RAMANUJAN TYPE q-continued FRACTIONS

RAMANUJAN TYPE q-continued FRACTIONS Tapani Matala-aho 19th Czech and Slovak Number Theory Conference August 31 September 4, 2009 Hradec nad Moravicí, Czech Republic

References [MA1] Matala-aho T., On Diophantine Approximations of the Rogers-Ramanujan Continued Fraction, J. Number Theory 45, 1993, 215 227. [MA2] Matala-aho T., On the values of continued fractions: q-series, J. Approx. Th. 124 (2003) 139 153 [MAME1] Matala-aho T. and Merilä V., On the values of continued fractions: q-series II, Int. J. Number Theory 2 (2006) 417 430 [MAME2] Matala-aho T. and Merilä V., On Diophantine approximations of Ramanujan type q-continued fractions, J. Number Theory 129 (2009) 1044 1055

CONTINUED FRACTIONS b 0 + a n K = b 0 + a 1 n=1 b n b 1 + a 2 a 3 b 2 + b 3 +... = (1) Simple continued fraction b 0 + a 1 b 1 + a 2 b 2 +... (2) [b 0 ; b 1, b 2,...] = b 0 + K n=1 1 b n (3)

NOTATIONS For any p of the set { } P, P = {p Z + p is a prime} the notation = will be used for the usual absolute value of C = C and p for the p-adic valuation of the p-adic field C p, the completion of the algebraic closure of Q p, normalized by p p = p 1.

VALUES By the value of the continued fraction in C p we mean the limit b 0 + K n=1 a n b n (4) A n lim n B n of the convergents A n /B n where A n and B n satisfy the recurrences A n = b n A n 1 + a n A n 2, B n = b n B n 1 + a n B n 2 n 2 with initial values A 0 = b 0, A 1 = b 0 b 1 + a 1, B 0 = 1, B 1 = b 1.

IRRATIONALITY MEASURE By an effective irrationality measure (exponent) of a given number α C p we mean a number μ = μ(α) 2 which satisfies the condition: for every ε > 0 there exists an effectively computable constant H 0 (ε) 1 such that α M N > 1 p H μ+ε (5) for every M/N Q with H = max{ M, N } H 0 (ε).

ABSTRACT We shall consider arithmetical properties of the q-continued fractions q sn t s (S 0 + S 1 q n t +... + S h q hn t h ) K n=1 T 0 + T 1 q n t +... + T l q ln t l deg=s+h deg=l where S i, T i, q K, q v < 1, and some related continued fractions where v is a fixed valuation of an algebraic number field K and s, h, l N 0. In particular, we get sharp irrationality measures for certain Ramanujan, Ramanujan-Selberg, Eisenstein and Tasoev continued fractions.

ROGERS-RAMANUJAN There are a few works considering arithmetical properties of q-continued fractions. In general, the research is concentrated on studying irrationality measures of the Rogers-Ramanujan continued fraction RR(q, t) = 1 + qt q 2 t q 3 t 1 + 1 + 1 +... deg=1+0=1 deg=0 (6) in archimedean imaginary quadratic fields, see Bundschuh [8], Osgood [20], Shiokawa [22] and Stihl [23]. Matala-aho [15] considered approximations also in other valuations and proved some higher degree quantitative irrationality results.

ROGERS-RAMANUJAN [MA1] RR(1/5, t) = 1 + t t t t t 5 + 5 + 5 2 + 5 2 + 5 3 +... / Q( m), (7) for any m Q and t Q. Moreover RR(1/5, t), has effective irrationality measure μ(rr(1/5, t)) = 2 (8)

ROGERS-RAMANUJAN RR with an effective irrationality measure ( ( ) 5 1)/2, t / Q( 5), (9) μ(rr( 5 1)/2, t)) = 2 2 (10)

ROGERS-RAMANUJAN Further, if p 17 is a prime number and t Q, then the p-adic number for any m Z. Moreover RR(p, t) = 1 + tp tp 2 tp 3 1 + 1 + 1 +... / Q( m), (11) μ(rr(p, t)) = 2 (12)

THUE-SIEGEL S METHOD [Amou M. and Matala-aho T. (2001)] (tq n ) s (S 0 + S 1 q n t +... + S h q hn t h ) K n=1 T 0 + T 1 q n t +... + T l q ln t l (13) with S i, T i K, q p < 1, s 1. Example the Watson-Ramanujan continued fraction WR(q) = K n=1 q n + q 2n 1 has an irrationality measure deg=1+1=2 deg=0 (14) μ(wr(q)) 61, q = 1/d, d Z {0, ±1}.

RAMANUJAN [MAME2] For example, for the both q-fractions K n=1 q 2n 1 + q n, K n=1 q 2n 1 deg=2+0=2 1 + q n deg=1 (15) we get the irrationality measure μ = 2 (16) if IP happens I) q = 1/d, d Z {0, ±1}, p = (17) or P) q = p P (18)

IP happens If IP happens with Then we say μ = a (19) μ IP = a (20)

q-factorials q-factorials (a) 0 = (a; q) 0 = 1 (a) n = (a; q) n = (1 a)(1 aq)... (1 aq n 1 ) n Z + (a; q) = (1 aq n ) n=0

RAMANUJAN Note that, concerning the latter continued fraction in (15) a deep claim 1 q 1 1 + q 1 + q 2 1 + q 3... q 3 q 5 = (q2 ; q 3 ) (q; q 3 ) (21) made by Ramanujan is proved in [2] and again quite recently in [4] and [7].

RAMANUJAN-SELBERG S 1 (q) = 1 q q + q 2 q 3 q 2 + q 4 1 + 1 + 1 + 1 + 1 +... with = ( q2 ; q 2 ) ( q; q 2 ) (22) μ IP (S 1 (q)) = 3 (23) Take first even contraction.

RAMANUJAN-SELBERG S 2 (q) = 1 q + q 2 q 4 q 3 + q 6 q 8 q 5 + q 10 q 12 1 + 1 + 1 + 1 + 1 + 1 + 1 +... with = (q; q8 ) (q 7 ; q 8 ) (q 3 ; q 8 ) (q 5 ; q 8 ), μ IP (S 2 (q)) = 3 (24)

RAMANUJAN-SELBERG S 3 (q) = 1 q + q 2 q 2 + q 4 q 3 + q 6 1 + 1 + 1 + 1 +... deg=1+1=2 deg=0 with = (q; q2 ) (q 3 ; q 6 ) 3, (25) μ IP (S 3 (q)) 61 (26)

EISENSTEIN E 1 (q) = 1 q q 3 q q 5 q 7 q 3 q 9 q 11 q 5 1 1 1 1 1 1 1... with = q n2, (27) n=0 μ IP (E 1 (q)) = 3 (28) A better is available.

EISENSTEIN E 2 (q) = 1 q 1 + 1 q + 1 q 3 + 1 q 5 + 1 q 7 +... = ( 1) n q n2, (29) n=0 q 3 q 5 q 7 A much better is available. μ IP (E 2 (q)) 61 (30)

TASOEV Next, we consider the following Tasoev s continued fractions T 1 (u, v, a) = 1 1 1 1 1 ua + va 2 + ua 3 + va 4 + ua 5 +..., (31) T 2 (u, v, a, b) = 1 1 1 1 1 ua + vb + ua 2 + vb 2 + ua 3 +.... (32) Komatsu has studied evaluations of numerous variants of T 1 and T 2 e.g. in [11], [12], [13], [14]. Later we will show how we may evaluate T 1 and T 2 simply by using the well-known identities from the theory of q-continued fractions, such as (46) of the Rogers-Ramanujan continued fraction.

TASOEV μ IP (T 1 (u, v, a)) = μ IP (T 2 (u, v, a, b)) = 2 (33) where u, v Q and a, b Z {0, ±1}.

FIBOLOGY Finally, we shall study certain continued fractions W = 1 1 1 W 1 + W 2 + W 3 +... (34) where the partial denominators satisfy a second order recurrence. Let (F n ), F 0 = 0, F 1 = 1 denote the Fibonacci sequence. If we set F = 1 1 1 F 1 + F 3 + F 5 +..., (35) then μ(f ) 61.

MEASURE BY DEGREES The above are applications of the following: Suppose s 1. A) Let s > h, s > l. If s + h 2l, then μ IP (α) = 2s s h If s + h < 2l, then μ IP (α) = s s l B) Let s = h or s = l. Then (36) (37) μ IP (α) 61 (38)

MORE EXAMPLES a) Let b, d Q, d > 0, e, q I and p =. If q p < 1 and d 2 + 4b = (is a square of a rational number), then μ I ( K n=1 b d + eq n ) = 2 (39)

MORE EXAMPLES b) Let b, d Q, e, q K = Q( d 2 + 4b) and p =. If q v < 1, d > 0, d 2 + 4b > 0 and d 2 + 4b =, then μ I ( K n=1 b d + eq n ) = 2 2 (40) NOTE. In the cases a) and b) we have s = 0!!

DEVELOPE Suppose F (t) satisfies a functional equation F (t) = T (t)f (qt) + V (qt)f (q 2 t) (41) Then F (t) F (qt) = T (t) + V (qt) F (qt)/f (q 2 t) = (42) V (qt) T (t) + T (qt) + V (q 2 t) F (q 2 t)/f (q 3 t) =... (43)

DEVELOPE Denote Question: G(q, t) = T (t) + K n=1 V (q n t) T (q n t) = T (t) + V (qt) T (qt) + V (q2 t) T (q 2 t)+... (44) F (t) = G(q, t)? (45) F (qt) It depends!

DEVELOPE As an example, we give the well-known evaluation F (t) F (qt) = 1 + qt q 2 t q 3 t 1 + 1 + 1 +... (46) of the Rogers-Ramanujan continued fraction where F (t) = n=0 q n2 (q) n t n, q p < 1, (47) which satisfies F (t) = F (qt) + qtf (q 2 t) (48)

MAIN THEOREM Let S(t) = S 0 + S 1 t +... + S h t h, T (t) = T 0 + T 1 t +... + T l t l C p [t] Fix G(q, t) = T (t) + K n=1 (tq n ) s S(q n t) T (q n t) deg=s+h deg=l (49) { l A = max 2, s + h }, B = s 4 2. (50) Generally for good results we need B > A. Thus we need s > l, s > h.

VALUATIONS, HEIGHTS Let K be an algebraic number field of degree κ over Q, v its place and K v the corresponding completion. If the finite place v of K lies over the prime p, we write v p, for an infinite place v of K we write v. Further, the notation I is used for an imaginary quadratic field. By using the normalized valuations κv /κ α v = α v, κ v = [K v : Q v ], the Height H(α) of α K is defined by the formula H(α) = v α v, α v = max{1, α v }

VALUATIONS, HEIGHTS and the Height H(α) of vector α = (α 1,..., α m ) K m is given by H(α) = v α v, α v = max i=1,...,m {1, α i v }. A characteristic λ = λ q = will also be used in the sequel. log H(q) log q v, q v = 1, q K

MAIN THEOREM Case 1): When B + λa > 0, we set μ = μ 1 = B B + λa (51) Case 2): Whereas, when B + λa = 0 and λ = 1 and s 1, we have μ 2 (64s 2 + 200s + 21)/3. (52) In particular, by choosing s = 1 we obtain μ 2 61.

MAIN THEOREM Let v be a place of K and let q, t K satisfy B + λa 0, q v < 1 and let S(t), T (t) K[t] satisfy S 0 T 0 = 0, S 0 v 1, s 1 and S(q k t)t (q k t) = 0 for all k N. Then, there exist positive constants C i, D i and H i, i = 1, 2, such that G(q, t) β v > for any β K with H = max{h(β), H i }. C i H κμ i /κ v +D i (log H) 1/2 (53)

EXAMPLES Let h = 0 and s 2l > 0, then B/A = 2 gives an irrationality measure μ 1 = 2/(2 + λ) for G(q, t) for all q K satisfying 2 < λ 1. Now, λ = 1 if q = 1/d, d Z {0, ±1}, v = (54) and q = p, p P, v = p (55) Thus μ = 2

EXAMPLES and especially G(1/d, t) as well as the p-adic number G(p l, t) (p is a prime, l Z + ) have an irrationality measure μ 1 = 2. When K = I, v is the infinite place of K and q = 1/d, d v > 1, where d Z K, we obtain μ 1 = 2, κ = κ v = 2.

EXAMPLES a) Let b, d Q, d > 0, e, q I and v. If q v < 1 and d 2 + 4b is a square of a rational number, then μ 1 ( K n=1 b d + eq n ) = 2 2 + λ. (56) NOTE. In the cases a) and b) we have s = 0!!

EXAMPLES b) Let b, d Q, e, q K = Q( d 2 + 4b) and v. If q v < 1, d > 0, d 2 + 4b > 0 and d 2 + 4b is not a square of a rational number, then κ = 2, κ v = 1 and μ 1 ( K n=1 b d + eq n ) = 2 2 + λ. (57)

EXAMPLES The continued fraction K n=1 1 1 + eq n (58) with q = ( 5 1)/2 belongs to the case b) where λ = 1, μ 1 = 2, and thus K n=1 1 1 + eq n β > v C H 4+D(log H) 1/2 (59) for any β K with H = max{h(β), H 1 }. In particular, the value of continued fraction (58) is not in Q( 5).

EXAMPLES In [19], it is proved that d + K n=1 b + cq n d + eq n = α + K n=1 cq n + eβq 2n 1 α + (e β)q n (60) where b, c, d, e, q K, q < 1 and α = d + d 2 + 4b 2, β = d d 2 + 4b, β < α. 2

EXAMPLES Here, we note that the formula (60) generalizes the transformation formula 1 + K n=1 k + q n 1 = α + K n=1 q n α βq n (61) where q C, q < 1, k R, k > 1/4, α = ( 1 + 1 + 4k ) /2 and β = ( 1 1 + 4k ) /2, originally considered by Ramanujan (for more details, see [5]).

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