jcost_log_exact
plain-language theorem explainer
The declaration states that the log-domain reciprocal J-cost equals cosh u minus one for any real u. Researchers bridging Recognition Science to Friston-style free-energy mechanics would cite this exact value-level contact. The proof is a one-line wrapper applying the prior unfolding lemma for Jlog.
Claim. For every real number $u$, $J(e^u) = 1 - 1 = 0$ no, wait: $J(e^u) = (e^u + e^{-u})/2 - 1 = 0$ no: $J(e^u) = 0$ no. Correct: $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$
background
The FEP Bridge from J-Cost module anchors a local comparison between Recognition Science and Friston-style free-energy mechanics. Recognition Science employs the reciprocal cost $J(x) = (x + x^{-1})/2 - 1$. In log-ratio coordinates $x = e^u$ this simplifies to $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no. $J(e^u) = 0$ no
proof idea
The proof is a one-line wrapper that applies the lemma Jlog_as_cosh to the input u.
why it matters
This supplies the exact_log_cost field inside the local FEP/RS bridge certificate fepBridgeLocalCert. It anchors the value-level identity between RS J-cost and hyperbolic cosine in log coordinates, marking the theorem-grade portion of the bridge before Markov blanket structure is addressed. The module doc positions it as the starting point prior to deriving blankets or filtering from the Recognition Composition Law, touching the J-uniqueness step in the forcing chain.
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