Lower bounds for linear forms of logarithms (a la Baker)? - MathOverflow most recent 30 from http://mathoverflow.net2013-06-18T21:26:48Zhttp://mathoverflow.net/feeds/question/11212http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/11212/lower-bounds-for-linear-forms-of-logarithms-a-la-bakerLower bounds for linear forms of logarithms (a la Baker)?Joel Ouaknine2010-01-09T03:32:30Z2010-01-09T03:32:30Z
<p>Let $\lambda_1$, $\lambda_2$, and $a$ be three fixed complex algebraic
numbers.</p>
<p>For a given integer $n$, write
$\Theta(n) = \arg(a \lambda_1^n + \lambda_2^n)$.</p>
<p>Assuming $\Theta(n)$ is not zero, I am looking for 'good' lower bounds on
$|\Theta(n)|$. By 'good' I mean that if $|\Theta(n)| > B(n)$,
then $1/B(n)$ should asymptotically grow slower that any exponential in $n$.</p>
<p>Is there a way to use one of Baker's theorems (which provide effective
lower bounds on linear combinations of logs of algebraic numbers) to achieve
this?</p>
<p>For example, writing instead $\Gamma(n) = \arg(a \lambda_1^n \lambda_2^n)$
(say), one can get <em>polynomial</em> bounds on $|\Gamma(n)|$: noting that
$\displaystyle{\Gamma(n) = \log\left(\frac{a \lambda_1^n \lambda_2^n}
{|a \lambda_1^n \lambda_2^n|}\right) =
\log a + n \log \lambda_1 + n \log \lambda_2 - \log |a| -
n \log |\lambda_1| - n \log |\lambda_2|}$
we can apply e.g. Baker-Wustholz (1993) to the above linear form
and get a lower bound $|\Gamma(n)| > C(n)$ (assuming that $|\Gamma(n)|$ is non-zero)
such that $1/C(n)$ is in fact bounded by a fixed polynomial in $n$.</p>
<p>The problem in getting a similar lower bound for $|\Theta(n)|$ is that, even though
$\Theta(n)$ can be written as a linear combination of logs of algebraic
numbers of constant degree, as for $\Gamma(n)$, the <em>height</em> of the
algebraic number $a \lambda_1^n + \lambda_2^n$ is potentially exponential in
$n$, and it does not seem that taking logs will help here.</p>
<p>The critical case is of course when $\lambda_1$ and $\lambda_2$ have the
same magnitude. In fact, I would be happy for an approach with even very
simple values of $a$, such as $a = 2$.</p>