A note on linear forms in two logarithms: the argument of an algebraic power

TL;DR

This paper improves the lower bounds for the argument of algebraic numbers with absolute value one that are not roots of unity, advancing understanding in transcendence theory and algebraic number theory.

Contribution

It provides an enhanced lower bound estimate for the argument of algebraic powers, refining previous results in the field.

Findings

New lower bound for the argument of algebraic numbers with absolute value one.

Enhanced understanding of linear forms in two logarithms.

Implications for transcendence and algebraic independence.

Abstract

In this note, we shall give an improved lower bound for the argument of a power of a given algebraic number which has absolute value one but is not a root of unity.

Equations

$$b_1\log {\alpha }_1+b_2\log {\alpha }_2+\cdots +b_n\log {\alpha }_n$$

$$\Lambda =b_2\log {\alpha }_2-b_1\log {\alpha }_1,$$

$$\mathrm{h}(\alpha )=\frac{1}{d}\left(\log \left|a\right|+\sum _{i=1}^d\log \max \{1,\left|{\alpha }^{(i)}\right|\}\right),$$

$${\Lambda }_0=b_2\log \alpha -b_1\pi i,$$

$$\begin{array}{rl}D=& [\mathbb{Q}(\alpha ):\mathbb{Q}]/2,\\ a=& \max \{20,10.98\left|\log \alpha \right|+2D\mathrm{h}(\alpha )\},\\ h=& \max \{17,\sqrt{D}/10,D(\log (b_1/2a+b_2/68.9)+2.35)+5.03\},\end{array}$$

$$\left|{\Lambda }_0\right|\ge -8.87a{h}^2.$$

$$\begin{array}{rl}R=& R_1+R_2-1,S=S_1+S_2-1,N=KL,g=\frac{1}{4}-\frac{N}{12RS},\\ \sigma =& \frac{1+2\mu -{\mu }^2}{2},b=\frac{(R-1)b_2+(S-1)b_1}{2}{\left(\prod _{k=1}^{K-1}k!\right)}^{-2/(K^2-K)}.\end{array}$$

$$a_i\ge \rho \left|\log {\alpha }_i\right|-\log {\alpha }_i+2D\mathrm{h}({\alpha }_i)$$

$$\begin{array}{rl}& \#\{{\alpha }_1^r{\alpha }_2^s:0\le r<R_1,0\le s<S_1\}\ge L,\\ & \#\{r{b}_2+s{b}_1:0\le r<R_2,0\le s<S_2\}\ge (K-1)L\end{array}$$

$$K(\sigma L-1)\log \rho -(D+1)\log N-D(K-1)\log b-gL(R{a}_1+S{a}_2)>ϵ(N),$$

$${\Lambda }^{\prime }=\Lambda \max \left\{\frac{LS{e}^{LS|\Lambda |/(2b_2)}}{2b_2},\frac{LR{e}^{LR|\Lambda |/(2b_1)}}{2b_1}\right\}.$$

$${\Lambda }_1=b_2\log \alpha -\frac{b_1\pi i}{2},$$

$$\begin{array}{rl}{b}^{\prime }=& \frac{b_1}{9.05\pi +2D\mathrm{h}(\alpha )}+\frac{b_2}{9.05\pi },\\ D=& [\mathbb{Q}(\alpha ):\mathbb{Q}]/2,\\ a=& 9.05\pi +2D\mathrm{h}(\alpha ),\\ h=& \max \{17,D,D(\log b^{\prime }+2.96)+0.01\}.\end{array}$$

$$\log \left|{\Lambda }_1\right|>-2.7704a{h}^2.$$

$$\log \left|\arg ({\alpha }^n)\right|>-2.7704a{h}^2.$$

$$\log \left|{\Lambda }_1\right|\ge -b^{\prime }D\mathrm{h}(\alpha )-D\log 2>-2a{h}^2-D\log 2>-2.7704a{h}^2.$$

$$\begin{array}{rl}{\delta }_0=& 0.01,{\delta }_1=0.044,\mu =0.59,\rho =18.1,\\ {\alpha }_1=& i,{\alpha }_2=\alpha ,a_1=\frac{\rho \pi }{2},a_2=\frac{\rho \pi }{2}+2D\mathrm{h}(\alpha )=a.\end{array}$$

$$\begin{array}{rl}\sigma & =\frac{1+2\mu -{\mu }^2}{2},\lambda =\sigma \log \rho ,H=\frac{h}{\lambda }+\frac{1}{\sigma },\\ L_0& =H+\sqrt{H^2+\frac{1}{4}},L=\lfloor L_0\rfloor +\frac{1}{2}\end{array}$$

$$v_0(x)=\frac{1}{4a_1}+\frac{4}{3a_2}+\frac{x}{12a_1},v_1(x)=\frac{x}{3},v_2(x)=\lambda (x-H).$$

$$\begin{array}{rl}K=& 1+\lfloor kL{a}_1{a}_2\rfloor ,R_1=4,S_1=\lfloor \frac{L+3}{4}\rfloor ,\\ R_2=& 1+\lfloor \sqrt{(K-1)L{a}_2/a_1}\rfloor ,S_2=1+\lfloor \sqrt{(K-1)L{a}_1/a_2}\rfloor .\end{array}$$

$$\frac{L^2}{L-H}\le \frac{(L^{\pm }{)}^2}{L^{\pm }-H}=2L_0,$$

$$\sqrt{k}>\frac{v_1(L)}{v_2(L)}>\frac{L}{3\lambda (L-H)}>\frac{L^{+}}{3\lambda (L^{+}-H)}>0.2432,$$

$$\sqrt{k}<\frac{v_1(L^{-})}{2v_2(L^{-})}+\sqrt{{\left(\frac{v_1(L^{-})}{2v_2(L^{-})}\right)}^2+\frac{v_0(L^{-})}{v_2(L^{-})}}<0.279$$

$$\begin{array}{rl}H>& 7.5,15\le L\le L_0+\frac{1}{2}<0.92h,\\ 0.91<& \frac{(L^{+}{)}^3}{(3\lambda (L^{+}-H){)}^2}<kL<\frac{(L^{-}{)}^3}{(3\lambda (L^{-}-H){)}^2}<0.99,\\ K>& kL{a}_1{a}_2>700.\end{array}$$

$$\sqrt{k}L\le \frac{2L_0}{3\lambda }+\sqrt{{\left(\frac{2L_0}{3\lambda }\right)}^2+\frac{2L_0}{\lambda }\left(\frac{4}{3a_2}+\frac{1}{4a_1}+\frac{L^{+}}{12a_1}\right)}<0.239h.$$

$$gL(R{a}_1+S{a}_2)\le \frac{L}{4}(R_1{a}_1+S_1{a}_2)+\frac{L^{3/2}\sqrt{(K-1)a_1{a}_2}}{2}-\frac{K{L}^2}{12}\left(\frac{a_1}{S}+\frac{a_2}{R}\right)$$

$$\begin{array}{rl}\frac{1}{R}>\frac{1}{R_1+\sqrt{(K-1)L{a}_2/a_1}}=& \frac{1}{\sqrt{(K-1)L{a}_2/a_1}}-\frac{R_1}{(K-1)L{a}_2/a_1}\\ & +\frac{16a_1}{(K-1)L{a}_2(R_1+\sqrt{(K-1)L{a}_2/a_1})}\end{array}$$

$$\begin{array}{rl}\frac{1}{S}>\frac{1}{4+\sqrt{(K-1)L{a}_1/a_2}}=& \frac{1}{\sqrt{(K-1)L{a}_1/a_2}}-\frac{4}{(K-1)L{a}_1/a_2}\\ & +\frac{a_2{S}^2}{(K-1)L{a}_1(4+\sqrt{(K-1)L{a}_1/a_2})}.\end{array}$$

$$\begin{array}{rl}& K{L}^2\left(\frac{a_1}{S}+\frac{a_2}{R}\right)>(K-1)L^2\left(\frac{a_1}{S}+\frac{a_2}{R}\right)\\ & >2L^{3/2}\sqrt{(K-1)a_1{a}_2}-L(R_1{a}_1+S_1{a}_2)+\frac{a_{1}L{S}_1^2}{S_1+\sqrt{(K-1)L{a}_1/a_2}}\\ & +\frac{16a_{1}L}{4+\sqrt{(K-1)L{a}_2/a_1}}.\end{array}$$

$$\begin{array}{rl}gL(R{a}_1+S{a}_2)<& \frac{L}{3}(R_1{a}_1+S_1{a}_2)+\frac{L^{3/2}\sqrt{(K-1)a_1{a}_2}}{3}\\ & -\frac{a_{2}L{S}_1^2}{S_1+\sqrt{(K-1)L{a}_1/a_2}}\\ & -\frac{16a_{1}L}{4+\sqrt{(K-1)L{a}_2/a_1}}.\end{array}$$

Full text

1. Introduction

Since Baker [1, 2] found lower bounds for linear forms in logarithms

[TABLE]

with $\alpha_{i}$ complex algebraic numbers and $b_{i}$ integers, many authors such as Matveev [8] have given improved lower bounds for linear forms in logarithms of algebraic numbers.

Lower bounds for linear forms in two logarithms

[TABLE]

with $\alpha_{1},\alpha_{2}$ two complex algebraic numbers and $b_{1},b_{2}$ two positive integers had already been given by Gel’fond [4] and several authors such as Laurent [5, 6] and Laurent, Mignotte and Nesterenko [7] have given improved lower bounds.

For any algebraic number $\alpha$ of degree $d$ over ${\mathbb{Q}}$, we define the absolute logarithmic height of $\alpha$ by

[TABLE]

where $a$ is the leading coefficient of the minimal polynomial of $\alpha$ over ${\mathbb{Z}}$ and $\alpha^{(i)}(i=1,\ldots,d)$ denote the conjugates of $\alpha$ in complex numbers.

As an application of their lower bound for linear forms in two logarithms, Laurent, Mignotte and Nesterenko [7] gave an lower bound for the special logarithmic form

[TABLE]

where $\alpha$ is an algebraic number of absolute value one but not a root of unity and $b_{1},b_{2}$ are positive integers. Putting

[TABLE]

we have

[TABLE]

Later, Laurent [6] obtained the stronger lower bound for general linear forms in two logarithms in the following form:

Theorem 1.1**.**

Let $\Lambda=b_{2}\log\alpha_{2}-b_{1}\log\alpha_{1}$ be a linear form of two logarithms with $b_{1},b_{2}$ positive integers and $\alpha_{1},\alpha_{2}$ complex algebraic number. Put $D=[{\mathbb{Q}}(\alpha_{1},\alpha_{2}):{\mathbb{Q}}]/[{\mathbb{R}}(\alpha_{1},\alpha_{2}):{\mathbb{R}}]$.

Let $K$ be an integer $\geq 2$ and $L,R_{1},R_{2},S_{1},S_{2}$ be positive integers. Let $\rho$ and $\mu$ be real numbers with $\rho>1$ and $1/3\leq\mu\leq 1$. Put

[TABLE]

Let $a_{1},a_{2}$ be positive real numbers such that

[TABLE]

for $i=1,2$. Assume that

[TABLE]

and

[TABLE]

where $\epsilon(N)=2\log(N!N^{-N+1}(e^{N}+(e-1)^{N}))/N$.

Then $\left\lvert\Lambda^{\prime}\right\rvert>\rho^{-\mu KL}$, where

[TABLE]

However, Laurent has not given an improved lower bound for the special logarithmic form $\Lambda_{0}$. The purpose of this note is to deduce an improved lower bound for the special logarithmic form $\Lambda_{0}$ from Theorem 1.1.

Theorem 1.2**.**

Let

[TABLE]

where $b_{1},b_{2}$ are positive integers and $\alpha$ is an complex algebraic number of absolute value one but not a root of unity. Put

[TABLE]

Then,

[TABLE]

We note that we work a slightly generalized form $\Lambda_{1}$ rather than $\Lambda_{0}$.

It immediately follow from Theorem 1.2 that if $\alpha$ is an complex algebraic number of absolute value one but not a root of unity, $n$ is a positive integer and $b_{1}$ is the nearest integer from $2n\left\lvert\arg\alpha\right\rvert/\pi$, then, under the same notation as in Theorem 1.2 with $b_{2}=n$,

[TABLE]

2. Preliminaries to the proof

If $d=\gcd(b_{1},b_{2})>1$, then we divide $b_{i}$’s by $d$ to have another logarithmic form $\Lambda_{1}/d=(b_{2}/d)\log\alpha-(b_{1}/d)\pi i/2$. If Theorem 1.2 holds for $\left\lvert\Lambda_{1}/d\right\rvert$, then this would give the desired lower bound for $\left\lvert\Lambda_{1}\right\rvert$. Thus we may assume that $\gcd(b_{1},b_{2})=1$.

Moreover, we may assume that ${b^{\prime}}>4h^{2}$. Indeed, if ${b^{\prime}}\leq 4h^{2}$, then Liouville’s inequality immediately gives

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We set

[TABLE]

Clearly $\alpha_{i}$’s and $a_{i}$’s satisfy the condition (8) in Theorem 1.1.

Let

[TABLE]

and $k=(v_{1}(L)+\sqrt{v_{1}(L)^{2}+4v_{0}(L)v_{2}(L)})^{2}/(2v_{2}(L))^{2}$ be the positive real number such that $\sqrt{k}$ satisfies the quadratic equation $v_{2}(L)k-v_{1}(L)\sqrt{k}-v_{0}(L)=0$, where

[TABLE]

Moreover, we set

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We see that

[TABLE]

where $L^{\pm}=L_{0}\pm 1/2$,

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[TABLE]

and

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Using (21), we have

[TABLE]

3. Confirmation of the conditions of Theorem 1.1

In this section, we shall confirm the conditions of Theorem 1.1.

In order to obtain an upper bound for $gL(Ra_{1}+Sa_{2})$, we follow the proof of Lemme 9 of [7]. We begin by quoting the upper bound

[TABLE]

provided by (5.19) of [7].

As in [7], using the identity $\frac{1}{x+y}=\frac{1}{x}-\frac{y}{x^{2}}+\frac{y^{2}}{x^{2}(x+y)}$, we obtain

[TABLE]

and

[TABLE]

These lower bounds yield that

[TABLE]

Now, (26) gives

[TABLE]

Recalling that $R_{1}=4,S_{1}=\left\lfloor(L+3)/4\right\rfloor\geq L/4$ and $K-1<kLa_{1}a_{2}$, we have

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Now we follow the proof of Lemme 10 of [7]. From (22) and (24), we see that

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and

[TABLE]

Hence, we have

[TABLE]

where

[TABLE]

(24) implies that $f_{1}(K)<f_{1}(600)<0.00084$. Moreover, it follows from (22) that $f_{2}(K):=f_{1}(K)+3/2+\log((1+\delta_{1})/2\sqrt{k})<2.96$ and therefore

[TABLE]

From (30) and (36), we see that the left of (10) is at least

[TABLE]

say.

We can easily see that $\Phi=v_{2}k-v_{1}\sqrt{k}-v_{0}=0$.

Now we would like to show that $\Theta>\epsilon(N)$. Our argument is similar to the argument in Section 3.2.2 of [6]. We observe that $\Theta=(D-1)\Theta_{0}+\Theta_{1}$, where

[TABLE]

We recall the assumption that ${b^{\prime}}>4h^{2}$ and we observe that $h/\lambda>6$ and $L\leq L^{+}<h$. Thus we obtain

[TABLE]

and

[TABLE]

On the other hand, (24) gives that $N=KL>10000$ and, using Feller’s version [3] of Stirling’s formula, we have $\epsilon(N)<\epsilon(10000)<0.004$. This implies that our settings of $k,L,R_{1},S_{1},R_{2},S_{2}$ satisfy (10).

Now we shall confirm (9). Since $\alpha_{2}=\alpha$ is not a root of unity, $\alpha_{1}^{r}\alpha_{2}^{s}(0\leq r\leq 3,0\leq s\leq S_{1}-1$ take $4S_{1}\geq L$ different values and therefore the former part of (9) holds.

It follows from (25) that $R_{2}-1<\sqrt{(K-1)La_{2}/a_{1}}<\sqrt{k}La_{2}<a_{2}h/4<2a_{2}h^{2}$ and, similarly, $S_{2}-1<a_{1}h/4<2a_{1}h^{2}$. Since we have assumed that ${b^{\prime}}>4h^{2}$, $b_{1}>2a_{2}h^{2}>R_{2}-1$ or $b_{2}>2a_{1}h^{2}>S_{2}-1$.

Thus we can see that $R_{2}-1<b_{1}$ or $S_{2}-1<b_{2}$. If we have $r_{1}b_{2}-s_{1}b_{1}=r_{2}b_{2}-s_{2}b_{1}$ for some integers $r_{1},r_{2},s_{1},s_{2}$ with $0\leq r_{1},r_{2}\leq R_{2}-1,0\leq s_{1},s_{2}\leq S_{2}-1$, then $(r_{1}-r_{2})b_{2}=(s_{1}-s_{2})b_{1}$ and $\left\lvert r_{1}-r_{2}\right\rvert\leq R_{2}-1,\left\lvert s_{1}-s_{2}\right\rvert\leq S_{2}-1$. Since we have assumed that $\gcd(b_{1},b_{2})=1$, $r_{1}\equiv r_{2}\pmod{b_{1}}$ and $s_{1}\equiv s_{2}\pmod{b_{2}}$. If $R_{2}-1<b_{1}$, then $r_{1}=r_{2}$. If $S_{2}-1<b_{2}$, then $s_{1}=s_{2}$. Hence, we must have $r_{1}=r_{2}$ and $s_{1}=s_{2}$. This yields that $rb_{2}-sb_{1}(0\leq r\leq R_{2}-1,0\leq s\leq S_{2}-1)$ take $R_{2}S_{2}>(K-1)L$ different values. Hence, the latter part of (9) also holds.

Thus we have confirmed that Theorem 1.1 holds with our settings.

4. Computation of the constants

Now we apply Theorem 1.1 to obtain $\log\left\lvert\Lambda_{1}^{\prime}\right\rvert>-\mu KL\log\rho$, where

[TABLE]

By (22), we have $KL<L(1+kLa_{1}a_{2})<kL^{2}a_{1}a_{2}(1+1/(kLa_{1}a_{2}))<1.00126kL^{2}a_{1}a_{2}$ and $\mu KL\log\rho<2.7701ah^{2}$. Now Theorem 1.1 gives

[TABLE]

We may assume that $\log\left\lvert\Lambda_{1}\right\rvert<-2.75ah^{2}$.

$R<(L+3)/4+\sqrt{k}La_{2}<0.75+0.92h+0.257ha_{2}<0.291ha_{2}\leq 0.291ah$ and $S<4+\sqrt{k}La_{1}<4+0.257ha_{1}<0.266ah$.

From $a\geq\rho\pi/2$ and $h\geq 17$, we see that $\log(ah^{2})/(ah^{2})<0.0011$. Thus, we obtain $RL,SL<0.268ah^{2}$ and therefore $\log\max\{RL,SL\}+\log\Lambda_{1}<-2.749ah^{2}$.

[TABLE]

This immediately gives that

[TABLE]

This proves Theorem 1.2.