Has arctangential scale ever introduced in maths before?

Question

There are different scales used in mathematics - particularly when it comes to plot. Two major ones are linear and logarithmic. But I was wondering how can one grasp the whole set of numbers in one graph on finite amount of space. Then I came to arctangential scale.

So, how do I define it?

Let there be function $f(x)$ and function $h(x)$ which would be an arctangetial mapping from function $f(x)$.

Then: $$X=\arctan(x)\cdot \frac 2 \pi$$

$$h(x)=\arctan(f(x))\cdot \frac 2 \pi$$

Function $h(X)$ is defined on $[-1; 1]$ domain which would be equal to the domain $[-\infty;\infty]$ for function $f(x)$. Range $[-1; 1]$ of function $h(X)$ correspons to the range $[-\infty;\infty]$ for function $f(x)$.

Have I invented something new or has this been used in mathematics before?

Answer 1

Stereographic projection is a well-studied mapping of the real line onto a circle; specifically, the map $$x = \frac{1 - t^2}{1 + t^2}, \quad y = \frac{2t}{1 + t^2}$$ maps points $t \in (-\infty,\infty)$ onto the unit circle $x^2 + y^2 = 1$ except for missing the point $(-1,0)$. (We can extend the definition to the extended real line by mapping $\infty$ to $(-1,0)$.) Intuitively, it is the point of intersection of the unit circle with the line that passes through $(-1,0)$ with slope $t$.

The relevance to your question is that in polar coordinates, stereographic projection maps $t$ to $$r = 1, \quad \theta = 2 \arctan t,$$ and this map between $t \in (\infty,\infty)$ and $\theta \in (-\pi,\pi)$ is the same as yours (with some scaling).

Stereographic projection has some nice properties:

• It is birational. That is, $x$ and $y$ are rational functions of $t$, and the inverse mapping $t = y/(x+1)$ is also a rational function of $x$ and $y$. In particular, if $t$ is rational, so are $x$ and $y$ - this gives a way of generating all Pythagorean triples.
• It generalizes nicely to higher dimensions: we can map $\mathbb{R}^n$ to the $n$-sphere $S^n = \{\mathbf p \in \mathbb{R}^{n+1} : \|\mathbf p\| = 1\}$ by $$\mathbf t = (t_1,\ldots,t_n) \mapsto \frac{(1 - \|\mathbf t\|^2, 2t_1, \ldots, 2t_n)}{1 + \|\mathbf t\|^2}.$$ This mapping is conformal, i.e. it preserves angles.