Thread: Busted functions View Single Post
2006-02-22, 12:09   #3
Dougy

Aug 2004
Melbourne, Australia

23×19 Posts

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.

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