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# a05a - CRN 87401 MAD 6206 Combinatorics I Fall 2011...

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Unformatted text preview: CRN: 87401 MAD 6206 Combinatorics I, Fall 2011 Assignment 5 – suggested solution Question 1. (Eulerian numbers and their generating function) Recall that the the Eulerian numbers (A(n, k ) = An,k ) are deﬁned as the number of permutations on [n] with k runs. We proved in class that they satisfy the following recurrence A(n, k ) = (n − k + 1)A(n − 1, k − 1) + kA(n − 1, k ), n ≥ 1, k ≥ 1, with boundary conditions A(n, 0) = δ0,n , A(0, k ) = δ0,k . Let A(n, k ) G(z, x) = n≥0 k≥0 z n xk . n! (a) Prove that (1 − zx) ∂ ∂ G(z, x) − x(1 − x) G(z, x) = xG(z, x). ∂z ∂x 1 (b) To solve the pde, we change the variable z as follows. Let u = x exp(z (1 − x)) (or z = 1−x log(u/x)), and by replacing z in G(z, x), we obtain a function Y (u, x) so that Y (u, x) = G(z, x). Prove that ∂ Y (u, x) = −Y /(1 − x), ∂x and hence solve for G(z, x). (c) Prove that k A(n, k ) = j =0 n+1 (k − j )n . j (−1)j Note. The end of the assignment contains a table of Eulerian numbers. Solution. (a) We shall obtain the given pde from the recurrence directly as follows. We multiply both sides of the recurrence relation by nz n−1 xk /n!, sum over n ≥ 1 and k ≥ 1, and then using the boundary conditions, we obtain nA(n, k ) n≥1 k≥1 z n−1 xk = n! nA(n, k ) n≥0 k≥0 z n−1 xk ∂ = G(z, x) n! ∂z and n≥1 k≥1 = n≥1 k≥1 nA(n − 1, k − 1) (n + 1)A(n, k ) = n≥0 k≥0 = n(n − k + 1)A(n − 1, k − 1) z n−1 xk − (n − 1)! z n xk+1 − n! ∂ ∂ (zxG(z, x)) − x2 G(z, x), ∂z ∂x z n−1 xk n! n≥1 k≥1 (k − 1)A(n − 1, k − 1) kA(n, k ) n≥0 k≥0 z n xk+1 n! z n−1 xk (n − 1)! and also n≥1 k≥1 nkA(n − 1, k ) z n−1 xk n! = n≥1 k≥0 = n≥1 k≥0 = nkA(n − 1, k ) kA(n − 1, k ) kA(n, k ) n≥0 k≥0 =x z n−1 xk n! z n−1 xk (n − 1)! z n xk n! ∂ G(z, x). ∂z Hence, we have ∂ ∂ ∂ ∂ G(z, x) = (zxG(z, x)) − x2 G(z, x) + x G(z, x). ∂z ∂z ∂x ∂z Rewriting gives ∂ ∂ G(z, x) − x(1 − x) G(z, x) = xG(z, x). ∂z ∂x (b) Note that with u = x exp(z (1 − x)) and G(z, x) = Y (u, x), we have (1 − zx) Gz (z, x) = x(1 − x) exp(z (1 − x))Yu (u, x) Gx (z, x) = (exp(z (1 − x)) − zx exp(z (1 − x)))Yu (u, x) + Yx (u, x) = (1 − zx) exp(z (1 − x))Yu (u, x) + Yx (u, x). The diﬀerential equation in Y now follows from (a). Upon solving for Y , we see that Y (u, x) = B (u)(1 − x) for some function B . Thus G(z, x) = B (x exp(z (1 − x)))(1 − x). But G(0, x) = 1 implies B (x) = 1/(1 − x). Hence, G(z, x) = B (x exp(z (1 − x)))(1 − x) = 1−x . 1 − x exp(z (1 − x)) (c) A(n, k ) = n![z n xk ]G(z, x) = n![z n xk ](1 − x) = n![z n xk ](1 − x) = [xk ] j ≥0 xj j n = [x ] j =0 k = j =0 j ≥0 xj j n tn (1 − x)n xj j n (1 − x)n+1 k k j ≥0 xj exp(jz (1 − x)) (−1)k−j j n n+1 (−x)k−j k−j n+1 . k−j 1 n! Question 2. (Eulerian numbers, Stirling numbers, and ﬁnite diﬀerence) induction to prove that for any x, A(n, k ) k x+n−k n = xn , Use the recurrence in Q1 and ∀n ≥ 0, where the sum is over all values of k and we assume that A(n, k ) = 0 for all k < 0. (b) Prove that for integers m, n ≥ 0, A(n, k ) k m x+n−k n−m = j =0 m (−1)j (x + m − j )n . j (c) Prove that A(n, k ) k n−k n−m = m! n . m Solution. (a) We use induction on n. For the basis step, we observe that when n = 0, LHS = A(0, 0) x 0 = 1 = x0 = RHS. For the induction step, let n ≥ 1, and assume as induction hypothesis that the formula holds for all smaller values of n. We want to prove the formula holds for n. Using the recurrence given in Q1, we have (with A(n, 0) = 0) that A(n, k ) LHS = k ≥0 = k ≥1 = k ≥1 = k ≥0 = k ≥0 x+n−k n ((n − k + 1)A(n, k − 1) + kA(n − 1, k )) (n − k + 1)A(n − 1, k − 1) (n − k )A(n − 1, k ) x+n−k + n x+n−k−1 + n A(n − 1, k ) (n − k ) k ≥0 x+n−k n k ≥1 kA(n − 1, k ) kA(n − 1, k ) x+n−k x+n−k−1 +k n n x+n−k n x+n−k n . But (n − k ) x+n−k−1 x+n−k +k n n =n x+n−k−1 x+n−k−1 +k n n−1 Thus, using the induction hypothesis, LHS = k A(n − 1, k ) · x · x+n−k−1 n−1 = x · xn−1 = xn . =x x+n−k−1 . n−1 This completes the induction step, and hence the induction proof. (b) We perform ∆m on the identity found in (a). Note that ∆m A(n, k ) k x+n−k n x+n−k n A(n, k )∆m = k A(n, k ) = k x+n−k . n−m Also, as ∆ = E − I , where E is the shift operator and I is the identity, we have m m n m m (−1)j E m−j (xn ) = j n ∆ (x ) = (E − I ) (x ) = j =0 m j =0 m (−1)j (x + m − j )n . j (c) We set x = 0 to get k n−k A(n, k ) n−m m = j =0 n m . (−1)j (m − j )n = m! m j Question 3. (The Γ function and the falling factorial) We deﬁned Γ(z ) by 1 = z eγz Γ(z ) ∞ n=1 {(1 + z/n) exp(−z/n)} , where γ denotes the Euler constant, and n γ = lim (Hn − log n) , n→∞ 1/k is the nth harmonic number. where Hn = k=1 (a) Show that Γ(z + 1) = z Γ(z ), z = 0, −1, −2, −3, . . . . (b) We deﬁned in class that z ! = Γ(z + 1), for z = −1, −2, . . ., and that for any z, n, (z )n = lim w→z Prove that for integer m ≥ 0, we have (z )−m = (c) Show that ′ Γ (z ) = −Γ(z ) · (d) Prove that for integer k ≥ 0, z! . (z − n)! m 1 i=1 z +i . 1 +γ+ z ∞ n=1 1 1 − n+z n Γ′ (k + 1) = k ! (Hk − γ ) . . Solution. (a) From the deﬁnition of Γ, 1 Γ(z + 1) m = (z + 1) exp(γ (z + 1)) lim m→∞ n=1 m ((1 + (z + 1)/n) exp(−(z + 1)/n)) = exp(γz ) exp(γ )(1 + z ) lim m→∞ n=1 n+1 n 1+ z n+1 exp − z+1 n m+1 = exp(γz ) lim m→∞ × lim m→∞ = exp(γz ) ∞ n=1 = n=1 exp ((1 + z/n) exp(−z/n)) z m+1 m exp − n=1 1 + log(m + 1) + γ n ((1 + z/n) exp(−z/n)) 1 . z Γ(z ) Since Γ(z + 1) and z Γ(z ) are both deﬁned for z = 0, −1, −2, −3, . . ., the result follows. (b) From the deﬁnition of z ! and from (a) above, we have z ! = z · (z − 1)!. Thus, for integer m ≥ 0, (z )−m = lim w→z w! = lim (w + m)! w→z m i=1 1 = w+i m i=1 1 . z+i Note that the limit in the deﬁnition is needed. If otherwise, we have for example (−10)2 = not deﬁned without the limit. (c) Notice that − ln Γ(z ) = ln z + γz + ∞ n=1 ln(n + z ) − ln(n) − (−10)! (−12)! which is z . n The given formula is obtained by diﬀerentiating with respect to z . (d) Take the formula in (c) and set z = k + 1. (Note. We must not break the sum ∞ 1 n=1 n+k − ∞1 n=1 n ∞ n=1 1 n+k because the individual sums do not converge.) Question 4. (Generalized Fibonacci numbers) Deﬁne for integer n ≥ 0, a polynomial Fn by n Fn (z ) = k=0 n−k k z. k We showed in class that if α, β are values of x satisfying x2 − x − z = 0, then Fn (z ) = 1 (αn+1 − β n+1 ). α−β (a) Verify that α and β can be written as r exp(±iθ), where √ 2r cos(θ) = 1, r = −z, − 1 n into and hence prove that rn sin((n + 1)θ) . sin(θ) Fn (z ) = In particular, check that n n−k 2(n+3)/2 (−2)k = √ sin((n + 1)θ), k 7 k=0 1 where θ = cos−1 ( √8 ) (b) By expressing r and z in terms of θ, prove that for any integer n ≥ 0, n sin((n + 1)θ) = sin(θ) k=0 n−k (−1)k (2 cos(θ))n−2k . k Solution. (a) Since α and β are solutions to a quadratic, they can be written as “complex conjugates” in the form of r exp(±iθ), where r2 = αβ = −z, 2r cos(θ) = α + β = 1. Note that r and θ can be complex numbers, with cos(x) = this, we have Fn (z ) = 1 2 eix + e−ix and sin(x) = 1 2 eix − e−ix . From 1 rn sin((n + 1)θ) 1 (αn+1 − β n+1 ) = (2rn+1 i sin((n + 1)θ) = . α−β 2ri sin(θ) sin(θ) The special case where z = −2 follows with r = (b) From (a), we have √ 2, cos(θ) = 2−3/2 and sin(θ) = n sin((n + 1)θ) = r−n sin(θ)Fn (z ) = r−n k=0 Note that in terms of θ, Since z = −r2 , and 7/8. n−k k z. k r = (2 cos(θ))−1 r−n z k = (−1)k r2k−n = (−1)k (2 cos(θ))n−2k , we have n sin((n + 1)θ) = sin(θ) k=0 n−k (−1)k (2 cos(θ))n−2k . k Question 5 (Finite diﬀerence) Recall that n ∆n f (x) = k=0 n (−1)n−k f (x + k ). k (a) Show that for integer n ≥ 0, n k=0 (−1)k n (n + 1)! = . k (x + k + 1)(x + k + 2) (x + 1)(x + 2) · · · (x + n + 2) (b) For integers m and n ≥ 0, and for any x, ﬁnd a closed form for the sum n Fm,n (x) = k=0 x+k n . (−1)n−k m k In particular, evaluate F5,10 (x) and Fn,n (x). Solution. (a) Let f (x) = (x)−2 . Then ∆n f (x) = (−2)n (x)−2−n = (−1)n (n + 1)! , (x + 1)(x + 2) · · · (x + n + 2) while f (x + k ) = (x + k )−2 and n n (−1)n−k (x + k )−2 = k k=0 n n (−1)n−k . k (x + k + 1)(x + k + 2) k=0 Therefore, n k=0 (−1)k n (n + 1)! = . k (x + k + 1)(x + k + 2) (x + 1)(x + 2) · · · (x + n + 2) As a concrete example, if we take n = 2, then the formula becomes 1 (x+1)(x+2)(x+3)(x+4) 1 6 = (b) We apply the ∆n = (E − I )n to f (x) = 1 (x+1)(x+2) x m ∆n − 2 (x+2)(x+3) + 1 (x+3)(x+4) . in two diﬀerent ways to get x m = x . m−n Also, ∆n x m n = k=0 x n (−1)n−k E k m k n = k=0 x+k n . (−1)n−k m k Thus, Fm,n (x) = ∆n f (x) = x . m−n As special cases, when n > m, the formula gives Fm,n (x) = 0 and Fn,n (x) = 1. Question 6. (Vandermonde’s convolution) Prove the following version of Vandermonde’s convolution: For integer ℓ ≥ 0 and integers m, n, k ℓ m+k s−m s+k . (−1)k = (−1)ℓ+m n−ℓ n Solution. We try the following. Since ℓ is a nonnegative integer, we have z ℓ = (1 − (1 − z ))ℓ = k (−1)m+k ℓ (1 − z )m+k , m+k and therefore, z ℓ (1 − z )s−m = Hence, by considering coeﬃcients of (−1)n−ℓ zn (−1)m+k k ℓ (1 − z )s+k . m+k on both sides, we obtain s−m n−ℓ (−1)m+k = k s+k ℓ , (−1)n n m+k which can be reduced to the required identity easily. Note that the sum in Q5(b) is a special case of this problem. Notes. As Joseph pointed out, we can also apply the result in Q5(b) as follows. In Q5(b), we showed that n x n−m n x+k (−1)n−k . k m = k=0 Replacing n with ℓ, and m with n, in the above identity, we have x ℓ−n x+k ℓ . (−1)ℓ−k k n = k Switching the summation index (replacing k with k + m gives x ℓ−n x+k+m ℓ . (−1)ℓ−k−m n k+m = k Finally, replace x with s − m to obtain s−m ℓ−n s+k ℓ , (−1)ℓ−k−m n k+m = k from which the required identity follows easily (by multiplying both sides with (−1)ℓ+m . Question 7. (Inversion, Stirling numbers, and Bell numbers) (a) Recall that k ! n equals the number of k distributions of n distinct balls to k distinct boxes so that no box is empty. Use a combinatorial argument to show that m n m n . k! m= k k k=0 Use inversion to show that n m m! m = k=0 m (−1)m−k k n . k (b) We showed in class that the Bell number Bn satisﬁes n Bn+1 = k=0 n Bk , k integer n ≥ 0. Use inversion to prove that n Bn = k=0 n (−1)n−k Bk+1 , k integer n ≥ 0. (c) Prove that n k=0 n (−1)n−k Bk = 1, k integer n ≥ 0. Solution. (a) To distribute n distinct balls to m distinct boxes (and there are mn such distributions), we ﬁrst decide which of the (say k ) boxes are not empty, and then distribution the n balls to these nonempty boxes. The number of ways to do this is m n m . k! k k mn = k=0 Since am,k = m and bm,k = k and vk = k n ) that since m k (−1)m−k are connecting coeﬃcients, we see that (by using uk = k ! n k n am,k uk , vm = k=0 we have, from um = m k=0 bm,k vk , that m m m! k = k=0 m (−1)m−k k n . k (b) Let uk = Bk and vk = Bk+1 . Then the given identity is simply vn = have vn = n=0 n (−1)n−k uk , which is k k n Bn = k=0 n n k=0 k uk . Upon inversion, we n (−1)n−k Bk+1 . k (c) Note that n xn = k=0 Thus, an,k = n k and bn,k = n k n n (x)k , k (x)n = k=0 n (−1)n−k xk . k (−1)n−k are connecting coeﬃcients. Next, it is clear that n Bn = k=0 n . k Thus, if we choose uk = 1 and vk = Bk , then we have from un = n 1= k=0 Table of Eulerian numbers n (−1)n−k Bk . k n k=0 bn,k vk that n 0 1 2 3 4 5 6 7 8 A(n, 0) 1 0 0 0 0 0 0 0 0 A(n, 1) A(n, 2) A(n, 3) A(n, 4) A(n, 5) A(n, 6) A(n, 7) A(n, 8) 1 1 1 1 1 1 1 1 1 4 11 26 57 120 247 1 11 66 302 1191 4293 1 26 302 2416 15619 1 57 1191 15619 1 120 4293 1 247 1 ...
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