Symbol baro

The passage around this formula

Summed over an n -step trajectory, that gives $C_n = \sum_{i=1}^{n} c_i = n\,p_{\text{in}}(\tau+\delta) + p_{\text{in}}\,\sigma\,\frac{n(n-1)}{2} + n\,p_{\text{out}}\,\bar{o}$. The middle term is the one that matters. It grows as n(n-1)/2 — quadratically in step count — while the other two terms grow only linearly. For a short loop, the quadratic term is negligible next to the fixed tax and the output cost. For a long one, it dominates completely, and no amount of shrinking $\tau$ or $\bar{o}$ changes that; the loop’s own length has become the largest line item, all by itself.

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