3

I've been trying to write a formal definition for a $k$-involutible function in that the function has to satisfy the following properties:

  1. $k$ is a positive integer.
  2. $f \in \mathbb{R}(x)$ (as in, $f$ is a rational function. I explicitly wrote out the implication in the first line of my definition because my target audience is unfamiliar with the notation $\mathbb{R}(x)$. Defining it just to write $\mathbb{R}(x)$ once seems a little excessive in my opinion.)
  3. $f$ satisfies $f^k(x) = x$, but does not satisfy $f(x) = x, f^2(x) = x, \dots, f^{k-1}(x) = x$. In this sense, a function can only satisfy this condition for one particular value of $k$. (I defined a "minimality condition" because I'll refer back to this multiple times)
  4. $f$ is not a piecewise function.
  5. Here's the... tricky part. I want $f$ to be defined over $\mathbb{R}$ except for when the denominator of $f$ is equal to $0$ (and thus making $f$ undefined). Does this equate to just saying "$f$ is defined over $\mathbb{R}$ except in places where it isn't" or is this condition simply not needed?

Below is my attempt at a proper and formal definition (I'm not too sure what people would call "formal" - I'm still a high school student, after all).

Attempt at Definition

We call a rational function $f = p/q$, where $p, q \in \mathbb{R}[x]$, $k$-involutible for $k \in \mathbb{Z}^+ \geq 2$ if and only if $f$ satisfies

$$ \underbrace{f(f(f( \dots f}_{k \text{ times}}( x ))) \dots ) = f^k(x) = x $$

and there does not exist $n \in \mathbb{Z}^+$ such that $1 \leq n < k$ yields $f^n(x) = x$ (referred to as the minimality condition). Moreover, as $f$ is a rational function, let

$$ S = \{x \mid q(x) = 0 \} $$

In order for a function to be $k$-involutible we mandate that $f$ is a non-piecewise function such that the domain of $f$ is defined over $\mathbb{R} \setminus S$. Moreover, we define a special case $k = 1$ where the only $1$-involutible that exists is the identity function $f(x) = x$.

Question

How can I improve this definition? Also, is it excessively convoluted / complicated? It reads fine to me but I'm not sure if I'm breaking some kind of mathematical convention / notation rules, or if it's not rigorous enough.

Also, is it acceptable to insert periods at the end of equations defined by $$ ... $$ (in the sense that the equation in display style would end in a period on the same line) where it would usually be grammatically required?

Arkyter
  • 85
  • sorry if this question seems trivial to people, or if it simply doesn't belong on SE. I'll remove it if needed! :) thanks in advance, though. – Arkyter Jan 18 '22 at 21:04
  • 2
    I suggest you motivate it: where and how do your functions appear? – FShrike Jan 18 '22 at 21:10
  • what do you mean by this? uhh.. I'm planning to describe how I can use a rotation matrix-esque method to generate $k$-involutible functions. It's an interesting method, I suppose :D – Arkyter Jan 18 '22 at 22:15
  • Well, a definition is considered good partly it if fits the context of its application. You can define whatever you want; in order for me to know that the name and conditions on it make sense, I should have an idea of context. I say “I” in a general sense; I am not and will pass on commenting on your definition’s... suitability. – FShrike Jan 18 '22 at 22:22
  • i'm doing a project on finding a function that generates $k$-involutibles. at the beginning, it's probably important to define what a $k$-involutible is, so i'm writing a formal definition for it given the things i wanted to define in there as outlined above. I'm really only looking if I'm breaking some sort of mathematical convention in this definition, but additional comments are cool! – Arkyter Jan 18 '22 at 22:41
  • You say $f(x)=x$ is the only 1-involutible function, but why wouldn't $g(x)=x(x-3)/(x-3)$ be another? Or is the point that for $k=1$ you make a special definition with the intent of excluding things like $g$? – Mark S. Jan 19 '22 at 13:20
  • I suppose you bring up a nice point - my intention was originally to look at how a $k$-involutible would look like / transform ($f^2(x), f^3(x), \dots$) when graphed on the cartesian plane, and I figured things like $g$ would basically be $f(x) = x$ excluding $x = 3$. Then again, you can always append things like $(x-3)/(x-3)$ to the end of fractions, right? It's cool, but not exactly what I'm trying to look for. Sorry. – Arkyter Jan 19 '22 at 16:01

0 Answers0