Jabotinsky's iterative logarithm bo198214 Administrator Posts: 1,624 Threads: 103 Joined: Aug 2007 05/21/2008, 06:00 PM (This post was last modified: 05/21/2008, 06:06 PM by bo198214.) Jabotinsky and also Ecalle define something they call iterative logarithm (which I here abbreviate as ilog). It is defined by $\text{ilog}(f)=\frac{\partial f^{\circ t}(x)}{\partial t}|_{t=0}$ Jabotinsky  mentiones the similarity to the convetional log: $\frac{\partial b^t}{\partial t}|_{t=0}=b^t \ln(b)|_{t=0}=\ln(b)$ And we can easily derive the equation $\text{ilog}(f^{\circ w})=w\;\text{ilog}(f)$ by: $\frac{\partial \left(f^{\circ w}\right)^{\circ t}}{\partial t}|_{t=0}= \frac{\partial f^{\circ w t}}{\partial t}|_{t=0}= w\;\frac{\partial f^{\circ w t}}{\partial w t}|_{t=0}=w\;\text{ilog}(f)$ I also can verify this via my powerseries package for the example $f(x)=x+x^2+x^3$. However Jabotinsky also claims that: $\text{ilog}(f\circ g)=\text{ilog}(f)+\text{ilog}(g)$ which I can neither derive nor which is confirmed by the powerseries package. When I set $f(x)=x+x^2$ and $g(x)=x+x^2+x^3$ then $\text{ilog}(f\circ g)-(\text{ilog}(f)+\text{ilog}(g))=-\frac{1}{2}x^4+\frac{5}{2}x^5-\frac{22}{3}x^6+\frac{29}{2}x^7+\dots$  Eri Jabotinsky, Analytic Iteration, Transactions of the American Mathematical Society, Vol. 108, No. 3 (Sep., 1963), pp. 457-477 Attached Files powerseries.py (Size: 13.39 KB / Downloads: 829) Ivars Long Time Fellow    Posts: 366 Threads: 26 Joined: Oct 2007 05/22/2008, 09:02 AM bo198214 Wrote:$\text{ilog}(f\circ g)=\text{ilog}(f)+\text{ilog}(g)$ Not that I would know it, but would it not be more logical that in general case: $\text{ilog}(f\circ g)=g*\text{ilog}(f)+f*\text{ilog}(g)$? Perhaps there is a value of t/definition of $\text{ilog}(f)$ where multipliers g and f disappear. Ivars bo198214 Administrator Posts: 1,624 Threads: 103 Joined: Aug 2007 05/22/2008, 10:48 AM Ivars Wrote:Not that I would know it, but would it not be more logical that in general case: $\text{ilog}(f\circ g)=g*\text{ilog}(f)+f*\text{ilog}(g)$? Perhaps there is a value of t/definition of $\text{ilog}(f)$ where multipliers g and f disappear. Dont know what you mean. This law is in correspondence with the normal logarithmic law: $\ln(a\cdot b)=\ln(a)+\ln(b)$ The iterative logarithm is supposed to behave similar by swapping multiplication with composition. Ivars Long Time Fellow    Posts: 366 Threads: 26 Joined: Oct 2007 05/22/2008, 01:13 PM Hmm To put it differently, does there exist t real or complex where Your powerseries package would give 0 in that exercise? Ivars andydude Long Time Fellow    Posts: 510 Threads: 44 Joined: Aug 2007 05/22/2008, 05:53 PM I tried using my finite power series methods with this, and here is the code: Code:def iter_log(expr, t, x):   if is_parabolic(expr, x, 0):     f = parabolic_flow(expr, t, x)   else:     f = hyperbolic_flow(expr, t, x)   return diff(f, t).subs(t=0) def cto(ch): return dict([('C'+str(k), var(ch+str(k))) for k in range(1,10)]) t = var('t') f = p_poly(x, 6).subs(cto('F')) g = p_poly(x, 6).subs(cto('G')) fog = taylor(f.subs(x=g).expand(), x, 0, 6) z1 = taylor(   ( iter_log(f, t, x)   + iter_log(g, t, x)   - iter_log(fog, t, x)   ).expand(), x, 0, 6) z2 = taylor(   ( g*iter_log(f, t, x)   + f*iter_log(g, t, x)   - iter_log(fog, t, x)   ).expand(), x, 0, 6) z3 = taylor(   ( iter_log(f, t, x)   + iter_log(f.subs(x=f), t, x)   - iter_log(f.subs(x=f).subs(x=f), t, x)   ).expand(), x, 0, 6) ... sage: z1 (F2*G3 - F2*G2^2 - F3*G2 + F2^2*G2)*x^4/2 sage: z2 (-G2 - F2)*x^2 + (-G3 + G2^2 + G2 - F3 + F2^2 + F2)*x^3 + ... sage: z3 0 So, Jabotinsky is wrong. If Jabotinsky is right, then z1 = 0, but according to this, $z1 = \frac{x^4}{2}((f_2g_3 - f_3g_2) + (f_2 - g_2)f_2g_2) + \cdots$ which is only true if $f = g$ or if g is an iterate of f. Sadly, the same is true of z2, only more so, because it displayes non-zero terms way before the 4th term, so it is more obvious that it is non-zero. Andrew Robbins bo198214 Administrator Posts: 1,624 Threads: 103 Joined: Aug 2007 05/22/2008, 06:48 PM Gosh, I should learn proper reading first. Jabotinsky states: If $F(G(z))=G(F(z))$ it can be shown  that $G(z)$ is some iterate of $F(z)$ and hence that [translated into our style]: $\text{ilog}(F\circ G)=\text{ilog}(F)+\text{ilog}(G)$  J. Hadamard, Two works on iteration, Bull. Amer. Math. Soc. 50 (1944), 67-75 Sorry, for the inconvenience, this answers also Ivars question. (And btw. it should be very rare to find so obvious mistakes in peer reviewed journals ...) andydude Long Time Fellow    Posts: 510 Threads: 44 Joined: Aug 2007 05/22/2008, 07:43 PM bo198214 Wrote:(And btw. it should be very rare to find so obvious mistakes in peer reviewed journals ...)Right, I should have read Jabotinsky before stating that, sorry. I have his "Analytic Iteration" (not the one with Erdos) and his "L-sequences for ..." papers, so I should have read them if either of them contain this theorem. Anyways, its still very interesting, and I just did a test to see if it applies to hyperbolic iteration as well, and it does: Code:g = h_poly(x, 6) il1 = iter_log(g, t, x) il2 = iter_log(g.subs(x=g), t, x) il3 = iter_log(g.subs(x=g).subs(x=g), t, x) sage: il1 + il2 - il3 0 Andrew Robbins andydude Long Time Fellow    Posts: 510 Threads: 44 Joined: Aug 2007 05/22/2008, 09:22 PM This must mean that: $\text{ilog}(f^{\circ t}) = t \text{ilog}(f)$ so $f^{\circ t} = \text{ilog}^{-1}(t \text{ilog}(f))$ At first I thought this was either an Abel or Schroeder function, but it seems that it is neither. Because I've also heard Abel functions referred to as "iterational logarithm" before (I think Peter Walker calls them this). So how is this related to Abel functions? What is this? Since it cannot be uniquely identified by the term "iterational logarithm", then can it be called a Hadamard function? a Jabotinsky function? I am having a lot of trouble comparing this to Abel and Schroeder functions. The best I can do is solve all of them for n: $n = \frac{\text{ilog}(f^n(x))}{\text{ilog}(x)} = \log_c\left(\frac{\sigma(f^n(x))}{\sigma(x)}\right) = \alpha(f^n(x)) - \alpha(x)$ Andrew Robbins Ivars Long Time Fellow    Posts: 366 Threads: 26 Joined: Oct 2007 05/23/2008, 07:05 AM (This post was last modified: 05/23/2008, 07:12 AM by Ivars.) andydude Wrote:This must mean that: $\text{ilog}(f^{\circ t}) = t \text{ilog}(f)$ so $f^{\circ t} = \text{ilog}^{-1}(t \text{ilog}(f))$ Andrew Robbins Would that be true also for imaginary t? t=I, like: $\text{ilog}(f^{\circ I}) = I \text{ilog}(f)$ $f^{\circ I} = \text{ilog}^{-1}(I \text{ilog}(f))$ The functions having ilog property ilog(fg)=ilog (f) + ilog(g) seem to be rather wide class, am I right? Would it be true for ( I am only writing down iterates of f): any 1/m, 1/n - are there any constraints on m, n like they must have common integer which they both divide? any 1/x, 1/y or just 1/x, 1-1/x = (x-1)/(x), or n/x , 1-n/x = (x-n)/x? purely complex conjugate iterates, like it, -it ? z and z*? Please tell me where I am first wrong in my conjecturing. Best regards, Ivars bo198214 Administrator Posts: 1,624 Threads: 103 Joined: Aug 2007 05/23/2008, 09:24 AM andydude Wrote:This must mean that: $\text{ilog}(f^{\circ t}) = t \text{ilog}(f)$ so $f^{\circ t} = \text{ilog}^{-1}(t \text{ilog}(f))$ At first I thought this was either an Abel or Schroeder function, but it seems that it is neither. The categories (non-mathematical) are different: $\text{ilog}$ maps a function to a function (or better a formal powerseries to a formal powerseries) while the Abel function maps values to values. And before writing $\text{ilog}^{-1}$ you should assure that it is invertible, which stronly seems not to be the case. To be shorter I adopt Ecalle's notation and write $f_\ast$ for $\text{ilog}(f)$ and write $f^{\ast}$ for the regular Abel function of $f$. Then the relation between them both is: $\frac{\partial f^\ast(x)}{\partial x}=1/f_\ast(x)$ This can easily deduced by: \begin{align*} f^\ast(f^{\circ t}(x))&=t+f^\ast(x)\\ \frac{\partial f^\ast(f^{\circ t})}{\partial t}&=1\\ \frac{\partial f^\ast(x)}{\partial x} \frac{\partial f^{\circ t}}{\partial t}&=1\\ \end{align*} In particular $1/f_\ast$ is a meromorphic function i.e. can be expressed as $c_{-n}x^{-n}+\dots+c_{-1}x^{-1}+c_0+c_1x+c_2x^2+\dots$ If you now integrate this expression $x^{-1}$ becomes $\log(x)$. As this is so important, Ecalle calls $c_{-1}$ the "residu iteratif" (which he introduces in his first theorem in "Theorie des Invariants Holomorphes") so that we have $f^\ast(z) = c_{-1} \log(z) + F(z)$ where $F(z)$ is a meromorphic function. I think similar considerations can be found in Szerkes' paper too. Quote:Because I've also heard Abel functions referred to as "iterational logarithm" before (I think Peter Walker calls them this). Ecalle calls the Abel function of $f$ "iterateur de $f$" and he calls $f_\ast$ "logarithme iteratif de $f$". « Next Oldest | Next Newest »

 Possibly Related Threads… Thread Author Replies Views Last Post Derivative of the Tetration Logarithm Catullus 1 876 07/03/2022, 07:23 AM Last Post: JmsNxn Iterated Hyperbolic Sine and Iterated Natural Logarithm Catullus 2 1,370 06/11/2022, 11:58 AM Last Post: tommy1729 Jabotinsky IL and Nixon's program: a first categorical foundation MphLee 10 7,089 05/13/2021, 03:11 PM Last Post: MphLee Is bugs or features for fatou.gp super-logarithm? Ember Edison 10 20,588 08/07/2019, 02:44 AM Last Post: Ember Edison Can we get the holomorphic super-root and super-logarithm function? Ember Edison 10 21,438 06/10/2019, 04:29 AM Last Post: Ember Edison True or False Logarithm bo198214 4 16,778 04/25/2012, 09:37 PM Last Post: andydude Base 'Enigma' iterative exponential, tetrational and pentational Cherrina_Pixie 4 16,642 07/02/2011, 07:13 AM Last Post: bo198214 Principal Branch of the Super-logarithm andydude 7 24,042 06/20/2011, 09:32 PM Last Post: tommy1729 Logarithm reciprocal bo198214 10 36,567 08/11/2010, 02:35 AM Last Post: bo198214 Kneser's Super Logarithm bo198214 16 73,453 01/29/2010, 06:43 AM Last Post: mike3

Users browsing this thread: 1 Guest(s) 