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This book chapter defines $CV$ as follows:

$$ CV^2 = 2\pi\nu^2\int^\frac{V_{th}-V_{ss}}{\sigma_V}_\frac{V_{r}-V_{ss}}{\sigma_V}dxe^{x^2}\int^x_{-\inf}dye^{y^2}(1 + erf(y)) $$

However, Figure 15.2 in this chapter shows that when $\nu$ (the firing rate) is 0, $CV$ is 1. How is this possible? Doesn't $\nu=0$ imply $CV=0$? It seems to me that if $\nu$ were 0, everything else on the rhs above would be multiplied by 0.

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  • $\begingroup$ A quick look doesn't confirm that $\nu$ is the same thing as the firing rate which is plotted in 15.2 A or C. Maybe it is, but Figure C is NOT a plot of the equation you have; it's a parametric plot of A and B. $\endgroup$
    – Bill N
    Aug 31 '15 at 1:52
  • $\begingroup$ @BillN hmm. $\nu$ is defined as "the steady state firing rate" and elsewhere in the text seems to be referred to with the shorthand "the firing rate." indeed, C is a parametric plot of A and B. but B is a plot of CV, as defined in my question, across different values of $\mu_c$, which determines $V_{ss}$. you might have a point that i'm missing, but i'm not sure what it is. $\endgroup$
    – dbliss
    Aug 31 '15 at 1:58
  • $\begingroup$ @BillN i think the issue is that $\nu$ is never exactly 0 -- it tends toward 0 at the limit to infinity, but is always non-zero. $\endgroup$
    – dbliss
    Aug 31 '15 at 2:01
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The premise of the question is false. $\nu$ is never exactly 0. It tends toward 0 as $\mu_c$ decreases. However, it is always non-zero and positive. Hence, the value of $CV$ depends on the other variables in its definition.

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