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Unformatted text preview: MAD 6206 Combinatorics I, Fall 2011 CRN: 87401 Assignment 4 – suggested solution Question 1 . (Catalan numbers and sequences of “+1”s and “ − 1”s) Assume that p,r are nonnegative integers. Let Ω p,r be the number of sequences { a i } i ≥ 1 , where a i = ± 1, with p “ − 1”s and p + r “+1”s (so that each sequence sums to r , that is, ∑ 2 p + r i =1 a i = r ). Let C p,r = r 2 p + r parenleftbigg 2 p + r p parenrightbigg , where C p, = δ p, . Note that C p, 1 = 1 2 p +1 ( 2 p +1 p ) = 1 p +1 ( 2 p p ) is the (standard) Catalan number defined in class. We showed in class that the generating function C ( z ) = ∑ p ≥ C p, 1 z p of Catalan numbers satisfies C ( z ) = 1 + z C ( z ) 2 . (1) Recall also that C p, 1 equals the number of sequences of p “ − 1”s and p “+1”s so that every partial sum is at most zero. Let f p,r be the in Ω p,r with the property that every partial sum is less than r . (That is, ∑ j i =1 a i < r for all j < 2 p + r .) We shall show that f p,r = C p,r . (a) Prove (using a simple argument, with our knowledge of C p, 1 given in the problem) that f p, 1 = C p, 1 . (b) Prove that f p,r = summationdisplay p 1 + p 2 + ··· + p r = p C p 1 , 1 C p 2 , 1 ··· C p r , 1 , where the sum is over all p i ≥ 0 satisfying p 1 + p 2 + ··· + p r = p . (c) Prove that summationdisplay p ≥ f p,r z p = C ( z ) r . (d) One (standard) method to find [ z p ] C ( z ) r is by the following theorem. Theorem 1 (Lagrange’s Inversion Formula) . Suppose that φ ( y ) is analytic at y = y and φ ( y ) negationslash = 0 . If y = y + zφ ( y ) , then an analytic function f ( y ) can be expanded as a power series in z by f ( y ) = f ( y ) + ∞ summationdisplay n =1 z n n ! d n − 1 dy n − 1 ( f ′ ( y ) φ ( y ) n ) y = y . Use the theorem to prove that f p,r = C p,r . (Note. In order to apply the Theorem to equation (1), we shall take y = C ( z ) ,φ ( y ) = y 2 ,y = 1 and f ( y ) = y r .) (e) As an example, find [ z n ] z 3 C ( z ) 6 , for n ≥ 3. Solution . (a) For each sequence ω ∈ Ω p, of p “ − 1”s and p “+1”s so that every partial sum is at most 0, we add a “+1” at the end to obtain a sequence ω ′ ∈ Ω p, 1 . It is clear that the partial sums (except the full sum) of ω ′ so obtained are always less than 1. Also, the map ω → ω ′ is clearly a bijection. Thus, the number of sequences in Ω p, 1 with partial sums less than 1 equals C p, 1 . (b) Since the partial sums can only be increased by “+1”, every sequence ω ∈ Ω p,r with partial sums less than r is the concatenation of r sequences ω 1 ,ω 2 ,...,ω r , where the partial sums in ω i ∈ Ω p i , 1 , for each i , are less than 1. Since p 1 + p 2 + ··· + p r = p , we see that C p,r is simply the rfold convolution of C p, 1 . The result hence follows. (c) The result follows directly from the fact that the “product of generating functions of sequences” equals the “generating function of the convolution of the sequences.”the “generating function of the convolution of the sequences....
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 Spring '11
 Suen
 Taylor Series, Summation, Generating function

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