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PS04-solutions

# PS04-solutions - MATH 681 Problem Set#4 1(10 points For...

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MATH 681 Problem Set #4 1. (10 points) For this problem, it will be helpful to note the following two power series expansions: e x + e - x 2 = 1 + x 2 2! + x 4 4! + x 6 6! + · · · e x - e - x 2 = x + x 3 3! + x 5 5! + x 7 7! + · · · (a) (5 points) Find an exponential generating function for the sequence { a n } of the number of ways to build an n -letter string consisting of As, Bs, Cs, and Ds such that there is at least one A, an even number of Bs, an odd number of Cs, and any number of Ds. The exponential generating function for selecting one or more As is x + x 2 2 + x 3 6 + x 4 24 + · · · = e x - 1. The EGF for selecting an even number of Bs is 1+ x 2 2 + x 4 24 + · · · = e x + e - x 2 . The EGF for selecting an odd number of Cs is x + x 3 6 + x 5 120 + · · · = e x - e - x 2 . The EGF for selecting an arbitrary number of Ds is 1+ x + x 2 2 + x 3 6 + x 4 24 + · · · = e x . Assemblnig all these into the EGF for the procedure as a whole: ( e x - 1) e x + e - x 2 e x - e - x 2 e x = e 4 x - e 3 x - 1 + e - 3 x 4 The expansion of the product is not strictly necessary here, but will be useful for the next part of this problem. (b) (5 points) Using your exponential generating function, find a formula for a n . e 4 x - e 3 x - 1 + e - 3 x 4 = 1 4 " X n =0 (4 x ) n n ! - X n =0 (3 x ) n n ! - 1 + X n =0 ( - 3 x ) n n ! # = - 1 4 + X n =0 4 n - 3 n + ( - 1) n 4 x n n ! So a 0 will be somewhat a special case, due to the hanging constant term in the EGF; a 0 = 4 0 - 3 0 +( - 1) 0 4 - 1 4 = 0, for every other value of n , a n = 4 n - 3 n +( - 1) n 4 . 2. (20 points) A string is called “excellent” if it consists of any number of As, any number of Bs, and exactly one C. Let a n represent the number of excellent strings of length n . (a) (5 points) Construct an exponential generating function g ( x ) = n =0 a n x n n ! . The EGFs for selecting As, Bs, and Cs are respectively e x , e x , and x . Thus g ( x ) = e x e x x = xe 2 x . (b) (5 points) Using casewise analysis on the first term of an excellent string, find a recurrence relation, with initial conditions, for a n . An excellent string of length n may begin with an A or B, in which case the n - 1 letters that follow must contain exactly one C; that is, an excellent string of length n can be produced by suffixing an A or B with an excellent string of length n - 1. There are 2 ways to choose the prefix and a n - 1 possible excellent Page 1 of 6 Due October 22, 2009

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MATH 681 Problem Set #4 strings which follow it, for a total of 2 a n - 1 excellent strings of length n generated this way.
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PS04-solutions - MATH 681 Problem Set#4 1(10 points For...

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