By Petkovsek M., Wilf H.S., Zeilberger D.
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"The nice e-book of nature," stated Galileo, "can be learn basically by means of those that understand the language within which it was once written. And this language is arithmetic. " within the Language of arithmetic, award-winning writer Keith Devlin unearths the important position arithmetic performs in our everlasting quest to appreciate who we're and the realm we are living in.
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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.
A) k n k n+a k n−a n−k (b) k n p k (c) k k k (−1) n−k 2. ) 3. Each of the following sums can be evaluated in a simple form. In each case first write the sum in standard hypergeometric notation. Then consult the list in this chapter to find a database member that has the given sum as a special case. Then use the right hand side of the database sum, suitably specialized, to find the simple form of the given sum. Then check your answer numerically for a few small values of the free parameters. n k 2n−1 k (a) n k=0 (b) k (c) k n (k+3a)!
For now we’re going to focus on how to use the key, rather than on how to find it. We’ll follow the standardized proof through, step by step. In Step 1, our term t(n, k) is nk , and the right hand side is rhs(n) = 2n . For Step 2, we divide through by 2n and find that the standardized summand is F (n, k) = nk 2−n , and we now want to prove that k F (n, k) = 1, for this F . In Step 3 we use the key. We take our rational function R(n, k) = k/(2(k −n− 1)), and we define a new function k n −n 2 2(k − n − 1) k kn!