### 0010: Article 0 (Pi Formulas 1)

 Back to Index       Go to Updates Page     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  (For an updated summary, also see Ramanujan's Pi Formulas and the Hypergeometric Function.)   I.  Class number h(-d) = 1.  It is quite well-known that,   eπ√7 ≈ 153 + 697 eπ√11 ≈ 323 + 738 eπ√19 ≈ 963 + 743 eπ√43 ≈ 9603 + 743.999… eπ√67 ≈ 52803 + 743.99999… eπ√163 ≈ 6403203 + 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,   eπ√7 ≈ 33(32-22)3 + 697 eπ√11 ≈ 43(32-1)3 + 738 eπ√19 ≈ 123(32-1)3 + 743 eπ√43 ≈ 123(92-1)3 + 744 eπ√67 ≈ 123(212-1)3 + 744 eπ√163 ≈ 123(2312-1)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,   eπ√11 ≈ x24 – 24;    (x3-2x2+2x-2 = 0) eπ√19 ≈ x24 – 24;    (x3-2x-2 = 0) eπ√43 ≈ x24 – 24;    (x3-2x2-2 = 0) eπ√67 ≈ x24 – 24;    (x3-2x2-2x-2 = 0) eπ√163 ≈ x24 – 24;   (x3-6x2+4x-2 = 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 x3-6x2+4x-2 = 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 eπ√163 gives an inkling of the reason why.)    These transcendental numbers can also be closely approximated by algebraic numbers of degree 4,   eπ√19 ≈ 35(3-√(2v1))-2 - 12.00006…    eπ√43 ≈ 35(9-√(2v2))-2  - 12.000000061…    eπ√67 ≈ 35(21-√(2v3))-2 - 12.00000000036…    eπ√163 ≈ 35(231-√(2v4))-2 - 12.00000000000000021…      (notice how the numbers {3, 9, 21, 231} appear again) and where the vi are,   v1 = -3 + 1√(3*19) v2 = -39 + 7√(3*43) v3 = -219 + 31√(3*67) v4 = -26679 + 2413√(3*163)   and that,   3*26(-32 + 3*19*12) = 962 3*26(-392 + 3*43*72) = 9602 3*26(-2192 + 3*67*312) = 52802 3*26(-266792 + 3*163*24132) = 6403202   which, with the appropriate fractional power, are precisely the j-invariants.  As well as for algebraic numbers of degree 6,   eπ√19 ≈ (5x)3 - 6.000010…    eπ√43 ≈ (5x)3 - 6.000000010…    eπ√67 ≈ (5x)3 - 6.000000000061…    eπ√163 ≈ (5x)3 - 6.000000000000000034…     where the x’s are given respectively by the appropriate root of the sextics,   5x6-96x5-10x3+1 = 0 5x6-960x5-10x3+1 = 0 5x6-5280x5-10x3+1 = 0 5x6-640320x5-10x3+1 = 0   with the j-invariants 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,   5x3 - 5(53+86φ)x2 + 5(8+13φ)x - (18+29φ) = 0 5x3 - 20(73+118φ)x2 - 20(21+34φ)x - (47+76φ) = 0 5x3 - 20(8849+14318φ)x2 + 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[5x3 - 5(53+86φ)x2 + 5(8+13φ)x - (18+29φ),  φ2-φ-1,  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,   Fn-1 + Fn φ = φn   A related one for Lucas numbers is,   Ln-1 + Ln φ = φn√5   Hence, the cubic factors simplify as,   5x3 - 5 (4φ8 + φ3) x2 + 5φ7 x - φ7√5 = 0 5x3 - 20 (2φ10 + φ6) x2 - 20φ9 x - φ9√5 = 0 5x3 - 20 (4√5 φ17 + φ9) x2 + 20φ15 x - φ15√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,   x1 = eπi/24 η(τ)/η(2τ) x2 = eπi/12 η(τ)/η(3τ) x3 = eπi/6  η(τ)/η(5τ)   leading to,   eπ√163 ≈ x124 - 24.0000000000000010… eπ√163 ≈ x212 - 12.00000000000000021…  eπ√163 ≈ x36  - 6.000000000000000034…   where the eta quotients xi are the algebraic numbers given above.  There are also nice Diophantine relationships for the approximants of degree 1,   7 (123 + 153) = 32 *632 11 (123 + 323) = 42 *154219 (123 + 963) = 122 *3422 43 (123 + 9603) = 122 *162542 67 (123 + 52803) = 122 *2617022 163 (123 + 6403203) = 122 *5451401342   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)/(153)n+1/2 1/π = 4 Σ c (154n+15)/(323)n+1/2 1/π = 12 Σ c (342n+25)/(963)n+1/2 1/π = 12 Σ c (16254n+789)/(9603)n+1/2 1/π = 12 Σ c (261702n+10177)/(52803)n+1/2 1/π = 12 Σ c (545140134n+13591409)/(6403203)n+1/2   where the sum Σ goes from n = 0 to ∞.  Beautiful, aren’t they?  The modular function responsible for this is the j-function which has the q-series expansion,   j(q) = 1/q + 744 + 196884q + 21493760q2 + 864299970q3 + ...   which explains why, especially for these discriminants, the number eπ√d has an excess close to the integer "744".  This is sequence A000521 of the OEIS and, other than the constant term, is the McKay-Thompson 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 eπ√m gets intriguingly close to an integer with a fixed excess "104",   eπ√5 ≈ 322 + 100 eπ√13 ≈ 2882 + 103.9… eπ√37 ≈ 141122 + 103.9999…   eπ√6 ≈ 482 - 106 eπ√10 ≈ 1442 - 104.2… eπ√22 ≈ 15842 - 104.001… eπ√58 ≈ 1568162 - 104.0000001…   Just like for d with class number h(-d) = 1, there are also nice Diophantine relationships between these numbers,   5 (44 + 322) = 42 *202 13 (44 + 2882) = 42 *2602 37 (44 + 141122) = 42 *214602   6 (-44 + 482) = 82 *3*82 10 (-44 + 1442) = 82 *2*402 22 (-44 + 15842) = 82 *11*2802 58 (-44 + 1568162) = 82 *2*1055602   So the sums for the first set are perfect squares, while the second set are almost-squares. 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)/(322)n+1/2 1/π = 4 Σ r (-1)n (260n+23)/(2882)n+1/2 1/π = 4 Σ r (-1)n (21460n+1123)/(141122)n+1/2   1/π = 8√3  Σ r (8n+1)/(482)n+1/2 1/π = 8√2  Σ r (40n+4)/(1442)n+1/2 1/π = 8√11  Σ r (280n+19)/(15842)n+1/2 1/π = 8√2  Σ r (105560n+4412)/(1568162)n+1/2   where n = 0 to ∞.  The modular function, call it r(q), that is responsible for this has the q-series expansion,   r(q) = 1/q + 104 + 4372q + 96256q2 + 1240002q3 + …   and explains why the number eπ√m has an excess close to the integer "104".  This is sequence A007267 and, excepting the first term, is the McKay-Thompson 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 eπ√(m/3) gets close to an integer with a fixed excess "42", though I will focus only on the three highest m,   eπ√(17/3) ≈ 123 + 41.6… eπ√(41/3) ≈ 483 + 41.99… eπ√(89/3) ≈ 3003 + 41.9999…   and three non-fundamental odd m = {9, 25, 49},   eπ√(9/3) ≈ 3*43 + 38.8… eπ√(25/3) ≈ 5*123 + 41.91… eπ√(49/3) ≈ 7*363 + 41.998…   Since these non-fundamental 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 (2233 + 123) = 62 *512 3*41 (2233 + 483) = 62 *6152 3*89 (2233 + 3003) = 62 *141512   3 (2233 + 3*43) = 62 *52 3 (2233 + 5*123) = 62 *272 3 (2233 + 7*363) = 62 *1652   so the sums are perfect squares.  Define r2 = (2233)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+n-1).  Equivalently, let r2 = (2n)! (3n)!/ (n!5), then,   1/π = 2 Σ r2 (-1)n (51n+7)/(123)n+1/2 1/π = 2 Σ r2 (-1)n (615n+53)/(483)n+1/2 1/π = 2 Σ r2 (-1)n (14151n+827)/(3003)n+1/2   1/π = 2*3 Σ r2 (-1)n (5n+1)/(3*43)n+1/2 1/π = 2*5 Σ r2 (-1)n (27n+3)/(5*123)n+1/2 1/π = 2*7 Σ r2 (-1)n (165n+13)/(7*363)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 Ramanujan-Type Series for 1/π" (2000).  The modular function responsible, call it h(q), has q-series expansion,   h(q) = 1/q + 42 + 783q + 8672q2 + 65367q3 + …   This is sequence A030197 and, perhaps not surprisingly, is the McKay-Thompson series of class 3A for the Monster group.  To summarize, there is this deep relationship between transcendental numbers of the form eπ√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.       ◄