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 2006-02-22, 08:01 #1 Dougy     Aug 2004 Melbourne, Australia 23·19 Posts Busted functions Definition: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function. Then if, for all $m \in \mathbb{N}$, one of the following is true: (i) $f^n(m)$ is not prime for some $n \in \mathbb{N}$. (where $f^1(m)=f(m)$ and $f^n(m)=f(f^{n-1}(m))$, i.e. function composition) (ii) $f^i(m)=f^j(m)$ for distinct $i,j \in \mathbb{N}$. then $f$ is called a busted function. Some functions are easily busted. E.g. $f(x)=a$ for some $a \in \mathbb{N}$, or $f(x)=x$. While some functions are not bustible. E.g. Any function satisfying $f(p_n)=p_{n+1}$, where $p_i$ is the $i^{th}$ prime. Question: What functions are busted? The motivation behind this conjecture is to disprove anyone who claims something like "if $x_1$ is a prime, and $x_n := f(x_{n-1})$, then $(x_n)$ is a sequence of primes". A function becomes 'busted' when it stops producing new primes. Has this been done before? Can anyone easily bust some functions? Last fiddled with by Dougy on 2006-02-22 at 08:03
2006-02-22, 09:23   #2
Orgasmic Troll
Cranksta Rap Ayatollah

Jul 2003

10100000012 Posts

Quote:
 Originally Posted by Dougy Definition: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function. Then if, for all $m \in \mathbb{N}$, one of the following is true: (i) $f^n(m)$ is not prime for some $n \in \mathbb{N}$. (where $f^1(m)=f(m)$ and $f^n(m)=f(f^{n-1}(m))$, i.e. function composition) (ii) $f^i(m)=f^j(m)$ for distinct $i,j \in \mathbb{N}$. then $f$ is called a busted function. Some functions are easily busted. E.g. $f(x)=a$ for some $a \in \mathbb{N}$, or $f(x)=x$. While some functions are not bustible. E.g. Any function satisfying $f(p_n)=p_{n+1}$, where $p_i$ is the $i^{th}$ prime. Question: What functions are busted? The motivation behind this conjecture is to disprove anyone who claims something like "if $x_1$ is a prime, and $x_n := f(x_{n-1})$, then $(x_n)$ is a sequence of primes". A function becomes 'busted' when it stops producing new primes. Has this been done before? Can anyone easily bust some functions?
condition II can be rewritten as $f^i(m)=m$ for some i

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.

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.

2006-02-22, 12:09   #3
Dougy

Aug 2004
Melbourne, Australia

15210 Posts

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.

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

2006-02-22, 12:50   #4
Dougy

Aug 2004
Melbourne, Australia

23·19 Posts

Quote:
 Originally Posted by Dougy ...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.
After I posted it I noticed that this was incorrect. So I'll give another example of a function $f$ which would not be included if $f:\mathbb{N} \rightarrow \mathbb{N}$. Eg. $f(x)=2^{x-2}+1$ has the property that $f(1)$ is not a natural number, while $f(x) \in \mathbb{N}$ when $x \in \mathbb{N}$ and $x>1$.

It is possible to create a related function $g:\mathbb{N} \rightarrow \mathbb{N}$ such that $g(x)=f(x)$ whenever $f(x) \in \mathbb{N}$, and $g(x)=1$ whenever $f(x)$ is not a natural number. But then this would be asking a different question.

Last fiddled with by Dougy on 2006-02-22 at 12:52

 2006-02-22, 20:59 #5 Orgasmic Troll Cranksta Rap Ayatollah     Jul 2003 641 Posts Oops. I think mine implies yours, but not the converse

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