# 02 - MAD 6206 Combinatorics I Fall 2011 CRN 87401 On...

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Unformatted text preview: MAD 6206 Combinatorics I, Fall 2011 CRN: 87401 On Gosper’s Algorithm The aim is to find a simple form for the sum S ( m ) = ∑ m n =0 t n where r ( n ) = t n +1 /t n is a rational function of n . That is, t n is a hypergeometric term in n and does not depend on m . 1 A Brief Description The idea is find a hypergeometric term z n such that z n +1 − z n = t n , (1) and the sum is therefore S ( m ) = z m +1 − z . To find z n , we write y ( n ) = z n t n = 1 z n +1 z n − 1 , where since z n is a hypergeometric term, we see that y is a rational fucntion, and the problem is turned into finding a rational function y ( n ) where from (1), y ( n + 1) r ( n ) − y ( n ) = 1 . (2) The main step is to find functions a ( n ) ,b ( n ) and c ( n ) (with some required properties to be stated later) so that r ( n ) = t n +1 t n = a ( n ) b ( n ) c ( n + 1) c ( n ) , (3) and (2) now becomes c ( n + 1) y ( n + 1) a ( n ) b ( n ) − y ( n ) c ( n ) = c ( n ) . Writing x ( n ) = c ( n ) y ( n ) b ( n − 1) , or y ( n ) = b ( n − 1) x ( n ) c ( n ) , the previous recurrence in y becomes a recurrence in a rational function x : a ( n ) x ( n + 1) − b ( n − 1) x ( n ) = c ( n ) Now the functions a,b,c in (3) are chosen so that x is a polynomial in n . The first order recurrence in x is then solved (if possible), and hence z n = y ( n ) t n = b ( n − 1) x ( n ) c ( n ) t n . (4) 2 The functions a, b, c Theorem 1. Let r be a rational function over a field K with characteristic zero. Then there are polynomials a,b,c over K such that b,c are monic and r ( n ) = a ( n ) b ( n ) c ( n + 1) c ( n ) , where (i) gcd( a ( n ) ,b ( n + h )) = 1 for all integer h ≥ , (ii) gcd( a ( n ) ,c ( n )) = 1 , (iii) gcd( b ( n ) ,c ( n + 1)) = 1 Proof. Suppose that r ( n ) = f ( n ) g ( n ) , where f,g are polynomials in n over K . We can assume that g is monic and that gcd( f ( n ) ,g ( n )) = 1. If there does not exist h > 0 such that gcd( f ( n ) ,g ( n + h )) = 1, then take a = f , b = g and c = 1 and we are done. Otherwise let h be the smallest positive integer such u ( n ) = gcd( f ( n ) ,g ( n + h )) is not a constant . Now u ( n ) ,u ( n − h ) respectively are divisors of f ( n ) and g ( n ), and we write f ( n ) = u ( n ) f 1 ( n ) , g ( n ) = u ( n − h ) g 1 ( n ) , c 1 ( n ) = h productdisplay i =1 u ( n − i ) , giving that r ( n ) = f ( n ) g ( n ) = f 1 ( n ) g 1 ( n ) × u ( n ) u ( n − 1) ··· u ( n − h + 1) u ( n − 1) n ( n − 2) ··· u ( n − h ) = f 1 ( n ) g 1 ( n ) c 1 ( n + 1) c 1 ( n ) . By the minimality of h , we see that gcd( f 1 ( n ) ,c 1 ( n )) = 1 , gcd( g 1 ( n ) ,c 1 ( n + 1)) = 1 . If gcd( f 1 ( n ) ,g 1 ( n )) = 1 then we are done, else we can repeat the above process for f 1 ( n ) and g 1 ( n )....
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02 - MAD 6206 Combinatorics I Fall 2011 CRN 87401 On...

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