Reference for notions of angular mean

Question

Intuitively I'd like to have a "well behaved" mean value M of a series of lines' angular coefficients $m_1$, $m_2$, .., $m_n$ such as for example

$$M=1/n (\sum_{k=1}^n m_k)$$

The problem is I'm not sure it is truly well behaved. Converting tangents to arctangents, I'd like to use a function:

$M: (-\pi/2,\pi/2)^n \to (-\pi/2,\pi/2)$ (limits not included) such that for all $x, y, z \in$ Domain:

1. $M(x, x)=x$ for every number of equal arguments. In particular $M(x)=x$
2. $M(x, y, z)=M(x,z,y)=M(y,x,z)=M(y,z,x)$, ..., etc for every permutation and number of arguments.
3. $ (x>y) \rightarrow (x>M(x,y)>y)$
4. $M$ isn't in general associative. That is $\lnot M(x,y,z)=M(x,M(y,z))$

Discounting the transformations cartesian cohordinates, radiants, polar, etc what problems might the previous formula for M have? Are there alternatives?
The notion of circular mean doesn't seem helpful as it includes the whole $(0,2\pi)$ and averages over $sin$/$cos$ values which doesn't seem equivalent to what I'm aiming for.
Any reference? Should I ask MO?

Thank you.

Answer 1

Given the slopes $m_k$ of lines passing through origin, not parallel to the $y$ axis, $$y = m_k x$$ there are two obvious definitions for their mean or average.

The first one is the mean slope as shown by OP, $$M = \frac{1}{N} \sum_{k=1}^{N} m_k \tag{1}\label{label1}$$ and the second one is the mean angle $\theta$ between each line and the $x$ axis that OP also mentioned (but did not show explicitly), $$M = \tan{\theta}, \quad \theta = \frac{1}{N} \sum_{k=1}^{N} \arctan(m_k) \tag{2}\label{label2}$$ i.e. $$M = \tan\left(\frac{1}{N}\sum_{k=1}^{N} \arctan(m_k)\right)$$ What is the difference between the two, then?

Consider a set of two lines with $m_1 = 0$ and $m_2 = 4$.
The first one yields $M = 2$, and the second one $M = 0.78$.
The angles of the two original lines are $0^\circ$ and about $76^\circ$. The angle of the first mean is about $63^\circ$ and the angle of the second mean is about $38^\circ$.

So, the first one yields the mean slope, whereas the second one the mean angle. Whichever is more proper for your use case, depends on what the lines and their slopes represent.

In simpler terms, if you consider the lines as the point set where they intersect $x = 1$, mean slope is equivalent to picking the point at $x = 1$ with the mean $y$ coordinate. For the mean angle, you pick points on say the positive $x$ half of the unit circle, and measure their angles $\varphi$ in radians, along the circle from the intersection with the $x$ axis, with the signed distances being positive on the positive $y$ side, negative on the other side.

Now, let's assume $f$ is a monotonically increasing continuous function over all reals, i.e. $f(x) \lt f(x + \delta)$ for $x \in \mathbb{R}$ and $0 \lt \delta \in \mathbb{R}$. Then, $f$ is invertible: $y = f(x) \iff f^{-1}(y) = x$. With this, we can define an $f$-mean for the lines passing through origin: $$M_f = f^{-1}\left(\frac{1}{N}\sum_{k=1}^{N} f(m_k)\right) \tag{3}\label{label3}$$ In the case of $f(x) = x$, this yields $\eqref{label1}$; in the case of $f(x) = \arctan(x)$, $f^{-1}(x) = \tan(x)$, this yields $\eqref{label1}$.

If we consider $\eqref{label1}$ to have a measurement line at $x = 1$, and $\eqref{label2}$ to have a measurement curve a circular arc (half-circle on the positive $x$ axis), then $f$ represents all other possible measurement curves.