Iterability of exp(x)-1 bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 08/11/2007, 09:33 PM (This post was last modified: 08/11/2007, 09:45 PM by bo198214.) Hm, this is strange. I just tried numerically the half iterate of $e^x-1$ and it looks quite convergent. However I read in the article of Erdös and Jabotinsky [1]: Quote:The function $e^z-1$ was shown by I. N. Baker [2] to have no real non-integer iterates. M. Levine [3] showed, using some results of the present paper, that this function and the functions $z+z^2$ and $\frac{z}{1-z^2}$ have no analytic iterate. Now I am confused. If someone wants to compare, I computed the half iterate of $f(x):=e^x-1$ to be $f^{\circ 1/2}(x)=x+{\frac {1}{4}}{x}^{2}+{\frac {1}{48}}{x}^{3}+{\frac {1}{3840}}{x}^{ 5}-{\frac {7}{92160}}{x}^{6}+{\frac {1}{645120}}{x}^{7}+{\frac {53}{ 3440640}}{x}^{8}-{\frac {281}{30965760}}{x}^{9}+O \left( {x}^{10} \right)$ I verified that indeed $f^{1/2}\circ f^{1/2}=f$ and it converges for example for $x=1.0$ Code:1.0,1.250000000,1.270833333,1.270833333,1.271093750,1.271017795,1.271019345,1.271034749,1.271025674,1.271025591,1.271029198,1.271027503 Where is the mistake? [1] P. Erdös and E. Jabotinksy, On analytic iteration, J. Analyse Math. 8, 1960/1961, 361-376. [2] I. N. Baker, Zusammensetzungen ganzer Funktionen, Math. Z 69, 1958, 121-163. [3] M. Levin, MSc. Thesis, Israel Institute of Technology, 1960. Daniel Fellow Posts: 69 Threads: 25 Joined: Aug 2007 08/11/2007, 09:49 PM (This post was last modified: 08/13/2007, 10:45 PM by Daniel.) exp(x)-1 definately has non-integer iterated; see heirarchies of height 1/2 at http://tetration.org/Combinatorics/Schro...index.html which is listed in the OEIS as A052122. I don't have access to my computer, but it looks like our results for heirarchies of height 1/2 agree. I also have the general solution which checks with OEIS entries for heirarchies of height -2, -1, 1/2, 1, 2, 3 and 4. jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/11/2007, 10:25 PM Hmm, looks like you beat me to the punch. I just got done posting this over in a separate discussion thread: http://math.eretrandre.org/tetrationforu...d=40#pid40 bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 08/11/2007, 11:42 PM Daniel Wrote:exp(x)-1 definately has non-integer iterated; see heirarchies of height 1/2 at http://tetration.org/Combinatorics/Schro...index.html which is listed in the OEIS as A052122. I don't have access to my computer, but it looks like our results for heirarchies of height 1/2 agree. I also have the general solution which checks with OEIS entries for heirarchies of height -2, -1, 1/2, 1, 2, 3 and 4. The general formula comes from the double binomial expansion, ${f^{\circ s}}_n=\sum_{i=0}^{n-1} (-1)^{n-1-i}\left(s\\i\right)\left(s-1-i\\n-1-i\right){f^{\circ i}}_{n}$ The formula is reliable, I just computed it for the case s=1/2, to exemplify convergence. However I just looked in Baker's Paper and indeed he states (as a German native I just translate it): Quote:Proposition 17. Let $F(z)=e^z-1$; for each real $\sigma$ let $F_\sigma(z)$ be the by $F(F_\sigma(z))=F_\sigma(F(z))$ uniquely determined formal series which has the form $F_\sigma(z)=z+\frac{\sigma}{2}z^2+\sum_{m=3}^\infty a_m(\sigma)z^m$. Then $F_\sigma(z)$ has a positive radius of convergence if and only if $\sigma$ is an integer number. $F_\sigma(z)$ is the $n$-th iterate of $F(z)$ for integer $\sigma=n>0$, hence an entire function. $F_\sigma(z)$ is the inverse series development of $F_{-n}$ for integer $\sigma=n<0$. So instead just of to numerically verify, can we prove that $(e^x-1)^{\circ 1/2}$ converges for some $x>0$? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 08/12/2007, 09:07 AM Inspired by JaydFox mentioning P. L. Walker (which I didnt have heard of before) I just read the abstract of his paper [1]: Quote:The author considers the Abel functional equation $g(\varphi(x))=g(z)+1$, where $\varphi$ is a given entire function and $g$ is an unknown entire function to be found. The inverse function $f=g^{-1}$ (if one exists) must satisfy (1) $f(w+1)=\varphi(f(w))$. The purpose of this paper is to show that for a wide class of entire functions, which includes $\varphi(z)=e^z-1$, equation (1) has a nonconstant entire solution. So I guess some results of I. N. Baker are indeed errournous. [1] P. L. Walker, A class of functional equations which have entire solutions, Bull. Austral. Math. Soc. 38 (198, no. 3, 351-356 jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/12/2007, 04:41 PM (This post was last modified: 08/12/2007, 04:46 PM by jaydfox.) bo198214 Wrote:Inspired by JaydFox mentioning P. L. Walker (which I didnt have heard of before) I just read the abstract of his paper [1]: Quote:The author considers the Abel functional equation $g(\varphi(x))=g(z)+1$, where $\varphi$ is a given entire function and $g$ is an unknown entire function to be found. The inverse function $f=g^{-1}$ (if one exists) must satisfy (1) $f(w+1)=\varphi(f(w))$. The purpose of this paper is to show that for a wide class of entire functions, which includes $\varphi(z)=e^z-1$, equation (1) has a nonconstant entire solution. So I guess some results of I. N. Baker are indeed errournous. [1] P. L. Walker, A class of functional equations which have entire solutions, Bull. Austral. Math. Soc. 38 (198, no. 3, 351-356I think that settles it. If $\varphi(z)=e^z-1$ has an entire solution, then the cheta function should define a unique solution for tetration. And with my exact formula for base conversion, we can solve all bases. I'll put together a consolidated post with all the necessarily formulae. Currently they are spread across separate posts (which is good, as this allows feedback on each portion of the solution). bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 08/13/2007, 08:54 PM (This post was last modified: 08/13/2007, 08:55 PM by bo198214.) Now I indeed had a look at Quote:[1] P. L. Walker, A class of functional equations which have entire solutions, Bull. Austral. Math. Soc. 38 (198, no. 3, 351-356 but things become more complicated! Lets starting with his theorem: Quote:Theorem 2. Let $\phi$ be an entire function of the form $\phi(z)=z+\sum_{n=1}^\infty c_n z^{n+1}$, where $c_1>0, c_n\ge 0$ for all $n$, and either (i) $c_2\neq c_1^2$ or (ii) $c_3. Then the sequence $(f_n)$ defined in Theorem 1 converges uniformly on every $\overline{S}(0,M)$ to a function $f$ which is an entire non-constant solution of (2). Where (2) is $f(w+1)=\phi(f(w))$. The also mentioned theorem 1 and sequence $f_n$ does not matter yet. What however really bothers me, that it seems not to be true: Let $\phi(x)=x+x^2$. This is a feasible function for theorem 2, with $c_1=1$ and $c_2=0\neq c_1^2$. Now I looked at the (unqiue) half iterate $\phi^{\circ 1/2}(x)=f(f^{-1}(x)+1/2)$ which should be entire too, for comparison some members of its series: $ x+{\frac {1}{2}}{x}^{2}-{\frac {1}{4}}{x}^{3}+{\frac {1}{4}}{x}^{4}-{ \frac {5}{16}}{x}^{5}+{\frac {27}{64}}{x}^{6}-{\frac {9}{16}}{x}^{7}+{ \frac {171}{256}}{x}^{8}-{\frac {69}{128}}{x}^{9}+O \left( {x}^{10} \right)$ and tested convergence at $x=5$ (an entire function has an infinite radius of convergence, so it should converge for every $x$) and what did I find? Divergence! Quote:0, 5.0, 17.50000000, -13.75000000, 142.5000000, -834.0625000, 5757.734375, -38187.57812, 222737.7149, -830118.7301, -3591005.876, 123831803.7, -1672085945. So it seems this proof is also not reliable... Boy that shakes my trust in professional mathematics. jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/13/2007, 09:06 PM (This post was last modified: 08/13/2007, 09:08 PM by jaydfox.) The logarithm is "entire" for reals > 0 (not quite "entire", but you know what I mean), but it's standard power series diverges outside the range (0, 2]. Analytic extension is used outside that range. So this function might only converge for x<=1. Besides, you do know what the formula equals at x=5, don't you? Well, for starters, at x=4, it's approximately 5.03481484682034616908489989276E+41. Bear in mind, this function is equal to the second iterated logarithm (base e) of my cheta function, shifted by a constant in the x direction. So it has tetrational growth. It's like trying to solve sin(x) using the power series, and then saying it appear divergent because you tried to solve the power series for x=100. So no worries. Just make sure it converges for x=3, which should equal 96.0223655650268799109865292599. ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/13/2007, 09:13 PM jaydfox Wrote:The logarithm is "entire" for reals > 0 (not quite "entire", but you know what I mean), but it's standard power series diverges outside the range (0, 2]. Analytic extension is used outside that range. So this function might only converge for x<=1. Besides, you do know what the formula equals at x=5, don't you? Well, for starters, at x=4, it's approximately 5.03481484682034616908489989276E+41. Bear in mind, this function is equal to the second iterated logarithm (base e) of my cheta function, shifted by a constant in the x direction. So it has tetrational growth. It's like trying to solve sin(x) using the power series, and then saying it appear divergent because you tried to solve the power series for x=100. So no worries. Just make sure it converges for x=3, which should equal 96.0223655650268799109865292599.Sorry, I got mixed up. I was thinking of the iterating function itself. I was thinking of solving for the 5th iterate. You meant solving the half-iterate for x=5. Big difference. Nevertheless, I would try various x values and see what the radius of convergence really is. It might be 1. But a radius of convergence of 1 is still far more than enough allow analytic extension. To be a valid test, you'll need far more than the first 10 terms, though. (And to be a proof, you need a formula for all of them, etc., which is what I'm working on, in the general case). ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/13/2007, 09:16 PM jaydfox Wrote:jaydfox Wrote:The logarithm is "entire" for reals > 0 (not quite "entire", but you know what I mean), but it's standard power series diverges outside the range (0, 2]. Analytic extension is used outside that range. So this function might only converge for x<=1. Besides, you do know what the formula equals at x=5, don't you? Well, for starters, at x=4, it's approximately 5.03481484682034616908489989276E+41. Bear in mind, this function is equal to the second iterated logarithm (base e) of my cheta function, shifted by a constant in the x direction. So it has tetrational growth. It's like trying to solve sin(x) using the power series, and then saying it appear divergent because you tried to solve the power series for x=100. So no worries. Just make sure it converges for x=3, which should equal 96.0223655650268799109865292599.Sorry, I got mixed up. I was thinking of the iterating function itself. I was thinking of solving for the 5th iterate. You meant solving the half-iterate for x=5. Big difference. Nevertheless, I would try various x values and see what the radius of convergence really is. It might be 1. But a radius of convergence of 1 is still far more than enough allow analytic extension. To be a valid test, you'll need far more than the first 10 terms, though. (And to be a proof, you need a formula for all of them, etc., which is what I'm working on, in the general case). Wait, now I'm really mixed up. You posted a different power series from the one we were looking at earlier. The previous one matched my calculations. What's this new one? Is this the series for the iterating function itself? If so, I stand by my first post about trying x=3. ~ Jay Daniel Fox « Next Oldest | Next Newest »