# MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 39, Monday, December 8.

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MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 39, Monday, December 8

Example 3: Compound Inhomogeneous Term. a n = 3a n-1 – 4n + 3 £ 2 n. a 1 = 8. a n = A3 n + B 1 n + B 0 + B £ 2 n. a n-1 = A3 n-1 + B 1 (n-1) + B 0 + B £ 2 n-1. A3 n + B 1 n + B 0 + B £ 2 n = A3 n + 3B 1 (n-1) + 3B 0 + 3B £ 2 n-1 –4n + 3 £ 2 n. B 1 = 3B 1 – 4. Hence B 1 = 2. B 0 = -6 + 3B 0. Hence B 0 = 3. 2B = 3B + 6. Hence B = -6. From a 1 = 8 we obtain A = 5. a n = 5 £ 3 n + 2n + 3 – 6 £ 2 n.

7.5. Solutions with Generating Functions Homework (MATH 310#12M): Read 8.1 Do 7.5: all odd numbered problems

Example 1: Summation Recurrence (See Example 3, 7.1, p. 275) Dividing the plane: a n = a n-1 + n, a 0 = 1. a n x n = a n-1 x n + nx n. g(x) – a 0 = xg(x) + x/(1-x) 2. g(x) = 1/(1-x) + x/(1-x) 3 = [1 + C(n+1,2)]x n.

Example 2: Fibonacci Relation a n = a n-1 + a n-2 a 0 = a 1 = 1. g(x) – a 0 – a 1 x = x[g(x) - a 0 ] + x 2 g(x). g(x) = 1/(1 – x – x 2 ) = A/(1-ax) + B/(1-bx). a n = Aa n + Bb n.

Example 3: Selection Without Repetition a n,k = a n-1,k + a n-1,k-1, a n,0 = a n,n = 1. g n (x) = a n,0 + a n,1 x +... + a n,n x n. g n (x) – 1 = g n-1 (x) – 1 + xg n-1 (x). g n (x) = (1+x)g n-1 (x). g 0 (x) = 1. g n (x) = (1+x) n. a n,k = C(n,k).

Example 4: Placing Parentheses a n = a 1 a n-1 + a 2 a n-2 +... + a n-2 a 2 + a n-1 a 1. a 0 = 0, a 1 = 1. g(x) – x = [g(x)] 2. g(x) = ½(1 § sqrt(1-4x)). Since g(0) = 0 g(x) = ½(1 - sqrt(1-4x)). sqrt(1-4x) = (1 – 4x) 1/2 = 1 –C(1/2,1)(4x) + C(1/2,2)(4x) 2 -... C(q,n) = q(q-1)(q-2) £... £ [q-(n-1)]/n! C(1/2,n)(-4) n = -(2/n)C(2n-2,n-1) a n = (1/n)C(2n-2,n-1), n ¸ 1. Catalan numbers.

Example 5: Simultaneous Recurrence Relations (see Example 11 from 7.1) a n = a n-1 + b n-1 + c n-1. b n = 3 n-1 – c n-1. c n = 3 n-1 – b n-1. a 1 = b 1 = c 1 = 1, a 0 = 1, b 0 = 0, c 0 = 0. A(x) – a 0 = xA(x) + xB(x) + xC(x). B(x) – b 0 =x(1-3x) -1 – xC(x). C(x) – c 0 = x(1 – 3x) -1 – xB(x). B(x) = C(x) = (1/4)/(1 – 3x) – (1/4)/(1 + x). A(x) = [xB(x) +xC(x) + 1]/(1-x) b n = c n = (1/4)[3 n – (-1) n ] a n = (1/4)(3 n +3), n even a n = (1/4)(3 n + 1), n odd.

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