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Consider the function $\sigma : \mathbb{Z} \to \mathbb{Z}$ where $\sigma = n - 3$. The orbits are

$\{3n : n \in \mathbb{Z} \}, \{3n + 1 : n \in \mathbb{Z} \}, \{ 3n + 2: n \mathbb{Z} \}$

What exactly are they really doing? I tried listing out a few numbers $\sigma (1), \sigma(2), \sigma(3)$ etc... but they just keep going on. How are they finding these? What happens if the function is more complicated like $\rho = n^2 + 1$? What strategy are they using? What is their thinking towards this problem?

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By the orbit of a number $n$, they mean the numbers $n,\sigma(n),\sigma(\sigma(n)),\sigma(\sigma(\sigma(n)))$ and so on; but also $\sigma^{-1}(n),\sigma^{-1}(\sigma^{-1}(n))$ and so on. When $\sigma(n)=n-3$, that's the numbers $n,n-3,n-6,n-9$ and so on, and also $n+3,n+6,n+9$ and so on. So, depending on the value of $n$, you get all the multiples of 3, or all the numbers one more than a multiple of 3, or all the numbers two more than a multiple of 3.

It's not always this easy --- there are some fairly simple functions for which it's notoriously difficult to find all the orbits --- have a look for the Collatz problem to see what I mean.

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But you obviously know that Collatz is not a permutation - which is what concerns the OP. –  alancalvitti Dec 22 '12 at 17:30
    
@alancalvitti, you are right. However, there are small variations on Collatz which are permutations, and which appear to be as hard to analyze as Collatz is. –  Gerry Myerson Dec 23 '12 at 4:56
    
Can you give an example please? I understand that group theory gets ever more complicated as time goes on (entire journals devoted to the subject) and all the arrows are iso's. but specific permutation on $\mathbb N$ or $\mathbb Z$? –  alancalvitti Dec 23 '12 at 6:40
    
@alancalvitti, consider the map given by $f(3n+1)=4n+1$, $f(3n-1)=4n-1$, $f(3n)=2n$. This map is a permutation on the positive integers. I believe it is unknown whether the orbit of 8 goes to infinity or is eventually periodic. It's discussed briefly in Guy's Unsolved Problems In Number Theory. –  Gerry Myerson Dec 23 '12 at 16:30
    
Interesting function, I plotted it w/ Mathematica. However, permutations are bijections, hence surjective, whereas your $f$ is not. For example, these numbers have no preimage: 1, 2, 3, 4, 8, 9, 10, 11... ($f$ is odd, so similarly for the negative counterparts) –  alancalvitti Dec 23 '12 at 18:22
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