How to prove $b \ll a (a^{2}b^2)^{2.1/k} $

Question

In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is

written on page $9$ that

It follows from (13) and Lemma 2 with $\eta=0.1 $ we have that, $$b \ll a (a^{2}b^2)^{2.1/k} $$

The image of page 9 :

How do we prove $b \ll a (a^{2}b^2)^{2.1/k} $?

My Confusion : Note, if $k$ increases then ${2.1/k}$ goes close to $0$, so it is confusing

how the inequality holds true.

Answer 1

This is a partial solution to what I think is your problem (namely, that you're having a tough time working through this paper), although not to the immediate question you asked; I'm suggesting that you consider writing to the author saying something like this:

Hello.

I'm a computer science student, and I've been trying to work my way through your paper, "Linear Forms in the Logarithms of ...". As you can see by the questions I've been asking on Math Stackexchange (https://math.stackexchange.com/users/257587/mike-sq), my mathematical background is not all that sophisticated, so some of the steps that you assumed readers could fill in are well beyond my filling-in skills. I'd appreciate it if you could help me in a few places.

I apologize for writing in English; I do not speak French well enough to write coherent questions about math in French.

Right now I'm stuck on the remark following Lemma 2 on page 9. You write ...

It's hard for me to imagine that you'd get no response at all to such a note, even if the response is something like "I think, judging by some of your earlier questions, that perhaps that this paper is a bit too advanced for your current level of mathematical understanding and sophistication. Perhaps you should complete at least the junior-level courses for mathematics majors at your university before trying to read it, or you'll have troubles with almost every sentence." The author, after all, is probably very well suited to judge what preparation is needed to read his paper.

What I anticipate is that the author will say something like "This statement holds under the hypotheses written above; perhaps I should have been clearer, but $a$ must be at least ..., and $k$ is no greater than ..., hence ..."

Answer 2

We know, $ \left|\frac{a+1}{a}- \left(\frac{xz}{y^2}\right)^k\right|\leq \frac{1}{b} \cdots (13)$

Here, $xz= ((a+1)(ab^2+1))^\frac{1}{k}, y^2= (ab+1)^\frac{2}{k}.$

So, $ \left|\frac{a+1}{a}- \frac{(xz)^k}{(ab+1)^2} \right|\leq \frac{1}{b} \cdots (a)$

Since, $(ab+1)>ab\implies(ab+1)^\frac{2}{k}>(ab)^\frac{2}{k}$

$\implies \frac{1}{(ab+1)^\frac{2}{k}}<\frac{1}{(ab)^\frac{2}{k}} \implies \frac{(xz)^k}{(ab+1)^\frac{2}{k}}<\frac{(xz)^k}{(ab)^\frac{2}{k}}$

$\implies -\frac{(xz)^k}{(ab+1)^\frac{2}{k}}> -\frac{(xz)^k}{(ab)^\frac{2}{k}}$ [since, the inequality sign

changes when both sides are multiplied by $-1$.]

$\implies-\frac{(xz)^k}{(ab)^\frac{2}{k}} < -\frac{(xz)^k}{(ab+1)^\frac{2}{k}}$ [changing sides]

$\implies (\frac{a+1}{a}) -\frac{(xz)^k}{(ab)^\frac{2}{k}} < (\frac{a+1}{a}) -\frac{(xz)^k}{(ab+1)^\frac{2}{k}}$

$\implies \left|\frac{a+1}{a}- \frac{(xz)^k}{(ab)^\frac{2}{k}} \right| \leq \left|\frac{a+1}{a}- \frac{(xz)^k}{(ab+1)^\frac{2}{k}} \right|$

Using inequality (a), we deduce-

$\left|\frac{a+1}{a}- \frac{(xz)^k}{(ab)^\frac{2}{k}} \right| \leq \frac{1}{b} \cdots (b)$

By inspection, we see- $ \left|(\frac{a+1}{a})^\frac{1}{k} \right |\leq \left|\frac{a+1}{a} \right | $

$\implies \left|(\frac{a+1}{a})^\frac{1}{k}- \frac{(xz)^k}{(ab)^\frac{2}{k}}\right |\leq \left|\frac{a+1}{a}- \frac{(xz)^k}{(ab)^\frac{2}{k}} \right | $

[subtracting $\frac{(xz)^k}{(ab)^\frac{2}{k}}$ on both sides]

Using inequality (b), we deduce-

$\left|(\frac{a+1}{a})^\frac{1}{k}- \frac{(xz)^k}{(ab)^\frac{2}{k}}\right |\leq \frac{1}{b}\ \cdots (c) $

To use Lemma 2, we need to re-parameterize $p, q$ of Lemma 2

according to inequality $(13)$, so let, $p= (xz)^k, q= (ab)^\frac{2}{k} $, then-

$ \left|(1+\frac{1}{a})^\frac{1}{k}- \left(\frac{p}{q}\right)\right| > \frac{c_5}{aq^{\epsilon +2 }} $ [ Lemma 2]

$\implies \left|(1+ \frac{1}{a})^\frac{1}{k}- \frac{(xz)^k}{(ab)^\frac{2}{k}} \right| > \frac{c_5}{a(ab)^{\frac{2(\epsilon +2 )}{k}}} $

$\implies \frac{1}{b} > \frac{c_5}{a(ab)^{\frac{2(\epsilon +2 )}{k}}} $ [ using inequality (c)]

$\implies a(a^2b^2)^{\frac{2.1}{k}} >bc_5 $ [$\epsilon =0.1$]

$\implies a(a^2b^2)^{\frac{2.1}{k}} >b \implies b \ll_k a(a^2b^2)^{\frac{2.1}{k}} $