Why $\cot^{-1}x$ is an odd function in Mathematica

Question

The function $f(x)=\cot^{-1} x$ is well known to be neither even nor odd because $\cot^{-1}(-x)=\pi-\cot^{-1} x$. it's domain is $(-\infty, \infty)$ and range is $(0, \pi)$. Today, I was surprised to notice that Mathematica treats it as an odd function, and yields its plot as given below:

How to reconcile this ? I welcome your comments.

Edit: I used: Plot[ArcCot[x], {x, -3, 3}] there to plot

Answer 1

We have that $\cot^{-1}(-x)$ is invertible only on suitable restrictions, in this case it seems Mathematica is considering the following definition

$$f(x)=\cot^{-1}(x): \mathbb R \to \left(-\frac \pi 2, \frac \pi 2\right)$$

that is also the definition used by Wolfram.

Answer 2

From Inverse Cotangent on Wolfram MathWorld:

There are at least two possible conventions for defining the inverse cotangent. This work follows the convention of Abramowitz and Stegun (1972, p. 79) and the Wolfram Language, taking $\cot^{-1}x$ to have range $(-\pi/2,\pi/2]$, a discontinuity at $x=0$, and the branch cut placed along the line segment $(-i,i)$.

This definition is also consistent, as it must be, with the Wolfram Language's definition of ArcTan, so ArcCot[z] is equal to ArcTan[1/z].

A different but common convention (e.g., Zwillinger 1995, p. 466; Bronshtein and Semendyayev, 1997, p. 70; Jeffrey 2000, p. 125) defines the range of $\cot^{-1}x$ as $(0,\pi)$, thus giving a function that is continuous on the real line $\Bbb R$.

The former definition is what Mathematica uses. Note that with that definition, $\cot^{-1}(0) = \pi/2$, so it is an odd function only if you exclude $x=0$ from the domain.

The latter definition satisfies $\cot^{-1}(-x)=\pi-\cot^{-1} x$ and is not an odd function.

Answer 3

I think the Range of $f(x)=\cot^{-1} x$ as $(0,\pi)$ and hence the mixed parity of this function gets preference as then $f(x)$ is continuous in the its domain $(-\infty, \infty)$, specially at $x=0$. Then $\cot^{-1}(-x)=\pi-\cot (x)$.

In problem solving, for students, teachers and examiners this convention is most welcome for the sake of consistency. According to this $\cot^{-1} x$ function should look as below: