Real-analytic tetration uniqueness criterion?

[mike3]

Yes of course you are right.
Sorry silly of me.

But the argument still remains ...

What is your opinion about the periodic functions that I tried ?
Did you try those ?

regards

tommy1729

Exactly what \( \theta(x) \) mapping did you try? What was its \( \sup\ |\theta(x)| \)? I tried \( \sin(6 \pi x)^6 \) (5 derivatives at \( \frac{n}{6} \) 0, i.e. \( |\theta^{(k)}(x)| = 0 \) for \( k < 6 \) and \( x = \frac{n}{6} \)) divided by 100,000 and it was in fact much easier to pick that up than to pick up the usual sin mapping.

EDIT: Just tried \( \theta(x) = \frac{\sin(6 \pi x)^{25}}{100000} \). Darn easy to pick that up (got it on the 3rd derivative!) and it meets all your criteria (first 24 derivatives at every \( \frac{n}{6} \) zero). I even tried it with \( 100000 \) replaced by 1 billion (so \( \sup\ |\theta(x)| = 10^{-9} \)) and was still able to pick it up.

I think I know why, too: any such function with lots of flat spots is going to have "steeper" (in a relative sense) peaks (the peaks are more "concentrated" in terms of the interval they cover regardless of their amplitude). That is, \( |\theta'(x)| \) (and higher derivatives) is going to achieve bigger values for such functions at a given amplitude. Pure \( \sin \) is probably the "softest" periodic function you can get, and so is actually the hardest one to detect.

EDIT 2: Just tried a \( \theta(x) = \frac{\sin(2\pi x) + \frac{1}{10} \sin(4 \pi x)}{K} \). I.e. like in my original post, only with an added harmonic. The added harmonic made it somewhat easier to pick up. So I conjecture that straight up \( \sin(2\pi x) \) (divided by a \( K \), of course) is the hardest periodic function to detect.

And yes, you have to try \( x < 1 \). It looks it may still be detectable for larger \( x \) but it would take a LOT more derivatives.