Lower bounds for linear forms of logarithms (a la Baker)? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T21:26:48Z http://mathoverflow.net/feeds/question/11212 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/11212/lower-bounds-for-linear-forms-of-logarithms-a-la-baker Lower bounds for linear forms of logarithms (a la Baker)? Joel Ouaknine 2010-01-09T03:32:30Z 2010-01-09T03:32:30Z <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>