By Marko Petkovsek, Herbert S. Wilf

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2(−n − 12 )! (− 12 )! A similar argument shows that (−2n − 12 )! (−1)n (n − 12 )! = , (−n − 12 )! (2n − 12 )! 2 2n . n (2n − 12 )! (− 12 )! But for every positive integer m, 1 1 3 1 1 (m − )! = (m − )(m − ) · · · ( )(− )! 2 2 2 2 2 (2m − 1)(2m − 3) · · · 1 1 = (− )! 2m 2 (2m)! 1 = m (− )!. 4 m! 6 Using the database 47 So we can simplify our answer all the way down to f(n) = labor is that we have found the identity (−1)k k 2n k 2k k 2n 2 . 3) we realize that it is a special case of Dixon’s identity, and we further realize that the “lookup” in the database was not quite a routine matter!

In which tk+1 P (k) = , tk Q(k) where P and Q are polynomials in k. In this case we will call the terms hypergeometric terms. /((k + 3)! (2k + 7)). Hypergeometric series are very important in mathematics. Many of the familiar functions of analysis are hypergeometric. These include the exponential, logarithmic, trigonometric, binomial, and Bessel functions, along with the classical orthogonal polynomial sequences of Legendre, Chebyshev, Laguerre, Hermite, etc. 3 How to identify a series as hypergeometric 35 metric, then identifying precisely which hypergeometric function it is, and finally by using known results about such functions.

The answers are formally correct, but we need some further analysis to transform them into readily useable form. /(−2n)!? Imagine, for a moment, that n is near a positive integer, but is not equal to a positive integer. Then we use the reflection formula for the Γ-function π Γ(z)Γ(1 − z) = sin πz once more, in the equivalent form π (−z)! = . (z − 1)! sin πz When n is near, but not equal to, a positive integer we find that (−n)! π (sin 2nπ)(2n − 1)! 2(2n − 1)! cos nπ = = . (−2n)! (sin nπ)(n − 1)!