Top step of the flip-flop #
For k = t ε_{m+1}^{β-γ}, t ≥ 1/2: log (KBar(k)/(√(9/80) ε_m^{β+γ})) equals
log (t √(80/9)) up to O(ε_m^δ).
For k = t ε_{m+1}^{β-γ}, t ≥ 1/2: log (KBar(k)/(√(9/80) ε_m^{β+γ})) equals
log (t √(80/9)) up to O(ε_m^δ).