Attempt to find a limit point but each step needs doubling the precision...

[tommy1729]

I wonder how the analogue pictures look like for exp(x).

regards

tommy1729

[Gottfried]

Gottfried,

Have you looked at this thread before?
http://math.eretrandre.org/tetrationforu...php?tid=89

Hi Jay, that were wonderful images; I forgot to take a look at them. Sometimes I think we should try to compose some "paedagogical" workout of what we've really achieved here in the forum and the main links to the articles...

I think, if you look at the pictures, that you will see that there are singularities, fanning out from the fixed point at real 4. To get to the real axis, you will have to navigate a course between the singularities. This is likely the cause of the slow convergence, as well as the need for high precision.

I've never actually tried to reach the real axis, and indeed I assumed it was impossible. But without actually trying, I can't be sure. If you do find a path to the real axis, I'd be curious to know the details.

Yes, I'm trying to get some more comprehension of this aspect. To get aware of singularities should help against wrong speculations...

Gottfried

[Gottfried]

Upps, I think I'd mystified myself... and the properties in question are actually well known.

Recap what I've done:
1) I chose some \( x_0 \) denoting some significant number below the lower fixpoint; one of the most natural might be \( x_0=1 \), then \( x_{-1} = 0, x_{-2} = - \infty \) .

2) I mapped this numbers by the imaginary height-iteration by \( \rho=\pi*I/ \log(u) \)(where \( u=\log(2) \) is the log of the lower fixpoint ) to the interval 2..4 on the real line. And gave accordingly indices \( y_0,y_{-1},y_{-2} \) and so on.

3) Now my question was, what is the number \( z_{-4} \), to where , for instance \( y_{-4} \), is mapped by that imaginary height? Something exceeding negative infinity? But seemingly to the right side f the upper fixpoint 4 ...

4) Well, what my procedures actually do (and what I did no more consider seriously) is (to allow use of the power series of the Schroeder-function) to iterate the value \( y_{-4} \) very near towards the lower fixpoint 2 (say by h=80 iterations), apply the Schroeder-mechanism for the imaginary height to map to the other side of the fixpoint, and iterate backwards h-times to find \( z_{-4} \).


Hmm... what I was missing was, that the mapping around the lower fixpoint does nothing else than to find a (real!) value below 2. So in fact I chose one value \( x_0 \), mapped it to the value \( y_0 \) and now simply mapped it back ... and applied the h-fold backwards iteration.
That calls for sarcasm... I simply could have used the original \( x_0 \) value and directly iterate backwards (without the twofold mapping!), say 4 times towards negative infinity! And this means simply to look at the \( \log(1+\epsilon) \) and \( \log(1-\epsilon) \), then at the log() of this, then again at the log() of this, and one more time at the log() of this. No need for 1600 digits precision.

We have that the interval \( (0..1] \) maps to \( (-\infty.. 0] \) by one backward-iteration, then that interval maps to the full line \( (-\infty+ w*\pi*I) .. + (\infty + w*\pi*I) \) where w is \( \exp(u)/u \), and that line maps to a shape of a distorted waterdrop with one sharp edge located outwards to \( +\infty \) at the real line.

This seems to be the simple answer; and my question, where the mapping of \( y_{-4} \) is located is answered by "towards real positive infinity".


If that reasoning is correct so far, then we could also complete the white spaces in Jay's images by neighboured drop-shapes, all with their edge at real-positive infinity.

If this is all correct, then... it was really so simple. ;-)

Gottfried

[tommy1729]

Hmm

Why does not every path become a loop ?

Afterall we consider iterations in the direction of the period ?

But for some reason we are talking about going to oo.


regards

tommy1729

[tommy1729]

I think the white space needs to be computed from the other 3 superfunctions.

It might be a uniqueness criterion if these new contours do not overlap with the older ones.

Switching branches fails because we have a different functional equation at another branch ( as discussed with sheldon ).

Its fascinating to apparantly see that the boundaries of the current pics are created by the higher fixpoints.

Now you should be able to complete the pictures !

regards

tommy1729

[tommy1729]

See the recently created related thread

http://math.eretrandre.org/tetrationforu...hp?tid=933

regards

tommy1729