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notes-090924

# notes-090924 - MATH 681 1 1.1 Notes Combinatorics and Graph...

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MATH 681 Notes Combinatorics and Graph Theory I 1 Generating functions, continued 1.1 Exponential generating functions and set-partitions At this point, we’ve come up with good generating-function discussions based on 3 of the 4 rows of our twelvefold way. Will our integer-partition tricks work with exponential generating functions to give us rules for putting distinct balls into indistinct boxes (a.k.a. set-partitions)? Recall that for the integer-partition problem our series of sequential procedures was to first select the number of boxes with one ball, then the number of boxes with 2 balls, then the number of boxes with 3 balls, etc. Would that work in this case? Let’s try it: Question 1: Find an exponential generating function for the number of ways to distribute n balls among unlabeled boxes. Answer 1: An exponential generating function for the number of ways to select boxes with 1 ball in is easy: there is only one way to fill zero boxes, one way to fill one box, etc., so the generating function would be: 1 + x + x 2 2! + · · · = X n =0 x n n ! = e x Filling boxes with two balls gets harder, however: there is one way to set up zero boxes with two balls each, one way to set up one box, three ways to set up two boxes, and fifteen ways to set up six, etc. This ends up being ugly pretty fast, and doesn’t lend itself to computation. However, since we already know this statistic (these are the Bell numbers), we shouldn’t have to go home empty-handed: we can actually work out what the generating function is! Recall that B n = n X k =1 S ( n, k ) = n X k =1 1 k ! k X i =0 ( - 1) k - i k i i n So the generating function we seek is the heinous expression: X n =0 B n x n n ! = X n =0 n X k =1 k X i =0 ( - 1) k - i ( k i ) i n n ! k ! x n This is actually pretty easy! We can find the generating function for S ( n, k ) from the known generating function for k ! S ( n, k ) : X n =0 k ! S ( n, k ) x n n ! = ( e x - 1) k so X n =0 S ( n, k ) x n n ! = ( e x - 1) k k ! and since B n = n k =0 S ( n, k ) , and we can include the zero terms when k > n to get B n = Page 1 of 9 September 24, 2009

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MATH 681 Notes Combinatorics and Graph Theory I k =0 S ( n, k ) , then: X n =0 B n x n n ! = X n =0 X k =0 S ( n, k ) x n n ! = X k =0 X n =0 S ( n, k ) x n n ! = X k =0 ( e x - 1) k k ! = e ( e x - 1) As far as I know, there is no “from first principles” argument to derive this same elegant and simple form. 1.2 Proving binomial identities with generating functions We can prove binomial identities with generating functions, if we know the GFs of the various binomial expressions. For instance, here’s one from the last problem set: Question 2: Prove that n i =0 i ( n i ) = n 2 n - 1 . Answer 2: If we can show that the generating functions n =0 n i =0 i ( n i ) x n and n =0 n 2 n - 1 x n are equal, then we’re done, since two GFs are equivalent if and only if they’re equal in all coefficients.
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