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I want to define a function $f$ to be "$n$-involutory" if it satisfies $$f(f(f\dots(x)))\dots=f^n(x)=x $$ where there does not exist $k$ such that $1 \leq k < n$ where $f^k(x) = x$. In general, is it acceptable to invent terms as long as I define them properly? (I'm still working on the definition for $n$-involutory, but I'm pretty sure this suffices for my question)

I've been looking around for a while around SE but I'm still relatively confused about it. Sorry if my question doesn't really belong on here :c

Arkyter
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  • Did you mean $1≤k<n$? In any case: as long as the term does not have a generally accepted definition already, it's perfectly reasonable to introduce a definition. Just make sure your definition is clear and complete. – lulu Dec 15 '21 at 17:23
  • fixed it!! thanks for pointing it out. also, what would you consider a clear and complete definition? I'm still in high school so are there specific things I should consider while writing these? – Arkyter Dec 15 '21 at 17:24
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    Worth remarking: generally speaking, an "involutory function" is one that satisifes $f\circ f(x)=x$. That would include the identity function while your notion of $2-$involutory would exclude the identity. That's ok, though I would point things like that out along with the definition. – lulu Dec 15 '21 at 17:26
  • alright! I saw the wikipedia article on an "involutory function" so I wanted to generalize it a little :) – Arkyter Dec 15 '21 at 17:26
  • I think what you wrote was clear (now that you fixed the typo). As I say, I'd point out that your notion doesn't exactly match words in standard use, as the standard terms tend not to be concerned with the minimal "involutory" order. – lulu Dec 15 '21 at 17:27
  • alright - thank you! – Arkyter Dec 15 '21 at 17:28
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    Minor technical note: as it stands, your definition is not clear when $n=1$ (as $1≤k<1$ isn't what you want). As the only $1-$involutory function is the identity, you might want to declare that one as a special case. – lulu Dec 15 '21 at 17:29
  • okie. stack exchange is telling me to avoid discussions in comments so I'll head off for now, but thank you so much for your notes! – Arkyter Dec 15 '21 at 17:30
  • Isn't this notion just the notion of finite order in groups? In this case, the group of bijections of a set. – lhf Dec 15 '21 at 17:31
  • huh? Can you explain? – Arkyter Dec 15 '21 at 19:48
  • See https://en.wikipedia.org/wiki/Order_(group_theory) – lhf Dec 16 '21 at 00:48

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