0013: Article 3 (Pi Formulas and the Monster)

 
 
Pi Formulas and the Monster Group

(Or How a Monster Can Bake a Pi)

 

by Tito Piezas III

 

 

Abstract:  The McKay-Thompson series of class 1A, 2A, 3A, and 4A for the Monster group, along with certain expressions involving Dedekind eta quotients, are used as a unifying framework for the four kinds of pi formulas found by Ramanujan, the Chudnovskys, and the Borweins.  Formulas for four kinds of singular moduli using the same eta quotients will also be given.

 

 

I. Introduction

II. The Dedekind eta function

III. The variable C: The j-function j(τ) and r-functions rp(τ).

IV. The variables A and B

V. Ramanujan and class number h(-d) = 1

VI. Formulas for Singular Moduli

VII. Going beyond

 

 

I. Introduction

 

Ramanujan gave many beautiful and unusual formulas for pi and, inspired by these, the Chudnovsky and Borwein brothers came up with similar ones.   First though, define the factorial quotients,

 

h1 = (6n)! / ((3n)! n!3)

h2 = (4n)! / (n!4)

h3 = (2n)!(3n)! / (n!5)

h4 = (2n)!3 / (n!6)

 

or, equivalently, as Pochhammer symbol products,

 

h1 = 1728n (1/2)n(1/6)n(5/6)n / (n!3)

h2 = 256n (1/2)n(1/4)n(3/4)n / (n!3)

h3 = 108n (1/2)n(1/3)n(2/3)n / (n!3)

h4 = 64n (1/2)n(1/2)n(1/2)n / (n!3)

 

where (a)n = (a)(a+1)(a+2)…(a+n-1).  Starting with n = 0, these generate the following integer sequences (with limiting ratio Lp between successive terms as n → ∞),

 

h1[n] = {1, 120, 83160, 81681600,…}            A001421    (L1 = 1728)

h2[n] = {1, 24, 2520, 369600,…}                    A008977    (L2 = 256)

h3[n] = {1, 12, 540, 33600,…}                        A184423    (L3 = 108)

h4[n] = {1, 8, 216, 8000,…}                            A002897    (L4 = 64)

 

As J. Guillera points out in his paper, "A class of conjectured series representations for 1/π", these sequences satisfy a first-order recurrence relation whose coefficients are 3rd degree polynomials in n,

 

n3 h1[n]-24(2n-1)(6n-1)(6n-5) h1[n-1] = 0

n3 h2[n]- 8(2n-1)(4n-1)(4n-3) h2[n-1] = 0

n3 h3[n]- 6(2n-1)(3n-1)(3n-2) h3[n-1] = 0

n3 h4[n]- 8(2n-1)(2n-1)(2n-1) h4[n-1] = 0

 

For example, the first sequence yields,

 

h1[1] = 120;  h1[2] = 83160;

 

Thus, n = 2 solves the cubic,

 

n3 (83160) - 24(2n-1)(6n-1)(6n-5) (120) = 0

 

Nice, isn't it?  One can also see the affinity between the hp (as Pochhammer symbols) and the recurrence relations, as well as the appearance of the limiting ratio Lp

 

Below is the complete list of 36 pi formulas using the four kinds of hp and three constants {A,B,C} where C is an integer, and {A,B} are either integers or have a square root. The formulas beginning in red are 16 of Ramanujan’s 17 pi formulas found in his 1914 paper, Modular Equations and Approximations to π (one had a radical C as it involved the golden ratio) .  The formula beginning in blue was also given by Ramanujan in his Notebooks

 

With sum Σ going from n = 0 to ∞, then,

 

1. j(τ)

 

Let h1 = (6n)! / ((3n)! n!3)  =  1728n (1/2)n(1/6)n(5/6)n / (n!3)

 

(none)                                                                                   eπ√303 + …

1/π = 3 Σ h1 (-1)n (63n+8)/(153)n+1/2                                    eπ√7153 + 696

1/π = 4 Σ h1 (-1)n (154n+15)/(323)n+1/2                                eπ√11323 + 738

1/π = 12 Σ h1 (-1)n (342n+25)/(963)n+1/2                              eπ√19963 + 743

1/π = 36 Σ h1 (-1)n (506n+31)/(3*1603)n+1/2                        eπ√273*1603 + 743.9…

1/π = 12 Σ h1 (-1)n (16254n+789)/(9603)n+1/2                      eπ√439603 + 743.99…

1/π = 12 Σ h1 (-1)n (261702n+10177)/(52803)n+1/2               eπ√6752803 + 743.99999…

1/π = 12 Σ h1 (-1)n (545140134n+13591409)/(6403203)n+1/2      eπ√1636403203 + 743.999999999…

 

(none)                                                                           eπ√4123 - …

1/π = 8 Σ h1 (28n+3)/(203)n+1/2,                                   eπ√8203 - 771

1/π = 72 Σ h1 (11n+1)/(2*303)n+1/2                              eπ√122*303 - 747

1/π = 24√2 Σ h1 (63n+5)/(663)n+1/2                              eπ√16663 - 744

1/π = 162 Σ h1 (133n+8)/(2553)n+1/2                            eπ√282553 -744.01…

 

2. r2(τ)

 

Let h2 = (4n)! / (n!4)  =  256n (1/2)n(1/4)n(3/4)n / (n!3)

 

1/π = 4 Σ h2 (-1)n (20n+3)/(322)n+1/2                            eπ√5322 + 100

1/π = 4 Σ h2 (-1)n (260n+23)/(2882)n+1/2                      eπ√132882 + 103.9…

1/π = 4 Σ h2 (-1)n (21460n+1123)/(141122)n+1/2          eπ√37141122 + 103.9999…

 

1/π = √7 Σ h2 (-1)n (65n+8)/(632)n+1/2                         eπ√7632 + 102.9...

1/π = 4*3 Σ h2 (-1)n (28n+3)/(3*642)n+1/2                    eπ√93*642 + 103...

1/π = 4*5 Σ h2 (-1)n (644n+41)/(5*11522)n+1/2            eπ√255*11522 + 103.999..

 

1/π = 4√2  Σ h2 (7n+1)/(2*182)n+1/2                             eπ√42*182 - 112

1/π = 8√3  Σ h2 (8n+1)/(482)n+1/2                                 eπ√6482 - 106

1/π = 32√2  Σ h2 (10n+1)/(124)n+1/2                             eπ√10124 - 104.2…

1/π = 48√3  Σ h2 (40n+3)/(7842)n+1/2                           eπ√187842 -104.00…

1/π = 8√11  Σ h2 (280n+19)/(15842)n+1/2                     eπ√2215842 - 104.001…

1/π = 32√2  Σ h2 (26390n+1103)/(3964)n+1/2               eπ√583964 - 104.00000…

 

3. r3(τ)

 

Let h3 = (2n)!(3n)! / (n!5)  = 108n (1/2)n(1/3)n(2/3)n / (n!3)

 

(none)                                                                            eπ√(5/3)33 +

1/π = 2 Σ h3 (-1)n (51n+7)/(123)n+1/2                            eπ√(17/3)123 + 41.6…

1/π = 2 Σ h3 (-1)n (615n+53)/(483)n+1/2                        eπ√(41/3)483 + 41.99…

1/π = 2 Σ h3 (-1)n (14151n+827)/(3003)n+1/2                eπ√(89/3)3003 + 41.9999…

 

1/π = 2*3 Σ h3 (-1)n (5n+1)/(3*43)n+1/2                        eπ√(9/3)3*43 + 38.8…

1/π = 2*5 Σ h3 (-1)n (27n+3)/(5*123)n+1/2                    eπ√(25/3)5*123 + 41.91…

1/π = 2*7 Σ h3 (-1)n (165n+13)/(7*363)n+1/2                eπ√(49/3)7*363 + 41.998…

 

(none)                                                                             eπ√(4/3)22(33) - …

1/π = 2√2 Σ h3 (6n+1)/(63)n+1/2,                                    eπ√(8/3)63  -  47

1/π = 4√2 Σ h3 (15n+2)/(2*93)n+1/2                               eπ√(16/3)2*93- 42.5…

1/π = 2√5 Σ h3 (33n+4)/(153)n+1/2                                 eπ√(20/3)153 - 42.2…

 

4. r4(τ)

 

Let h4 = (2n)!3 / (n!6)  = 64n (1/2)n(1/2)n(1/2)n / (n!3)

 

(none)                                                                             eπ√123 + 15.1…

1/π = 4 Σ h4 (-1)n (4n+1)/(26)n+1/2                                eπ√226 + 21.0…

1/π = 8 Σ h4 (-1)n (6n+1)/(29)n+1/2                                eπ√429 + 23.5…

1/π = 4 Σ h4 (6n+1)/(28)n+1/2                                        eπ√328 - 25.2…

1/π = 4 Σ h4 (42n+5)/(212)n+1/2                                     eπ√7212 - 24…

 

The functions j(τ) and rp(τ) will be discussed in a short while.  (Surprisingly, Ramanujan missed two of the r2(τ) which were later found by Berndt, Guillera, et al.)  The associated power of Gelfond’s constant eπ is given at the right side for comparison.  The “excess” {744, 104, 42, 24} is also significant, as this can be a clue to what modular form is being used.  In his 1914 paper, Ramanujan gave little explanation on how he came up with the formulas other than saying that there were “corresponding theories”.  He apparently wasn’t aware of it, but he was in fact exploring certain natural consequences of the Monster group (predicted sixty years later in 1973 by Fischer and Griess, and constructed by Griess in 1980).  The general form of these pi formulas F is,

 

1/π = Σ hp (An+B) / Cn

 

where {A,B,C} are algebraic numbers, and n = 0 to ∞ (as in the rest of this paper).  There is an infinite number of F, a special few use just integers while some involve involve familiar irrational constants.  For example, one found by Ramanujan involved the golden ratio φ = (1+√5)/2,

 

1/π = (1/16) Σ h4 ((-6+42φ)n + (-3+5φ)) / (212φ8)n

 

Another one by J. Guillera, simplified from its original form in Mathworld's Pi Formulas [117] is, 

 

1/π = (10/51/4) Σ h4 (6(18φ-29)n + (47φ-76)) / (26φ24)n

 

For this article, the author tweaked the numerators so that they are polynomials in φ and found that the second involve four consecutive Lucas numbers (in blue) as {2, 1, 3, 4, 7, 11, 18, 29, 47, 76,…}!  Why that is so, I have no idea. 


(Update, 5/23/11)  Qiaochu Yuan from MathStackExchange gave the identity for Fibonacci numbers,

 

Fn-1 + Fn φ = φn

 

A related one for Lucas numbers is,

 

Ln φ - Ln+1 = (1/φn-1) √5

 

Hence, Guillera's pi formula simplifies as,

 

1/π = (10/51/4) Σ h4 (6√5/φ6n + √5/φ8) / (26φ24)n

 

or just simply,

 

1/π = 10(51/4)/φ6  Σ h4 (6n + 1/φ2) / (26φ24)   

 

One mystery solved.  (End update)  


Surprisingly, the tribonacci constant T which is the limiting ratio of the tribonacci numbers (in the same manner as the golden ratio is for the Fibonacci and Lucas numbers) can also be used in a pi formula. The constant T is the real root of x3-x2-x-1 = 0.  Define v = 1/T,  

 

A = (1/2)(1+2v)(1+4v)

B = (1/4)(-1+2v+4v2)

C = (v+1)24

h4 = (2n)!3 / (n!6)     

 

then 1/π = Σ h4 (An+B)/Cn.

 

See Article 2, “A Tale of Four Constants”, which discusses the golden ratio, silver ratio, tribonacci constant, and the plastic constant, including their surprising appearance in pi formulas.  So how do we find these kinds of formulas?  An easy approach is, given the factorial quotient hp, if any two of {A,B,C} is known, then one can solve for the third using enough terms of the summation, easily done by computer algebra systems like Mathematica.  But first of all, what are the {A,B,C}’s?

 
 

II. The Dedekind eta function, η(τ)

 

The Dedekind eta function is a modular form usually defined in terms of the infinite product,

 

η(τ) = q1/24 Π (1-qm)

 

where m = 1 to infinity, q = e2πiτ = exp(2πiτ), and τ is the half-period ratio.  To point out again, it involves a 24th root (which “explains” why the integer 24 plays such a significant role in certain contexts).  This can be calculated in Mathematica as,

 

η(τ) = N[DedekindEta[τ], n]

 

for arbitrary n decimal places.

 

 

III. The variable C: The j-function j(τ) and r-functions rp(τ).

 

First define,

 

qp = e2πiτ = exp(2πiτ)

 

It is easily seen that for imaginary values τ = (1+√-d)/2, or τ = √-m, (and similar ones) where d or m is a positive real, then qp is a real number.

 

Proof:

 

Case 1.  τ = (1+√-d)/2    (Associated with odd discriminant d

 

qp = e2πi τ = e2πi (1+√-d)/2 = eπi (1+√-d) = eπi eπi√-d = (-1) e-π√d = -1/(eπ√d)

 

Case 2.  τ = √-m      (Associated with even discriminant d = 4m) 

 

qp = e2πi τ = e2πi√-m = e-2π√m = 1/(e2π√m)

 

and other cases.  Thus as d increases, qp becomes a very small real number, negative for the first case, and positive for the second.  See also Mathworld’s j-function for a similar derivation. (End proof.)

 

Then using qp = e2πiτ = exp(2πiτ), define the following functions j(τ) and rp(τ),

 

1. j(τ) = (f224 + 16)3 / f224  = 1/q + 744 + 196884q + 21493760q2 + 864299970q3 + … (A000521, A007240)

 

2. r2(τ) = (f224 + 64)2 / f224  = 1/q + 104 + 4372q + 96256q2 + 1240002q3 + …  (A007267)

 

3. r3(τ) = (f312 + 27)2 / f312  = 1/q + 42 + 783q + 8672q2 + 65367q3 + … (A030197)

 

4. r4(τ) = (f48 + 16)2 / f48  = 1/q + 24 + 276q + 2048q2 + 11202q3 + … (A097340A107080)

 

and perhaps also relevant,

 

5. r7(τ) = (f74 + 7)2 / f74  = 1/q + 10 + 51q + 204q2 + 681q3 + …   (A030183)

 

The rp(τ) have a common form.  Let k = 24/(p-1), and we get k = {24, 12, 8, 4}.  Then,

 

rp(τ) = (fpk + pk/4)2 / fpk 

 

Furthermore, the constant term c = {104, 42, 24, 10} of their q-series expansion also has a common form in terms of {p, k} as,

 

c = 2pk/4-k

 

Using appropriate Dedekind eta quotient fp, then j(τ) and rp(τ) are real algebraic numbers.  The first is the well-known j-function and its q-expansion is the McKay-Thompson series of class 1A for the Monster group, while the r-functions rp(τ) involve class pA for p = {2, 3, 4, 7}.  

 

Examples:

 

Bearing in mind that qp = e2πiτ = exp(2πiτ):
 
1. For j(τ), let p = 1, τ = (1+√-163)/2, hence q = -e-π√163.  Substituting this q into the j-function's q-series above, we get,

 

j((1+√-163)/2) = -6403203

 

2. Let p = 2, and τ = 1+√-37, hence q2 = e-2π√37.  By FTA*, this has 2 roots, namely q = ±e-π√37.  Using the negative root q = -e-π√37,

 

r2(1+√-37) = -141122 = -(84√2)4

 

Note:  Fundamental discriminant d = -4∙37 has class number h(d) = 2.

 

3. Let p = 3, and τ = (1+√-267)/2, hence q3 = -e-π√267.  By FTA, this has 3 roots.  Using the real root, we have q = (-e-π√267)1/3 = -e-π√(89/3), hence, 

 

r3((1+√-267)/2) = -3003

 

*Note:  With FTA as Fundamental Theorem of Algebra.  When verifying these identities using computer algebra software (CAS) like Mathematica or Maple, one should be careful since Mathematica gives the cube root of a negative real number r, or N[(-r)1/3], as a complex value.  Of course, by FTA this is still valid but becomes a problem when the real root, like q above, is desired.
 
4. Let p = 4, and τ = 2+√-16, hence q4 = e-2π√16.  This has 4 roots, but using the real negative root, q = -e-2π, 

 

r4(2+√-16) = -29

 

5. Let p = 7, and τ = (1+√-427)/2, hence q7 = -e-π√427.  Again using the real root, we have q = (-e-π√427)1/7 = -e-π√(61/7), hence, 

 

r7((1+√-427)/2) = -7(392)

 

Alternatively, instead of using the q-series expansion, one can employ the Dedekind eta quotients fp.  (Note: In Mathematica, the syntax [] is simply used to introduce a function’s argument.)  Recalling that η[τ] is the Dedekind eta function, then,

 

1. j(τ) = (f224 + 16)3 / f224,    where f2 = η[τ/2] / η[τ],  for τ = √-m, or τ = (1+√-d)/2.

 

2. r2(τ) = (f224 + 64)2 / f224,   where f2 = η[τ/2] / η[τ],  for τ = √-m, or τ = 1+√-m.

 

3. r3(τ) = (f312 + 27)2 / f312,   

 

a) f3 = η[τ/3] / η[τ],  for τ = √-m

b) f3 = (-1)1/12 η[(τ+1)/3] / η[τ],  for τ = (1+√-d)/2

 

4. r4(τ) = (f48 + 16)2 / f48,     where f4 = η[τ/4] / η[τ],  for τ = √-m, or τ = 2+√-

 

5. r7(τ) = (f74 + 7)2 / f74,

 

a) f7 = η[τ/7] / η[τ],  for τ = √-m

b) f7 = (-1)1/4 η[(τ+3)/7] / η[τ],  for τ = (1+√-d)/2

 

(Notice the qualitative difference of fp for odd and even p.)

 

Mathematica examples:

 

1. j(τ)

 

j(τ) = N[(f224 + 16)3 / f224 /. f2 → DedekindEta[τ/2]/DedekindEta[τ] /. τ → √-4, 1000] = 663

j(τ) = N[(f224 + 16)3 / f224 /. f2 → DedekindEta[τ/2]/DedekindEta[τ] /. τ → (1+√-7)/2, 1000] = -153

 

2. r2(τ)

 

r2(τ) = N[(f224 + 64)2 / f224 /. f2 → DedekindEta[τ/2]/DedekindEta[τ] /. τ → √-58, 1000] = 3964

r2(τ) = N[(f224 + 64)2 / f224 /. f2 → DedekindEta[τ/2]/DedekindEta[τ] /. τ → 1+√-37, 1000] = -141122 = -(84√2)4

 

3. r3(τ)

 

r3(τ) = N[(f312 + 27)2 / f312  /. f3 → DedekindEta[τ/3]/DedekindEta[τ] /. τ → √-6, 1000] = 63

r3(τ) = N[(f312 + 27)2 / f312  /. f3 → (-1)1/12DedekindEta[(τ+1)/3]/DedekindEta[τ] /. τ → (1+√-267)/2, 1000] = -3003

 

4. r4(τ)

 

r4(τ) = N(f48 + 16)2 / f48  /. f4 → DedekindEta[τ/4]/DedekindEta[τ] /. τ → √-28, 1000] = 212

r4(τ) = N(f48 + 16)2 / f48  /. f4 → DedekindEta[τ/4]/DedekindEta[τ] /. τ → 2+√-16, 1000] = -29

 

5. r7(τ)

 

r7(τ) = N[(f74 + 7)2 / f74  /. f7 → DedekindEta[τ/7]/DedekindEta[τ] /. τ → √-28, 1000] = 14(32)

r7(τ) = N[(f74 + 7)2 / f74  /. f7 → (-1)1/4DedekindEta[(τ+3)/7]/DedekindEta[τ] /. τ → (1+√-427)/2, 1000] = -7(392)

 

 
So what is the variable C in the pi formula?  It turns out that Cp = j(τ) or rp(τ)!  For more about the r-functions rp(τ), see also Article 1, “The j-function and Its Cousins”.
 
  

IV. The variables A and B

 

Given the eqn,

 

1/π = Σ hp (An + B) / Cn,  

 

Conveniently, A can be expressed in terms of C as,

 

1. for h1 and j-function j(τ):   A1 = √a1,  a1 = d(1 – 123/C1);  and C1 = j(τ).

2. for hp and r-functions rp(τ):  Ap = (1/p)√ap,  ap = d(1 – 4pv/Cp);  v = 6/(p-1);  and Cp = rp(τ).

 

hence,

 

A1 = (1/1)√(d(1 – 1728/C1))

A2 = (1/2)√(d(1 – 256/C2))

A3 = (1/3)√(d(1 – 108/C3))

A4 = (1/4)√(d(1 – 64/C4))

 

The discriminant d depends on the form of τ:

 

Case 1:  If τ = a+√-m,  for a = {0,1,2},  then d = 4m.

Case 2:  If τ = (1+√-d)/2,  then d = d.

 

Examples.  Using the τ’s from the previous section:

 

Let τ = {√-4,  √-58,  √-6,  √-28},

 

C1 = j(τ) = 663,        a1 = 4*4(1 – 1728/C1),    so A1 = (1/1)√a1 = (84/121)√33

C2 = r2(τ) = 3964,    a2 = 4*58(1 – 256/C2),    so A2 = (1/2)√a2 = (52780/9801)√2

C3 = r3(τ) = 63,        a3 = 4*6(1 – 108/C3),      so A3 = (1/3)√a3 = (2/3)√2

C4 = r4(τ) = 212,       a4 = 4*28(1 – 64/C4),      so A4 = (1/4)√a4 = 21/8

 

Let τ = {(1+√-7)/2,  1+√-37,  (1+√-267)/2,  2+√-16},

 

C1 = j(τ) = -153,            a1 = 7(1 – 1728/C1),        so A1 = (1/1)√a1 = (21/25)√15

C2 = r2(τ) = -141122,    a2 = 4*37(1 – 256/C2),    so A2 = (1/2)√a2 = 5365/882

C3 = r3(τ) = -3003,        a3 = 267(1 – 108/C3),      so A3 = (1/3)√a3 = (4717/1500)√3

C4 = r4(τ) = -29,            a4 = 4*16(1 – 64/C4),      so A4 = (1/4)√a4 = (3/2)√2

 

We now have A and C.  Unfortunately, B is a complicated function in terms of C, and beyond the scope of this paper.  To find B, one can just numerically solve for it using enough terms of the summation.  If C is an algebraic number of degree n, then B should be of degree either n or 2n, and an integer relations algorithm can then find its minimal polynomial.  For example, using the last entry with C4 = -29,

 

x = NSolve[Sum[h4(An+B)/Cn /. A →(3/2)√2  /. C→ -29 /. h4 → (2n)!3 / n!6, {n, 0, 50}] = 1/Pi, B, 30] = 0.353553390593…

 

The Mathematica add-on Recognize[x, k, v] tries to recognize the real number x as a root of an kth degree equation in some variable v.  Applying the command on x for k = 2 yields the simple equation 1-8B2 = 0, so B = (1/4)√2, hence,

 

1/π = Σ h4 ((3/2)√2n + (1/4)√2) / (-29)n
 
simplified as,
 
1/π = (1/4)√2 Σ h4 (6n+1) / (-29)n
 
Equivalently,

 

1/π = 8 Σ h4 (-1)n (6n+1) / (29)n+1/2

 

The fact that A applies a square root on an expression involving C explains the “n+1/2” exponent in the denominator though, if C is negative, then (-1)n should be factored out as in the above example.

 

  

V.  Ramanujan and Class number h(-d) = 1

 

Why didn’t Ramanujan find the Chudnovsky family using the j-function j(τ) and class number h(-d) = 1?  He almost did.  To see this, first define v1 = (1/2)n(1/6)n(5/6)n / (n!3) where (a)n is the Pochhammer symbol.  The original form of two of his 17 formulas were,

 

1/π = (2√3)/(5√5)  Σ v1 (11n+1)(4/125)n         

1/π = (18√3)/(85√85)  Σ v1 (133n+8)/(4/85)3n

 

However, this can be transformed to,

 

1/π = 72 Σ h1 (11n+1)/(2*303)n+1/2,                              eπ√122*303 - 747…

1/π = 162 Σ h1 (133n+8)/(2553)n+1/2,                            eπ√282553 -744.01…

 

where, as before, h1 = (6n)! / ((3n)! n!3).  Thus, Ramanujan also found formulas using non-fundamental discriminants d = {12, 28} with class number h(-d) = 1!  In fact, in his Lost Notebook, he wrote down some of his results on d = 11, 19, 43, 67, 163, fundamental d which have h(-d) = 1.  I’m sure that had he only lived long enough (he died at age 32), he would have found back in the 1920s the formulas discovered by the Chudnovsky brothers only in 1987.

 

 

VI.  Formulas for Singular Moduli

 

Other than pi formulas, the same eta quotients fp are also useful for calculating singular moduli. To recall, the hp can also be defined as Pochhammer symbol products,

 

h1 = 1728n (1/2)n(1/6)n(5/6)n / (n!3)

h2 = 256n (1/2)n(1/4)n(3/4)n / (n!3)

h3 = 108n (1/2)n(1/3)n(2/3)n / (n!3)

h4 = 64n (1/2)n(1/2)n(1/2)n / (n!3)

 

Expressed in this way, the connection to four kinds of singular moduli become more transparent.  First, given the hypergeometric function 2F1,

 

2F1(a,b;c;z):= Σ g zn/n!   

 

where,

 

g = (a)n(b)n / (c)n

 

with |z| < 1, and n = 0  to ∞.  In Mathematica, this is Hypergeometric2F1[a,b,c,z].  Define,

 

m1 = 2F1(a, b; 1; 1-αN)

m2 = 2F1(a, b; 1; αN)

 

Then, for positive numbers {a,b} such that a+b = 1, the singular modulus αN is the unique positive real number 0 < αN < 1 satisfying,

 

m1/m2 = √N

 

To find αN, it can be observed that,

 

1. Let {a,b} = {1/6, 5/6}. Then p = 1, αN = (1-√(1-u1))/2, and u1 = 1728 / j(τ)

2. Let {a,b} = {1/4, 3/4}. Then p = 2, αN = 1/(u2+1), and u2 = (1/26) f224   

3. Let {a,b} = {1/3, 2/3}. Then p = 3, αN = 1/(u3+1), and u3 = (1/33) f312

4. Let {a,b} = {1/2, 1/2}. Then p = 4, αN = 1/(u4+1), and u4 = (1/42) f48

 

where fp are the Dedekind eta quotients,

 

f2 = η[τ/2] / η[τ]

f3 = η[τ/3] / η[τ]

f4 = η[τ/4] / η[τ]

 

and τ = √(-pn). 
 
Example.  In his second letter to Hardy, Ramanujan gave the amazing radical,

 

α210 = (8-3√7)2 (2-√3)2 (6-√35)2 (√15-√14)2 (1-√2)4 (3-√10)4 (4-√15)4 (√7-√6)4 = 2.7066… x 10(-19)

 

This applies to the case {a,b} = {1/2, 1/2}, and N = 210.  To confirm this, set p = 4, so,

 

α210 = 1/(u4+1),  where u4 = (1/42) f48 = (1/42)(η[τ/4]/η[τ])8 
 
and τ = √(-210p) = √(-210*4), and we find that indeed,

 

α210 = 2.7066… x 10(-19)

 

such that,
 

m1/m2 = √210

 
To express αN as a product of simple radicals (in fact, the factors are units in some quadratic field) just like what Ramanujan ingeniously did is another matter entirely, and requires much mathematical dexterity.

 

 

VII.  Going Beyond

 

The natural generalization of Ramanujan’s pi formulas is,

 

1/πk = Σ h  P(n) / Cn       (eq.1)

 

where h is some factorial quotient and the numerator P(n) is a polynomial in n of degree k. Ramanujan’s was just the case k = 1. A semi-trivial solution to eq.1 is taking kth powers of known formulas for k = 1, with the resulting P(n) as a complicated mess.  However, without squaring, Jesús Guillera has found many simple and elegant examples for k = 2 using various h, and Boris Gourevitch has found a single instance of k = 3.  To give examples, first define,

 

v2 = (2n)!/(n!2) 

 

then, again for n = 0 to ∞,

 

1. 1/π = 22 Σ v23 (6n+1) / (28)n+1/2,

2. 1/π = 22 Σ v23 (42n+5) / (212)n+1/2

3. 1/π2 = 23 Σ v25 (-1)n  (20n2+8n+1) / (212)n+1/2

4. 1/π2 = 23 Σ v25 (-1)n  (820n2+180n+13) / (220)n+1/2

5. 1/π3 = 25 Σ v27 (168n3+76n2+14n+1) / (220)n+1/2

6. 1/π4 = 27 Σ v29  P(n)(?) / (C)n+1/2

 

The first two were found by Ramanujan, the next two by Guillera (correcting two typos in Mathworld), and the fifth is by Gourevitch.  Does the pattern go on for all 1/πk using polynomials of degree k?  If so, is there a function that generates the variable C just like for (1) and (2)?  This phenomenon is simply begging for an explanation.   

 

Anybody knows how to find (6), where P(n) is a 4th deg polynomial and C is a power of 2?
 
 
P.S.  Digressing a bit, here is the introductory scene of The Last Unicorn, one of my favorite animated films when I was young.  The song is by the legendary band, America.  The theme song is beautiful and the movie as well:
 

  

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© Sept 2010 (modified May 2011), Tito Piezas III

You can email author at tpiezas@gmail.com
 
 
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