Valeri's second point measured: the similarity function belongs to the PLACE it is read in, not to the method. False position's three-century stall is a similarity read at order one on an approach of order m: with sigma = |f(far)| the residual ratio is (e/E)^m, so raising it to the measured 1/m recovers the distance ratio exactly. Paired inside each cell, the correction is cheaper in 7 of 7 safeguards for S = 1/(1+z) - the one function that makes that place exact inverse interpolation - and in only 29 of 60 cells over all nine functions, i.e. a coin. It does NOT beat Chandrupatla (2884 against 2683 and 2423), it HARMS the slope reading (2806 -> 2854), and it does not cure false position: 127/154 and 10676 become 128/154 and 10043, where Illinois buys 153/154 and 7045 with one constant. Estimator bias measured and stated: a pure cube reads 2.830, not 3

Also fixes a real regression from cc2c499: the derived3_cap branch was inserted in front of the derived3_m note and swallowed its division, so derived3_m silently became derived3 (2806 against 2806) and a refutation recorded in the paper stopped being reproducible. The full pin caught it; a cheap pin now guards it too. And \label{sec:method} never existed while the table on page 7 referenced it.

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

article/sim_numerics.tex

@@ -213,6 +213,7 @@ the evaluations.
 \end{corollary}
 
 \section{The method}
+\label{sec:method}
 
 Corollaries~\ref{cor:chosen}--\ref{cor:derived} fix every choice in the method
 before any benchmark is consulted. We record them here as a prediction, and
@@ -563,6 +564,66 @@ The readings do not compose. Taken together, (2) and (3) give $2884$ and
 $2859$ --- worse than either alone. Two selections that are each an improvement
 interfere, which is a reason to measure composition rather than assume it.
 
+\subsection{The similarity function belongs to the place, not to the method}
+\label{sec:perplace}
+
+Everything above varies \emph{where} the similarity is read while keeping one
+function $S$ for the whole method. That is itself an assumption, and it is
+measurable. Proposition~\ref{prop:falsi} says the two-point normalisation is
+false position; what it does not say is \emph{why} false position stalls. Write
+$|f|\sim K e^m$ near the current point, $e=|x-r|$. Then the residual ratio
+against the far end needs no slope at all,
+\begin{equation}
+\label{eq:orderz}
+z \;=\; \frac{|f(x)|}{|f(x_{\mathrm{far}})|} \;=\; \Bigl(\frac{e}{E}\Bigr)^{m},
+\qquad\text{so}\qquad z^{1/m} \;=\; \frac{e}{E},
+\end{equation}
+and under $S=1/(1+z)$ the step fraction is $z/(1+z)$, which is exactly inverse
+linear interpolation. False position therefore reads a dimensionless ratio
+\emph{at order one} on an approach of order $m$: its three-century-old stall is
+a mismatch between the similarity function and the place it is read in. The
+exponent $1/m$ is not a constant --- $m=1/(1-ff''/f'^2)$ comes from three
+evaluations already paid for, and is $1$ whenever the power model does not
+apply.
+
+Measured over the $252$ cells of the grid, the correction behaves exactly as
+that reading predicts and not at all as a universal improvement. Paired inside
+each (function, safeguard) cell of the far reading it is cheaper in $29$ of
+$60$ cells --- indistinguishable from a coin. But in the one pairing where the
+far reading \emph{is} exact inverse interpolation, $S=1/(1+z)$, it is cheaper
+in all seven safeguards:
+
+\begin{table}[h]
+\centering
+\begin{tabular}{lrrrr}
+\toprule
+safeguard & APS $m{=}1$ & APS measured $m$ & CHA $m{=}1$ & CHA measured $m$\\
+\midrule
+both    & 4522 & \textbf{3675} & 3387 & \textbf{3216}\\
+bracket & 4606 & \textbf{3542} & 3571 & \textbf{3303}\\
+credit  & 5190 & \textbf{4686} & 3013 & \textbf{2884}\\
+project & 6932 & \textbf{6364} & 4794 & \textbf{4559}\\
+safe    & 5987 & \textbf{5229} & 4108 & \textbf{3923}\\
+step    & 6435 & \textbf{5551} & 4257 & \textbf{4022}\\
+none    & 10676 & \textbf{10043} & 8265 & \textbf{7794}\\
+\bottomrule
+\end{tabular}
+\caption{The measured order in the step place, on the far reading. Cheaper in
+$7$ of $7$ safeguards for the similarity that matches the place, and in $29$ of
+$60$ cells over all nine functions.}
+\end{table}
+
+Two limits must be stated with it. The estimator is \emph{biased low}: on a
+pure cube, with no noise at all, it returns $2.830$ rather than $3$, because a
+parabola through three points carries only the second difference. The bias is
+towards $m=1$, that is towards the uncorrected reading, which is the harmless
+direction, but it is a bias. And the correction does not rescue the bracket
+reading as a method: the best cell it reaches on the wide-bracket set is
+$2884$, against $2683$ for reading~(3) above and $2423$ for Chandrupatla. In
+the \emph{slope} reading the same exponent is harmful, $2806\to2854$, which is
+the same statement read from the other side --- a function that does not belong
+to a place does not become right by being applied there.
+
 \subsection{Six problems of the two sets are scored unsolved for every method}
 \label{sec:fun7}
 
@@ -693,6 +754,18 @@ $3288$ 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{Reading the similarity at the measured order does not cure false
+position.} Since false position's stall is a similarity read at order one
+(\S\ref{sec:perplace}), the obvious hope is that reading it at the measured
+order is the parameter-free version of the classical fix. It is not. Unguarded,
+the correction takes the bracket reading from $127/154$ and $10676$ to $128/154$
+and $10043$ --- one problem and six per cent --- where Illinois, at the price of
+one constant (halve the stale ordinate), reaches $153/154$ and $7045$. A sixth
+of the gap is not the gap. The correction is real and it is in the right place;
+it is simply much smaller than a constant that was chosen for exactly this
+failure. \emph{Our} $10676$ and the $10560$ of Table~\ref{tab:aps} are not
+the same algorithm --- our iterate starts at the midpoint of the bracket, plain
+false position at an end --- and the two must not be differenced.
 \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