Séminaire Lotharingien de Combinatoire 60 (2008), Article B60b

A CONTINUED FRACTION EXPANSION FOR A q-TANGENT

FUNCTION: AN ELEMENTARY PROOF

HELMUT PRODINGER

Abstract. We prove a continued fraction expansion for a certain q-tangent function that was conjectured by the present writer, then proved by Fulmek, now in a completely elementary way.

1. Introduction

In [3], the present writer 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}.

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

For the q-tangent function tan_{q}(z) = \frac{sin_{q}(z)}{cos_{q}(z)}, the following continued fraction expansion was conjectured in [3]:

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} − ⋯}}}}.

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

In [1], this statement was proven using heavy machinery from q-analysis.

Happily, after about 8 years, I was now successful to provide a complete elementary proof that I will present in the next section.

Key words and phrases. q-tangent, continued fraction.

2. The proof

We write

\frac{z sin_{q}(z)}{cos_{q}(z)} = \frac{z^{2}}{N_{0}} = \frac{z^{2}}{C_{1} − \frac{z^{2}}{N_{1}}} = \frac{z^{2}}{C_{1} − \frac{z^{2}}{C_{2} − \frac{z^{2}}{N_{2}}}} = …,

and set

N_{i} = \frac{a_{i}}{b_{i}} .

This means that

N_{i} = C_{i+1} − \frac{z^{2}}{N_{i+1}}

or

\frac{z^{2}}{N_{i+1}} = C_{i+1} − N_{i}

and

\frac{b_{i+1}z^{2}}{a_{i+1}} = C_{i+1} − \frac{a_{i}}{b_{i}} = \frac{C_{i+1}b_{i} − a_{i}}{b_{i}} .

Therefore we may identify numerators and denominators, and put a_{i} = b_{i−1} and

b_{i+1}z^{2} = C_{i+1}b_{i} − b_{i−1}.

The initial conditions are

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

The constants C_{i} guarantee that all the b_{i} are power series, i.e., they make the constant term in C_{i+1}b_{i}−b_{i−1} disappear. Our goal is to show that C_{i} = [2i−1]_{q}q^{(−1)^{i−1}i(i−1)/2−i+1} are the (unique) numbers that do this. We are proving the claim by proving the following explicit formula for b_{i}:

b_{i} = ∑_{n≥0} \frac{(−1)^{n}z^{2n}}{[2n + 2i + 1]_{q}!}( ∏_{j=1}^{i}[2n + 2j]_{q})q^{(n+⌊ \frac{i+1}{2} ⌋)^{2}+[i odd]\binom{i+1}{2}}.

Note that the C_{i} are uniquely determined by the imposed condition, and since the b_{i} are power series, we are done once we prove this formula by induction. The first two instances satisfy this, and we do the induction step now:

C_{i+1}b_{i} − b_{i−1}

= [2i + 1]_{q}q^{(−1)^{i}\binom{i+1}{2}−i} ∑_{n≥0} \frac{(−1)^{n}z^{2n}}{[2n + 2i + 1]_{q}!}( ∏_{j=1}^{i}[2n + 2j]_{q})q^{(n+⌊ \frac{i+1}{2} ⌋)^{2}+[i odd]\binom{i+1}{2}}

− ∑_{n≥0} \frac{(−1)^{n}z^{2n}}{[2n + 2i − 1]_{q}!}( ∏_{j=1}^{i−1}[2n + 2j]_{q})q^{(n+⌊ \frac{i}{2} ⌋)^{2}+[i−1 odd]\binom{i}{2}}

= ∑_{n≥0} \frac{(−1)^{n}z^{2n}}{[2n + 2i + 1]_{q}!}([2i + 1]_{q}( ∏_{j=1}^{i}[2n + 2j]_{q})q^{(n+⌊ \frac{i+1}{2} ⌋)^{2}+[i odd]\binom{i+1}{2}+(−1)^{i}\binom{i+1}{2}−i}

− [2n + 2i + 1]_{q}( ∏_{j=1}^{i}[2n + 2j]_{q})q^{(n+⌊ \frac{i}{2} ⌋)^{2}+[i−1 odd]\binom{i}{2}})

= \frac{1}{1 − q} ∑_{n≥0} \frac{(−1)^{n}z^{2n}}{[2n + 2i + 1]_{q}!}( ∏_{j=1}^{i}[2n + 2j]_{q})

× ((1 − q^{2i+1})q^{(n+⌊ \frac{i+1}{2} ⌋)^{2}+[i even]\binom{i+1}{2}−i} − (1 − q^{2n+2i+1})q^{(n+⌊ \frac{i}{2} ⌋)^{2}+[i even]\binom{i}{2}}).

The last bracket in this expression can be simplified for i even:

−q^{(n+ \frac{i}{2} )^{2}+\binom{i}{2}+2i+1}(1 − q^{2n})

and for i odd:

−q^{(n+ \frac{i−1}{2} )^{2}}(1 − q^{2n}).

Putting everything together, we arrive at

C_{i+1}b_{i} − b_{i−1} = ∑_{n≥0} \frac{(−1)^{n−1}z^{2n}}{[2n + 2i + 1]_{q}!}( ∏_{j=0}^{i}[2n + 2j]_{q})q^{(n+⌊ \frac{i}{2} ⌋)^{2}+[i even]\binom{i+2}{2}}.

Notice that the constant term vanishes, whence

b_{i+1} = ∑_{n≥1} \frac{(−1)^{n−1}z^{2n−2}}{[2n + 2i + 1]_{q}!}( ∏_{j=0}^{i}[2n + 2j]_{q})q^{(n+⌊ \frac{i}{2} ⌋)^{2}+[i even]\binom{i+2}{2}}

= ∑_{n≥0} \frac{(−1)^{n}z^{2n}}{[2n + 2(i + 1) + 1]_{q}!}( ∏_{j=1}^{i+1}[2n + 2j]_{q})q^{(n+⌊ \frac{i+2}{2} ⌋)^{2}+[i + 1 odd]\binom{i+2}{2}},

which is the announced formula.

References

1. M. Fulmek, A continued fraction expansion for a q-tangent function, Sém. Loth. Comb., 45,

(2000), [B45b] (5 pages).

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. H. Prodinger, Combinatorics of geometrically distributed random variables: new q-tangent and

q-secant numbers, Int. J. Math. Math. Sci., 24, (2000), 825–838.

Helmut Prodinger, Department of Mathematics, University of Stellenbosch, 7602 Stellenbosch, South Africa

E-mail address: hproding@sun.ac.za