Help : Deriving an inequality related to Lucas Sequence

Aug 2019
Hi, I am studying a paper by Yann Bugeaud (click here), on page 13 there is an inequality (16) as given below-

$\Lambda:=\left|\left(\frac{\alpha(\gamma-\delta)}{\gamma(\alpha-\beta)}\right)\left(\frac{\gamma^{s}}{\alpha^{r}}\right)^{-d}-1\right| \ll \alpha^{-\eta r d}$

which is obtained from -

Let $\theta>1$ and $d>1$ be an integer. We suppose that (15) with $\operatorname{gcd}(m-1, n-1)=d$ and $(m-1) /(n-1) \leq \theta$ is satisfied and we put
m-1=d r, \quad n-1=d s
where $r$ and $s$ are positive integers. We assume that $y \geq 7$. We re-write (15) as
\frac{\alpha}{\alpha-\beta} \alpha^{r d}-\frac{\gamma}{\gamma-\delta} \gamma^{s d}=\frac{\beta^{1+r d}}{\alpha-\beta}-\frac{\delta^{1+s d}}{\gamma-\delta}

,on page 12.

How the inequality (16) is derived? I couldn't figure it out. However one of my forum member tried but it has two problems (problems are marked as "how?"), it is given below-

One of the member tried to prove it but failed, his answer is given as below-

If we divide both sides of equation (15) of page 12 by $\frac{\gamma}{\gamma-\delta} \gamma^{s d}$, we get$\frac{\alpha(\gamma-\delta)}{\gamma(\alpha-\beta)}\left(\frac{\gamma^{s}}{\alpha^{\prime}}\right)^{-d}-1=\frac{\gamma-\delta}{\alpha-\beta} \cdot \frac{\beta^{1+\alpha}}{\gamma^{1+\alpha}}-\left(\frac{\delta}{\gamma}\right)^{1+s d}$

Then, we have $\left.\left|\frac{\alpha(\gamma-\delta)}{\gamma(\alpha-\beta)}\left(\frac{\gamma^{\prime}}{\alpha^{\prime}}\right)^{-d}-1\right|=\left|\frac{\gamma-\delta}{\alpha-\beta} \cdot \frac{\beta^{1+\pi \alpha}}{\gamma^{1+\alpha d}}-\left(\frac{\delta}{\gamma}\right)^{1+s d}\right| \text { since } \frac{\delta}{\gamma} \approx 0 \text { (i.e. } 0<\frac{\delta}{\gamma}<1\right)$ from $\gamma \geq|\delta|^{1+\eta} \gg|\delta|$, we have $\left|\frac{\gamma-\delta}{\alpha-\beta} \cdot \frac{\beta^{1+\gamma d}}{\gamma^{1+\alpha}}-\left(\frac{\delta}{\gamma}\right)^{1+s d}\right| \approx\left|\frac{\gamma-\delta}{\alpha-\beta} \cdot \frac{\beta^{1+\alpha d}}{\gamma^{1+\alpha}}\right|$

Note that as the values of $\gamma$ become larger and larger, $\frac{\delta}{\gamma}$ become smaller and smaller and become close to 0 so, we ignored $\frac{\delta}{\gamma}$ by writing $\frac{\delta}{\gamma}=0$

Similarly, $\frac{\delta}{\gamma} \approx 0$ and $\frac{\beta}{\alpha} \approx 0$ from $\alpha \geq|\beta|^{1+\eta} \gg|\beta|,$ we have, $\frac{\gamma-\delta}{\alpha-\beta}=\frac{1-\frac{\delta}{\eta}}{1-\frac{\beta}{a}} \cdot \frac{\gamma}{\alpha} \approx \frac{1-0}{1-0} \cdot \frac{\gamma}{\alpha}=\frac{\gamma}{\alpha}$ and $\operatorname{so}\left|\frac{\gamma-\delta}{\alpha-\beta} \cdot \frac{\beta^{1+\alpha}}{\gamma^{1+\alpha}}\right| \approx\left|\frac{\gamma}{\alpha} \cdot \frac{\beta^{1+\alpha d}}{\gamma^{1+\alpha d}}\right|$

Since $\alpha^{r} \approx \gamma^{s}$ (how?), we have $\left|\frac{\gamma}{\alpha} \cdot \frac{\beta^{1+\alpha}}{\gamma^{1+\alpha d}}\right| \approx\left|\frac{\gamma}{\alpha} \cdot \frac{\beta^{1+\gamma d}}{\gamma \alpha^{\prime 2}}\right|=\frac{\gamma}{\alpha} \cdot \frac{|\beta|^{1+\gamma d}}{\gamma \alpha^{\prime d}}=\frac{|\beta|^{1+\alpha}}{\alpha^{1+\alpha} d}$

By inspection $\alpha \ll \alpha^{1+(1-\eta)}$ and from $\alpha \geq|\beta|^{1+\eta} \gg|\beta| \Longrightarrow \alpha^{r d} \gg|\beta|^{r d}$ so we get, $\frac{\left.|\beta|^{1} \beta\right|^{r d}}{\alpha^{1+r d}} \ll \frac{\alpha^{1}|\beta|^{r d}}{\alpha^{1+r d}}$

$\Longrightarrow \frac{\left.|\beta|^{1} \beta\right|^{\pi d}}{\alpha^{1+\mu d}} \ll \frac{\alpha^{1}|\beta|^{(l-\eta) \pi d}}{\alpha^{1+r d}}(\text { how? }) \Longrightarrow \frac{|\beta|^{1+\alpha d}}{\alpha^{1+\alpha d}} \ll \frac{\alpha^{1+(l-\nu) \nu t}}{\alpha^{1+\alpha d}}=\alpha^{-\eta r d}$

This answer is taken from another member which is not complete. It is not clear how those two questions would be resolved. Can any one show the derivation of inequality (16)?

Thanks in Advance.


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