notes-090917 - MATH 681 Notes Combinatorics and Graph...

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Unformatted text preview: MATH 681 Notes Combinatorics and Graph Theory I 1 Generating functions, continued 1.1 Generating functions and partitions We can make use of generating functions to answer some questions a bit more restrictive than weve done so far: Question 1: Find a generating function for the number of ways to distribute n balls among 3 boxes if the first box can contain any number of balls, the second box contains an odd number of balls, and the third contains a multiple of 5 less than 15. Answer 1: The procedure for filling the first box is known to be represented by the g.f. 1+ x + x 2 + x 3 + = 1 1- x . The second, however, would be x + x 3 + x 5 + x 7 + = x (1+ x 2 + x 4 + x 6 + ) = x 1- x 2 , and the third would be 1+ x 5 + x 10 = 1- x 15 1- x 5 . In total, we would have x- x 15 (1- x )(1- x 2 )(1- x 5 ) , which is not tremendously amenable to simplification, but we could actually do it with partial fractions! (this one was done with computer assistance) 3 2(1- x ) 2- 33 4(1- x )- 1 4(1 + x ) + 7 + 6 x + 5 x 2 + 4 x 3 + 3 x 4 + 2 x 5 + 2 x 6 + x 7 + x 8 The three terms at the beginning of this expansion expand to n =0 3 2 ( n +2- 1 2- 1 ) x n- 33 4 x n- 1 4 (- x ) n So for x > 8 , the coefficient of x n is 3 2 ( n + 1)- 33 4- (- 1) n 4 . Note that we couldve phrased this another way: Question 2: Find a generating function for the number of nonnegative integer solutions to x 1 + (2 x 2 + 1) + 5 x 3 = n , with 5 x 3 < 30 . which means we can now produce generating functions if not simple answers to the number of ways to divide any n into linear combinations. Heres one such question which has a certain amount of everyday relevance: x 1 + 5 x 2 + 10 x 3 + 25 x 4 + 50 x 5 + 100 x 6 = n A nonnegative integer solution to this is equivalent to a way of making change for n cents in standard American change (including the half-dollar and dollar coin). The generating function for this can be worked out to be 1 (1- x )(1- x 5 )(1- x 10 )(1- x 25 )(1- x 50 )(1- x 100 ) Trying to actually calculate this, even with partial fractions, is not recommended. However, using partial sums and computer algebra systems, we can find values for individual entries: Question 3: How many ways are there to make change for a dollar? Answer 3: From the above, we can get an approximation sufficient for worikng out the first hundred terms of the sequence: (1+ x + x 2 + + x 100 )(1+ x 5 + x 10 + + x 100 )(1+ x 10 + x 20 + + x 100 )(1+ x 25 + x 50 + x 75 + x 100 )(1+ x 50 + x 100 )(1+ x 100 ) which gives us a 600th-degree polynomial: 1 + x + x 2 + x 3 + x 4 + 2 x 5 + + 293 x 100 + + x 600 . We can thus see that there are 293 ways to make change for a dollar....
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notes-090917 - MATH 681 Notes Combinatorics and Graph...

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