\section{Choosing the endpoints}\label{sec:endpoints}

\subsection{A choice valid for every starting index and count}

To balance the $m$th power in the numerator, we also make the
pole exponent $S$ proportional to $m$. Take
\[
 r=7,\qquad L=18,\qquad u=\frac1{50},\qquad
 S_0=s_B+1,\qquad S=mS_0.
\]
The integer $S$ is odd, and the resulting endpoint is
$S-1=m(s_B+1)-1$. Let $f_0,U_0$ denote the expressions
\eqref{eq:f}--\eqref{eq:U} with multiplicity $1$, derivative
order $0$ and denominator exponent $S_0$. Then
\begin{equation}\label{eq:scaled-criterion}
 f=f_0^m,\qquad U=U_0^m e_0^{4(m-1)}.
\end{equation}
Thus $f_0<1$ will give the first condition for every $m$.
For the second, we need enough decay in $U_0$ to pay the extra
derivative cost $e_0^{4(m-1)}$.

To see how $S_0$ is chosen, put
\[
 A=\frac{11\log11-7\log7}{4}-\log4-\log e_0.
\]
From \eqref{eq:K}, the inequality
$\lceil18/(p-1)\rceil\le18$ gives
\[
 A_d\le d^{36},\qquad K\le B^{36D}.
\]
Consequently
\begin{align*}
 \frac14\log U_0
 &=\log K+18D\log(18+u)-AS_0\\
 &\le36D\log B+18D\log(18+u)-AS_0.
\end{align*}
The first two terms account for the denominator normalization and
the numerator growth. The negative term is the decay gained from
the pole exponent. We will verify the elementary bounds
\begin{equation}\label{eq:constants}
 \log(18+u)<\frac{29}{10},\qquad A>\frac45
\end{equation}
below. They give
\[
 \frac14\log U_0
 <36D\log B+\frac{261}{5}D-\frac45S_0.
\]
The coefficient $45$ in $S_0$ cancels the logarithmic cost because
$(4/5)45=36$. The coefficient $66$ leaves a further margin because
$(4/5)66=264/5>261/5$. These choices account for $33$ and $45/2$
in \eqref{eq:closed-endpoint}. Indeed, that definition gives
\[
 S_0\ge66D+45D\log B+1>18D,
 \qquad
 \frac14\log U_0<-\frac35D-\frac45.
\]
This margin will absorb the cost of differentiation.

To verify the bounds just used and the first condition $f_0<1$,
we use three rational inequalities:
\begin{align}
 901^4\left(\frac{351}{551}\right)^{66}&<1,
                                                    \label{eq:f66}\\
 (901/50)^{10}&<(163/60)^{29},\label{eq:log-bound}\\
 \left((4e_0)^4\frac{7^7}{11^{11}}\right)^5e_0^{16}&<1.
                                                    \label{eq:slope}
\end{align}
Here $f_0=901^{4D}(351/551)^{S_0}$, so $S_0\ge66D$ and
\eqref{eq:f66} give $f_0<1$. Since $18+u=901/50$ and
$163/60<\ep$, \eqref{eq:log-bound} proves the first bound in
\eqref{eq:constants}. The logarithm of the left side of
\eqref{eq:slope} is $-20A+16\log e_0$. Since $\log e_0>1$,
it gives $A>4/5$, the second bound. Finally,
\begin{equation}\label{eq:log-e0}
 e_0<\sum_{j=0}^5\frac{(11/10)^j}{j!}<\ep^{11/10},
 \qquad\hbox{so }\log e_0<\frac{11}{10}.
\end{equation}
All the rational comparisons are checked in the accompanying code.

Using \eqref{eq:scaled-criterion}, \eqref{eq:log-e0} and $D\ge1$,
we obtain
\[
 \frac14\log U
 <-m\left(\frac35D+\frac45\right)+(m-1)\frac{11}{10}
 \le-\frac3{10}m-\frac{11}{10}<0.
\]
Thus $f<1$ and $U<1$ for every positive odd $m$ and every $B\ge1$.
Proposition~\ref{prop:criterion} proves Theorem~\ref{thm:intervals}.
At $B=1$ we have $D_1=1$ and $s_1=66$, giving $67m-1$.

\subsection{Sharper individual intervals}

The criterion can be checked directly without the upper bound
$K\le B^{36D}$. The following rational choices satisfy
$S>mDL$, $f(u)<1$ and $U<1$:
\begin{center}
\begin{tabular}{@{}rrrrrr@{}}
\toprule
$m$&$B$&$s=S-1$&$r$&$u$&$U$(approx)\\
\midrule
3&1&198&7&$21/2000$&0.00275\\
5&1&330&7&$103/10\,000$&0.00659\\
7&1&462&7&$13/1250$&0.01585\\
19&2&3080&7&$19/10\,000$&0.00183\\
19&3&6950&7&$59/100\,000$&0.02403\\
\bottomrule
\end{tabular}
\end{center}
The displayed decimals illustrate the margin; the inequalities
are verified exactly over $\Q$. In the row with $m=19$ and $B=2$,
the forms are supported on $\beta(20),\beta(22),\ldots,\beta(3080)$,
and the criterion guarantees at least two irrational members.
These choices prove the intervals listed in the introduction.