On restricted approximation measures of Jacobi’s triple product

Abstract

We obtain rational approximations for Jacobi’s triple product

$$\begin{aligned} \Pi _q(t):= \prod _{m=1}^\infty (1-q^{2m})(1+q^{2m-1}t)(1+q^{2m-1}t^{-1}), \end{aligned}$$

when $t=a/b\in {\mathbb {Q}}$ is non-zero and $q=1/d$ with $d\in {\mathbb {Z}}{\setminus }\{0, \pm 1 \}$. Especially we give effective and restricted approximation for the values of Jacobi’s triple product and for the values of Euler’s infinite product.

1 Introduction and results

In the following $\Vert x\Vert$ denotes the distance of a real number x to the nearest integer. Let $\xi$ be an irrational real number. Then the irrationality exponent $\mu (\xi )$ of $\xi$ is defined by setting $\mu ( \xi ) =v(\xi )+1$, where $v(\xi )$ is the infimum of the real numbers u for which the inequality

$$\begin{aligned} \left\| N\xi \right\| >N^{-u} \end{aligned}$$

holds for every sufficiently large positive integer N. By restricting the set of positive integers N in the above definition to a certain infinite subset of positive integers, we get the definition of so-called restricted irrationality exponents. Let now d be an integer, $|d|\ge 2$. We will follow Bennett and Bugeaud [3] by defining $v_d(\xi )$ to be the infimum of the real numbers u for which the inequality

$$\begin{aligned} \left\| d^s\xi \right\| >|d|^{-su} \end{aligned}$$

holds for every sufficiently large positive integer s. Likewise, $v_d^{\text {eff}}(\xi )$ denotes the infimum of the real numbers u for which there exists a computable constant $c(\xi ,d)$ such that the condition

$$\begin{aligned} \left\| d^s\xi \right\| >c(\xi ,d)|d|^{-su} \end{aligned}$$

holds for every sufficiently large positive integer s. Further, we call $v_d(\xi )+1$ and $v_d^{\text {eff}}(\xi )+1$ restricted irrationality exponents of $\xi$.

Amou and Bugeaud [1] noted that $v_d(\xi ) \ge 0$ for all irrational real numbers $\xi$, and furthermore $v_d(\xi )=0$ for almost all irrational real numbers $\xi$, provided that $d\ge 2$. However, if $\xi$ is a classical mathematical constant like $\sqrt{2}$, e or $\pi$, we do not even know whether $v_d(\xi )=0$ for any d. On the other hand, for certain explicit numbers there are already results which give upper bounds for restricted irrationality exponents. Namely, Rivoal [10] proved that $v_d(\log r)$ is arbitrarily close to 0 for certain integers d, when $r\in {\mathbb {Q}}$ is sufficiently close to 1. See also Dubickas [6]. Recently, Bennett and Bugeaud [3] proved that there exists an effectively computable positive constant $\tau _1=\tau _1(p)$ such that

$$\begin{aligned} v_p^{\text {eff}}\left( \sqrt{p^2+1} \right) \le 1-\tau _1 \end{aligned}$$

for every prime number p. They also noted that one can deduce the existence of an effectively computable positive constant $\tau _2=\tau _2(d,k)$ such that

$$\begin{aligned} v_d^{\text {eff}}\left( \sqrt{d^{2k}+1}\right) \le 1-\tau _2 \end{aligned}$$

for every positive integer k and $d\ge 2$.

In the present work, we investigate restricted rational approximations for the values of Jacobi’s triple product

$$\begin{aligned} \Pi _q(t):= \prod _{m=1}^\infty (1-q^{2m})(1+q^{2m-1}t)(1+q^{2m-1}t^{-1}) \end{aligned}$$

at $t=a/b\in {\mathbb {Q}}\setminus \{0\}$ and $q=1/d$, where $d\in {\mathbb {Z}}{\setminus }\{0, \pm 1 \}$. Particularly, we consider determining effective exponents $v_d^{\text {eff}}(\Pi _{\frac{1}{d}}(a/b))$. Furthermore, we obtain that $v_d(\Pi _{\frac{1}{d}}(a/b))=0$.

Theorem 1

Let $t=a/b\in {\mathbb {Q}}\setminus \{0\}$, $\gcd (a,b)= 1$, $d\in {\mathbb {Z}}$ and   $\max \{|a|,|b|\}<|d|$. Then for all $s, M\in {\mathbb {Z}}$ with $s\ge C$ we have

$$\begin{aligned} \left| \Pi _{\frac{1}{d}}(t) - \frac{M}{d^s}\right| > \frac{1}{2|d|^{s(1+\varepsilon _1(s))}},\quad \varepsilon _1(s)=\frac{6}{\sqrt{s}}+\frac{8}{s}, \end{aligned}$$

where $C=\left( 3\max \{|a|,|b|\}-1\right) ^2/4$. Consequently, $v_d(\Pi _{\frac{1}{d}}(t))= 0$.

It is remarkable that (as far as we know) the only irrationality measure results for Jacobi’s triple product at arbitrary rational $t\ne 0$ are outcomes of the linear independence results for (the right-hand side of) Jacobi’s Theta function

$$\begin{aligned} \Theta (q,t):= \sum _{n=0}^\infty q^{n^2}t^n,\quad |q|<1. \end{aligned}$$

Namely, because $\Pi _{\frac{1}{d}}(t)=-1+\Theta (1/d,t)+\Theta (1/d,t^{-1})$, the result of Bundschuh and Shiokawa in [5] implies the estimate

$$\begin{aligned} \mu \left( \Pi _{\frac{1}{d}}(t) \right) \le \frac{5+\sqrt{17}}{2}=4.5615\ldots \end{aligned}$$

for $d\in {\mathbb {Z}}\setminus \{0, \pm 1 \}$ and $t\in {\mathbb {Q}}\setminus \{0\}$.

We also study the restricted approximations for Euler’s infinite product

$$\begin{aligned} \pi _q(t):=\prod _{n=1}^\infty (1 - q^{n}t) \end{aligned}$$

at $t=1$, when $q=1/d$, $d\in {\mathbb {Z}}\setminus \{0, \pm 1 \}$ and $q= (1-\sqrt{5})/(1+\sqrt{5})$. Jacobi’s triple product has a q-expansion, given by the well-known Jacobi’s triple product identity

$$\begin{aligned} \prod _{m=1}^\infty (1-q^{2m})(1+q^{2m-1}t)(1+q^{2m-1}t^{-1})= \sum _{n=-\infty }^{\infty } t^n q^{n^2}, \end{aligned}$$

(1)

see e.g. [2, p. 498]. Our proof of Theorem 1 will be based on this identity. By replacing q with $q^{3/2}$ and t with $-q^{-1/2}$ in (1), we obtain, after simplification, that

$$\begin{aligned} \prod _{m=1}^\infty (1-q^{3m})(1-q^{3m-2})(1-q^{3m-1})= \sum _{n=-\infty }^{\infty } (-1)^n q^{(3n^2-n)/2}. \end{aligned}$$

This can be rewritten as

$$\begin{aligned} \pi _q:=\prod _{n=1}^\infty (1-q^n)= 1 + \sum _{n=1}^\infty (-1)^n \left( q^{n(3n-1)/2} +q^{n(3n+1)/2} \right) , \end{aligned}$$

(2)

which is the famous Euler’s pentagonal formula, see e.g. [2, p. 500]. On the basis of the above consideration Euler’s infinite product $\pi _q$ seems to be a special case of Jacobi’s triple product. But because of the square root substitutions we can not obtain our results for Euler’s product $\pi _q$ from Theorem 1. Therefore, we investigate separately the product $\pi _q$ at $q=1/d$, $d\in {\mathbb {Z}}\setminus \{0, \pm 1 \}$.

Theorem 2

Let $d\in {\mathbb {Z}}\setminus \{0, \pm 1 \}$, $M\in {\mathbb {Z}}$ and $s\in {\mathbb {Z}}_+$. Then

$$\begin{aligned} \left| \pi _{\frac{1}{d}}(1)- \frac{M}{d^s}\right| > \frac{1}{2 |d|^{s(1+\varepsilon _2(s))}},\quad \varepsilon _2(s) = \frac{3+\sqrt{1+24s}}{2s}. \end{aligned}$$

Consequently, $\ v_d(\pi _{\frac{1}{d}}(1))=0.$

Theorem 2 improves considerably the earlier results concerning this special case. Recently, Leinonen et al. [7] obtained that $v_d(\pi _{\frac{1}{d}}(t))=1.1547\ldots$ with arbitrary $t\in {\mathbb {Q}}\setminus \{0\}$. It should be noted that there are more general results available which consider the irrationality exponent of Euler’s product. Already in 1969 Bundschuh [4] proved that the irrationality exponent of the product $\pi _{\frac{1}{d}}(t)$ satisfies the inequality $\mu (\pi _{\frac{1}{d}}(t))\le 7/3$, for $|d|\in {\mathbb {Z}}_{\ge 2}$ and $t\in {\mathbb {Q}}\setminus \{0\}$. This is still the best known upper bound for $\mu (\pi _{\frac{1}{d}}(t))$. For a more extensive overview on the arithmetical properties of Euler’s infinite product $\pi _q(t)$, see e.g. [7].

The next theorem is inspired by the work [8], where the authors investigated the distances between Fibonomials. Therefore we consider restricted approximations over the number field ${\mathbb {K}}={\mathbb {Q}}(\sqrt{5})$, only. In the following, the notation ${\mathbb {Z}}_{{\mathbb {K}}}$ denotes the ring of integers of ${\mathbb {K}}$ and ${\overline{A}}:=a-b\sqrt{5}$ denotes the field conjugate of $A=a+b\sqrt{5}\in {\mathbb {K}}$.

Theorem 3

Let ${\mathbb {K}}={\mathbb {Q}}(\sqrt{5})$, $q=(1-\sqrt{5})/(1+\sqrt{5})$, $\alpha = (1+\sqrt{5})/2$ and $s\in {\mathbb {Z}}_+$. Let $M\in {\mathbb {Z}}_{\mathbb {K}}{\setminus }\{0\}$ be such that $\left| {\overline{M}}\right| \le |M|$. Then

$$\begin{aligned} \left| \pi _q(1)- \frac{M}{\alpha ^s}\right| > \frac{1}{2\alpha ^{s(2 +\varepsilon _3(s))}},\quad \varepsilon _3(s)= \frac{17+2\sqrt{1+24s}}{s}. \end{aligned}$$

(3)

The lower bound in (3) is an improvement to the result proved in [8], where the corresponding approximation exponent is $s(3 +\varepsilon (s))$ and $M=(\sqrt{5})^l$, $l\in {\mathbb {Z}}_+$. There are more general approximation results for Euler’s infinite product and related q-series over number fields, see e.g. [9]. The results in [9] imply that there exist positive constants $\Gamma$ and $H_0$ such that

$$\begin{aligned} \left| \pi _q(1)- \frac{M}{N}\right| > \frac{1}{H^{14/3 +\varepsilon (H)}},\quad \varepsilon (H)=\frac{\Gamma }{\sqrt{\log H}}, \end{aligned}$$

for all $M/N\in {\mathbb {Q}}(\sqrt{5})$, where $q=(1-\sqrt{5})/(1+\sqrt{5})$, $M,N\in {\mathbb {Z}}_{{\mathbb {Q}}(\sqrt{5})}$, $N\ne 0$ and $H=\max \{|M|,|N|,|{\overline{M}}|,|{\overline{N}}|\}\ge H_0$. On the other hand, in the Remark section of this paper we prove that for all $\tau \in {\mathbb {R}}\setminus {\mathbb {Q}}(\sqrt{5})$ there exists an infinite sequence of fractions $M/N\in {\mathbb {Q}}(\sqrt{5})$, where $M,N\in {\mathbb {Z}}_{{\mathbb {Q}}(\sqrt{5})}$ and $N\ne 0$, such that

$$\begin{aligned} \left| \tau - \frac{M}{N}\right| < \frac{9+4\sqrt{5}}{|N|^4}. \end{aligned}$$

2 Proof of Theorem 1

We suppose that $N\in {\mathbb {Z}}_+$. By Jacobi’s triple product identity (1) we have

$$\begin{aligned} \Pi _{\frac{1}{d}}(t)=\sum _{n=-\infty }^{\infty } d^{-n^2}t^n. \end{aligned}$$

Because $t=a/b$, we obtain that

$$\begin{aligned} (ab)^Nd^{N^2}\Pi _{\frac{1}{d}}(t)= A_N(d,t) + R_N(d,t), \end{aligned}$$

(4)

where

$$\begin{aligned} A_N(d,t) = (ab)^Nd^{N^2} + (ab)^N\sum _{n=1}^N d^{N^2-n^2}\left( t^n+\frac{1}{t^n}\right) \in {\mathbb {Z}}\end{aligned}$$

and

$$\begin{aligned} R_N(d,t)= (ab)^N\sum _{n=N+1}^\infty \frac{t^n+\frac{1}{t^n}}{d^{n^2-N^2}} \in {\mathbb {Q}}\left[ \left[ t, \frac{1}{t}, \frac{1}{d}\right] \right] . \end{aligned}$$

We write $n=N+1+k$. Since $n^2-N^2\ge 2N+1+(2N+3)k\$ for all $k\in {\mathbb {Z}}_{\ge 0}$, we get that

$$\begin{aligned} |R_N(d,t)|&\le \frac{|a|^{2N+1}}{|b||d|^{2N+1}}\sum _{k=0}^\infty \left( \frac{|a|}{|b||d|^{2N+3}}\right) ^k+ \frac{|b|^{2N+1}}{|a||d|^{2N+1}}\sum _{k=0}^\infty \left( \frac{|b|}{|a||d|^{2N+3}}\right) ^k\\&= \frac{|a|^{2N+1}}{|b||d|^{2N+1}-|a|/|d|^2}+\frac{|b|^{2N+1}}{|a||d|^{2N+1}-|b|/|d|^2}. \end{aligned}$$

Further, our assumption $\max \{|a|,|b|\}+1\le |d|$ implies that

$$\begin{aligned} |R_N(d,t)|&\le \frac{|a|^{2N+1}}{(|a|+1)^{2N+1}-1}+\frac{|b|^{2N+1}}{(|b|+1)^{2N+1}-1}\\&\le \frac{|a|^{2N+1}}{|a|^{2N+1}+(2N+1)|a|^{2N}}+\frac{|b|^{2N+1}}{|b|^{2N+1}+(2N+1)|b|^{2N}}. \end{aligned}$$

We choose ${\hat{N}}:=(3\max \{|a|,|b|\}-1)/2$. Then

$$\begin{aligned} |R_N(d,t)| \le \frac{1}{2} \end{aligned}$$

(5)

for all $N\ge {\hat{N}}$.

Let us denote

$$\begin{aligned} \Lambda := \Pi _{\frac{1}{d}}(t)- \frac{M}{d^s}. \end{aligned}$$

By using (4) we obtain that

$$\begin{aligned} (ab)^N d^{N^2}\Lambda = A_N(d,t) - (ab)^N M d^{N^2-s} + R_N(d,t), \end{aligned}$$

where the main term

$$\begin{aligned} \Delta _N(t):= A_N(d,t)- (ab)^N Md^{N^2-s} \end{aligned}$$

is a rational integer, assuming that $N \ge \sqrt{s}$. Because the determinant

$$\begin{aligned}&\left| \begin{matrix} A_N(d,t) &{} (ab)^N d^{N^2}\\ A_{N+1}(d,t) &{} (ab)^{N+1}d^{(N+1)^2} \end{matrix}\right| \\&= (ab)^{N+1}d^{(N+1)^2}A_N(d,t) - (ab)^Nd^{N^2}A_{N+1}(d,t)\\&= (ab)^{2N+1}d^{N^2}d^{(N+1)^2} \left( \sum _{n=1}^N d^{-n^2}\left( t^n+\frac{1}{t^n}\right) - \sum _{n=1}^{N+1}d^{ - n^2}\left( t^n+\frac{1}{t^n}\right) \right) \\&=-(ab)^{2N+1}d^{N^2} \left( t^{N+1}+\frac{1}{t^{N+1}}\right) \ne 0, \end{aligned}$$

we get that $\Delta _N(t)\ne 0$ or $\Delta _{N+1}(t) \ne 0$. We let now N be such that $\hat{N}\le \sqrt{s}\le N< \sqrt{s}+2$ and $\Delta _{N}(t)\in {\mathbb {Z}}\setminus \{0\}$. Hence,

$$\begin{aligned} 1\le |\Delta _N(t)|=|(ab)^Nd^{N^2}\Lambda -R_N(d,t)|\le |ab|^N|d|^{N^2}|\Lambda |+|R_N(d,t)|. \end{aligned}$$

By (5) we get the approximation

$$\begin{aligned} 1 \le 2|ab|^{N}|d|^{N^2}|\Lambda |< 2|d|^{s\left( 1+6/\sqrt{s}+8/s\right) }|\Lambda |, \end{aligned}$$

which completes the proof of Theorem 1.

3 Proof of Theorem 2

We suppose that $N\in {\mathbb {Z}}_+$. By Euler’s pentagonal formula (2) we have

$$\begin{aligned} \pi _{1/d}(1)= 1 + \sum _{n=1}^\infty (-1)^n \left( d^{-n(3n-1)/2} +d^{-n(3n+1)/2} \right) . \end{aligned}$$

Hence, we can write

$$\begin{aligned} d^{N(3N+1)/2}\pi _{1/d}(1)= A_N(d)+ R_N(d), \end{aligned}$$

(6)

where

$$\begin{aligned} A_N(d)= & {} d^{N(3N+1)/2}\\{} & {} + \sum _{n=1}^N (-1)^n\left( d^{N(3N+1)/2 - n(3n-1)/2} + d^{N(3N+1)/2 - n(3n+1)/2}\right) \in {\mathbb {Z}}[d], \end{aligned}$$

and

$$\begin{aligned} R_N(d)&=\sum _{n=N+1}^\infty (-1)^n\left( \frac{1}{d^{ n(3n-1)/2 -N(3N+1)/2}} + \frac{1}{d^{n(3n+1)/2 -N(3N+1)/2}}\right) \\&\in {\mathbb {Z}}[[1/d]]. \end{aligned}$$

By noting that $(N+1)(3(N+1)-1)/2 -N(3N+1)/2= 2N+1$ we deduce that

$$\begin{aligned} |R_N(d)| \le \frac{1}{|d|^{2N+1}}\sum _{n=0}^\infty \frac{1}{|d|^n} \le \frac{2}{|d|^{2N+1}}\le \frac{1}{4}. \end{aligned}$$

(7)

Let us denote

$$\begin{aligned} \Lambda := \pi _{1/d}(1)-\frac{M}{d^s}, \end{aligned}$$

(8)

where $M \in {\mathbb {Z}}$, $s\in {\mathbb {Z}}_+$. By using (6) and (8), we get that

$$\begin{aligned} d^{N(3N+1)/2} \Lambda = A_N(d) - Md^{N(3N+1)/2-s} + R_N(d), \end{aligned}$$

where the term

$$\begin{aligned} \Delta _N:= A_N(d) - Md^{N(3N+1)/2-s} \end{aligned}$$

is an integer if $N(3N+1)/2-s >0$. Additionally,

$$\begin{aligned} d \not \mid \Delta _N = (-1)^N + (-1)^N d^{N} + \cdots + d^{N(3N+1)/2}- Md^{N(3N+1)/2 -s}. \end{aligned}$$

Hence, $\Delta _N \ne 0$ and further

$$\begin{aligned} 1 \le |\Delta _N| = |d^{N(3N+1)/2}\Lambda - R_N(d)|\le |d|^{N(3N+1)/2}|\Lambda | + |R_N(d)|. \end{aligned}$$

By (7), we obtain that

$$\begin{aligned} 1< 2|\Lambda | |d|^{N(3N+1)/2}. \end{aligned}$$

In particular, this lower bound holds when N is such that

$$\begin{aligned} (N-1)(3N-2)/2\le s<N(3N+1)/2. \end{aligned}$$

In this case,

$$\begin{aligned} N(3N+1)/2&=(N-1)(3N-2)/2+3N-1\\&\le s\left( 1+\frac{3+\sqrt{1+24s}}{2s}\right) , \end{aligned}$$

and we obtain Theorem 2.

4 Proof of Theorem 3

We suppose that $N\in {\mathbb {Z}}_+$. By (2), we have

$$\begin{aligned} \begin{aligned} \pi _q(1)&= 1+ \sum _{n=1}^\infty (-1)^n \left( \left( \frac{1-\sqrt{5}}{1+\sqrt{5}}\right) ^{n(3n-1)/2} + \left( \frac{1-\sqrt{5}}{1+\sqrt{5}}\right) ^{n(3n+1)/2}\right) \\&=1 + \sum _{n=1}^\infty \left( \frac{(-1)^{n + n(3n-1)/2}}{\alpha ^{n(3n-1)}} + \frac{(-1)^{n+ n(3n+1)/2}}{\alpha ^{n(3n+1)}} \right) .\\ \end{aligned} \end{aligned}$$

Hence, we can write

$$\begin{aligned} \alpha ^{N(3N+1)}\pi _q = A_N(\alpha )+ R_N(\alpha ), \end{aligned}$$

(9)

where

$$\begin{aligned} \begin{aligned} A_N(\alpha ) = \alpha ^{N(3N+1)} + \sum _{n=1}^N (&(-1)^{n+ n(3n-1)/2}\alpha ^{N(3N+1) - n(3n-1)} \\ +&(-1)^{n+n(3n+1)/2}\alpha ^{N(3N+1) - n(3n+1)} ) \in {\mathbb {Z}}[\alpha ], \end{aligned} \end{aligned}$$

(10)

and

$$\begin{aligned} R_N(\alpha ) = \sum _{n=N+1}^\infty \left( \frac{(-1)^{n+ n(3n-1)/2}}{\alpha ^{ n(3n-1) -N(3N+1)}} + \frac{(-1)^{n + n(3n+1)/2}}{\alpha ^{n(3n+1)-N(3N+1)}}\right) \in {\mathbb {Z}}[[1/\alpha ]]. \end{aligned}$$

By noting that $(N+1)(3(N+1)-1) - N(3N+1) =4N+2$, we deduce that

$$\begin{aligned} |R_N(\alpha )| \le \frac{1}{\alpha ^{4N+2}} \sum _{n=0}^\infty \frac{1}{\alpha ^n} = \frac{1}{\alpha ^{4N+1}(\alpha -1)}=\frac{1}{\alpha ^{4N}}. \end{aligned}$$

(11)

We denote

$$\begin{aligned} \Lambda :=\pi _q(1) -\frac{M}{\alpha ^s}. \end{aligned}$$

(12)

From Eqs. (9) and (12) we get that

$$\begin{aligned} \alpha ^{N(3N+1)} \Lambda = A_N(\alpha ) - M\alpha ^{N(3N+1)-s} + R_N(\alpha ). \end{aligned}$$

Because $A_N(\alpha )\in {\mathbb {Z}}[\alpha ]$ and $M, \alpha , 1/\alpha \in {\mathbb {Z}}_{\mathbb {K}}$, we have

$$\begin{aligned} \Delta _N:= A_N(\alpha ) - M\alpha ^{N(3N+1)-s}\in {\mathbb {Z}}_{\mathbb {K}}. \end{aligned}$$

Since the determinant

$$\begin{aligned} \begin{aligned} \left| \begin{matrix} A_N(\alpha ) &{} \alpha ^{N(3N+1)} \\ A_{N+1}(\alpha ) &{} \alpha ^{(N+1)(3(N+1)+1)}\end{matrix}\right| = \alpha ^{(N+1)(3(N+1)+1)}A_N(\alpha ) - \alpha ^{N(3N+1)}A_{N+1}(\alpha )\\ =\alpha ^{N(3N+1)}(-1)^{N+ (N+1)(3N+2)/2}\left( \alpha ^{2(N+1)} +(-1)^{N+1} \right) \end{aligned} \end{aligned}$$

is non-zero. We have that $\Delta _N\ne 0$ or $\Delta _{N+1}\ne 0$. So, we can choose N such that $\Delta _{N} \in {\mathbb {Z}}_{\mathbb {K}}\setminus \{0\}$. Hence, we have

$$\begin{aligned} 1 \le |N_{{\mathbb {K}}/{\mathbb {Q}}}(\Delta _{N})| = |\Delta _{N}||\overline{\Delta _{N}}| = |\alpha ^{N(3N+1)}\Lambda - R_{N}(\alpha )||\overline{\Delta _{N}}|. \end{aligned}$$

(13)

Let us bound from above the absolute value of the conjugate $\overline{\Delta _{N}}$. First, we note that

$$\begin{aligned} \begin{aligned} |\overline{\Delta _{N}}|&= |\overline{A_{N}(\alpha )- M\alpha ^{N(3N+1) -s}} |= |A_{N}({\overline{\alpha }}) - {\overline{M}}{\overline{\alpha }}^{N(3N+1)-s}|\\&\le |A_{N}({\overline{\alpha }})| + |{\overline{M}}||{\overline{\alpha }}|^{N(3N+1)-s}.\end{aligned} \end{aligned}$$

By using (10), we get that

$$\begin{aligned} \begin{aligned} |A_{N}({\overline{\alpha }} )|&< \sum _{n=0}^{N(3N+1)} |{\overline{\alpha }}|^{n} < \sum _{n=0}^\infty |{\overline{\alpha }}|^n = \frac{1}{1+{\overline{\alpha }}}=\alpha ^2. \end{aligned} \end{aligned}$$

We can restrict the approximation to such numbers $M/\alpha ^s$ that

$$\begin{aligned} \left| \pi _q(1)-\frac{M}{\alpha ^s}\right| \le 1. \end{aligned}$$

Because

$$\begin{aligned} \pi _q(1)\sim 1.226742\ldots , \end{aligned}$$

it is enough to consider numbers $M/\alpha ^s$ satisfying

$$\begin{aligned} 0<M/\alpha ^s\le 2.226742\ldots <\alpha ^2. \end{aligned}$$

Now we suppose that $N(3N+1)\ge 2s$. Since $|{\overline{M}}| \le |M|$, by our assumption, we obtain that

$$\begin{aligned} |{\overline{M}}||{\overline{\alpha }}|^{N(3N+1)-s} \le |M|\left| \frac{-1}{\alpha }\right| ^{N(3N+1)-s} = |M|\alpha ^{s-N(3N +1)}\le \frac{M}{\alpha ^s} < \alpha ^2. \end{aligned}$$

Hence,

$$\begin{aligned} |\overline{\Delta _N}|< 2\alpha ^2. \end{aligned}$$

(14)

Inequalities (13) and (14) imply now that

$$\begin{aligned} \frac{1}{2\alpha ^2} < \alpha ^{N(3N+1)}|\Lambda | + |R_{N}(\alpha )|. \end{aligned}$$

By (11), we have

$$\begin{aligned} |R_{N}(\alpha )| \le \frac{1}{\alpha ^4}, \end{aligned}$$

which implies

$$\begin{aligned} 1< 2|\Lambda |\alpha ^5 \alpha ^{N(3N +1)}. \end{aligned}$$

We fix an integer ${\hat{N}}$ such that ${\hat{N}}(3{\hat{N}}+1)<2s\le ({\hat{N}}+1)(3({\hat{N}}+1)+1)$. We can now suppose that N is ${\hat{N}}+1$ or ${\hat{N}}+2$. Hence,

$$\begin{aligned} N(3N+1)&\le ({\hat{N}}+2)(3({\hat{N}}+2)+1) ={\hat{N}}(3{\hat{N}}+1)+12{\hat{N}}+14\\&<s\left( 2+\frac{12+2\sqrt{1+24s}}{s}\right) , \end{aligned}$$

and we obtain Theorem 3.

5 Remark

When the approximations $M/N\in {\mathbb {Q}}(\sqrt{5})$ are not restricted, then there are better approximations for general $\tau \in {\mathbb {R}}{\setminus } {\mathbb {Q}}(\sqrt{5})$. The ring of integers ${\mathbb {Z}}_{{\mathbb {Q}}(\sqrt{5})}= {\mathbb {Z}}[\omega ]$, where $\omega =\frac{1+\sqrt{5}}{2}$. We call the fraction

$$\begin{aligned} \frac{a+b\omega }{c+d\omega },\quad a,b,c,d\in {\mathbb {Z}} \end{aligned}$$

primitive whenever the vector (a, b, c, d) is primitive, meaning $\gcd (a,b,c,d)=1$.

Lemma 1

Let $\tau \in {\mathbb {R}}\setminus {\mathbb {Q}}(\sqrt{5})$. Then there exists an infinite sequence of primitive fractions $M/N\in {\mathbb {Q}}(\sqrt{5})$, where $M,N\in {\mathbb {Z}}_{{\mathbb {Q}}(\sqrt{5})}$, $N\ne 0$, such that

$$\begin{aligned} \left| \tau - \frac{M}{N}\right| < \frac{9+4\sqrt{5}}{|N|^4}. \end{aligned}$$

(15)

Proof

Let $\tau \in {\mathbb {R}}\setminus {\mathbb {Q}}(\sqrt{5})$. Because $1,\omega ,\tau$ and $\tau \omega$ are linearly independent over ${\mathbb {Q}}$, there exists an infinite sequence of primitive integer 4–tuples $(a,b,c,d)\in {\mathbb {Z}}^4\setminus \{(0,0,0,0)\}$ such that

$$\begin{aligned} \left| a+b\omega +c\tau +d\tau \omega \right| < \frac{1}{H^3}, \end{aligned}$$

(16)

where $H=\max \{|b|,|c|,|d|\}\ge 1$ (see Corollary 1D in [11, p. 27]). If $c=d=0$, then $H=|b|\ge 1$ and

$$\begin{aligned} 0\ne \left| a+b\omega \right| < \frac{1}{|b|^3}. \end{aligned}$$

Thus,

$$\begin{aligned} 1\le \left| (a+b\omega )(a+b{\overline{\omega }})\right| < \frac{\left| a+b{\overline{\omega }}\right| }{|b|^3}\le \frac{\left| \sqrt{5}\right| }{|b|^2}+\frac{1}{|b|^6} \end{aligned}$$

implying $|b|= 1$ and so $|a|\le 2$, contradicting the fact that there are infinitely many (a, b, c, d). Hence there exists an infinite sequence of integer 4–tuples $(a,b,c,d)\in {\mathbb {Z}}^4\setminus \{(0,0,0,0)\}$ satisfying (16) with $(c,d)\ne (0,0)$. We also note that $|c+d\omega |\le (1+\omega )H$. Consequently,

$$\begin{aligned} \left| \tau - \frac{a+b\omega }{c+d\omega }\right| < \frac{1}{H^3|c+d\omega |} \le \frac{(1+\omega )^3}{|c+d\omega |^4}= \frac{9+4\sqrt{5}}{|c+d\omega |^4}, \end{aligned}$$

which completes the proof . $\square$

The above bound (15) is a variation of the fundamental result presented in e.g. [11, p. 253].