Define $ f: \mathbb{R}^{4} \rightarrow \mathbb{R}^{4} $ by \begin{equation} \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|). \end{equation}
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!
