Does the logistic function have a relation with $\arctan(x)$?

Question

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?

Answer 1

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

Answer 2

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.

Answer 3

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 $x^{-1}$: $$ \eqalign{\arctan(x) \sim \frac{\pi}{2} - \frac{1}{x} & \ \text{as $x \to +\infty$}\cr \arctan(x) \sim -\frac{\pi}{2} - \frac{1}{x} & \ \text{as $x \to -\infty$}\cr}$$