Research Article Parametric Evaluations of the Rogers-Ramanujan Continued Fraction

International Mathematics and Mathematical Sciences Volume 011, Article ID 940839, 11 pages doi:10.1155/011/940839 Research Article Parametric Evaluations of the Rogers-Ramanujan Continued Fraction Nikos Bagis Department of Informatics, Aristotle University of Thessaloniki, Thessaloniki, Greece Correspondence should be addressed to Nikos Bagis, bagkis@hotmail.com Received 1 May 011; Accepted 9 August 011 Academic Editor: B. N. Mandal Copyright q 011 Nikos Bagis. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. In this paper with the help of the inverse function of the singular moduli we evaluate the Rogers- Ranmanujan continued fraction and its first derivative. 1. Introductory Definitions and Formulas For q < 1, the Rogers-Ramanujan continued fraction RRCF see 1 is defined as R q ) : q1/5 1 q 1 q q 3 1 1 1. 1.1 We also define n 1 a; q )n : 1 aq k), k0 1. f q ) : 1 q n ) q; q ). n1 Ramanujan give the following relations which are very useful: 1 R q ) 1 R q ) f ) q 1/5 q 1/5 f q 5), 1.3

International Mathematics and Mathematical Sciences 1 R 5 q ) 11 R5 q ) f ) 6 q qf 6 q 5). 1.4 From the theory of elliptic functions see 1 3, Kx π/ 0 1 1 x sin t dt, 1.5 is the elliptic integral of the first kind. It is known that the inverse elliptic nome k, k r 1 k r is the solution of the equation Kk r Kk r, 1.6 where r R. When r is rational then the are algebraic numbers. We can also write the function f using elliptic functions. It holds see 3 andalsoholds f q ) 8 8/3 π 4 q 1/3 /3 ) 8/3Kkr 4, 1.7 f q ) 6 k rk 3. 1.8 π 3 q 1/ From 4 it is known that R q ) 1 5 q 5/6 f q ) 4 R q ) 6 R q ) 5 11 R q ) 5. 1.9 Consider now for every 0 <x<1 the equation x, 1.10 which has solution r k 1 x. 1.11 Hence for example k 1 1 ) 1. 1.1 With the help of k 1 function we evaluate the Rogers Ramanujan continued fraction.

International Mathematics and Mathematical Sciences 3. Propositions The relation between k 5r and is see 1 page 80 k 5r k rk 5r 41/3 k 5r k rk 5r) 1/3 1..1 For to solve.1 we give the following. Proposition.1. The solution of the equation x 6 x 3 16 10x ) w 15x 4 w 0x 3 w 3 15x w 4 x 10 16x ) w 5 w 6 0.. when one knows w is given by y 1/ w1/ w1/ x 1/ 1 4 3 L 1/6 4M1/6 M1/6 L 1/6 ) 1 3 L 1/6 4M1/6 M1/6 L 1/6 ),.3 where L18 L w 664 3L < 1,.4 M 18 L 64 3L..5 If it happens that x and y k 5r,thenr k 1 x and w k 5r, w k 5r k r. Proof. The relation.3 can be found using Mathematica. See also 5. Proposition.. If q e πr and a a r k r k 5r ) M 5 r 3, k 5r.6 then a r R q ) 5 11 R 5 q ),.7 where M 5 r is root of 5x 1 5 1 x 56 k r x.

4 International Mathematics and Mathematical Sciences Proof. Suppose that N n μ, where n is positive integer and μ is positive real then it holds that [ ] K n ) [ ] μ M n μ K μ,.8 where Kμ Kk μ The following formula for M 5 r is known: 5M 5 r 1 5 1 M 5 r 56 k r) M5 r..9 Thus, if we use 1.4 and 1.7 and the above consequence of the theory of elliptic functions, we get: R 5 q ) 11 R 5 q ) f 6 q ) qf 6 q 5) a a r k r k 5r ) M 5 r 3. k 5r.10 See also 4, 5. 3. The Main Theorem From Proposition. and relation w k 5r we get w 5 kr w k3 r k r 1 ) a r M 5 r. 3 3.1 Combining. and 3.1,weget [ ] [ ] 10kr 4 6kr 6 a r M 5 r 3 kr 6 16kr 8 kr 3 6a r M 5 r 3 kr 3 kr 5 6a r M 5 r 3 kr 5 w [ ] a r M 5 r 3 kr 15a r M 5 rkr 4 w 0a r M 5 r 3 kr 3 w 3 15a r M 5 r 3 kr w 4 0. 3. Solving with respect to a r M 5 r 3,weget a r M 5 r 3 16k 6 r 6k 4 r wk 3 r 10k r w k 4 r 6k 3 r w 0k 3 r w 15w k r 6 w 15w 4 w. 3.3 Also we have Kk 5r K M 5r 1 m k 5r k 5r k 5r k 5r 1 w w k r ) 1 ww. 3.4

International Mathematics and Mathematical Sciences 5 The above equalities follow from 1 page 80 Entry 13-xii and the definition of w. Notethat m is the multiplier. Hence for given 0 < w < 1, we find L R and we get the following parametric evaluation for the Rogers Ramanujan continued fraction R e π ) 5 rl 11 R e π ) 5 rl 16kr 6 6kr 4 wkr 3 10kr w a r kr 4 6kr 3 w 0 w 3 15w kr 6 w 15w 4 w w w k r ) 3 ww. 3.5 Thus for a given w we find L and M from.4 and.5. Setting the values of M, L, w in.3 we get the values of x and y see Proposition.1. Hence from 3.5 if we find k 1 x r we know Re πr. The clearer result is as follows. Main Theorem. When w is a given real number, one can find x from.3. Then for the Rogers- Ramanujan continued fraction the following holds: R e π ) k 1 5 x 11 R e π ) k 1 5 x 16x 6 6x 4 wx 3 10x wx a r x 4 6x 3 w 0xw 3 15w x 6xw 15w 4 w w x w ww ) 3 1 x x. 1 x 3.6 Theorem 3.1. the first derivative). One has R e π k 1 x ) 4/3 x 1/ 1 x ) 5w 1/6 w /3 w x w ww ) 1/ 1 x x 1 x R e π ) k 1 x K xe π k 1 x. π 3.7 Proof. Combining 1.7 and 1.9 and Proposition. we get the proof. We will see now how the function k 1 x plays the same role in other continued fractions. Here we consider also the Ramanujan s Cubic fraction see 5, which is completely solvable using. Define the function Gx x x 3x x 3/ x 1 3 x 4x 3x 3/ x. 3.8 Set for a given 0 <w 3 < 1 x Gw 3. 3.9

6 International Mathematics and Mathematical Sciences Then as in Main Theorem, for the Cubic continued fraction V q, the following holds see 5: t V e π ) ) k 1 x 1 x 1/3 w 1/4 3 1/3 x 1/3 1 w 3. 3.10 Observe here that again we only have to know k 1 x. If x, for a certain r, then k 9r w 3, 3.11 and if we set T 1 8V q ) 3, 3.1 then the follwing holds: x 1 T3 T3 1 T3 T 3, 3.13 which is solvable always in radicals quartic equation. When we know w 3 we can find x from x Gw 3 and hence t. The inverse also holds: if we know t V q we can find T and hence x. Thew 3 can be found by the degree 3 modular equation which is always solvable in radicals: 3.14 k 9r k 9r 1. Let now V q ) z q V 1 z, 3.15 if V i t : 1 ) 3 1 8t 3 3 1 8t 1 3 1 8t 3 3, 3.16 1 8t 3 then V i V )) e π x k x, 3.17

International Mathematics and Mathematical Sciences 7 or ) V e π r V 1 i, V e π ) k 1 x V 1 i x, 3.18 or e π k 1x V 1 ) V 1 i x V i V 1 x, k 1 V i V q ))) 1 π log q ) r. 3.19 Setting now values into 3.19 we get values for k 1. The function V i is an algebraic function. 4. Evaluations of the Rogers-Ramanujan Continued Fraction Note that if x, r Q, then we have the classical evaluations with and k 5r. Evaluations 1 We have R e π) 1 5 5 5, R e π) 8 9 5 5 50 ) e π ) 5 4 5 5 π Γ. 3 4 4.1 Assume that x 1/, hence k 1 1/ 1. From.5 which for this x can be solved in radicals, with respect to w,wefind w 4 ) 5 1 1 7 5 15. 4. Hence from w 1 w4 1 x, 4.3 x

8 International Mathematics and Mathematical Sciences we get w 1 1 30 14 5 9 150 70 5 ) 1/4. 4.4 Setting these values to 3.6 we get the value of a r and then Rq in radicals. The result is R e π) 5 11 R e π ) 5 1 8 1 5 3 5 30 14 ) 5 30 14 5 3/8 3 5 30 14 ) 5 1 1 30 14 5 9 150 70 ) 1/4 5 [ 1574 704 5 655 30 14 5 93 150 70 1 5]. 4.5 3 Set w 1/64 and a 1359863889, b 36855, then x 9 ab ) [ 5/6 4915 6 1/3 ab ) 1/6 ) 5/6 960 ab 6 /3 ab ) 3/ 384 /3 3 1/6[ 64 ab ) 3 1/6 /3 45387963 b a ) /3 819 6 1/3 ab ) 1/3 185 6 /3 a b ) ] /3 6 5/6 a b ) 1/6 [ 4096 1/3 3 5/6 45387963 a b ) 1/3 36855 /3 3 1/6 45387963 a b ) /3 150958080.6 1/3 ab ) 1/3 45305819 6 /3 ab ) /3 19 45305819 185 a )]] 1. 4.6 4 For w 77 108 13 385, 4.7 108

International Mathematics and Mathematical Sciences 9 we get 77/1 13 ) 385 /1 x 4 7. 4.8 Hence R exp 77/1 13 ) 1/ 385 /1 π k 1 4 7 8071 18 1075 55 18 1 18 55740148 3470530 55) 1/5. 4.9 5 Set q e π r 0, then from ) V e π r 0 V 1 i 0 V 0, V q 1/3) ) 1 V q ) V q ) 3 V q 1 V q ) 4V q ). 4.10 We can evaluate all V q 0 n ) b 0 n Algebraic function of r 0, 4.11 where q 0 n e π r 0 /3 n, V i V q0 n )) V i b 0 n 0 /9 n, 4.1 hence k 1 V i b 0 n r 0 9 n. 4.13

10 International Mathematics and Mathematical Sciences An example for r 0 is ) V e π 3 1, V e ) π /3 1 1/3 3 1 ) V e π /9 ρ 1/3 3, 1/3, 4.14 where ρ 3 can be evaluated in radicals but for simplicity we give the polynomial form 1 7x 6408x 50048x 3 5164x 4 4608x 5 51x 6 0. 4.15 Then, respectively, we get the values k 49 1 35 4 3 99 70 ) ) 9, k 1 V i ρ 1/3 3 )) 81, 4.16. Hence k 1 V i b 0 n r 0 9 n. 4.17 Also it holds that R e π r 0 /3 n) 5 11 R e π r 0 /3 n) 5 16xn 6 6xn 4 w n xn 3 10xn w n x n xn 4 6xnw 3 n 0x n wn 3 15wnx n 6x n w n 15wn 4 wn w n w n x n 1 xn w n w n x n 1 x n 3, 4.18 where x n V i b 0 n known. The w n are given from. in this case we do not find a way to evaluate w n in radicals.

International Mathematics and Mathematical Sciences 11 Theorem 4.1. Set then w 108 144a 4 4a 8 a 1 11664 108 144a 4 4a 8 a 1) 6 3, 4.19 108 a 4 1 a 4) a 4 6 a 4) 1 a 4) 36 18a4 a 8) x 3 3 a 4 a ), 6 a 4 R e π ) k 1 5 x 11 R e π ) k 1 5 x Aa, 4.0 where w 1 w4 1 x. 4.1 x The Aa is a known algebraic function of a and can calculated from the Main Theorem. References 1 B. C. Berndt, Ramanujan s Notebooks. Part III, Springer, New York, NY, USA, 1991. I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products, Academic Press, New York, NY, USA, 1980. 3 E. T. Whittaker and G. N. Watson, A Course of Modern Analysis, Cambridge University Press, Cambridge, UK, 1996. 4 N. Bagis and M. L. Glasser, Integrals related with Rogers Ramanujan continued fraction and q- products, http://arxiv.org/abs/0904.1641. 5 N. Bagis, The complete evaluation of Rogers Ramanujan and other continued fractions with elliptic functions, http://arxiv.org/abs/1008.1304.

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