Pi Formulas and the Monster Group
(Or How a Monster Can Bake a Pi)
by Tito Piezas III
Abstract: The McKayThompson 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 jfunction j(τ) and rfunctions r_{p}(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,
h_{1} = (6n)! / ((3n)! n!^{3}) h_{2} = (4n)! / (n!^{4}) h_{3} = (2n)!(3n)! / (n!^{5}) h_{4} = (2n)!^{3} / (n!^{6})
or, equivalently, as Pochhammer symbol products,
h_{1} = 1728^{n} (1/2)_{n}(1/6)_{n}(5/6)_{n} / (n!^{3}) h_{2} = 256^{n} (1/2)_{n}(1/4)_{n}(3/4)_{n} / (n!^{3}) h_{3} = 108^{n} (1/2)_{n}(1/3)_{n}(2/3)_{n} / (n!^{3}) h_{4} = 64^{n} (1/2)_{n}(1/2)_{n}(1/2)_{n} / (n!^{3})
where (a)_{n} = (a)(a+1)(a+2)…(a+n1). Starting with n = 0, these generate the following integer sequences (with limiting ratio L_{p} between successive terms as n → ∞),
h_{1}[n] = {1, 120, 83160, 81681600,…} A001421 (L_{1} = 1728) h_{2}[n] = {1, 24, 2520, 369600,…} A008977 (L_{2} = 256) h_{3}[n] = {1, 12, 540, 33600,…} A184423 (L_{3} = 108) h_{4}[n] = {1, 8, 216, 8000,…} A002897 (L_{4} = 64)
As J. Guillera points out in his paper, "A class of conjectured series representations for 1/π", these sequences satisfy a firstorder recurrence relation whose coefficients are 3rd degree polynomials in n,
n^{3 }h_{1}[n]24(2n1)(6n1)(6n5) h_{1}[n1] = 0 n^{3 }h_{2}[n] 8(2n1)(4n1)(4n3) h_{2}[n1] = 0 n^{3 }h_{3}[n] 6(2n1)(3n1)(3n2) h_{3}[n1] = 0 n^{3 }h_{4}[n] 8(2n1)(2n1)(2n1) h_{4}[n1] = 0
For example, the first sequence yields,
h_{1}[1] = 120; h_{1}[2] = 83160;
Thus, n = 2 solves the cubic,
n^{3 }(83160)  24(2n1)(6n1)(6n5) (120) = 0
Nice, isn't it? One can also see the affinity between the h_{p} (as Pochhammer symbols) and the recurrence relations, as well as the appearance of the limiting ratio L_{p}.
Below is the complete list of 36 pi formulas using the four kinds of h_{p} 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 h_{1} = (6n)! / ((3n)! n!^{3}) = 1728^{n} (1/2)_{n}(1/6)_{n}(5/6)_{n} / (n!^{3})
(none) e^{π√3} ≈ 0^{3} + … 1/π = 3 Σ h_{1} (1)^{n} (63n+8)/(15^{3})^{n+1/2} e^{π√7} ≈ 15^{3} + 696 1/π = 4 Σ h_{1} (1)^{n} (154n+15)/(32^{3})^{n+1/2} e^{π√11} ≈ 32^{3} + 738 1/π = 12 Σ h_{1} (1)^{n} (342n+25)/(96^{3})^{n+1/2} e^{π√19} ≈ 96^{3} + 743 1/π = 36 Σ h_{1} (1)^{n} (506n+31)/(3*160^{3})^{n+1/2} e^{π√27} ≈ 3*160^{3} + 743.9… 1/π = 12 Σ h_{1} (1)^{n} (16254n+789)/(960^{3})^{n+1/2} e^{π√43} ≈ 960^{3} + 743.99… 1/π = 12 Σ h_{1} (1)^{n} (261702n+10177)/(5280^{3})^{n+1/2} e^{π√67} ≈ 5280^{3} + 743.99999… 1/π = 12 Σ h_{1} (1)^{n} (545140134n+13591409)/(640320^{3})^{n+1/2} e^{π√163} ≈ 640320^{3} + 743.999999999…
(none) e^{π√4} ≈ 12^{3}  … 1/π = 8 Σ h_{1} (28n+3)/(20^{3})^{n+1/2}, e^{π√8} ≈ 20^{3}  771 1/π = 72 Σ h_{1} (11n+1)/(2*30^{3})^{n+1/2} e^{π√12} ≈ 2*30^{3}  747 1/π = 24√2 Σ h_{1} (63n+5)/(66^{3})^{n+1/2} e^{π√16} ≈ 66^{3}  744 1/π = 162 Σ h_{1} (133n+8)/(255^{3})^{n+1/2} e^{π√28} ≈ 255^{3} 744.01…
2. r_{2}(τ)
Let h_{2} = (4n)! / (n!^{4}) = 256^{n} (1/2)_{n}(1/4)_{n}(3/4)_{n} / (n!^{3})
1/π = 4 Σ h_{2} (1)^{n }(20n+3)/(32^{2})^{n+1/2} e^{π√5} ≈ 32^{2} + 100 1/π = 4 Σ h_{2} (1)^{n }(260n+23)/(288^{2})^{n+1/2} e^{π√13} ≈ 288^{2} + 103.9… 1/π = 4 Σ h_{2} (1)^{n }(21460n+1123)/(14112^{2})^{n+1/2} e^{π√37} ≈ 14112^{2} + 103.9999…
1/π = √7 Σ h_{2} (1)^{n }(65n+8)/(63^{2})^{n+1/2} e^{π√7} ≈ 63^{2} + 102.9... 1/π = 4*3 Σ h_{2} (1)^{n }(28n+3)/(3*64^{2})^{n+1/2} e^{π√9} ≈ 3*64^{2} + 103... 1/π = 4*5 Σ h_{2} (1)^{n }(644n+41)/(5*1152^{2})^{n+1/2} e^{π√25} ≈ 5*1152^{2} + 103.999..
1/π = 4√2 Σ h_{2} (7n+1)/(2*18^{2})^{n+1/2} e^{π√4} ≈ 2*18^{2}  112 1/π = 8√3 Σ h_{2} (8n+1)/(48^{2})^{n+1/2} e^{π√6} ≈ 48^{2}  106 1/π = 32√2 Σ h_{2} (10n+1)/(12^{4})^{n+1/2} e^{π√10} ≈ 12^{4}  104.2… 1/π = 48√3 Σ h_{2} (40n+3)/(784^{2})^{n+1/2} e^{π√18} ≈ 784^{2} 104.00… 1/π = 8√11 Σ h_{2} (280n+19)/(1584^{2})^{n+1/2} e^{π√22} ≈ 1584^{2}  104.001… 1/π = 32√2 Σ h_{2} (26390n+1103)/(396^{4})^{n+1/2} e^{π√58} ≈ 396^{4}  104.00000…
3. r_{3}(τ)
Let h_{3} = (2n)!(3n)! / (n!^{5}) = 108^{n} (1/2)_{n}(1/3)_{n}(2/3)_{n} / (n!^{3})
(none) e^{π√(5/3)} ≈ 3^{3} + 1/π = 2 Σ h_{3} (1)^{n }(51n+7)/(12^{3})^{n+1/2} e^{π√(17/3)} ≈ 12^{3} + 41.6… 1/π = 2 Σ h_{3} (1)^{n }(615n+53)/(48^{3})^{n+1/2} e^{π√(41/3)} ≈ 48^{3} + 41.99… 1/π = 2 Σ h_{3} (1)^{n }(14151n+827)/(300^{3})^{n+1/2} e^{π√(89/3)} ≈ 300^{3} + 41.9999…
1/π = 2*3 Σ h_{3} (1)^{n }(5n+1)/(3*4^{3})^{n+1/2} e^{π√(9/3)} ≈ 3*4^{3} + 38.8… 1/π = 2*5 Σ h_{3} (1)^{n }(27n+3)/(5*12^{3})^{n+1/2} e^{π√(25/3)} ≈ 5*12^{3} + 41.91… 1/π = 2*7 Σ h_{3} (1)^{n }(165n+13)/(7*36^{3})^{n+1/2} e^{π√(49/3)} ≈ 7*36^{3} + 41.998…
(none) e^{π√(4/3)} ≈ 2^{2}(3^{3})  … 1/π = 2√2 Σ h_{3} (6n+1)/(6^{3})^{n+1/2}, e^{π√(8/3)} ≈ 6^{3}  47 1/π = 4√2 Σ h_{3} (15n+2)/(2*9^{3})^{n+1/2} e^{π√(16/3)} ≈ 2*9^{3} 42.5… 1/π = 2√5 Σ h_{3} (33n+4)/(15^{3})^{n+1/2} e^{π√(20/3)} ≈ 15^{3}  42.2…
4. r_{4}(τ)
Let h_{4} = (2n)!^{3} / (n!^{6}) = 64^{n} (1/2)_{n}(1/2)_{n}(1/2)_{n} / (n!^{3})
(none) e^{π√1} ≈ 2^{3} + 15.1… 1/π = 4 Σ h_{4} (1)^{n} (4n+1)/(2^{6})^{n+1/2} e^{π√2} ≈ 2^{6} + 21.0… 1/π = 8 Σ h_{4} (1)^{n} (6n+1)/(2^{9})^{n+1/2} e^{π√4} ≈ 2^{9} + 23.5… 1/π = 4 Σ h_{4} (6n+1)/(2^{8})^{n+1/2} e^{π√3} ≈ 2^{8}  25.2… 1/π = 4 Σ h_{4} (42n+5)/(2^{12})^{n+1/2} e^{π√7} ≈ 2^{12}  24… The functions j(τ) and r_{p}(τ) will be discussed in a short while. (Surprisingly, Ramanujan missed two of the r_{2}(τ) which were later found by Berndt, Guillera, et al.) The associated power of Gelfond’s constant e^{π} 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 h_{p} (An+B) / C^{n}
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 h_{4} ((6+42φ)n + (3+5φ)) / (2^{12}φ^{8})^{n}
Another one by J. Guillera, simplified from its original form in Mathworld's Pi Formulas [117] is,
1/π = (10/5^{1/4}) Σ h_{4} (6(18φ29)n + (47φ76)) / (2^{6}φ^{24})^{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,
F_{n1} + F_{n} φ = φ^{n}
A related one for Lucas numbers is,
L_{n} φ  L_{n+1} = (1/φ^{n1}) √5
Hence, Guillera's pi formula simplifies as,
1/π = (10/5^{1/4}) Σ h_{4} (6√5/φ^{6}n + √5/φ^{8}) / (2^{6}φ^{24})^{n}
or just simply,
1/π = 10(5^{1/4})/φ^{6} Σ h_{4} (6n + 1/φ^{2}) / (2^{6}φ^{24})^{n }
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 x^{3}x^{2}x1 = 0. Define v = 1/T,
A = (1/2)(1+2v)(1+4v) B = (1/4)(1+2v+4v^{2}) C = (v+1)^{24} h_{4} = (2n)!^{3} / (n!^{6})
then 1/π = Σ h_{4} (An+B)/C^{n}.
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 h_{p}, 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) = q^{1/24} Π (1q^{m})
where m = 1 to infinity, q = e^{2πiτ} = exp(2πiτ), and τ is the halfperiod 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 jfunction j(τ) and rfunctions r_{p}(τ).
First define,
q^{p} = e^{2πiτ} = 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 q^{p} is a real number.
Proof:
Case 1. τ = (1+√d)/2 (Associated with odd discriminant d)
q^{p} = e^{2πi τ} = e^{2πi (1+√d)/2} = e^{πi (1+√d)} = e^{πi} e^{πi√d} = (1) e^{π√d} = 1/(e^{π√d})
Case 2. τ = √m (Associated with even discriminant d = 4m)
q^{p} = e^{2πi τ} = e^{2πi√m} = e^{2π√m} = 1/(e^{2π√m})
and other cases. Thus as d increases, q^{p} becomes a very small real number, negative for the first case, and positive for the second. See also Mathworld’s jfunction for a similar derivation. (End proof.)
Then using q^{p} = e^{2πiτ} = exp(2πiτ), define the following functions j(τ) and r_{p}(τ),
1. j(τ)_{ }= (f_{2}^{24} + 16)^{3} / f_{2}^{24} = 1/q + 744 + 196884q + 21493760q^{2} + 864299970q^{3} + … (A000521, A007240)
2. r_{2}(τ) = (f_{2}^{24} + 64)^{2} / f_{2}^{24} = 1/q + 104 + 4372q + 96256q^{2} + 1240002q^{3} + … (A007267)
3. r_{3}(τ) = (f_{3}^{12} + 27)^{2} / f_{3}^{12} = 1/q + 42 + 783q + 8672q^{2} + 65367q^{3} + … (A030197)
4. r_{4}(τ) = (f_{4}^{8} + 16)^{2} / f_{4}^{8} = 1/q + 24 + 276q + 2048q^{2} + 11202q^{3} + … (A097340, A107080)
and perhaps also relevant,
5. r_{7}(τ) = (f_{7}^{4} + 7)^{2} / f_{7}^{4} = 1/q + 10 + 51q + 204q^{2} + 681q^{3} + … (A030183)
The r_{p}(τ) have a common form. Let k = 24/(p1), and we get k = {24, 12, 8, 4}. Then,
r_{p}(τ) = (f_{p}^{k} + p^{k/4})^{2} / f_{p}^{k}
Furthermore, the constant term c = {104, 42, 24, 10} of their qseries expansion also has a common form in terms of {p, k} as,
c = 2p^{k/4}k
Using appropriate Dedekind eta quotient f_{p}, then j(τ) and r_{p}(τ) are real algebraic numbers. The first is the wellknown jfunction and its qexpansion is the McKayThompson series of class 1A for the Monster group, while the rfunctions r_{p}(τ) involve class pA for p = {2, 3, 4, 7}.
Examples:
Bearing in mind that q^{p} = e^{2πiτ} = exp(2πiτ):
1. For j(τ), let p = 1, τ = (1+√163)/2, hence q = e^{π√163}. Substituting this q into the jfunction's qseries above, we get,
j((1+√163)/2) = 640320^{3}
2. Let p = 2, and τ = 1+√37, hence q^{2} = e^{2π√37}. By FTA*, this has 2 roots, namely q = ±e^{π√37}. Using the negative root q = e^{π√37},
r_{2}(1+√37) = 14112^{2} = (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 q^{3} = 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,
r_{3}((1+√267)/2) = 300^{3}
*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 q^{4} = e^{2π√16}. This has 4 roots, but using the real negative root, q = e^{2π},
r_{4}(2+√16) = 2^{9}
5. Let p = 7, and τ = (1+√427)/2, hence q^{7} = e^{π√427}. Again using the real root, we have q = (e^{π√427})^{1/7} = e^{π√(61/7)}, hence,
r_{7}((1+√427)/2) = 7(39^{2})
Alternatively, instead of using the qseries expansion, one can employ the Dedekind eta quotients f_{p}. (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)_{ }= (f_{2}^{24} + 16)^{3} / f_{2}^{24}, where f_{2} = η[τ/2] / η[τ], for τ = √m, or τ = (1+√d)/2.
2. r_{2}(τ) = (f_{2}^{24} + 64)^{2} / f_{2}^{24}, where f_{2} = η[τ/2] / η[τ], for τ = √m, or τ = 1+√m.
3. r_{3}(τ) = (f_{3}^{12} + 27)^{2} / f_{3}^{12},
a) f_{3} = η[τ/3] / η[τ], for τ = √m b) f_{3} = (1)^{1/12} η[(τ+1)/3] / η[τ], for τ = (1+√d)/2
4. r_{4}(τ) = (f_{4}^{8} + 16)^{2} / f_{4}^{8}, where f_{4} = η[τ/4] / η[τ], for τ = √m, or τ = 2+√m
5. r_{7}(τ) = (f_{7}^{4} + 7)^{2} / f_{7}^{4},
a) f_{7} = η[τ/7] / η[τ], for τ = √m b) f_{7} = (1)^{1/4} η[(τ+3)/7] / η[τ], for τ = (1+√d)/2
(Notice the qualitative difference of f_{p} for odd and even p.)
Mathematica examples:
1. j(τ)
j(τ) = N[(f_{2}^{24} + 16)^{3} / f_{2}^{24} /. f_{2} → DedekindEta[τ/2]/DedekindEta[τ] /. τ → √4, 1000] = 66^{3} j(τ) = N[(f_{2}^{24} + 16)^{3} / f_{2}^{24} /. f_{2} → DedekindEta[τ/2]/DedekindEta[τ] /. τ → (1+√7)/2, 1000] = 15^{3}
2. r_{2}(τ)
r_{2}(τ) = N[(f_{2}^{24} + 64)^{2} / f_{2}^{24} /. f_{2} → DedekindEta[τ/2]/DedekindEta[τ] /. τ → √58, 1000] = 396^{4} r_{2}(τ) = N[(f_{2}^{24} + 64)^{2} / f_{2}^{24} /. f_{2} → DedekindEta[τ/2]/DedekindEta[τ] /. τ → 1+√37, 1000] = 14112^{2} = (84√2)^{4}
3. r_{3}(τ)
r_{3}(τ) = N[(f_{3}^{12} + 27)^{2} / f_{3}^{12} /. f_{3} → DedekindEta[τ/3]/DedekindEta[τ] /. τ → √6, 1000] = 6^{3} r_{3}(τ) = N[(f_{3}^{12} + 27)^{2} / f_{3}^{12} /. f_{3} → (1)^{1/12}DedekindEta[(τ+1)/3]/DedekindEta[τ] /. τ → (1+√267)/2, 1000] = 300^{3}
4. r_{4}(τ)
r_{4}(τ) = N(f_{4}^{8} + 16)^{2} / f_{4}^{8} /. f_{4} → DedekindEta[τ/4]/DedekindEta[τ] /. τ → √28, 1000] = 2^{12} r_{4}(τ) = N(f_{4}^{8} + 16)^{2} / f_{4}^{8} /. f_{4} → DedekindEta[τ/4]/DedekindEta[τ] /. τ → 2+√16, 1000] = 2^{9}
5. r_{7}(τ)
r_{7}(τ) = N[(f_{7}^{4} + 7)^{2} / f_{7}^{4} /. f_{7} → DedekindEta[τ/7]/DedekindEta[τ] /. τ → √28, 1000] = 14(3^{2}) r_{7}(τ) = N[(f_{7}^{4} + 7)^{2} / f_{7}^{4} /. f_{7} → (1)^{1/4}DedekindEta[(τ+3)/7]/DedekindEta[τ] /. τ → (1+√427)/2, 1000] = 7(39^{2})
So what is the variable C in the pi formula? It turns out that C_{p} = j(τ) or r_{p}(τ)! For more about the rfunctions r_{p}(τ), see also Article 1, “The jfunction and Its Cousins”.
IV. The variables A and B
Given the eqn,
1/π = Σ h_{p} (An + B) / C^{n},
Conveniently, A can be expressed in terms of C as,
1. for h_{1} and jfunction j(τ): A_{1} = √a_{1}, a_{1} = d(1 – 12^{3}/C_{1}); and C_{1} = j(τ). 2. for h_{p} and rfunctions r_{p}(τ): A_{p} = (1/p)√a_{p}, a_{p} = d(1 – 4p^{v}/C_{p}); v = 6/(p1); and C_{p} = r_{p}(τ).
hence,
A_{1} = (1/1)√(d(1 – 1728/C_{1})) A_{2} = (1/2)√(d(1 – 256/C_{2})) A_{3} = (1/3)√(d(1 – 108/C_{3})) A_{4} = (1/4)√(d(1 – 64/C_{4}))
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},
C_{1} = j(τ) = 66^{3}, a_{1} = 4*4(1 – 1728/C_{1}), so A_{1} = (1/1)√a_{1} = (84/121)√33 C_{2} = r_{2}(τ) = 396^{4}, a_{2} = 4*58(1 – 256/C_{2}), so A_{2} = (1/2)√a_{2} = (52780/9801)√2 C_{3} = r_{3}(τ) = 6^{3}, a_{3} = 4*6(1 – 108/C_{3}), so A_{3} = (1/3)√a_{3} = (2/3)√2 C_{4} = r_{4}(τ) = 2^{12}, a_{4} = 4*28(1 – 64/C_{4}), so A_{4} = (1/4)√a_{4} = 21/8
Let τ = {(1+√7)/2, 1+√37, (1+√267)/2, 2+√16},
C_{1} = j(τ) = 15^{3}, a_{1} = 7(1 – 1728/C_{1}), so A_{1} = (1/1)√a_{1} = (21/25)√15 C_{2} = r_{2}(τ) = 14112^{2}, a_{2} = 4*37(1 – 256/C_{2}), so A_{2} = (1/2)√a_{2} = 5365/882 C_{3} = r_{3}(τ) = 300^{3}, a_{3} = 267(1 – 108/C_{3}), so A_{3} = (1/3)√a_{3} = (4717/1500)√3 C_{4} = r_{4}(τ) = 2^{9}, a_{4} = 4*16(1 – 64/C_{4}), so A_{4} = (1/4)√a_{4} = (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 C_{4} = 2^{9},
x = NSolve[Sum[h_{4}(An+B)/C^{n} /. A →(3/2)√2 /. C→ 2^{9} /. h_{4} → (2n)!^{3} / n!^{6}, {n, 0, 50}] = 1/Pi, B, 30] = 0.353553390593…
The Mathematica addon 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 18B^{2} = 0, so B = (1/4)√2, hence,
1/π = Σ h_{4} ((3/2)√2n + (1/4)√2) / (2^{9})^{n}
simplified as,
1/π = (1/4)√2 Σ h_{4} (6n+1) / (2^{9})^{n}
Equivalently,
1/π = 8 Σ h_{4} (1)^{n} (6n+1) / (2^{9})^{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 jfunction j(τ) and class number h(d) = 1? He almost did. To see this, first define v_{1} = (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 v_{1} (11n+1)(4/125)^{n} 1/π = (18√3)/(85√85) Σ v_{1} (133n+8)/(4/85)^{3n}
However, this can be transformed to,
1/π = 72 Σ h_{1} (11n+1)/(2*30^{3})^{n+1/2}, e^{π√12} ≈ 2*30^{3}  747… 1/π = 162 Σ h_{1} (133n+8)/(255^{3})^{n+1/2}, e^{π√28} ≈ 255^{3} 744.01…
where, as before, h_{1} = (6n)! / ((3n)! n!^{3}). Thus, Ramanujan also found formulas using nonfundamental 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 f_{p} are also useful for calculating singular moduli. To recall, the h_{p} can also be defined as Pochhammer symbol products,
h_{1} = 1728^{n} (1/2)_{n}(1/6)_{n}(5/6)_{n} / (n!^{3}) h_{2} = 256^{n} (1/2)_{n}(1/4)_{n}(3/4)_{n} / (n!^{3}) h_{3} = 108^{n} (1/2)_{n}(1/3)_{n}(2/3)_{n} / (n!^{3}) h_{4} = 64^{n} (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 _{2}F_{1},
_{2}F_{1}(a,b;c;z):= Σ g z^{n}/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,
m_{1} = _{2}F_{1}(a, b; 1; 1α_{N}) m_{2} = _{2}F_{1}(a, b; 1; α_{N})
Then, for positive numbers {a,b} such that a+b = 1, the singular modulus α_{N} is the unique positive real number 0 < α_{N} < 1 satisfying,
m_{1}/m_{2} = √N
To find α_{N}, it can be observed that,
1. Let {a,b} = {1/6, 5/6}. Then p = 1, α_{N} = (1√(1u_{1}))/2, and u_{1} = 1728 / j(τ) 2. Let {a,b} = {1/4, 3/4}. Then p = 2, α_{N} = 1/(u_{2}+1), and u_{2} = (1/2^{6}) f_{2}^{24} 3. Let {a,b} = {1/3, 2/3}. Then p = 3, α_{N} = 1/(u_{3}+1), and u_{3} = (1/3^{3}) f_{3}^{12} 4. Let {a,b} = {1/2, 1/2}. Then p = 4, α_{N} = 1/(u_{4}+1), and u_{4} = (1/4^{2}) f_{4}^{8}
where f_{p} are the Dedekind eta quotients,
f_{2} = η[τ/2] / η[τ] f_{3} = η[τ/3] / η[τ] f_{4} = η[τ/4] / η[τ]
and τ = √(pn).
Example. In his second letter to Hardy, Ramanujan gave the amazing radical,
α_{210} = (83√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,
α_{210} = 1/(u_{4}+1), where u_{4} = (1/4^{2}) f_{4}^{8} = (1/4^{2})(η[τ/4]/η[τ])^{8}
and τ = √(210p) = √(210*4), and we find that indeed,
α_{210} = 2.7066… x 10^{(19)}
such that,
m_{1}/m_{2} = √210 To express α_{N} 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/π^{k} = Σ h P(n) / C^{n} (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 semitrivial 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,
v_{2} = (2n)!/(n!^{2})
then, again for n = 0 to ∞,
1. 1/π = 2^{2} Σ v_{2}^{3} (6n+1) / (2^{8})^{n+1/2}, 2. 1/π = 2^{2} Σ v_{2}^{3} (42n+5) / (2^{12})^{n+1/2} 3. 1/π^{2} = 2^{3} Σ v_{2}^{5} (1)^{n }(20n^{2}+8n+1) / (2^{12})^{n+1/2} 4. 1/π^{2} = 2^{3} Σ v_{2}^{5} (1)^{n }(820n^{2}+180n+13) / (2^{20})^{n+1/2} 5. 1/π^{3} = 2^{5} Σ v_{2}^{7} (168n^{3}+76n^{2}+14n+1) / (2^{20})^{n+1/2} 6. 1/π^{4} = 2^{7} Σ v_{2}^{9} 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/π^{k} 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
