Busted functions

Dougy · 2006-02-22, 08:01

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?

Orgasmic Troll · 2006-02-22, 09:23

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.

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.

Dougy · 2006-02-22, 12:50

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.

Orgasmic Troll · 2006-02-22, 20:59

Oops. I think mine implies yours, but not the converse

2006-02-22, 08:01	#1
Dougy Aug 2004 Melbourne, Australia 152₁₀ 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

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2006-02-22, 20:59	#5
Orgasmic Troll Cranksta Rap Ayatollah Jul 2003 1010000001₂ Posts	Oops. I think mine implies yours, but not the converse