Touchard Polynomials, Stirling Numbers and Random Permutations
Ross Pinsky
Department of Mathematics, Technion 32000 Haifa, ISRAEL
September 3, 2018
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations |Sn| = n! 1 2 3 4 5 6 S 3 σ = = (124)(36)(5), 6 1 2 4 6 1 5 3 1 2 3 4 5 6 S 3 σ = = (136245) = (624513) 6 2 3 4 6 5 1 2
σ2 is a 6-cycle in S6. There are 5! different 6-cycles in S6.
There are (n − 1)! different n-cycles in Sn.
Notation: [n] = {1, 2, ··· , n}
Sn = permutations of [n]
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations 1 2 3 4 5 6 S 3 σ = = (124)(36)(5), 6 1 2 4 6 1 5 3 1 2 3 4 5 6 S 3 σ = = (136245) = (624513) 6 2 3 4 6 5 1 2
σ2 is a 6-cycle in S6. There are 5! different 6-cycles in S6.
There are (n − 1)! different n-cycles in Sn.
Notation: [n] = {1, 2, ··· , n}
Sn = permutations of [n]
|Sn| = n!
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations 1 2 3 4 5 6 S 3 σ = = (136245) = (624513) 6 2 3 4 6 5 1 2
σ2 is a 6-cycle in S6. There are 5! different 6-cycles in S6.
There are (n − 1)! different n-cycles in Sn.
Notation: [n] = {1, 2, ··· , n}
Sn = permutations of [n]
|Sn| = n! 1 2 3 4 5 6 S 3 σ = = (124)(36)(5), 6 1 2 4 6 1 5 3
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations σ2 is a 6-cycle in S6. There are 5! different 6-cycles in S6.
There are (n − 1)! different n-cycles in Sn.
Notation: [n] = {1, 2, ··· , n}
Sn = permutations of [n]
|Sn| = n! 1 2 3 4 5 6 S 3 σ = = (124)(36)(5), 6 1 2 4 6 1 5 3 1 2 3 4 5 6 S 3 σ = = (136245) = (624513) 6 2 3 4 6 5 1 2
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations There are (n − 1)! different n-cycles in Sn.
Notation: [n] = {1, 2, ··· , n}
Sn = permutations of [n]
|Sn| = n! 1 2 3 4 5 6 S 3 σ = = (124)(36)(5), 6 1 2 4 6 1 5 3 1 2 3 4 5 6 S 3 σ = = (136245) = (624513) 6 2 3 4 6 5 1 2
σ2 is a 6-cycle in S6. There are 5! different 6-cycles in S6.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Notation: [n] = {1, 2, ··· , n}
Sn = permutations of [n]
|Sn| = n! 1 2 3 4 5 6 S 3 σ = = (124)(36)(5), 6 1 2 4 6 1 5 3 1 2 3 4 5 6 S 3 σ = = (136245) = (624513) 6 2 3 4 6 5 1 2
σ2 is a 6-cycle in S6. There are 5! different 6-cycles in S6.
There are (n − 1)! different n-cycles in Sn.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations n X k ≡ ank x , x ∈ R, n ≥ 1. (1) k=1 Unsigned Stirling Numbers of the First Kind: |s(n, k)| := the number of permutations in Sn with exactly k cycles, 1 ≤ k ≤ n Recurrence relation |s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1; (2) |s(n + 1, k)| = n|s(n, k)| + |s(n, k − 1)|, 2 ≤ k ≤ n.
It is easy to check that the coefficients {ank } in (1) also satisfy (2); thus, ank = |s(n, k)|. Therefore n X x(n) = |s(n, k)|xk . (3) k=1
Rising Factorials x(n) := x(x +1) ··· (x +n−1)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Unsigned Stirling Numbers of the First Kind: |s(n, k)| := the number of permutations in Sn with exactly k cycles, 1 ≤ k ≤ n Recurrence relation |s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1; (2) |s(n + 1, k)| = n|s(n, k)| + |s(n, k − 1)|, 2 ≤ k ≤ n.
It is easy to check that the coefficients {ank } in (1) also satisfy (2); thus, ank = |s(n, k)|. Therefore n X x(n) = |s(n, k)|xk . (3) k=1
Rising Factorials
n (n) X k x := x(x +1) ··· (x +n−1) ≡ ank x , x ∈ R, n ≥ 1. (1) k=1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Recurrence relation |s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1; (2) |s(n + 1, k)| = n|s(n, k)| + |s(n, k − 1)|, 2 ≤ k ≤ n.
It is easy to check that the coefficients {ank } in (1) also satisfy (2); thus, ank = |s(n, k)|. Therefore n X x(n) = |s(n, k)|xk . (3) k=1
Rising Factorials
n (n) X k x := x(x +1) ··· (x +n−1) ≡ ank x , x ∈ R, n ≥ 1. (1) k=1 Unsigned Stirling Numbers of the First Kind: |s(n, k)| := the number of permutations in Sn with exactly k cycles, 1 ≤ k ≤ n
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations It is easy to check that the coefficients {ank } in (1) also satisfy (2); thus, ank = |s(n, k)|. Therefore n X x(n) = |s(n, k)|xk . (3) k=1
Rising Factorials
n (n) X k x := x(x +1) ··· (x +n−1) ≡ ank x , x ∈ R, n ≥ 1. (1) k=1 Unsigned Stirling Numbers of the First Kind: |s(n, k)| := the number of permutations in Sn with exactly k cycles, 1 ≤ k ≤ n Recurrence relation |s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1; (2) |s(n + 1, k)| = n|s(n, k)| + |s(n, k − 1)|, 2 ≤ k ≤ n.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Therefore
n X x(n) = |s(n, k)|xk . (3) k=1
Rising Factorials
n (n) X k x := x(x +1) ··· (x +n−1) ≡ ank x , x ∈ R, n ≥ 1. (1) k=1 Unsigned Stirling Numbers of the First Kind: |s(n, k)| := the number of permutations in Sn with exactly k cycles, 1 ≤ k ≤ n Recurrence relation |s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1; (2) |s(n + 1, k)| = n|s(n, k)| + |s(n, k − 1)|, 2 ≤ k ≤ n.
It is easy to check that the coefficients {ank } in (1) also satisfy (2); thus, ank = |s(n, k)|.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Rising Factorials
n (n) X k x := x(x +1) ··· (x +n−1) ≡ ank x , x ∈ R, n ≥ 1. (1) k=1 Unsigned Stirling Numbers of the First Kind: |s(n, k)| := the number of permutations in Sn with exactly k cycles, 1 ≤ k ≤ n Recurrence relation |s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1; (2) |s(n + 1, k)| = n|s(n, k)| + |s(n, k − 1)|, 2 ≤ k ≤ n.
It is easy to check that the coefficients {ank } in (1) also satisfy (2); thus, ank = |s(n, k)|. Therefore n X x(n) = |s(n, k)|xk . (3) k=1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Substituting −x for x in (4), we have n (−x)n = −x(−x −1) ··· (−x −n+1) = (−1) x(x +1) ··· (x +n−1) = (−1)nx(n). Substituting above −x for x, we have n n (n) from (3) n X k (x)n = (−1) (−x) = (−1) |s(n, k)|(−x) = k=1 n (5) X (−1)n−k |s(n, k)|xk . k=1 Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called the Stirling Numbers of the First Kind. From (5) we have n X k (x)n = s(n, k)x . (6) k=1
Falling Factorials
(x)n := x(x − 1) ··· (x − n + 1), x ∈ R, n ≥ 1. (4)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations (−1)nx(x +1) ··· (x +n−1) = (−1)nx(n). Substituting above −x for x, we have n n (n) from (3) n X k (x)n = (−1) (−x) = (−1) |s(n, k)|(−x) = k=1 n (5) X (−1)n−k |s(n, k)|xk . k=1 Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called the Stirling Numbers of the First Kind. From (5) we have n X k (x)n = s(n, k)x . (6) k=1
Falling Factorials
(x)n := x(x − 1) ··· (x − n + 1), x ∈ R, n ≥ 1. (4) Substituting −x for x in (4), we have
(−x)n = −x(−x −1) ··· (−x −n+1) =
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations = (−1)nx(n). Substituting above −x for x, we have n n (n) from (3) n X k (x)n = (−1) (−x) = (−1) |s(n, k)|(−x) = k=1 n (5) X (−1)n−k |s(n, k)|xk . k=1 Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called the Stirling Numbers of the First Kind. From (5) we have n X k (x)n = s(n, k)x . (6) k=1
Falling Factorials
(x)n := x(x − 1) ··· (x − n + 1), x ∈ R, n ≥ 1. (4) Substituting −x for x in (4), we have n (−x)n = −x(−x −1) ··· (−x −n+1) = (−1) x(x +1) ··· (x +n−1)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Substituting above −x for x, we have n n (n) from (3) n X k (x)n = (−1) (−x) = (−1) |s(n, k)|(−x) = k=1 n (5) X (−1)n−k |s(n, k)|xk . k=1 Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called the Stirling Numbers of the First Kind. From (5) we have n X k (x)n = s(n, k)x . (6) k=1
Falling Factorials
(x)n := x(x − 1) ··· (x − n + 1), x ∈ R, n ≥ 1. (4) Substituting −x for x in (4), we have n (−x)n = −x(−x −1) ··· (−x −n+1) = (−1) x(x +1) ··· (x +n−1) = (−1)nx(n).
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations =
n (5) X (−1)n−k |s(n, k)|xk . k=1 Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called the Stirling Numbers of the First Kind. From (5) we have n X k (x)n = s(n, k)x . (6) k=1
Falling Factorials
(x)n := x(x − 1) ··· (x − n + 1), x ∈ R, n ≥ 1. (4) Substituting −x for x in (4), we have n (−x)n = −x(−x −1) ··· (−x −n+1) = (−1) x(x +1) ··· (x +n−1) = (−1)nx(n). Substituting above −x for x, we have n n (n) from (3) n X k (x)n = (−1) (−x) = (−1) |s(n, k)|(−x) k=1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called the Stirling Numbers of the First Kind. From (5) we have n X k (x)n = s(n, k)x . (6) k=1
Falling Factorials
(x)n := x(x − 1) ··· (x − n + 1), x ∈ R, n ≥ 1. (4) Substituting −x for x in (4), we have n (−x)n = −x(−x −1) ··· (−x −n+1) = (−1) x(x +1) ··· (x +n−1) = (−1)nx(n). Substituting above −x for x, we have n n (n) from (3) n X k (x)n = (−1) (−x) = (−1) |s(n, k)|(−x) = k=1 n (5) X (−1)n−k |s(n, k)|xk . k=1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations From (5) we have n X k (x)n = s(n, k)x . (6) k=1
Falling Factorials
(x)n := x(x − 1) ··· (x − n + 1), x ∈ R, n ≥ 1. (4) Substituting −x for x in (4), we have n (−x)n = −x(−x −1) ··· (−x −n+1) = (−1) x(x +1) ··· (x +n−1) = (−1)nx(n). Substituting above −x for x, we have n n (n) from (3) n X k (x)n = (−1) (−x) = (−1) |s(n, k)|(−x) = k=1 n (5) X (−1)n−k |s(n, k)|xk . k=1 Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called the Stirling Numbers of the First Kind.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Falling Factorials
(x)n := x(x − 1) ··· (x − n + 1), x ∈ R, n ≥ 1. (4) Substituting −x for x in (4), we have n (−x)n = −x(−x −1) ··· (−x −n+1) = (−1) x(x +1) ··· (x +n−1) = (−1)nx(n). Substituting above −x for x, we have n n (n) from (3) n X k (x)n = (−1) (−x) = (−1) |s(n, k)|(−x) = k=1 n (5) X (−1)n−k |s(n, k)|xk . k=1 Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called the Stirling Numbers of the First Kind. From (5) we have n X k (x)n = s(n, k)x . (6) k=1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Example: S(4, 3) = 5: {1, 2}, {3}, {4} {1, 3}, {2}, {4} {1, 4}, {2}, {3} {2, 3}, {1}, {4} {2, 4}, {1}, {3}
Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations {1, 2}, {3}, {4} {1, 3}, {2}, {4} {1, 4}, {2}, {3} {2, 3}, {1}, {4} {2, 4}, {1}, {3}
Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets. Example: S(4, 3) = 5:
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets. Example: S(4, 3) = 5: {1, 2}, {3}, {4} {1, 3}, {2}, {4} {1, 4}, {2}, {3} {2, 3}, {1}, {4} {2, 4}, {1}, {3}
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Theorem.
n n X x = S(n, k)(x)k , (7) k=1
Proof. Consider functions f :[n] → X , where |X | = x ∈ N. How many such functions are there? The answer by direct count: xn. An alternative indirect count: For k = 1, 2, ··· , n, let ck denote the number of such functions with |Im(f )| = k. (If x < n, then ck = 0 for x < k ≤ n.) Pn So the answer by this indirect count is k=1 ck . n Pn Thus, x = k=1 ck . To complete the proof we now show that ck = S(n, k)(x)k .
Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Proof. Consider functions f :[n] → X , where |X | = x ∈ N. How many such functions are there? The answer by direct count: xn. An alternative indirect count: For k = 1, 2, ··· , n, let ck denote the number of such functions with |Im(f )| = k. (If x < n, then ck = 0 for x < k ≤ n.) Pn So the answer by this indirect count is k=1 ck . n Pn Thus, x = k=1 ck . To complete the proof we now show that ck = S(n, k)(x)k .
Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets. Theorem.
n n X x = S(n, k)(x)k , (7) k=1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations How many such functions are there? The answer by direct count: xn. An alternative indirect count: For k = 1, 2, ··· , n, let ck denote the number of such functions with |Im(f )| = k. (If x < n, then ck = 0 for x < k ≤ n.) Pn So the answer by this indirect count is k=1 ck . n Pn Thus, x = k=1 ck . To complete the proof we now show that ck = S(n, k)(x)k .
Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets. Theorem.
n n X x = S(n, k)(x)k , (7) k=1
Proof. Consider functions f :[n] → X , where |X | = x ∈ N.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations The answer by direct count: xn. An alternative indirect count: For k = 1, 2, ··· , n, let ck denote the number of such functions with |Im(f )| = k. (If x < n, then ck = 0 for x < k ≤ n.) Pn So the answer by this indirect count is k=1 ck . n Pn Thus, x = k=1 ck . To complete the proof we now show that ck = S(n, k)(x)k .
Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets. Theorem.
n n X x = S(n, k)(x)k , (7) k=1
Proof. Consider functions f :[n] → X , where |X | = x ∈ N. How many such functions are there?
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations An alternative indirect count: For k = 1, 2, ··· , n, let ck denote the number of such functions with |Im(f )| = k. (If x < n, then ck = 0 for x < k ≤ n.) Pn So the answer by this indirect count is k=1 ck . n Pn Thus, x = k=1 ck . To complete the proof we now show that ck = S(n, k)(x)k .
Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets. Theorem.
n n X x = S(n, k)(x)k , (7) k=1
Proof. Consider functions f :[n] → X , where |X | = x ∈ N. How many such functions are there? The answer by direct count: xn.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations For k = 1, 2, ··· , n, let ck denote the number of such functions with |Im(f )| = k. (If x < n, then ck = 0 for x < k ≤ n.) Pn So the answer by this indirect count is k=1 ck . n Pn Thus, x = k=1 ck . To complete the proof we now show that ck = S(n, k)(x)k .
Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets. Theorem.
n n X x = S(n, k)(x)k , (7) k=1
Proof. Consider functions f :[n] → X , where |X | = x ∈ N. How many such functions are there? The answer by direct count: xn. An alternative indirect count:
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Pn So the answer by this indirect count is k=1 ck . n Pn Thus, x = k=1 ck . To complete the proof we now show that ck = S(n, k)(x)k .
Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets. Theorem.
n n X x = S(n, k)(x)k , (7) k=1
Proof. Consider functions f :[n] → X , where |X | = x ∈ N. How many such functions are there? The answer by direct count: xn. An alternative indirect count: For k = 1, 2, ··· , n, let ck denote the number of such functions with |Im(f )| = k. (If x < n, then ck = 0 for x < k ≤ n.)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations n Pn Thus, x = k=1 ck . To complete the proof we now show that ck = S(n, k)(x)k .
Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets. Theorem.
n n X x = S(n, k)(x)k , (7) k=1
Proof. Consider functions f :[n] → X , where |X | = x ∈ N. How many such functions are there? The answer by direct count: xn. An alternative indirect count: For k = 1, 2, ··· , n, let ck denote the number of such functions with |Im(f )| = k. (If x < n, then ck = 0 for x < k ≤ n.) Pn So the answer by this indirect count is k=1 ck .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations To complete the proof we now show that ck = S(n, k)(x)k .
Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets. Theorem.
n n X x = S(n, k)(x)k , (7) k=1
Proof. Consider functions f :[n] → X , where |X | = x ∈ N. How many such functions are there? The answer by direct count: xn. An alternative indirect count: For k = 1, 2, ··· , n, let ck denote the number of such functions with |Im(f )| = k. (If x < n, then ck = 0 for x < k ≤ n.) Pn So the answer by this indirect count is k=1 ck . n Pn Thus, x = k=1 ck .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets. Theorem.
n n X x = S(n, k)(x)k , (7) k=1
Proof. Consider functions f :[n] → X , where |X | = x ∈ N. How many such functions are there? The answer by direct count: xn. An alternative indirect count: For k = 1, 2, ··· , n, let ck denote the number of such functions with |Im(f )| = k. (If x < n, then ck = 0 for x < k ≤ n.) Pn So the answer by this indirect count is k=1 ck . n Pn Thus, x = k=1 ck . To complete the proof we now show that ck = S(n, k)(x)k .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations For f with |Im(f )| = k, let Im(f ) = {x1, ··· , xk }. Then −1 −1 −1 f ({x1}), f ({x2}), ··· , f ({xk }) is a partition of [n] into k non-empty sets, which we will denote by {B1, ··· , Bk }. Now: 1. There are S(n, k) ways to choose the sets {B1, ··· , Bk }; x 2. There are k was to choose the {x1, ··· , xk }; 3. There are k! ways to make the correspondence between −1 −1 −1 f ({x1}), f ({x2}), ··· , f ({xk }) and {B1, ··· , Bk }.
x x! Thus ck = S(n, k) × k × k! = S(n, k) (x−k)! = S(n, k)(x)k .
Consider functions f :[n] → X , where |X | = x ∈ N. Let ck denote the number of such functions with |Im(f )| = k. Proof that ck = S(n, k)(x)k :
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Then −1 −1 −1 f ({x1}), f ({x2}), ··· , f ({xk }) is a partition of [n] into k non-empty sets, which we will denote by {B1, ··· , Bk }. Now: 1. There are S(n, k) ways to choose the sets {B1, ··· , Bk }; x 2. There are k was to choose the {x1, ··· , xk }; 3. There are k! ways to make the correspondence between −1 −1 −1 f ({x1}), f ({x2}), ··· , f ({xk }) and {B1, ··· , Bk }.
x x! Thus ck = S(n, k) × k × k! = S(n, k) (x−k)! = S(n, k)(x)k .
Consider functions f :[n] → X , where |X | = x ∈ N. Let ck denote the number of such functions with |Im(f )| = k. Proof that ck = S(n, k)(x)k :
For f with |Im(f )| = k, let Im(f ) = {x1, ··· , xk }.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Now: 1. There are S(n, k) ways to choose the sets {B1, ··· , Bk }; x 2. There are k was to choose the {x1, ··· , xk }; 3. There are k! ways to make the correspondence between −1 −1 −1 f ({x1}), f ({x2}), ··· , f ({xk }) and {B1, ··· , Bk }.
x x! Thus ck = S(n, k) × k × k! = S(n, k) (x−k)! = S(n, k)(x)k .
Consider functions f :[n] → X , where |X | = x ∈ N. Let ck denote the number of such functions with |Im(f )| = k. Proof that ck = S(n, k)(x)k :
For f with |Im(f )| = k, let Im(f ) = {x1, ··· , xk }. Then −1 −1 −1 f ({x1}), f ({x2}), ··· , f ({xk }) is a partition of [n] into k non-empty sets, which we will denote by {B1, ··· , Bk }.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations x 2. There are k was to choose the {x1, ··· , xk }; 3. There are k! ways to make the correspondence between −1 −1 −1 f ({x1}), f ({x2}), ··· , f ({xk }) and {B1, ··· , Bk }.
x x! Thus ck = S(n, k) × k × k! = S(n, k) (x−k)! = S(n, k)(x)k .
Consider functions f :[n] → X , where |X | = x ∈ N. Let ck denote the number of such functions with |Im(f )| = k. Proof that ck = S(n, k)(x)k :
For f with |Im(f )| = k, let Im(f ) = {x1, ··· , xk }. Then −1 −1 −1 f ({x1}), f ({x2}), ··· , f ({xk }) is a partition of [n] into k non-empty sets, which we will denote by {B1, ··· , Bk }. Now: 1. There are S(n, k) ways to choose the sets {B1, ··· , Bk };
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations 3. There are k! ways to make the correspondence between −1 −1 −1 f ({x1}), f ({x2}), ··· , f ({xk }) and {B1, ··· , Bk }.
x x! Thus ck = S(n, k) × k × k! = S(n, k) (x−k)! = S(n, k)(x)k .
Consider functions f :[n] → X , where |X | = x ∈ N. Let ck denote the number of such functions with |Im(f )| = k. Proof that ck = S(n, k)(x)k :
For f with |Im(f )| = k, let Im(f ) = {x1, ··· , xk }. Then −1 −1 −1 f ({x1}), f ({x2}), ··· , f ({xk }) is a partition of [n] into k non-empty sets, which we will denote by {B1, ··· , Bk }. Now: 1. There are S(n, k) ways to choose the sets {B1, ··· , Bk }; x 2. There are k was to choose the {x1, ··· , xk };
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations x x! Thus ck = S(n, k) × k × k! = S(n, k) (x−k)! = S(n, k)(x)k .
Consider functions f :[n] → X , where |X | = x ∈ N. Let ck denote the number of such functions with |Im(f )| = k. Proof that ck = S(n, k)(x)k :
For f with |Im(f )| = k, let Im(f ) = {x1, ··· , xk }. Then −1 −1 −1 f ({x1}), f ({x2}), ··· , f ({xk }) is a partition of [n] into k non-empty sets, which we will denote by {B1, ··· , Bk }. Now: 1. There are S(n, k) ways to choose the sets {B1, ··· , Bk }; x 2. There are k was to choose the {x1, ··· , xk }; 3. There are k! ways to make the correspondence between −1 −1 −1 f ({x1}), f ({x2}), ··· , f ({xk }) and {B1, ··· , Bk }.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations x! = S(n, k) (x−k)! = S(n, k)(x)k .
Consider functions f :[n] → X , where |X | = x ∈ N. Let ck denote the number of such functions with |Im(f )| = k. Proof that ck = S(n, k)(x)k :
For f with |Im(f )| = k, let Im(f ) = {x1, ··· , xk }. Then −1 −1 −1 f ({x1}), f ({x2}), ··· , f ({xk }) is a partition of [n] into k non-empty sets, which we will denote by {B1, ··· , Bk }. Now: 1. There are S(n, k) ways to choose the sets {B1, ··· , Bk }; x 2. There are k was to choose the {x1, ··· , xk }; 3. There are k! ways to make the correspondence between −1 −1 −1 f ({x1}), f ({x2}), ··· , f ({xk }) and {B1, ··· , Bk }.
x Thus ck = S(n, k) × k × k!
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations = S(n, k)(x)k .
Consider functions f :[n] → X , where |X | = x ∈ N. Let ck denote the number of such functions with |Im(f )| = k. Proof that ck = S(n, k)(x)k :
For f with |Im(f )| = k, let Im(f ) = {x1, ··· , xk }. Then −1 −1 −1 f ({x1}), f ({x2}), ··· , f ({xk }) is a partition of [n] into k non-empty sets, which we will denote by {B1, ··· , Bk }. Now: 1. There are S(n, k) ways to choose the sets {B1, ··· , Bk }; x 2. There are k was to choose the {x1, ··· , xk }; 3. There are k! ways to make the correspondence between −1 −1 −1 f ({x1}), f ({x2}), ··· , f ({xk }) and {B1, ··· , Bk }.
x x! Thus ck = S(n, k) × k × k! = S(n, k) (x−k)!
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Consider functions f :[n] → X , where |X | = x ∈ N. Let ck denote the number of such functions with |Im(f )| = k. Proof that ck = S(n, k)(x)k :
For f with |Im(f )| = k, let Im(f ) = {x1, ··· , xk }. Then −1 −1 −1 f ({x1}), f ({x2}), ··· , f ({xk }) is a partition of [n] into k non-empty sets, which we will denote by {B1, ··· , Bk }. Now: 1. There are S(n, k) ways to choose the sets {B1, ··· , Bk }; x 2. There are k was to choose the {x1, ··· , xk }; 3. There are k! ways to make the correspondence between −1 −1 −1 f ({x1}), f ({x2}), ··· , f ({xk }) and {B1, ··· , Bk }.
x x! Thus ck = S(n, k) × k × k! = S(n, k) (x−k)! = S(n, k)(x)k .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations S(n, k) (Stirling number of the second kind) is the number of ways to partition [n] into k nonempty subsets |s(n, k)| (unsigned Stirling number of the first kind) is the number of permutations of [n] which have k cycles s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind) Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 by defining S(n, k) = s(n, k) = 0, for k > n. ∞ Let S = {Snk }n,k=1 denote the ∞ × ∞ matrix with entries Snk = S(n, k). ∞ Let s = {snk }n,k=1 denote the ∞ × ∞ matrix with entries snk = s(n, k). Corollary. Ss = sS = Id. ( 1, if n = m; That is, P∞ S(n, j)s(j, m) = j=1 0, if n 6= m.
n Pn x = k=1 S(n, k)(x)k , n ≥ 1 Pn k (x)n = k=1 s(n, k)x , n ≥ 1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 by defining S(n, k) = s(n, k) = 0, for k > n. ∞ Let S = {Snk }n,k=1 denote the ∞ × ∞ matrix with entries Snk = S(n, k). ∞ Let s = {snk }n,k=1 denote the ∞ × ∞ matrix with entries snk = s(n, k). Corollary. Ss = sS = Id. ( 1, if n = m; That is, P∞ S(n, j)s(j, m) = j=1 0, if n 6= m.
n Pn x = k=1 S(n, k)(x)k , n ≥ 1 Pn k (x)n = k=1 s(n, k)x , n ≥ 1 S(n, k) (Stirling number of the second kind) is the number of ways to partition [n] into k nonempty subsets |s(n, k)| (unsigned Stirling number of the first kind) is the number of permutations of [n] which have k cycles s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations ∞ Let S = {Snk }n,k=1 denote the ∞ × ∞ matrix with entries Snk = S(n, k). ∞ Let s = {snk }n,k=1 denote the ∞ × ∞ matrix with entries snk = s(n, k). Corollary. Ss = sS = Id. ( 1, if n = m; That is, P∞ S(n, j)s(j, m) = j=1 0, if n 6= m.
n Pn x = k=1 S(n, k)(x)k , n ≥ 1 Pn k (x)n = k=1 s(n, k)x , n ≥ 1 S(n, k) (Stirling number of the second kind) is the number of ways to partition [n] into k nonempty subsets |s(n, k)| (unsigned Stirling number of the first kind) is the number of permutations of [n] which have k cycles s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind) Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 by defining S(n, k) = s(n, k) = 0, for k > n.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations ∞ Let s = {snk }n,k=1 denote the ∞ × ∞ matrix with entries snk = s(n, k). Corollary. Ss = sS = Id. ( 1, if n = m; That is, P∞ S(n, j)s(j, m) = j=1 0, if n 6= m.
n Pn x = k=1 S(n, k)(x)k , n ≥ 1 Pn k (x)n = k=1 s(n, k)x , n ≥ 1 S(n, k) (Stirling number of the second kind) is the number of ways to partition [n] into k nonempty subsets |s(n, k)| (unsigned Stirling number of the first kind) is the number of permutations of [n] which have k cycles s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind) Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 by defining S(n, k) = s(n, k) = 0, for k > n. ∞ Let S = {Snk }n,k=1 denote the ∞ × ∞ matrix with entries Snk = S(n, k).
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Corollary. Ss = sS = Id. ( 1, if n = m; That is, P∞ S(n, j)s(j, m) = j=1 0, if n 6= m.
n Pn x = k=1 S(n, k)(x)k , n ≥ 1 Pn k (x)n = k=1 s(n, k)x , n ≥ 1 S(n, k) (Stirling number of the second kind) is the number of ways to partition [n] into k nonempty subsets |s(n, k)| (unsigned Stirling number of the first kind) is the number of permutations of [n] which have k cycles s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind) Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 by defining S(n, k) = s(n, k) = 0, for k > n. ∞ Let S = {Snk }n,k=1 denote the ∞ × ∞ matrix with entries Snk = S(n, k). ∞ Let s = {snk }n,k=1 denote the ∞ × ∞ matrix with entries snk = s(n, k).
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations ( 1, if n = m; That is, P∞ S(n, j)s(j, m) = j=1 0, if n 6= m.
n Pn x = k=1 S(n, k)(x)k , n ≥ 1 Pn k (x)n = k=1 s(n, k)x , n ≥ 1 S(n, k) (Stirling number of the second kind) is the number of ways to partition [n] into k nonempty subsets |s(n, k)| (unsigned Stirling number of the first kind) is the number of permutations of [n] which have k cycles s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind) Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 by defining S(n, k) = s(n, k) = 0, for k > n. ∞ Let S = {Snk }n,k=1 denote the ∞ × ∞ matrix with entries Snk = S(n, k). ∞ Let s = {snk }n,k=1 denote the ∞ × ∞ matrix with entries snk = s(n, k). Corollary. Ss = sS = Id.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations n Pn x = k=1 S(n, k)(x)k , n ≥ 1 Pn k (x)n = k=1 s(n, k)x , n ≥ 1 S(n, k) (Stirling number of the second kind) is the number of ways to partition [n] into k nonempty subsets |s(n, k)| (unsigned Stirling number of the first kind) is the number of permutations of [n] which have k cycles s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind) Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 by defining S(n, k) = s(n, k) = 0, for k > n. ∞ Let S = {Snk }n,k=1 denote the ∞ × ∞ matrix with entries Snk = S(n, k). ∞ Let s = {snk }n,k=1 denote the ∞ × ∞ matrix with entries snk = s(n, k). Corollary. Ss = sS = Id. ( 1, if n = m; That is, P∞ S(n, j)s(j, m) = j=1 0, if n 6= m.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Consider the real vector space V of polynomials of the Pn k form k=1 ck x , n ≥ 1, ck ∈ R. . Of course, 2 3 B1 := {x, x , x , ···} is a basis for V .
By (*), B2 := {(x)1, (x)2, (x)3, ···} is also a basis for V . By (*) and (**), the matrices S and s transform between the two basis, and thus Ss = sS = I .
n Pn x = k=1 S(n, k)(x)k , n ≥ 1 (*) Pn k (x)n = k=1 s(n, k)x , n ≥ 1 (**) Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 by defining S(n, k) = s(n, k) = 0, for k > n. ∞ Let S = {Snk }n,k=1 denote the ∞ × ∞ matrix with entries Snk = S(n, k). ∞ Let s = {snk }n,k=1 denote the ∞ × ∞ matrix with entries snk = s(n, k). Corollary. Ss = sS = Id. Proof.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations . Of course, 2 3 B1 := {x, x , x , ···} is a basis for V .
By (*), B2 := {(x)1, (x)2, (x)3, ···} is also a basis for V . By (*) and (**), the matrices S and s transform between the two basis, and thus Ss = sS = I .
n Pn x = k=1 S(n, k)(x)k , n ≥ 1 (*) Pn k (x)n = k=1 s(n, k)x , n ≥ 1 (**) Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 by defining S(n, k) = s(n, k) = 0, for k > n. ∞ Let S = {Snk }n,k=1 denote the ∞ × ∞ matrix with entries Snk = S(n, k). ∞ Let s = {snk }n,k=1 denote the ∞ × ∞ matrix with entries snk = s(n, k). Corollary. Ss = sS = Id. Proof. Consider the real vector space V of polynomials of the Pn k form k=1 ck x , n ≥ 1, ck ∈ R.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations By (*), B2 := {(x)1, (x)2, (x)3, ···} is also a basis for V . By (*) and (**), the matrices S and s transform between the two basis, and thus Ss = sS = I .
n Pn x = k=1 S(n, k)(x)k , n ≥ 1 (*) Pn k (x)n = k=1 s(n, k)x , n ≥ 1 (**) Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 by defining S(n, k) = s(n, k) = 0, for k > n. ∞ Let S = {Snk }n,k=1 denote the ∞ × ∞ matrix with entries Snk = S(n, k). ∞ Let s = {snk }n,k=1 denote the ∞ × ∞ matrix with entries snk = s(n, k). Corollary. Ss = sS = Id. Proof. Consider the real vector space V of polynomials of the Pn k form k=1 ck x , n ≥ 1, ck ∈ R. . Of course, 2 3 B1 := {x, x , x , ···} is a basis for V .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations By (*) and (**), the matrices S and s transform between the two basis, and thus Ss = sS = I .
n Pn x = k=1 S(n, k)(x)k , n ≥ 1 (*) Pn k (x)n = k=1 s(n, k)x , n ≥ 1 (**) Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 by defining S(n, k) = s(n, k) = 0, for k > n. ∞ Let S = {Snk }n,k=1 denote the ∞ × ∞ matrix with entries Snk = S(n, k). ∞ Let s = {snk }n,k=1 denote the ∞ × ∞ matrix with entries snk = s(n, k). Corollary. Ss = sS = Id. Proof. Consider the real vector space V of polynomials of the Pn k form k=1 ck x , n ≥ 1, ck ∈ R. . Of course, 2 3 B1 := {x, x , x , ···} is a basis for V .
By (*), B2 := {(x)1, (x)2, (x)3, ···} is also a basis for V .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations n Pn x = k=1 S(n, k)(x)k , n ≥ 1 (*) Pn k (x)n = k=1 s(n, k)x , n ≥ 1 (**) Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 by defining S(n, k) = s(n, k) = 0, for k > n. ∞ Let S = {Snk }n,k=1 denote the ∞ × ∞ matrix with entries Snk = S(n, k). ∞ Let s = {snk }n,k=1 denote the ∞ × ∞ matrix with entries snk = s(n, k). Corollary. Ss = sS = Id. Proof. Consider the real vector space V of polynomials of the Pn k form k=1 ck x , n ≥ 1, ck ∈ R. . Of course, 2 3 B1 := {x, x , x , ···} is a basis for V .
By (*), B2 := {(x)1, (x)2, (x)3, ···} is also a basis for V . By (*) and (**), the matrices S and s transform between the two basis, and thus Ss = sS = I .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Example: n = 8, m = 4: P∞ 0 = j=1 S(8, j)s(j, 4) = S(8, 4)−S(8, 5)|s(5, 4)|+S(8, 6)|s(6, 4)|−S(8, 7)|s(7, 4)|+|s(8, 4)|.
S(n, k) (Stirling number of the second kind) is the number of ways to partition [n] into k nonempty subsets |s(n, k)| (unsigned Stirling number of the first kind) is the number of permutations of [n] which have k cycles s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind) Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 by defining S(n, k) = s(n, k) = 0, for k > n. ∞ Let S = {Snk }n,k=1 denote the ∞ × ∞ matrix with entries Snk = S(n, k). ∞ Let s = {snk }n,k=1 denote the ∞ × ∞ matrix with entries snk = s(n, k). Corollary. Ss = sS = Id ( 1, if n = m; That is, P∞ S(n, j)s(j, m) = j=1 0, if n 6= m.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations S(n, k) (Stirling number of the second kind) is the number of ways to partition [n] into k nonempty subsets |s(n, k)| (unsigned Stirling number of the first kind) is the number of permutations of [n] which have k cycles s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind) Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 by defining S(n, k) = s(n, k) = 0, for k > n. ∞ Let S = {Snk }n,k=1 denote the ∞ × ∞ matrix with entries Snk = S(n, k). ∞ Let s = {snk }n,k=1 denote the ∞ × ∞ matrix with entries snk = s(n, k). Corollary. Ss = sS = Id ( 1, if n = m; That is, P∞ S(n, j)s(j, m) = j=1 0, if n 6= m. Example: n = 8, m = 4: P∞ 0 = j=1 S(8, j)s(j, 4) = S(8, 4)−S(8, 5)|s(5, 4)|+S(8, 6)|s(6, 4)|−S(8, 7)|s(7, 4)|+|s(8, 4)|.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations n nth moment: µn;λ := EX
Recalling the Poisson distribution Let X be a random variable with the distribution Poiss(λ), λ > 0:
−λ λj P(X = j) = e j! , j = 0, 1, 2, ··· .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Recalling the Poisson distribution Let X be a random variable with the distribution Poiss(λ), λ > 0:
−λ λj P(X = j) = e j! , j = 0, 1, 2, ··· . n nth moment: µn;λ := EX
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations =
∞ min(j,n) X λj X = e−λ S(n, k)(j) because (j) = 0, k > j) j! k k j=0 k=1 n ∞ X X λj = e−λ S(n, k) (j) . j! k k=1 j=k
P∞ λj P∞ λj k P∞ λj−k k P∞ λl But j=k j! (j)k = j=k (j−k)! = λ j=k (j−k)! = λ l=0 l! = λk eλ. Thus, n X k µn;λ = S(n, k)λ . k=1
Let X be a random variable with the distribution Poiss(λ), λ > 0:
−λ λj P(X = j) = e j! , j = 1, 2, ··· . n nth moment: µn;λ := EX
∞ j ∞ j n X λ (7) X λ X µ = e−λ jn = e−λ S(n, k)(j) n;λ j! j! k j=0 j=0 k=1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations because (j)k = 0, k > j)
n ∞ X X λj = e−λ S(n, k) (j) . j! k k=1 j=k
P∞ λj P∞ λj k P∞ λj−k k P∞ λl But j=k j! (j)k = j=k (j−k)! = λ j=k (j−k)! = λ l=0 l! = λk eλ. Thus, n X k µn;λ = S(n, k)λ . k=1
Let X be a random variable with the distribution Poiss(λ), λ > 0:
−λ λj P(X = j) = e j! , j = 1, 2, ··· . n nth moment: µn;λ := EX
∞ j ∞ j n X λ (7) X λ X µ = e−λ jn = e−λ S(n, k)(j) = n;λ j! j! k j=0 j=0 k=1 ∞ min(j,n) X λj X = e−λ S(n, k)(j) j! k j=0 k=1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations n ∞ X X λj = e−λ S(n, k) (j) . j! k k=1 j=k
P∞ λj P∞ λj k P∞ λj−k k P∞ λl But j=k j! (j)k = j=k (j−k)! = λ j=k (j−k)! = λ l=0 l! = λk eλ. Thus, n X k µn;λ = S(n, k)λ . k=1
Let X be a random variable with the distribution Poiss(λ), λ > 0:
−λ λj P(X = j) = e j! , j = 1, 2, ··· . n nth moment: µn;λ := EX
∞ j ∞ j n X λ (7) X λ X µ = e−λ jn = e−λ S(n, k)(j) = n;λ j! j! k j=0 j=0 k=1 ∞ min(j,n) X λj X = e−λ S(n, k)(j) because (j) = 0, k > j) j! k k j=0 k=1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations P∞ λj P∞ λj k P∞ λj−k k P∞ λl But j=k j! (j)k = j=k (j−k)! = λ j=k (j−k)! = λ l=0 l! = λk eλ. Thus, n X k µn;λ = S(n, k)λ . k=1
Let X be a random variable with the distribution Poiss(λ), λ > 0:
−λ λj P(X = j) = e j! , j = 1, 2, ··· . n nth moment: µn;λ := EX
∞ j ∞ j n X λ (7) X λ X µ = e−λ jn = e−λ S(n, k)(j) = n;λ j! j! k j=0 j=0 k=1 ∞ min(j,n) X λj X = e−λ S(n, k)(j) because (j) = 0, k > j) j! k k j=0 k=1 n ∞ X X λj = e−λ S(n, k) (j) . j! k k=1 j=k
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations k P∞ λj−k k P∞ λl = λ j=k (j−k)! = λ l=0 l! = λk eλ. Thus, n X k µn;λ = S(n, k)λ . k=1
Let X be a random variable with the distribution Poiss(λ), λ > 0:
−λ λj P(X = j) = e j! , j = 1, 2, ··· . n nth moment: µn;λ := EX
∞ j ∞ j n X λ (7) X λ X µ = e−λ jn = e−λ S(n, k)(j) = n;λ j! j! k j=0 j=0 k=1 ∞ min(j,n) X λj X = e−λ S(n, k)(j) because (j) = 0, k > j) j! k k j=0 k=1 n ∞ X X λj = e−λ S(n, k) (j) . j! k k=1 j=k
P∞ λj P∞ λj But j=k j! (j)k = j=k (j−k)!
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations k P∞ λl = λ l=0 l! = λk eλ. Thus, n X k µn;λ = S(n, k)λ . k=1
Let X be a random variable with the distribution Poiss(λ), λ > 0:
−λ λj P(X = j) = e j! , j = 1, 2, ··· . n nth moment: µn;λ := EX
∞ j ∞ j n X λ (7) X λ X µ = e−λ jn = e−λ S(n, k)(j) = n;λ j! j! k j=0 j=0 k=1 ∞ min(j,n) X λj X = e−λ S(n, k)(j) because (j) = 0, k > j) j! k k j=0 k=1 n ∞ X X λj = e−λ S(n, k) (j) . j! k k=1 j=k
P∞ λj P∞ λj k P∞ λj−k But j=k j! (j)k = j=k (j−k)! = λ j=k (j−k)!
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations = λk eλ. Thus, n X k µn;λ = S(n, k)λ . k=1
Let X be a random variable with the distribution Poiss(λ), λ > 0:
−λ λj P(X = j) = e j! , j = 1, 2, ··· . n nth moment: µn;λ := EX
∞ j ∞ j n X λ (7) X λ X µ = e−λ jn = e−λ S(n, k)(j) = n;λ j! j! k j=0 j=0 k=1 ∞ min(j,n) X λj X = e−λ S(n, k)(j) because (j) = 0, k > j) j! k k j=0 k=1 n ∞ X X λj = e−λ S(n, k) (j) . j! k k=1 j=k
P∞ λj P∞ λj k P∞ λj−k k P∞ λl But j=k j! (j)k = j=k (j−k)! = λ j=k (j−k)! = λ l=0 l!
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Thus, n X k µn;λ = S(n, k)λ . k=1
Let X be a random variable with the distribution Poiss(λ), λ > 0:
−λ λj P(X = j) = e j! , j = 1, 2, ··· . n nth moment: µn;λ := EX
∞ j ∞ j n X λ (7) X λ X µ = e−λ jn = e−λ S(n, k)(j) = n;λ j! j! k j=0 j=0 k=1 ∞ min(j,n) X λj X = e−λ S(n, k)(j) because (j) = 0, k > j) j! k k j=0 k=1 n ∞ X X λj = e−λ S(n, k) (j) . j! k k=1 j=k
P∞ λj P∞ λj k P∞ λj−k k P∞ λl But j=k j! (j)k = j=k (j−k)! = λ j=k (j−k)! = λ l=0 l! = λk eλ.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Let X be a random variable with the distribution Poiss(λ), λ > 0:
−λ λj P(X = j) = e j! , j = 1, 2, ··· . n nth moment: µn;λ := EX
∞ j ∞ j n X λ (7) X λ X µ = e−λ jn = e−λ S(n, k)(j) = n;λ j! j! k j=0 j=0 k=1 ∞ min(j,n) X λj X = e−λ S(n, k)(j) because (j) = 0, k > j) j! k k j=0 k=1 n ∞ X X λj = e−λ S(n, k) (j) . j! k k=1 j=k
P∞ λj P∞ λj k P∞ λj−k k P∞ λl But j=k j! (j)k = j=k (j−k)! = λ j=k (j−k)! = λ l=0 l! = λk eλ. Thus, n X k µn;λ = S(n, k)λ . k=1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Pn k Touchard Polynomial: Tn(x) := k=1 S(n, k)x
µn;λ = Tn(λ) (Compare Tn with (6).) ∞ n X λj X e−λ jn = EX n = µ = T (λ) := S(n, k)λk j! n;λ n j=0 k=1 When λ = 1: ∞ n 1 X jn X = EX n = µ = T (1) := S(n, k) := B , e j! n;1 n n j=0 k=1
where Bn is the nth Bell number: Bn = the total number of partitions of [n].
Let X be a random variable with the distribution Poiss(λ), λ > 0: −λ λj P(X = j) = e j! , j = 1, 2, ··· . n P∞ −λ λj n nth moment: µn;λ := EX = j=0 e j! j Pn k µn;λ = k=1 S(n, k)λ
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations µn;λ = Tn(λ) (Compare Tn with (6).) ∞ n X λj X e−λ jn = EX n = µ = T (λ) := S(n, k)λk j! n;λ n j=0 k=1 When λ = 1: ∞ n 1 X jn X = EX n = µ = T (1) := S(n, k) := B , e j! n;1 n n j=0 k=1
where Bn is the nth Bell number: Bn = the total number of partitions of [n].
Let X be a random variable with the distribution Poiss(λ), λ > 0: −λ λj P(X = j) = e j! , j = 1, 2, ··· . n P∞ −λ λj n nth moment: µn;λ := EX = j=0 e j! j Pn k µn;λ = k=1 S(n, k)λ Pn k Touchard Polynomial: Tn(x) := k=1 S(n, k)x
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations (Compare Tn with (6).) ∞ n X λj X e−λ jn = EX n = µ = T (λ) := S(n, k)λk j! n;λ n j=0 k=1 When λ = 1: ∞ n 1 X jn X = EX n = µ = T (1) := S(n, k) := B , e j! n;1 n n j=0 k=1
where Bn is the nth Bell number: Bn = the total number of partitions of [n].
Let X be a random variable with the distribution Poiss(λ), λ > 0: −λ λj P(X = j) = e j! , j = 1, 2, ··· . n P∞ −λ λj n nth moment: µn;λ := EX = j=0 e j! j Pn k µn;λ = k=1 S(n, k)λ Pn k Touchard Polynomial: Tn(x) := k=1 S(n, k)x
µn;λ = Tn(λ)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations ∞ n X λj X e−λ jn = EX n = µ = T (λ) := S(n, k)λk j! n;λ n j=0 k=1 When λ = 1: ∞ n 1 X jn X = EX n = µ = T (1) := S(n, k) := B , e j! n;1 n n j=0 k=1
where Bn is the nth Bell number: Bn = the total number of partitions of [n].
Let X be a random variable with the distribution Poiss(λ), λ > 0: −λ λj P(X = j) = e j! , j = 1, 2, ··· . n P∞ −λ λj n nth moment: µn;λ := EX = j=0 e j! j Pn k µn;λ = k=1 S(n, k)λ Pn k Touchard Polynomial: Tn(x) := k=1 S(n, k)x
µn;λ = Tn(λ) (Compare Tn with (6).)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations When λ = 1: ∞ n 1 X jn X = EX n = µ = T (1) := S(n, k) := B , e j! n;1 n n j=0 k=1
where Bn is the nth Bell number: Bn = the total number of partitions of [n].
Let X be a random variable with the distribution Poiss(λ), λ > 0: −λ λj P(X = j) = e j! , j = 1, 2, ··· . n P∞ −λ λj n nth moment: µn;λ := EX = j=0 e j! j Pn k µn;λ = k=1 S(n, k)λ Pn k Touchard Polynomial: Tn(x) := k=1 S(n, k)x
µn;λ = Tn(λ) (Compare Tn with (6).) ∞ n X λj X e−λ jn = EX n = µ = T (λ) := S(n, k)λk j! n;λ n j=0 k=1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations := Bn,
where Bn is the nth Bell number: Bn = the total number of partitions of [n].
Let X be a random variable with the distribution Poiss(λ), λ > 0: −λ λj P(X = j) = e j! , j = 1, 2, ··· . n P∞ −λ λj n nth moment: µn;λ := EX = j=0 e j! j Pn k µn;λ = k=1 S(n, k)λ Pn k Touchard Polynomial: Tn(x) := k=1 S(n, k)x
µn;λ = Tn(λ) (Compare Tn with (6).) ∞ n X λj X e−λ jn = EX n = µ = T (λ) := S(n, k)λk j! n;λ n j=0 k=1 When λ = 1: ∞ n 1 X jn X = EX n = µ = T (1) := S(n, k) e j! n;1 n j=0 k=1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Let X be a random variable with the distribution Poiss(λ), λ > 0: −λ λj P(X = j) = e j! , j = 1, 2, ··· . n P∞ −λ λj n nth moment: µn;λ := EX = j=0 e j! j Pn k µn;λ = k=1 S(n, k)λ Pn k Touchard Polynomial: Tn(x) := k=1 S(n, k)x
µn;λ = Tn(λ) (Compare Tn with (6).) ∞ n X λj X e−λ jn = EX n = µ = T (λ) := S(n, k)λk j! n;λ n j=0 k=1 When λ = 1: ∞ n 1 X jn X = EX n = µ = T (1) := S(n, k) := B , e j! n;1 n n j=0 k=1 where Bn is the nth Bell number: Bn = the total number of partitions of [n]. Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Let X be a random variable with the distribution Poiss(λ), λ > 0: −λ λj P(X = j) = e j! , j = 1, 2, ··· . n nth moment: µn;λ := EX Pn k µn;λ = k=1 S(n, k)λ Pn k Touchard Polynomial: Tn(x) := k=1 S(n, k)x
µn;λ = Tn(λ)
∞ n X λj X e−λ jn = EX n = µ = T (λ) := S(n, k)λk j! n;λ n j=0 k=1
When λ = 1: ∞ n 1 X jn X = EX n = µ = T (1) := S(n, k) := B , (8) e j! n;λ n n j=0 k=1 where Bn is the nth Bell number: Bn = the total number of partitions of [n]. (8) is known as Dobi´nski’sformula.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations 1 Uniform probability measure on Sn: Pn(σ) := n! , σ ∈ Sn
We now define a family of non-uniform probability measures on Sn:
Let kn(σ) = the number of cycles in σ.
For θ > 0, define the following probability measure on Sn: kn(σ) (θ) θ P kn(σ) Pn (σ) = , where Nn,θ = θ is the normalizing Nn,θ σ∈Sn constant. (3) We have N = P θkn(σ) = Pn θk |s(n, k)| = θ(n). n,θ σ∈Sn k=1 k (σ) Thus, P(θ)(σ) = θ n , σ ∈ S . n θ(n) n θ > 1: the measure favors permutations with many cycles θ ∈ (0, 1): the measure favors permutations with few cycles θ = 1: uniform measure
Random Permutations
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations We now define a family of non-uniform probability measures on Sn:
Let kn(σ) = the number of cycles in σ.
For θ > 0, define the following probability measure on Sn: kn(σ) (θ) θ P kn(σ) Pn (σ) = , where Nn,θ = θ is the normalizing Nn,θ σ∈Sn constant. (3) We have N = P θkn(σ) = Pn θk |s(n, k)| = θ(n). n,θ σ∈Sn k=1 k (σ) Thus, P(θ)(σ) = θ n , σ ∈ S . n θ(n) n θ > 1: the measure favors permutations with many cycles θ ∈ (0, 1): the measure favors permutations with few cycles θ = 1: uniform measure
Random Permutations 1 Uniform probability measure on Sn: Pn(σ) := n! , σ ∈ Sn
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Let kn(σ) = the number of cycles in σ.
For θ > 0, define the following probability measure on Sn: kn(σ) (θ) θ P kn(σ) Pn (σ) = , where Nn,θ = θ is the normalizing Nn,θ σ∈Sn constant. (3) We have N = P θkn(σ) = Pn θk |s(n, k)| = θ(n). n,θ σ∈Sn k=1 k (σ) Thus, P(θ)(σ) = θ n , σ ∈ S . n θ(n) n θ > 1: the measure favors permutations with many cycles θ ∈ (0, 1): the measure favors permutations with few cycles θ = 1: uniform measure
Random Permutations 1 Uniform probability measure on Sn: Pn(σ) := n! , σ ∈ Sn
We now define a family of non-uniform probability measures on Sn:
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations For θ > 0, define the following probability measure on Sn: kn(σ) (θ) θ P kn(σ) Pn (σ) = , where Nn,θ = θ is the normalizing Nn,θ σ∈Sn constant. (3) We have N = P θkn(σ) = Pn θk |s(n, k)| = θ(n). n,θ σ∈Sn k=1 k (σ) Thus, P(θ)(σ) = θ n , σ ∈ S . n θ(n) n θ > 1: the measure favors permutations with many cycles θ ∈ (0, 1): the measure favors permutations with few cycles θ = 1: uniform measure
Random Permutations 1 Uniform probability measure on Sn: Pn(σ) := n! , σ ∈ Sn
We now define a family of non-uniform probability measures on Sn:
Let kn(σ) = the number of cycles in σ.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations (3) We have N = P θkn(σ) = Pn θk |s(n, k)| = θ(n). n,θ σ∈Sn k=1 k (σ) Thus, P(θ)(σ) = θ n , σ ∈ S . n θ(n) n θ > 1: the measure favors permutations with many cycles θ ∈ (0, 1): the measure favors permutations with few cycles θ = 1: uniform measure
Random Permutations 1 Uniform probability measure on Sn: Pn(σ) := n! , σ ∈ Sn
We now define a family of non-uniform probability measures on Sn:
Let kn(σ) = the number of cycles in σ.
For θ > 0, define the following probability measure on Sn: kn(σ) (θ) θ P kn(σ) Pn (σ) = , where Nn,θ = θ is the normalizing Nn,θ σ∈Sn constant.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations (3) = θ(n).
k (σ) Thus, P(θ)(σ) = θ n , σ ∈ S . n θ(n) n θ > 1: the measure favors permutations with many cycles θ ∈ (0, 1): the measure favors permutations with few cycles θ = 1: uniform measure
Random Permutations 1 Uniform probability measure on Sn: Pn(σ) := n! , σ ∈ Sn
We now define a family of non-uniform probability measures on Sn:
Let kn(σ) = the number of cycles in σ.
For θ > 0, define the following probability measure on Sn: kn(σ) (θ) θ P kn(σ) Pn (σ) = , where Nn,θ = θ is the normalizing Nn,θ σ∈Sn constant.
We have N = P θkn(σ) = Pn θk |s(n, k)| n,θ σ∈Sn k=1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations k (σ) Thus, P(θ)(σ) = θ n , σ ∈ S . n θ(n) n θ > 1: the measure favors permutations with many cycles θ ∈ (0, 1): the measure favors permutations with few cycles θ = 1: uniform measure
Random Permutations 1 Uniform probability measure on Sn: Pn(σ) := n! , σ ∈ Sn
We now define a family of non-uniform probability measures on Sn:
Let kn(σ) = the number of cycles in σ.
For θ > 0, define the following probability measure on Sn: kn(σ) (θ) θ P kn(σ) Pn (σ) = , where Nn,θ = θ is the normalizing Nn,θ σ∈Sn constant. (3) We have N = P θkn(σ) = Pn θk |s(n, k)| = θ(n). n,θ σ∈Sn k=1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations θ > 1: the measure favors permutations with many cycles θ ∈ (0, 1): the measure favors permutations with few cycles θ = 1: uniform measure
Random Permutations 1 Uniform probability measure on Sn: Pn(σ) := n! , σ ∈ Sn
We now define a family of non-uniform probability measures on Sn:
Let kn(σ) = the number of cycles in σ.
For θ > 0, define the following probability measure on Sn: kn(σ) (θ) θ P kn(σ) Pn (σ) = , where Nn,θ = θ is the normalizing Nn,θ σ∈Sn constant. (3) We have N = P θkn(σ) = Pn θk |s(n, k)| = θ(n). n,θ σ∈Sn k=1 k (σ) Thus, P(θ)(σ) = θ n , σ ∈ S . n θ(n) n
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Random Permutations 1 Uniform probability measure on Sn: Pn(σ) := n! , σ ∈ Sn
We now define a family of non-uniform probability measures on Sn:
Let kn(σ) = the number of cycles in σ.
For θ > 0, define the following probability measure on Sn: kn(σ) (θ) θ P kn(σ) Pn (σ) = , where Nn,θ = θ is the normalizing Nn,θ σ∈Sn constant. (3) We have N = P θkn(σ) = Pn θk |s(n, k)| = θ(n). n,θ σ∈Sn k=1 k (σ) Thus, P(θ)(σ) = θ n , σ ∈ S . n θ(n) n θ > 1: the measure favors permutations with many cycles θ ∈ (0, 1): the measure favors permutations with few cycles θ = 1: uniform measure
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations (n) Let Cj (σ) = the number of j-cycles in σ ∈ Sn, j = 1, ··· , n. (n) n Then for each θ > 0, we can think of {Cj }j=1 as random (θ) variables on the probability space (Sn, Pn ). (n) (θ) Theorem. For any j, the distribution of Cj under Pn converges θ weakly to the distribution Poiss( j ) as n → ∞. That is, θ m (θ) (n) θ ( ) − j j limn→∞ Pn (Cj = m) = e m! , for m = 0, 1, ··· . (n) Remark. C1 (σ) is the number of fixed points of σ: (n) (θ) C1 = |{k ∈ [n]: σk = k}|. So under Pn , the distribution of the number of fixed points in a permutation in Sn converges as n → ∞ to the Poisson distribution with parameter θ. In particular, under the uniform measure (θ = 1), it converges to the Poisson distribution with parameter 1.
k (σ) P(θ)(σ) = θ n , σ ∈ S . n θ(n) n θ > 1: the measure favors permutations with many cycles θ ∈ (0, 1): the measure favors permutations with few cycles θ = 1: uniform measure
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations (n) n Then for each θ > 0, we can think of {Cj }j=1 as random (θ) variables on the probability space (Sn, Pn ). (n) (θ) Theorem. For any j, the distribution of Cj under Pn converges θ weakly to the distribution Poiss( j ) as n → ∞. That is, θ m (θ) (n) θ ( ) − j j limn→∞ Pn (Cj = m) = e m! , for m = 0, 1, ··· . (n) Remark. C1 (σ) is the number of fixed points of σ: (n) (θ) C1 = |{k ∈ [n]: σk = k}|. So under Pn , the distribution of the number of fixed points in a permutation in Sn converges as n → ∞ to the Poisson distribution with parameter θ. In particular, under the uniform measure (θ = 1), it converges to the Poisson distribution with parameter 1.
k (σ) P(θ)(σ) = θ n , σ ∈ S . n θ(n) n θ > 1: the measure favors permutations with many cycles θ ∈ (0, 1): the measure favors permutations with few cycles θ = 1: uniform measure (n) Let Cj (σ) = the number of j-cycles in σ ∈ Sn, j = 1, ··· , n.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations (n) (θ) Theorem. For any j, the distribution of Cj under Pn converges θ weakly to the distribution Poiss( j ) as n → ∞. That is, θ m (θ) (n) θ ( ) − j j limn→∞ Pn (Cj = m) = e m! , for m = 0, 1, ··· . (n) Remark. C1 (σ) is the number of fixed points of σ: (n) (θ) C1 = |{k ∈ [n]: σk = k}|. So under Pn , the distribution of the number of fixed points in a permutation in Sn converges as n → ∞ to the Poisson distribution with parameter θ. In particular, under the uniform measure (θ = 1), it converges to the Poisson distribution with parameter 1.
k (σ) P(θ)(σ) = θ n , σ ∈ S . n θ(n) n θ > 1: the measure favors permutations with many cycles θ ∈ (0, 1): the measure favors permutations with few cycles θ = 1: uniform measure (n) Let Cj (σ) = the number of j-cycles in σ ∈ Sn, j = 1, ··· , n. (n) n Then for each θ > 0, we can think of {Cj }j=1 as random (θ) variables on the probability space (Sn, Pn ).
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations That is, θ m (θ) (n) θ ( ) − j j limn→∞ Pn (Cj = m) = e m! , for m = 0, 1, ··· . (n) Remark. C1 (σ) is the number of fixed points of σ: (n) (θ) C1 = |{k ∈ [n]: σk = k}|. So under Pn , the distribution of the number of fixed points in a permutation in Sn converges as n → ∞ to the Poisson distribution with parameter θ. In particular, under the uniform measure (θ = 1), it converges to the Poisson distribution with parameter 1.
k (σ) P(θ)(σ) = θ n , σ ∈ S . n θ(n) n θ > 1: the measure favors permutations with many cycles θ ∈ (0, 1): the measure favors permutations with few cycles θ = 1: uniform measure (n) Let Cj (σ) = the number of j-cycles in σ ∈ Sn, j = 1, ··· , n. (n) n Then for each θ > 0, we can think of {Cj }j=1 as random (θ) variables on the probability space (Sn, Pn ). (n) (θ) Theorem. For any j, the distribution of Cj under Pn converges θ weakly to the distribution Poiss( j ) as n → ∞.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations (n) Remark. C1 (σ) is the number of fixed points of σ: (n) (θ) C1 = |{k ∈ [n]: σk = k}|. So under Pn , the distribution of the number of fixed points in a permutation in Sn converges as n → ∞ to the Poisson distribution with parameter θ. In particular, under the uniform measure (θ = 1), it converges to the Poisson distribution with parameter 1.
k (σ) P(θ)(σ) = θ n , σ ∈ S . n θ(n) n θ > 1: the measure favors permutations with many cycles θ ∈ (0, 1): the measure favors permutations with few cycles θ = 1: uniform measure (n) Let Cj (σ) = the number of j-cycles in σ ∈ Sn, j = 1, ··· , n. (n) n Then for each θ > 0, we can think of {Cj }j=1 as random (θ) variables on the probability space (Sn, Pn ). (n) (θ) Theorem. For any j, the distribution of Cj under Pn converges θ weakly to the distribution Poiss( j ) as n → ∞. That is, θ m (θ) (n) θ ( ) − j j limn→∞ Pn (Cj = m) = e m! , for m = 0, 1, ··· .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations k (σ) P(θ)(σ) = θ n , σ ∈ S . n θ(n) n θ > 1: the measure favors permutations with many cycles θ ∈ (0, 1): the measure favors permutations with few cycles θ = 1: uniform measure (n) Let Cj (σ) = the number of j-cycles in σ ∈ Sn, j = 1, ··· , n. (n) n Then for each θ > 0, we can think of {Cj }j=1 as random (θ) variables on the probability space (Sn, Pn ). (n) (θ) Theorem. For any j, the distribution of Cj under Pn converges θ weakly to the distribution Poiss( j ) as n → ∞. That is, θ m (θ) (n) θ ( ) − j j limn→∞ Pn (Cj = m) = e m! , for m = 0, 1, ··· . (n) Remark. C1 (σ) is the number of fixed points of σ: (n) (θ) C1 = |{k ∈ [n]: σk = k}|. So under Pn , the distribution of the number of fixed points in a permutation in Sn converges as n → ∞ to the Poisson distribution with parameter θ. In particular, under the uniform measure (θ = 1), it converges to the Poisson distribution with parameter 1.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Method of moments: It is enough to show that for all m, the (n) mth moment of Cj converges to the mth moment of the 1 distribution Poiss( j ). That is, it is enough to show that
(1) (n)m 1 lim En Cj = µm; 1 = Tm . n→∞ j j
(n) (θ) Theorem. For any j, the distribution of Cj under Pn converges θ weakly to the distribution Poiss( j ) as n → ∞. That is, θ θ m (θ) (n) ( j ) − m limn→∞ Pn (Cj = m) = e m! , for m = 0, 1, ··· . Proof for θ = 1 (the same method works with virtually no extra work for all θ).
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations That is, it is enough to show that
(1) (n)m 1 lim En Cj = µm; 1 = Tm . n→∞ j j
(n) (θ) Theorem. For any j, the distribution of Cj under Pn converges θ weakly to the distribution Poiss( j ) as n → ∞. That is, θ θ m (θ) (n) ( j ) − m limn→∞ Pn (Cj = m) = e m! , for m = 0, 1, ··· . Proof for θ = 1 (the same method works with virtually no extra work for all θ). Method of moments: It is enough to show that for all m, the (n) mth moment of Cj converges to the mth moment of the 1 distribution Poiss( j ).
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations (n) (θ) Theorem. For any j, the distribution of Cj under Pn converges θ weakly to the distribution Poiss( j ) as n → ∞. That is, θ θ m (θ) (n) ( j ) − m limn→∞ Pn (Cj = m) = e m! , for m = 0, 1, ··· . Proof for θ = 1 (the same method works with virtually no extra work for all θ). Method of moments: It is enough to show that for all m, the (n) mth moment of Cj converges to the mth moment of the 1 distribution Poiss( j ). That is, it is enough to show that
(1) (n)m 1 lim En Cj = µm; 1 = Tm . n→∞ j j
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations We will show that (1) (n)m 1 En Cj = µm; 1 = Tm , for n ≥ mj. j j For D ⊂ [n] with |D| = j, define ( 1, if σ has a cycle consisting of the elements of D; 1D (σ) = 0, otherwise. (n) P Then Cj (σ) = D⊂[n]:|D|=j 1D (σ) and (n) m X X Cj (σ) = 1D1 (σ) ··· 1Dm (σ) = D1⊂[n]:|D1|=j Dm⊂[n]:|Dm|=j m X Y 1Dl (σ). l=1 D1,D2,··· ,Dm :|Dl |=···=|Dm|=j So (1) (n)m P (1) Qm En Cj = Pn l=1 1Dl = 1 . D1,D2,··· ,Dm :|Dl |=···=|Dm|=j
(1) (n)m 1 Proof that limn→∞ En C = µ 1 = Tm . j m; j j
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations For D ⊂ [n] with |D| = j, define ( 1, if σ has a cycle consisting of the elements of D; 1D (σ) = 0, otherwise. (n) P Then Cj (σ) = D⊂[n]:|D|=j 1D (σ) and (n) m X X Cj (σ) = 1D1 (σ) ··· 1Dm (σ) = D1⊂[n]:|D1|=j Dm⊂[n]:|Dm|=j m X Y 1Dl (σ). l=1 D1,D2,··· ,Dm :|Dl |=···=|Dm|=j So (1) (n)m P (1) Qm En Cj = Pn l=1 1Dl = 1 . D1,D2,··· ,Dm :|Dl |=···=|Dm|=j
(1) (n)m 1 Proof that limn→∞ En C = µ 1 = Tm . j m; j j We will show that (1) (n)m 1 En Cj = µm; 1 = Tm , for n ≥ mj. j j
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations (n) P Then Cj (σ) = D⊂[n]:|D|=j 1D (σ) and (n) m X X Cj (σ) = 1D1 (σ) ··· 1Dm (σ) = D1⊂[n]:|D1|=j Dm⊂[n]:|Dm|=j m X Y 1Dl (σ). l=1 D1,D2,··· ,Dm :|Dl |=···=|Dm|=j So (1) (n)m P (1) Qm En Cj = Pn l=1 1Dl = 1 . D1,D2,··· ,Dm :|Dl |=···=|Dm|=j
(1) (n)m 1 Proof that limn→∞ En C = µ 1 = Tm . j m; j j We will show that (1) (n)m 1 En Cj = µm; 1 = Tm , for n ≥ mj. j j For D ⊂ [n] with |D| = j, define ( 1, if σ has a cycle consisting of the elements of D; 1D (σ) = 0, otherwise.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations and
(n) m X X Cj (σ) = 1D1 (σ) ··· 1Dm (σ) = D1⊂[n]:|D1|=j Dm⊂[n]:|Dm|=j m X Y 1Dl (σ). l=1 D1,D2,··· ,Dm :|Dl |=···=|Dm|=j So (1) (n)m P (1) Qm En Cj = Pn l=1 1Dl = 1 . D1,D2,··· ,Dm :|Dl |=···=|Dm|=j
(1) (n)m 1 Proof that limn→∞ En C = µ 1 = Tm . j m; j j We will show that (1) (n)m 1 En Cj = µm; 1 = Tm , for n ≥ mj. j j For D ⊂ [n] with |D| = j, define ( 1, if σ has a cycle consisting of the elements of D; 1D (σ) = 0, otherwise. (n) P Then Cj (σ) = D⊂[n]:|D|=j 1D (σ)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations So (1) (n)m P (1) Qm En Cj = Pn l=1 1Dl = 1 . D1,D2,··· ,Dm :|Dl |=···=|Dm|=j
(1) (n)m 1 Proof that limn→∞ En C = µ 1 = Tm . j m; j j We will show that (1) (n)m 1 En Cj = µm; 1 = Tm , for n ≥ mj. j j For D ⊂ [n] with |D| = j, define ( 1, if σ has a cycle consisting of the elements of D; 1D (σ) = 0, otherwise. (n) P Then Cj (σ) = D⊂[n]:|D|=j 1D (σ) and (n) m X X Cj (σ) = 1D1 (σ) ··· 1Dm (σ) = D1⊂[n]:|D1|=j Dm⊂[n]:|Dm|=j m X Y 1Dl (σ). l=1 D1,D2,··· ,Dm :|Dl |=···=|Dm|=j
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations (1) (n)m 1 Proof that limn→∞ En C = µ 1 = Tm . j m; j j We will show that (1) (n)m 1 En Cj = µm; 1 = Tm , for n ≥ mj. j j For D ⊂ [n] with |D| = j, define ( 1, if σ has a cycle consisting of the elements of D; 1D (σ) = 0, otherwise. (n) P Then Cj (σ) = D⊂[n]:|D|=j 1D (σ) and (n) m X X Cj (σ) = 1D1 (σ) ··· 1Dm (σ) = D1⊂[n]:|D1|=j Dm⊂[n]:|Dm|=j m X Y 1Dl (σ). l=1 D1,D2,··· ,Dm :|Dl |=···=|Dm|=j So (1) (n)m P (1) Qm En Cj = Pn l=1 1Dl = 1 . D1,D2,··· ,Dm :|Dl |=···=|Dm|=j
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Qm Now l=1 1Dl 6≡ 0, (that is, there exists some σ ∈ Dn such that Qm l=1 1Dl (σ) 6= 0),
if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 or
Di1 ∩ Di2 = ∅. That is, if and only if there exist a k ∈ [m] and disjoint sets k m k {Ai }i=1 such that {Dl }l=1 = {Ai }i=1. Qm Qk In this case, l=1 1Dl = i=1 1Ai , and k (1) Qm (1) Qk (j−1)! (n−kj)! Pn l=1 1Dl = 1 = Pn i=1 1Ai = 1 = n! . (Here we use the assumption that n ≥ mj, which insures that n ≥ kj.)
m (1) (n)m X (1) Y En Cj = Pn 1Dl = 1 . l=1 D1,D2,··· ,Dm :|Dl |=···=|Dm|=j
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations (that is, there exists some σ ∈ Dn such that Qm l=1 1Dl (σ) 6= 0),
if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 or
Di1 ∩ Di2 = ∅. That is, if and only if there exist a k ∈ [m] and disjoint sets k m k {Ai }i=1 such that {Dl }l=1 = {Ai }i=1. Qm Qk In this case, l=1 1Dl = i=1 1Ai , and k (1) Qm (1) Qk (j−1)! (n−kj)! Pn l=1 1Dl = 1 = Pn i=1 1Ai = 1 = n! . (Here we use the assumption that n ≥ mj, which insures that n ≥ kj.)
m (1) (n)m X (1) Y En Cj = Pn 1Dl = 1 . l=1 D1,D2,··· ,Dm :|Dl |=···=|Dm|=j
Qm Now l=1 1Dl 6≡ 0,
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 or
Di1 ∩ Di2 = ∅. That is, if and only if there exist a k ∈ [m] and disjoint sets k m k {Ai }i=1 such that {Dl }l=1 = {Ai }i=1. Qm Qk In this case, l=1 1Dl = i=1 1Ai , and k (1) Qm (1) Qk (j−1)! (n−kj)! Pn l=1 1Dl = 1 = Pn i=1 1Ai = 1 = n! . (Here we use the assumption that n ≥ mj, which insures that n ≥ kj.)
m (1) (n)m X (1) Y En Cj = Pn 1Dl = 1 . l=1 D1,D2,··· ,Dm :|Dl |=···=|Dm|=j
Qm Now l=1 1Dl 6≡ 0, (that is, there exists some σ ∈ Dn such that Qm l=1 1Dl (σ) 6= 0),
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations That is, if and only if there exist a k ∈ [m] and disjoint sets k m k {Ai }i=1 such that {Dl }l=1 = {Ai }i=1. Qm Qk In this case, l=1 1Dl = i=1 1Ai , and k (1) Qm (1) Qk (j−1)! (n−kj)! Pn l=1 1Dl = 1 = Pn i=1 1Ai = 1 = n! . (Here we use the assumption that n ≥ mj, which insures that n ≥ kj.)
m (1) (n)m X (1) Y En Cj = Pn 1Dl = 1 . l=1 D1,D2,··· ,Dm :|Dl |=···=|Dm|=j
Qm Now l=1 1Dl 6≡ 0, (that is, there exists some σ ∈ Dn such that Qm l=1 1Dl (σ) 6= 0), if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 or
Di1 ∩ Di2 = ∅.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Qm Qk In this case, l=1 1Dl = i=1 1Ai , and k (1) Qm (1) Qk (j−1)! (n−kj)! Pn l=1 1Dl = 1 = Pn i=1 1Ai = 1 = n! . (Here we use the assumption that n ≥ mj, which insures that n ≥ kj.)
m (1) (n)m X (1) Y En Cj = Pn 1Dl = 1 . l=1 D1,D2,··· ,Dm :|Dl |=···=|Dm|=j
Qm Now l=1 1Dl 6≡ 0, (that is, there exists some σ ∈ Dn such that Qm l=1 1Dl (σ) 6= 0), if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 or
Di1 ∩ Di2 = ∅. That is, if and only if there exist a k ∈ [m] and disjoint sets k m k {Ai }i=1 such that {Dl }l=1 = {Ai }i=1.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations and k (1) Qm (1) Qk (j−1)! (n−kj)! Pn l=1 1Dl = 1 = Pn i=1 1Ai = 1 = n! . (Here we use the assumption that n ≥ mj, which insures that n ≥ kj.)
m (1) (n)m X (1) Y En Cj = Pn 1Dl = 1 . l=1 D1,D2,··· ,Dm :|Dl |=···=|Dm|=j
Qm Now l=1 1Dl 6≡ 0, (that is, there exists some σ ∈ Dn such that Qm l=1 1Dl (σ) 6= 0), if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 or
Di1 ∩ Di2 = ∅. That is, if and only if there exist a k ∈ [m] and disjoint sets k m k {Ai }i=1 such that {Dl }l=1 = {Ai }i=1. Qm Qk In this case, l=1 1Dl = i=1 1Ai ,
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations (Here we use the assumption that n ≥ mj, which insures that n ≥ kj.)
m (1) (n)m X (1) Y En Cj = Pn 1Dl = 1 . l=1 D1,D2,··· ,Dm :|Dl |=···=|Dm|=j
Qm Now l=1 1Dl 6≡ 0, (that is, there exists some σ ∈ Dn such that Qm l=1 1Dl (σ) 6= 0), if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 or
Di1 ∩ Di2 = ∅. That is, if and only if there exist a k ∈ [m] and disjoint sets k m k {Ai }i=1 such that {Dl }l=1 = {Ai }i=1. Qm Qk In this case, l=1 1Dl = i=1 1Ai , and k (1) Qm (1) Qk (j−1)! (n−kj)! Pn l=1 1Dl = 1 = Pn i=1 1Ai = 1 = n! .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations m (1) (n)m X (1) Y En Cj = Pn 1Dl = 1 . l=1 D1,D2,··· ,Dm :|Dl |=···=|Dm|=j
Qm Now l=1 1Dl 6≡ 0, (that is, there exists some σ ∈ Dn such that Qm l=1 1Dl (σ) 6= 0), if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 or
Di1 ∩ Di2 = ∅. That is, if and only if there exist a k ∈ [m] and disjoint sets k m k {Ai }i=1 such that {Dl }l=1 = {Ai }i=1. Qm Qk In this case, l=1 1Dl = i=1 1Ai , and k (1) Qm (1) Qk (j−1)! (n−kj)! Pn l=1 1Dl = 1 = Pn i=1 1Ai = 1 = n! . (Here we use the assumption that n ≥ mj, which insures that n ≥ kj.)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations The number of ways to construct k disjoint (ordered) sets (A1, ··· , Ak ), each with j elements from [n], is n n−j n−(m−1)j n! j j ··· j = (j!)k (n−jk)! . k m Given {Ai }i=1, the number of ways to choose {Dl }l=1 so that m k {Dl }l=1 = {Ai }i=1 is equal to S(m, k). So m k (1) (n) m X (j − 1)! (n − kj)! n! E C = × × S(m, k) = n j n! (j!)k (n − jk)! k=1 m X 1k 1 S(m, k) = Tm( ) = µm; 1 . j j j k=1
(1) (n)m P (1) Qm En Cj = Pn l=1 1Dl = 1 . D1,D2,··· ,Dm :|Dl |=···=|Dm|=j Qm Now l=1 1Dl 6≡ 0 if and only if there exist a k ∈ [m] and disjoint k m k sets {Ai }i=1 such that {Dl }l=1 = {Ai }i=1. In this case, Qm Qk l=1 1Dl = i=1 1Ai , and k (1) Qm (1) Qk (j−1)! (n−kj)! Pn l=1 1Dl = 1 = Pn i=1 1Ai = 1 = n! .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations k m Given {Ai }i=1, the number of ways to choose {Dl }l=1 so that m k {Dl }l=1 = {Ai }i=1 is equal to S(m, k). So m k (1) (n) m X (j − 1)! (n − kj)! n! E C = × × S(m, k) = n j n! (j!)k (n − jk)! k=1 m X 1k 1 S(m, k) = Tm( ) = µm; 1 . j j j k=1
(1) (n)m P (1) Qm En Cj = Pn l=1 1Dl = 1 . D1,D2,··· ,Dm :|Dl |=···=|Dm|=j Qm Now l=1 1Dl 6≡ 0 if and only if there exist a k ∈ [m] and disjoint k m k sets {Ai }i=1 such that {Dl }l=1 = {Ai }i=1. In this case, Qm Qk l=1 1Dl = i=1 1Ai , and k (1) Qm (1) Qk (j−1)! (n−kj)! Pn l=1 1Dl = 1 = Pn i=1 1Ai = 1 = n! . The number of ways to construct k disjoint (ordered) sets (A1, ··· , Ak ), each with j elements from [n], is n n−j n−(m−1)j n! j j ··· j = (j!)k (n−jk)! .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations So
m k (1) (n) m X (j − 1)! (n − kj)! n! E C = × × S(m, k) = n j n! (j!)k (n − jk)! k=1 m X 1k 1 S(m, k) = Tm( ) = µm; 1 . j j j k=1
(1) (n)m P (1) Qm En Cj = Pn l=1 1Dl = 1 . D1,D2,··· ,Dm :|Dl |=···=|Dm|=j Qm Now l=1 1Dl 6≡ 0 if and only if there exist a k ∈ [m] and disjoint k m k sets {Ai }i=1 such that {Dl }l=1 = {Ai }i=1. In this case, Qm Qk l=1 1Dl = i=1 1Ai , and k (1) Qm (1) Qk (j−1)! (n−kj)! Pn l=1 1Dl = 1 = Pn i=1 1Ai = 1 = n! . The number of ways to construct k disjoint (ordered) sets (A1, ··· , Ak ), each with j elements from [n], is n n−j n−(m−1)j n! j j ··· j = (j!)k (n−jk)! . k m Given {Ai }i=1, the number of ways to choose {Dl }l=1 so that m k {Dl }l=1 = {Ai }i=1 is equal to S(m, k).
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations =
m X 1k 1 S(m, k) = Tm( ) = µm; 1 . j j j k=1
(1) (n)m P (1) Qm En Cj = Pn l=1 1Dl = 1 . D1,D2,··· ,Dm :|Dl |=···=|Dm|=j Qm Now l=1 1Dl 6≡ 0 if and only if there exist a k ∈ [m] and disjoint k m k sets {Ai }i=1 such that {Dl }l=1 = {Ai }i=1. In this case, Qm Qk l=1 1Dl = i=1 1Ai , and k (1) Qm (1) Qk (j−1)! (n−kj)! Pn l=1 1Dl = 1 = Pn i=1 1Ai = 1 = n! . The number of ways to construct k disjoint (ordered) sets (A1, ··· , Ak ), each with j elements from [n], is n n−j n−(m−1)j n! j j ··· j = (j!)k (n−jk)! . k m Given {Ai }i=1, the number of ways to choose {Dl }l=1 so that m k {Dl }l=1 = {Ai }i=1 is equal to S(m, k). So m k (1) (n) m X (j − 1)! (n − kj)! n! E C = × × S(m, k) n j n! (j!)k (n − jk)! k=1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations 1 = Tm( ) = µm; 1 . j j
(1) (n)m P (1) Qm En Cj = Pn l=1 1Dl = 1 . D1,D2,··· ,Dm :|Dl |=···=|Dm|=j Qm Now l=1 1Dl 6≡ 0 if and only if there exist a k ∈ [m] and disjoint k m k sets {Ai }i=1 such that {Dl }l=1 = {Ai }i=1. In this case, Qm Qk l=1 1Dl = i=1 1Ai , and k (1) Qm (1) Qk (j−1)! (n−kj)! Pn l=1 1Dl = 1 = Pn i=1 1Ai = 1 = n! . The number of ways to construct k disjoint (ordered) sets (A1, ··· , Ak ), each with j elements from [n], is n n−j n−(m−1)j n! j j ··· j = (j!)k (n−jk)! . k m Given {Ai }i=1, the number of ways to choose {Dl }l=1 so that m k {Dl }l=1 = {Ai }i=1 is equal to S(m, k). So m k (1) (n) m X (j − 1)! (n − kj)! n! E C = × × S(m, k) = n j n! (j!)k (n − jk)! k=1 m X 1 k S(m, k) j k=1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations = µ 1 . m; j
(1) (n)m P (1) Qm En Cj = Pn l=1 1Dl = 1 . D1,D2,··· ,Dm :|Dl |=···=|Dm|=j Qm Now l=1 1Dl 6≡ 0 if and only if there exist a k ∈ [m] and disjoint k m k sets {Ai }i=1 such that {Dl }l=1 = {Ai }i=1. In this case, Qm Qk l=1 1Dl = i=1 1Ai , and k (1) Qm (1) Qk (j−1)! (n−kj)! Pn l=1 1Dl = 1 = Pn i=1 1Ai = 1 = n! . The number of ways to construct k disjoint (ordered) sets (A1, ··· , Ak ), each with j elements from [n], is n n−j n−(m−1)j n! j j ··· j = (j!)k (n−jk)! . k m Given {Ai }i=1, the number of ways to choose {Dl }l=1 so that m k {Dl }l=1 = {Ai }i=1 is equal to S(m, k). So m k (1) (n) m X (j − 1)! (n − kj)! n! E C = × × S(m, k) = n j n! (j!)k (n − jk)! k=1 m X 1 k 1 S(m, k) = T ( ) j m j k=1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations (1) (n)m P (1) Qm En Cj = Pn l=1 1Dl = 1 . D1,D2,··· ,Dm :|Dl |=···=|Dm|=j Qm Now l=1 1Dl 6≡ 0 if and only if there exist a k ∈ [m] and disjoint k m k sets {Ai }i=1 such that {Dl }l=1 = {Ai }i=1. In this case, Qm Qk l=1 1Dl = i=1 1Ai , and k (1) Qm (1) Qk (j−1)! (n−kj)! Pn l=1 1Dl = 1 = Pn i=1 1Ai = 1 = n! . The number of ways to construct k disjoint (ordered) sets (A1, ··· , Ak ), each with j elements from [n], is n n−j n−(m−1)j n! j j ··· j = (j!)k (n−jk)! . k m Given {Ai }i=1, the number of ways to choose {Dl }l=1 so that m k {Dl }l=1 = {Ai }i=1 is equal to S(m, k). So m k (1) (n) m X (j − 1)! (n − kj)! n! E C = × × S(m, k) = n j n! (j!)k (n − jk)! k=1 m X 1k 1 S(m, k) = Tm( ) = µm; 1 . j j j k=1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations