Pi Formulas
I. Class number h(d) = 1 II. Class number h(d) = 2, where d = 4m
III. Class number h(d) = 2, where d = 3m
I. Class number h(d) = 1. It is quite wellknown that,
e^{π√7} ≈ 15^{3} + 697
e^{π√11} ≈ 32^{3} + 738
e^{π√19} ≈ 96^{3} + 743
e^{π√43} ≈ 960^{3} + 743.999… e^{π√67} ≈ 5280^{3} + 743.99999… e^{π√163} ≈ 640320^{3} + 743.99999999999…
with the last also known as the Ramanujan constant. The discriminants d under the square root are the six highest of the nine Heegner numbers. As I pointed out in a 2008 sci.math.research post, it is not so commonly known there is another internal structure,
e^{π√7} ≈ 3^{3}(3^{2}2^{2})^{3} + 697
e^{π√11} ≈ 4^{3}(3^{2}1)^{3} + 738
e^{π√19} ≈ 12^{3}(3^{2}1)^{3} + 743
e^{π√43} ≈ 12^{3}(9^{2}1)^{3} + 744 e^{π√67} ≈ 12^{3}(21^{2}1)^{3} + 744 e^{π√163} ≈ 12^{3}(231^{2}1)^{3} + 744
The reason for the squares of {3, 9, 21, 231} is due to a certain Eisenstein series, and they will appear again below. Before going to the pi formulas, it can be pointed out that these transcendental numbers, in addition to being closely approximated by integers (which are simply algebraic numbers of degree 1), can also be closely approximated by algebraic numbers of degree 3,
e^{π√11} ≈ x^{24} – 24; (x^{3}2x^{2}+2x2 = 0) e^{π√19} ≈ x^{24} – 24; (x^{3}2x2 = 0) e^{π√43} ≈ x^{24} – 24; (x^{3}2x^{2}2 = 0) e^{π√67} ≈ x^{24} – 24; (x^{3}2x^{2}2x2 = 0) e^{π√163} ≈ x^{24} – 24; (x^{3}6x^{2}+4x2 = 0)
The corresponding cubic equations are extremely simple and have a very similar form. In fact, x can be exactly expressed in terms of the Dedekind eta function η(τ), a modular function which involves a 24th root, and explains the 24 in the approximations. (Note: As Rich Schroeppel points out, John Brillhart, in the early days of computers, looked at continued fractions of cubic irrationalities and found a few had some interesting properties, one of which was x^{3}6x^{2}+4x2 = 0. If you look at the first 200 terms, it is curiously peppered with large terms. Of course, the fact that it is related to e^{π√163} gives an inkling of the reason why.)
These transcendental numbers can also be closely approximated by algebraic numbers of degree 4,
e^{π√19} ≈ 3^{5}(3√(2v_{1}))^{2}  12.00006… e^{π√43} ≈ 3^{5}(9√(2v_{2}))^{2}  12.000000061… e^{π√67} ≈ 3^{5}(21√(2v_{3}))^{2}  12.00000000036… e^{π√163} ≈ 3^{5}(231√(2v_{4}))^{2}  12.00000000000000021…
(notice how the numbers {3, 9, 21, 231} appear again) and where the v_{i }are,
v_{1} = 3 + 1√(3*19) v_{2} = 39 + 7√(3*43) v_{3} = 219 + 31√(3*67) v_{4} = 26679 + 2413√(3*163)
and that,
3*2^{6}(3^{2 }+ 3*19*1^{2}) = 96^{2} 3*2^{6}(39^{2 }+ 3*43*7^{2}) = 960^{2} 3*2^{6}(219^{2 }+ 3*67*31^{2}) = 5280^{2} 3*2^{6}(26679^{2 }+ 3*163*2413^{2}) = 640320^{2}
which, with the appropriate fractional power, are precisely the jinvariants. As well as for algebraic numbers of degree 6,
e^{π√19} ≈ (5x)^{3}  6.000010… e^{π√43} ≈ (5x)^{3}  6.000000010… e^{π√67} ≈ (5x)^{3}  6.000000000061… e^{π√163} ≈ (5x)^{3}  6.000000000000000034…
where the x’s are given respectively by the appropriate root of the sextics,
5x^{6}96x^{5}10x^{3}+1 = 0 5x^{6}960x^{5}10x^{3}+1 = 0 5x^{6}5280x^{5}10x^{3}+1 = 0 5x^{6}640320x^{5}10x^{3}+1 = 0
with the jinvariants appearing again. Another interesting feature of these sextics is that they are not only algebraic, but are also solvable in radicals. These factor into two cubics (with the exception of the first which has a quadratic factor) over the extension Q(φ) where φ = (1+√5)/2 or the golden ratio. The last three, respectively, then have the cubic factors,
5x^{3 } 5(53+86φ)x^{2 }+ 5(8+13φ)x  (18+29φ) = 0 5x^{3 } 20(73+118φ)x^{2 } 20(21+34φ)x  (47+76φ) = 0 5x^{3 } 20(8849+14318φ)x^{2 }+ 20(377+610φ)x  (843+1364φ) = 0
Amazingly, pairs of consecutive Fibonacci numbers F(n) = {0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610,…} appear in the x term, as well as Lucas numbers L(n) = {2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364,…} in the constant term. Why this is so I have no idea as it came as a surprise while I was looking at their factorization over Q(φ). If you have Mathematica, the factorization is easily verified by the Resultant[] function,
Resultant[5x^{3 } 5(53+86φ)x^{2 }+ 5(8+13φ)x  (18+29φ), φ^{2}φ1, φ]
which eliminates the variable φ between the two equations and recovers the original sextic. (Likewise for the other two.) (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_{n1} + L_{n} φ = φ^{n}√5
Hence, the cubic factors simplify as,
5x^{3 } 5 (4φ^{8 }+ φ^{3}) x^{2 }+ 5φ^{7 }x  φ^{7}√5 = 0 5x^{3 } 20 (2φ^{10 }+ φ^{6}) x^{2 } 20φ^{9 }x  φ^{9}√5 = 0 5x^{3 } 20 (4√5 φ^{17 }+ φ^{9}) x^{2 }+ 20φ^{15 }x  φ^{15}√5 = 0
or the coefficients neatly contain powers of φ. (End update) These algebraic approximants of degree 3,4,6 can be exactly expressed as Dedekind eta η(τ) quotients. For example, let τ = (1+√163)/2, then,
x_{1} = e^{πi/24} η(τ)/η(2τ) x_{2} = e^{πi/12 }η(τ)/η(3τ) x_{3} = e^{πi/6 }η(τ)/η(5τ)
leading to,
e^{π√163} ≈ x_{1}^{24}  24.0000000000000010… e^{π√163} ≈ x_{2}^{12}  12.00000000000000021… e^{π√163} ≈ x_{3}^{6}  6.000000000000000034…
where the eta quotients x_{i} are the algebraic numbers given above. There are also nice Diophantine relationships for the approximants of degree 1,
7 (12^{3} + 15^{3}) = 3^{2} *63^{2} 11 (12^{3} + 32^{3}) = 4^{2} *154^{2} 19 (12^{3} + 96^{3}) = 12^{2} *342^{2}43 (12^{3} + 960^{3}) = 12^{2} *16254^{2} 67 (12^{3} + 5280^{3}) = 12^{2} *261702^{2} 163 (12^{3} + 640320^{3}) = 12^{2} *545140134^{2}
So the sums are perfect squares. And where else do the blue and red numbers appear? Of all places, in formulas for π. Let c = (1)^{n}(6n)! / ((n!)^{3}(3n)!), then,
1/π = 3 Σ c (63n+8)/(15^{3})^{n+1/2} 1/π = 4 Σ c (154n+15)/(32^{3})^{n+1/2} 1/π = 12 Σ c (342n+25)/(96^{3})^{n+1/2} 1/π = 12 Σ c (16254n+789)/(960^{3})^{n+1/2} 1/π = 12 Σ c (261702n+10177)/(5280^{3})^{n+1/2} 1/π = 12 Σ c (545140134n+13591409)/(640320^{3})^{n+1/2}
where the sum Σ goes from n = 0 to ∞. Beautiful, aren’t they? The modular function responsible for this is the jfunction which has the qseries expansion,
j(q) = 1/q + 744 + 196884q + 21493760q^{2} + 864299970q^{3} + ...
which explains why, especially for these discriminants, the number e^{π√d} has an excess close to the integer "744". This is sequence A000521 of the OEIS and, other than the constant term, is the McKayThompson series class 1A of the Monster simple group. These pi formulas were discovered by the Chudnovsky brothers inspired by Ramunujan’s formulas given in the next section (which uses d with class number 2). For more details, see “Pi formulas, Ramanujan, and the Baby Monster Group”, by this author (File 07).
II. Class number h(d) = 2, where d = 4m. There are exactly 18 negative fundamental discriminants d with class number h(d) = 2, seven of which are even, namely d = 4m, for m = {5, 13, 37} and m = {6, 10, 22, 58}. As m increases, the expression e^{π√m} gets intriguingly close to an integer with a fixed excess "104",
e^{π√5} ≈ 32^{2} + 100 e^{π√13} ≈ 288^{2} + 103.9… e^{π√37} ≈ 14112^{2} + 103.9999…
e^{π√6} ≈ 48^{2}  106 e^{π√10} ≈ 144^{2}  104.2… e^{π√22} ≈ 1584^{2}  104.001… e^{π√58} ≈ 156816^{2}  104.0000001…
Just like for d with class number h(d) = 1, there are also nice Diophantine relationships between these numbers,
5 (4^{4} + 32^{2}) = 4^{2} *20^{2} 13 (4^{4} + 288^{2}) = 4^{2} *260^{2} 37 (4^{4} + 14112^{2}) = 4^{2} *21460^{2}
6 (4^{4} + 48^{2}) = 8^{2} *3*8^{2} 10 (4^{4} + 144^{2}) = 8^{2} *2*40^{2} 22 (4^{4} + 1584^{2}) = 8^{2} *11*280^{2} 58 (4^{4} + 156816^{2}) = 8^{2} *2*105560^{2}
So the sums for the first set are perfect squares, while the second set are almostsquares. The blue and red numbers again appear in pi formulas courtesy of, who else, but Ramanujan. Let r = (4n)! / (n!^{4}), then,
1/π = 4 Σ r (1)^{n }(20n+3)/(32^{2})^{n+1/2} 1/π = 4 Σ r (1)^{n }(260n+23)/(288^{2})^{n+1/2} 1/π = 4 Σ r (1)^{n }(21460n+1123)/(14112^{2})^{n+1/2}
1/π = 8√3 Σ r (8n+1)/(48^{2})^{n+1/2} 1/π = 8√2 Σ r (40n+4)/(144^{2})^{n+1/2} 1/π = 8√11 Σ r (280n+19)/(1584^{2})^{n+1/2} 1/π = 8√2 Σ r (105560n+4412)/(156816^{2})^{n+1/2}
where n = 0 to ∞. The modular function, call it r(q), that is responsible for this has the qseries expansion,
r(q) = 1/q + 104 + 4372q + 96256q^{2} + 1240002q^{3} + …
and explains why the number e^{π√m} has an excess close to the integer "104". This is sequence A007267 and, excepting the first term, is the McKayThompson series of Class 2A for the Monster group.
III. Class number h(d) = 2, where d = 3m. There are exactly 5 discriminants of this form, namely d = 3m, for m = {5, 8, 17, 41, 89}. This time, the expression e^{π√(m/3)} gets close to an integer with a fixed excess "42", though I will focus only on the three highest m,
e^{π√(17/3)} ≈ 12^{3} + 41.6… e^{π√(41/3)} ≈ 48^{3} + 41.99… e^{π√(89/3)} ≈ 300^{3} + 41.9999…
and three nonfundamental odd m = {9, 25, 49},
e^{π√(9/3)} ≈ 3*4^{3} + 38.8… e^{π√(25/3)} ≈ 5*12^{3} + 41.91… e^{π√(49/3)} ≈ 7*36^{3} + 41.998…
Since these nonfundamental m are the squares of the first 3 odd primes, the approximations look nice, being the product of that prime and a cube. As usual, there are nice Diophantine relationships,
3*17 (2^{2}3^{3} + 12^{3}) = 6^{2} *51^{2} 3*41 (2^{2}3^{3} + 48^{3}) = 6^{2} *615^{2} 3*89 (2^{2}3^{3} + 300^{3}) = 6^{2} *14151^{2}
3 (2^{2}3^{3} + 3*4^{3}) = 6^{2} *5^{2} 3 (2^{2}3^{3} + 5*12^{3}) = 6^{2} *27^{2} 3 (2^{2}3^{3} + 7*36^{3}) = 6^{2} *165^{2}
so the sums are perfect squares. Define r_{2} = (2^{2}3^{3})^{n} (1/2)_{n}(1/3)_{n}(2/3)_{n} / (n!^{3}) where (a)_{n} is the rising factorial, or Pochhammer symbol, such that (a)_{n} = (a)(a+1)(a+2)…(a+n1). Equivalently, let r_{2} = (2n)! (3n)!/ (n!^{5}), then,
1/π = 2 Σ r_{2} (1)^{n }(51n+7)/(12^{3})^{n+1/2} 1/π = 2 Σ r_{2} (1)^{n }(615n+53)/(48^{3})^{n+1/2} 1/π = 2 Σ r_{2} (1)^{n }(14151n+827)/(300^{3})^{n+1/2}
1/π = 2*3 Σ r_{2} (1)^{n }(5n+1)/(3*4^{3})^{n+1/2} 1/π = 2*5 Σ r_{2} (1)^{n }(27n+3)/(5*12^{3})^{n+1/2} 1/π = 2*7 Σ r_{2} (1)^{n }(165n+13)/(7*36^{3})^{n+1/2}
where n = 0 to ∞. These have been derived by this author (On Ramanujan's Other Pi Formulas, file 08), though Chan and Liaw gave a different form for the first one in their paper, "Cubic Modular Equations and New RamanujanType Series for 1/π" (2000). The modular function responsible, call it h(q), has qseries expansion,
h(q) = 1/q + 42 + 783q + 8672q^{2} + 65367q^{3} + …
This is sequence A030197 and, perhaps not surprisingly, is the McKayThompson series of class 3A for the Monster group. To summarize, there is this deep relationship between transcendental numbers of the form e^{π√d}, the modular functions j(q), r(q), h(q), pi formulas, and the Monster group. (End update.)
P.S. Digressing for a moment, just a little Youtube post that I like. This is Simon Zimmer's excellent unplugged cover version of Christopher Cross' Sailing. The guitar playing is brilliant, and the deeper timbre of this artist's voice is a refreshing contrast to the original. Wouldn't it be great to go sailing?
You can email author at tpiezas@gmail.com.
