November 28, Western Washington University. Ramanujan s Congruences. Bryan Clark, Lesley Lowery, Nhan Nguyen. Introduction. History.

Size: px

Start display at page:

Download "November 28, Western Washington University. Ramanujan s Congruences. Bryan Clark, Lesley Lowery, Nhan Nguyen. Introduction. History."

Myles French
5 years ago
Views:

1 Western Washington University November 28, 2012

2 Srinivasa Ramanujan Srinivasa Ramanujan (22 December April 1920)

3 Srinivasa Ramanujan Considered to have no formal training in pure mathematics

4 Srinivasa Ramanujan Considered to have no formal training in pure mathematics Education Government Arts College Pachaiyappa s College

5 Srinivasa Ramanujan Considered to have no formal training in pure mathematics Education Government Arts College Pachaiyappa s College Influenced by G. H. Hardy

6 Srinivasa Ramanujan Considered to have no formal training in pure mathematics Education Government Arts College Pachaiyappa s College Influenced by G. H. Hardy Known for

7 Srinivasa Ramanujan Considered to have no formal training in pure mathematics Education Government Arts College Pachaiyappa s College Influenced by G. H. Hardy Known for Landau - Ramanujan constant lim x N(x) ln(x) x

8 Srinivasa Ramanujan Considered to have no formal training in pure mathematics Education Government Arts College Pachaiyappa s College Influenced by G. H. Hardy Known for Landau - Ramanujan constant N(x) ln(x) lim x x Ramanujan theta function f (a, b) = a n(n+1)/2 b n(n 1)/2 n=

9 Srinivasa Ramanujan Considered to have no formal training in pure mathematics Education Government Arts College Pachaiyappa s College Influenced by G. H. Hardy Known for Landau - Ramanujan constant N(x) ln(x) lim x x Ramanujan theta function f (a, b) = a n(n+1)/2 b n(n 1)/2 n= Rogers - Ramanujan identities

10 Srinivasa Ramanujan Considered to have no formal training in pure mathematics Education Government Arts College Pachaiyappa s College Influenced by G. H. Hardy Known for Landau - Ramanujan constant N(x) ln(x) lim x x Ramanujan theta function f (a, b) = a n(n+1)/2 b n(n 1)/2 n= Rogers - Ramanujan identities Ramanujan conjecture

11 Srinivasa Ramanujan Considered to have no formal training in pure mathematics Education Government Arts College Pachaiyappa s College Influenced by G. H. Hardy Known for Landau - Ramanujan constant N(x) ln(x) lim x x Ramanujan theta function f (a, b) = a n(n+1)/2 b n(n 1)/2 n= Rogers - Ramanujan identities Ramanujan conjecture q sum c q (n) = a=1 (a,q)=1 e 2πi a q n

12 The of - Ramanujan first postulates his congruences, inspired when he examined a table of the values of p(n) for values of n from 1 to 200.

13 The of - Ramanujan first postulates his congruences, inspired when he examined a table of the values of p(n) for values of n from 1 to 200. of mod 5, mod 7 congruences Uses Theta functions, f (a, b) = a n(n+1)/2 b n(n 1)/2, n= to sketch proofs of the congruences: p(25n 1) 0 (mod 25) p(49n 2) 0 (mod 49).

14 The of - After death in 1920, G. H. Hardy edits and publishes one of manuscripts which contains new proofs.

15 The of - After death in 1920, G. H. Hardy edits and publishes one of manuscripts which contains new proofs. First published proof of the mod 11 congruence s make use of Eisenstein series

16 Statement of the Theorem Theorem Let n N then p(5n + 4) 0 (mod 5)

17 . We are looking for p(5n + 4) so we are interested in [z 5n+4] p(n)z n n 0 So it suffices to look at [z ( ) 5n] z i 1 (1 zi ) (*)

18 Continued. To get at star, look at 3 z i 1(1 z i ) (1 z i ) i 1 Through some algebra we find that this equals: ( 1) r+s r(r+1) 1+ + (2r + 1)z s(3s+1) 2 2 ( ) r 0 s Z

19 Continued. If the exponent of ( ) is a multiple of 5 We see that Therefore 1 + r(r + 1) 2 + s(3s + 1) 2 0 (mod 5) (2r + 1) 2 + 2(s + 1) 2 0 (mod 5) (2r + 1) 0 (mod 5)

20 Continued. From the last slide [z 5n ] z i 1(1 z i ) 0 (mod 5) (**)

21 Continued. To get ( ) let s now consider [z 5n] We multiply by 1, [z 5n] z i 1 (1 zi ) (1 z 5i ). 4 z i 1(1 z i ) i 1 i 1 (1 z5i ) ( i 1 ((1 zi ) 5. )

22 Continued. By binomial expansion (1 z i ) 5 (1 z 5i ) (mod 5) z i 1(1 i ) 5 (1 z 5i ) (mod 5) i 1 i 1 (1 z5i ) i 1 (1 1 (mod 5). ( ) zi ) 5

23 Continued. From ( ) and ( ) we have [z 5n] 4 z i 1(1 z i ) [z 5n] i 1 (1 z5i ) ( i 1 ((1 zi ) 5 0 (mod 5) ) z i 1 (1 zi ) (1 z 5i ) 0 (mod 5) i 1

24 Continued. Finally [z ( ) 5n] z i 1 (1 zi ) (*)

25 (mod 7) Case Theorem Let n N then p(7n + 5) 0 (mod 7)

26 . Similar to the (mod 5) proof

27 (mod 11) Case Theorem Let n N then p(11n + 6) 0 (mod 11)

28 We will show that the style of proof used to prove the (mod 5) and the (mod 7) cases will not work for the (mod 11) case.

29 We will show that the style of proof used to prove the (mod 5) and the (mod 7) cases will not work for the (mod 11) case.

30 Attempt. We are looking for p(11n + 6) so we are interested in [z 11n+6] p(n)z n n 0 It would be sufficient to look at [z ( 11n] z 5 ) i 1 (1 zi ) ( )

31 Continued. To get at, look at z 5 9 i 1(1 z i ) (1 z i ) i 1 Through some algebra we find that this equals: ( 1) r+s+t+u (2r + 1)(2s + 1)(2t + 1)z α Where α = 5 + r(r+1)+s(s+1)+t(t+1)+u(3u+1) 2 The summations extending form 0 < r <, 0 < s <, 0 < t <, and u Z ( )

32 continued. If the exponent of ( ) is a multiple of r(r + 1) 2 We see that + s(s + 1) 2 + t(t + 1) 2 + u(3u + 1) 2 0 (mod 11) (2r + 1) 2 + (2s + 1) 2 + (2t + 1) 2 + (u + 2) 2 0 (mod 11)

33 continued. (2r + 1) 2 0, 1, 3, 4, 5, 9 (mod 11) (2s + 1) 2 0, 1, 3, 4, 5, 9 (mod 11) (2t + 1) 2 0, 1, 3, 4, 5, 9 (mod 11) (u + 2) 2 0, 1, 3, 4, 5, 9 (mod 11) This implies that there are multiple ways to make (2r + 1) 2 + (2s + 1) 2 + (2t + 1) 2 + (u + 2) 2 0 (mod 11) thus we can not proceed with this style of proof.

35 New Directions: Dyson - Freeman Dyson, then an undergraduate at Cambridge, looks for a combinatorial proof of the congruences. To do this, he defines the rank r of a partition π as follows: Let l(π) be the largest part of π. Let ν(π) be the number of parts of π. Then r(π) := l(π) ν(π).

36 New Directions: Dyson - Freeman Dyson, then an undergraduate at Cambridge, looks for a combinatorial proof of the congruences. To do this, he defines the rank r of a partition π as follows: Let l(π) be the largest part of π. Let ν(π) be the number of parts of π. Then r(π) := l(π) ν(π). Rank can be easily visualized using Ferrers Diagrams.

37 New Directions: Dyson Dyson also comes up with generating functions related to the rank, by defining N(m, n) as the number of partitions of n with rank m, and N(m, q, n) as the number of partitions of n with a rank congruent to m (mod q).

38 New Directions: Dyson Dyson also comes up with generating functions related to the rank, by defining N(m, n) as the number of partitions of n with rank m, and N(m, q, n) as the number of partitions of n with a rank congruent to m (mod q). Dyson claims that for n = 5k + 4, N(0, 5, 5k + 4) = N(1, 5, 5k + 4) = N(2, 5, 5k + 4) = N(3, 5, 5k + 4) = N(4, 5, 5k + 4) = p(5k + 4), 5 and that a similar relationship exists for the mod 7 congruence.

39 New Directions: Dyson Dyson also comes up with generating functions related to the rank, by defining N(m, n) as the number of partitions of n with rank m, and N(m, q, n) as the number of partitions of n with a rank congruent to m (mod q). Dyson claims that for n = 5k + 4, N(0, 5, 5k + 4) = N(1, 5, 5k + 4) = N(2, 5, 5k + 4) = N(3, 5, 5k + 4) = N(4, 5, 5k + 4) = p(5k + 4), 5 and that a similar relationship exists for the mod 7 congruence. He creates tables of calculated values to back up his claims, but does not provide a proof.

40 Let n = 4. n = 5(0) + 4. The partitions of n are:

41 Let n = 4. n = 5(0) + 4. The partitions of n are: : r(π) = 3 2 (mod 5) : r(π) = 1 4 (mod 5) : r(π) = 0 0 (mod 5) : r(π) = 1 1 (mod 5) 4 : r(π) = 3 3 (mod 5)

42 Let n = 4. n = 5(0) + 4. The partitions of n are: : r(π) = 3 2 (mod 5) : r(π) = 1 4 (mod 5) : r(π) = 0 0 (mod 5) : r(π) = 1 1 (mod 5) 4 : r(π) = 3 3 (mod 5) We have one partition in each group, modulo 5! Great!

43 Let n = 4. n = 5(0) + 4. The partitions of n are: : r(π) = 3 2 (mod 5) : r(π) = 1 4 (mod 5) : r(π) = 0 0 (mod 5) : r(π) = 1 1 (mod 5) 4 : r(π) = 3 3 (mod 5) We have one partition in each group, modulo 5! Great! Exercise: Try this with n = 5 = 7(0) + 5.

44 Let n = 6. n = 11(0) + 6. The partitions of n are: : r(π) = 5 6 (mod 11) : r(π) = 3 8 (mod 11) : r(π) = 2 9 (mod 11) : r(π) = 1 10 (mod 11) : r(π) = 1 10 (mod 11) : r(π) = 0 0 (mod 11) : r(π) = 1 1 (mod 11) : r(π) = 1 1 (mod 11) : r(π) = 2 2 (mod 11) : r(π) = 3 3 (mod 11) 6 : r(π) = 5 5 (mod 11)

45 Let n = 6. n = 11(0) + 6. The partitions of n are: : r(π) = 5 6 (mod 11) : r(π) = 3 8 (mod 11) : r(π) = 2 9 (mod 11) : r(π) = 1 10 (mod 11) : r(π) = 1 10 (mod 11) : r(π) = 0 0 (mod 11) : r(π) = 1 1 (mod 11) : r(π) = 1 1 (mod 11) : r(π) = 2 2 (mod 11) : r(π) = 3 3 (mod 11) 6 : r(π) = 5 5 (mod 11)

46 Let n = 6. n = 11(0) + 6. The partitions of n are: : r(π) = 5 6 (mod 11) : r(π) = 3 8 (mod 11) : r(π) = 2 9 (mod 11) : r(π) = 1 10 (mod 11) : r(π) = 1 10 (mod 11) : r(π) = 0 0 (mod 11) : r(π) = 1 1 (mod 11) : r(π) = 1 1 (mod 11) : r(π) = 2 2 (mod 11) : r(π) = 3 3 (mod 11) 6 : r(π) = 5 5 (mod 11)

47 Let n = 6. n = 11(0) + 6. The partitions of n are: : r(π) = 5 6 (mod 11) : r(π) = 3 8 (mod 11) : r(π) = 2 9 (mod 11) : r(π) = 1 10 (mod 11) : r(π) = 1 10 (mod 11) : r(π) = 0 0 (mod 11) : r(π) = 1 1 (mod 11) : r(π) = 1 1 (mod 11) : r(π) = 2 2 (mod 11) : r(π) = 3 3 (mod 11) 6 : r(π) = 5 5 (mod 11) So rank doesn t give us 11 equal groupings!

49 New Directions: Dyson Once again, the mod 11 case proves to be difficult!

50 New Directions: Dyson Once again, the mod 11 case proves to be difficult! Dyson postulates that some other statistic of a partition exists that will divide the 5k + 4, 7k + 5, and 11k + 6 partitions into equally-sized classes, and thus prove the congruences combinatorially. He names this statistic the crank, since he believes it is similar to the rank.

51 New Directions: Dyson Once again, the mod 11 case proves to be difficult! Dyson postulates that some other statistic of a partition exists that will divide the 5k + 4, 7k + 5, and 11k + 6 partitions into equally-sized classes, and thus prove the congruences combinatorially. He names this statistic the crank, since he believes it is similar to the rank. He leaves the finding of the crank, and the proof of his rank postulates to others...

52 New Directions: Atkin, Swinerton-Dyer - A. O. L. Atkin and P. Swinnerton-Dyer prove Dyson s rank postulate for the mod 5 and mod 7 congruences works. Several new congruences and partition identities come from their work.

53 New Directions: Andrews, Garvan - Andrews and Garvan find the elusive crank!

54 New Directions: Andrews, Garvan - Andrews and Garvan find the elusive crank! Crank is similar to rank, as Dyson thought. Crank divides p(5k + 4) into classes, based on its residue mod 5 - just like rank did.

55 New Directions: Andrews, Garvan - Andrews and Garvan find the elusive crank! Crank is similar to rank, as Dyson thought. Crank divides p(5k + 4) into classes, based on its residue mod 5 - just like rank did. Andrews and Garvan prove that by using the crank, the classes of partitions are equal in size - for all three congruences!

56 Crank definition Let π be a partition of n. Let l(π) be the largest part of π. Let ω(π) be the number of ones in π. Let µ(π) be the number of parts of (π) which are > ω(π).

57 Crank definition Let π be a partition of n. Let l(π) be the largest part of π. Let ω(π) be the number of ones in π. Let µ(π) be the number of parts of (π) which are > ω(π). Then the crank is defined as { l(π), if ω(π) = 0 c(π) = µ(π) ω(π), if ω(π) > 0

58 Let n = 4. n = 5(0) + 4. The partitions of n are:

59 Let n = 4. n = 5(0) + 4. The partitions of n are: : c(π) = 4 1 (mod 5) : c(π) = 2 3 (mod 5) : c(π) = 2 2 (mod 5) : c(π) = 0 0 (mod 5) 4 : c(π) = 4 4 (mod 5)

60 Let n = 4. n = 5(0) + 4. The partitions of n are: : c(π) = 4 1 (mod 5) : c(π) = 2 3 (mod 5) : c(π) = 2 2 (mod 5) : c(π) = 0 0 (mod 5) 4 : c(π) = 4 4 (mod 5) So crank gives us one partition in each group, modulo 5, just like rank did. So far, so good.

61 Let n = 4. n = 5(0) + 4. The partitions of n are: : c(π) = 4 1 (mod 5) : c(π) = 2 3 (mod 5) : c(π) = 2 2 (mod 5) : c(π) = 0 0 (mod 5) 4 : c(π) = 4 4 (mod 5) So crank gives us one partition in each group, modulo 5, just like rank did. So far, so good. Exercise: Try this with n = 5 = 7(0) + 5.

62 Let n = 6. n = 11(0) + 6. The partitions of n are: : c(π) = 6 5 (mod 11) : c(π) = 4 7 (mod 11) : c(π) = 2 9 (mod 11) : c(π) = 2 2 (mod 11) : c(π) = 3 8 (mod 11) : c(π) = 1 1 (mod 11) : c(π) = 3 3 (mod 11) : c(π) = 1 10 (mod 11) : c(π) = 4 4 (mod 11) : c(π) = 0 0 (mod 11) 6 : c(π) = 6 6 (mod 11)

63 Let n = 6. n = 11(0) + 6. The partitions of n are: : c(π) = 6 5 (mod 11) : c(π) = 4 7 (mod 11) : c(π) = 2 9 (mod 11) : c(π) = 2 2 (mod 11) : c(π) = 3 8 (mod 11) : c(π) = 1 1 (mod 11) : c(π) = 3 3 (mod 11) : c(π) = 1 10 (mod 11) : c(π) = 4 4 (mod 11) : c(π) = 0 0 (mod 11) 6 : c(π) = 6 6 (mod 11) So crank works for the troublesome mod 11 congruence!

64 Q : Why do motorcycle gang members use their motorcycles to get to work?

65 Q : Why do motorcycle gang members use their motorcycles to get to work? A : Because members of cyclical groups commute.

66 for listening. Thanks for a fun quarter, Stephanie!! Have a great break.

67 Aigner, Martin. A Course in enumeration. Berlin: Springer, Andrews, George E., and Garvan, F. G., Dyson s crank of a partition. Bulletin of the AMS., 18 (April 1988), no. 2, Atkin, A.O.L. and Swinnerton-Dyer, P., Some properties of partitions. Proc. London Math. Soc., (3) 4 (), F. J. Dyson, Some guesses in the theory of partitions. Eureka (Cambridge). 8 (), Ramanujan, Srinivasa, Congruence properties of partitions. In Collected Papers of Srinivasa Ramanujan. G. H. Hardy, P.V. Seshu Aiyar, and B. M. Wilson, Eds. Providence, RI: AMS Chelsea, , Some properties of p(n), the number of partitions of n. In Collected Papers of Srinivasa Ramanujan. G. H. Hardy, P.V. Seshu Aiyar, and B. M. Wilson, Eds. Providence, RI: AMS Chelsea, Srinivasa Ramanujan. 20 November 2012, 23:37 UTC. In Wikipedia: The Free Encyclopedia. Wikimedia Foundation Inc. Encyclopedia on-line. Available from ramanujan. Internet. Retrieved 25 November 2012.

50. ON THE EXPANSION OF RAMANUJAN S CONTINUED FRACTION. Dedicated to George E. Andrews on the occasion of his 60th Birthday.

50. ON THE EXPANSION OF RAMANUJAN S CONTINUED FRACTION. Dedicated to George E. Andrews on the occasion of his 60th Birthday. 50 ON THE EXPANSION OF RAMANUJAN S CONTINUED FRACTION Dedicated to George E Andrews on the occasion of his 60th Birthday Introduction The continued fraction R( = 1 + 2 3 1+ 1+ 1+ has become known as Ramanujan