Partition Numbers. Kevin Y.X. Wang. September 5, School of Mathematics and Statistics The University of Sydney
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1 Partition Numbers Kevin Y.X. Wang School of Mathematics and Statistics The University of Sydney September 5, 2017
2 Outline 1 Introduction 2 Generating function of partition numbers 3 Jacobi s Triple Product Identity 4 Recurrence formula of partition numbers 5 Proof of JTPI
3 Find the pattern (Christmas trees?)
4 Find the pattern (Christmas trees?)
5 Partition Number Definition Partition number of n Z +, written as p(n), is a way of writing n as sums of positive integers.
6 Partition Number Definition Partition number of n Z +, written as p(n), is a way of writing n as sums of positive integers. Example Take example of n = 4: So p(4) = 5.
7 Some observations p(n) 1. Strictly increasing. p(n) is hard to compute for a given n.
8 Some observations p(n) 1. Strictly increasing. p(n) is hard to compute for a given n. Not too crazy to think that p(n) can be related to p(n 1), p(n 2),... p(1). i.e. recursive formula! But it is not clear how we should add up these numbers to get p(n).
9 Obvious observations Leonhard Euler: obviously p(n) x n 1 = 1 x k. n=1 k=1
10 Obvious observations Leonhard Euler: obviously p(n) x n 1 = 1 x k. n=1 k=1 Leonhard Euler: again, obviously p(n) = p(n 1) + p(n 2) p(n 5) p(n 7) + p(n 12) + p(n 15). These numbers are pentagonal numbers, g k = 1 2 k(3k 1)
11 Obvious observations Leonhard Euler: obviously p(n) x n 1 = 1 x k. n=1 k=1 Leonhard Euler: again, obviously p(n) = p(n 1) + p(n 2) p(n 5) p(n 7) + p(n 12) + p(n 15). These numbers are pentagonal numbers, g k = 1 2k(3k 1) Srinivasa Ramanujan: obviously p (5n + 4) 0 (mod 5).
12 Obviously not obvious results 1. How should we explore the structure of this sequence? 2. Why and how does the pentagonal numbers come up in the study of partition numbers? 3. Sufficient conditions for divisibility of p(n). 4. An unified theorem that explains all of the above.
13 Outline 1 Introduction 2 Generating function of partition numbers 3 Jacobi s Triple Product Identity 4 Recurrence formula of partition numbers 5 Proof of JTPI
14 GF of partition numbers What is the coefficient of x 4 in following expansion? (1 + x)(1 + x 2 )(1 + x 3 )(1 + x 4 )(1 + x 5 )(1 + x 6 )(1 + x 7 )... Equal to the number of ways of getting x 4, since all coefficients are 1.
15 GF of partition numbers What is the coefficient of x 4 in following expansion? (1 + x)(1 + x 2 )(1 + x 3 )(1 + x 4 )(1 + x 5 )(1 + x 6 )(1 + x 7 )... Equal to the number of ways of getting x 4, since all coefficients are 1. Consider each bracket, and choose a single term from each.
16 GF of partition numbers What is the coefficient of x 4 in following expansion? (1 + x)(1 + x 2 )(1 + x 3 )(1 + x 4 )(1 + x 5 )(1 + x 6 )(1 + x 7 )... Equal to the number of ways of getting x 4, since all coefficients are 1. Consider each bracket, and choose a single term from each. Anything beyond (1 + x 4 ) will be irrelevant, so: 1, 1, 1, x 4 x, 1, x 3, 1 Thus coefficient of x 4 is 2
17 GF of partition numbers What is the coefficient of x 4 in following expansion? (1 + x + x 2 )(1 + x 2 + x 4 )(1 + x 3 + x 6 )(1 + x 4 + x 8 )... =(1 + x 1 + x 1+1 )(1 + x 2 + x 2+2 )(1 + x 3 + x 3+3 )(1 + x 4 + x 4+4 )...
18 GF of partition numbers What is the coefficient of x 4 in following expansion? =(1 + x 1 + x 1+1 )(1 + x 2 + x 2+2 )(1 + x 3 + x 3+3 )(1 + x 4 + x 4+4 )... x 4 x 3 x 1 x 2+2 x 2 x 1+1
19 GF of partition numbers What is the coefficient of x 4 in following expansion? =(1 + x 1 + x 1+1 )(1 + x 2 + x 2+2 )(1 + x 3 + x 3+3 )(1 + x 4 + x 4+4 )... x 4 x 3 x 1 x 2+2 x 2 x 1+1
20 GF of partition numbers What is the coefficient of x 4 in following expansion? =(1 + x 1 + x 1+1 )(1 + x 2 + x 2+2 )(1 + x 3 + x 3+3 )(1 + x 4 + x 4+4 )... x 4 x 3 x 1 x 2+2 x 2 x 1+1
21 GF of partition numbers What is the coefficient of x 4 in following expansion? =(1 + x 1 + x 1+1 )(1 + x 2 + x 2+2 )(1 + x 3 + x 3+3 )(1 + x 4 + x 4+4 )... x 4 x 3 x 1 x 2+2 x 2 x 1+1
22 GF of partition numbers What is the coefficient of x 4 in following expansion? =(1 + x 1 + x 1+1 )(1 + x 2 + x 2+2 )(1 + x 3 + x 3+3 )(1 + x 4 + x 4+4 )... x 4 x 3 x 1 x 2+2 x 2 x 1+1 Each of these terms directly correspond to a partition of 4, with the restriction each part appears at most twice.
23 GF of partition numbers What is the coefficient of x 4 in following expansion? (1+x+x 2 + +x m )(1+x 2 +x 4 + +x 2m )(1+x 3 +x 6 + +x 3m )(1+x 4 +x 8 + +x 4m )... Using the same idea, each of these terms directly correspond to a partition of 4, with the restriction each part appears at most m times.
24 GF of partition numbers What is the coefficient of x 4 in following expansion? (1+x+x 2 + +x m )(1+x 2 +x 4 + +x 2m )(1+x 3 +x 6 + +x 3m )(1+x 4 +x 8 + +x 4m )... Using the same idea, each of these terms directly correspond to a partition of 4, with the restriction each part appears at most m times. Coefficient of x n is equal to partition of n, with the restriction each part appears at most m times. Let m, we would remove this restriction and obtain the unrestricted partition number, p(n) as the coefficient of x n
25 GF of partition numbers p(n) x n = (1 + x + x )(1 + x 2 + x )(1 + x 3 + x )..., n=1 = 1 1 x = k=1 1 1 x x k. 1 1 x 3...,
26 GF of partition numbers p(n) x n = (1 + x + x )(1 + x 2 + x )(1 + x 3 + x )..., n=1 Euler is happy = 1 1 x = k=1 1 1 x x k. 1 1 x 3...,
27 GF of partition numbers p(n) x n = (1 + x + x )(1 + x 2 + x )(1 + x 3 + x )..., n=1 Euler is happy = 1 1 x = k=1 1 1 x x k. 1 1 x 3..., GF often allow extra insights into sequences
28 GF of partition numbers p(n) x n = (1 + x + x )(1 + x 2 + x )(1 + x 3 + x )..., n=1 = 1 1 x = k=1 1 1 x x k. 1 1 x 3..., Euler is happy GF often allow extra insights into sequences This is not quite as satisfying, how about this? ( ) ( ) 1 = p(n) x n 1 x k n=1 k=1 (1)
29 Outline 1 Introduction 2 Generating function of partition numbers 3 Jacobi s Triple Product Identity 4 Recurrence formula of partition numbers 5 Proof of JTPI
30 How to break down the infinite product 1 hour lecture of contents. Euler s Pentagonal Number Theorem breaks down this product as:
31 How to break down the infinite product 1 hour lecture of contents. Euler s Pentagonal Number Theorem breaks down this product as: p(n) q n ( = 1 q k ) = ( 1) n q 1 2 n(3n 1). (2) The proof of EPNT is very easy n=1 k=1 n=
32 How to break down the infinite product 1 hour lecture of contents. Euler s Pentagonal Number Theorem breaks down this product as: p(n) q n ( = 1 q k ) = ( 1) n q 1 2 n(3n 1). (2) n=1 k=1 n= The proof of EPNT is very easy... Provided that you can prove something stronger Jacobi s Triple Product Identity ( 1 + xq k ) ( 1 + x 1 q k 1) ( 1 q k) = q 1 2 n(n+1) x n. (3) k=1 n=
33 Proof of EPNT Assume JTPI is true: ( 1 + xq k ) ( 1 + x 1 q k 1) ( 1 q k) = q 1 2 n(n+1) x n k=1 n= Substituting q q 3 and x q 1 (1 q 3k 2 )(1 q 3k 1 )(1 q 3k ) = ( 1) n q 1 2 n(3n 1). k=1 n= But the LHS has consecutive exponents, ( 1 q k ) = ( 1) n q 1 2 n(3n 1). Done! k=1 n=
34 Alert!
35 Outline of proof for JTPI Jacobi s Triple Product Identity ( 1 + xq k ) ( 1 + x 1 q k 1) ( 1 q k) = k=1 n= q 1 2 n(n+1) x n.
36 Outline of proof for JTPI Jacobi s Triple Product Identity ( 1 + xq k ) ( 1 + x 1 q k 1) ( 1 q k) = k=1 We will first define: i=1 n= q 1 2 n(n+1) x n. ( f(x) := 1 + xq i ) ( 1 + x 1 q j 1). (4) j=1 The proof then follows this simple structure: 1. Collapsing f(x) into a 0 (q) RHS
37 Outline of proof for JTPI Jacobi s Triple Product Identity ( 1 + xq k ) ( 1 + x 1 q k 1) ( 1 q k) = k=1 We will first define: i=1 n= q 1 2 n(n+1) x n. ( f(x) := 1 + xq i ) ( 1 + x 1 q j 1). (4) j=1 The proof then follows this simple structure: 1. Collapsing f(x) into a 0 (q) RHS 2. a 0 = m=0 b m q m
38 Outline of proof for JTPI Jacobi s Triple Product Identity ( 1 + xq k ) ( 1 + x 1 q k 1) ( 1 q k) = k=1 We will first define: i=1 n= q 1 2 n(n+1) x n. ( f(x) := 1 + xq i ) ( 1 + x 1 q j 1). (4) j=1 The proof then follows this simple structure: 1. Collapsing f(x) into a 0 (q) RHS 2. a 0 = m=0 b m q m 3. Realise the formation of a 0 require cancellation of powers of q, which forms a special partition of m
39 Outline of proof for JTPI Jacobi s Triple Product Identity ( 1 + xq k ) ( 1 + x 1 q k 1) ( 1 q k) = k=1 We will first define: i=1 n= q 1 2 n(n+1) x n. ( f(x) := 1 + xq i ) ( 1 + x 1 q j 1). (4) j=1 The proof then follows this simple structure: 1. Collapsing f(x) into a 0 (q) RHS 2. a 0 = m=0 b m q m 3. Realise the formation of a 0 require cancellation of powers of q, which forms a special partition of m 4. Create a pictorial bijection between p(n) and b m for all n, m
40 Outline of proof for JTPI Jacobi s Triple Product Identity ( 1 + xq k ) ( 1 + x 1 q k 1) ( 1 q k) = k=1 We will first define: i=1 n= q 1 2 n(n+1) x n. ( f(x) := 1 + xq i ) ( 1 + x 1 q j 1). (4) j=1 The proof then follows this simple structure: 1. Collapsing f(x) into a 0 (q) RHS 2. a 0 = m=0 b m q m 3. Realise the formation of a 0 require cancellation of powers of q, which forms a special partition of m 4. Create a pictorial bijection between p(n) and b m for all n, m 5. Thus, f(x) can be expressed as two infinite product as well as the power series of p(n) (inverse of the third infinite product on LHS) and RHS.
41 Outline 1 Introduction 2 Generating function of partition numbers 3 Jacobi s Triple Product Identity 4 Recurrence formula of partition numbers 5 Proof of JTPI
42 Deriving the recurrence formula 1 1 q k = p(n) q n k=1 n=1 ( ) ( ) 1 = p(n) q n 1 q k n=1 k=1 ( ) ( 1 = p(n) q n n=1 n= ( ) ( 1 = p(n) q n 1 + n=1 ( 1) n q 1 2 n(3n 1) ) m=1 (by applying EPNT) [ ] ) ( 1) m q m(3m 1) 2 + q m(3m+1) 2.
43 Deriving the recurrence formula 1 1 q k = p(n) q n k=1 n=1 ( ) ( ) 1 = p(n) q n 1 q k n=1 k=1 ( ) ( 1 = p(n) q n n=1 n= ( ) ( 1 = p(n) q n 1 + n=1 But multiplying power series is easy! ( ) ( a n q n n=0 m=0 ( 1) n q 1 2 n(3n 1) ) m=1 b m q m ) = (by applying EPNT) [ ] ) ( 1) m q m(3m 1) 2 + q m(3m+1) 2. ( t ) a t i b i q t. t=0 i=0
44 Deriving the recurrence formula ( ) ( 1 = p(n) q n 1 + n=1 m=1 [ ] ) ( 1) m q m(3m 1) 2 + q m(3m+1) 2. By multiplication of power series, and equating the coefficient for q t, t > 0: 0 = p (t) + { ( ( 1) m p t m 1 ) ( m(3m 1) + p t 2 This equation becomes well-defined if we set p(t) = 0 for all t < 0. )} m(3m + 1). 2
45 Deriving the recurrence formula ( ) ( 1 = p(n) q n 1 + n=1 m=1 [ ] ) ( 1) m q m(3m 1) 2 + q m(3m+1) 2. By multiplication of power series, and equating the coefficient for q t, t > 0: 0 = p (t) + { ( ( 1) m p t m 1 ) ( m(3m 1) + p t 2 )} m(3m + 1). 2 This equation becomes well-defined if we set p(t) = 0 for all t < 0. E.g. Consider t = 6, p(6) = {p(6 1) + p(6 2)} {p(6 5) + p(6 7)} = p(5) + p(4) p(1) p( 1) = = 11
46 Sure... That was obvious...
47 Sketch of proof for Ramanujan s congruences Start with JTPI, make specific substitutions like we did with EPNT ] The coefficient of x 5m+5 in x is divisible by 5 by modular arithmetic [ (1 x 5 )(1 x 10 )... (1 x)(1 x 2 )...
48 Sketch of proof for Ramanujan s congruences Start with JTPI, make specific substitutions like we did with EPNT ] The coefficient of x 5m+5 in x is divisible by 5 by modular arithmetic [ (1 x 5 )(1 x 10 )... (1 x)(1 x 2 )... This forces the coefficient of x 5m+5 in x k=1 1 1 x k = x (1 x)(1 x 2 )... [ (1 x 5 )(1 x 10 )... = x (1 x)(1 x 2 )... [ (1 x 5 )(1 x 10 )... = x (1 x)(1 x 2 )... ] 1 1 (1 x 5 ) (1 x 10 )... ] ( 1 + x 5 + x ) ( 1 + x 10 + x )...
49 Sketch of proof for Ramanujan s congruences Start with JTPI, make specific substitutions like we did with EPNT ] The coefficient of x 5m+5 in x is divisible by 5 by modular arithmetic [ (1 x 5 )(1 x 10 )... (1 x)(1 x 2 )... This forces the coefficient of x 5m+5 in x k=1 1 1 x k = x (1 x)(1 x 2 )... [ (1 x 5 )(1 x 10 )... = x (1 x)(1 x 2 )... [ (1 x 5 )(1 x 10 )... = x (1 x)(1 x 2 )... ] 1 1 (1 x 5 ) (1 x 10 )... ] ( 1 + x 5 + x ) ( 1 + x 10 + x )...
50 Sketch of proof for Ramanujan s congruences Start with JTPI, make specific substitutions like we did with EPNT ] The coefficient of x 5m+5 in x is divisible by 5 by modular arithmetic [ (1 x 5 )(1 x 10 )... (1 x)(1 x 2 )... This forces the coefficient of x 5m+5 in x x k=1 1 1 x k = x (1 x)(1 x 2 )... [ (1 x p(n) x n 5 )(1 x 10 )... = x (1 x)(1 x 2 )... n=1 [ (1 x 5 )(1 x 10 )... = x (1 x)(1 x 2 )... ] 1 1 (1 x 5 ) (1 x 10 )... ] ( 1 + x 5 + x ) ( 1 + x 10 + x )...
51 Sketch of proof for Ramanujan s congruences Start with JTPI, make specific substitutions like we did with EPNT ] The coefficient of x 5m+5 in x is divisible by 5 by modular arithmetic [ (1 x 5 )(1 x 10 )... (1 x)(1 x 2 )... This forces the coefficient of x 5m+5 in x n=1 k=1 1 1 x k = x (1 x)(1 x 2 )... [ (1 x x p(n) x n 5 )(1 x 10 )... = x (1 x)(1 x 2 )... n=1 [ (1 x p(n) x n+1 5 )(1 x 10 )... = x (1 x)(1 x 2 )... ] 1 1 (1 x 5 ) (1 x 10 )... ] ( 1 + x 5 + x ) ( 1 + x 10 + x )...
52 Sketch of proof for Ramanujan s congruences Start with JTPI, make specific substitutions like we did with EPNT ] The coefficient of x 5m+5 in x is divisible by 5 by modular arithmetic [ (1 x 5 )(1 x 10 )... (1 x)(1 x 2 )... This forces the coefficient of x 5m+5 in x n=1 k=1 1 1 x k = x (1 x)(1 x 2 )... [ (1 x x p(n) x n 5 )(1 x 10 )... = x (1 x)(1 x 2 )... n=1 [ (1 x p(n) x n+1 5 )(1 x 10 )... = x (1 x)(1 x 2 )... ] 1 1 (1 x 5 ) (1 x 10 )... ] ( 1 + x 5 + x ) ( 1 + x 10 + x )... Equating the power of x gives 5m + 5 = n + 1 = n = 5m + 4. So, p (5n + 4) 0 (mod 5)
53 Final remarks Theorem (Ramanujan s Congruences) For n N p (5n + 4) 0 (mod 5) p (7n + 5) 0 (mod 7) p (11n + 6) 0 (mod 11)
54 Final remarks Theorem (Ramanujan s Congruences) For n N p (5n + 4) 0 (mod 5) p (7n + 5) 0 (mod 7) p (11n + 6) 0 (mod 11) p( k + 237) 0 (mod 13)
55 Final remarks Theorem (Ramanujan s Congruences) For n N p (5n + 4) 0 (mod 5) p (7n + 5) 0 (mod 7) p (11n + 6) 0 (mod 11) p( k + 237) 0 (mod 13) In 2001, Ken Ono proved such congruence relation exist for every prime greater than 3 In 2006, Ono further proved if N is a integer coprime to 6, then there are integers a and b such that: p(am + b) 0 (mod N). (5) Partition numbers and their approximations can be used in quantum physics.
56 Outline 1 Introduction 2 Generating function of partition numbers 3 Jacobi s Triple Product Identity 4 Recurrence formula of partition numbers 5 Proof of JTPI
57 Outline of proof for JTPI Jacobi s Triple Product Identity ( 1 + xq k ) ( 1 + x 1 q k 1) ( 1 q k) = q 1 2 n(n+1) x n. k=1 n= We will first define: ( f(x) := 1 + xq i ) ( 1 + x 1 q j 1). (6) i=1 j=1 The proof then follows this simple structure 1. Collapsing f(x) into just one term 2. Staring at that one term very hard and apply the previous counting exponent idea
58 Symmetry of f(x) ( f(x) = 1 + xq i ) ( 1 + x 1 q j 1) i=1 j=1 = (1 + xq)(1 + xq 2 )(1 + xq 3 )... (1 + x 1 )(1 + x 1 q)(1 + x 1 q 2 )...
59 Symmetry of f(x) ( f(x) = 1 + xq i ) ( 1 + x 1 q j 1) i=1 j=1 = (1 + xq)(1 + xq 2 )(1 + xq 3 )... (1 + x 1 )(1 + x 1 q)(1 + x 1 q 2 )... Via substitution, f(xq) = (1 + xq 2 )(1 + xq 3 )(1 + xq 4 )... (1 + x 1 q 1 )(1 + x 1 )(1 + x 1 q)... (7)
60 Symmetry of f(x) Via substitution, ( f(x) = 1 + xq i ) ( 1 + x 1 q j 1) i=1 j=1 = (1 + xq)(1 + xq 2 )(1 + xq 3 )... (1 + x 1 )(1 + x 1 q)(1 + x 1 q 2 )... f(xq) = (1 + xq 2 )(1 + xq 3 )(1 + xq 4 )... (1 + x 1 q 1 )(1 + x 1 )(1 + x 1 q)... = (1 + x 1 q 1 (1 + xq) ) (1 + xq) (1 + xq2 )(1 + xq 3 )(1 + xq 4 )... (1 + x 1 )(1 + x 1 q)... (7)
61 Symmetry of f(x) Via substitution, ( f(x) = 1 + xq i ) ( 1 + x 1 q j 1) i=1 j=1 = (1 + xq)(1 + xq 2 )(1 + xq 3 )... (1 + x 1 )(1 + x 1 q)(1 + x 1 q 2 )... f(xq) = (1 + xq 2 )(1 + xq 3 )(1 + xq 4 )... (1 + x 1 q 1 )(1 + x 1 )(1 + x 1 q)... = (1 + x 1 q 1 (1 + xq) ) (1 + xq) (1 + xq2 )(1 + xq 3 )(1 + xq 4 )... (1 + x 1 )(1 + x 1 q)... = (1 + x 1 q 1 ) (1 + xq) f(x) (7)
62 Symmetry of f(x) Via substitution, ( f(x) = 1 + xq i ) ( 1 + x 1 q j 1) i=1 j=1 = (1 + xq)(1 + xq 2 )(1 + xq 3 )... (1 + x 1 )(1 + x 1 q)(1 + x 1 q 2 )... f(xq) = (1 + xq 2 )(1 + xq 3 )(1 + xq 4 )... (1 + x 1 q 1 )(1 + x 1 )(1 + x 1 q)... = (1 + x 1 q 1 (1 + xq) ) (1 + xq) (1 + xq2 )(1 + xq 3 )(1 + xq 4 )... (1 + x 1 )(1 + x 1 q)... = (1 + x 1 q 1 ) (1 + xq) = x 1 q 1 f(x) f(x) (7)
63 f(x) as a formal Laurent series What is f(x) if not just a bunch of x n with coefficients at the front? f(x) = n= a n x n.
64 f(x) as a formal Laurent series What is f(x) if not just a bunch of x n with coefficients at the front? f(x) = n= a n x n. Substituting x for xq, and using f(xq) = x 1 q 1 f(x), we have: f(xq) = a n q n x n n=
65 f(x) as a formal Laurent series What is f(x) if not just a bunch of x n with coefficients at the front? f(x) = n= a n x n. Substituting x for xq, and using f(xq) = x 1 q 1 f(x), we have: f(xq) = a n q n x n = n= n= a n q 1 x n 1.
66 f(x) as a formal Laurent series What is f(x) if not just a bunch of x n with coefficients at the front? f(x) = n= a n x n. Substituting x for xq, and using f(xq) = x 1 q 1 f(x), we have: f(xq) = a n q n x n = n= n= a n q 1 x n 1. Equating coefficient of x n 1, we have:
67 f(x) as a formal Laurent series What is f(x) if not just a bunch of x n with coefficients at the front? f(x) = n= a n x n. Substituting x for xq, and using f(xq) = x 1 q 1 f(x), we have: f(xq) = a n q n x n = n= n= a n q 1 x n 1. Equating coefficient of x n 1, we have: a n = q n a n 1.
68 JTPI step 1 proven a n = q n a n 1 = q n (q n 1 a n 2 ) = q n q n 1 (q n 2 a n 3 ). = q n+(n 1)+(n 2)+ +1 a 0 = q 1 2 n(n+1) a 0
69 JTPI step 1 proven a n = q n a n 1 = q n (q n 1 a n 2 ) = q n q n 1 (q n 2 a n 3 ). = q n+(n 1)+(n 2)+ +1 a 0 = q 1 2 n(n+1) a 0 Thus, f(x) = a n x n n= = a 0 q 1 2 n(n+1) x n. n=
70 Sketch of Step 2 a 0 is simply the coefficient of x 0 in the expansion of ( f(x) = 1 + xq i ) ( 1 + x 1 q j 1) i=1 j=1 = (1 + xq)(1 + xq 2 )(1 + xq 3 )(1 + xq 4 )... (1 + x 1 )(1 + x 1 q)(1 + x 1 q 2 )(1 + x 1 q 3 )...
71 Sketch of Step 2 a 0 is simply the coefficient of x 0 in the expansion of ( f(x) = 1 + xq i ) ( 1 + x 1 q j 1) i=1 j=1 = (1 + xq)(1 + xq 2 )(1 + xq 3 )(1 + xq 4 )... (1 + x 1 )(1 + x 1 q)(1 + x 1 q 2 )(1 + x 1 q 3 )... As long as we have an equal number of terms from the first and second infinite product, we are guaranteed the cancellation of powers of x
72 Sketch of Step 2 a 0 is simply the coefficient of x 0 in the expansion of ( f(x) = 1 + xq i ) ( 1 + x 1 q j 1) i=1 j=1 = (1 + xq)(1 + xq 2 )(1 + xq 3 )(1 + xq 4 )... (1 + x 1 )(1 + x 1 q)(1 + x 1 q 2 )(1 + x 1 q 3 )... As long as we have an equal number of terms from the first and second infinite product, we are guaranteed the cancellation of powers of x A power series in q will capture different selection of terms: a 0 = b m q m. (8) m=0
73 Sketch of Step 2 ( f(x) = 1 + xq i ) ( 1 + x 1 q j 1) i=1 j=1 = (1 + xq)(1 + xq 2 )(1 + xq 3 )(1 + xq 4 )... (1 + x 1 )(1 + x 1 q)(1 + x 1 q 2 )(1 + x 1 q 3 )... Consider m = 4. Keep in mind we need an equal number of terms from the first and second product in f(x), the only possible ways to obtain x 0 q 4 are: (1 + xq) (1 + x 1 q 3 ) (1 + xq 2 ) (1 + x 1 q 2 ) (1 + xq 3 ) (1 + x 1 q) (1 + xq 4 ) (1 + x 1 ) (1 + xq)(1 + xq 2 ) (1 + x 1 )(1 + x 1 q).
74 A special partition In particular, the number m is partitioned into the sum of distinct elements from the positive integers plus an equal number of distinct elements from the natural numbers (including 0), i.e. m = (i 1 + i i n ) + (j 1 + j j n ); (9) where: i 1 < i 2 < < i n Z +, and j 1 < j 2 < < j n N. The number of ways of achieving this partition for a fixed m is the same as p(m).
75 Ferrers Diagram Figure: 19 = = ( ) + ( ) Strictly decreasing sequence is guaranteed by the diagonal Equal number of parts is guaranteed by the use of 0 in the lower half
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