Séminaire Lotharingien de Combinatoire 45 (2001), Article B45b

A CONTINUED FRACTION EXPANSION FOR A q-TANGENT FUNCTION

MARKUS FULMEK

Abstract. We prove a continued fraction expansion for a certain q–tangent function that was conjectured by Prodinger.

1. Introduction

In [4], Prodinger defined the following q–trigonometric functions

sin_{q}(z) = ∑_{n=0}^{∞} \frac{(−1)^{n}z^{2n+1}}{[2n + 1]_{q}!} q^{n^{2}},

cos_{q}(z) = ∑_{n=0}^{∞} \frac{(−1)^{n}z^{2n}}{[2n]_{q}!} q^{n^{2}}.

Here, we use standard q–notation:

[n]_{q} := \frac{1 − q^{n}}{1 − q} , [n]_{q}! := [1]_{q}[2]_{q} … [n]_{q},

(a; q)_{n} := (1 − a)(1 − aq) … (1 − aq^{n−1}).

These q–functions are variations of Jackson’s [2] q-sine and q-cosine functions.

For the q–tangent function tan_{q} = \begin{matrix}sin & q \\ cos & q \end{matrix}, Prodinger conjectured the following continued fraction expansion (see [4, Conjecture 10]):

−z tan_{q}(z) = −\frac{z^{2}}{[1]_{q}q^{0} − \frac{z^{2}}{[3]_{q}q^{−2} − \frac{z^{2}}{[5]_{q}q^{1} − \frac{z^{2}}{[7]_{q}q^{−9} − ⋯}}}}. (1)

Here, the powers of q are of the form (−1)^{n−1}n(n − 1)/2 − n + 1.

The purpose of this note is to prove this statement. In our proof, we make use of the polynomials (see [3, §2, (11)]) A_{n}(z) and B_{n}(z), which are given recursively by

Date: January 25, 2001.

A_{n}(z) = b_{n}A_{n−1}(z) − z^{2}A_{n−2}(z),  (2)
B_{n}(z) = b_{n}B_{n−1}(z) − z^{2}B_{n−2}(z);  (3)

with initial conditions (see [3, §2, (12)])

A_{−1} = 1, B_{−1} = 0, A_{0} = b_{0}, B_{0} = 1,

where b_{0} = 0, b_{n} = [2n − 1]_{q}q^{(−1)^{n−1}n(n−1)/2−n+1}. As is well known (see [3, §2]), the continued fraction terminated after the term b_{n} is equal to \frac{A_{n}}{B_{n}}, whence (1) follows from the assertion

A_{n} cos_{q} +zB_{n} sin_{q} = O(z^{2n+1}),  (4)

i.e., the leading 2n coefficients of z vanish in (4).

In Section 2 we give a proof of (4) (and thus of (1)).

2. The proof

Both A_{n} and B_{n} are polynomials in z^{2}:

A_{n}(z) = ∑_{j} c_{n,j}z^{2j}, B_{n}(z) = ∑_{j} d_{n,j}z^{2j}.

Observe that from the recursions (2) and (3) we obtain immediately

c_{n,k} = b_{n}c_{n−1,k} − c_{n−2,k−1} and d_{n,k} = b_{n}d_{n−1,k} − d_{n−2,k−1}, (5)

with initial conditions

c_{0,k} = d_{0,k} = c_{−1,k} = d_{−1,k} = c_{n,0} = d_{n,−1} = 0, c_{1,1} = −1, d_{0,0} = 1.

Given this notation, we have to prove the following assertion for the coefficients of z^{2k} in (4): For n ≥ 1, 0 ≤ k ≤ n, there holds

∑_{i=0}^{k} c_{n,i} \frac{(−1)^{k−i}}{[2k − 2i]_{q}!}q^{(k−i)^{2}} + ∑_{i=0}^{k} d_{n,i} \frac{(−1)^{k−i−1}}{[2k − 2i − 1]_{q}!}q^{(k−i−1)^{2}} = 0. (6)

In fact, we shall state and prove a slightly more general assertion: Lemma 1. Given the above definitions, we have for all n ≥ 1, k ≥ 0:

∑_{i=0}^{k−1}(−1)^{i} \frac{q^{(k−i−1)^{2}}}{[2k − 2i − 2]_{q}!} (c_{n,i+1} + \frac{d_{n,i}}{[2k − 2i − 1]_{q}} ) =

(−1)^{n}q^{(5+3(−1)^{n}−12k−4(−1)^{n}k+8k^{2}+8n−8kn+4n^{2}−2(−1)^{n}n^{2})/8}

× \frac{∏_{s=k−n}^{k}[2s]_{q}}{[2k]_{q}!} . (7)

Note that the left hand side of (7) is the same as in (6), and the right hand side of (7) vanishes for 0 ≤ k ≤ n. Hence (6) (and thus Prodinger’s conjecture) is an immediate consequence of Lemma 1.

Proof. We perform an induction on k for arbitrary n.

The case k = 0 is immediate. For the case k = 1, observe that

−c_{n,1} = d_{n,0} = ∏_{s=1}^{n} b_{n}.

For the inductive step (k − 1) → k, we shall rewrite the recursions (5) in the following way:

c_{n,k} = − ∑_{i=0}^{n−2} (c_{i,k−1} ∏_{j=i+3}^{n} b_{j}) , d_{n,k} = − ∑_{i=0}^{n−2} (d_{i,k−1} ∏_{j=i+3}^{n} b_{j}) .

Substitution of these recursions into (7) and interchange of summations transform the identity into

\frac{q^{(k−1)^{2}}(1 − [2k − 1]_{q}) ∏_{s=1}^{n} b_{s}}{[2k − 1]_{q}!} +∑_{i=0}^{n−2} (rhs(i, k − 1) ∏_{j=i+3}^{n} b_{j}) = rhs(n, k),

where rhs(n, k) denotes the right hand side of (7).

Now we use the induction hypothesis. As it turns out, factorization of powers of q from (rhs(i, k − 1) ∏_{j=i+3}^{n} b_{j}) yields the same power for 2i and 2i + 1, whence we can group these terms together. After several steps of simplification we arrive at the following identity:

( ∑_{j=0}^{⌈ \frac{n−4}{2} ⌉} \frac{(q^{−2k+6}; q^{4})_{j}(q^{−2k+4}; q^{4})_{j}(q^{17/2−k}; q^{4})_{j}(−q^{17/2−k}; q^{4})_{j}}{(q^{7}; q^{4})_{j}(q^{9}; q^{4})_{j}(q^{9/2−k}; q^{4})_{j}(−q^{9/2−k}; q^{4})_{j}} q^{(2k−1)j})

× \frac{(q; q^{2})_{n}q(1 − q^{2k−1})(1 − q^{2k−2})(1 − q^{9−2k})}{(1 − q)(1 − q^{3})(1 − q^{5})} − (1 − q^{2k−1})(q^{3}; q^{2})_{n−1}

− (−1)^{n}q^{(−1+(−1)^{n}+2k−2(−1)^{n}k+2n−4kn+4n^{2})/4}(q^{2k−2n}; q^{2})_{n}

+ (q; q^{2})_{n} + χ(n)(1 − q^{2k−1})q^{n(2n−2k+1)/2}(q^{2k−2n+2}; q^{2})_{n−1} = 0, (8)

where χ(n) = 1 for n even and 0 for n odd.

The sum can be evaluated by means of the very–well–poised _{6}ϕ_{5} summation formula [1, (2.7.1); Appendix (II.20)]:

∑_{j=0}^{∞} \frac{(a; q)_{j}(\sqrt{a}q; q)_{j}(−\sqrt{a}q; q)_{j}(b; q)_{j}(c; q)_{j}(d; q)_{j}}{(q; q)_{j}(\sqrt{a}; q)_{j}(−\sqrt{a}; q)_{j}(\frac{aq}{b}; q)_{j}(\frac{aq}{c}; q)_{j}(\frac{aq}{d}; q)_{j}} (\frac{aq}{bcd})^{j}

= \frac{(aq; q)_{∞}(\frac{aq}{bc}; q)_{∞}(\frac{aq}{bd}; q)_{∞}(\frac{aq}{cd}; q)_{∞}}{(\frac{aq}{b}; q)_{∞}(\frac{aq}{c}; q)_{∞}(\frac{aq}{d}; q)_{∞}(\frac{aq}{bcd}; q)_{∞}} . (9)

The sum we are actually interested in does not extend to infinity, so we rewrite is as follows:

∑_{j=0}^{⌈ \frac{n−4}{2} ⌉} s(n, k, j) = ∑_{j=0}^{∞} s(n, k, j) − ∑_{j=⌈ \frac{n−2}{2} ⌉}^{∞} s(n, k, j)

= ∑_{j=0}^{∞} s(n, k, j) − s(n, k, ⌈\frac{n − 2}{2}⌉) ∑_{j=0}^{∞} \frac{s(n, k, j + \frac{n−2}{2})}{s(n, k, ⌈\frac{n−2}{2}⌉)},

where s(n, k, j) denotes the summand in (8). Now, replacing q by q^{4}, a by q^{−2k+8a+9}, b by q^{−2k+4a+4}, c by q^{−2k+4a+6} and d by q^{4} in the summand of (9) gives \frac{s(n,k,j+a)}{s(n,k,a)} times the fraction \frac{(q^{−2k+8a+9};q^{4})_{j}(q^{4};q^{4})_{j}}{(q^{−2k+8a+9};q^{4})_{j}(q^{4};q^{4})_{j}}, which cancels. So we obtain after some simplification:

∑_{j=0}^{x} \frac{(q^{−2k+6}; q^{4})_{j}(q^{−2k+4}; q^{4})_{j}(q^{17/2−k}; q^{4})_{j}(−q^{17/2−k}; q^{4})_{j}}{(q^{7}; q^{4})_{j}(q^{9}; q^{4})_{j}(q^{9/2−k}; q^{4})_{j}(−q^{9/2−k}; q^{4})_{j}} q^{(2k−1)j}

= \frac{(1 − q^{3})(1 − q^{5})}{(1 − q^{2k−1})(1 − q^{9−2k})} (1 − q^{(2k−1)(x+1)} \frac{(q^{−2k+4}; q^{2})_{2x+2}}{(q^{3}; q^{2})_{2x+2}} ) .

Substitution of this evaluation in (8) and simplification yield for both cases n even (n = 2N) and odd (n = 2N − 1) the same equation

(q^{−2k+2}; q^{2})_{2N−1} = −q^{−2(k−N)(2N−1)}(q^{2k−4N+2}; q^{2})_{2N−1},

which, of course, is true. This finishes the proof. □

References

[1] G. Gasper and M. Rahman, Basic hypergeometric series, Encyclopedia of Mathematics and its Applications 35, Cambridge University Press, Cambridge, 1990.

[2] F.H. Jackson, A basic–sine and cosine with symbolic solutions of certain differential equations, Proc. Edinburgh Math. Soc., 22, (1904), 28–39.

[3] O. Perron, Die Lehre von den Kettenbrüchen, 1. Band, B.G. Teubner, Stuttgart, 1977.

[4] H. Prodinger, Combinatorics of geometrically distributed random variables: New q-tangent an q-secant numbers, Int. J. Math. Math. Sci. (to appear).

Institut für Mathematik der Universität Wien, Strudlhofgasse 4, A-1090 Wien, Austria

E-mail address: Markus.Fulmek@Univie.Ac.At
WWW: http://www.mat.univie.ac.at/~mfulmek