Touchard Polynomials, Stirling Numbers and Random Permutations Touchard Polynomials, Stirling...

download Touchard Polynomials, Stirling Numbers and Random Permutations Touchard Polynomials, Stirling Numbers

of 115

  • date post

    22-Jul-2020
  • Category

    Documents

  • view

    0
  • download

    0

Embed Size (px)

Transcript of Touchard Polynomials, Stirling Numbers and Random Permutations Touchard Polynomials, Stirling...

  • 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

  • Notation:

    [n] = {1, 2, · · · , n}

    Sn = permutations of [n]

    |Sn| = n!

    S6 3 σ1 = (

    1 2 3 4 5 6 2 4 6 1 5 3

    ) = (124)(36)(5),

    S6 3 σ2 = (

    1 2 3 4 5 6 3 4 6 5 1 2

    ) = (136245) = (624513)

    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

  • Notation:

    [n] = {1, 2, · · · , n}

    Sn = permutations of [n]

    |Sn| = n!

    S6 3 σ1 = (

    1 2 3 4 5 6 2 4 6 1 5 3

    ) = (124)(36)(5),

    S6 3 σ2 = (

    1 2 3 4 5 6 3 4 6 5 1 2

    ) = (136245) = (624513)

    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

  • Notation:

    [n] = {1, 2, · · · , n}

    Sn = permutations of [n]

    |Sn| = n!

    S6 3 σ1 = (

    1 2 3 4 5 6 2 4 6 1 5 3

    ) = (124)(36)(5),

    S6 3 σ2 = (

    1 2 3 4 5 6 3 4 6 5 1 2

    ) = (136245) = (624513)

    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

  • Notation:

    [n] = {1, 2, · · · , n}

    Sn = permutations of [n]

    |Sn| = n!

    S6 3 σ1 = (

    1 2 3 4 5 6 2 4 6 1 5 3

    ) = (124)(36)(5),

    S6 3 σ2 = (

    1 2 3 4 5 6 3 4 6 5 1 2

    ) = (136245) = (624513)

    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

  • Notation:

    [n] = {1, 2, · · · , n}

    Sn = permutations of [n]

    |Sn| = n!

    S6 3 σ1 = (

    1 2 3 4 5 6 2 4 6 1 5 3

    ) = (124)(36)(5),

    S6 3 σ2 = (

    1 2 3 4 5 6 3 4 6 5 1 2

    ) = (136245) = (624513)

    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

  • Notation:

    [n] = {1, 2, · · · , n}

    Sn = permutations of [n]

    |Sn| = n!

    S6 3 σ1 = (

    1 2 3 4 5 6 2 4 6 1 5 3

    ) = (124)(36)(5),

    S6 3 σ2 = (

    1 2 3 4 5 6 3 4 6 5 1 2

    ) = (136245) = (624513)

    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

  • Rising Factorials

    x (n) := x(x +1) · · · (x +n−1)

    ( ≡

    n∑ k=1

    ankx k ) , x ∈ R, n ≥ 1. (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; |s(n + 1, k)| = n|s(n, k)|+ |s(n, k − 1)|, 2 ≤ k ≤ n.

    (2)

    It is easy to check that the coefficients {ank} in (1) also satisfy (2); thus, ank = |s(n, k)|. Therefore

    x (n) = n∑

    k=1

    |s(n, k)|xk . (3)

    Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

  • Rising Factorials

    x (n) := x(x +1) · · · (x +n−1) ( ≡

    n∑ k=1

    ankx k ) , x ∈ R, n ≥ 1. (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; |s(n + 1, k)| = n|s(n, k)|+ |s(n, k − 1)|, 2 ≤ k ≤ n.

    (2)

    It is easy to check that the coefficients {ank} in (1) also satisfy (2); thus, ank = |s(n, k)|. Therefore

    x (n) = n∑

    k=1

    |s(n, k)|xk . (3)

    Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

  • Rising Factorials

    x (n) := x(x +1) · · · (x +n−1) ( ≡

    n∑ k=1

    ankx k ) , x ∈ R, n ≥ 1. (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; |s(n + 1, k)| = n|s(n, k)|+ |s(n, k − 1)|, 2 ≤ k ≤ n.

    (2)

    It is easy to check that the coefficients {ank} in (1) also satisfy (2); thus, ank = |s(n, k)|. Therefore

    x (n) = n∑

    k=1

    |s(n, k)|xk . (3)

    Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

  • Rising Factorials

    x (n) := x(x +1) · · · (x +n−1) ( ≡

    n∑ k=1

    ankx k ) , x ∈ R, n ≥ 1. (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; |s(n + 1, k)| = n|s(n, k)|+ |s(n, k − 1)|, 2 ≤ k ≤ n.

    (2)

    It is easy to check that the coefficients {ank} in (1) also satisfy (2); thus, ank = |s(n, k)|. Therefore

    x (n) = n∑

    k=1

    |s(n, k)|xk . (3)

    Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

  • Rising Factorials

    x (n) := x(x +1) · · · (x +n−1) ( ≡

    n∑ k=1

    ankx k ) , x ∈ R, n ≥ 1. (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; |s(n + 1, k)| = n|s(n, k)|+ |s(n, k − 1)|, 2 ≤ k ≤ n.

    (2)

    It is easy to check that the coefficients {ank} in (1) also satisfy (2); thus, ank = |s(n, k)|.

    Therefore

    x (n) = n∑

    k=1

    |s(n, k)|xk . (3)

    Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

  • Rising Factorials

    x (n) := x(x +1) · · · (x +n−1) ( ≡

    n∑ k=1

    ankx k ) , x ∈ R, n ≥ 1. (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; |s(n + 1, k)| = n|s(n, k)|+ |s(n, k − 1)|, 2 ≤ k ≤ n.

    (2)

    It is easy to check that the coefficients {ank} in (1) also satisfy (2); thus, ank = |s(n, k)|. Therefore

    x (n) = n∑

    k=1

    |s(n, k)|xk . (3)

    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 (−x)n = −x(−x−1) · · · (−x−n+1) = (−1)nx(x+1) · · · (x+n−1) = (−1)nx (n). Substituting above −x for x , we have

    (x)n = (−1)n(−x)(n) from (3)

    = (−1)n n∑

    k=1

    |s(n, k)|(−x)k =

    n∑ k=1

    (−1)n−k |s(n, k)|xk . (5)

    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

    (x)n = n∑

    k=1

    s(n, k)xk . (6)

    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 (−x)n = −x(−x−1) · · · (−x−n+1) =

    (−1)nx(x+1) · · · (x+n−1) = (−1)nx (n). Substituting above −x for x , we have

    (x)n = (−1)n(−x)(n) from (3)

    = (−1)n n∑

    k=1

    |s(n, k)|(−x)k =

    n∑ k=1

    (−1)n−k |s(n, k)|xk . (5)

    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

    (x)n = n∑

    k=1

    s(n, k)xk . (6)

    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 (−x)n = −x(−x−1) · · · (−x−n+1) = (−1)nx(x+1) · · · (x+n−1)

    = (−1)nx (n). Substituting above −x for x , we have

    (x)n = (−1)n(−x)(n) from (3)

    = (−1)n n∑

    k=1

    |s(n, k)|(−x)k =

    n∑ k=1

    (−1)n−k |s(n, k)|xk . (5)

    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

    (x)n = n∑

    k=1

    s(n, k)xk . (6)

    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 (−x)n = −x(−x−1) · · · (−x−n+1) = (−1)nx(x+1) · · · (x+n−1) = (−1)nx (n).

    Substituting above −x for x , we have

    (x)n = (−1)n(−x)(n) from (3)

    = (−1)n n∑

    k=1

    |s