TITLE: Does the logistic function have a relation with $\arctan(x)$?
QUESTION [6 upvotes]: The logistic function is: $$f(x)=\frac{L}{1+e^{-k(x-x_0)}}+B.$$
It's plot looks similar to the plot of $\arctan(x)$. Therefore, I was wondering whether there is a relationship between these two functions.
Can one transform the logistic function in such a way that it equals $\arctan(x)$? For example by giving the constants certain values?

REPLY [9 votes]: One of the differences between a logistic function and the arctan is that the logistic function approaches its asymptotes exponentially, i.e. (if $k > 0$)
$$\eqalign{f(x) \sim L + B - L \exp(k x_0) \exp(-k x) & \ \text{as $x \to +\infty$}\cr
f(x) \sim B + L \exp(-k x_0) \exp(k x) & \ \text{as $x \to -\infty$}}$$
while the arctan approaches its asymptotes much more slowly, like $x^{-1}$:
$$ \eqalign{\arctan(x) \sim \frac{\pi}{2} - \frac{1}{x} & \ \text{as $x \to +\infty$}\cr
    \arctan(x) \sim -\frac{\pi}{2} - \frac{1}{x} & \ \text{as $x \to -\infty$}\cr}$$