Instantaneous zero-derivative condition for C

Read the equation

Setting the C derivative to zero produces this relation between E, C, L*, and time. If those starred values are constants, the oscillatory term generally still changes with time, so a single constant balance cannot hold for all times unless its amplitude vanishes or another term compensates. It should not be treated as a constant equilibrium without those conditions.

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Original context

The Fibonacci ratio $\fib{n}/\fib{n-1}$ converges to $\golden$ as $n \to \infty$. The modulation term $\sin(\golden t) \cdot \fib{n}/\fib{n-1}$ therefore oscillates with amplitude approaching $\golden$.

At equilibrium, $dC/dt = 0$, giving:
\begin{equation}
\alpha(E^* - C^*) + \beta \cdot L^* \cdot \sin(\golden t) \cdot \golden = 0
\end{equation}

Time-averaging eliminates the oscillatory term (since $\langle \sin(\golden t) \rangle = 0$), yielding $E^* = C^*$.

Similarly, $dE/dt = 0$ at equilibrium gives:
\begin{equation}
-\gamma C^* + I^* \cdot (1 + \cos(\golden t \cdot \fib{n})) = 0

\alpha(E^* - C^*) + \beta \cdot L^* \cdot \sin(\golden t) \cdot \golden = 0

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