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1 Named numbres Ngày 25 tháng 11 năm 2011 () Named numbres Ngày 25 tháng 11 năm / 7

2 Fibonacci, Catalan, Stirling, Euler, Bernoulli Many sequences are famous. 1 1, 2, 3, 4,... the integers. () Named numbres Ngày 25 tháng 11 năm / 7

3 Fibonacci, Catalan, Stirling, Euler, Bernoulli Many sequences are famous. 1 1, 2, 3, 4,... the integers. 2 1, 3, 5, 7, 9... the odd integers. () Named numbres Ngày 25 tháng 11 năm / 7

4 Fibonacci, Catalan, Stirling, Euler, Bernoulli Many sequences are famous. 1 1, 2, 3, 4,... the integers. 2 1, 3, 5, 7, 9... the odd integers. 3 2, 3, 5, 7, 11, 13,... () Named numbres Ngày 25 tháng 11 năm / 7

5 Fibonacci, Catalan, Stirling, Euler, Bernoulli Many sequences are famous. 1 1, 2, 3, 4,... the integers. 2 1, 3, 5, 7, 9... the odd integers. 3 2, 3, 5, 7, 11, 13,... () Named numbres Ngày 25 tháng 11 năm / 7

6 Fibonacci, Catalan, Stirling, Euler, Bernoulli Many sequences are famous. 1 1, 2, 3, 4,... the integers. 2 1, 3, 5, 7, 9... the odd integers. 3 2, 3, 5, 7, 11, 13,... the prime numbers. 4 1, 1, 2, 3, 5, 8, 13,... () Named numbres Ngày 25 tháng 11 năm / 7

7 Fibonacci, Catalan, Stirling, Euler, Bernoulli Many sequences are famous. 1 1, 2, 3, 4,... the integers. 2 1, 3, 5, 7, 9... the odd integers. 3 2, 3, 5, 7, 11, 13,... the prime numbers. 4 1, 1, 2, 3, 5, 8, 13,... () Named numbres Ngày 25 tháng 11 năm / 7

8 Fibonacci, Catalan, Stirling, Euler, Bernoulli Many sequences are famous. 1 1, 2, 3, 4,... the integers. 2 1, 3, 5, 7, 9... the odd integers. 3 2, 3, 5, 7, 11, 13,... the prime numbers. 4 1, 1, 2, 3, 5, 8, 13,... Fibonacci numbers. ( n ) 5 () Named numbres Ngày 25 tháng 11 năm / 7

9 Fibonacci, Catalan, Stirling, Euler, Bernoulli Many sequences are famous. 1 1, 2, 3, 4,... the integers. 2 1, 3, 5, 7, 9... the odd integers. 3 2, 3, 5, 7, 11, 13,... the prime numbers. 4 1, 1, 2, 3, 5, 8, 13,... Fibonacci numbers. ( n ) 5 () Named numbres Ngày 25 tháng 11 năm / 7

10 Fibonacci, Catalan, Stirling, Euler, Bernoulli Many sequences are famous. 1 1, 2, 3, 4,... the integers. 2 1, 3, 5, 7, 9... the odd integers. 3 2, 3, 5, 7, 11, 13,... the prime numbers. 4 1, 1, 2, 3, 5, 8, 13,... Fibonacci numbers. ( 5 n ) binomials. ( 6 1 2n ) n+1 n () Named numbres Ngày 25 tháng 11 năm / 7

11 Fibonacci, Catalan, Stirling, Euler, Bernoulli Many sequences are famous. 1 1, 2, 3, 4,... the integers. 2 1, 3, 5, 7, 9... the odd integers. 3 2, 3, 5, 7, 11, 13,... the prime numbers. 4 1, 1, 2, 3, 5, 8, 13,... Fibonacci numbers. ( 5 n ) binomials. ( 6 1 2n ) n+1 n () Named numbres Ngày 25 tháng 11 năm / 7

12 Fibonacci, Catalan, Stirling, Euler, Bernoulli Many sequences are famous. 1 1, 2, 3, 4,... the integers. 2 1, 3, 5, 7, 9... the odd integers. 3 2, 3, 5, 7, 11, 13,... the prime numbers. 4 1, 1, 2, 3, 5, 8, 13,... Fibonacci numbers. ( 5 n ) binomials. ( 6 1 2n ) n+1 n the Catalan numbers. () Named numbres Ngày 25 tháng 11 năm / 7

13 Fibonacci, Catalan, Stirling, Euler, Bernoulli Many sequences are famous. 1 1, 2, 3, 4,... the integers. 2 1, 3, 5, 7, 9... the odd integers. 3 2, 3, 5, 7, 11, 13,... the prime numbers. 4 1, 1, 2, 3, 5, 8, 13,... Fibonacci numbers. ( 5 n ) binomials. ( 6 1 2n ) n+1 n the Catalan numbers. The reason they are named is because they appear in many forms in mathematics and other sciences. () Named numbres Ngày 25 tháng 11 năm / 7

14 Stirling Numbers Stirling Numbers are named after the Scottish mathematician James Stirling who introduced them in the 18 th century. There are two inds of Stirling numbers (with various notation): [ ] n Stirling numbers of the first ind: = c(n, ) { n Stirling numbers of the second ind: } = S(n, ) Both numbers describe combinatorial counting that lead to a triangular recurrence relation similar to the binomial coefficients. () Named numbres Ngày 25 tháng 11 năm / 7

15 Stirling numbers of the second ind: Question In how many ways can you partition the set {a, b, c, d, e} into two non-empty subsets? () Named numbres Ngày 25 tháng 11 năm / 7

16 Stirling numbers of the second ind: Question In how many ways can you partition the set {a, b, c, d, e} into two non-empty subsets? Answer We can just list the subsets: {{a}, {, b, c, d, e}}, {{b}, {a, c, d, e}}..., {{a, b}, {c, d, e}},... {{d, e}, {, a, b, c}}. For a total of: () Named numbres Ngày 25 tháng 11 năm / 7

17 Stirling numbers of the second ind: Question In how many ways can you partition the set {a, b, c, d, e} into two non-empty subsets? Answer We can just list the subsets: {{a}, {, b, c, d, e}}, {{b}, {a, c, d, e}}..., {{a, b}, {c, d, e}},... {{d, e}, {, a, b, c}}. For a total of: () Named numbres Ngày 25 tháng 11 năm / 7

18 Stirling numbers of the second ind: Question In how many ways can you partition the set {a, b, c, d, e} into two non-empty subsets? Answer We can just list the subsets: {{a}, {, b, c, d, e}}, {{b}, {a, c, d, e}}..., {{a, b}, {c, d, e}},... {{d, e}, {, a, b, c}}. For a total of: = 15 different partitions. Definition { n } The Stirling number of the second ind is the number of ways to partition an n-set into non-empty subsets. () Named numbres Ngày 25 tháng 11 năm / 7

19 From the example we can derive a recurrence relation for these numbers: () Named numbres Ngày 25 tháng 11 năm / 7

20 From the example we can derive a recurrence relation for these numbers: Let B be an n set. We can divide the patitions into two sets A: partitions that include a fixed singleton {x 0 } and B: the rest. () Named numbres Ngày 25 tháng 11 năm / 7

21 From the example we can derive a recurrence relation for these numbers: Let B be an n set. We can divide the patitions into two sets A: partitions that include a fixed singleton {x 0 } and B: the rest. Clearly, there are S(n 1, 1) distinct partitions in A. () Named numbres Ngày 25 tháng 11 năm / 7

22 From the example we can derive a recurrence relation for these numbers: Let B be an n set. We can divide the patitions into two sets A: partitions that include a fixed singleton {x 0 } and B: the rest. Clearly, there are S(n 1, 1) distinct partitions in A. As for the set B for every partition B \ {x 0 } into subset we can add x 0 to any one of the parttions yielding: { } { } { } n n 1 n 1 = + 1 () Named numbres Ngày 25 tháng 11 năm / 7

23 From the example we can derive a recurrence relation for these numbers: Let B be an n set. We can divide the patitions into two sets A: partitions that include a fixed singleton {x 0 } and B: the rest. Clearly, there are S(n 1, 1) distinct partitions in A. As for the set B for every partition B \ {x 0 } into subset we can add x 0 to any one of the parttions yielding: { } { } { } n n 1 n 1 = + 1 This triangular relation is very similar to Pascal s identity for binomials. () Named numbres Ngày 25 tháng 11 năm / 7

24 An application The polynomials of degree n form a vector space over the field R, and so do the polynomials {1, x, x(x 1), x(x 1)(x 2)...}. A common notation for x(x 1)... (x j + 1) is x j This means that the polynomials x can be expressed as linear combination of these polynomials: x = a i x i i=0 () Named numbres Ngày 25 tháng 11 năm / 7

25 An application The polynomials of degree n form a vector space over the field R, and so do the polynomials {1, x, x(x 1), x(x 1)(x 2)...}. A common notation for x(x 1)... (x j + 1) is x j This means that the polynomials x can be expressed as linear combination of these polynomials: What are the coefficients a i? x = a i x i i=0 () Named numbres Ngày 25 tháng 11 năm / 7

26 Claim: x n = n =0 { n } x Chứng minh. () Named numbres Ngày 25 tháng 11 năm / 7

27 Claim: x n = n =0 { n } x Chứng minh. 1 a. x x = x +1 + x. () Named numbres Ngày 25 tháng 11 năm / 7

28 Claim: x n = n =0 { n } x Chứng minh. 1 a. x x = x +1 + x. 2 n 1 { x x n 1 n 1 = x =0 n 1 { n 1 =0 } x = n 1 =0 } { x n Follows from the triangular relation. { n 1 } x = } (x +1 + x ) = n =0 { n } x () Named numbres Ngày 25 tháng 11 năm / 7

29 The Stirling numbers of the first ind are defined by a closely related relation: It counts in how many ways you can arrange n objects into disjoint cycles. So for example, the partitions {[1, 3][2, 5, 4]} and {[1, 3][2, 4, 5]} are distinct but {[3, 1], [5, 4, 2]} is the same as the first partition. () Named numbres Ngày 25 tháng 11 năm / 7

30 The Stirling numbers of the first ind are defined by a closely related relation: It counts in how many ways you can arrange n objects into disjoint cycles. So for example, the partitions {[1, 3][2, 5, 4]} and {[1, 3][2, 4, 5]} are distinct but {[3, 1], [5, 4, 2]} is the same as the first partition. We leave it to you to show that: 1 [ n ] [ n 1 = (n 1) ] [ n ] () Named numbres Ngày 25 tháng 11 năm / 7

31 The Stirling numbers of the first ind are defined by a closely related relation: It counts in how many ways you can arrange n objects into disjoint cycles. So for example, the partitions {[1, 3][2, 5, 4]} and {[1, 3][2, 4, 5]} are distinct but {[3, 1], [5, 4, 2]} is the same as the first partition. We leave it to you to show that: 1 [ n ] [ n 1 = (n 1) 2 Let x n = x(x + 1)... (x + n 1) ] [ n ] () Named numbres Ngày 25 tháng 11 năm / 7

32 The Stirling numbers of the first ind are defined by a closely related relation: It counts in how many ways you can arrange n objects into disjoint cycles. So for example, the partitions {[1, 3][2, 5, 4]} and {[1, 3][2, 4, 5]} are distinct but {[3, 1], [5, 4, 2]} is the same as the first partition. We leave it to you to show that: 1 [ n ] [ n 1 = (n 1) 2 Let x n = x(x + 1)... (x + n 1) 3 x n = n x =0 ] [ n ] () Named numbres Ngày 25 tháng 11 năm / 7

33 The Stirling numbers of the first ind are defined by a closely related relation: It counts in how many ways you can arrange n objects into disjoint cycles. So for example, the partitions {[1, 3][2, 5, 4]} and {[1, 3][2, 4, 5]} are distinct but {[3, 1], [5, 4, 2]} is the same as the first partition. We leave it to you to show that: 1 [ n ] [ n 1 = (n 1) 2 Let x n = x(x + 1)... (x + n 1) 3 x n = n 4 x =0 n =0 [ n ] = n! ] [ n ] () Named numbres Ngày 25 tháng 11 năm / 7

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