Section 12 4 October 2026

Rotor wave and velocity-weighted error

Jeremy Rodgers · Independent Researcher
doi:10.5281/zenodo.23131081

Original PDF ↗Section Markdown Full Markdown All contents

Reading position 14 of 37

12 Rotor wave and velocity-weighted error

In one sector write $z=(X-o)/h-i/2-d_i$ , $\omega_i=\sqrt{g_i/\alpha}$ , and $b_i=\cos(\omega_i t)$ . Let $a_{\eta_t}$ be the unnormalized quintic taper: one on $|z|\le1/2-2\eta_t$ , zero on $|z|\ge1/2-\eta_t$ , and a $S_5$ transition between. The harmonic compact comparison carries amplitude $a_{\eta_t}(z/b_i)/\sqrt{2b_i}$ and phase $-\alpha K\omega_i\tan(\omega_it)z^2/2$ in each cell.

The physical initial wave is $1/\sqrt2$ , so its initial error is not zero. Exact integrals give

d_0^2=\frac{643\eta_t}{231},\quad A_1^2=\frac{20}{7\eta_t},\quad A_2^2=\frac{240}{7\eta_t^3},\quad A_3^2=\frac{1440}{\eta_t^5}.

The comparison lies in the quadratic potential core whenever it is used. Its residual norm is $A_2/(2\alpha Kb_i^2)$ .

Lemma 12.1 (Finite rotor comparison)

Through capture time $t_b=1.28$ , let $r_i$ be the exact rigid-sector wave. Then

\norm{r_i-\Phi_i}\le d_0+\epsilon_R=:d, \qquad \epsilon_R=\frac{2A_2}{.999K},

\frac{\norm{p_X(r_i-\Phi_i)}}{mh} \le \frac{A_1}{.999K}+.501d_0+\frac{4.02A_3}{K^2} +1.002\epsilon_R=:q.

For the selected row, $d<D=.007233$ and $q<Q=.003983$ .

Proof

The residual norm integral uses $\omega_it<1.3$ and $\tan1.3<4$ . The static nonnegative rotor Hamiltonian propagates its own square-root norm. Multiplying that estimate by $\sqrt{2/(\alpha K)}$ gives the initial derivative contribution $A_1/(\alpha K)+\omega_i d_0/2$ . Differentiating the residual introduces $A_3/(2\alpha^2K^2b_i^3)$ and its harmonic phase derivative; the potential part adds the same $\omega_i/2$ residual bound. Integrate with $\int_0^{1.3}\sec^3\theta\,d\theta<8$ , $\omega_i<1.002$ , and $4/(\alpha^2\omega_i)<4.02$ . Weak $H^3$ taper derivatives suffice; no fourth derivative of the sharp taper is invoked. The outward rational square-root calculation in Appendix A proves the final decimals.

□

At $t_a=1.22$ , each $\Phi_i$ is supported in its own outcome region even with $|d_i|\le.01$ . The wrong-side wave probability is at most $D^2$ . For the isolated spinor rotor, the guidance map commutes with $X\mapsto X+h$ . The initial preimage of the classifier is therefore periodic, including its circle circulation. Lemma 11.1 and the wrong-side estimate yield the actual isolated-label error

|P(\ell(X^{\rm iso}_{t_a})=1)-p|\le\frac1{2N}+D^2.