Bummer! - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Bummer! (/showthread.php?tid=69) Pages: 1 2 3 4 5 6 regular iteration at two different fixed points - bo198214 - 06/19/2009 I just discovered that this was already researched in 1968 [1]. Karlin and Mcgregor showed that: If $f$ is a function holomorphic and single valued on the complement of a closed countable set in the extended complex plane. Let $s_1\neq s_2$ two fixed points of $f$ such that $|f'(s_0)|,|f'(s_1)|\neq 0,1$ and $f([s_1,s_2])\subseteq [s_1,s_2]$. Then the regular iterations at $s_1$ and $s_2$ are equal if and only if $f$ is a fractional linear function. [1] Karlin, S., & Mcgregor, J. (1968). Embedding iterates of analytic functions with two fixed points into continuous groups. Trans. Am. Math. Soc., 132, 137–145. RE: regular iteration at two different fixed points - Kouznetsov - 06/19/2009 (06/19/2009, 08:51 AM)bo198214 Wrote: If $f$ is a function holomorphic and single valued on the complement of a closed countable set in the extended complex plane. Let $s_1\neq s_2$ two fixed points of $f$ such that $|f'(s_0)|,|f'(s_1)|\neq 0,1$ and $f([s_1,s_2])\subseteq [s_1,s_2]$. Then the regular iterations at $s_1$ and $s_2$ are equal if and only if $f$ is a fractional linear function. ...Example: for positive b, let f(z)=z^b. It has fixed points 0 and 1. The superfunction for such f is F(z)=exp(b^z) Does it correspond to some fractional linear Schroeder? RE: regular iteration at two different fixed points - bo198214 - 06/19/2009 (06/19/2009, 11:41 AM)Kouznetsov Wrote: Example: for positive b, let f(z)=z^b. It has fixed points 0 and 1. The superfunction for such f is F(z)=exp(b^z) Does it correspond to some fractional linear Schroeder?It does not fit the preconditions as $f'(0)=0$. It has no regular iteration at 0. For most b it is not even holomorphic at 0. RE: Bummer! - nuninho1980 - 05/29/2011 (10/06/2007, 07:05 AM)bo198214 Wrote: x of max |y| may be 3.08853227...(we remember that this new number succeed "Euler") RE: Bummer! - bo198214 - 05/29/2011 (05/29/2011, 09:37 PM)nuninho1980 Wrote: your picture with graphic is missed!?!?! (your post #4) No, its not missing. Perhaps your browser has a caching problem. You also can find it under the direct URL: http://math.eretrandre.org/tetrationforum/attachment.php?aid=91 RE: Bummer! - nuninho1980 - 05/30/2011 (05/29/2011, 10:11 PM)bo198214 Wrote: No, its not missing. Perhaps your browser has a caching problem. You also can find it under the direct URL: http://math.eretrandre.org/tetrationforum/attachment.php?aid=91I edited my previous post. thank! RE: Bummer! - sheldonison - 05/30/2011 (05/29/2011, 09:37 PM)nuninho1980 Wrote: x of max |y| may be 3.08853227...(we remember that this new number succeed "Euler") Nuninho, you'll have to explain a little more before I can understand. The graph is the difference between the two superfunctions of sqrt(2), which are nearly identical going from f(x)=4 downto f(x)=2. RE: Bummer! - bo198214 - 05/31/2011 (05/30/2011, 03:25 PM)sheldonison Wrote: (05/29/2011, 09:37 PM)nuninho1980 Wrote: x of max |y| may be 3.08853227...(we remember that this new number succeed "Euler") Nuninho, you'll have to explain a little more before I can understand. The graph is the difference between the two superfunctions of sqrt(2), which are nearly identical going from f(x)=4 downto f(x)=2. I think I understood now what he meant! Like the Euler number is the argument x where b^x = x for the b where b^x has exactly one fixpoint - in the transition from two fixpoints to no real fixpoint (i.e. $b=e^{1/e}$), the "succeeding Euler number" is the argument x where b[4]x = x for the b where b[4]x has exactly one fixpoint - in the transition from two fixpoints to no real fixpoint (on x>0, because there is always a fixpoint between -2 and 0 for the tetrational.) This b is around 1.635.... And it looks in the graph as if the maximum of the difference is achieved exactly at this new Euler number. PS: We also have the analogon to the function $x^{1/x}$, this function has its maximum at x=e and the maximum is e^(1/e). Same with the self-tetra-root it has its maximum at the "succeeding Euler number" 3.08853227... and its value is around 1.635... as Mike found out in this thread. PS2: Ya and it also has the analogon to the maximum b such that b[4]oo exists (which is e^(1/e)[4]oo = e). So the maximum base for which b[5]oo exists is $1.635[5]\infty\approx 3.0885$. Andrew/Nuninho pointed it out in this thread. RE: Bummer! - JmsNxn - 05/31/2011 (05/31/2011, 09:05 AM)bo198214 Wrote: Like the Euler number is the argument x where b^x = x for the b where b^x has exactly one fixpoint - in the transition from two fixpoints to no real fixpoint (i.e. $b=e^{1/e}$), the "succeeding Euler number" is the argument x where b[4]x = x for the b where b[4]x has exactly one fixpoint - in the transition from two fixpoints to no real fixpoint (on x>0, because there is always a fixpoint between -2 and 0 for the tetrational.) This b is around 1.635.... And it looks in the graph as if the maximum of the difference is achieved exactly at this new Euler number. PS: We also have the analogon to the function $x^{1/x}$, this function has its maximum at x=e and the maximum is e^(1/e). Same with the self-tetra-root it has its maximum at the "succeeding Euler number" 3.08853227... and its value is around 1.635... as Mike found out in this thread. PS2: Ya and it also has the analogon to the maximum b such that b[4]oo exists (which is e^(1/e)[4]oo = e). So the maximum base for which b[5]oo exists is $1.635[5]\infty\approx 3.0885$. Andrew/Nuninho pointed it out in this thread. All of this was so fascinating, to see such a connection between exponentiation, tetration and pentation. Truly fascinating. RE: Bummer! - sheldonison - 06/01/2011 (05/31/2011, 09:05 AM)bo198214 Wrote: (05/29/2011, 09:37 PM)nuninho1980 Wrote: x of max |y| may be 3.08853227...(we remember that this new number succeed "Euler") .... I think I understood now what he meant! Like the Euler number is the argument x where b^x = x for the b where b^x has exactly one fixpoint - in the transition from two fixpoints to no real fixpoint (i.e. $b=e^{1/e}$), the "succeeding Euler number" is the argument x where b[4]x = x for the b where b[4]x has exactly one fixpoint - in the transition from two fixpoints to no real fixpoint (on x>0, because there is always a fixpoint between -2 and 0 for the tetrational.) This b is around 1.635.... And it looks in the graph as if the maximum of the difference is achieved exactly at this new Euler number.... I refreshed my memory on Henryk's graph, and on the Tetra-Euler constant. Nuinho really like's that number! Last year, I posted my own tetra-Euler results and graph. But I think in this particular case, the connection between Henryk's graph and the Tetra-Euler constant is probably a coincidence. For base= $\sqrt{2}$, the lower fixed point=2, and the upper fixed point=4. As I understand it, the range of Henryk's graph goes from the lower fixed point to the upper fixed point. He is graphing $f^{-1}(z) - g^{-1}(z) - k$, where f(z) is the upper entire $\text{SuperFunction}_{\sqrt{2}}(z)$, and g(z) is the lower superfunction $\text{sexp}_{\sqrt{2}}(z)$, and k is the average of the difference between the two. Going from 4=upper fixed point at -infinity, down to 2=lower fixed point at +infinity, there is a small 1-periodic wobble that defines the difference between these two functions, where $\text{sexp}_{\sqrt{2}}(z)=\text{SuperFunction}_{\sqrt{2}}(z+k+\theta(z))$. The problem is that as the base gets closer and closer to eta, the upper fixed point can get arbitrarily close to e, so the maximum, can't occur at the Tetra-Euler number, ~3.0885322718067176544821807826411. For example, for B=1.4409, the upper fixed point=3.07763784998, and the lower fixed point=2.42379627885. So, then the maximum can't occur at Tetra-Euler, since Tetra-Euler is outside the range of the graph, right? - Sheldon