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Dougy · 2006-02-22, 12:09

Thanks for your response.

Quote:

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.

Quote:

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..

Quote:

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.