Deriving an inequality from a paper

In summary: The assertion is that ##0 \ll \dfrac{\alpha (\gamma - \delta)}{\gamma (\alpha -\beta)} \cdot \dfrac{\alpha^{dr}}{\gamma^{ds}}\ll 2##.
  • #1
Andrew_99
4
0
Hi, I am studying a paper by Yann Bugeaud:

http://irma.math.unistra.fr/~bugeaud/travaux/ConfMumbaidef.pdf
on page 13 there is an inequality (16) as given below-

attachment.php


which is obtained from -

attachment.php


, 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-

attachment.php


It is not clear how those two questions would be resolved.

Can anyone show the derivation of inequality (16)?

Thanks in Advance.
 
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  • #2
Are you missing the latex or images for the equations of interest in your post?

I see gaps but no equations.

POSTSCRIPT: I also had to edit your post to make the link more explicit. We sometimes get posts like this that are actually spam with hidden links.
 
  • #3
Multiplying it leads to the question whether
$$
\left|\dfrac{\beta^m}{\gamma^n} \cdot \dfrac{\gamma-\delta}{\alpha-\beta}- \dfrac{\delta^n}{\gamma^n} \right| \ll \dfrac{1}{\alpha^{\eta(m-1)}}
$$
given ##\alpha \geq \gamma \geq |\delta|^{1+\eta}\, , \,\alpha \geq |\beta|^{1+\eta}##.

Are there additional relations by the fact that ##(\alpha,\beta)## and ##(\gamma,\delta)## are Lucas pairs?
An idea could be to write the left hand side as a power series in ##\alpha## and estimate the coefficients.
 
  • #4
jedishrfu said:
Are you missing the latex or images for the equations of interest in your post?

I see gaps but no equations.

POSTSCRIPT: I also had to edit your post to make the link more explicit. We sometimes get posts like this that are actually spam with hidden links.
I used Image from paper.
 
  • #5
fresh_42 said:
Are there additional relations by the fact that ##(\alpha,\beta)## and ##(\gamma,\delta)## are Lucas pairs?
An idea could be to write the left hand side as a power series in ##\alpha## and estimate the coefficients.
I tried, two other persons tried, see the last image, but failed to get the result, Plz help.
 
  • #6
Andrew_99 said:
I tried, two other persons tried, see the last image, but failed to get the result, Plz help.
fresh_42 said:
Are there additional relations by the fact that ##(\alpha,\beta)## and ##(\gamma,\delta)## are Lucas pairs?

As far as I can see, there is no restriction on ##\eta## except ##0<\eta <\frac{1}{2}.## Let us assume that ##\eta## is close to zero. Then the assertion is that ##0 \ll \dfrac{\alpha (\gamma - \delta)}{\gamma (\alpha -\beta)} \cdot \dfrac{\alpha^{dr}}{\gamma^{ds}}\ll 2##, i.e. the quotient is close to ##1## in this case. This can be written as ##\alpha^m \gamma +\gamma^n \beta \approx \gamma^n\alpha + \alpha^m \delta##.

As this condition only depends on the given Lucas pairs ##u_m(\alpha,\beta) = v_n(\gamma,\delta)##, I assume that both equations are related; which I don't know since I haven't read the paper. If so, then I would start here and look whether we can go backwards to
$$
\left| \dfrac{\alpha (\gamma - \delta)}{\gamma (\alpha -\beta)} \cdot \dfrac{\alpha^{dr}}{\gamma^{ds}} -1 \right| \ll \alpha^{-\eta rd}
$$
 
  • #7
fresh_42 said:
As this condition only depends on the given Lucas pairs ##u_m(\alpha,\beta) = v_n(\gamma,\delta)##, I assume that both equations are related; which I don't know since I haven't read the paper.
Read page 12 and 13 (at most, page 10, 11) of below pdf file, for all information required for my query, no need to read the whole paper.
http://irma.math.unistra.fr/~bugeaud/travaux/ConfMumbaidef.pdf

The problem is algebraic derivation which I could not figure out.

You wrote-
fresh_42 said:
the assertion is that ## \dfrac{\alpha (\gamma - \delta)}{\gamma (\alpha -\beta)} \cdot \dfrac{\alpha^{dr}}{\gamma^{ds}}\ll 2##,
do you think
##\alpha^{-\eta rd} \approx 2## ? also please see our attempt in the last image of fist post.
anyway, please inform me if find anything, thanks.
 

Related to Deriving an inequality from a paper

1. How do you derive an inequality from a paper?

To derive an inequality from a paper, you need to carefully read and understand the information presented in the paper. Look for key equations, definitions, and assumptions that can help you formulate an inequality. You may also need to use mathematical techniques such as algebra, calculus, or statistics to manipulate the information and arrive at the desired inequality.

2. What are the steps involved in deriving an inequality from a paper?

The steps involved in deriving an inequality from a paper may vary depending on the specific paper and its content. However, some general steps include: identifying the relevant information in the paper, understanding the context and assumptions, formulating an inequality based on the information, and using mathematical techniques to manipulate the information and arrive at the desired inequality.

3. Can anyone derive an inequality from a paper?

In theory, anyone with a basic understanding of mathematics and the subject matter of the paper can attempt to derive an inequality. However, it is recommended to have a strong background in the subject and a good understanding of mathematical techniques to ensure accuracy and validity of the derived inequality.

4. What are some common mistakes to avoid when deriving an inequality from a paper?

Some common mistakes to avoid when deriving an inequality from a paper include misinterpreting the information presented in the paper, using incorrect mathematical techniques, and making assumptions that are not supported by the paper. It is important to carefully read and understand the paper and double-check all calculations and assumptions to avoid errors.

5. How can I verify the accuracy of my derived inequality?

To verify the accuracy of your derived inequality, you can compare it with the original paper and check if it aligns with the information and assumptions presented. You can also consult with other experts in the field or perform additional calculations and analyses to confirm the validity of your derived inequality.

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