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The source replaces the time-dependent oscillation with its mean and treats L as constant in that step. If this averaging is justified and the sine mean is zero, the relation reduces to E=C* when α≠0. Correlation between a changing L(t) and the sinusoid must be addressed before extending that conclusion to the original trajectory.
Terms & definitions
| Term | Meaning in this source |
|---|---|
| C_star | Candidate mean C |
| E_star | Candidate mean E |
| L_star | Assumed constant or averaged L |
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Original context
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The error is $O(\golden^{-2n})$, converging exponentially.
\textbf{Equilibrium Analysis.} At equilibrium, all time derivatives vanish. From Equation~\ref{eq:dC}:
\begin{equation}
0 = \alpha(E^* - C^*) + \beta L^* \langle \sin(\golden t) \rangle \cdot \golden
\end{equation}
The time-average $\langle \sin(\golden t) \rangle = 0$ over complete periods, yielding $E^* = C^*$.
From Equation~\ref{eq:dE}:
\begin{equation}
0 = -\gamma C^* + I^* \cdot (1 + \langle \cos(\golden t \cdot \fib{n}) \rangle)Exact original notation
0 = \alpha(E^* - C^*) + \beta L^* \langle \sin(\golden t) \rangle \cdot \golden
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Sources & provenance
0 = \alpha(E^* - C^*) + \beta L^* \langle \sin(\golden t) \rangle \cdot \golden
0 = \alpha(E^* - C^*) + \beta L^* \langle \sin(\golden t) \rangle \cdot \golden
0 = \alpha(E^* - C^*) + \beta L^* \langle \sin(\golden t) \rangle \cdot \golden
0 = \alpha(E^* - C^*) + \beta L^* \langle \sin(\golden t) \rangle \cdot \golden