# Continuous Collatz Conjecture

Has anyone studied the real function $$f(x) = \frac{ 2 + 7x - ( 2 + 5x )\cos{\pi x}}{4}$$ (and $f(f(x))$ and $f(f(f(x)))$ and so on) with respect to the Collatz conjecture?

It does what Collatz does on integers, and is defined smoothly on all the reals.

I looked at $$\frac{ \overbrace{ f(f(\cdots(f(x)))) }^{\text{n times}} }{x}$$ briefly, and it appears to have bounds independent of $n$.

Of course, the function is very wiggly, so Mathematica's graph is probably not $100\%$ accurate.

-
Please review the formulas because I'm not sure if that's exactly what you wanted to write. –  Adrián Barquero Apr 2 '11 at 19:32
@Adrian Thanks for the cleanup; yes, that's what I meant. –  barrycarter Apr 2 '11 at 19:54
add comment

## 1 Answer

The analogue of the Collatz conjecture would be false, since the image of an interval of length $1$, say, contains an interval of length $1$ strictly to the right. That means you can find a point whose images keep moving to the right.

-
Nice observation. I don't understand your conclusion, though. I can see that for every $n$, you can find a point that moves to the right $n$ times, but how do you find a point that keeps moving (and doesn't reach a limit, either)? –  Sebastian Reichelt Apr 2 '11 at 22:01
Here is something to consider, though: There are lots of $x$ that map to themselves, the lowest around $1.736$. I guess that does make the analogue of the Collatz conjecture false, whatever that analogue might be exactly. –  Sebastian Reichelt Apr 2 '11 at 22:14
You use the nested interval theorem: A nested sequence of closed intervals has nonempty intersection. –  Douglas Zare Apr 2 '11 at 23:07
add comment