Talk:Wilson–Cowan model

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$S({\bar {N}})=\int _{\frac {\theta }{{\bar {N}}(t)}}^{\infty }C(w)dw$

Where does this integral come from? I cannot find it in the original paper and I doubt its correctness. Example: If the excitation threshold $\theta$ is twice the average excitation ${\bar {N}}$ , then this would integrate over all neurons that have $\omega \geq 2$ afferent synapses. However, in order for a neuron to belong to the excited subpopulation $S({\bar {N}})$ , it should have at least twice the average no. of afferent nerve fibers, not just two afferent synapses (or more). — Preceding unsigned comment added by LoSchizzatore (talk • contribs) 12:53, 19 July 2018 (UTC)[reply]