%! TEX root = DA.tex
% vim: tw=50
% 15/11/2024 12PM

\begin{proposition*}[1]
  Let $S = (T_0 + 1) T_1$ be non-negative
  integers. Let $w_1, \ldots, w_{T_1}$ and $\xi_1,
  \ldots, \xi_S$ be  two sets of distinct real
  numbers.

  Then
  \[
    \det[\xi_S^\tau \exp(w_t
    \xi_S)]_{\substack{\tau, t \\ s}}
    \neq 0
  ,\]
  with: $\tau = 0, \ldots, T_0$, $t = 1, \ldots,
  T_1$, $s = 0, \ldots, S$.
\end{proposition*}

alternant / interpolation determinant

\begin{proposition*}[2]
  Let $T \in \Zbb_{\ge 1}$, let $w_1, \ldots, w_T$
  be distinct real numbers. Let $P_1, \ldots, P_T
  \in \Rbb[X]$ be non-zero. Then the function
  \[
    F(x) = P_1(x) e^{w_1 x} + \cdots + P_T(x)
    e^{w_T x}
  \]
  has at most $\deg P_1 + \cdots + \deg P_T + T -
  1$ real zeroes counting multiplicities.
\end{proposition*}

\begin{proof}[Proposition (2) $\implies$
  Proposition (1)]
  Suppose to the contrary that $\det = 0$. Then
  there exists $a_{\tau, t} \in \Rbb$ not all $0$
  such that
  \[
    \sum a_{\tau, t} x^\tau \exp(w_t x)
  \]
  vanishes for all $x = \xi_1, \ldots, \xi_S$.
  This is a function of the type in Proposition
  (2). Each polynomil is of degree $\le T_0$, and
  there are $T_1$ many of them, so there can be no
  more than $T_0 \cdot T_1 + T_1 - 1 < S$ zeroes.
\end{proof}

\begin{lemma}
  Let $f$ be a $C^\infty$ function on $\Rbb$ with
  $N$ real zeroes. Then $f'$ has at least $N - 1$
  zeroes.
\end{lemma}

Corollary of Rolle's Theorem.

\begin{proof}[Proof of Proposition (2)]
  By induction on $N \defeq \deg P_1 + \cdots +
  \deg P_T + T - 1$. If $N = 0$, then $T = 1$ and
  $\deg P_1 = 0$. So $F(x) = a \cdot \exp(w_1 x)$
  for some $a \neq 0$. This indeed has no zeroes.

  Suppose $N > 0$ and the claim holds for $N - 1$.

  We assume as we may that $w_1 = 0$ (if not, then
  replace $w_j$ by $w_j - w_1$, which has the
  effect of replacing $F$ by $F \cdot e^{-w_1
  \cdot x}$).

  Then by the lemma, $F$ has at most one more zero
  than
  \[
    F' = \ub{P_1(x)'}_{\deg P_1 - 1}
    + \ub{(P_2'(x) + P_2(x) w_w) e^{w_2 x}}_{\deg
    P_2} + \cdots
  \]
  By the induction hypothesis, $F'$ has at most $N
  - 1$ zeroes, so $F$ has at most $N$ zeroes.
\end{proof}

Now we return to proving \nameref{gs}.

Let $z_1, z_2 \in \Rbb_{\neq = 0}$ such that
$\alpha_j = e^{\lambda_j} \in \ol{\Qbb}$ for $j =
1, 2$.

We aim for a contradiction. We have integers $L,
T_0, T_1, S$ such that
\[
  L = (T_0 + 1)(2T_1 + 1) = (2S + 1)^2
.\]
Let
\[
  \Delta = \det[(s_1 + \beta s_2)^\tau
  \exp(\lambda_1 t(s_1 + \beta s_2))]_{\substack{\tau, t \\ s_1, s_2}}
.\]
Last time:
\[
  \log |\Delta| \le -cL^2 \log E + CLT_0 \log ES +
  CLT_1 ES
\]
where $E \in \Rbb_{> 1}$ arbitrary.

Apply Proposition (1) with $\xi_S = (s_1 + \beta
s_2)$ with some enumeration of $s_1, s_2$ and $w_t
= \lambda_1 t$. Then $\Delta \neq 0$.

Recall:
\[
  \Delta = \det[(s_1 + \beta s_2)^\tau \alpha_1^{t
  s_1} \alpha_2^{t s_2}]_{\substack{\tau, t
  \\ s_1, s_2}}
.\]
Then
\[
  \Delta = P(\beta, \alpha_1, \alpha_2)
\]
for some $P \in \Zbb[X, Y, Z]$. So:
\[
  H(\Delta) \le \mathcal{L}(P) \cdot H(\beta)^{T_0
  \cdot L} \cdot H(\alpha_1)^{T_1 S \cdot L}
  H(\alpha_2)^{T_1 SL}
\]
using
\[
  \mathcal{L}(P_1, P_2) \le \mathcal{L}(P_1)
  \mathcal{L}(P_2)
\]
and
\[
  \mathcal{L} \left( \sum P_j \right) \le \sum
  \mathcal{L}(P_j)
\]
we get
\[
  \mathcal{L}(P) \le L! \cdot (2S)^{T_0 L}
.\]
Liouville bound:
\[
  \log |\Delta|
  > -C(\log L! + T_0 L \log S)
.\]
Take: $E = 10$.

Then we have a contradiction if
\[
  -cL^2 + CL T_0 \log S + CL T_1 S
  < -C(L \cdot \log L + T_0 L \log S + T_1 LS)
.\]
I want:
\[
  L^2 > C(T_0 L \log S + LT_1 S)
.\]
Take: $S \approx L^{\half}$, $T_0 \approx L^{1 -
\eps}$, $T_1 \approx L^\eps$.