Read the equation
The source averages the cosine in the energy equation. With a zero cosine mean and nonzero γ, the averaged balance gives C=I/γ. This conclusion belongs to the averaged model under the stated forcing assumptions; it does not by itself solve the oscillating original equation.
Terms & definitions
| Term | Meaning in this source |
|---|---|
| C_star | Candidate averaged C |
| I_star | Constant input |
| γ | Consumption rate |
How it works
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Original context
Read the surrounding source passage
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)
\end{equation}
For large $n$, the cosine term averages to zero (since $\golden \cdot \fib{n}$ is irrational for all $n$), giving:
\begin{equation}
C^* = \frac{I^*}{\gamma}
\end{equation}Exact original notation
0 = -\gamma C^* + I^* \cdot (1 + \langle \cos(\golden t \cdot \fib{n}) \rangle)Related builds
Sources & provenance
0 = -\gamma C^* + I^* \cdot (1 + \langle \cos(\golden t \cdot \fib{n}) \rangle)
0 = -\gamma C^* + I^* \cdot (1 + \langle \cos(\golden t \cdot \fib{n}) \rangle)
0 = -\gamma C^* + I^* \cdot (1 + \langle \cos(\golden t \cdot \fib{n}) \rangle)
0 = -\gamma C^* + I^* \cdot (1 + \langle \cos(\golden t \cdot \fib{n}) \rangle)