Multiple exp^[1/2](z) by same sexp ? - 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: Multiple exp^[1/2](z) by same sexp ? (/showthread.php?tid=854) Pages: 1 2 Multiple exp^[1/2](z) by same sexp ? - tommy1729 - 04/29/2014 Consider the analytic solution sexp(z) that has the minimal amount of singularities => only singularities on the negative real axis. ( uniqueness has been shown , related is TPID 4 : http://math.eretrandre.org/tetrationforum/showthread.php?tid=747) Now let sexp(A) = u where Re(A) > 0,Im(A) > 0 and exp^[v](u) = u for real positive v. Notice that most u have such a v because of the chaotic behaviour of the iterations of exp. ( we ignore branches and functional equations in this sentense ) Now if such A,u are defined on the fundamental branch of sexp such that we have sexp(z+1)=exp(sexp(z)) then we have to conclude something special : exp^[v](u)= u = sexp(A) = sexp(A+v) Now if all if for w and v : 0 < w < 1 << v exp^w(u) = sexp(A + w) = sexp(A + v + w) = sexp(A+ 2v + w) THEN WE MUST CONCLUDE BY INDUCTION AND ANALYTIC CONTINUATION : -------------------------------------- sexp is periodic with period v !! -------------------------------------- But this contradicts the nonperiodicity of sexp and the branches of sexp !! Hence we must have for some u : set w = 1/2 then sexp(A+1/2 + v) =/= sexp(A+1/2) THUS ----------------------------------------------------------------------------- exp^[1/2](u) IS NOT DEFINED UNIQUELY BY THE SAME BRANCH OF SEXP THAT INCLUDES A ----------------------------------------------------------------------------- EUREKA ! this is surprising not ? this reminds me of the recent threads such as "slog(sexp(z)" AND MUCH MORE of a thread by gottfried : http://math.eretrandre.org/tetrationforum/showthread.php?tid=499 I bet you can see the relevance. regards tommy1729 RE: Multiple exp^[1/2](z) by same sexp ? - sheldonison - 04/29/2014 (04/29/2014, 12:19 PM)tommy1729 Wrote: ----------------------------------------------------------------------------- exp^[1/2](u) IS NOT DEFINED UNIQUELY BY THE SAME BRANCH OF SEXP THAT INCLUDES A -----------------------------------------------------------------------------I wasn't able to follow your logic -- too many variable names that seem to depend on each other. You have two previous threads on exp^{1/2}, which discussed the branch singularity at L, and the branch singularity at sexp(-2.5)=~-0.36+Pi i. Can you compare this result with the previous result, that showed exp^{1/2} has branches? http://math.eretrandre.org/tetrationforum/showthread.php?tid=849 http://math.eretrandre.org/tetrationforum/showthread.php?tid=544 I do find it interesting that it seems there are no known entire functions with fractional exponential growth. I would define the exponential growth by the following equation, converging to a number bigger than zero, and less than one, for any arbitrary function f. For example, f=exp^{0.5} would have a growth rate of 0.5. The gamma function has a growth rate of 1, as does exponentiation to any base. The double exponential has a growth rate of 2. iterating x^2 or any finite polynomial has a growth rate of zero. Iterating a super-exponential function (there are many entire examples) would have an infinite growth rate. $\text{growth}_f = \lim_{n \to\infty}\frac{\text{slog}(f^{o n})}{n}$ Perhaps this entire function growth rate question should be a conjecture? It seems it would be fairly easy to falsify, if one could find a counter-example. RE: Multiple exp^[1/2](z) by same sexp ? - tommy1729 - 04/29/2014 (04/29/2014, 07:02 PM)sheldonison Wrote: (04/29/2014, 12:19 PM)tommy1729 Wrote: ----------------------------------------------------------------------------- exp^[1/2](u) IS NOT DEFINED UNIQUELY BY THE SAME BRANCH OF SEXP THAT INCLUDES A -----------------------------------------------------------------------------I wasn't able to follow your logic -- too many variable names that seem to depend on each other. You have two previous threads on exp^{1/2}, which discussed the branch singularity at L, and the branch singularity at sexp(-2.5)=~-0.36+Pi i. Can you compare this result with the previous result, that showed exp^{1/2} has branches? http://math.eretrandre.org/tetrationforum/showthread.php?tid=849 http://math.eretrandre.org/tetrationforum/showthread.php?tid=544I do not feel those threads are strongly related. Quote:I do find it interesting that it seems there are no known entire functions with fractional exponential growth. I would define the exponential growth by the following equation, converging to a number bigger than zero, and less than one, for any arbitrary function f. For example, f=exp^{oz} would have a growth rate of 0.5. The gamma function has a growth rate of 1, as does exponentiation to any base. The double exponential has a growth rate of 2. iterating x^2 or any finite polynomial has a growth rate of zero. Iterating a super-exponential function (there are many entire examples) would have an infinite growth rate. $\text{growth}_f = \lim_{n \to\infty}\frac{\text{slog}(f^{o n})}{n}$ Perhaps this entire function growth rate question should be a conjecture? It seems it would be fairly easy to falsify, if one could find a counter-example. Lets call that "conjecture entire 1". Your intuition is correct. And also it is strongly related to another recent conjecture of you ( or a repost of it ) : $\lim_{z \to \infty} \text{slog}(f^z)=\text{slog}(f^{z+1})-1$ (made in the slog_b(sexp_b(z)) thread) Lets call that " conjecture entire 2" Im very optimistic about both conjectures and dare to say a proof is 99% complete. I have believed them for over 25 years. As for " conjecture entire 1" I feel forced to notice http://en.wikipedia.org/wiki/Weierstrass_product Yes the famous weierstrass factorization theorem. Together with induction that should be a strong tool in a proof. Notice that the product expansions given in the wiki as an example ( sin and cos ) have MULTIPLE product expansions ! This might complicate the proof. if every entire f(z) = z^a exp(g(z)) (1-z/a_n) (product over positive integer n , a_n integers not necc all distinct , g(z) another entire function ) That would be simpler. But its not exactly like that ... Maybe there is already a proof of it in the literature ? " conjecture entire 2" has also been considered for proof by me. The issue here is the nondifferentiable nature of certain aspects. one more thing about " conjecture entire 1" In a recent thread I started , I asked for an entire function that grows like tetration. Lets say F(z) By analogue you ask for an entire function that grows like exp^[1/2]. How about F(F^[-1](z)+1/2) as a solution to your problem ? Clearly F^[-1] cannot be entire ( it must have branches , being the inverse of a nontrivial entire function ). Hence F(F^[-1](z)+1/2) is NOT the solution we want. Maybe that helps. regards tommy1729 RE: Multiple exp^[1/2](z) by same sexp ? - sheldonison - 04/30/2014 (04/29/2014, 10:04 PM)tommy1729 Wrote: [quote='sheldonison' pid='6885' dateline='1398794535'] Quote:..... I do find it interesting that it seems there are no known entire functions with fractional exponential growth. I would define the exponential growth by the following equation.... $\text{growth}_f = \lim_{n \to\infty}\frac{\text{slog}(f^{o n})}{n}$ ..... Lets call that "conjecture entire 1". Your intuition is correct. And also it is strongly related to another recent conjecture of you ( or a repost of it ) : $\lim_{z \to \infty} \text{slog}(f^z)=\text{slog}(f^{z+1})-1$ (made in the slog_b(sexp_b(z)) thread) Lets call that " conjecture entire 2" Im very optimistic about both conjectures and dare to say a proof is 99% complete. I have believed them for over 25 years. ..... As for " conjecture entire 1" I feel forced to notice http://en.wikipedia.org/wiki/Weierstrass_product Yes the famous weierstrass factorization theorem. Together with induction that should be a strong tool in a proof. .... you ask for an entire function that grows like exp^[1/2]. How about F(F^[-1](z)+1/2) as a solution to your problem ? Clearly F^[-1] cannot be entire ( it must have branches , being the inverse of a nontrivial entire function ). Hence F(F^[-1](z)+1/2) is NOT the solution we want. Maybe that helps. regards tommy1729Tommy, Interesting, thanks for your response... - Sheldon RE: Multiple exp^[1/2](z) by same sexp ? - tommy1729 - 04/30/2014 (04/30/2014, 10:47 AM)sheldonison Wrote: (04/29/2014, 10:04 PM)tommy1729 Wrote: [quote='sheldonison' pid='6885' dateline='1398794535'] Quote:..... I do find it interesting that it seems there are no known entire functions with fractional exponential growth. I would define the exponential growth by the following equation.... $\text{growth}_f = \lim_{n \to\infty}\frac{\text{slog}(f^{o n})}{n}$ ..... Lets call that "conjecture entire 1". Your intuition is correct. And also it is strongly related to another recent conjecture of you ( or a repost of it ) : $\lim_{z \to \infty} \text{slog}(f^z)=\text{slog}(f^{z+1})-1$ (made in the slog_b(sexp_b(z)) thread) Lets call that " conjecture entire 2" Im very optimistic about both conjectures and dare to say a proof is 99% complete. I have believed them for over 25 years. ..... As for " conjecture entire 1" I feel forced to notice http://en.wikipedia.org/wiki/Weierstrass_product Yes the famous weierstrass factorization theorem. Together with induction that should be a strong tool in a proof. .... you ask for an entire function that grows like exp^[1/2]. How about F(F^[-1](z)+1/2) as a solution to your problem ? Clearly F^[-1] cannot be entire ( it must have branches , being the inverse of a nontrivial entire function ). Hence F(F^[-1](z)+1/2) is NOT the solution we want. Maybe that helps. regards tommy1729Tommy, Interesting, thanks for your response... - Sheldon Even more intresting, I have a proof of "conjecture entire 1". ON the other side its a shame we missed it because its almost written before here. Its a bit sketchy but the proof goes something like this : Lets say exp*(z) is a function with order 1 and 2 conjugate repelling fixpoints. sexp*(z) is its superfunction. If we want an entire function close to exp^[1/2](z) then it should be exp*^[1/2](z). So exp*^[1/2] is not entire because sexp*(z) is not entire. sexp*(z) is not entire => sexp*(slog*(z)+1/2) is not entire. So for the same reasons exp^[1/2](z) is not entire , exp*^[1/2](z) is not entire. we know sexp*(z) is not entire because it has to have log branches or algebraic branches ( coming from sexp*(z-1) = log*(sexp*(z)) ). regards tommy1729 RE: Multiple exp^[1/2](z) by same sexp ? - sheldonison - 04/30/2014 (04/30/2014, 12:39 PM)tommy1729 Wrote: .... So exp*^[1/2] is not entire because sexp*(z) is not entire. sexp*(z) is not entire => sexp*(slog*(z)+1/2) is not entire. So for the same reasons exp^[1/2](z) is not entire , exp*^[1/2](z) is not entire. we know sexp*(z) is not entire because it has to have log branches or algebraic branches ( coming from sexp*(z-1) = log*(sexp*(z)) ). regards tommy1729 What about f(z)=sexp(2*z)? We can generate the function $g(z) = f(f^{-1}(z)+1/2)=\exp(z)$, which is entire... I guess that's not fair since the multiple branches as you circle the fixed point of L all agree at integer values of n. But what if the fixed point is real? Can we say that $g(z) = f(f^{-1}(z)+1/2)=\exp(z)$ is never entire? RE: Multiple exp^[1/2](z) by same sexp ? - tommy1729 - 04/30/2014 (04/30/2014, 01:08 PM)sheldonison Wrote: What about f(z)=sexp(2*z)? We can generate the function $g(z) = f(f^{-1}(z)+1/2)=\exp(z)$, which is entire... I guess that's not fair since the multiple branches as you circle the fixed point of L all agree at integer values of n. Your f(z) has order 2 and your g(z) has order 1. In my example it was order 1 and order 1/2. The clue is that a point on sexp*(a) that is analytic cannot be mapped to a singularity at sexp*(a+b) while assuming that exp*^[b] is entire. Quote:But what if the fixed point is real? Can we say that $g(z) = f(f^{-1}(z)+1/2)=\exp(z)$ is never entire? If the only fixpoint is real then it resembles iterations of eta^z (or exp(z)-1). But eta^z (or exp(z)-1) has a parabolic fixpoint ( you are long enough into tetration to know what that implies ). --- It seems that things get complicated if we assume more than 1 fixpoint pair. However this means sexp* has more branches than sexp. And It seems therefore we get the same result. This lacks some formal formulation I admit. But I think you understand what I mean. --- regards tommy1729 RE: Multiple exp^[1/2](z) by same sexp ? - tommy1729 - 05/05/2014 as for the " conjecture entire 1 " things get intresting knowing that the entire functions grow like exp or faster despite being slower on the real line. For instance fake logs or fake sqrt. examples sinh(sqrt(x)). (= fake of exp(sqrt(x)) ) sum x^n/(2^n!) or this fake log here : http://math.eretrandre.org/tetrationforum/showthread.php?tid=821 f(z) = ln( (2sinh(z) - z) / z ) --- edit --- What Im trying to say here is this : Despite entire superfunctions like that of 2sinh , " conjecture entire 1 " is still not formally decided. I mentioned the weierstrass product , but however it gives imho a slight illusion. The weierstrass product gives the impression that most entire functions that grow slower than exp(exp(x)) behave very much like exp(g(x)) p(x) where g(x) and p(x) are very close to polynomials. ( i sometimes call that conjecture 0 but that term has also been used for other conjectures ) The point of these " false functions " is to show this is not true !! sinh(sqrt(z)) Always grows slower than about abs(exp(sqrt(abs(z)))) for abs(z). This follows from absolute convergeance ideas. ln( (2sinh(z) - z) / z ) Always grows slower than about abs(ln( (2sinh(abs(z)) - abs(z)) / abs(z) )) for abs(z). This follows from absolute convergeance ideas. So there is a " false function " for every elementary function. ( I still find this counterintuitive when I see weierstrass , sometimes even a proof is not convincing ? Maybe someone can enlighten me beyond proofs/examples ) SO why is there perhaps no " false function " for exp^[1/2] ? Despite the arguments above, we still lack a proof , disproof or example. So lets continue : How - very sketchy - should such a " candidate function " look like ? These candidate functions need to be between polynomial and exponential. More precisely they need to be polynomial << f(z) << exp(ln(a) x^b). So for instance f(z) ( the candidate ) cannot grow like exp(sqrt(x)-1). We know from the previous posts that taking the half-iterate of a function close to exp makes a nice argument, but not a good example or counterexample. ( e.g. 2sinh^[1/2] ) 2 Possibilities remain : 1) A power tower. Something like ln(x)^ln(ln(x))^... But thats even hard to prove being analytic. 2) sum z^n/(2^n!) As a first remark I note that its not immediately clear - without alot of work - if a Taylor series is faster or slower than a power tower. Second remark : x^ln(x) grows to slow : the growth = 0. ( actually the " fake x^ln(x) " to be exact ) This has already been demonstrated by me before. This leads us back to remark 1 actually how do 1) , 2) , x^ln(x) relate ? For the power tower case , I refer to the Original thread : http://math.eretrandre.org/tetrationforum/showthread.php?tid=799 I will update that thread with a new idea. ( actually an old , but new here ) ------------------------------------------------------------------------------------- I conjecture a false exp^[1/2](z) = f(z) by chosing a suitable positive real A for f(z)=0.5 + z + ( sexp(slog(1!)+A) )^-1 z^2 + ( sexp(slog(2!)+A) )^-1 z^3 + ... + ( sexp(slog((n-1)!)+A) )^-1 z^n where n goes to +oo. ------------------------------------------------------------------------------------- I think that expressed way better how I think about the subject then the post before. So I edited. regards tommy1729 " Reality is that what does not go away when you stop believing in it " RE: Multiple exp^[1/2](z) by same sexp ? - tommy1729 - 05/05/2014 A big edit has been made to the previous post. Also included are 2 links that are imho intresting. There is also a new post in one of those links. regards tommy1729 " We have to prove with nonsense that nonsense is nonsense " RE: Multiple exp^[1/2](z) by same sexp ? - sheldonison - 05/06/2014 (05/05/2014, 12:25 PM)tommy1729 Wrote: .... We know from the previous posts that taking the half-iterate of a function close to exp makes a nice argument, but not a good example or counterexample. ( e.g. 2sinh^[1/2] ) 2 Possibilities remain : 1) A power tower. Something like ln(x)^ln(ln(x))^... But thats even hard to prove being analytic. 2) sum z^n/(2^n!) ....The Taylor series of half exponential's should behave similarly as real(z) increases since the exponential approximations behave similarly as real(z) increases. Can this be used to build an entire half iterate? What is the pattern? The half exponential's I have worked most with are Kneser's tetration, which has singularities at -0.362+/-Pi i, plus a "quiet" singularity at L/L*, many others. But as z gets larger, it behaves very nicely. The other half exponential I've worked with is 2sinh^{1/2}(z), which has a singularity at 2i, as well as all of the iterates of 2i. I've also worked a little bit with the half exponential of exp(z)-1, and with sinh^{1/2}(z) both of which are generated from the parabolic case with a singularity at zero. For exp(x) itself, if half is a Taylor series at x0, the recursive definition is $\text{half}(\exp(x0)+x) = \exp(\text{half}(x0+\log(1+\frac{x}{\exp(x0)})))$ All three of these half iterates behave very nicely as real(z) gets larger, especially compared with tetration, and all three have similar Taylor series patterns, but I don't know what the pattern is. Here is the Taylor series for 2sinh^{1/2} at z0~=179.3912939581, whose half iterate is ~= e^^3, as well as the similar Kneser half iterate, and the half iterate of exp(z)-1. Can we quantify how much faster the derivatives decay, than exp(x+e^^2)? This might help in building an entire function, which approximates the half iterate as real(z) increases, or it may show why it may not be possible to generate an entire half exponential approximation; I will post more later; the equation above may prove very useful. All three of these half exponential functions have positive Taylor series coefficients up to around x^14 or x^15th. Code:2sinh^{1/2)(179.3912939581+x) ~=        3814279.104760 +x^ 1*  116894.1861623 +x^ 2*  1545.365235620 +x^ 3*  11.55206923902 +x^ 4*  0.05384577308168 +x^ 5*  0.0001631177103376 +x^ 6*  0.0000003259720865567 +x^ 7*  0.000000000430018019731 +x^ 8*  3.722580437964 E-13 +x^ 9*  2.100799565010 E-16 +x^10*  7.704363816259 E-20 +x^11*  1.839961959607 E-23 +x^12*  2.873242189147 E-27 +x^13*  3.114122886329 E-31 +x^14*  7.902808213762 E-35 +x^15* -8.028657785155 E-37 KneserHalf(179.1155195732+x) ~=        3814279.104760 +x^ 1*  117195.7788544 +x^ 2*  1553.468579782 +x^ 3*  11.64401876124 +x^ 4*  0.05442019863822 +x^ 5*  0.0001652812729209 +x^ 6*  0.0000003310497888663 +x^ 7*  0.0000000004374835751370 +x^ 8*  3.790743244215 E-13 +x^ 9*  2.138878310517 E-16 +x^10*  7.832144844649 E-20 +x^11*  1.865015582946 E-23 +x^12*  2.896225937469 E-27 +x^13*  3.309923535497 E-31 +x^14* -1.731956033634 E-35 +x^15* -3.215662487204 E-37 HalfExpz-1(178.0013796735+x) ~=        3814279.104760 +x^ 1*  118320.9653987 +x^ 2*  1583.539131951 +x^ 3*  11.98264489348 +x^ 4*  0.05651283677712 +x^ 5*  0.0001730408420441 +x^ 6*  0.0000003488473243786 +x^ 7*  0.0000000004627633828955 +x^ 8*  4.009594804049 E-13 +x^ 9*  2.251097354280 E-16 +x^10*  8.155859990453 E-20 +x^11*  1.910315886523 E-23 +x^12*  2.897614845893 E-27 +x^13*  3.466010556218 E-31 +x^14* -1.279953070466 E-34 +x^15*  1.112984355720 E-37