combinatorics type, or as a learn complement for the professional in enumerative combinatorics.
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Extra resources for Combinatorial Identities for Stirling Numbers: The Unpublished Notes of H W Gould
I=0 (x − y)k (k) f (y). k! We record this result as Taylor’s Polynomial Expansion: Let f (x) = n f (x) = k=0 (x − y)k (k) f (y), k! n i=0 ai xi . 5) where f (k) (y) denotes the k th derivative of f with respect to x evaluated at y. page 22 September 15, 2015 12:0 ws-book9x6 Combinatorial Identities for Stirling Nu... 5) appears in many guises. 5) becomes n f (x + y) = k=0 xk (k) f (y). k! 6) set y = 0 to obtain n f (x) = k=0 xk (k) f (0), k! 7) a finite polynomial form of the Maclaurin series. 1 Two Summation Interchange Formulas Not all iterative series have independent indices.
22) k=n+1 i=k+1−n We claim that n n 2n−1 k k−⌊ n ⌋(k−n) ai,j = i=1 j=1 ai,k−i+1 . 22), we need to analyze some inequalities. First, let = 0 = nk . Hence, i = 1+ k−1 (k−n) = 1 and 1 ≤ k ≤ n−1. 22). 22). To obtain the third sum on the right side of = 1 = nk . 22), we let n + 1 ≤ k ≤ 2n − 1. Notice that k−1 n Hence, i = 1 + k−1 (k − n) = 1 + 1(k − n) = k + n − 1 and k − nk (k − n) n = k − 1(k − n) = n. 23) to obtain 2n−1 n n ϕ(j) = f (i) i=1 k k−⌊ n ⌋(k−n) j=1 f (i)ϕ(k − i + 1), n ≥ 1. 24) k=1 i=1+⌊ k−1 ⌋(k−n) n Since many combinatorial identities have indices beginning at zero, it would n n be beneficial to have a convolution formula for i=0 f (i) j=0 ϕ(j).
11) holds for n ≥ 0. 8) implies that n n n 2n 2n 2n + 1 22n = + = . 12) are basic applications of the binomial theorem. 2), we demonstrate just one more which is related to Melzak’s theorem. Let n be n xk , where x is a nonzero a nonnegative integer. Define Sn := k=0 nk k+1 complex number. 2) implies that n Sn = k=0 = = 1 n xk = n+1 k k+1 1 n+1 n+1 k=1 1 x(n + 1) n k=0 n+1 k x k+1 1 n + 1 k−1 x = x(n + 1) k n+1 k=0 n+1 k=1 n+1 k x k (x + 1)n+1 − 1 n+1 k . 13) page 15 September 15, 2015 12:0 ws-book9x6 16 Combinatorial Identities for Stirling Nu...