TAILIEUCHUNG - Báo cáo toán học: "Continued fractions related to (t, q)-tangents and variants"

Tuyển tập các báo cáo nghiên cứu khoa học ngành toán học tạp chí Department of Mathematic dành cho các bạn yêu thích môn toán học đề tài: Continued fractions related to (t, q)-tangents and variants. | Continued fractions related to t q -tangents and variants Helmut Prodinger Department of Mathematics 7602 Stellenbosch South Africa hproding@ Submitted May 16 2011 Accepted Aug 3 2011 Published Aug 19 2011 Mathematics Subject Classifications 05A30 05A10 To Doron Zeilberger who turned me into an addict of creative guessing Abstract For the q-tangent function introduced by Foata and Han this volume we provide the continued fraction expansion by creative guessing and a routine verification. Then an even more recent q-tangent function due to Cieslinski is also expanded. Lastly a general version is considered that contains both versions as special cases. 1 Foata and Han s tangent function Foata and Han 3 defined sin r u 1 n q q 2n 1 u2n 1 i q n 0 q q 2n i cos r u nn qr q 2n Un cosq u A 1 u n 0 q q 2n tan r u q cosqr u Here we use the classic notation where we assume q 1 x q n 1 x 1 xq . 1 xqn-1 . Note that for r TO we obtain the classic q-tangent function of Jackson s 5 . THE ELECTRONIC JOURNAL OF COMBINATORICS 18 2 2011 P18 1 In this paper we compute the continued fraction expansion of this new q-tangent function. In the spirit of Zeilberger the coefficients in it ak in the sequel were obtained first by guessing them. After that some additional power series sk z were also guessed using the recursion that later will be proved . Once one has them the proof of a recursion for sk z is routine and turns immediately into the continued fraction expansion. In a sense this is the most elementary approach possible. Set for k 1 z . X 1 V qr-k q2 k n 1 qr k 1 q Sk z q N q q2 k n 1 q2 q2 z znH Sk-1 z akSk zH q and for k 0 a qr 1-k q2 k 1 q2k 1 qr-k q2 k 1qk Note that for r X we obtain 1 q2k 1 ak k qk which are the well-known coefficients for the classic q-tangent function. Now we compute 2 qr 1-k q2 k n qr k q2 n q q2 k n q2 q2 n _ qr 1-k q2 k 1 q2k 1 fc 1 qr-k q2 k n 1 qr k 1 q2 n qr-k q2 k 1qk q q q2 k n 1 q2 q2 n q 2 qr-k q2 k n 1 qr k 1 q2 n q q2 k n 1 q2 q2 n qr 1-k .

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