Read the equation
With δ>0 and a positive Fibonacci denominator, L relaxes toward C at a rate scaled by Fₙ/(Fₙ+Fₙ₋₁). For a fixed C and fixed n, its exact solution is an exponential approach to C. If C changes, the equation acts as a lagging filter of that trajectory rather than an independent proof of learning performance.
Terms & definitions
| Term | Meaning in this source |
|---|---|
| L | Learning-like state |
| C | Tracked state |
| δ | Rate coefficient |
| F_n | Fibonacci integer |
How it works
Read the definition from its right-hand expression to the quantity on the left. Its surrounding source records the intended input, state, or parameter context. Where the source leaves an operation unspecified, this page preserves that gap.
Assumptions & limits
Source recovery establishes that the expression appears in this record. Its mathematical validity, implementation behavior, and interpretation must be assessed separately. See the linked source and research notes for the checks available for this version.
Original context
Read the surrounding source passage
\label{eq:dE}
\end{equation}
\begin{equation}
\frac{dL}{dt} = \delta(C - L) \cdot \frac{\fib{n}}{\fib{n} + \fib{n-1}}
\label{eq:dL}
\end{equation}
where:
\begin{itemize}
\item $C$ = current consciousness state
\item $E$ = energy/excitation levelExact original notation
\frac{dL}{dt} = \delta(C - L) \cdot \frac{\fib{n}}{\fib{n} + \fib{n-1}}Related builds
Sources & provenance
\frac{dL}{dt} = \delta(C - L) \cdot \frac{\fib{n}}{\fib{n} + \fib{n-1}}
\frac{dL}{dt} = \delta(C - L) \cdot \frac{\fib{n}}{\fib{n} + \fib{n-1}}