Read the equation
The first difference shows whether an update raises or lowers the current state; the second shows whether it lies above or below one. For 1/5≤z<1, both expressions imply z<F(z)≤1. These sign relations support a monotone convergence argument within that invariant positive interval.
Terms & definitions
| Term | Meaning in this source |
|---|---|
| F | Average recurrence map |
| z | Current positive state |
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
6.3 Proposition 3: every positive averaged trajectory converges to one Two factorizations establish the result: Fz−z=1−z2z+17z, Fz−1=5z−1z−17z. For z>1, these show 1<Fz<z. The sequence decreases while bounded below by one. For 1/5≤z<1, they show z<Fz≤1, so the sequence increases while bounded above by one. For 0<z<1/5, the next value is greater than one and thereafter belongs to the first case. A bounded monotone sequence has a limit; continuity and the positive fixed-point equation force that limit to be one. For z0=2, local computation gives: Iteration
Exact original notation
F\left(z\right)-z=\frac{\left(1-z\right)\left(2z+1\right)}{7z}, F\left(z\right)-1=\frac{\left(5z-1\right)\left(z-1\right)}{7z}.Sources & provenance
F\left(z\right)-z=\frac{\left(1-z\right)\left(2z+1\right)}{7z}, F\left(z\right)-1=\frac{\left(5z-1\right)\left(z-1\right)}{7z}.
F\left(z\right)-z=\frac{\left(1-z\right)\left(2z+1\right)}{7z}, F\left(z\right)-1=\frac{\left(5z-1\right)\left(z-1\right)}{7z}.