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Old 2006-02-22, 08:01   #1
Dougy
 
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Default 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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Old 2006-02-22, 09:23   #2
Orgasmic Troll
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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.
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Old 2006-02-22, 12:09   #3
Dougy
 
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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
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Old 2006-02-22, 12:50   #4
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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
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Old 2006-02-22, 20:59   #5
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Oops. I think mine implies yours, but not the converse
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