0
$\begingroup$

Consider a Cobweb diagram on the following function in the interval $(\frac12,1]$:

COllatz Conjecture Fractal

The function is clearly arranged as a fractal, in fact its most obvious symmetry is $x\cong \dfrac{x+2}4$

By this symmetry alone, the function has infinitely many points which map to any given point. And in fact, this is the only symmetry by which the function does not surject. On other words, in the quotient by this symmetry, it surjects onto any given choice of representatives.

It seems to me to follow that the right hand part of the diagram $(\frac34,1]$ should be all we need in order to create the cobweb diagram.

You may notice that this entire right hand segment of the function is below the $y=x$ line (which I have not drawn on). (apologies, the $x$ axis is inadvertently scaled to be over-square, with the effect that the $y=x$ line wouldn't run at 45 degrees.)

Anyway. Am I right in thinking that this latter fact, i.e. the function lying entirely below the 45 degree line in this representative interval, is sufficient to guarantee the function has no periodic points?

If so, this is sufficient to prove Lagarias' periodicity conjecture, since this graph is topologically conjugate to the Collatz conjecture function.

$\endgroup$
2
  • $\begingroup$ If I understand what you are asking, the answer is 'no'. From your graph, $f(x)<x$ for $x\in (\frac{3}{4},1]$. Potentially, there is a period $2$ point $x\in (\frac{3}{4},1]$ such that $f(x)\in (\frac{1}{2},\frac{3}{4}]$ and $f(f(x))=x$. Possibly, there are periodic points that bounce back and forth between the two intervals. I haven't studied the graph closely, but the answer to your general question seems to be 'no'. $\endgroup$ Commented 2 days ago
  • $\begingroup$ @TheOtherTerry if we can restrict it to a period 2 point that would be fine! In fact that would still be sufficient to prove the periodicity conjecture. $\endgroup$ Commented yesterday

0

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.