Symbol C_t

The passage around this formula

…The crossover, where 1/($f_1$$\delta^2$) = $\ln$ S/2$\pi$ , sits at $\delta^\ast$ $\approx$ 0.188\,$\mathrm{rad}$ , about eleven degrees, for this hole and this choice of $f_1$ . A fourth, trivial floor closes the set of checks: for $C_t$ smaller than the input systems themselves, or for t before scrambling has had time to run at all, neither $\epsilon_L(t)$ nor $\operatorname{Var}$$\hat g$ meets any nontrivial bound — nothing has left yet, which is simply causality, not a result.

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