Does the logistic function have a relation with $\arctan(x)$? 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?
 A: 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 $x^{-1}$:
$$ \eqalign{\arctan(x) \sim \frac{\pi}{2} - \frac{1}{x} & \ \text{as $x \to +\infty$}\cr
    \arctan(x) \sim -\frac{\pi}{2} - \frac{1}{x} & \ \text{as $x \to -\infty$}\cr}$$
A: Lets have a look at the inverse functions of the logistic function and arctan
Consider the sigmoid
$$sigmoid(x)=\frac{1}{1+e^x}$$
It's inverse function is
$$logit(x)=\log(\frac{x}{1-x})$$
Now the comparison becomes much clearer.
$tan$ is periodic ($tan(x)=tan(x+2k\pi)$), while $logit$ is not.
The $logit$ is a bijection, while $tan$ is not
A: There's a nice comparison of sigmoid functions that you can find here: Does the logistic function have a relation with arctan(x)
specifically this:
https://en.wikipedia.org/wiki/Sigmoid_function#/media/File:Gjl-t(x).svg
You probably want to compare your sigmoid function on equal footing to arctangent, in both domain and range so you'd want something like $1+\frac{2}{\pi}\arctan(kx)$
They do not share any real relationship other than both being "sigmoidal functions", and become more similar as $k$ goes to 0 or infinity.
