Valeri's objection measured: the principle is right (a similarity step should be large far away and small near), and our sigma = |f'|D defeats it — at x=1e4 on a 2e4 bracket the normalised residual is 0.167, so S=0.833 'similar' and the step is small five thousand step-lengths from the root. Selecting sigma = min(|f'|D, max|f| at the ends) — no new constant — improves 403 to 390 on eight hand-picked wide problems but is WORSE over the grid: APS 3288 vs 2726, CHA 3111 vs 2683, so the mode is rejected by its pre-registered rule. Its extra solves are two fun7 problems landing inside the flat zero region, i.e. the scoring artefact and not a capability. Written into Negative results as the answer to the natural objection

Co-Authored-By: Claude Opus 5 <noreply@anthropic.com>

article/sim_numerics.tex

@@ -677,6 +677,22 @@ canonical set against $2527$ --- a difference of $1922$ evaluations for one
 choice of criterion. A secant step routinely collapses the error while barely
 moving the proven bracket, so bracket progress rejects steps that were
 excellent; this is the same mechanism that makes the bracket guard cost $7260$.
+\item \textbf{Making the scale independent of the bracket does not help either.}
+The natural objection to the derived scale is that it defeats the very principle
+of a similarity step: with $\sigma=|f'|D$ the normalised residual is
+$|f|/(|f'|D)$, which on the $2\times10^4$-wide bracket of $x^3-2x-5$ at
+$x=10^4$ equals $0.167$, so $S=0.833$ --- ``similar'' --- and the step is small
+although the root is some five thousand step-lengths away. The similarity is
+reading what fraction of the \emph{bracket} the Newton step is, not how far the
+iterate is from the root. Selecting the scale between the two quantities already
+computed, $\sigma=\min(|f'|D,\max(|f(A)|,|f(B)|))$ --- no new constant, a
+\texttt{min} over what is in hand --- does give a full-bracket step far away and
+Newton's step near the root, and on eight hand-picked wide problems it improves
+$403$ evaluations to $390$. Over the whole grid it is worse on both sets:
+$3288$ on the canonical set against $2726$, and $3111$ on the wide-bracket set
+against $2683$. Its apparent extra solves ($194$ instead of $193$) are two
+\texttt{fun7} problems where it lands inside the flat zero region close enough to
+satisfy the harness --- the artefact of \S\ref{sec:fun7}, not a capability.
 \item \textbf{A logarithmic change of variable does not buy the wide brackets.}
 Working in $u=\operatorname{sign}(x-c)\log(1+|x-c|/\varepsilon_x)$ makes the
 step multiplicative, which is the right diagnosis of the flat families --- CHA