The Rogers-Ramanujan Functions and Computer Algebra

Transcription

1 Ramanujan Congruences / 1 Joint Mathematics Meetings AMS Special Session in honor of Dennis Stanton s 65th Birthday San Diego, Jan 10 13, 2018 The Rogers-Ramanujan Functions and Computer Algebra Peter Paule (joint work with S. Radu) Johannes Kepler University Linz Research Institute for Symbolic Computation (RISC)

2 Ramanujan Congruences / Prelude: Felix Klein s icosahedron 2 Prelude: Felix Klein s icosahedron

3 Ramanujan Congruences / Prelude: Felix Klein s icosahedron 3 G := {3D rotations g : g fixes the icosahedron 1 } 1

4 Ramanujan Congruences / Prelude: Felix Klein s icosahedron 3 G := {3D rotations g : g fixes the icosahedron 1 } By stereographic projection: rotation g G αz + β γz + δ Moeb G 1

5 Ramanujan Congruences / Prelude: Felix Klein s icosahedron 3 G := {3D rotations g : g fixes the icosahedron 1 } By stereographic projection: K := { f C(z) : f rotation g G αz + β γz + δ Moeb G ( ) αz + β = f(z) for all γz + δ } αz + β γz + δ Moeb G 1

6 Ramanujan Congruences / Prelude: Felix Klein s icosahedron 4 K := { f C(z) : f ( ) αz + β = f(z) for all γz + δ } αz + β γz + δ Moeb G Felix Klein ( ) found:

7 Ramanujan Congruences / Prelude: Felix Klein s icosahedron 4 K := { f C(z) : f ( ) αz + β = f(z) for all γz + δ } αz + β γz + δ Moeb G Felix Klein ( ) found: where K = C(J(z))

8 Ramanujan Congruences / Prelude: Felix Klein s icosahedron 4 K := { f C(z) : f ( ) αz + β = f(z) for all γz + δ } αz + β γz + δ Moeb G Felix Klein ( ) found: where K = C(J(z)) J(z) = (z20 228z z z 5 + 1) 3 z 5 (z z 5 1) 5.

9 Ramanujan Congruences / An excercise from a combinatorics course 5 An excercise from a combinatorics course

10 Ramanujan Congruences / An excercise from a combinatorics course Write a program which lists all partitions of n a) into parts which are congruent to 1 or 4 mod 5, and b) into parts whose differences are at least two. Investigate your data. If a) is replaced by 2 or 3 mod 5, can you find an appropriate b)?

11 Ramanujan Congruences / An excercise from a combinatorics course 7 Consider a 1,a 2 1 a 1 a 2 2 x a 1 1 xa 2 2 qa 1+a 2.

12 Ramanujan Congruences / An excercise from a combinatorics course 7 Consider a 1,a 2 1 a 1 a 2 2 x a 1 1 xa 2 2 qa 1+a 2. In[5]:= << RISC Omega Omega Package version 2.49 written by Axel Riese (in cooperation with George E. Andrews and Peter Paule) c RISC-JKU In[6]:= OR[OSum[x1 a1 x2 a2 q a1+a2, {a1 a2 2, a2 1}, λ]] Out[6]= x1 3 x2 q 4 (1 x1 q)(1 x1 x2 q 2 ) In[7]:= Series[%, {q, 0, 9}]

13 Ramanujan Congruences / An excercise from a combinatorics course 7 Consider a 1,a 2 1 a 1 a 2 2 x a 1 1 xa 2 2 qa 1+a 2. In[9]:= << RISC Omega Omega Package version 2.49 written by Axel Riese (in cooperation with George E. Andrews and Peter Paule) c RISC-JKU In[10]:= OR[OSum[x1 a1 x2 a2 q a1+a2, {a1 a2 2, a2 1}, λ]] Out[10]= x1 3 x2 q 4 (1 x1 q)(1 x1 x2 q 2 ) In[11]:= Series[%, {q, 0, 9}] Out[11]= x1 3 x2 q 4 + x1 4 x2 q 5 + ( x1 5 x2 + x1 4 x2 2) q 6 + ( x1 6 x2 + x1 5 x2 2) q 7 + ( x1 7 x2 + x1 6 x2 2 + x1 5 x2 3) q 8 + ( x1 8 x2 + x1 7 x2 2 + x1 6 x2 3) q 9 + O[q] 10 In[12]:= %/.{x1 1, x2 1} Out[12]= q 4 + q 5 + 2q 6 + 2q 7 + 3q 8 + 3q 9 + O[q] 10

14 Ramanujan Congruences / An excercise from a combinatorics course 8 Consider a 1,a 2,a 3 1 a 1 a 2 2,a 2 a 3 2 x a 1 1 xa 2 2 xa 3 3 qa 1+a 2 +a 3.

15 Ramanujan Congruences / An excercise from a combinatorics course 8 Consider a 1,a 2,a 3 1 a 1 a 2 2,a 2 a 3 2 x a 1 1 xa 2 2 xa 3 3 qa 1+a 2 +a 3. In[15]:= OR[OSum[x1 a1 x2 a2 x3 a3 q a1+a2+a3, a1 a2 2, a2 a3 2, a3 1, l]] Out[15]= q 9 x1 5 x2 3 x3 (1 qx1) (1 q 2 x1x2) (1 q 3 x1x2x3)

16 Ramanujan Congruences / An excercise from a combinatorics course 8 Consider a 1,a 2,a 3 1 a 1 a 2 2,a 2 a 3 2 x a 1 1 xa 2 2 xa 3 3 qa 1+a 2 +a 3. In[17]:= OR[OSum[x1 a1 x2 a2 x3 a3 q a1+a2+a3, a1 a2 2, a2 a3 2, a3 1, l]] Out[17]= q 9 x1 5 x2 3 x3 (1 qx1) (1 q 2 x1x2) (1 q 3 x1x2x3) In[18]:= Series[%, {q, 0, 11}] Out[18]= x1 5 x2 3 x3 q 9 + x1 6 x2 3 x3 q 10 + ( x1 7 x2 3 x3 + x1 6 x2 4 x3 ) q 11 + O[q] 12

17 Ramanujan Congruences / An excercise from a combinatorics course 8 Consider a 1,a 2,a 3 1 a 1 a 2 2,a 2 a 3 2 x a 1 1 xa 2 2 xa 3 3 qa 1+a 2 +a 3. In[19]:= OR[OSum[x1 a1 x2 a2 x3 a3 q a1+a2+a3, a1 a2 2, a2 a3 2, a3 1, l]] Out[19]= q 9 x1 5 x2 3 x3 (1 qx1) (1 q 2 x1x2) (1 q 3 x1x2x3) In[20]:= Series[%, {q, 0, 11}] Out[20]= x1 5 x2 3 x3 q 9 + x1 6 x2 3 x3 q 10 + ( x1 7 x2 3 x3 + x1 6 x2 4 x3 ) q 11 + O[q] 12 For the counting functions; i.e., 1 = x 1 = x 2 = x 3 =..., 1, q 1 1 q, q 4 (1 q) (1 q 2 ), q 9 (1 q) (1 q 2 ) (1 q 3 ), a.s.o.

18 Ramanujan Congruences / An excercise from a combinatorics course 9 Consider a 1,a 2,a 3 2 a 1 a 2 2,a 2 a 3 2 x a 1 1 xa 2 2 xa 3 3 qa 1+a 2 +a 3.

19 Ramanujan Congruences / An excercise from a combinatorics course 9 Consider a 1,a 2,a 3 2 a 1 a 2 2,a 2 a 3 2 x a 1 1 xa 2 2 xa 3 3 qa 1+a 2 +a 3. In[22]:= OR[OSum[x1 a1 x2 a2 x3 a3 q a1+a2+a3, a1 a2 2, a2 a3 2, a3 2, l]] Out[22]= q 12 x1 6 x2 4 x3 2 (1 qx1) (1 q 2 x1x2) (1 q 3 x1x2x3)

20 Ramanujan Congruences / An excercise from a combinatorics course 9 Consider a 1,a 2,a 3 2 a 1 a 2 2,a 2 a 3 2 x a 1 1 xa 2 2 xa 3 3 qa 1+a 2 +a 3. In[23]:= OR[OSum[x1 a1 x2 a2 x3 a3 q a1+a2+a3, a1 a2 2, a2 a3 2, a3 2, l]] Out[23]= q 12 x1 6 x2 4 x3 2 (1 qx1) (1 q 2 x1x2) (1 q 3 x1x2x3) For the counting functions; i.e., 1 = x 1 = x 2 = x 3 =..., 1, q1+1 1 q, q 4+2 (1 q) (1 q 2 ), q 9+3 (1 q) (1 q 2 ) (1 q 3 ), a.s.o.

21 Ramanujan Congruences / An excercise from a combinatorics course 10 Consider a 1,a 2,a 3 3 a 1 a 2 2,a 2 a 3 2 x a 1 1 xa 2 2 xa 3 3 qa 1+a 2 +a 3.

22 Ramanujan Congruences / An excercise from a combinatorics course 10 Consider a 1,a 2,a 3 3 a 1 a 2 2,a 2 a 3 2 x a 1 1 xa 2 2 xa 3 3 qa 1+a 2 +a 3. In[25]:= OR[OSum[x1 a1 x2 a2 x3 a3 q a1+a2+a3, a1 a2 2, a2 a3 2, a3 3, l]] Out[25]= q 15 x1 7 x2 5 x3 3 (1 qx1) (1 q 2 x1x2) (1 q 3 x1x2x3)

23 Ramanujan Congruences / An excercise from a combinatorics course 10 Consider a 1,a 2,a 3 3 a 1 a 2 2,a 2 a 3 2 x a 1 1 xa 2 2 xa 3 3 qa 1+a 2 +a 3. In[26]:= OR[OSum[x1 a1 x2 a2 x3 a3 q a1+a2+a3, a1 a2 2, a2 a3 2, a3 3, l]] Out[26]= q 15 x1 7 x2 5 x3 3 (1 qx1) (1 q 2 x1x2) (1 q 3 x1x2x3) For the counting functions; i.e., 1 = x 1 = x 2 = x 3 =..., 1, q1+2 1 q, q 4+4 (1 q) (1 q 2 ), q 9+6 (1 q) (1 q 2 ) (1 q 3 ), a.s.o.

24 Ramanujan Congruences / An excercise from a combinatorics course 11 This gives: The counting (generating) function of partitions into parts with minimal difference two and smallest part m + 1 is k=0 q k2 +mk (q; q) k = R(q m ), where

25 Ramanujan Congruences / An excercise from a combinatorics course 11 This gives: The counting (generating) function of partitions into parts with minimal difference two and smallest part m + 1 is k=0 q k2 +mk (q; q) k = R(q m ), where R(z) := k=0 q k2 z k (q; q) k. Rogers-Ramanujan functions: R(1) and R(q).

26 Ramanujan Congruences / q-shift Equations: Automatic Derivation 12 q-shift Equations: Automatic Derivation

27 Ramanujan Congruences / q-shift Equations: Automatic Derivation 13 Recall R(z) = k=0 f k (z) with f k (z) := qk2 z k (q; q) k. TASK. Determine rational functions r j = r j (z, q) such that r 0 R(z) + r 1 R(q z) + r 2 R(q 2 z) = 0.

28 Ramanujan Congruences / q-shift Equations: Automatic Derivation 13 Recall R(z) = k=0 f k (z) with f k (z) := qk2 z k (q; q) k. TASK. Determine rational functions r j = r j (z, q) such that r 0 R(z) + r 1 R(q z) + r 2 R(q 2 z) = 0. NOTE. The summand is q-hypergeometric: hence also f k+1 (z) f k (z) = q2k+1 z 1 q k+1 Q(z, q)(qk ), r 0 f k (z) + r 1 f k (q z) + r 2 f k (q 2 z) ( f k (q z) = f k (z) r 0 + r 1 f k (z) + r f k (q 2 ) z) 2 f k (z) ( ) = f k (z) r 0 + r 1 q k + r 2 q 2k.

29 Ramanujan Congruences / q-shift Equations: Automatic Derivation 14 TASK. Determine rational functions r j = r j (z, q) such that r 0 R(z) + r 1 R(q z) + r 2 R(q 2 z) = 0. NEW TASK. Determine rational functions r j = r j (z, q) and a q-hypergeometric g k (z) such that r 0 f k (z) + r 1 f k (qz) + r 2 f k (q 2 z) = g k+1 (z) g k (z). Then

30 Ramanujan Congruences / q-shift Equations: Automatic Derivation 14 TASK. Determine rational functions r j = r j (z, q) such that r 0 R(z) + r 1 R(q z) + r 2 R(q 2 z) = 0. NEW TASK. Determine rational functions r j = r j (z, q) and a q-hypergeometric g k (z) such that r 0 f k (z) + r 1 f k (qz) + r 2 f k (q 2 z) = g k+1 (z) g k (z). Then r 0 R(z) + r 1 R(q z) + r 2 R(q 2 z) = r 0 f k (z) + r 1 k=0 k=0 = g (z) g 0 (z). f k (qz) + r 2 k=0 f k (q 2 z)

31 Ramanujan Congruences / q-shift Equations: Automatic Derivation 15 NEW TASK modified: We compute r 0 N f k (z) + r 1 N f k (qz) + r 2 N k=0 k=0 k=0 and send N to : f k (q 2 z) = g N+1 (z) g 0 (z),

32 Ramanujan Congruences / q-shift Equations: Automatic Derivation 15 NEW TASK modified: We compute r 0 N f k (z) + r 1 N f k (qz) + r 2 N k=0 k=0 k=0 and send N to : f k (q 2 z) = g N+1 (z) g 0 (z), In[29]:= << RISC qzeil Package q-zeilberger version 4.50 written by Axel Riese c RISC- JKU [ q k 2 z k { In[30]:= qtelescope, {k, 0, N}, qparameterized 1, q k, q 2k}] (q; q) k

33 Ramanujan Congruences / q-shift Equations: Automatic Derivation 15 NEW TASK modified: We compute r 0 N f k (z) + r 1 N f k (qz) + r 2 N k=0 k=0 k=0 and send N to : f k (q 2 z) = g N+1 (z) g 0 (z), In[31]:= << RISC qzeil Package q-zeilberger version 4.50 written by Axel Riese c RISC- JKU [ q k 2 z k { In[32]:= qtelescope, {k, 0, N}, qparameterized 1, q k, q 2k}] (q; q) k Out[32]= Sum [ F 0(k) + F 1(k) + q z F 2(k), {k, 0, N}] = qn2 +2N+1 z N+1 (q; q) N

34 Ramanujan Congruences / q-shift Equations: Automatic Derivation 15 NEW TASK modified: We compute r 0 N f k (z) + r 1 N f k (qz) + r 2 N k=0 k=0 k=0 and send N to : f k (q 2 z) = g N+1 (z) g 0 (z), In[33]:= << RISC qzeil Package q-zeilberger version 4.50 written by Axel Riese c RISC- JKU [ q k 2 z k { In[34]:= qtelescope, {k, 0, N}, qparameterized 1, q k, q 2k}] (q; q) k Out[34]= Sum [ F 0(k) + F 1(k) + q z F 2(k), {k, 0, N}] = qn2 +2N+1 z N+1 (q; q) N NOTE. F 0 (k) = f k (z), F 1 (k) = q k f k (z) = f k (qz), and F 2 (k) = q 2k f k (z) = f k (q 2 z).

35 Ramanujan Congruences / q-shift Equations: Automatic Derivation 16 RECALL. We computed N N N ( 1) f k (z) + 1 f k (qz) + q z f k (q 2 z) = q(n+1)2 z N+1. (q; q) N k=0 k=0 k=0 For this implies R(z) = k=0 q k2 z k (q; q) k = f k (z), k=0

36 Ramanujan Congruences / q-shift Equations: Automatic Derivation 16 RECALL. We computed N N N ( 1) f k (z) + 1 f k (qz) + q z f k (q 2 z) = q(n+1)2 z N+1. (q; q) N k=0 k=0 k=0 For this implies R(z) = k=0 q k2 z k (q; q) k = f k (z), k=0 R(z) + R(qz) + q z R(q 2 z) = 0.

37 Ramanujan Congruences / q-shift Equation: Application 1 17 q-shift Equation: Application 1

38 Ramanujan Congruences / q-shift Equation: Application 1 18 RECALL: q z R(q 2 z) = R(z) R(qz) and R(q m ) = qr(q 2 ) = R(1) R(q), k=0 q k2 +m k (q; q) k

39 Ramanujan Congruences / q-shift Equation: Application 1 18 RECALL: q z R(q 2 z) = R(z) R(qz) and R(q m ) = qr(q 2 ) = R(1) R(q), q 3 R(q 3 ) = q ( R(q) R(q 2 ) ) = R(1) + (1 + q)r(q), k=0 q k2 +m k (q; q) k

40 Ramanujan Congruences / q-shift Equation: Application 1 18 RECALL: q z R(q 2 z) = R(z) R(qz) and R(q m ) = qr(q 2 ) = R(1) R(q), q 3 R(q 3 ) = q ( R(q) R(q 2 ) ) = R(1) + (1 + q)r(q), q 6 R(q 4 ) = q 3 ( R(q 2 ) R(q 3 ) ) = (1 + q 2 )R(1) (1 + q + q 2 )R(q), k=0 q k2 +m k (q; q) k

41 Ramanujan Congruences / q-shift Equation: Application 1 18 RECALL: q z R(q 2 z) = R(z) R(qz) and R(q m ) = qr(q 2 ) = R(1) R(q), q 3 R(q 3 ) = q ( R(q) R(q 2 ) ) = R(1) + (1 + q)r(q), q 6 R(q 4 ) = q 3 ( R(q 2 ) R(q 3 ) ) = (1 + q 2 )R(1) (1 + q + q 2 )R(q), q 10 R(q 5 ) = q 6 ( R(q 3 ) R(q 4 ) ) k=0 q k2 +m k (q; q) k = (1 + q 2 + q 3 )R(1)+(1 + q + q 2 + q 3 + q 4 )R(q),

42 Ramanujan Congruences / q-shift Equation: Application 1 18 RECALL: q z R(q 2 z) = R(z) R(qz) and R(q m ) = qr(q 2 ) = R(1) R(q), q 3 R(q 3 ) = q ( R(q) R(q 2 ) ) = R(1) + (1 + q)r(q), q 6 R(q 4 ) = q 3 ( R(q 2 ) R(q 3 ) ) = (1 + q 2 )R(1) (1 + q + q 2 )R(q), q 10 R(q 5 ) = q 6 ( R(q 3 ) R(q 4 ) ) k=0 q k2 +m k (q; q) k = (1 + q 2 + q 3 )R(1)+(1 + q + q 2 + q 3 + q 4 )R(q), q 15 R(q 6 ) = q 10 ( R(q 4 ) R(q 5 ) ) = (1 + q 2 + q 3 + q 4 + q 6 )R(1) (1 + q + q 2 + q 3 + 2q 4 + q 5 + q 6 )R(q)

43 Ramanujan Congruences / q-shift Equation: Application 1 18 RECALL: q z R(q 2 z) = R(z) R(qz) and R(q m ) = qr(q 2 ) = R(1) R(q), q 3 R(q 3 ) = q ( R(q) R(q 2 ) ) = R(1) + (1 + q)r(q), q 6 R(q 4 ) = q 3 ( R(q 2 ) R(q 3 ) ) = (1 + q 2 )R(1) (1 + q + q 2 )R(q), q 10 R(q 5 ) = q 6 ( R(q 3 ) R(q 4 ) ) k=0 q k2 +m k (q; q) k = (1 + q 2 + q 3 )R(1)+(1 + q + q 2 + q 3 + q 4 )R(q), q 15 R(q 6 ) = q 10 ( R(q 4 ) R(q 5 ) ) = (1 + q 2 + q 3 + q 4 + q 6 )R(1) (1 + q + q 2 + q 3 + 2q 4 + q 5 + q 6 )R(q) Polynomial coefficients: any pattern?

44 Ramanujan Congruences / q-shift Equation: Application 1 19 RECALL: q (m 2 ) R(q m ) = ( 1) m d n (q) R(1)+( 1) m e n (q) R(q). (GIS)

45 Ramanujan Congruences / q-shift Equation: Application 1 19 RECALL: q (m 2 ) R(q m ) = ( 1) m d n (q) R(1)+( 1) m e n (q) R(q). (GIS) In[37]:= << RISC qgeneratingfunctions qgeneratingfunctions Package version written by Christoph Koutschan, c RISC-JKU In[38]:= QREGuess[{1, 0, 1, 1, 1 + q 2, 1 + q 2 + q 3, 1 + q 2 + q 3 + q 4 + q 6 }, d[n]]

46 Ramanujan Congruences / q-shift Equation: Application 1 19 RECALL: q (m 2 ) R(q m ) = ( 1) m d n (q) R(1)+( 1) m e n (q) R(q). (GIS) In[39]:= << RISC qgeneratingfunctions qgeneratingfunctions Package version written by Christoph Koutschan, c RISC-JKU In[40]:= QREGuess[{1, 0, 1, 1, 1 + q 2, 1 + q 2 + q 3, 1 + q 2 + q 3 + q 4 + q 6 }, d[n]] Out[40]= { d[n 2]q n + q 2 d[n 1] q 2 d[n] = 0, d[0] = 1, d[1] = 0 }

47 Ramanujan Congruences / q-shift Equation: Application 1 19 RECALL: q (m 2 ) R(q m ) = ( 1) m d n (q) R(1)+( 1) m e n (q) R(q). (GIS) In[41]:= << RISC qgeneratingfunctions qgeneratingfunctions Package version written by Christoph Koutschan, c RISC-JKU In[42]:= QREGuess[{1, 0, 1, 1, 1 + q 2, 1 + q 2 + q 3, 1 + q 2 + q 3 + q 4 + q 6 }, d[n]] Out[42]= { d[n 2]q n + q 2 d[n 1] q 2 d[n] = 0, d[0] = 1, d[1] = 0 } d n+2 (q) = d n+1 (q) + q n d n (q), n 0; d 0 (q) = 1, d 1 (q) = 0. NOTE. T. Garrett, M. Ismail, and D. Stanton proved (GIS) by evaluating an integral involving q-hermite polynomials in two different ways and equating the results [Adv. Appl. Math., 1999].

48 Ramanujan Congruences / q-shift Equation: Application 2 20 q-shift Equation: Application 2

49 Ramanujan Congruences / q-shift Equation: Application 2 21 RECALL R(qz) + q z R(q 2 z) = R(z).

50 Ramanujan Congruences / q-shift Equation: Application 2 21 RECALL R(qz) + q z R(q 2 z) = R(z). R(qz) R(z) + qz R(q2 z) = 1 R(z)

51 Ramanujan Congruences / q-shift Equation: Application 2 21 RECALL R(qz) + q z R(q 2 z) = R(z). R(qz) R(z) + qz R(q2 z) R(z) ( 1 + qz R(q2 z) R(qz) = 1 ) R(qz) R(z) = 1

52 Ramanujan Congruences / q-shift Equation: Application 2 21 RECALL R(qz) + q z R(q 2 z) = R(z). R(qz) R(z) + qz R(q2 z) R(z) ( 1 + qz R(q2 z) R(qz) R(qz) R(z) = qz R(q2 z) R(qz) = 1 ) R(qz) R(z) = 1

53 Ramanujan Congruences / q-shift Equation: Application 2 21 RECALL R(qz) + q z R(q 2 z) = R(z). R(qz) R(z) + qz R(q2 z) R(z) ( 1 + qz R(q2 z) R(qz) R(qz) R(z) = 1 = 1 + qz R(q2 z) R(qz) 1 + = 1 ) R(qz) R(z) = 1 1 qz 1 + q 2 z R(q3 z) R(q 2 z) =...

54 Ramanujan Congruences / q-shift Equation: Application 2 22 For z = 1 and q = q(τ) = exp(2πiτ) this gives: q 1 5 r(τ) := R(q) R(1) = q q q3 1 +

55 Ramanujan Congruences / q-shift Equation: Application 2 22 For z = 1 and q = q(τ) = exp(2πiτ) this gives: q 1 5 r(τ) := R(q) R(1) = q q q3 1 + NOTE. The continued fraction converges for τ H and also for some τ R.

56 Ramanujan Congruences / q-shift Equation: Application 2 22 For z = 1 and q = q(τ) = exp(2πiτ) this gives: q 1 5 r(τ) := R(q) R(1) = q q q3 1 + NOTE. The continued fraction converges for τ H and also for some τ R. E.g., for τ = 0 and τ = 1/2 one has:

57 Ramanujan Congruences / q-shift Equation: Application 2 23 and r(0) = =

58 Ramanujan Congruences / q-shift Equation: Application 2 23 and r(0) = e πi 5 r(1/2) = = =

59 Ramanujan Congruences / q-shift Equation: Application 2 24 For τ H Ramanujan found several beautiful evaluations; e.g., r(i) = e 2π 5 = e 2π e 4π e 6π

60 Ramanujan Congruences / q-shift Equation: Application 2 24 For τ H Ramanujan found several beautiful evaluations; e.g., r(i) = e 2π 5 = e 2π e 4π e 6π [These formulas] defeated me me completely. I had never seen anything in the least like this before. A single look at them is enough to show that they could only be written down by a mathematician of the highest class. They must be true because no one would have the imagination to invent them. [G.H. Hardy]

61 Ramanujan Congruences / q-shift Equation: Application 2 25 RECALL r(i) = e 2π 5 = e 2π e 4π e 6π

62 Ramanujan Congruences / q-shift Equation: Application 2 25 RECALL r(i) = e 2π 5 = e 2π e 4π e 6π EXPLANATION. r(τ) = q 1/5 R(q)/R(1) satisfies j(τ) = (r(τ)20 228r(τ) r(τ) r(τ) 5 + 1) 3 r(τ) 5 (r(τ) r(τ) 5 1) 5. Now setting τ = i, using j(i) = 1728, gives the following equation for r = r(i):

63 Ramanujan Congruences / q-shift Equation: Application r 5 (r r 5 1) 5 +(r r r r 5 +1) 3 = 0. The polynomial factors as follows:

64 Ramanujan Congruences / q-shift Equation: Application r 5 (r r 5 1) 5 +(r r r r 5 +1) 3 = 0. The polynomial factors as follows: In[45]:= 1728r 5 ( 1+11r 5 +r 10 ) 5 +(1+228r r r 15 +r 20 ) 3 //Factor Out[45]= ( r ) 2 ( r 4 + 2r 3 6r 2 2r + 1 ) 2 ( r 8 r 6 + r 4 r ) 2 ( r 8 6r r 6 18r r r r 2 + 6r + 1 ) 2 ( r 8 + 4r r r 5 + 5r 4 22r r 2 4r + 1 ) 2

65 Ramanujan Congruences / q-shift Equation: Application r 5 (r r 5 1) 5 +(r r r r 5 +1) 3 = 0. The polynomial factors as follows: In[47]:= 1728r 5 ( 1+11r 5 +r 10 ) 5 +(1+228r r r 15 +r 20 ) 3 //Factor Out[47]= ( r ) 2 ( r 4 + 2r 3 6r 2 2r + 1 ) 2 ( r 8 r 6 + r 4 r ) 2 ( r 8 6r r 6 18r r r r 2 + 6r + 1 ) 2 Consider ( r 8 + 4r r r 5 + 5r 4 22r r 2 4r + 1 ) 2 In[48]:= Solve[r 4 + 2r 3 6r 2 2r + 1 == 0, r] Out[48]= { { r { ( r ( 5 5) } { (, r ( ) )}, 2 2 ( 5 + 5) )} { (, r 1 1 ( ) )} } 5 2

66 Ramanujan Congruences / Back to Felix Klein 27 Back to Felix Klein

67 Ramanujan Congruences / Back to Felix Klein 28 RECALL: j(τ) = (r(τ)20 228r(τ) r(τ) r(τ) 5 + 1) 3 r(τ) 5 (r(τ) r(τ) 5 1) 5. NOTE 1. For the rational functions fixed by the icosahedral group, { ( ) } αz + β αz + β K := f C(z) : f = f(z) for all γz + δ γz + δ Moeb G, Felix Klein found K = C(J(z)), where J(z) = (z20 228z z z 5 + 1) 3 z 5 (z z 5 1) 5.

68 Ramanujan Congruences / Back to Felix Klein 29 RECALL: j(τ) = (r(τ)20 228r(τ) r(τ) r(τ) 5 + 1) 3 r(τ) 5 (r(τ) r(τ) 5 1) 5. NOTE 2. This relation can be derived (and thus proved) algorithmically in a straight-forward fashion.

69 Ramanujan Congruences / Back to Felix Klein 29 RECALL: j(τ) = (r(τ)20 228r(τ) r(τ) r(τ) 5 + 1) 3 r(τ) 5 (r(τ) r(τ) 5 1) 5. NOTE 2. This relation can be derived (and thus proved) algorithmically in a straight-forward fashion. SETTING: j(τ) and r(τ) 5 are modular functions for {( ) ( ) ( ) } a b a b 1 Γ 1 (5) = SL 2 (Z) : (mod 5) c d c d 0 1 [P. and Radu. Rogers-Ramanujan Functions, Modular Functions, and Computer Algebra, 2017].

70 Ramanujan Congruences / Back to Felix Klein 29 RECALL: j(τ) = (r(τ)20 228r(τ) r(τ) r(τ) 5 + 1) 3 r(τ) 5 (r(τ) r(τ) 5 1) 5. NOTE 2. This relation can be derived (and thus proved) algorithmically in a straight-forward fashion. SETTING: j(τ) and r(τ) 5 are modular functions for {( ) ( ) ( ) } a b a b 1 Γ 1 (5) = SL 2 (Z) : (mod 5) c d c d 0 1 [P. and Radu. Rogers-Ramanujan Functions, Modular Functions, and Computer Algebra, 2017]. NOTE 3. To prove that r(τ) 5 is a modular function use the Rogers-Ramanujan identities to obtain r(τ) 5 (1 q 5m+1 ) 5 (1 q 5m+4 ) 5 = q (1 q 5m+2 ) 5 (1 q 5m+3 ) 5. m=0

71 Ramanujan Congruences / Back to Felix Klein 29 RECALL: j(τ) = (r(τ)20 228r(τ) r(τ) r(τ) 5 + 1) 3 r(τ) 5 (r(τ) r(τ) 5 1) 5. NOTE 2. This relation can be derived (and thus proved) algorithmically in a straight-forward fashion. SETTING: j(τ) and r(τ) 5 are modular functions for {( ) ( ) ( ) } a b a b 1 Γ 1 (5) = SL 2 (Z) : (mod 5) c d c d 0 1 [P. and Radu. Rogers-Ramanujan Functions, Modular Functions, and Computer Algebra, 2017]. NOTE 4. Source of inspiration for the Felix Klein connection : W. Duke [Continued Fractions and Modular Functions].

72 Ramanujan Congruences / Conclusion 30 Conclusion

73 Ramanujan Congruences / Conclusion 31 Back to the Rogers-Ramanujan exercise: where is it taken from?

74 Ramanujan Congruences / Conclusion 31 Back to the Rogers-Ramanujan exercise: where is it taken from?

75 Ramanujan Congruences / Conclusion 32

76 Ramanujan Congruences / Conclusion 33

77 Ramanujan Congruences / Conclusion 34 Key ingredients (cont d) for proving the (j(τ), r(τ) 5 ) relation: x(τ) := r(τ) 5, y(τ) := 1 x(τ), j(τ) M(Γ 1(5)). cusps of Γ 1 (5): = 1 0, 0 1, 1 2, 2 5 ; widths 1, 5, 5, 1. j(τ) has poles at all these cusps of orders 1, 5, 5, 1, resp. x(τ) has a zero of order 1 at ; the q γ -expansions of x(τ) at 0 and 1 2 have both order 0, and x(τ) has a simple pole at 2 5.

78 Ramanujan Congruences / Conclusion 34 Key ingredients (cont d) for proving the (j(τ), r(τ) 5 ) relation: x(τ) := r(τ) 5, y(τ) := 1 x(τ), j(τ) M(Γ 1(5)). cusps of Γ 1 (5): = 1 0, 0 1, 1 2, 2 5 ; widths 1, 5, 5, 1. j(τ) has poles at all these cusps of orders 1, 5, 5, 1, resp. x(τ) has a zero of order 1 at ; the q γ -expansions of x(τ) at 0 and 1 2 have both order 0, and x(τ) has a simple pole at 2 5. j(τ) (x(τ) x(0))5 (x(τ) x(1/2)) 5 x(τ) 11 M (Γ 1 (5)).

79 Ramanujan Congruences / Conclusion 35 RECALL. x(0) = r(0) 5 = φ 5 and x(1/2) = r(1/2) 5 = 1 φ 5 This gives f(τ) = j(τ) (x(τ) x(0))5 (x(τ) x(1/2)) 5 x(τ) 11 M (Γ 1 (5)) = j(τ) (x(τ)2 + 11x(τ) 1) 5 x(τ) 11 = q12 q11 q q 9...

80 Ramanujan Congruences / Conclusion 35 RECALL. x(0) = r(0) 5 = φ 5 and x(1/2) = r(1/2) 5 = 1 φ 5 This gives f(τ) = j(τ) (x(τ) x(0))5 (x(τ) x(1/2)) 5 x(τ) 11 M (Γ 1 (5)) = j(τ) (x(τ)2 + 11x(τ) 1) 5 x(τ) 11 = q12 q11 q q 9... We reduce this with y(τ) = 1 x(τ) = 1 q q + 5q2 15q 3 24q q

81 Ramanujan Congruences / Conclusion 35 RECALL. x(0) = r(0) 5 = φ 5 and x(1/2) = r(1/2) 5 = 1 φ 5 This gives f(τ) = j(τ) (x(τ) x(0))5 (x(τ) x(1/2)) 5 x(τ) 11 M (Γ 1 (5)) = j(τ) (x(τ)2 + 11x(τ) 1) 5 x(τ) 11 = q12 q11 q q 9... We reduce this with y(τ) = 1 x(τ) = 1 q q + 5q2 15q 3 24q q First reduction step: f(τ) + y(τ) 12

82 Ramanujan Congruences / Conclusion 36

83 Ramanujan Congruences / Conclusion 37 etc., until one reaches

84 Ramanujan Congruences / Conclusion 38

85 Ramanujan Congruences / Conclusion 38

86 Ramanujan Congruences / Conclusion 39 Summarizing, since y(τ) = 1 Klein s icosohedral equation: x(τ) we discovered and proved j x(x x 1) 5 + (x 4 228x x x + 1) 3 = 0.

87 Ramanujan Congruences / Conclusion 39 Summarizing, since y(τ) = 1 Klein s icosohedral equation: x(τ) we discovered and proved j x(x x 1) 5 + (x 4 228x x x + 1) 3 = 0. Recall that x(τ) = r(τ) 5. Setting x = z 5 gives Klein s absolute invariant for the icosaeder: j = (z20 228z z z 5 + 1) 3 z 5 (z z 5 1) 5.