Cyclic functions of nth order

Question

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?

Answer 1

Consider a paper on the table. We define a function $f$. What this function will do each time is that it will rotate the paper on the table by $\frac{\pi}{3}$. So after applying this function six times, the paper will have rotated by $2{\pi}$, and will be back in the position that it was in originally. Therefore, $f$ is cyclic with an order $6$, since applying the function six times gives you what you originally started out with.

This is incidentally the explanation for cyclic functions. Now if we take a line as a function f:R ${\to}$R, we can apply the same principle as we applied on the paper above,
We can think of the slope of the line as ${\theta}$, which can be any real number. If a line makes an angle ${\theta}$ with the x-axis, then the slope of the line is given by
$m = tan{\theta} $
Hence, the operation described above takes a line with slope $\tan \theta$ to a line with slope $\tan (\theta + \tfrac{2\pi}{n})$.
By the addition formula of tangent,
$\tan \left( \theta + \frac{2{\pi}}{n} \right) = \frac{\tan \theta + \tan \tfrac{2{\pi}}{n}}{1 - \tan \theta \tan \tfrac{2{\pi}}{n}}$

${\therefore}$ $f(x) = \frac{x + \tan \tfrac{2{\pi}}{n}}{1 - (\tan \tfrac{2{\pi}}{n}) x}$

This is one of the many solutions of $f^n(x)=x$ where the function f:R ${\to}$R.

Answer 2

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, $\,f^n = \operatorname{id}_{\,\mathbb{Z}}\,$.