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 using sinh(x) ? tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 07/04/2011, 10:56 PM (07/02/2011, 10:32 PM)tommy1729 Wrote: in particular i still dont know why 1.729 i ^1.729 i ^ ... 0.5 + 0.5 i gives a circle and that bothers me. nobody is telling me ... anyway you might be intrested in the formula for the assumed invariant : for real z > 0 tommysexp_e(z,x) = tommysexp_e(z,inv(x)) then inv_e(z,x) = tommyQ_e^[2](z,x) tommyQ_e(z,x) = tommysexp_e( - z, - tommysexp_e(z,x)) sheldonison Long Time Fellow Posts: 630 Threads: 22 Joined: Oct 2008 07/05/2011, 01:09 AM (This post was last modified: 07/05/2011, 01:53 AM by sheldonison.) (07/04/2011, 10:56 PM)tommy1729 Wrote: (07/02/2011, 10:32 PM)tommy1729 Wrote: in particular i still dont know why 1.729 i ^1.729 i ^ ... 0.5 + 0.5 i gives a circle and that bothers me. nobody is telling me ... I can't follow your latest tommysexp equations right now, but the part about 1.729i is an approximation for b=1.712936...i, which is the base on the Shell Thron boundary region with real(b)=0. All bases on the Shell Thron boundary with an irrational period are going to give a circle (as long as the irrational period is a Brjuno number). The definition is the abs(log(fixed_point))=1, which leads to a neutral fixed point. Look up Siegel Disc on Wikipedia, for some background. Also, recent discussion has some really good papers referenced in it (on the Brjuno number). Each iteration circles around the h(z) function (inverse Schroder function), so how circular the path is depends on the magnitude of the initial point (0.5+0.5i)-L, which for this case is L~= 0.390842217 + 0.4586753i, so 0.5+0.5i is pretty close to the fixed point, so the path is fairly circular. 1.729*I is a very very slightly repelling fixed point, that will eventually diverge, abs(log(L))=1.00244, but it would take a long time. - Sheldon tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 07/05/2011, 12:06 PM (07/05/2011, 01:09 AM)sheldonison Wrote: (07/04/2011, 10:56 PM)tommy1729 Wrote: (07/02/2011, 10:32 PM)tommy1729 Wrote: in particular i still dont know why 1.729 i ^1.729 i ^ ... 0.5 + 0.5 i gives a circle and that bothers me. nobody is telling me ... I can't follow your latest tommysexp equations right now, but the part about 1.729i is an approximation for b=1.712936...i, which is the base on the Shell Thron boundary region with real(b)=0. All bases on the Shell Thron boundary with an irrational period are going to give a circle (as long as the irrational period is a Brjuno number). The definition is the abs(log(fixed_point))=1, which leads to a neutral fixed point. Look up Siegel Disc on Wikipedia, for some background. Also, recent discussion has some really good papers referenced in it (on the Brjuno number). Each iteration circles around the h(z) function (inverse Schroder function), so how circular the path is depends on the magnitude of the initial point (0.5+0.5i)-L, which for this case is L~= 0.390842217 + 0.4586753i, so 0.5+0.5i is pretty close to the fixed point, so the path is fairly circular. 1.729*I is a very very slightly repelling fixed point, that will eventually diverge, abs(log(L))=1.00244, but it would take a long time. - Sheldon ah , so its not a perfect circle ? it just approximates one ? didnt gottfried say it was a perfect circle ? i find 0.5 + 0.5 i - L still relatively large to give such a result , but it must be visual illusion and just a circle approximation then - if i understand you -. thanks. tommy1729 tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 11/18/2012, 11:41 PM (This post was last modified: 11/18/2012, 11:43 PM by tommy1729.) For those who are confused , let me explain the next big step about this. If we want a superfunction sexp(z) that is analytic for Re(z) > C >> 0 for some C then we need tommysexp(z) or sheldontommysexp(z) to have the same property. Lets call this C property. tommysexp(z) and sheldontommysexp(z) agree on the reals (converge speed may differ though ). Now tommysexp(z) seems to be only defined for the real z so it requires analytic continuation ( assuming it is possible ). However sheldontommysexp(z) IS periodic and might be analytic in that period. However to go beyond that period we will need analytic continuation because the C property forbids nonreal periodicity. ( again assuming it is possible ). The analytic continuation of tommysexp(z) and sheldontommysexp(z) are identical function. Hence either/both tommysexp and sheldontommysexp have property C. To prove property C of tommysexp(z) = sheldontommysexp(z) one can choose one of those 2 and investigate its analytic continuation/singularities/discontinu on the complex plane within the periodic strip for Re(z) > C. If it turns out there are no singularities in the periodic semistrip near the reals , then both functions have property C and are equal. If there are singularities , property C is broken however it might still have regions of analyticity. A natural boundary proves nowhere analytic ( because exp has fixpoints ). The reason why tommysexp and sheldontommysexp are equivalent is because of a base change like trick explained by sheldon in this thread. The periodicity is also explained here by him. Notice that a log(f(z)) with f(z) periodic remains periodic. Also notice that the C property is a uniqueness criterion ! To see this think of (my proof of) TPID 4. Or the 1 period function wobble of sexp(z). regards tommy1729 sheldonison Long Time Fellow Posts: 630 Threads: 22 Joined: Oct 2008 11/20/2012, 12:01 AM (This post was last modified: 11/21/2012, 04:04 PM by sheldonison.) (11/18/2012, 11:41 PM)tommy1729 Wrote: tommysexp(z) and sheldontommysexp(z) agree on the reals (converge speed may differ though ). Now tommysexp(z) seems to be only defined for the real z so it requires analytic continuation ( assuming it is possible ). However sheldontommysexp(z) IS periodic and might be analytic in that period. However to go beyond that period we will need analytic continuation because the C property forbids nonreal periodicity. ( again assuming it is possible ). $\text{tommysexp}_3(z)=\log^{[3]}(\text{2sinh}^{[z]}(\exp^{[3]}(1)))$ This is tommysexp generated with three iterated logarithms. This function is reasonably well behaved at the origin, with a radius of convegence of approximately 0.457. For reference, the taylor series coefficients are listed below. $\text{tommysexp}_4(z)=\log^{[4]}(\text{2sinh}^{[z]}(\exp^{[4]}(1)))$ But the next step, for n=4, is not so well behaved in the complex plane. The radius of convergence in the complex plane is about 0.03469, and by my count, there are 587507 singularities within a radius of 0.035. And its not just a matter of extending the function beyond the wall of singularities, because the function misbehaves inside the wall of singularities too. Here is where singularities occur, due to the log(0). $\text{2sinh}^{[z]}=n\pi i$ $\text{2sinh}^{[z+1]}=0$ The other interesting thing about the singularities is how quickly they fall off. At 99.999% of the singularity radius, $|\text{tommysexp}_4(z)-\text{tommysexp}_3(z)|<10^{-50}$. Its almost as if one can totally ignore the effect of the fourth iteration. And in fact the 4th iteration has no measurable effect on any taylor series coefficient one would use in normal computation. But, somewhere around the 1352620th taylor series term there is an abrupt transition. The $\text{tommysexp}_4(z)$ function starts behaving radically differently than the $\text{tommysexp}_3(z)$, as the taylor series terms finally become dominated by the wall of nearby singularities. By the way, estimating the 1.35 millionth taylor series coefficent for a function is a very delicate calculation involving algorithms to make approximations cauchy integrals at carefully picked radius's of $s(z)-\log(e^{s(z)}-e^{-s(z)})=\sum_{n\to\infty}-e^{-2n s(z)}$, where $s(z)=\text{2sinh}^{[z]}$. After picking an appropriate approximation radius, I can estimate the log of the desired taylor series coefficient. It helps that usually the different values of "n" in the sum can be treated independently. I have a pari-gp program, but haven't posted the details, which are complicated. The algorithm works for both the 2sinh(z) superfunction, as well as the exp(z-1) superfunction. One final thought. $\text{tommysexp}_3(z)$ has a radius of convergence of ~0.457 which can be seen in the taylor series coefficients below. It is not as well behaved as sexp(z) in the complex plane. The interesting thing is that $\text{tommysexp}_4(z)$ has the same first 40 taylor series coefficients listed below, accurate to millions of decimal digits. This would also be true of the first million or so terms, until the transition. So, in addition to misbehaving at the 1.35 millionth taylor series term, $\text{tommysexp}_4(z)$ still has all smaller taylor series coefficients dominated by the singularities in $\text{tommysexp}_3(z)$. $\text{tommysexp}_5(z)$ would have a radius of convergence of less than 10^-7, that would effect uncountably large taylor series terms due to a wall of uncountably many nearby singularities. There's nothing particularly special about doing the approximations for tommysexp(0), and similar misbehavior would be expected for any value of tommysexp(z). I could post the methods I used for these approximations which might lead to a rigorous proof, but the general problem is complicated. You need to show the taylor series terms eventually grow faster than any radius of convergence. The difficulty is coming up with a rigorous language to express appropriate approximations of superexponentially large taylor series terms. Even so, I am convinced that tommysexp(z) is nowhere analytic, even though it is infinitely differentiable at the real axis. - Sheldon added pretty graph showing accuracy of tommysexp3 taylor series terms, and switchover to tommmysexp4 taking over near 1.35 millionth taylor series term     Code:Taylor series coefficients for tommysexp(0) a0=   1.00000000000 a1=   1.09146536077 a2=   0.273334906394 a3=   0.215218479242 a4=   0.0652715037680 a5=   0.0391656564309 a6=   0.0171521314068 a7=   0.0117058806325 a8=   0.00471958861559 a9=   0.00123667589678 a10= -0.00226288150336 a11=  0.00321559868096 a12= -0.00820154271015 a13=  0.00218777707040 a14=  0.0477372146016 a15= -0.117247563749 a16= -0.0788849220733 a17=  0.883038307996 a18= -0.610166815881 a19= -5.10470797278 a20=  7.81612106347 a21=  28.6479580355 a22= -60.0481521891 a23= -173.314801252 a24=  382.323334609 a25=  1156.14068365 a26= -2075.65103517 a27= -8124.15769733 a28=  8589.21772341 a29=  56026.7180961 a30= -10973.8915911 a31= -353646.843999 a32= -264336.726813 a33=  1880963.01782 a34=  3669078.97541 a35= -7012765.98983 a36= -30701470.8277 a37=  1548864.00646 a38=  184151139.583 a39=  254566511.140 tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 11/20/2012, 10:53 PM (11/20/2012, 12:01 AM)sheldonison Wrote: $\text{2sinh}^{[z]}=n\pi i$ What is wrong with that ? We can compute log of that ? sheldonison Long Time Fellow Posts: 630 Threads: 22 Joined: Oct 2008 11/20/2012, 11:01 PM (This post was last modified: 11/20/2012, 11:03 PM by sheldonison.) (11/20/2012, 10:53 PM)tommy1729 Wrote: (11/20/2012, 12:01 AM)sheldonison Wrote: $\text{2sinh}^{[z]}=n\pi i$ What is wrong with that ? We can compute log of that ?if $\text{2sinh}^{[z]}=n\pi i$ then $\text{2sinh}^{[z+1]}=e^{n\pi i}-e^{-n\pi i}=0$ and therefore there is a singularity. $\log(\text{2sinh}^{[z+1]})=\log(0)$ - Sheldon tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 11/20/2012, 11:32 PM I think the same applies to the base change ... I do not immediately know a way around it. Although those singularities maybe vanish from analytic continuation if they were in an undefined region before analytic continuation. sheldonison Long Time Fellow Posts: 630 Threads: 22 Joined: Oct 2008 11/20/2012, 11:46 PM (This post was last modified: 11/20/2012, 11:53 PM by sheldonison.) (11/20/2012, 11:32 PM)tommy1729 Wrote: I think the same applies to the base change ... I do not immediately know a way around it.Yes! The exact same thing applies to the base change. Iterating $s(z)=\exp(z-1)$ is exactly analogous to iterating tetration base $\eta=\exp(1/e)$. Then, taking the logarithm as the first step in the base change results in singularities whenever $s^{[z]}=2n\pi i$, where $\log(s^{[z+1]})=\log(e^{2n\pi i}-1)=\log(0)$ Actually, for every singularity in the base change, there are two singularities for 2sinh(z), and the singularities for $2n\pi i$ for both functions are very nearly in the same place in the complex plane! - Sheldon tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 11/28/2012, 06:59 PM (This post was last modified: 11/28/2012, 07:40 PM by tommy1729.) I would like to point out that tommysexp for bases $>= e^2$ is problemfree for real values. The reason is that for bases $>=e^2$ there is a unique (entire) superfunction for the 2sinh ( example base $e^2$ thus $e^{2x} - e^{-2x}$). And that is true because 2sinh(2x) has only one complex fixpoint : 0. So for these bases tommysexp is valid , has uniqueness and existance and satisfies all desired properties with the possible exception of analytic. However it is Coo. While doing basechange I prefer to use tommysexp base exp(2). Lets call that basechangetommy. I suspect basechangetommy and tommysexp for bases larger than exp(2) to have similar properties. I even considered them to be equal once , although I now keep it as a sort of vague closeness idea. I am considering properties and might even call it my new pet ideas in tetration. I believe in intresting properties for these functions. Im also considering replacing the logs with functions that are asymptotic to logs but entire. ( an idea that I came up with together with mick , the guy who posts on MSE ) However that is quite complicated and not well understood at the moment. For instance it is unknown if this is the sought of analytic continuation or a totally different function ? « Next Oldest | Next Newest »

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