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Old 2006-02-22, 12:09   #3
Dougy's Avatar
Aug 2004
Melbourne, Australia

23×19 Posts

Thanks for your response.

Originally Posted by TravisT
condition II can be rewritten as f^i(m)=m for some i
I thought of this, however consider, for example, a function that satisfies f(2)=5, f(5)=7, f(7)=5. Then if m=2, f^n(m)=5,7,5,7,5,7,... for n=1,2,3,.... Informally, a loop might occur, but it might not loop right back to the beginning.

Originally Posted by TravisT
are you sure you mean that f: \mathbb{R} \rightarrow \mathbb{R}? if I have f(x) = ax + \pi, a \in \mathbb{N} then that's a busted function.
I certainly do mean f: \mathbb{R} \rightarrow \mathbb{R}. The other possibility f: \mathbb{N} \rightarrow \mathbb{N} would not include all of the relevant functions, for example, f(x)=\frac{3}{2}x+\frac{1}{2}, if m=3 then f^n(m) will still always be a natural number.

Your example is a busted function, however a rather trivial one. It would perhaps be possible to define f: \mathbb{N} \rightarrow \mathbb{R}, but then f(f(x)) is not well defined. Basically the functions of any real interest are smooth functions, like polynomials, exponentations, etc..

Originally Posted by TravisT
what does bustible mean? Either a function is busted or it is not, bustible sounds like we can do something to the function to make sure it is busted.
Yes I shouldn't have used the term 'bustible'. This is a human word which I used to mean "it is possible to prove that this function is busted." In fact lets make it a definition.

Definition: Let f: \mathbb{R} \rightarrow \mathbb{R} be a function. Then if there exists a proof that f is busted, we say f is bustible.

Last fiddled with by Dougy on 2006-02-22 at 12:15
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