# Tetration Forum

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@ Sheldon's root 2 postings:

Hmm, this is all very interesting. I myself am interested in $\text{sexp}_{\sqrt{2}}(z)$; since I think this may be the natural base semi-operators work in. Therefore I have a few questions:

I understand $\text{Lsexp}_{\sqrt{2}}(\text{Lsexp}_{\sqrt{2}}^{-1}(z) + \sigma) = \exp_{\sqrt{2}}^{\circ \sigma}(z)\,\,;\,\,\R(z) < 2$
and
$\text{Usexp}_{\sqrt{2}}(\text{Usexp}_{\sqrt{2}}^{-1}(z) + \sigma) = \exp_{\sqrt{2}}^{\circ \sigma}(z)\,\,;\,\,\R(z) > 4$
Therefore how do we generate $\exp_{\sqrt{2}}^{\circ \sigma}(z)\,\,;\,\,\R(z) \in (2, 4)$?
Do we create a middle super function?

Secondly, why doesn't $\text{Usexp}_{\sqrt{2}}(0) = 8$? Is there a reason it isn't like this? because if it was centered at 8 it would give the beautiful result:
$\text{Usexp}_{\sqrt{2}}(\sigma) = 4\,\,\bigtriangleup_{\sigma}^{\small{\sqrt{2}}}\,\,4\,\,:\,\,\R(\sigma) \le 2$

It would also be consistent with the cheta function, where it's centered at 2 times the fix point.
(06/09/2011, 06:20 PM)JmsNxn Wrote: [ -> ]Hmm, this is all very interesting. I myself am interested in $\text{sexp}_{\sqrt{2}}(z)$; since I think this may be the natural base semi-operators work in. Therefore I have a few questions:

Here is some introduction into the topic.
(06/09/2011, 06:20 PM)JmsNxn Wrote: [ -> ]@ Sheldon's root 2 postings:

Hmm, this is all very interesting. I myself am interested in $\text{sexp}_{\sqrt{2}}(z)$; since I think this may be the natural base semi-operators work in. Therefore I have a few questions:
....
Therefore how do we generate $\exp_{\sqrt{2}}^{\circ \sigma}(z)\,\,;\,\,\R(z) \in (2, 4)$?
Do we create a middle super function?
Hey James,
$\exp_{\sqrt{2}}^{\circ \sigma}(z)=\text{Usexp_{\sqrt{2}}(\text{Uslog_{\sqrt{2}}(z)+\sigma)$
The short answer, is that these functions are imaginary periodic. Usexp(z) has a period of approximately 19.236i. USexp is real valued at the real axis going from 4+delta to infinity. But at exactly half that period, the Usexp(z) function is also real valued from -infinity to infinity, gently making a transition from 4-delta to 2+delta. Lsexp(z) has a period of about 17.143i. And at exactly half that period, the Lsexp(z) function is real valued from -infinity to infinity, also gently making a transition from 4-delta to 2+delta. In going from 4-delta to 2+delta, these two functions can be lined up, so that they are nearly identical, but they differ by a tiny amount!

So, for $z \in (2, 4)\,\; \Im(\text{Uslog_{\sqrt{2}}(z))\approx9.62i
$
. Hope that helps.
- Sheldon
Here are some more links (to go on wiki page?)
http://math.eretrandre.org/tetrationforu...96#pid3296
http://math.eretrandre.org/tetrationforu...534#pid534
Thanks sheldon that was really helpful. Is there any code yet generating these two functions?

And my second question still stands, though;
why exactly does $\text{Usexp}_{\sqrt{2}}(0) = 5.767053...$, is this point arbitrary? If it was shifted to 8 it wouldn't affect its status as a super function of root 2 would it? It's just a horizontal shift right?
(06/09/2011, 11:04 PM)JmsNxn Wrote: [ -> ]Thanks sheldon that was really helpful. Is there any code yet generating these two functions?

And my second question still stands, though;
why exactly does $\text{Usexp}_{\sqrt{2}}(0) = 5.767053...$, is this point arbitrary? If it was shifted to 8 it wouldn't affect its status as a super function of root 2 would it? It's just a horizontal shift right?

Horizontal shift -- yes and no. It is what comes out of the limit equation, that generates Usexp(0), limit as n->infinity. I'll need to dig that equation out. But even if you do a horizontal shift, it gets cancelled out since Uslog is the inverse of Usexp. I do have the lower level primitives, "superf(z)" and a "isuperf(z)" functions in kneser.gp. Type init(sqrt(2)), and those two functions are available. For bases<eta, which also have a lower superfunction, I have also implemented superf2(z), and isuperf2(z).
- Sheldon
(06/09/2011, 06:20 PM)JmsNxn Wrote: [ -> ]Therefore how do we generate $\exp_{\sqrt{2}}^{\circ \sigma}(z)\,\,;\,\,\R(z) \in (2, 4)$?
Do we create a middle super function?

Beginning at x=1 we use the lower fixpoint (at 2) and the iterates are always between -infty and 2. if we begin at x=3 the iterates are always between 2 and 4. However, we can connect the two areas. If we begin at x=1 and iterate with the complex height h=0 + 2*Pi*i/log(log(2)) then we get exactly one value in the 2..4 interval. That is that we just switch the sign of the value of the schrÃ¶der-function from positive to negative (one half round in the complex plane). That makes it also possible to define a "norm"-height for that values in the 2..4-interval. We set the real part of the height = 0 where x=1 was mapped to.

Gottfried