Login

***bo198214*** · 03/29/2009, 11:23 AM

As it is well-known we have for $b<e^{1/e}$
the regular superexponential at the lower fixed point.

This can be obtained by computing the Schroeder function at the fixed point $a$ of $F(x)=b^x$ .

More precisely we set
$G(x)=F(x+a)-a = b^{x+a}-a=b^a b^x -a = a b^x -a = a(b^x-1)$
This is a function with fixed point at 0, it is the function $F$ shifted that its fixed point is at 0.

We compute the Schroeder function $\chi$ of $G$ , i.e. the solution of:
$\chi(G(x))=c\chi(x)$ where $c=G'(0)=a\ln(b)=\ln(b^a)=\ln(a)$ .
This has a unique analytic solution with $\chi'(0)=1$ .

Then we get the super exponential by
$\operatorname{sexp}_b(t)=a+\chi^{-1}(c^x \chi(y_0)$
$y_0$ is adjusted such that
$1=\operatorname{sexp}_b(0)=a+\chi^{-1}(\chi(y_0))=a+y_0$
i.e. $y_0=1-a$ .

This procedure can be applied to any fixed point $a$ of $b^x$ .
The normal regular superexponential is obtained by applying it to the lower fixed point.

Now the upper regular superexponential $\operatorname{usexp}$ is the one obtained at the upper fixed point of $b^x$ .
For this function we have however always $\operatorname{usexp}(x)>a$ ,
so the condition $\operatorname{usexp}(0)=1$ can not be met.
Instead we normalize it by $\operatorname{usexp}(0)=a+1$ , which gives the formula:
$\operatorname{usexp}_b(t)=a+\chi^{-1}\left(\ln(a)^x \chi(1)\right)$

The interesting difference to the normal regular superexponential is that upper on is entire, while the normal one has a singularity at -2 and is no more real for $x<-2$ .

It is entire because the inverse Schroeder function $\chi^{-1}$ is entire, it can be continued from an initial small disk of radius r around 0 By the equation
$\chi^{-1}(c^n x)=G^{[n]}(\chi(x))$
We know that $c>1$ thatswhy we cover the whole complex plane with $c^nx$ , $x$ from the initial disc around 0, and we know that $G^{[n}]$ is entire.

Here are some pictures of $\operatorname{sexp}$ that are computed via the regular schroeder function as powerseries for our beloved base $b=\sqrt{2}$ , $a=2,4$ :

Filename: lower_upper_sexp.png Size: 12.14 KB Less than 1 minute ago

and here the upper super exponential base 2 alone:

Filename: upper_sexp.png Size: 9.49 KB Less than 1 minute ago

andydude · 03/31/2009, 04:29 AM

wow, how bizarre...

***bo198214*** · 03/31/2009, 08:26 AM

Ansus Wrote:But this does not satisfy the functional equation of tetration, yes?

It satisfies all except $f(0)=1$ .

***bo198214*** · 03/31/2009, 08:31 AM

Ansus Wrote:So it is iterated exponential rather than tetration? Does it have asymptote?

Yes $y=4$ for $x\to -\infty$

sheldonison · (This post was last modified: 04/03/2009, 03:11 PM by sheldonison.)

bo198214 Wrote:....
Instead we normalize it by $\operatorname{usexp}(0)=a+1$ , which gives the formula:
$\operatorname{usexp}_b(t)=a+\chi^{-1}\left(\ln(a)^x \chi(1)\right)$

The interesting difference to the normal regular superexponential is that upper on is entire, while the normal one has a singularity at -2 and is no more real for $x<-2$ .
....

Does this upper super expoonential equation also hold for b= $e^{1/e}$ ?
Is this "chi" the same as the "Chi distribution" used in probability? Any links to a definition for
$\chi$ and $\chi^{-1}$

***bo198214*** · 04/03/2009, 04:22 PM

sheldonison Wrote:Does this upper super expoonential equation also hold for b= $e^{1/e}$ ?

Interesting question. Unfortunately the convergence gets quite bad for $b$ approaching $e^{1/e}$ , so I could not really check numerically.
On the other hand Walker describes also two solutions for $b=e^{1/e}$ in "On the solutions of an Abelian equation". I did not really read this article, but I think he also showed that these solutions are not the limit of approaching $e^{1/e}$ .

Quote:Is this "chi" the same as the "Chi distribution" used in probability?

No, not at all. Its just somewhat similar to "Sch" in Schroeder.

Quote: Any links to a definition for
$\chi$ and $\chi^{-1}$

Ya, for example in the thread regular slog.
Literature is: Szekeres "Regular iteration of real and complex functions."

sheldonison · (This post was last modified: 04/05/2009, 04:35 PM by sheldonison.)

bo198214 Wrote:....
Unfortunately the convergence gets quite bad for $b$ approaching $e^{1/e}$ , so I could not really check numerically.
On the other hand Walker describes also two solutions for $b=e^{1/e}$ in "On the solutions of an Abelian equation". I did not really read this article, but I think he also showed that these solutions are not the limit of approaching $e^{1/e}$ .
....
in the thread regular slog.
Literature is: Szekeres "Regular iteration of real and complex functions."

Kouznetsov has graphs of the lower super exponential for $b=e^{1/e}$ in the citizendium wiki. He says "the function approaches its limiting value e, almost everywhere". I haven't seen any graphs for the upper superexponential though.

For $b>e^{1/e}$ , the function exponentially decays to its limiting value in the complex plane at +/- i $\infty$ . This is probably also true for the upper super exponential for $b=e^{1/e}$ , as the value at the real axis increases ...

tommy1729 · 04/05/2009, 07:05 PM

bo198214 Wrote:As it is well-known we have for $b<e^{1/e}$
the regular superexponential at the lower fixed point.

This can be obtained by computing the Schroeder function at the fixed point $a$ of $F(x)=b^x$ .

More precisely we set
$G(x)=F(x+a)-a = b^{x+a}-a=b^a b^x -a = a b^x -a = a(b^x-1)$
This is a function with fixed point at 0, it is the function $F$ shifted that its fixed point is at 0.

We compute the Schroeder function $\chi$ of $G$ , i.e. the solution of:
$\chi(G(x))=c\chi(x)$ where $c=G'(0)=a\ln(b)=\ln(b^a)=\ln(a)$ .
This has a unique analytic solution with $\chi'(0)=1$ .

Then we get the super exponential by
$\operatorname{sexp}_b(t)=a+\chi^{-1}(c^x \chi(y_0)$
$y_0$ is adjusted such that
$1=\operatorname{sexp}_b(0)=a+\chi^{-1}(\chi(y_0))=a+y_0$
i.e. $y_0=1-a$ .

This procedure can be applied to any fixed point $a$ of $b^x$ .
The normal regular superexponential is obtained by applying it to the lower fixed point.

Now the upper regular superexponential $\operatorname{usexp}$ is the one obtained at the upper fixed point of $b^x$ .
For this function we have however always $\operatorname{usexp}(x)>a$ ,
so the condition $\operatorname{usexp}(0)=1$ can not be met.
Instead we normalize it by $\operatorname{usexp}(0)=a+1$ , which gives the formula:
$\operatorname{usexp}_b(t)=a+\chi^{-1}\left(\ln(a)^x \chi(1)\right)$

The interesting difference to the normal regular superexponential is that upper on is entire, while the normal one has a singularity at -2 and is no more real for $x<-2$ .

It is entire because the inverse Schroeder function $\chi^{-1}$ is entire, it can be continued from an initial small disk of radius r around 0 By the equation
$\chi^{-1}(c^n x)=G^{[n]}(\chi(x))$
We know that $c>1$ thatswhy we cover the whole complex plane with $c^nx$ , $x$ from the initial disc around 0, and we know that $G^{[n}]$ is entire.

Here are some pictures of $\operatorname{sexp}$ that are computed via the regular schroeder function as powerseries for our beloved base $b=\sqrt{2}$ , $a=2,4$ :

[attachment=467]

and here the upper super exponential base 2 alone:
[attachment=468]

nice post.

thanks.

regards

tommy1729

***bo198214*** · (This post was last modified: 04/06/2009, 06:37 AM by bo198214.)

sheldonison Wrote:Kouznetsov has graphs of the lower super exponential for $b=e^{1/e}$ in the citizendium wiki. He says "the function approaches its limiting value e, almost everywhere". I haven't seen any graphs for the upper superexponential though.

I guess that the upper exponential for $b\uparrow e^{1/e}$ converges pointwise to the constant function $e$ (which of course also a solution of $f(x+1)=\left(e^{1/e}\right)^{f(x)}$ ).

sheldonison · (This post was last modified: 04/22/2009, 05:34 PM by sheldonison.)

bo198214 Wrote:As it is well-known we have for $b<e^{1/e}$
the regular superexponential at the lower fixed point.

This can be obtained by computing the Schroeder function at the fixed point $a$ of $F(x)=b^x$ .
.....
Now the upper regular superexponential $\operatorname{usexp}$ is the one obtained at the upper fixed point of $b^x$ .
For this function we have however always $\operatorname{usexp}(x)>a$ ,
so the condition $\operatorname{usexp}(0)=1$ can not be met.
Instead we normalize it by $\operatorname{usexp}(0)=a+1$ , which gives the formula:
$\operatorname{usexp}_b(t)=a+\chi^{-1}\left(\ln(a)^x \chi(1)\right)$

The "upper/lower" properties of these two sexp solutions are very interesting, especially being able to convert one to the other. The "upper" solution approaches the larger fixed point at -infinity, and the lower solution approaches the smaller fixed point at +infinity.

Can this be applied to Kneser's fixed point solution for bases larger than (e^(1/e))? For base e, Kneser's solution, has complex values at the real number line, and the function approaches the fixed point as x grows towards +infinity. But the desired solution has real values for all x>-2, and complex values for all x<-2 (except for the singularities). Moreover, the desired solution approaches the fixed point, as real x approaches -infinity.

This has probably already been done, but can Kneser's base e solution, approaching a complex fixed point at +infinity, be converted it to another solution, approaching the fixed point at -infinity, with real values at the real number line, for all x>-2? Perhaps this line of reasoning isn't applicable because the resulting solution, approaching the fixed point at -infinity, probably would not have imaginary values of zero for for real all x>-2.
- Sheldon

Login
Username:
Password:	Lost Password?
	Remember me

Possibly Related Threads...
Thread		Author	Replies	Views	Last Post
	Why the beta-method is non-zero in the upper half plane	JmsNxn	0	463	09/01/2021, 01:57 AM Last Post: JmsNxn
	Growth of superexponential	Balarka Sen	7	15,120	03/06/2013, 11:55 PM Last Post: tommy1729
	Nowhere analytic superexponential convergence	sheldonison	14	30,304	02/10/2011, 07:22 AM Last Post: sheldonison
	superexponential below -2	bo198214	10	17,617	05/27/2008, 01:09 PM Last Post: GFR