The upper superexponential bo198214 Administrator Posts: 1,616 Threads: 102 Joined: Aug 2007 03/29/2009, 11:23 AM As it is well-known we have for $b 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 $b=\sqrt{2}$, $a=2,4$:     and here the upper super exponential base 2 alone:     andydude Long Time Fellow Posts: 510 Threads: 44 Joined: Aug 2007 03/31/2009, 04:29 AM wow, how bizarre... bo198214 Administrator Posts: 1,616 Threads: 102 Joined: Aug 2007 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 Administrator Posts: 1,616 Threads: 102 Joined: Aug 2007 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 Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 04/03/2009, 03:06 PM (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 Administrator Posts: 1,616 Threads: 102 Joined: Aug 2007 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 Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 04/05/2009, 12:45 PM (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 Ultimate Fellow Posts: 1,742 Threads: 382 Joined: Feb 2009 04/05/2009, 07:05 PM bo198214 Wrote:As it is well-known we have for $b 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 Administrator Posts: 1,616 Threads: 102 Joined: Aug 2007 04/06/2009, 06:35 AM (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 Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 04/22/2009, 05:02 PM (This post was last modified: 04/22/2009, 05:34 PM by sheldonison.) bo198214 Wrote:As it is well-known we have for $b 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 « Next Oldest | Next Newest »

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