%! TEX root = DA.tex
% vim: tw=50
% 21/10/2024 12PM

\begin{theorem*}
  Let $\alpha_1, \ldots, \alpha_n \in
  \ol{\Qbb}_{\neq 0}$. Let $\log \alpha_j$ be a
  choice of their logarithms. Let $b_1, \ldots,
  b_n \in \Zbb$. Let
  \begin{align*}
    A_j
    &= \max(\height(f(\alpha_1)), \ldots,
    H(f(\alpha_n)), \exp(|log \alpha_1|), \ldots,
    \exp(|\log \alpha_n|), 10) \\
    B^*
    &= \max \left( \frac{|b_1|}{\log A_n}, \ldots,
    \frac{|b_{n - 1}|}{\log A_n}, |b_n|, 10) \right) \\
    \Lambda
    &= b_1 \log \alpha_1 + \cdots + b_n \log
    \alpha_n
    \qquad \qquad \text{$\leftarrow$ homogeneous}
  \end{align*}
  Then there is an effective constant $C$ that
  depends only on $n$ and the degree of
  $\Qbb(\alpha_1, \ldots, \alpha_n)$ such that
  \[
    |\Lambda| > \exp(-C(\log A_1), \ldots, (\log
    A_n)(\log B^*)) \qquad \text{or} \qquad
    \Lambda = 0
  .\]
\end{theorem*}

Observe
\[
  \exp(\Re \log \alpha_j) = |\alpha_j| \le
  \height(f(\alpha_j))
.\]
Recall
\[
  B = \max(|b_1|, \ldots, |b_n|, \log A_1, \ldots,
  \log A_n, 10)
.\]
Typical scenario: $\alpha_1, \ldots, \alpha_{n -
1}$ fixed numbers, $b_n = 1$, $b_j \sim \log A_n$.

In the setting of Diophantine approximations, it
is possible to show
\[
  \left| \alpha - \frac{p}{q} \right|  > c(\alpha)
  \cdot \frac{1}{q^{d - \eps(\alpha)}}
,\]
with $c(\alpha)$ and $\eps(\alpha)$ being
effective constants.

\begin{proposition*}
  There is an effective absolute constant $C$ such
  that for all $p, q, k, m$:
  \[
    |p 2^k - q 3^m| > \frac{\max(2^k,
    3^m)}{\max(p, q, 10)^{-C\log (\max(k, m) /
    \log \max(p, q, 10) + 10)}}
  ,\]
  or $p2^k = q3^m$.
\end{proposition*}

\begin{proof}
  Suppose $3^m > 2^k$.
  \[
    \Lambda = k\log 2 - m \log 3 + 1 \cdot \log (p
    / q)
  .\]
  $A_2 = A_1 = 10$, $A_3 = \max(p, q, 10)$
  \[
    B^* = \frac{\max(k, m)}{\log A_3} + 1
  .\]
  Then:
  \[
    |\Lambda| > \exp(-C \log A_3 \log B^*) =
    A_3^{-C \log B^*}
  .\]
  $|\exp(\Lambda) - 1| > \frac{1}{10}|\Lambda|$
  \[
    \exp(\Lambda) - 1 = \left| 2^k \cdot ^{-m}
    \cdot \frac{p}{q} -1  \right| \ge
    A_3^{-\tilde{C} \log B^*}
  \]
  Multiply by $q \cdot 3^m$.
\end{proof}

Before:
\[
  |p2^k - q3^m| > C\frac{\max(2^k, 3^m)^{1 -
  \eps}}{\max(p, q)}
.\]
The new bound wins when $\max(p, q) < \max(2^k,
3^m)^{o(1)}$.

In particular, when $p = q = 1$:
\[
  |2^k - 3^m| > \frac{\max(2^k, 3^m)}{\max(k, m)^{C}}
  \qquad \text{vs} \qquad
  |2^k - 3^m| > C 2^{1 -\eps k}
.\]

$p_1 2^{k_1} + p_2 3^{k_2} + p_3 5^{k_3}$ for $k_1,
k_2, k_3 \in \Zbb_{> 0}$, $p_1, p_2, p_3 \in
\Zbb$.

Recall: $N(a, b)$ is the number of non-zero digits
in the base $b$ expansion of $a$.

\begin{theorem*}[Stewart]
  There is an effective absolute constant $C$ such
  that
  \[
    N(a, 2) + N(a, 3) \ge \frac{\log \log a}{\log
    \log \log a + C} - 1
  ,\]
  for $a \in \Zbb_{\ge 0}$.
\end{theorem*}

Digit expansion of $a$
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/56d63fb6ef574116.png}
\end{center}
$a = p2^{k_1} + e_1$.

We need $p^K e_1 < 2^{k_1}$ where $K = C \log
\log_2 a$ (this is an upper bound for the exponent
of $\max(p, q, 10)$ in the proposition).
Previously we have $p e_1 < 2^{k_1 (1 - \eps)}$.

Alternative to heights of minimal polynomials (is
better behaved under operations like addition):

\begin{fcdefn}[Mahler measure]
  \glsnoundefn{mahler}{Mahler measure}{Mahler
  measures}%
  \glssymboldefn{mahler}%
  Let $P \in \Cbb[X]$
  \begin{align*}
    P(X)
    &= a_d X^d + a_{d - 1} X^{d - 1} + \cdots
    + A_0 \\
    &= a_d(X - \alpha_1) \cdots (X - \alpha_d)
  \end{align*}
  Then we define
  \[
    M(P) = |a_d| \cdot \prod_{j = 1}^{d} \max(1,
    |\alpha_j|)
  .\]
\end{fcdefn}

We could define the height of an algebraic number
$\alpha$ as
\[
  H(\alpha) = \mahler(f_\alpha)^{\frac{1}{\deg
  f_\alpha}}
,\]
but instead we will define it in a different (but
equivalent) way.
