Very cryptic stuff

By Jörg Arndt <arndt@jjj.de> This document is online at http://www.jjj.de/

26-November-1996

This file is here for the sole purpose of giving the nice explicit formulas for 1/π (3c for n=1555, 3b for n=190, 1 for n=862, 3a for n=163). For more formulas and explanation cf. the paper ’More Ramanujan-type Series for 1/π’ in the proceedings ’Ramanujan Revisited’, Proceedings of the Centenary Conference, Univ. of Illinois at Urbana-Champaign June 1-5, 1987.

The explicit formulas were made 1994 using Mathematica (for numerical computation of the quantities in the general formulas and for finding the minimal polynomials) and MapleV (for solving the polynomials and numerical verification of the results).

(\frac{2 K}{π} (k))^{2} = m(k) F (Φ(k))
= \frac{1}{1 + k^{2}} _{3}F_{2} (\frac{1}{4}, \frac{3}{4}, \frac{1}{2}; 1, 1; (\frac{g^{12} + g^{−12}}{2} )^{−2})

= \frac{1}{k^{′2} − k^{2}} _{3}F_{2} (\frac{1}{4}, \frac{3}{4}, \frac{1}{2}; 1, 1; − (\frac{G^{12} − G^{−12}}{2} )^{−2})
= \frac{1}{\sqrt{1 − k^{2} k^{′2}}} _{3}F_{2} (\frac{1}{6}, \frac{5}{6}, \frac{1}{2}; 1, 1; J^{−1})

1. Series in x_{N} N ≥ 3:

\frac{1}{π} = ∑_{n=0}^{∞} \frac{(\frac{1}{4})n (\frac{2}{4})n (\frac{3}{4})n}{n!^{3}} d_{n}(N) x_{N}^{2n+1}

x_{N} := (\frac{g_{N}^{12} + g_{N}^{−12}}{2} )^{−1} = \frac{4 k(N) k^{′2}(N)}{(1 + k^{2}(N))^{2}}

d_{n} := (\frac{α(N) x_{N}^{−1}}{1 + k^{2}(N)} − \frac{\sqrt{N}}{4} g_{N}^{−12}) + n \sqrt{N} (\frac{g_{N}^{12} − g_{N}^{−12}}{2} )

2. Series in y_{N} N ≥ 4:

\frac{1}{π} = ∑_{n=0}^{∞} (−1)^{n} \frac{(\frac{1}{4})n (\frac{2}{4})n (\frac{3}{4})n}{n!^{3}} e_{n}(N) y_{N}^{2n+1}

y_{N} := (\frac{G_{N}^{12} − G_{N}^{−12}}{2} )^{−1} = \frac{4 k(N) k′(N)}{1 − (2 k(N) k′(N))^{2}}

e_{n} := (\frac{α(N) y_{N}^{−1}}{k^{′2}(N) − k^{2}(N)} + \frac{\sqrt{N}}{2} k^{2}(N) G_{N}^{12}) + n \sqrt{N} (\frac{G_{N}^{12} + G_{N}^{−12}}{2} )

3. Series in J_{N}^{−1} N ≥ 2:

\frac{1}{π} = ∑_{n=0}^{∞} \frac{(\frac{1}{6})n (\frac{3}{6})n (\frac{5}{6})n}{n!^{3}} f_{n}(N) (J_{N}^{−1/2})^{2n+1}

J_{N}^{−1} := \frac{27 G_{N}^{24}}{(4 G_{N}^{24} − 1)^{3}} = \frac{27 g_{N}^{24}}{(4 g_{N}^{24} + 1)^{3}}

f_{n} := \frac{1}{3\sqrt{3}} (\sqrt{N} \sqrt{1 − G_{N}^{−24}} + 2 (α(N) − \sqrt{N} k^{2}(N)) (4G_{N}^{24} − 1)) +

+n \sqrt{N} \frac{2}{3\sqrt{3}} ((8G_{N}^{24} + 1) \sqrt{1 − G_{N}^{−24}})

\frac{1}{π} = \frac{1}{\sqrt{−12 J}} ∑_{n=0}^{∞} \frac{(\frac{1}{6})n (\frac{3}{6})n (\frac{5}{6})n}{n!^{3}} \frac{A + n B}{J^{n}}

(type 3c) for n=1555 we get:

A := 5280419026080999965452185 + 2361475178400070170568800 \sqrt{5} +

+32 \sqrt{5}(10891728551171178200467436212395209160385656017 +

+4870929086578810225077338534541688721351255040 \sqrt{5})^{1/2}

B := 654159204458052267524145750 + 292548889855077669080467200 \sqrt{5} +

+209664 \sqrt{3110}(6260208323789001636993322654444020882161 +

+2799650273060444296577206890718825190235 \sqrt{5})^{1/2}

J := −[17897749588626020 + 8004116944887336 \sqrt{5} +

+108 \sqrt{5}(10985234579463550323713318473 +

+4912746253692362754607395912 \sqrt{5})^{1/2}]^{3}

(each term adds 50 correct digits)

0 = 91056965337194438815073158624225 + 9214187265360390391808927003100 A +

+672035320036821921804675631270 A^{2} − 21121676104323999861808740 A^{3} + A^{4}

0 = 17514180018137387326565131389045795600 + 193756947585743300725193322013380000 B +

+2063419786805410130433556462222680 B^{2} − 2616636817832209070096583000 B^{3} + B^{4}

0 = 27192565672854630400 + 49698245345181030400 U +

+22885453089727782720 U^{2} − 71590998354504080 U^{3} + U^{4} where J = −U^{3}

\frac{1}{π} = \frac{1}{\sqrt{3J}} ∑_{n=0}^{∞} \frac{(\frac{1}{6})n (\frac{3}{6})n (\frac{5}{6})n}{n!^{3}} \frac{A + n B}{J^{n}}

(type 3b) for n=190 we get:

A := 21242668516504965 + 15020834958518500\sqrt{2} +

+2\sqrt{5}(45125096427586568251645610141659 + 31908261685643312902173585434250\sqrt{2})^{1/2}

B := 1839779353703421900 + 1300920456890691000\sqrt{2} +

+24337404\sqrt{10}(1142912476713024496667 + 808161162586491705750\sqrt{2})^{1/2}

J := [71864175655 + 22725423252\sqrt{10} + 2808\sqrt{5}(261993316778681 + 82849561276216\sqrt{10})^{1/2}]^{3}

(each term adds 34 digits)

0 = 18983882886895192207942622025 − 2233154457185835655186373700 A +

+5704998902295029443240990 A^{2} − 84970674066019860 A^{3} + A^{4}

0 = 11316047287507303785105891917318400 − 143564046791790430632439232928000 B +

+21396235898865291113024998560 B^{2} − 7359117414813687600 B^{3} + B^{4}

0 = 1860185517864501025 − 1262383694834359900 U +

+44498697145120230 U^{2} − 287456702620 U^{3} + U^{4} where J = −U^{3}

\frac{1}{π} = ∑_{n=0}^{∞} \frac{(\frac{1}{4})n (\frac{2}{4})n (\frac{3}{4})n}{n!^{3}} \frac{A + n B}{X^{2n+1}}

(type 1) for n=862 we get:

A := [4521962731044058367634998271455136035/4 + 799377627848523458605912125112563234\sqrt{2} +

+12(17750127552909235203012377369182079345275390781190873870656491261057219 + +1255123555958829884236839904476079251826408616374387198187634303258534\sqrt{2})^{1/2}]^{1/2}

B := [9617761395088953485915444091307636106000 + 6800784302301588686616253973429782154400\sqrt{2}

+52003425600\sqrt{2}(34204566586722903151731072537516469136640672047198830592963 + +24186280981018566606552309811255775851849456510216830399522\sqrt{2})^{1/2}]^{1/2}

X := 1670141896514232075 + 1180968660568974600\sqrt{2} +

+2736\sqrt{2}(372627201865017746341791564603 + 263487221293322577155951514850\sqrt{2})^{1/2} (each term adds 37 digits)

\frac{1}{π} = \frac{6541681608}{\sqrt{640320}^{3}} ∑_{k=0}^{∞} (\frac{13591409}{545140134} + k) (\frac{(6k)!}{(3k)! (k!)^{3}} \frac{(−1)^{k}}{640320^{3k}} )

Chudnovsky’s formula (type 3a for n=163)

Elliptic integral of the first kind:

K(k) := \frac{π}{2} _{2}F_{1} (\frac{1}{2} , \frac{1}{2} ; 1; k^{2}) = \frac{π}{2} ∑_{i=0}^{∞} (\frac{(2 i − 1)!!}{2^{i} i!})^{2} k^{2i}
= ∫_{0}^{1} \frac{dt}{\sqrt{(1 − t^{2})(1 − k^{2} t^{2})}} = ∫_{0}^{π/2} \frac{dΘ}{\sqrt{1 − k^{2} sin^{2}(Θ)}}
= \frac{π}{2 AGM(1, k′)}
= \frac{π}{2} Θ_{3}^{2}(q) q = e^{−πK′(k)/K(k)}

Elliptic integral of the second kind:

E(k) := \frac{π}{2} _{2}F_{1} (−\frac{1}{2} , \frac{1}{2} ; 1; k^{2}) = \frac{π}{2} (1 − ∑_{i=0}^{∞} (\frac{(2 i − 1)!!}{2^{i} i!})^{2} \frac{k^{2i}}{2i − 1})
= ∫_{0}^{1} \begin{matrix}√ & \overline{1 − k^{2} t^{2}} \\ √ & \overline{1 − t^{2}} \end{matrix} dt = ∫_{0}^{π/2} \sqrt{1 − k^{2} sin^{2}(Θ)} dΘ

Derivatives:

\frac{dK}{dk} = \frac{E − k^{′2} K}{k k′} \frac{dE}{dk} = \frac{E − K}{k}

Differential equation for K and E:

(k^{3} − k) \frac{d^{2}y}{dk^{2}} + (3 k^{2} − 1) \frac{dy}{dk} + k y = 0

( also satisfied by AGM(1, k)^{−1} and AGM(1 + k, 1 − k)^{−1} ) k′ := \sqrt{1 − k^{2}} K′(k) := K(k′), E′(k) := E(k′)

Θ_{2}(q) := ∑_{n=−∞}^{∞} q^{(n+1/2)^{2}}  Θ_{3}(q) := ∑_{n=−∞}^{∞} q^{n^{2}}  Θ_{4}(q) := ∑_{n=−∞}^{∞} (−1)^{n} q^{n^{2}}
k(q) := k = \frac{Θ_{2}^{2}(q)}{Θ_{3}^{2}(q)}  k′(q) := k′ = \frac{Θ_{4}^{2}(q)}{Θ_{3}^{2}(q)}  q = e^{− \frac{K′(k)}{K(k)}}

E = k^{′2} K + k k^{′2} \frac{dK}{dk}

E′ K + K′ E − K′ K = \frac{π}{2}

singular value function:

k(N) : \frac{K′}{K} (k(N)) = \sqrt{N} k(N) := (\frac{Θ_{2}(q)}{Θ_{3}(q)})^{2} , with q := e^{−π \sqrt{N}}

k(0) = ∞, k(1) = \frac{1}{\sqrt{2}} , k(∞) = 0 k(N) algebraic for N rational

l(N) : \frac{K′}{K} (l(N)) = \frac{L′}{L} = \frac{1}{\sqrt{N}} l(N) := (\frac{Θ_{2}(r)}{Θ_{3}(r)})^{2} , with r := e^{−π/\sqrt{N}} = q^{1/N}

\frac{K′}{K} = N \frac{L′}{L}

u := k^{1/4} v := l^{1/4}

singular value function of the second kind:

α(N) := \frac{E′}{K} − \frac{π}{4 K^{2}} (where k := k(N))
= \sqrt{N} \frac{E′}{K′} − \frac{N π}{4 K^{′2}} = \frac{π}{4 K^{2}} − \sqrt{N} (\frac{E}{K} − 1)
= \frac{π^{−1} − \sqrt{N} 4 q \frac{Θ̇_{4}}{Θ_{4}}}{Θ_{3}^{4}} (where q = e^{−π \sqrt{N}})

α(1) = \frac{1}{2} , α(∞) = \frac{1}{π}

recursions:

α(4N) = \frac{4 α(N) − 2 \sqrt{N} k^{2}(N)}{(1 + k′(N))^{2}} = (1 + y^{2}) α(4N) − 2\sqrt{N}y  where y := k(4N) = \frac{1 − k′(N)}{1 + k′(N)}
α(16N) = (1 + y)^{4}α(N) − 4\sqrt{N}y(1 + y + y^{2})  where y := k(16N) = (\frac{1 − ^{4} \sqrt{1 − k^{2}(n)}}{1 + ^{4} \sqrt{1 − k^{2}(n)}})^{2}
α(N^{−1}) = \frac{\sqrt{N} − α(N)}{N}

G,g:

G := (2k k′)^{−1/12} , g := (\frac{2k}{k^{′2}} )^{−1/12}

recursions:

9 = (1 + 2\sqrt{2} \frac{G_{9n}^{3}}{G_{n}^{9}} ) (1 + 2\sqrt{2} \frac{G_{n}^{3}}{G_{9n}^{9}} ) 9 = (1 + 2\sqrt{2} \frac{g_{9n}^{3}}{g_{n}^{9}} ) (1 − 2\sqrt{2} \frac{g_{n}^{3}}{g_{9n}^{9}} )

G_{81n}^{3} = G_{9n} \frac{\sqrt{2}G_{9n} + G_{n}^{3}}{\sqrt{2}G_{n}^{3} − G_{9n}} g_{81n}^{3} = −g_{9n} \frac{\sqrt{2}g_{9n} + g_{n}^{3}}{\sqrt{2}g_{n}^{3} − g_{9n}}

J := \frac{(4 G^{24} − 1)^{3}}{27 G^{24}} = \frac{(4 g^{24} + 1)^{3}}{27 g^{24}} = \frac{4}{27} \frac{(1 − k^{2}(N) + k^{4}(N))^{3}}{k^{4}(N) (1 − k^{2}(N))^{2}} = \frac{4}{27} \frac{(1 − k^{2}(N)k^{′2}(N))^{3}}{k^{4}(N)k^{′4}(N)}

j := 1728 J multiplier:

M_{N}(l, k) := \frac{Θ_{3}^{2}(q)}{Θ_{3}^{2}(q^{1/N})} = \frac{K}{L}

\frac{1}{π} = \sqrt{N} k k^{′2} \frac{4KK̇}{π^{2}} + (α(N) − \sqrt{N} k^{2}) \frac{4K^{2}}{π^{2}} \frac{1}{K} = \sqrt{N} k k^{′2} \frac{4K̇}{π} + (α(N) − \sqrt{N} k^{2}) \frac{4K}{π}

9

Θ_{1}(z, q) := Θ_{1}(z, t) := 2q^{1/4} sin z − 2q^{9/4} sin 3z + 2q^{25/4} sin 5z − 2q^{49/4} sin 7z + ...
Θ_{2}(z, q) := Θ_{2}(z, t) := 2q^{1/4} cos z + 2q^{9/4} cos 3z + 2q^{25/4} cos 5z + 2q^{49/4} cos 7z + ...
Θ_{3}(z, q) := Θ_{3}(z, t) := 1 + 2q cos 2z + 2q^{4} cos 4z + 2q^{9} cos 6z + 2q^{16} cos 8z + ...
Θ_{4}(z, q) := Θ_{4}(z, t) := 1 − 2q cos 2z + 2q^{4} cos 4z − 2q^{9} cos 6z + 2q^{16} cos 8z − ...
where q := e^{πit}
Θ_{1}(z, q) := Θ_{1}(z, t) := ∑_{n∈Z}(−1)^{n}q^{((2n+1)/2)^{2}} sin (2n + 1)z
Θ_{2}(z, q) := Θ_{2}(z, t) := ∑_{n∈Z} q^{((2n+1)/2)^{2}} cos (2n + 1)z
Θ_{3}(z, q) := Θ_{3}(z, t) := ∑_{n∈Z} q^{(2n)^{2}} cos (2n)z
Θ_{4}(z, q) := Θ_{4}(z, t) := ∑_{n∈Z}(−1)^{n}q^{(2n)^{2}} cos (2n)z

Θ_{1}(z) = −Θ_{2}(z + \frac{π}{2} ) = −iMΘ_{3}(z + \frac{π}{2} + \frac{πt}{2}) = −iMΘ_{4}(z + \frac{πt}{2})
Θ_{2}(z) = MΘ_{3}(z + \frac{πt}{2}) = MΘ_{4}(z + \frac{π}{2} + \frac{πt}{2}) = Θ_{1}(z + \frac{π}{2} )
Θ_{3}(z) = Θ_{4}(z + \frac{π}{2} ) = MΘ_{1}(z + \frac{π}{2} + \frac{πt}{2}) = MΘ_{2}(z + \frac{πt}{2})
Θ_{4}(z) = −iMΘ_{1}(z + \frac{πt}{2}) = iMΘ_{2}(z + \frac{π}{2} + \frac{πt}{2}) = Θ_{3}(z + \frac{π}{2} )
where M := q^{1/4}e^{iz}
Θ_{1}(z) = Θ_{1}(z + π) = −qe^{2iz}Θ_{1}(z + πt)
Θ_{2}(z) = Θ_{2}(z + π) = +qe^{2iz}Θ_{2}(z + πt)
Θ_{3}(z) = Θ_{3}(z + π) = +qe^{2iz}Θ_{3}(z + πt)
Θ_{4}(z) = Θ_{4}(z + π) = −qe^{2iz}Θ_{4}(z + πt)

\frac{∂^{2}Θ(z, t)}{∂z^{2}} = \frac{4i}{π} \frac{∂Θ(z, t)}{∂t}

Θ_{3}^{2} = 1 + 4 ∑_{n=1}^{∞} \frac{q^{n}}{1 + q^{2n}}

Let T (x, q) := ∑_{−∞}^{∞} x^{n}q^{n^{2}} then:

T (x, q^{a})T (y, q^{b}) = ∑_{m,n} x^{m}y^{n}q^{am^{2}+bn^{2}} = ∑_{k=0}^{a+b−1} y^{k}q^{bk^{2}}T (xyq^{2bk}, q^{a+b})T (y^{a}x^{−b}q^{2abk}, q^{ab(a+b)})

Q_{0} := ∏_{n=1}^{∞} 1 − q^{2n} Q_{1} := ∏_{n=1}^{∞} 1 + q^{2n} Q_{2} := ∏_{n=1}^{∞} 1 + q^{2n−1} Q_{3} := ∏_{n=1}^{∞} 1 − q^{2n−1}

Q_{0}Q_{1} = Q_{0}(q^{2}) Q_{0}Q_{3} = Q_{0}(q^{1/2}) Q_{2}Q_{3} = Q_{3}(q^{2}) Q_{1}Q_{2} = Q_{1}(q^{1/2})

∑_{n=−∞}^{∞} (±1)^{n}q^{kn^{2}+ln} = ∏_{n=0}^{∞} (1 ± q^{2kn+k−l})(1 ± q^{2kn+k+l})(1 − q^{2kn+2k})
e.g. :1 = Q_{1}Q_{2}Q_{3}  Θ_{3}(q) = Q_{0}Q_{2}^{2} Θ_{4}(q) = Q_{0}Q_{3}^{2} Θ_{2}(q) = 2q^{1/4}Q_{0}Q_{1}^{2}

k = 3/2 and l = 1/2 gives:

∏_{n=1}^{∞} (1 − q^{n}) = ∑_{−∞}^{∞} (−1)^{n}q^{(3n+1)n/2}

triple-product identity:

∏_{n=1}^{∞} (1 − q^{2n})(1 + xq^{2n−1})(1 + x^{−1}q^{2n−1}) = ∑_{n=−∞}^{∞} x^{n}q^{n^{2}}

quintuple-product identity:

∏_{n=1}^{∞} (1 − q^{n})(1 − xq^{n})(1 − x^{−1}q^{n−1})(1 − x^{2}q^{2n−1})(1 − x^{−2}q^{2n−1}) =

= ∑_{n=−∞}^{∞} (x^{3n} − x^{−3n−1})q^{n(3n+1)/2}

equivalent:

∏_{n=1}^{∞} (1 − q^{2n})(1 − xq^{2n−1})(1 − x^{−1}q^{2n−1})(1 − x^{2}q^{4n−4})(1 − x^{−2}q^{4n−4}) =

= ∑_{n=−∞}^{∞} q^{3n^{2}−2n}[(x^{3n} + x^{−3n}) − (x^{3n−2} + x^{−(3n−2)})]