The Hardy-Ramanujan-Rademacher Expansion of p(n)

Transcription

1 The Hardy-Ramanujan-Rademacher Expansion of p(n) Definition 0. A partition of a positive integer n is a nonincreasing sequence of positive integers whose sum is n. We denote the number of partitions of n by p(n). For example, p(4) = 5 since the partitions of 4 are: The partition function has the following generating function: P (q) = p(n)q n = ( q n ) n=0 In this discussion we will focus on the exact formula for the partition function p(n), given by Rademacher: Theorem 0.2 (Rademacher) [ A (n) 2 where p(n) = 2 = A (n) = h d dx n= sinh((/)( 2 3 (x /24)) ] 2 ) (mod ),(h,)= (x /24) 2 ω h, e 2inh/ with ω h, a certain 24th root of unity defined as follows. In order to investigate the proof of the above theorem, we need to introduce some tools. x=n Definition 0.3. For example,. F N := { } h : h, N, (h, ) = [0, ] F 3 = {0, 3, 2, 23 }, Theorem 0.4 (Farey fractions) If h h are successive terms in F N, then the rational number with the least denominator lying strictly between h h is h+h + (mediant). 0-

2 Definition 0.5 { ( ) ω h, = h e ( i( 4 (2 h h)+ 2 ( )(2h h +h 2 h )) if h odd, ( h ) e ( i( 4 ( )+ 2 ( )(2h h +h 2 h )) if odd with (a/b) the Legendre symbol. Theorem 0.6 (Knopp, 970) If Rez > 0 h is a solution to hh (mod ), P (e 2i(h+iz)/ ) = ω h, z 2 e [(z z)/2] P (e [2i(h +iz )/] ). Bac to the exact formula for the partition function, let P (q) = p(n)q n = ( q n ). n=0 Each partial product N n= ( xn ) has a pole of order i at x = e. As z 0 (Rez > 0), the term e 2i(h +iz )/ 0. By theorem 0.6, n= P (e 2ih/ 2z/ ) ω h, z 2 e (z z ) 2. Goal: We want to divide the circle of integration into segments depending on which rational point e 2ih/ we are near. We can loo at the discreet set of those rational points e 2ih/ with 0 < N, a fixed positive integer, i.e the set of proper Farey fractions of order N. The idea is we want to use the mediants as the end points for an intervals to get a natural dissection of C. If h 0 / 0, h/, h / are three successive terms in F N, we write: θ 0, = N + ; By the residue theorem, θ h, = h h 0 + h 0 + h, h > 0; θ h, = h + h + h h. p(n) = P (x) dx. 2i c xn+ Let x = ρe 2iφ change the integral into polar coordinates p(n) = ρ n P (ρe 2ρ )(e 2inφ )dφ 0 0-2

3 Choose ρ = e 2. Hence = ρ n =,(h,)= =,(h,)= h, θ h, P (ρe 2ih h, θ h, Let z = iφ apply theorem 0.6 =,(h,)= h, ω h, θ h, +2iφ )e 2inh 2inφ dφ. P (e 2ih 2 ( iφ) )e ( 2inφ) dφ. z (z z 2 e ) 2 P (e 2i( h +iz ) )e 2inφ dφ. As z 0 with Rez > 0, e 2i(h +iz )/ 0 rapidly. Replace the integr P (x) by ( + (P (x) )) we have =,(h,)= h, ω h, θ h, z (z z 2 e ) 2 e 2inφ dφ +e 2n Let Σ 2 = e 2n =,(h,)= Σ = e 2n =,(h,)= h, ω h, =,(h,)= θ h, z (z z 2 e ) 2 h, ω h, h, ω h, θ h, θ h, z (z z 2 e ) 2 We ll show the contribution of Σ 2 is negligible. We have the bound for z = N 2 iφ: P (e 2i( h +iz ) )e 2inφ dφ. z (z z 2 e ) 2 e 2inφ dφ P (e 2i( h +iz ) )e 2inφ dφ. z (z z 2 e ) 2 We have each of θ h, P (e 2i( h +iz ) )e 2inφ z 2 e 2 Re(z ) = m= p(m)e 2Re(z )( m /24 ). θ h, satisfies 2N θ h, N, θ h, φ θ h,. N 2 2 (N 4 + φ 2 ) > N 2 2 N 4 + N 2 = + 2 N

4 z 2 = ( 2 N φ 2 ) 4 < ( 2 N 4 + N 2 ) N 2. Σ 2 e ( 2n ) e ( 2n ) e ( =,(h,)= 2 ) 2 4 N N 2 e ( 2 ) m= p(m)e (m 24 ) m= CN 2 e 2n. p(m)e (m 24 ) h, =,(h,)= h, This approaches to 0 as N where C is a constant. In the integral Σ, we change variables to ω, where ω = N 2 iφ Σ = e 2n 2 i N 2 iθ h, N 2 +iθ h, g(ω)dω θ h, θ h, where g(ω) = ω 2 e 2(n 24 )ω+ 2 2 ω. The integr is single valued analytic in the complex plane, so by Cauchy s theorem we may rewrite it as dφ. dφ. Σ = e 2n 2 i ( 0+ iθ h, N 2 iθ h, +iθ h, iθ h, N 2 +iθ h, +iθ h, Σ = e 2n 2 i (L I I 2 I 3 I 4 I 5 I 6 ). The integrals are demonstrated in the figure below, where L is the loop integral along the contour L. )g(ω)dω. 0-4

5 I 2, I 3, I 4, I 5 can be shown to be negligible. The integrals I I 6 are not negligible; however, I +I 6 = u e i 2 e 2 2 u +2(n 24 )u du+ = 2i ɛ t 2 e 2(n 24 )t = 2iH. u e i 2 e 2 2 u +2(n 24 )u du 2 2 t dt Combining the results of Σ Σ 2 we get an expression for p(n) Set ψ (n) = 2 i L H, then p(n) = N = A (n)ψ (n) + O[N 2 e 2n ] + O(N 2 ). Note that the error terms here are all 0 as N. We state the following theorem, which completes the proof Theorem 0.7 ψ (n) = 2 2 [ d dx sinh((/)( 2 3 (x /24)) ] 2 ) (x /24) 2 x=n. 0-5