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(06/23/2010, 09:11 PM)tommy1729 Wrote: to go towards the taylor series you simply take the derivates you need of \( \ln^{[n]} (\2sinh^{[z]}(exp^{[n]}(x))) \)
and create the taylor series with real coefficients.
then just plug in complex Z instead of x and your done.
however the question is , what is the radius of that taylor series expanded at x = 0 or even elsewhere.
if that radius is nonzero , we could probably extend further by using analytic continuation , or mittagleffler expansion.
and that is the way we go to the complex plane.
perhaps an intresting note is that if the ROC is large , we could check if that taylor series has the same period as exp(x) ( 2pi i ).
if it is indeed large enough and
1) the period is indeed 2pi i then the limit formula might hold for the complex plane and be equal to the taylor of the limit formula for the reals.
2) the period is not 2pi i then the limit formula will NOT hold for the complex plane and NOT be equal to the taylor of the limit formula for the reals.
also , if the radius is 0 everywhere , despite unlikely , then if the limit converges for all complex , then the function must have some local or global fractal or semifractal properties.
regards
tommy1729
perhaps constructing a fourier series is 'better' , assuming we have the period property of course.
'better ' in the sense of potentially easier to compute numerically , easier to compute 'symbolicly' ( four coeff ) and easier to prove related statements conjectures and properties.
this makes me doubt if taylor = four , if the period really exists and if the result only converges for x > 0 ... for reasons not yet explained ...
just saying that fourier expansion might be intresting imho.
if valid...
slightly off topic but i often think there should be a new type of series expansion designed for tetration ...
regards
tommy1729
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also worth mentioning i think :
let the base be a^(1/b) > sqrt(e) so that we can compute
the superfunction of f(x) = a^(1/b)^x with my method.
then consider t(x) = b(x + c)
and its inverse m(x) = x/b  c
m(f(t(x))) = m(a^(x+c)) = (1/b) a^(x+c)  c = (a^c / b) a^x  c
if a , b and c are chosen such that (a^c / b) a^x  c > x
we can compute the superfunction of (a^c / b) a^x  c by computing m(f^[z](t(x))).
regards
tommy1729
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(06/23/2010, 10:44 PM)tommy1729 Wrote: also worth mentioning i think :
let the base be a^(1/b) > sqrt(e) so that we can compute
the superfunction of f(x) = a^(1/b)^x with my method.
then consider t(x) = b(x + c)
and its inverse m(x) = x/b  c
m(f(t(x))) = m(a^(x+c)) = (1/b) a^(x+c)  c = (a^c / b) a^x  c
if a , b and c are chosen such that (a^c / b) a^x  c > x
we can compute the superfunction of (a^c / b) a^x  c by computing m(f^[z](t(x))).
regards
tommy1729
if (a^c / b) a^x  c = x we get the intresting case of yet another fixpoint.
associating functions without fixpoint with functions with 1 or more fixpoints is a intresting but complicated idea ...
it raises questions. can we determine the number of superfunctions by that ? can we define uniqueness in some way ?
for instance we associate g , f , k with g no fixpoint , f one fixpoint and k 2 fixpoints.
how many solutions ? 1 ? 2 ? 3 ? 4 ? 5 ? 6 ? oo ?
keep in mind that solution might be equal to eachother.
e.g. expanding f at its fixpoint = expanding k at its second fixpoint.
the key might be to notice that if a function with no fixp has the same superf as the associated with 1 , it may be unique.
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(06/24/2010, 12:29 PM)tommy1729 Wrote: (06/23/2010, 10:44 PM)tommy1729 Wrote: also worth mentioning i think :
let the base be a^(1/b) > sqrt(e) so that we can compute
the superfunction of f(x) = a^(1/b)^x with my method.
then consider t(x) = b(x + c)
and its inverse m(x) = x/b  c
m(f(t(x))) = m(a^(x+c)) = (1/b) a^(x+c)  c = (a^c / b) a^x  c
if a , b and c are chosen such that (a^c / b) a^x  c > x
we can compute the superfunction of (a^c / b) a^x  c by computing m(f^[z](t(x))).
regards
tommy1729
if (a^c / b) a^x  c = x we get the intresting case of yet another fixpoint.
associating functions without fixpoint with functions with 1 or more fixpoints is a intresting but complicated idea ...
it raises questions. can we determine the number of superfunctions by that ? can we define uniqueness in some way ?
for instance we associate g , f , k with g no fixpoint , f one fixpoint and k 2 fixpoints.
how many solutions ? 1 ? 2 ? 3 ? 4 ? 5 ? 6 ? oo ?
keep in mind that solution might be equal to eachother.
e.g. expanding f at its fixpoint = expanding k at its second fixpoint.
the key might be to notice that if a function with no fixp has the same superf as the associated with 1 , it may be unique.
however i believe in ' conservation of fixed points ' for analytic solutions.
for instance (a^c / b) a^x  c = x has no solution in nonnegative real x and real a,b,c > 1.
thus we linked the superfunction of a function without a real fixpoint to another without a real fixpoint.
a similar thing happens with bases below eta.
we then linked the superfunction of a function with 2 real fixpoints to another with 2 real fixpoints.
as you can see the amount of fixpoints remains constant hence ' conservation of fixed points '.
also intresting might be another example of substitution
exp(x^2 / 2) ^[z] => sqrt( exp^[z] (x^2))
and notice exp(x^2 / 2) = sqrt(exp(x^2)) = sqrt(e)^(x^2)
which leads me to the ' gaussian question '
exp( x^2 ) ^[z] = ?? however we have a fixpoint there
maybe intresting for statistics and combinatorics ...
regards
tommy1729
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(06/24/2010, 07:22 PM)tommy1729 Wrote: as you can see the amount of fixpoints remains constant hence ' conservation of fixed points '.
I am not really sure how you come to this conlcusion. Take
Base function: b*x and its superfunction: b^x: The base function has 1 fixed point (0), while the superfunction has fixed points depending on the value of b. Either two real, 1 real, or no real fixed point.
I would "conservation of fixed points" rather to noninteger iterates, as the demand that the noninteger iterates should have the same fixed points as the base function.
PS: Again the reminder: Dont quote whole posts! Always pick the particular quote you are replying to, or no quote at all, if it is a general reply. People are able to read the previous post, no need to repeat it for them.
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no quote
well Bo , i meant the conservation of the amount of important fixed points when linking one superfunction to another.
e.g. the link between e^x  1 and eta^x , both have only 1 important fixpoint. the link didnt change the amount of important fixpoints.
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06/27/2010, 03:31 AM
(This post was last modified: 06/29/2010, 02:02 AM by sheldonison.)
speaking of fixed points, the superfunction of 2sinh has an attracting fixed point of 0 + 1.895494239i, on the imaginary axis. I'm not sure what the exact region of convergence is (probably fractal), but any time real(superfunction(z))=0, then the superfunction(z+1, +2, +3 ... +n) will converge to this attracting fixed point. This is because if real(z)=0, then real(sexp2(z)) will also equal zero, which helps in understand why there is an attracting fixed point.
Regions where real(SuperFunction(z))=0 occur whenever imag(SuperFunction(z))=i*0.5*pi/ln(2), or i*(0.5+n)*pi/ln(2). I started to sketch out where the contours are. The SuperFunction(i*0.5*pi/ln(2)+x) converges to this fixed attracting point as x (real valued) increases. Other imaginary values of z close to 0.5*pi/ln(2) also converge to the attracting fixed point.
This adds complication to the "base change" converting the super function of sexp2 to sexp_e in those regions where real(superfunction(z)) approaches zero, since 2sinh has this attracting fixed point, but exp_e doesn't. Moreover, such regions approach the real axis as z increases.
 Sheldon
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(06/21/2010, 09:14 PM)tommy1729 Wrote: (06/09/2010, 12:18 PM)tommy1729 Wrote: \( \operatorname{TommySexp_e}(z,x)= \lim_{n \to \infty } \ln^{[n]} (\2sinh^{[z]}(exp^{[n]}(x))) \)
Ok, first let us verify that it is indeed an iteration of exp, i.e that \( f^z(x)=\text{TommySexp_e}(z,x) \) indeed satisfies:
\( f^{v+w}(x)=f^{v}(f^{w}(x)) \) and \( f^1(x)=\exp(x) \).
neglecting some rules of properly evaluating limits we get
\( f^v(f^w(x))=\lim_{n\to\infty} \ln^{[n]} (2\sinh^{[u]}(\exp^{[n]}(\ln^{[n]}(2\sinh^{[v]}(\exp^{[n]}(x))))))=\lim_{n\to\infty} \ln^{[n]}(2\sinh^{u+v}(\exp^{[n]}(x))=f^{u+v}(x) \).
and \( f^1(x)=\lim_{n\to\infty} \ln^{[n]}(2\sinh(\exp_^{[n]}(x))=\exp(x) \)
because towards infinity \( 2\sinh \) gets arbitrarily close to \( \exp \).
Basically thats the iteration equivalent of the Abel function Lévy proposes:
\( \beta(x) = \lim_{n\to\infty} \alpha(\exp^{[n]}(x))  \alpha(\exp^{[n]}(x_0)) \)
where \( \alpha \) is the Abel function of \( 2\sinh \) (or in Lévy's case \( \exp(x)1 \)).
The superfunction \( \sigma \) is then (the inverse of \( \beta \)):
\( \lim_{n\to\infty} \alpha(\exp^{[n]}(x))  \alpha(\exp^{[n]}(x_0))=y \)
\( \sigma(y)=x=\lim_{n\to\infty} \log^{[n]}(\alpha^{1}(y+\alpha(\exp^{[n]}(x_0)))) \)
\( \sigma(y)=\lim_{n\to\infty} \log^{[n]}(2\sinh^{[y]}(\exp^{[n]}(x_0)))) \)
which is the same as Tommy's superfunction.
We can do something similar with not only \( 2\sinh \) or \( \exp(x)1 \) but with any function that does not deviate too much from exp at infinity (i.e. all functions \( h \) such that \( \log^{[n]}(h(\exp^{[n]}(x)))\to \exp(x) \)).
Quote:as for the ROC i assume a plot for increasing n says more than a thousand words.
plot n = 1 > 100. z = 1/2
\( \ln^{[n]} (\2sinh^{[z]}(exp^{[n]}(x))) \)
I think you were not attentive when proposing to compute \( \exp^{[100]} \), already \( \exp^{[6]}(0) \) can not be computed in even sage's multiple precision arithmetic.
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(06/27/2010, 03:31 AM)sheldonison Wrote: speaking of fixed points, the superfunction of 2sinh has an attracting fixed point of 0 + 1.895494239i, on the imaginary axis. I'm not sure what the exact region of convergence is (probably fractal), but any time real(superfunction(z))=0, then the superfunction(z+1, +2, +3 ... +n) will converge to this attracting fixed point. This is because if real(z)=0, then real(sexp2(z)) will also equal zero, which helps in understand why there is an attracting fixed point.
Regions where real(SuperFunction(z))=0 occur whenever imag(SuperFunction(z))=i*0.5*pi/ln(2), or i*(0.5+n)*pi/ln(2). I started to sketch out where the contours are. The SuperFunction(i*0.5*pi/ln(2)+x) converges to this fixed attracting point as x (real valued) increases. Other imaginary values of 0 close to 0.5*pi/ln(2) also converge to the attracting fixed point.
This adds complication to the "base change" converting the super function of sexp2 to sexp_e in those regions where real(superfunction(z)) approaches zero, since 2sinh has this attracting fixed point, but exp_e doesn't. Moreover, such regions approach the real axis as z increases.
 Sheldon
no , we get rid of the fixed point of 2 sinh by the iterates of exp in my formula.
regards
tommy1729
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06/29/2010, 03:18 AM
(This post was last modified: 06/29/2010, 05:05 PM by sheldonison.)
(06/28/2010, 11:23 PM)tommy1729 Wrote: (06/27/2010, 03:31 AM)sheldonison Wrote: ....This adds complication to the "base change" converting the super function of sexp2 to sexp_e in those regions where real(superfunction(z)) approaches zero, since 2sinh has this attracting fixed point, but exp_e doesn't. Moreover, such regions approach the real axis as z increases.
 Sheldon
no , we get rid of the fixed point of 2 sinh by the iterates of exp in my formula.
regards
tommy1729 How is the fixed point removed? Starting with your equation for TommySexp, only interested in the "z" component, then let x=0,
\( \operatorname{TommySexp_e}(z)= \lim_{n \to \infty } \ln^{[n]} (\operatorname{2sinh}^{[z]}(\exp^{[n]}(0))) \)
This function gives the exact same results for all values of n as the following equivalent equation:
\( \operatorname{TommySexp_e}(z)= \lim_{n \to \infty } \ln^{[n]} (\operatorname{2sinh}^{[z+\operatorname{SuperFunction^{1}(\exp^{[n]}(0))]}} ) \)
Let \( k=\operatorname{SuperFunction^{1}(\exp^{[n]}(0))n \), then the following equation is also exactly equivalent:
\( \operatorname{TommySexp_e}(z)= \lim_{n \to \infty } \ln^{[n]} (\operatorname{SuperFunction(z+k+n)) \) As n goes to infinity, k converges quickly to the approximate value, k=0.067838366 (for n=0: k=0.0734181, for n=1: k=0.0663658, for n>=2: k=0.067838366)
Next, this equation can be used to compute TommySexp at z=0.5i*pi/ln(2), in the region where the attracting fixed point is, for increasing values of n
\( \ln^{[n]}(\operatorname{SuperFunction}(0.5i*\pi/\ln(2)+k+n)) \) as n increases. My guess is that everywhere in the region around the i=0.5*pi/ln(2) value, where the Superfunction converges to the attracting fixed point, TommySexp will converge to the repelling fixed point of e^z. with all derivatives going to zero.
For example, consider TommySexp for n=0, n=2, and n=10 at i0.5pi/ln(2). For n=0, real(x)=0, img(f)=i0.905, and for real(x)=1,img(f)=i1.572. This is a well defined analytic function. The function flattens out for n=2, and by the time n=10, it is converging towards the fixed point of "e"=0.318+i1.337 (hence my guess that all derivatives go to zero). Notice how the graph of TommySexp, for n=0, n=2 and n=10 flattens out and converges to the fixed point of "e" as n increases.
 Sheldon
