# Does the following function on $\mathbb{R}^{4}$ iterate to $(0,0,0,0)$ after infinitely many steps?

Define $f: \mathbb{R}^{4} \rightarrow \mathbb{R}^{4}$ by $$\forall (a,b,c,d) \in \mathbb{R}^{4}: \quad f(a,b,c,d) \stackrel{\text{def}}{=} (|a - b|,|b - c|,|c - d|,|d - a|).$$

For many of the 'obvious' $(a,b,c,d) \in \mathbb{R}^{4}$ that you start with, you will obtain ${f^{n}}(a,b,c,d) = (0,0,0,0)$ for all $n \in \mathbb{N}$ large enough.

There is also an example where ${f^{n}}(a,b,c,d) \neq (0,0,0,0)$ for all $n \in \mathbb{N}$ but still $\displaystyle \lim_{n \rightarrow \infty} {f^{n}}(a,b,c,d) = (0,0,0,0)$. The example is constructed as follows. Let $\alpha$ be the real root of the quartic polynomial $x^{4} - 2 x^{3} + 1 = 0$ that lies in the interval $(1,2)$. Then \begin{align} f(1,\alpha,\alpha^{2},\alpha^{3}) &= (|1 - \alpha|,|\alpha - \alpha^{2}|,|\alpha^{2} - \alpha^{3}|,|\alpha^{3} - 1|) \\ &= (\alpha - 1,\alpha^{2} - \alpha,\alpha^{3} - \alpha^{2},\alpha^{3} - 1) \\ &= (\alpha - 1) \cdot (1,\alpha,\alpha^{2},\alpha^{3}), \end{align} where the last equality is obtained by observing that $(\alpha - 1) \alpha^{3} = \alpha^{3} - 1$, which, in turn, is obtained from the quartic polynomial above. It follows that ${f^{n}}(1,\alpha,\alpha^{2},\alpha^{3}) = (\alpha - 1)^{n} \cdot (1,\alpha,\alpha^{2},\alpha^{3})$ for all $n \in \mathbb{N}$. As $\alpha - 1 \in (0,1)$, we see that $\displaystyle \lim_{n \rightarrow \infty} {f^{n}}(1,\alpha,\alpha^{2},\alpha^{3}) = (0,0,0,0)$, but clearly, no term is equal to $(0,0,0,0)$.

Is there an example where ${f^{n}}(a,b,c,d)$ does not converge to $(0,0,0,0)$? Thanks!

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Thanks! However, I'm looking at the case where $f$ is applied to $\mathbb{R}^{4}$ and not $\mathbb{Z}^{4}$. I've briefly scanned the article, but I don't see anything that helps to answer my question about the existence of a starting $4$-tuple that does not get iterated to $(0,0,0,0)$. – Haskell Curry Oct 14 '12 at 22:33
The first reference in Wikipedia says the example given is the only one that does not converge to $(0,0,0,0)$ in a finite number of steps. – Ross Millikan Oct 14 '12 at 23:05