aa r X i v : . [ m a t h . N T ] A p r Some onje tural ontinued fra tionsThomas Baru hel(cid:3)September 30, 2018
Method and notation
All results presented below have been found by experimental means. Millions ofrandom ontinued fra tions have been omputed at very high speed ( ode waswritten in C and double-pre ision (cid:13)oating numbers were used for the greatestpart of the omputation despite their relatively low pre ision). Relations weredete ted by mat hing values against various databases with a fast implementa-tion of the famous PSLQ algorithm. Of ourse, all expressions have been later he ked with a mu h higher pre ision, but they are not proved and they are opied here as mere onje tures.Several databases of interesting onstants were used; one of them was builtfrom the book of Mi hael Shamos. Later ad ho databases were built in orderto fo us more a urately on spe i(cid:12) al onstants.For onvenien e reasons, the notation being used below for the generalized on-tinued fra tions is the old one used by Gauss: K1n=1 bnan . It has to be rememberedhowever that the expression at the right of the K operator is not a ompletefra tion that ould be an elled down.a0 + 1Kn=1 bnan = a0 + b1a1 + b2a2 + b3a3 + . . . (cid:0) ((cid:24) + 1=2) =(cid:0) ((cid:24))A ontinued fra tion for the quotient (cid:0) ((cid:24) + 1=2) =(cid:0) ((cid:24)) involving simple polyno-mial fun tions was onje tured to exist and therefore an algorithm was writtenfor gathering as many onstants related to the (cid:0) fun tion as possible. Manyidentities were dete ted for parti ular values and a general expression was found:1Kn=1 ((cid:11)n + 1) ((cid:0)2(cid:11)n + 2(cid:11)(cid:24) (cid:0) 1)3(cid:11)n + 2 + (cid:11) = (cid:0)1 (cid:0) (cid:11) + (cid:0) (cid:18)(cid:24) + 12 (cid:19) (cid:0) (cid:18) 12(cid:11) (cid:19)(cid:0) ((cid:24)) (cid:0) (cid:18) (cid:11) + 12(cid:11) (cid:19) (1)(cid:3)baru helriseup.net 1
An identity involving the lemniscate constant
By looking for values related to the lemnis ate onstant in randomly generated ontinued fra tions made of integer terms, a general expression ould be found:1Kn=1 (cid:0) (2n + (cid:11)) (4n (cid:0) 4(cid:24) + 2(cid:11) (cid:0) 1)6n + 2(cid:11) + 2= (cid:0)2 (cid:0) (cid:11) + 2p2 (cid:0) (cid:18) 14 (cid:19) (cid:0) (cid:18) 34 (cid:19) (cid:0) (cid:16)1 + (cid:11)4 (cid:17) (cid:0) (cid:18) 34 (cid:0) (cid:11)4 + (cid:24)(cid:19)(cid:25) (cid:0) (cid:18) 12 + (cid:11)4 (cid:19) (cid:0) (cid:18) 14 (cid:0) (cid:11)4 + (cid:24)(cid:19) (2)where the ase (cid:11) = 1 here overs the ase (cid:11) = 2 in the previous formula (1).
The fun tion g is de(cid:12)ned as a generalized ontinued fra tiong (k; x) = 1Kn=1 (n + 1) (k (cid:0) n (cid:0) k=x) =x(n + 1) (1 + 1=x)then the following relation stands:2 + g ((cid:11); (cid:24)) + g (cid:18)(cid:11); (cid:24)(cid:24) (cid:0) 1 (cid:19)= (cid:11) ((cid:24) (cid:0) 1)(cid:11)=(cid:24)(cid:0)1 (cid:18) (cid:24)(cid:24) (cid:0) 1 (cid:19)(cid:11)(cid:0)2 B ((cid:11)=(cid:24); (cid:11) (cid:0) (cid:11)=(cid:24)) (3)where B is the beta fun tion.An easy ase o urs when (cid:11) = (cid:24)= ((cid:24) (cid:0) 1), whi h leads to g ((cid:11); (cid:24)) = 0. Then, hoosing (cid:24) = (cid:30) (the golden ratio) gives the ni e identity:(cid:30)(cid:30) = 2 + 1Kn=1 (n + 1) (1 (cid:0) n=(cid:30)) =(cid:30)(n + 1) (2 (cid:0) 1=(cid:30)) (4)while hoosing (cid:24) = (cid:30)2 gives:(cid:30)2=(cid:30) = 2 + 1Kn=1 (n + 1) (1 (cid:0) (n + 1) =(cid:30))(n + 1) (cid:30) (5)and similarly for other onstants like p3, 3p4, 4p5, et . The general expression oming from this ase (cid:11) = (cid:24)= ((cid:24) (cid:0) 1) being:(cid:18) xx (cid:0) 1 (cid:19)x(cid:0)1 = 2 + 1Kn=1 (n + 1) (x (cid:0) n (cid:0) 1) =x(n + 1) (x + 1) =x (6)Building a similar identity relying on g (2(cid:24); (cid:24)= ((cid:24) (cid:0) 1)) is not very diÆ ult, butthe resulting expression is not as simple as previously:x (cid:0) 12x (cid:0) 1 (cid:18) xx (cid:0) 1 (cid:19)2x(cid:0)1 (cid:0) 1! = 2 + 1Kn=1 (n + 1) (2x (cid:0) n (cid:0) 2) =x(n + 1) (x + 1) =x (7)2
A continued fraction for tanh zWhile trying to dete t mis ellaneous onstants by building ontinued fra tionswith simple polynomials, several di(cid:11)erent values related to the hyperboli tan-gent fun tion were dete ted. Generalizing happened to be very easy and theresulting ontinued fra tion seems to be di(cid:11)erent from the usual one:tanh z = 1Kn=1 1 + (cid:16)(n (cid:0) 1) (cid:25)4 z(cid:0)1(cid:17)2(2n (cid:0) 1) (cid:25)4 z(cid:0)1 (8)Convergen e is slower however than the lassi al ontinued fra tion for tanh.On the other hand, this version produ es ni e and simple ontinued fra tionswhen used with multiples of (cid:25).
Though Euler's ontinued fra tion formula gives an easy ontinued fra tion forthe following sum of produ ts, the ontinued fra tion below seems to have aqui ker onvergen e: 1Xn=1 1Qnk=0 (cid:11)k + 1 = 1Kn=1 ((cid:11)n)(cid:0)11 (9)Interesting values o ur for several values of (cid:11):(cid:0)q (cid:25)2e er(cid:12) 1p2 for (cid:11) = (cid:0)2 e2 (cid:0) 52 for (cid:11) = 12e (cid:0) 2 for (cid:11) = 1 (cid:0)1 + q e(cid:25)2 erf 1p2 for (cid:11) = 2
References [BA1991℄ D. H. Bailey and H. R. P. Ferguson, A polynomial time, numeri allystable integer relation algorithm, SRC Te hni al Report SRC-TR-92-066; RNR Te hni al Report RNR-91-032 (16 De ember 1991; 14 July1992), 1{14.[BA1999℄ D. H. Bailey, H. R. P. Ferguson and S. Arno, Analysis of PSLQ,an integer relation (cid:12)nding algorithm, Mathemati s of Computation,vol. 68, 351{369, 1999.[SH2001℄ M. Shamos, Catalogue of the real numbers, 2011; available online at http://euro.ecom.cmu.edu/shamos.htmlhttp://euro.ecom.cmu.edu/shamos.html