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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!

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1 Answer 1

up vote 1 down vote accepted

You are asking about Ducci sequences. Wikipedia is a good place to start.

EDIT: The precise problem in question here seems to have been studied in M. Lotan, A problem in difference sets, American Mathematical Monthly, 56 (1949) 535–541.

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Never knew that they had a name; thanks! –  Brian M. Scott Oct 14 '12 at 22:26
    
Though the Wikipedia article is about integers, and Haskell's example is not consistent with the statement "If n is a power of 2 every Ducci sequence eventually reaches the n-tuple (0,0,...,0) in a finite number of steps." –  Henry Oct 14 '12 at 22:31
    
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
    
Haskell, I've given you the name. I thought with a little work you'd be able to use that name to find more suitable references. I guess not. –  Gerry Myerson Oct 14 '12 at 22:54
    
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

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