Nowhere analytic superexponential convergence  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: Nowhere analytic superexponential convergence (/showthread.php?tid=578) Pages:
1
2

Nowhere analytic superexponential convergence  sheldonison  01/20/2011 I found some surprisingly close similarities in the behavior of tommysexp and the base change function, and may have made some progress on why both are probably nowhere analytic functions. In particular, my hypothesis is that is also a nowhere analytic function, whose convergence in some sense tracks the base change, and tommysexp function. I'm still working out the details, some of which are included today, and some of which will be in a future post. I calculated the taylor series and error terms and nearest singularities for the base change eta equations, see post, and for the tommysexp equations, see my recent post. Both iterate logarithms from one real valued superexponential superfunction, to generate a particular different real valued base(e) sexp(z) superexponential. The working assumption is that the resulting sexp(e) base change is nowhere analytic. But the intermediate approximations are defined in the complex plane. So here, I generated the series, for both approximations. The similarities between the two are somewhat striking, and lead to some new ideas to explore about nowhere analytic superexponential convergence. Recall, . So the first two logarithms in the base change are sort of trivial. For the following, I'm comparing five logarithms for the base change, with three logarithms for the 2sinnh superfunction.
The next iteration delta is defined as Let z=sexp(x). Using the chain rule, and first order derivatives, one can derive that the the next iteration convergence delta equations for cheta, and for 2insh superf are approximately the following, which has also been verified empirically. nextdelta_cheta_e(z)= nextdelta_tommysexp(z)= The tommysexp next iteration delta term is roughly the square of the cheta base change next iteration delta term. But that squaring becomes negligible compared to supperexponential converging terms. For both, I suspect this superexponential convergence at the real axis makes it impossible to model the nextdelta functions with a taylor series, which matches the singularities approaching arbitrarily close to the real axis. Below, you can see how very very tiny these next iteration deltas are, with a number like 10^165653 for the next iteration delta at z=0. For z>0.05, parigp overflows, and can't print the the next iteration delta in scientific form. This may also lead to a way to prove that these functions are nowhere analytic, since they are both very close to an analytic function with a radius of 0.46, but adding the tiny nextdelta dramatically reduces the taylor series convergence, because the nextdelta term has such quickly growing derivatives. And for the nextdelta after that, the nextdelta is tinier still, approximately 1/sexp(z+5), with a correspondingly faster growith in the derivatives. This leads to a natural question. Is this superexponential reciprocal summation also , but nowhere analytic? Interestingly, the intermediate terms for this summation do not have any actual singularities at radius 0.46, and then at radius 0.035, but it nonetheless behaves like it has a singularities at those radii. More comments comming in a future post! And it has some of the same convergence properties, since for large enough values of sexp(z), one can say that in some sense all of the following are approximately equal, . Then in that same sense, for tomysexp, and for the base change, and for the superexponential reciprocal summation, the nextdelta is approximately the same. nextdelta(n+4) approximation: 1/sexp(z+4) nextdelta(n+5) approximation: 1/sexp(z+5) nextdelta(n+6) approximation: 1/sexp(z+6) nextdelta(n+7) approximation: 1/sexp(z+7) Here are some computed results, with the tommysexp function on the left, and the base change function on the right. Results are included for closest singularities, next iteration delta, taylor series, and taylor series accuracy. The Taylor series for both was sampled around at a radius of 0.4 around about the origin, z=0, with 200 samples. Both have a singularity at a radius of 0.46. For the superfunction of 2sinh, the singularities occur whenever the superfunction is first equal to n*2*Pi*i. For the base change, the singularities occur where the superfunction of is first equal to e*(1+n*2*Pi*i).  Sheldon Code: tommysexp=superfunction(2sinh) cheta base change, upper sexp(eta) RE: Nowhere analytic superexponential convergence  tommy1729  01/21/2011 i will explain some things later on ... RE: Nowhere analytic superexponential convergence  sheldonison  01/26/2011 (01/20/2011, 05:23 PM)sheldonison Wrote: .... In particular, my hypothesis is that is also a nowhere analytic function...The goal of this post is to show that f(x), the superexponential reciprocal sum is not analytic on the complex plane, even though it converges superexponentially fast at the real axis. This post assumes that f(x) is , and that all of the derivatives of f(x) converge via the limit equation. First, assume that the sum is analytic. Then, around any point "x" along the real axis, there must exist r, which is the radius of a circle, for which all points inside that circle, such that , for the nth approximation term, 1/sexp(x+n+z)<1. After choosing any radius, no matter how small, we can choose a value of n, such that the approximation of f(x) has n terms, and show that for the nth term, , there exists a counter example, with z<r, and . Moreover, for the n+1 term, the radius will be superexponentially smaller still. Even though the superexponential reciprocal summation converges superexponentially quickly at the real axis, it does not converge in the complex plane, which means the function is only defined at the real axis. Also, each individual approximation is analytic, and there are no singularities, so long as . But as n increases, gets arbitrarily small, and the function converges at the real axis. But, as n increases, the radius for which holds also gets arbitrarily small, which is the exact opposite from what would be expected if the function were defined in the complex plane. Here I give an empirical (unproven) estimate of what the radius of z is. There exists a value of z, where , and where , as long as x>3.5 or so. This gives a way to calculate the upper bounds for the integer value of n to use for any particular radius r. n>slog(1/r)+2x will work, as long as slog(1/r)>=1.5. Summation terms with n greater than this value will not be stable in the complex plane within the radius r, because the term will be greater than 1, which is contrary to the assumption. Moreover, for any value of r, no matter how small, there is a value of n for which the summation not converging in the complex plane. Now, a few graphs. The first graph is a graph of the superexponential reciprocal summation, and its first three derivatives. For x>2.5, f(x)<2E78. Beginning with the third derivatives, oscillation is visible. For higher derivatives, the oscillation becomes more and more pronounced, and larger in magnitude. [attachment=845] The next chart, is a contour plot of where where sexp(z+1)=1. If sexp(z+1)=1 then it is also true that the absolute value of the reciprocal=1, 1/sexp(z+1)=1. This contour is also the contour where . This chart has . This contour is also where , where . If z is chosen from a point on this curve, then . Between this curve and the real axis, 1/sexp(z+1)<1. For every point on the real axis, if the radius is bigger than some value, the circle tangentially intersects the curve, and the radius of that circle gets arbitrarily small as x increases. [attachment=846] Then, going back to the superexponential reciprocal summation, in the complex plane the individual terms don't converge, because there isn't a radius for which all of the approximation terms of . This curve tracks very closely to the singularity points for the base change sexp(z), which tracks the 2sinh(z) singularities, and I could post some similarities later. I suspect there is some way to show that all three nowhere analytic functions behave in some underlying similar manner, and that perhaps this can help to prove that they are all nowhere analytic. As an example, consider f(0.5). The n=3 approximation term is 1/sexp(3.5+z). If z=0, 1/sexp(3.5)=1.6E78, which is very small, and convergence looks very good at the real axis. But, in the complex plane, the curve 1/sexp(3.5+z), is not so well behaved. In particular 1/sexp(3.5794+0.163i)=1.0050.174i, which is>1. This is approximately the closest point. The radius is given by r=abs(3.5794+0.163i3.5)=0.1813. The slog(1/r)+2 approximation equation gives 3.538, which is pretty good, compared to 3.5. Here is a graph of the contour of the n=3 term, a circle of radius r=0.181, for f(z)=1/sexp(3.5+z). For abs(z)<0.181, the 1/sexp(3.5+z)<1. First, I graph the circle from z=r*exp(0) to z=r*exp(Pi). Then I zoom in on z=r*exp(0.9...1.4). Finally, I graph the absolute value, 1/sexp(x+z) where z=r*exp(0.9..1.4). These are contour graphs, where red=real and green=imaginary. The third graph shows the approximately Gaussian decay of the 1/sexp(3.5+z), on the contour of the circle. The abs(1/sexp(3.5+z) quickly decays to 1E10, and will continue to decay from there. In some ways, the rapid growth from 1/sexp(3.5)=1E78 resembles the way a singularity might behave. As 1/sexp(z) superexponentially decays, the Gaussian distribution falls over a smaller and smaller region of the circle. [attachment=847] Also, here are the first 50 derivatives, for 1/sexp(2.5+z) compared with 1/sexp(3.5+z). For the first 43 derivatives, the 1/sexp(3.5+z) terms are smaller, but after that, the 1/sexp(3.5+z) terms take over. Notice that 1/sexp(3.5) is only 1.6E78 at the real axis; which is a very small correction. Code: n 1/sexp(2.5) 1/sexp(3.5) RE: Nowhere analytic superexponential convergence  tommy1729  01/26/2011 i would like to point out : i agree that the base change and my sinh method function are probably  without extensions  Coo but not complex analytic. but the point is my sinh method is Coo and REAL  analytic. Look : since log log ... exp exp ... (z) is only REAL  analytic and not complex analytic , but we can extend log log ... exp exp ... (z) simply to id(z) BECAUSE it is REAL analytic , and then it BECOMES complex  analytic. RE: Nowhere analytic superexponential convergence  bo198214  01/28/2011 (01/20/2011, 05:23 PM)sheldonison Wrote: I found some surprisingly close similarities in the behavior of tommysexp and the base change function, and may have made some progress on why both are probably nowhere analytic functions. Ya, actually this method would also work for a lot of other functions than or , I guess all these are nowhere analytic on the real line (and produce superexponentials). (01/26/2011, 11:17 PM)tommy1729 Wrote: i agree that the base change and my sinh method function are probably  without extensions  Coo but not complex analytic. Tommy, it seems you are not familiar with the definitions. "Realanalytic" means analytic at a certain interval of the real axis and the function returning real values there. "Analytic" at a point means there is a powerseries development with a nonzero convergence radius. If this is the case then there is disk around this point in the complex plane where the function is analytic/holomorphic. A term like "complexanalytic" does not exist. When one says "analytic" there must be including a statement of the domain, where it is analytic. A function that is analytic in the whole complex plane is called entire. So "nowhere analytic" on the real axis means: all the powerseries developments at the real axis have 0 convergence radius, which indicates a really strange function. RE: Nowhere analytic superexponential convergence  tommy1729  01/29/2011 ofcourse Bo ! i just wrote *complex*  analytic to informally point out the difference with realanalytic. of course i know : " "Realanalytic" means analytic at a certain interval of the real axis and the function returning real values there. "Analytic" at a point means there is a powerseries development with a nonzero convergence radius. If this is the case then there is disk around this point in the complex plane where the function is analytic/holomorphic " i thought it was clear i was speaking informally ... kind of embarrassing for me :/ now knowing that , you might wanna read my post again ? as i understood sheldon , he means " nowhereanalytic " rather than " nowhereanalytic " on the real axis. tommysexp is realanalytic. RE: Nowhere analytic superexponential convergence  bo198214  01/29/2011 (01/29/2011, 12:03 AM)tommy1729 Wrote: i just wrote *complex*  analytic to informally point out the difference with realanalytic. Sorry, tommy your post makes no sense then. You say its not analytic, but realanalytic. Its not analytic where? If it is real analytic then it is analytic on a neighborhood of the real axis/interval. Quote:as i understood sheldon , he means " nowhereanalytic " rather than " nowhereanalytic " on the real axis. The function is only defined on the real axis, of course he speaks about the real axis. Quote:tommysexp is realanalytic. I dont think so. Everything is pointing to the opposite: nowhere analytic on the real axis. This is because the sequence of functions gets singularities nearer and denser around the real axis (convergence radius shrinking to 0). This is no proof, but makes it quite unlikely that the limit function has somewhere nonzero convergence radius. RE: Nowhere analytic superexponential convergence  tommy1729  01/29/2011 well im sorry too Bo. but it seems you missed the point. lim n > oo log^[n] ( exp^[n] (x) ) also doesnt have a somewhere nonzero convergence radius. but clearly on the real line f(x) = lim n > oo log^[n] ( exp^[n] (x) ) is realanalytic because it simply reduces to f(x) = x for real x , which is clearly realanalytic. RE: Nowhere analytic superexponential convergence  mike3  01/29/2011 (01/29/2011, 03:26 PM)tommy1729 Wrote: well im sorry too Bo. No, it does have a nonzero radius, at every step of the way, since on the real line for , it equals . RE: Nowhere analytic superexponential convergence  tommy1729  01/29/2011 well according to Bo it is not realanalytic and it doesnt have a radius , and according to mike it does have a radius. a radius is usually considered a radius of a circle. for reals we usually talk about intervals. so opinions and/or terminology differs , i hope at least everyone agrees on : on the real line f(x) = lim n > oo log^[n] ( exp^[n] (x) ) is realanalytic because it simply reduces to f(x) = x for real x , which is clearly realanalytic. and for realanalytic i prefer to say : analytic on an interval , rather than a radius , for imho a nonzeroradius is for analytic taylor series  analytic on on a disk on the complex plane with nonzeroradius. 