Evaluations of a continued fraction of Ramanujan

Evaluations of a continued fraction of Ramanujan

N.D. BAGIS(*) - M.L. GLASSER(**)

ABSTRACT- We study the properties of a general continued fraction of Ramanujan.

In certain cases we evaluate it completely.

MATHEMATICSSUBJECTCLASSIFICATION(2010). 11A55; 33E05.

KEYWORDS. Continued Fractions; Ramanujan; Evaluations

1. Introduction Let

a;q

… †_k:ˆY^{k 1}

nˆ0

(1 aqⁿ) …1†

Then the Ramanujan eta function is

f( q):ˆ(q;q)₁ …2†

and the Weber function is

F( q):ˆ( q;q)₁: …3†

Also we denote by

K(x)ˆ Z^p=2

0

1 1 x²sin²(t)

p dt

…4†

the elliptic integral of the first kind.

(*) Indirizzo dell'A.: Department of Informatics, Aristotele University, Pane- pistimioupolis, Thessaloniki, Greece.

E-mail: [email protected]

(**) Indirizzo dell'A.: Department of Physics, Clarkson University Potsdam, NY 13699-5820, New York, United States.

E-mail: [email protected]

The functionkris defined as the solution to the equation (see [2], [7]) K(k⁰_r)

K(kr)ˆ 

pr …5†

whereris positive,qˆe ^pp_r

andk⁰ˆ 

1 k²

p . It is known that wheneverr is positive rational,k_ris an algebraic number.

In Berndt's book [5] pg. 21 and in [6], one can find the following ex- pansion

THEOREM 1.1. Suppose that q;a and b are complex numbers with q

j j<1, or that q;a, and b are complex numbers with aˆbq^m for some integer m. Then

UˆU(a;b;q)ˆ( a;q)₁(b;q)₁ (a;q)₁( b;q)₁ ( a;q)₁(b;q)₁‡(a;q)₁( b;q)₁ˆ …6†

ˆ a b 1 q‡

(a bq)(aq b) 1 q³‡

q(a bq²)(aq² b) 1 q⁵‡

q²(a bq³)(aq³ b)

1 q⁷‡ . . .

Our main concern in the present article is to simplify and evaluateUfor some specific cases ofa,bandq. For this purpose we define

Xˆ( a;q)₁(b;q)₁ (a;q)₁( b;q)₁; …7†

then

X 1 X‡1ˆU …8†

2. Propositions

PROPOSITION2.1. Set

f(q)ˆ X¹

nˆ 1

qⁿ² …9†

then

f(q) 1 f(q)‡1ˆ q

1‡q‡

q³ 1‡q³‡

q⁵ 1‡q⁵‡

q⁷ 1‡q⁷‡. . . …10†

PROOF. Take q!q² in (6) and then seta!q andb!q². Using [5]

chapter 16 Entry 22, we get the result (10).

The expansion (10) can also be found independently and directly by using Euler's general continued fraction expansion [8]. p

PROPOSITION2.2.

F( q) f( q) F( q)‡f( q)ˆ q

1 q‡

q³ 1 q³‡

q⁵ 1 q⁵‡

q⁷ 1 q⁷‡. . . …11†

PROOF. Set bˆ0 in (6) and thenaˆq. The result follows imediately from the definitions off andFby relations (2) and (3). p

PROPOSITION2.3.

f( q) F( q)

f( q)‡F( q)ˆf( q) 1 f( q)‡1 …12†

PROOF. This follows from Propositions 2.1 and 2.2 p We continue by setting

u₀(a;q):ˆ 2a 1 q‡

a²(1‡q)² 1 q³‡

a²q(1‡q²)² 1 q⁵‡

a²q²(1‡q³)²

1 q⁷‡ . . .

…13†

and

Pˆ ( a;q)₁ (a;q)1

₂

: …14†

Then

P 1

P‡1ˆu₀(a;q);

…15†

On solving (15) with respect toP, we get COROLLARY2.4. Ifjqj<1, then

…16† ( a;q)₁ (a;q)₁

₂

ˆ 1‡ 2 1

2a 1 q‡

a²(1‡q)² 1 q³‡

a²q(1‡q²)² 1 q⁵‡

a²q²(1‡q³)²

1 q⁷‡ . . .

COROLLARY2.5. Ifjqj<1andja=qj<1then 4X¹

nˆ0

a^2n‡1

(2n‡1)(1 q^2n‡1)ˆlog 1‡ 2 1 u₀(a;q)

…17†

PROOF. Take the logarithm on both sides of (16) and expand the two products in Taylor series. Then the double sums are easily rearranged to

get the desired result. p

From Corollaries 2.4 and 2.5, we proceed to the more general formula:

PROPOSITION2.6.

2X¹

nˆ0

a^2n‡1 b^2n‡1

(2n‡1)(1 q^2n‡1)ˆlog 1‡ 2 1 U(a;b;q)

…18†

and hence

1‡ 2

1 U(a;b;q)

₂

ˆ

1‡ 2

1 u0(a;q)

1‡ 2

1 u₀(b;q)

…19†

From relation (19) it is clear that to studyU we have only to examine u₀(a;q).

One can find many useful results on internet sites, such as http://pi.physik.uni-bonn.de/ dieckman/InfProd/InfProd.html In some cases the fractionu0(a;q) can calculated in terms of elliptic func- tions. For example

1‡ 2

1 u₀(q;q)ˆ p 2k⁰_rK(k_r): …20†

More generally, 4X¹

nˆ0

q^n(2n‡1)

(2n‡1)(1 q^2n‡1)ˆ 4X^{n 1}

jˆ1

arctanh(q^j) log 2k⁰_rK(kr) p

…21†

from which we obtain:

PROPOSITION2.7. Letnbe positive integer, then …22† 1‡ 2

1 u0(qⁿ;q)ˆ 1‡ 2

1 2qⁿ 1 q‡

q²ⁿ(1‡q)² 1 q³‡

q^2n‡1(1‡q²)² 1 q⁵‡

q^2n‡2(1‡q³)²

1 q⁷‡ . . .ˆ

ˆ p

2k⁰_rK(kr)exp 4X^{n 1}

jˆ1

arctanh(q^j)

" #

LEMMA2.8. Letn₁,n₂be positive integers, then …23† 1‡ 2

1 U(qⁿ¹;qⁿ²;q)ˆexp 4 ⁿX^{1 1}

j1ˆ1

arctanh(q^j1) X^{n2 1}

j2ˆ1

arctanh(q^j2)

!

" #

PROOF. The proof follows easily from Propositions 2.6 and 2.7. p PROPOSITION2.9.

X¹

nˆ0

qⁿ

1 a²q²ⁿˆ 1 1 q‡

a²(1 q)² 1 q³‡

qa²(1 q²)² 1 q⁵‡

q²a²(1 q³)²

1 q⁷‡ . . .

…24†

PROOF. Solve relation (18) forU(a;b;q), divide bya band then take

the limitb!a. p

PROPOSITION2.10. If qˆe ^pp_r , then K(kr)

2p ‡1

4ˆ 1

1 q‡

(1 q)² 1 q³‡

q(1 q²)² 1 q⁵‡

q²(1 q³)²

1 q⁷‡ . . .

…25†

PROOF. Set in (24)aˆi,qˆe ^p 

pr

and use the identity X¹

nˆ0

qⁿ

1‡q²ⁿ ˆK(kr) 2p ‡1

4 p

It is also known that whennis an integer,

1‡ 2

1 u0(q^n‡1=2;q)ˆexp 4X¹

nˆ0

q(2n‡1)(n‡1=2)

(2n‡1)(1 q^2n‡1)

" #

…26†

ˆexp 4X^{n 1}

jˆ0

arctanh(q^j‡1=2)‡arctanh(k_r)

" #

Hence

krˆtanh 4X^{n 1}

jˆ0

arctanh(q^j‡1=2)‡log 1‡ 2 1 u₀(q^n‡1=2;q)

" #

For every positive integer n. By this means we get a continued fraction expansion for the singular moduluskr

k⁰_r

1 k_rˆ 1‡ 2

1 u0(q¹⁼²;q) …27†

From Propositions 2.6 and 2.7 and Lemma 2.8, we have

PROPOSITION2.11. If c is positive real andn1,n2are positive integers then:

1‡ 2

1 U(q^n1‡c;q^n2‡c;q)ˆ …28†

ˆexp 2 X^{n1 1}

j1ˆ1

arctanh(q^j1‡c) X^{n2 1}

j2ˆ1

arctanh(q^j2‡c)

!

" #

: Moreover:

If UˆU(a;b;q), where qˆe ^pp_r and cˆ log(a)

p 

pr

ˆ log(b) p 

pr

…29†

then

1‡ 2

1 U(a;b;q)ˆ

ˆexp 2 X _log(a)

p^p_r

1 j1ˆ1

arctanh(q^j1‡c) X _log(b)

p^p_r

1 j2ˆ1

arctanh(q^j2‡c) 0

B@

1 CA 2

64

3 75

where xf g is the fractional part of x and x‰ Š is the largest integer not exceeding x.

REMARK2.12. Observe that forn₁ˆn₂ˆn

1‡ 2

1 U(q^n‡c;q^n‡c;q)ˆ1

Also, we observe that

1‡ 2

1 U(a; b;q)

ˆ 1‡ 2 1 u₀(a;q)

1‡ 2

1 u₀(b;q)

…30†

This relation is similar to (19).

PROPOSITION2.13.

…31†log 1‡ 2 1 U(aq;b;q)

log 1‡ 2

1 U(a;b;q)

ˆ 2arctanh(a)

PROOF. From (20)

log 1‡ 2

1 U(aq;b;q)

‡2X¹

nˆ0

b^2n‡1

(2n‡1)(1 q^2n‡1)ˆ2X¹

nˆ0

q^2n‡1a^2n‡1 (2n‡1)(1 q^2n‡1) or

log 1‡ 2

1 U(aq;b;q)

‡2X¹

nˆ0

b^2n‡1

(2n‡1)(1 q^2n‡1) 2X¹

nˆ0

a^2n‡1

(2n‡1)(1 q^2n‡1)ˆ

ˆ2X¹

nˆ0

q^2n‡1a^2n‡1

(2n‡1)(1 q^2n‡1) 2X¹

nˆ0

a^2n‡1

(2n‡1)(1 q^2n‡1)ˆ 2X¹

nˆ0

a^2n‡1 2n‡1

p PROPOSITION2.14.

…32† log 1‡ 2 1 U(a;b;q)

ˆ2X¹

j1ˆ1

arctanh(aq^j1) 2X¹

j2ˆ1

arctanh(bq^j2)

PROOF. In (31) replaceawithaqand add the result to (31) to arrive at

log 1‡ 2

1 U(0;b;q)

log 1‡ 2

1 U(a;b;q)

ˆ 2X¹

nˆ0

arctanh(aqⁿ)

By doing the same with

log 1‡ 2

1 U(0;b;q)

ˆ 2X¹

nˆ0

b^2n‡1

(2n‡1)(1 q^2n‡1); using the fact that

log 1‡ 2

1 U(0;0;q)

ˆ0

we obtain (32). p

REMARK2.15. From (32) one can show that whena,b,sare positive reals and

b a

s ˆnˆn1 n22N …33†

then for

xˆa andyˆa(b‡n2) bn1 a² a b …34†

andn1ˆ[y],n2ˆ[x], we can evaluate

log 1‡ 2

1 U(q^a;q^b;q^s)

ˆlog 1‡ 2

1 U(q^{s(x‡y [y])};q^{s(x‡y [x])};q^s)

ˆ

in the closed form

2^{[x] 1}X

jˆ1

arctanhq^{s(j‡f g‡x f g)y}

‡2^{[y] 1}X

jˆ1

arctanhq^{s(j‡f g‡x f g)y} …35†

From the relations (see [2])

sn(u)ˆsn(q;u)ˆ 2p Kk

X¹

nˆ0

q^n‡1=2sin 2K

p (2n‡1)u

1 q^2n‡1 …36†

cn(u)ˆcn(q;u)ˆ 2p Kk

X¹

nˆ0

q^n‡1=2cos 2K

p (2n‡1)u

1‡q^{2n 1} …37†

whereuˆ2Kz=pand Z

sn(u)duˆk_r¹

2 log 1 krcd(u) 1‡k_rcd(u)

ˆ 1

2klog w3 w2 

pk w3‡w2 

pk

! …38†

and from X¹

nˆ0

q^n‡1=2

(2n‡1)(1 q^2n‡1)ˆ i 4 arctan



m( q)

1 m( q)

s !

…39†

withm(q)ˆk²_rand (17) we get next

PROPOSITION2.16. If qˆe ^pp_r

, r>0andu2R, then

Re log 1‡ 2

1 u0(q¹⁼²e^iu;q)

ˆ …40†

ˆ 1

2 log w3 w2 

kr

p w₃‡w₂ 

k_r p

u

0 iarctan



m( q)

1 m( q)

s !

wherew_2;3are the Jacobi theta functions:

w2(q;z)ˆ2X¹

nˆ0

q^(n‡1=2)2cos((2n‡1)z) …41†

w₃(q;z)ˆ1‡2X¹

nˆ1

qⁿ²cos(2nz) …42†

PROPOSITION2.17. Ifu₁;u₂are real numbers andjqj<1, then …43† Re log 1‡ 2

1 U(q¹⁼²e^iu1;q¹⁼²e^iu2;q)

ˆ1

4 log w₃ w₂ 

kr

p w3‡w2 

kr

p

u2

u1

More generally,

cc(u)ˆcc(q;u):ˆ 2p Kk

X¹

nˆ0

q^n‡1=2cos (2n‡1)2Ku p

1‡q^2n‡1 ˆ

…44†

ˆtan 4Ku p

ss(u)‡q ¹sec 4Ku p

cn(u)

2ppqcos 2Ku p

sec 4Ku p

Kk(1‡q) where

ss(u)ˆss(q;u):ˆ 2p Kk

X¹

nˆ0

q^n‡1=2sin (2n‡1)2Ku p

1‡q^2n‡1 …45†

But cc(q;u)ˆ isign(q)cd( q;u), so by solving (44) for the function ss we get

…46†ss(q;u)ˆ

cn(q;u)csc 4Ku p

q ‡

ppqcsc 2Ku p

Kkr(1‡q) icd( q;u) cot 4Ku p

which leads to

PROPOSITION2.18. Ifuis inCsuch thatjq¹⁼²e^2iuK=pj<1and0<q<1, then

p² 4K²k

d

dt log 1‡ 2

1 u₀(iq¹⁼²e^2itK=p;q)

tˆuˆ …47†

ˆ

ippqcsc 2Ku p

kK(1 q) ‡ 1‡icot 4Ku p

cd(q;u)‡

csc 4Ku p

cn( q;u) q

PROOF. From (17), the definitions of cc, ss ande^ixˆ cos(x)‡i sin(x), we have

p² 4K²k

d

du log 1‡ 2

1 u0(iq¹⁼²e^2iuK=p;q)

ˆss( q;u) icc( q;u)ˆ

ˆss( q;u)‡cd(q;u)

Using (46), we get (47). p

REMARK2.19. The function ss(q;u) plays very important role here. If one finds its generalized integral, then the problem of the evaluation ofU will be solved.

REFERENCES

[1] M. ABRAMOWITZ and I.A. STEGUN, Handbook of Mathematical Functions, Dover Publications, 1972.

[2] J.V. ARMITAGE and W. F. EBERLEIN, Elliptic Functions, Cambridge Uni- versity Press, 2006.

[3] B.C. BERNDT,Ramanujan`s Notebooks Part I, Springer Verlag, New York, 1985.

[4] B.C. BERNDT,Ramanujan`s Notebooks Part II, Springer Verlag, New York, 1989.

[5] B.C. BERNDT,Ramanujan`s Notebooks Part III, Springer Verlag, New York, 1991.

[6] L. LORENTZENand H. WAADELAND,Continued Fractions with Applications, Elsevier Science Publishers B. V., North Holland, 1992.

[7] E.T. WHITTAKER and G. N. WATSON,A course on Modern Analysis, Cam- bridge U. P., 1927.

[8] H.S. WALL, Analytic Theory of Continued Fractions, Chelsea Publishers, New York, 1948.

Manoscritto pervenuto in redazione il 15 gennaio 2013.