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I was facing difficulty in calculating an nth order cyclic function for $f:N\rightarrow N$ I tried and found out some of them which are $\frac{1}{x}, x, n-x.$ But can't find anymore. Is there any general form for calculating an nth order cyclic function?

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  • $\begingroup$ Could you clarify on what you mean by cyclic of order $n$? $\endgroup$
    – qualcuno
    May 25, 2017 at 11:50
  • $\begingroup$ Yes, by cyclic of order n, I mean f∘^n (x)=x ie. f(f(f...n times f(x)))).. For eg cyclic function of two order would be f(f(x))=x. You can also see this [link] (math.stackexchange.com/questions/1879137/…) by @Anonymous. $\endgroup$
    – Bingming
    May 26, 2017 at 0:58
  • $\begingroup$ $f(x) = \frac{x + \tan \tfrac{2{\pi}}{n}}{1 - (\tan \tfrac{2{\pi}}{n}) x}$ This is one of the many solutions of $f^n(x)=x$ where the function f:R ${\to}$R which comes when we take f as a line $\endgroup$
    – Bingming
    Aug 19, 2017 at 4:57
  • $\begingroup$ Sorry, I can't post answer. Answer ban forbids me to help anyone. $\endgroup$
    – Bingming
    Aug 19, 2017 at 4:58

1 Answer 1

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Hint:  let $\,A=\{0,1,\cdots,n-1\}\,$ and note that $\,g : A \to A\,$ defined by $\,g(m) = (m+1) \bmod n\,$ is "cyclic" in the sense that $\,g^n = \operatorname{id}_A \,$ (where $\,g^n\,$ denotes composition $\,n\,$ times) because $\,n\,$ repeated iterations complete the cycle $\,m\to m+1 \to \cdots \to n-2 \to n-1 \to 0 \to 1 \cdots \to m-1 \to m\,$.

Now consider $\,\displaystyle\mathbb{Z} = \bigcup_{k \in \mathbb{Z}} \{kn, kn+1,\cdots,kn+n-1\}\,$ and define $\,f:\mathbb{Z}\to\mathbb{Z}\,$ as a $\,g$-like function on each of the subsets of $n$ consecutive numbers:

$$ f(m) \;=\; \left\lfloor \frac{m}{n} \right\rfloor \cdot n \;+\; (m+1) \bmod n $$

Then, by the same argument as before, $\,f^n = \operatorname{id}_{\,\mathbb{Z}}\,$.

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