1 Introduction
For any complex number q with
$|q|> 1$
, the q-analogue of the exponential function is defined by the absolutely convergent series
$$ \begin{align*} E_q(x) := 1+ \sum_{n=1}^{\infty} \frac{x^n}{[n]_{q} !}, \end{align*} $$
where
$[n]_{q} = q^n-1$
and
$[n]_{q}! = (q^n-1)(q^{n-1}-1) \cdots (q-1)$
. Similarly, the q-analogue of the logarithm is given by
$$ \begin{align*} L_q(x) := \sum_{n=1}^{\infty} \frac{x^n}{[n]_{q}} \quad \text{for } |x| < |q|. \end{align*} $$
The analogy between the classical functions and their q-analogues is driven by the limit
$$ \begin{align*} \lim_{q \rightarrow 1^+} \frac{q^n - 1}{q-1} = n. \end{align*} $$
Unlike the classical exponential and logarithm functions, their q-counterparts are related by the differential relation
$$ \begin{align*} L_q(x) = x \frac{E'_q(-x)}{E_q(-x)} \quad \mbox{for } |x| < |q|. \end{align*} $$
For more details, we refer the reader to [Reference Murty, Prasad, Rajan, Sankaranarayanan and Sengupta7, Section 6]. These functions appear in various contexts in combinatorics and number theory and are interesting functions in their own right.
The value at
$x=1$
of the q-logarithm function is of particular importance, as
$L_q(1) = \zeta _q(1)$
, where
$$ \begin{align*} \zeta_q(s) := \sum_{n=1}^{\infty} \frac{n^{s-1}}{[n]_{q}}, \end{align*} $$
is the q-analogue of the Riemann zeta-function (see [Reference Krattenthaler, Rivoal and Zudilin6]). The value
$\zeta _q(1)$
,
$$ \begin{align*} \zeta_q(1) = \sum_{n=1}^{\infty} \frac{1}{q^n -1}, \end{align*} $$
is often referred to as the q-harmonic series.
We examine the arithmetic nature and linear independence properties of certain special values of these functions. Recall that a real algebraic integer
$\omega $
is said to be a Pisot–Vijayaraghavan number (abbreviated to PV number) if
$\omega> 1$
and
$|\omega ^{(j)}| < 1$
for all other Galois conjugates
$\omega ^{(j)}$
of
$\omega $
. Immediate examples of PV numbers are positive integers greater than one. A nontrivial example is obtained by considering
$\beta $
, the real root of
$x^4 -x^3 -2x^2 +1$
with
$\beta> 1$
. Pisot [Reference Pisot8] showed that, in every real algebraic number field, there exist PV numbers that generate the field. These numbers make a fundamental appearance in Diophantine approximation and have been studied extensively.
Fix an algebraic integer
$q \neq 0$
and let
$n_q = [\mathbb {Q}(q) : \mathbb {Q}]$
. Let
$ \sigma _1$
,
$\sigma _2$
,
$\ldots ,\ \sigma _{n_q}$
denote the embeddings of
$\mathbb {Q}(q)$
into
$\mathbb {C}$
, with
$\sigma _1$
being identity. Let
$\mathcal {O}_{q}$
be the ring of integers of
$\mathbb {Q}(q)$
. For any algebraic number
$\alpha \in \mathbb {Q}(q)$
, the q-relative height of
$\alpha $
,
$H_q(\alpha )$
, is
$$ \begin{align*} H_q(\alpha) := \prod_{l=1}^{n_q} \max \{ 1, | \sigma_l(\alpha) |\}. \end{align*} $$
Thus, if q is a PV number, then
$H_q(q) = q$
.
Our first theorem concerns the linear independence of values of derivatives of a certain generalised q-exponential function. Let
$P(X) \in \mathbb {Z}[X]$
be a nonconstant polynomial such that
$P(q^t) \neq 0$
for all
$t \in \mathbb {N}$
. Then the generalised q-exponential function with respect to P is given by
$$ \begin{align*} E_{q,P}(x) = 1 + \sum_{n=1}^{\infty} \frac{x^n}{\prod_{t=1}^n P(q^t)}. \end{align*} $$
If
$P(X) = X-1$
, then
$E_{q,P}(x) = E_q(x)$
, the q-exponential function. Note that
$E_{q,P}(x)$
is a basic hypergeometric series, as defined in [Reference Gasper and Rahman5].
With this notation, we can state our first result.
Theorem 1.1. Assume that q or
$-q$
is a PV number. Let
$P(X)= L_{D} X^{D}+\cdots + c_{d} X^{d}\in \mathbb {Z}[X]$
be a nonconstant polynomial with
$P(q^t) \neq 0$
for
$t \geq 1$
,
$d \leq D$
and
$L_D c_d \neq 0$
. Let
$\alpha _1,\ldots ,\alpha _m$
be nonzero algebraic integers in
$\mathbb {Q}(q)$
satisfying
$$ \begin{align} \hspace{-8pt}|c_d|^{n_q-1} \max\{|\alpha_1|, |\alpha_2|, \ldots, |\alpha_m| \} \prod_{l=2}^{n_q} \max \{ 1, | \sigma_l(\alpha_1) |, | \sigma_l(\alpha_2) |, \ldots, | \sigma_l(\alpha_m) | \} < |q|^{D}. \end{align} $$
Suppose that
$\alpha _{k_1}/\alpha _{k_2}$
is not a root of unity for
$1 \leq k_1, k_2 \leq m$
and
$k_1 \neq k_2$
. Then the numbers in the set
$$ \begin{align*} S := \{ E_{q, P}^{(j)} ( \alpha_k ) : 1 \leq k \leq m, 0 \leq j \leq M \} \cup \{ 1 \} \end{align*} $$
are linearly independent over the field
$\mathbb {Q}(q)$
.
The following result is an immediate corollary of this theorem.
Corollary 1.2. Assume that q or
$-q$
is a PV number. Let
$\alpha _1,\ldots ,\alpha _m$
be nonzero algebraic integers in
$\mathbb {Q}(q)$
satisfying
$$ \begin{align*} \max\{|\alpha_1|, |\alpha_2|, \ldots, |\alpha_m| \} \prod_{l=2}^{n_q} \max \{ 1, | \sigma_l(\alpha_1) |, | \sigma_l(\alpha_2) |, \ldots, | \sigma_l(\alpha_m) | \} < |q|. \end{align*} $$
Suppose that
$\alpha _{k_1}/\alpha _{k_2}$
is not a root of unity for
$1 \leq k_1, k_2 \leq m$
and
$k_1 \neq k_2$
. Then the numbers in the set
$$ \begin{align*} S := \{ E_q^{(j)}(\alpha_k) : 1 \leq k \leq m, 0 \leq j \leq M \} \cup \{ 1 \} \end{align*} $$
are linearly independent over the field
$\mathbb {Q}(q)$
.
In particular, this gives the following result about the special functions discussed earlier.
Corollary 1.3. Assume that q or
$-q$
is a PV number and that
$\alpha \in \mathcal {O}_{q}$
satisfies
$$ \begin{align*} 0 < \min \{1, | \alpha| \} H_q ( \alpha) < |q|. \end{align*} $$
Then
$E_q(\alpha ), L_q(\alpha ) \not \in \mathbb {Q}(q)$
. In particular,
$\zeta _q(1)$
is irrational.
The irrationality and linear independence of the values of the q-logarithm function have been studied extensively. We refer the reader to [Reference Zudilin10] for a comprehensive history of the problem and an investigation of the values of a generalisation of the q-logarithm function. The irrationality of
$\zeta _q(1)$
when q is an integer was first obtained by Erdős [Reference Erdős4]. More recently, Tachiya [Reference Tachiya9, Theorem 2] proved that
$\zeta _q(1) \not \in \mathbb {Q}(q)$
when q is a PV number, which also follows from Corollary 1.3.
A special function closely related to the q-exponential function is the Tschakaloff function, given by
$$ \begin{align*} T_q(x) := 1 + \sum_{n=1}^{\infty} \frac{x^n}{q^{{n(n+1)}/{2}}}. \end{align*} $$
In our notation,
$T_q(x) = E_{q,I}(x)$
, where
$I(x) = x$
. Thus, Theorem 1.1 implies the following result.
Corollary 1.4. Assume that q or
$-q$
is a PV number. Suppose that
$ \alpha \in \mathcal {O}_{q}$
satisfies
$$ \begin{align*} 0 < { \min \{1, |\alpha| \} } {H_q(\alpha)} < |q|. \end{align*} $$
Then the numbers
$1, T_q(\alpha ), T_q^{(1)}(\alpha ), \ldots , T_q^{(m)}(\alpha )$
are linearly independent over
$\mathbb {Q}(q)$
.
It was brought to our notice by the referee that Theorem 1.1 follows from [Reference Amou, Matala-Aho and Väänänen1, Corollaries 5.1 and 5.2], which require a much weaker condition on the
$\alpha _k$
than in Theorem 1.1. In [Reference Amou, Matala-Aho and Väänänen1], Amou et al., prove a general result regarding linear independence of values of solutions to q-difference equations. The techniques necessary to prove this result are involved, whereas our proof of Theorem 1.1 follows from relatively elementary considerations.
The statements so far were concerned with the independence of values of a single function and its derivatives at several arguments. We now address the question of independence of different cognate functions at the same argument. For any
$M \in \mathbb {N}$
and any q with
$|q|>1$
, we define an arithmetic progression analogue,
$E_{q,M}(x)$
, of
$E_q(x)$
by
$$ \begin{align*} E_{q,M}(x) := 1 + \sum_{n=1}^{\infty} \frac{x^n}{[Mn]_q!}. \end{align*} $$
This is an entire function. Clearly,
$E_{q,1}(x) = E_q(x)$
and
$$ \begin{align*} E_{q,M}(x^M) = 1 + \sum_{\substack{n=1 \\ n \equiv 0 \bmod M}}^{\infty} \frac{x^n}{[n]_q !}. \end{align*} $$
Note that
$E_{q,M}$
is not a basic hypergeometric function.
For these special functions, we prove the following theorem.
Theorem 1.5. Assume that q or
$-q$
is a PV number and that
$a_1<\cdots <a_k$
are distinct positive integers. Let
$\alpha \in \mathcal {O}_{q}$
be such that
$1 \leq |\alpha |$
and
$$ \begin{align} H_q(\alpha) < |q|^{a_1}. \end{align} $$
Then the numbers
$$ \begin{align} 1, E_{q,a_1}(\alpha),\ldots, E_{q, a_k}(\alpha) \end{align} $$
are linearly independent over the field
$\mathbb {Q}(q)$
.
The approach in this paper is an adaptation of the proof of [Reference Murty, Prasad, Rajan, Sankaranarayanan and Sengupta7, Theorem 1.1], which is a modification of the argument by Duverney [Reference Duverney3]. In essence, it is similar to Fourier’s proof of the irrationality of the number e. The proof of Theorem 1.1 relies on a Diophantine lemma, which is a consequence of the Skolem–Mahler–Lech theorem. The proof of Theorem 1.5 is completed using a recursive elimination argument.
2 Proof of the theorems
An important ingredient in the proofs is the following particular case of the Skolem–Mahler–Lech theorem [Reference Corvaja and Zannier2, Theorem 4.3, page 124].
Theorem 2.1. Let
$\alpha _1,\ldots ,\alpha _k$
be nonzero algebraic numbers such that
$\alpha _i/\alpha _j$
is not a root of unity for
$1\leq i< j\leq k$
. Let
$P_1(x),\ldots , P_k(x)$
be nonzero polynomials with algebraic coefficients. Then there are only finitely many natural numbers n satisfying
$$ \begin{align*} P_1(n) \alpha^n_1 + \cdots + P_k(n) \alpha^n_k=0. \end{align*} $$
This is immediate from the Skolem–Mahler–Lech theorem since the sequence
$P_1(n) \alpha ^n_1 + \cdots + P_k(n) \alpha ^n_k$
is a nondegenerate recurrence sequence if none of the
$\alpha _i/\alpha _j$
(
$1\leq i< j\leq k$
) is a root of unity.
2.1 Proof of Theorem 1.1
Let
$f_j(x) := x^j E_{q,P}^{(j)}(x)$
for
$0 \leq j \leq M$
. Observe that the result follows if we show that
$1$
and the values
$f_j(\alpha _k)$
are
$\mathbb {Q}(q)$
-linearly independent for
$0 \leq j \leq M$
and
$1 \leq k \leq m$
. Indeed, suppose that
$\xi _0$
and
$\xi _{j,k}$
are algebraic numbers in
$\mathbb {Q}(q)$
for
$1 \leq k \leq m$
and
$0 \leq j \leq M$
, not all zero, such that
$$ \begin{align*} \xi_0 + \sum_{j=0}^M \sum_{k=1}^m \xi_{j,k} E_{q,P}^{(j)}(\alpha_k) = 0. \end{align*} $$
Then we obtain the nontrivial linear relation
$$ \begin{align*} \xi_0 + \sum_{j=0}^M \sum_{k=1}^m \frac{\xi_{j,k}}{\alpha_k^j} f_j(\alpha_k) = 0, \end{align*} $$
which again has coefficients in
$\mathbb {Q}(q)$
. Thus, it suffices to establish the linear independence of the
$f_j(\alpha _k)$
over
$\mathbb {Q}(q)$
.
Let
$r_0(X) = 1$
and
$r_j(X) := X (X-1) \cdots (X - j + 1)$
for
$1 \leq j \leq M$
. Then
$$ \begin{align*} f_j(x) = \sum_{n=j}^{\infty} \frac{r_j(n) x^n}{ \prod_{t=1}^n P(q^t)} = \sum_{n=1}^{\infty} \frac{r_j(n) x^n}{ \prod_{t=1}^n P(q^t)}, \end{align*} $$
since
$r_j(n) = 0$
for
$0 \leq n \leq j-1$
. Now, suppose that
$\lambda _0$
and
$\lambda _{j,k} \in \mathbb {Q}(q)$
are such that
$$ \begin{align*} \lambda_0 + \sum_{j=0}^M \sum_{k=1}^m \lambda_{j,k} f_j(\alpha_k) = 0. \end{align*} $$
Without loss of generality, we can assume that
$\lambda _0$
and the
$\lambda _{j,k}$
are algebraic integers. For
$1 \leq k \leq m$
, let
$A_k(X) := \sum _{j=0}^M \lambda _{j,k} r_j(X)$
. Then, from the definition of
$E_{q,P}(x)$
,
$$ \begin{align*} \widetilde{\lambda_0} + \sum_{n=1}^{\infty} \frac{\sum_{k=1}^m A_k(n) \alpha_k^n}{\prod_{t=1}^n P(q^t)} = 0, \end{align*} $$
where
$\widetilde {\lambda _0} = \lambda _0 + \sum _{k=1}^m \lambda _{0,k}$
.
Let N be a sufficiently large positive integer. We truncate the infinite sum above at N and clear denominators to obtain
$$ \begin{align} X_N:= \widetilde{\lambda_0} \prod_{t=1}^{N} P(q^t) + \sum_{n=1}^N \bigg(\sum_{k=1}^m A_k(n) \alpha_k^n \bigg) \prod_{t=n+1}^{N} P(q^t) = - \prod_{t=1}^N P(q^t) \sum_{n=N+1}^{\infty} \frac{\sum_{k=1}^m A_k(n)\alpha_k^n }{\prod_{t=1}^n P(q^t)}. \end{align} $$
Then
$X_N \in \mathcal {O}_{q}$
. Moreover, the right-hand side of (2.1) can be written as
$$ \begin{align} \prod_{t=1}^N &P(q^t) \sum_{n=N+1}^{\infty} \frac{\sum_{k=1}^m A_k(n) \alpha_k^n }{\prod_{t=1}^n P(q^t)} \nonumber \\ & = \frac{\sum_{k=1}^m A_k(N+1) \alpha_k^{N+1}}{P(q^{N+1})} + \frac{1}{P(q^{N+1})} \sum_{n=2}^{\infty} \frac{\sum_{k=1}^m A_k(N+n) \alpha_k^{N+n} }{\prod_{t=N+2}^{N+n} P(q^t)}. \end{align} $$
For simplicity of notation, let
$$ \begin{align*} \boldsymbol{\alpha} := \max \{|\alpha_1|, |\alpha_2|, \ldots, |\alpha_m| \}. \end{align*} $$
From the triangle inequality and the fact that each
$A_k(X)$
is a polynomial of degree M, for all
$\nu>0$
,
$$ \begin{align*} \bigg|\sum_{k=1}^m A_k(\nu) \alpha_k^{\nu} \bigg| \leq {\boldsymbol{\alpha}}^{\nu} \sum_{k=1}^m |A_k(\nu)| \ll {\nu}^M {\boldsymbol{\alpha}}^{\nu}. \end{align*} $$
Also, since
$|P(q^t)| \sim |q|^{tD}$
for t sufficiently large, the second term on the right-hand side of (2.2) is
$$ \begin{align*} \ll \frac{{\boldsymbol{\alpha}}^{N+1}}{|P(q^{N+1})|} \sum_{n=2}^{\infty} {{(n+N)}^M} \cdot {\bigg(\frac{\boldsymbol{\alpha}}{|q|^{DN}} \bigg)}^{n-1} \cdot {|q|}^{{- D( n^2+n-2 )}/{2}}. \end{align*} $$
This infinite series converges absolutely as
$|q|>1$
and the terms decay exponentially. Applying these bounds to the expression in (2.2) gives
$$ \begin{align} | X_N | \ll {\frac{\boldsymbol{\alpha}^{N+1}}{|P(q^{N+1})|} }N^M, \end{align} $$
where the implied constant depends on q, the
$\alpha _k$
and the coefficients
$\lambda _{j,k}$
.
We now estimate the size of conjugates of
$X_N$
. Since
$\pm q$
is a PV number,
$|\sigma _l(q)| < 1$
for
$2 \leq l \leq n_q$
. From the expression for
$X_N$
in (2.1), for all
$n \geq 0$
,
$$ \begin{align*} \sigma_l(X_N) = \sigma_l(\widetilde{\lambda_0}) \prod_{t=1}^{N} P ( {\sigma_l(q)}^t ) + \sum_{n=1}^N \bigg(\sum_{k=1}^m \sigma_l(A_k(n)) \sigma_l{(\alpha_k)}^n \bigg)\prod_{t=n+1}^{N} P ( {\sigma_l(q)}^t ). \end{align*} $$
Observe that
$$ \begin{align*} \bigg|\prod_{t=n+1}^N P ( \sigma_l(q^t) )\bigg|= \bigg|c_d\bigg(\prod_{t=n+1}^N \sigma_l(q^t)\bigg)^{d}\bigg|^{N-n} \prod_{t=n+1}^N \bigg| 1+\cdots+\frac{L_D}{c_d} ( \sigma_l(q^t) )^{D-d} \bigg|. \end{align*} $$
Since
$|\sigma _l(q)| < 1$
for
$2 \leq l \leq n_q$
, the series
$\sum _{t=1}^{\infty } {( \sigma _l ( q^t ) )}^s$
is absolutely convergent for
$1 \leq s \leq D-d$
. Thus, the infinite product
$$ \begin{align*} \prod_{t=1}^\infty \bigg| 1+\cdots+\frac{L_D}{c_d} ( \sigma_l(q^t) )^{D-d} \bigg| \end{align*} $$
is convergent and
$$ \begin{align*} \bigg|\prod_{t=n+1}^N P ( {\sigma_l(q)}^t ) \bigg| \ll |c_d|^{N-n} \prod_{t =n+1}^N |( \sigma_l(q^t))^{d(N-n)} | \ll |c_d|^{N-n}, \end{align*} $$
again since
$|\sigma _l(q)|<1$
for
$2\leq l \leq n_q$
. By these observations,
$$ \begin{align*} | \sigma_l(X_N)| \ll|c_d|^{N} \bigg( 1 + \sum_{n=1}^N |c_d|^{-n} \sum_{k=1}^m |\sigma_l(A_k(n))| |\sigma_l{(\alpha_k)}|^n \bigg). \end{align*} $$
Note that
$c_d \in \mathbb {Z}$
so that
$|c_d| \geq 1$
. Now,
$\sigma _l ( A_k(n)) = \sum _{j=0}^M \sigma _l ( \lambda _{j,k} ) r_j(n)$
, which is again a polynomial of degree M in n. Putting these bounds together, we deduce that
$$ \begin{align} | \sigma_l(X_N) | \ll N^{M+2} |c_d|^{N} {(\max \{ 1, | \sigma_l(\alpha_1) |, \ldots, | \sigma_l(\alpha_m) | \})}^N. \end{align} $$
As before, the implied constant only depends on q, the
$\alpha _k$
and the
$\lambda _{j,k}$
.
Multiplying the absolute values of all the conjugates of
$X_N$
and the corresponding bounds in (2.3) and (2.4) gives
$$ \begin{align*} \prod_{l=1}^{n_q} | \sigma_l(X_N) | &\ll \frac{N^{n_q(M+2)-2} |c_d|^{(n_q-1)N} {\boldsymbol{\alpha}}^{N}}{|P(q^{N+1})|} \bigg( \prod_{l=2}^{n_q} \max\{ 1, | \sigma(\alpha_1) |, \ldots, | \sigma(\alpha_m) | \} \bigg)^N \\ &\ll N^{n_q(M+2)-2} \bigg(\frac{{\boldsymbol{\alpha}} |c_d|^{(n_q-1)}{\prod_{l=2}^{n_q} \max \{ 1, | \sigma(\alpha_1)|, \ldots, | \sigma(\alpha_m) | \}}}{|q|^{D}}\bigg)^N. \end{align*} $$
By the hypothesis (1.1), the last bound tends to zero as
$N \rightarrow \infty $
. In particular,
$$ \begin{align*} \bigg| \prod_{l=1}^{n_q} \sigma_l(X_N) \bigg| < 1 \end{align*} $$
for all N sufficiently large. Here, the left-hand side is a power of the norm of an algebraic integer (noting that
$\mathbb {Q}(X_N)$
may be a strict subfield of
$\mathbb {Q}(q)$
). Thus,
$ \prod _{l=1}^{n_q} \sigma _l(X_N)$
must be a rational integer for all
$N>0$
. This is only possible if
$X_N = 0$
for all N sufficiently large.
Therefore, there exists a natural number
$N_0$
such that, for all
$N \geq N_0$
,
$$ \begin{align*} \frac{X_N}{\prod_{t=1}^N P(q^t)} = \widetilde{\lambda_0} + \sum_{n=1}^N \sum_{k=1}^m A_k(n) \alpha_k^n = 0. \end{align*} $$
Thus, considering the expression
$$ \begin{align*} \frac{X_{N+1}}{\prod_{t=1}^{N+1} P(q^t)} - \frac{X_N}{\prod_{t=1}^N P(q^t)}, \end{align*} $$
which equals zero for
$N> N_0$
, we obtain
$$ \begin{align*} A_1(N) \alpha_1^N + \cdots + A_m(N) \alpha^N_m = 0 \end{align*} $$
for all
$N> N_0$
. As
$\alpha _{k_1}/\alpha _{k_2}$
is not a root of unity, it follows from Theorem 2.1 that
$A_k(N) = 0$
for
$1 \leq k \leq m$
and all
$N> N_0$
. Thus, the polynomials
$A_k(X)$
are identically zero. Recall that
$$ \begin{align*} A_k(X) = \sum_{j=0}^M \lambda_{j,k} r_j(X), \end{align*} $$
and
$\deg r_j(X) = j$
. Since the
$r_j(X)$
have distinct degrees,
$A_k(X)$
is identically zero if and only if
$\lambda _{j,k} = 0$
for
$0 \leq j \leq M$
and
$1 \leq k \leq m$
. This completes the proof of the theorem.
2.2 Proof of Theorem 1.5
We begin along the same lines as in the proof of Theorem 1.1.
Suppose that the numbers in (1.3) are linearly dependent over
$\mathbb {Q}(q)$
. Then there exist algebraic integers
$\lambda _0, \lambda _1, \ldots , \lambda _k \in \mathcal {O}_q$
, not all zero, such that
$$ \begin{align*} \lambda_0+\lambda_1 E_{q,a_1}(\alpha)+\cdots+ \lambda_k E_{q,a_k}(\alpha)=0. \end{align*} $$
Without loss of generality, we can assume that
$\lambda _1 \neq 0$
. Otherwise, we can change the notation to replace
$a_j$
by
$a_1$
for the smallest
$j \leq k$
for which
$\lambda _j \neq 0$
and follow the argument below.
From the definition of the q-exponential function,
$$ \begin{align} \widetilde{\lambda_0} + \sum_{n=1}^{\infty} \frac{\lambda_1\alpha^n}{[a_1 n]_{q !}} +\cdots+ \sum_{n=1}^{\infty} \frac{\lambda_k \alpha^n}{[a_k n]_{q}!}= 0, \end{align} $$
where
$\widetilde {\lambda _0} = \lambda _0 + \lambda _1 + \cdots + \lambda _k$
. Set
$d=\mbox {lcm}\{a_1,\ldots ,a_k\}$
and
$d_i=d/a_i$
. Choose a large positive integer N and set
$N_i=N d_i$
for
$i=1, 2,\ldots ,k$
. With these choices of
$N_i$
,
$$ \begin{align*} a_1 N_1 = a_2 N_2 = \cdots = a_k N_k = d N. \end{align*} $$
Furthermore, for all
$i=1, 2, 3,\ldots ,k$
,
$$ \begin{align} \frac{[dN]_{q}! }{[a_i(N_i+1)]_{q}!} &= \frac{[a_i N_i]_{q}! }{[a_i(N_i+1)]_{q}!} \nonumber\\ &=\frac{(q^{a_i N_i}-1)\cdots(q-1)}{(q^{a_i N_i+a_i}-1)\cdots(q-1)}=\frac{1}{(q^{N d+a_i}-1)\cdots(q^{N d+1}-1)}. \end{align} $$
Now truncate the ith infinite sum in (2.5) at
$N_i$
and multiply by
$[dN]_{q}!$
to get
$$ \begin{align} X_N & := [dN]_{q}!\bigg(\widetilde{\lambda_0} + \sum_{n=1}^{N_1} \frac{\lambda_1 \alpha^n}{[a_1 n]_{q} !} +\cdots+ \sum_{n=1}^{N_k} \frac{\lambda_k \alpha^n}{[a_k n]_{q} !} \bigg) \nonumber \\ &\ =-[dN]_{q}!\bigg(\sum_{n=N_1+1}^{\infty} \frac{\lambda_1 \alpha^n}{[a_1 n]_{q} !} +\cdots+ \sum_{n=N_k+1}^{\infty} \frac{\lambda_k \alpha^n}{[a_k n]_{q} !}\bigg). \end{align} $$
Since
$[dN]_q! = [a_iN_i]_q!$
for
$1 \leq i \leq k$
,
$X_N$
is an algebraic integer in
$\mathcal {O}_q$
. We now estimate the right-hand side of (2.7). By an argument similar to the one in Theorem 1.1 and using (2.6), we deduce that
$$ \begin{align*} \bigg|[dN]_q! \sum_{n = N_j+1}^{\infty} \frac{\alpha^n}{[a_jn]_q!} \bigg| \ll \frac{{|\alpha|}^{N_j}}{|q|^{a_jdN}} \ll {\bigg( \bigg| \frac{\alpha^{d_j}}{q^{a_jd}} \bigg| \bigg)}^N, \end{align*} $$
since
$N_j = Nd_j$
. As
$a_1 < a_2 < \cdots < a_k$
,
$d_1> d_2 > \cdots > d_k$
and
$|\alpha | \geq 1$
,
$$ \begin{align} |X_N| \ll {\bigg( \bigg| \frac{\alpha^{d_1}}{q^{a_1d}} \bigg| \bigg)}^N. \end{align} $$
By the same argument as in the proof of Theorem 1.1, we can estimate the conjugates of
$X_{N}$
by
$$ \begin{align} |\sigma_l(X_N)|\ll N_1 {( \max\{1, |\sigma_l(\alpha)|\} )}^{N_1}. \end{align} $$
As before, the implied constant depends only on q, the
$a_i$
and the
$\lambda _{j,k}$
. Multiplying the bounds (2.8) and (2.9) for the absolute values of all the conjugates of
$X_N$
and noting that
$|\alpha | \geq 1 $
, we derive
$$ \begin{align*} \prod_{l=1}^{n_q} | \sigma_l(X_N) | \ll N_1^{n_q-1} { \bigg( \frac{|\alpha| \prod_{l=2}^{n_q} \max\{ 1, | \sigma_l(\alpha)| \} }{|q|^{a_1^2}} \bigg) }^{d_1N}. \end{align*} $$
By (1.2), the right-hand side tends to zero as
$N \rightarrow \infty $
. However, the left-hand side is a rational integer since it is a power of the norm of an algebraic integer. Therefore, there exists a natural number
$N_0$
such that
$X_N = 0$
for all
$N> N_0$
, which, in turn, implies that
$X_N=X_{N+1}=0$
. Consequently,
$$ \begin{align*} \widetilde{\lambda_0} + \sum_{n=1}^{N d_1} \frac{\lambda_1 \alpha^n}{[a_1 n]_{q}!} +\cdots+ \sum_{n=1}^{N d_k} \frac{\lambda_k \alpha^n}{[a_k n]_{q}!}=0 \end{align*} $$
and
$$ \begin{align*} \widetilde{\lambda_0} + \sum_{n=1}^{N d_1+d_1} \frac{\lambda_1 \alpha^n}{[a_1 n]_{q}!} +\cdots+ \sum_{n=1}^{N d_k+d_k} \frac{\lambda_k \alpha^n}{[a_k n]_{q}!}=0 \end{align*} $$
for all
$N> N_0$
. Subtracting these two relations gives
$$ \begin{align} \lambda_1 \sum_{n=N d_1+1}^{N d_1+d_1} \frac{\alpha^n}{[a_1 n]_{q}!} +\cdots+ \lambda_k \sum_{n=N d_k+1}^{N d_k+d_k} \frac{\alpha^n}{[a_k n]_{q}!}=0 \end{align} $$
for all
$N> N_0$
. Note that, for
$1 \leq j \leq k$
, we have
$Nd+a_j \leq a_j n \leq Nd + d$
in the above sums. Therefore,
$$ \begin{align*} \frac{\alpha^{Nd_k+1}}{[Nd+a_1]_q!} \end{align*} $$
divides each term in (2.10). By extracting this factor, we obtain
$$ \begin{align} & \lambda_1 \bigg(\alpha^{N(d_1 - d_k)} + \frac{\alpha^{N(d_1 - d_k)+1}}{(q^{Nd+2a_1}-1)\cdots(q^{Nd+a_1 + 1}-1)} + \cdots + \frac{\alpha^{N(d_1 - d_k)+d_1 -1}}{(q^{Nd+d}-1)\cdots(q^{Nd+a_1 + 1}-1)} \bigg) \nonumber \\ &\quad+ \lambda_2 \bigg(\frac{\alpha^{N(d_2 - d_k)}}{(q^{Nd+a_2}-1)\cdots(q^{Nd+a_1 + 1}-1)} + \cdots + \frac{\alpha^{N(d_2 - d_k)+d_2 -1}}{(q^{Nd+d}-1)\cdots(q^{Nd+a_1 + 1}-1)}\bigg) \nonumber \\ &\quad + \cdots \nonumber \\ &\quad + \lambda_k \bigg( \frac{1}{(q^{Nd+a_k}-1)\cdots(q^{Nd+a_1 + 1}-1)} + \cdots + \frac{\alpha^{d_k-1}}{(q^{Nd+d}-1)\cdots(q^{Nd+a_1 + 1}-1)}\bigg) = 0. \end{align} $$
Now, for
$1 \leq j \leq k$
and
$0 \leq l \leq d_j-1$
, the absolute value of the general term is
$$ \begin{align*} \bigg| \frac{\alpha^{N(d_j - d_k) + l}}{(q^{Nd+(l+1)a_j}-1)\cdots(q^{Nd+a_1+1}-1)}\bigg| \ll {\bigg| \frac{\alpha^{d_1 - d_k}}{q^{\delta d}} \bigg|}^N \end{align*} $$
except for
$j=1$
and
$l=0$
, with
$\delta = \min \{a_1, a_2 - a_1\}$
. Since
$1 \leq \delta $
, this implies that each term in (2.11) is
$\ll {|\alpha ^{d_1 - d_k}/q^d|}^N$
. By (1.2), this quotient is less than
$1$
, as
$1 \leq |\alpha | < |q|^{a_1}$
. Hence, taking the limit as
$N \rightarrow \infty $
, all terms in (2.11) tend to zero except the first, that is,
$\alpha ^{N(d_1-d_k)}$
. This implies that
$\lambda _1 = 0$
, which is a contradiction. This completes the proof.
Acknowledgments
This work was initiated when the second author was visiting Chennai Mathematical Institute during July 2022. He extends heartfelt gratitude to the institute for its hospitality and support. We also thank Prof. Ram Murty for helpful suggestions that improved the exposition of the paper. We are especially grateful to the referees for their careful reading and insightful comments about the earlier draft of this paper and thank them for making us aware of the results in [Reference Amou, Matala-Aho and Väänänen1].
