• 3 Vote(s) - 4.33 Average
• 1
• 2
• 3
• 4
• 5
 Searching for an asymptotic to exp[0.5] sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 05/16/2014, 07:27 AM (This post was last modified: 05/16/2014, 08:25 AM by sheldonison.) (05/15/2014, 09:49 PM)tommy1729 Wrote: .... As for f(z) and f(f(z))/exp(z) being entire ... I repeat my questions/remarks : What does the weierstrass product of those look like ?? ..... Notice f(z) = exp(g(z)) ( 1 + a_1 z)(1 + a_2 z) ... implies f(z) has no positive real zero's but an infinite amount of negative zero's ( the a_n ). Also f(z) = exp(g(z)) ( 1 + a_1 z)(1 + a_2 z) ... could be wrong ! The weierstrass product form could be more complicated ! Is it ? Similar questions for f(f(z)/exp(z). Since f(f(z))/exp(z) must be close to 1 for positive real z , it follows it must go to oo for other part of the complex plane. This post is just some pretty pictures, with all graphs from from -20 real to +20 real, and -5 imag to +20 imag with grid lines every 5 units. The first graph is the graph or the ratio that I wanted to create, which looks like I expected it would. Let's call the asymptotic half iterate function f (it needs a better name). This graph is f(f(z))/exp(z)-1, showing the very sharp switch from the region of good convergence, for positive reals, where f(f(z))/exp(z)-1 is very nearly zero, which is black. Here, I used the best approximation I have so far, which turns out to match the exponential function by a little better than 99% in the region of interest where the image is black. The boundary is pretty sharp. The conjecture is that there is a well defined boundary is exactly where f(z) takes on negative values, with imag(f(z))=0. At that boundary, the conjecture is that |f(f(z))/exp(z)-1|~=1. The 99.2% accuracy in this region may be as good as f(f(z))/exp(z) can be. With a 300 term Taylor series, accuracy peaks at 99.95% at z=2800, beyond which I would need to generate more Taylor series terms for convergence. f(f(pi i)=-1.021 + 0.053i, vs the exp(Pi*I)=-1.     The next graph is f(z), the asymptotic half iterate itself, using the same grid coordinates. You can see zeros for f on the real axis, as black dots, at -0.71, -4.26, and -15.21. The pattern goes on forever, as f grows at the negative real axis, oscillating between positive and negative.     The next graph is f(f(z)), same coordinates. You can clearly see the parts of the complex plane where f(f(z)) mimics the exponential, and the parts of the complex plane where it doesn't mimic the exponential function, which explains the first graph. And you can see the zeros, of f(f(z)), which are black dots. I'm pretty sure all the zeros of f(z) are on the negative real axis, and exp(z) has no zeros, so if you connect the black dots not on the real axis, this is where f(f(z))/exp(z) has to start to diverge from its asymptotic value of 1, and this is where f(z) is at the negative real axis. btw; thanks for your comments. The Weierstrass product form sounds interesting; I haven't worked with it before, so it might be a lot of learning before I could generate any useful results. - Sheldon     tommy1729 Ultimate Fellow Posts: 1,645 Threads: 369 Joined: Feb 2009 05/16/2014, 09:36 PM I was wondering about another asymptotic to exp^[0.5](x). For all n : a_n >=0. f(x) is an asymptotic. f(x) = a_0 + a_1 x - a_2 x^2 - a_3 x^3 + a_4 x^4 + a_5 x^5 - ... Where the periodic pattern is {+,+,-,-}. And also in the behaviour of f(-x) + f(x). The theory of " series multisections " has a lot to say about this I assume. Need further investigation. Yeah Im annoying, I know regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,645 Threads: 369 Joined: Feb 2009 05/16/2014, 10:16 PM I havent given this much consideration yet but let lim x->+oo : C = lim ( 1 - ( D exp_a^[b](x) / exp_a^[b](x) ) ) / ( 1 - ( D exp_c^[d](x) / exp_c^[d](x) ) ). For a,c > eta ( = exp(1/e)) [a and c are bases] For 0.25 < b,d < 0.5. (a-c)^2 + (b-d)^2 > 0. C > 0. (limit exists and is larger than 0). Is that true ? Do we know numerical examples of such a,b,c,d,C ? It seems - at first sight - that l'hopitals rule cannot be used. regards tommy1729 sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 05/18/2014, 06:14 PM (This post was last modified: 05/18/2014, 10:12 PM by sheldonison.) All of the zeros of the asymptotic exp[0.5] are roughly in correspondence with -exp(hn), on the negative part of the real axis. A better approximation might be the real crossings of the branch of the Kneser half exponential, that we usually don't look at, at the negative real axis, where this branch is arrived by by circling counter clockwise around L, and this branch is not real valued at the real axis. So, anyway, lets say we have these zeros of the asymptotic half iterate in an infinite list. Can we recover the asymptotic half iterate with the Weiestrass factorization theorem? - Sheldon JmsNxn Ultimate Fellow Posts: 895 Threads: 110 Joined: Dec 2010 05/22/2014, 12:16 AM (05/18/2014, 06:14 PM)sheldonison Wrote: So, anyway, lets say we have these zeros of the asymptotic half iterate in an infinite list. Can we recover the asymptotic half iterate with the Weiestrass factorization theorem? - Sheldon If you have some more to add on top of the theorem. All it will give you is the half iterate upto multiplication by an entire never zero function. These are usually solved using additional properties of the function. I'm sure there must be a way. sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 05/22/2014, 07:08 AM (05/22/2014, 12:16 AM)JmsNxn Wrote: (05/18/2014, 06:14 PM)sheldonison Wrote: So, anyway, lets say we have these zeros of the asymptotic half iterate in an infinite list. Can we recover the asymptotic half iterate with the Weiestrass factorization theorem? - Sheldon If you have some more to add on top of the theorem. All it will give you is the half iterate upto multiplication by an entire never zero function. These are usually solved using additional properties of the function. I'm sure there must be a way.Thanks James, It looks like the "multiplication by an entire never zero function" is exp(g(z)), where g(z) is any entire function. For our purposes, g(z)=k might work, if that is a legal choice. That would just be multiplication by a constant. Multiplication by exp of anything else would probably grow faster than the half iterate of exponentiation, which by definition grows slower than exp(z). - Sheldon tommy1729 Ultimate Fellow Posts: 1,645 Threads: 369 Joined: Feb 2009 05/22/2014, 08:31 AM (This post was last modified: 05/22/2014, 08:34 AM by tommy1729.) (05/22/2014, 07:08 AM)sheldonison Wrote: (05/22/2014, 12:16 AM)JmsNxn Wrote: (05/18/2014, 06:14 PM)sheldonison Wrote: So, anyway, lets say we have these zeros of the asymptotic half iterate in an infinite list. Can we recover the asymptotic half iterate with the Weiestrass factorization theorem? - Sheldon If you have some more to add on top of the theorem. All it will give you is the half iterate upto multiplication by an entire never zero function. These are usually solved using additional properties of the function. I'm sure there must be a way.Thanks James, It looks like the "multiplication by an entire never zero function" is exp(g(z)), where g(z) is any entire function. For our purposes, g(z)=k might work, if that is a legal choice. That would just be multiplication by a constant. Multiplication by exp of anything else would probably grow faster than the half iterate of exponentiation, which by definition grows slower than exp(z). - Sheldon "probably" is not so logical as you might think. Remember the fake log for instance. And there are many other " counter-intuitive " functions. Im carefull with the word counter-intuitive because i do not " believe " in that word ; everybody has a different intuition. Another " counter-intuitive " (entire) function is this one : $f(z)=\int_0^\infty \frac{e^{zt}}{t^t}\mathrm{d}t$ this entire f(z) is bounded outside the strip $Im(z)=< \pi$. ( this can be shown with contour integration or substitution ) So its almost everywhere constant. And one could take the fake log of this f(z) once or twice too. Or f(fake log(z)). The fact that a function has no 0 does not limit exp(g(z)) much. So exp(g(z)) can be very different from exp(polynomial). Therefore the "probably " idea is not very solid. On the other hand : if the truncated Taylor series can be approximated well with a truncated product (1+b_n x) then its likely that g(z) is indeed a constant. This is also likely because of the simplification that all derivates are positive and it might be possible to find a recursion to solve for b_n ... So I think truncation and induction will show us the way. Thinking ... regards tommy1729 sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 05/22/2014, 10:16 PM (This post was last modified: 05/23/2014, 04:24 AM by sheldonison.) (05/22/2014, 08:31 AM)tommy1729 Wrote: .... Another " counter-intuitive " (entire) function is this one : $f(z)=\int_0^\infty \frac{e^{zt}}{t^t}\mathrm{d}t$ this entire f(z) is bounded outside the strip $Im(z)=< \pi$. ( this can be shown with contour integration or substitution ) So its almost everywhere constant. .... The fact that a function has no 0 does not limit exp(g(z)) much. So exp(g(z)) can be very different from exp(polynomial). ... On the other hand : if the truncated Taylor series can be approximated well with a truncated product (1+b_n x) then its likely that g(z) is indeed a constant. This is also likely because of the simplification that all derivates are positive and it might be possible to find a recursion to solve for b_n ... So I think truncation and induction will show us the way. Thinking ... regards tommy1729Hey Tommy, Much to learn. I bought Conway's graduate complex analysis book, which includes the Weierstrass factorization theorem. I will add it to my large and growing collection of graduate math books, to supplement my somewhat weak formal math education (BSEE). For the truncated product equation, where z_n are the zeros of the Asymptotic half exponential, is this correct? $(1+b_n x)$ $b_n=\frac{-1}{z_n}$ update I tried Tommy's factor, which works surprisingly well. So I guess this is the Weierstrass factorization of the asymptotic half exponential. $f(z)=\text{half}(0)\times\prod_{n=1}^{\infty}(1-\frac{x}{z_n})$ Here are the first 10 zeros of the asymptotic half exponential. I also have an approximation for the zeros of the asymptotic function, in terms of Kneser's half exponential. Code:1 -0.71176762728441566602682009931906 2 -4.2615192715738731444168590500003 3 -15.214306922947794235707847543680 4 -43.768867332590888558785594556450 5 -109.77901963743164514158613984269 6 -229.50542893029538607241128473700 7 -458.89117411149970236796071783155 8 -861.01099146084982202350824423190 9 -1534.9231313380922901088132019252 10 -2623.2160464901874760015847745797- Sheldon tommy1729 Ultimate Fellow Posts: 1,645 Threads: 369 Joined: Feb 2009 05/23/2014, 10:53 PM Yes yes that is what I meant. I anticipated your approximation for the zeros of the asymptotic function, in terms of Kneser's half exponential. The Weierstrass product form does not only need a computation but also a proof ofcourse. Too illustrate why, I gave those counterintuitive functions as examples. A vague possible conjecture could be that exp(g(z)) is the " integer growth rate " for " most " entire functions. Im afraid to conjecture it , but the idea fascinates me and perhaps plays a role in a proof. I remind you that 2 posts are still completely unadressed, although ofcourse you might not have the time or habit of doing more than one thing at a time. As for your education , the fun thing is that " research is a hard - but open book - exam ". And you are certainly a respected member here. Another intresting/counterintuitive remark about product forms ( weierstrass or other ) for entire functions is the fact that for instance exp(x) can be written in the form : (1+c_n x^n) valid for abs(x)<1. That seems surprising and fascinating considering that exp(x) has a very different ( and dull ) weierstrass product form. This can all be generalized like almost forever , but the point is to warn about expansions that might only work locally and are thus not the weierstrass expansion ! Another "counter-intuitive" result. Btw first mentioned here by Gottfried and written about on his page : " dream of a sequence " link : http://go.helms-net.de/math/musings/drea...quence.pdf - This also reminds me of the classic pentagonal theorem of euler ofcourse - which might turn into an argument against truncation ... ... and also continued fractions ... id better stop here. But some day you will see continued fractions regards tommy1729 sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 05/25/2014, 03:00 PM (This post was last modified: 05/25/2014, 03:40 PM by sheldonison.) (05/23/2014, 10:53 PM)tommy1729 Wrote: I anticipated your approximation for the zeros of the asymptotic function, in terms of Kneser's half exponential. The Weierstrass product form does not only need a computation but also a proof ofcourse. Too illustrate why, I gave those counterintuitive functions as examples.A proof requires an analytic function, with a definition. Right now, all I have is a series of constructive algorithms to generate an asymptotic half Taylor series, along with some estimates for the error terms. Perhaps one can formalize a definition in terms of how well f(f(z)) converges to exp(z), or to the Kneser half iterate. The best algorithm is Version IV, which uses a Cauchy circle integral approximation of a function with a fairly large discontinuity at the negative real axis, which causes us to carefully choose where to evaluate the Cauchy integral to minimize the numeric impact of the jump discontinuity at the negative real axis. A picture showing the discontinuity at the negative real axis is at the end of this post. $a_n = \frac{\exp( \exp^{0.5}(h_n) - n h_n)}{2\pi } \int_{-\pi i}^{\pi i} \exp ( b_2 x^2 + b_3 x^3 + b_4 x^4 + ... ) dx \;\;$ this a_n is the most accurate entire asymptotic Taylor series! b_n are the coefficients of the Kneser $\exp^{0.5}(x)$ Taylor series at h_n, and for the definite integral, x=h_n Now from the Kneser function itself, one can generate a Weiestrass zero approximation half iterate, which represents yet another version, it appears to be not quiet as good as version IV, but this is primarily because even though the approximation for the zeros, as compared to version IV, is very good, for the larger zeros, is is not for the smaller zeros. This approximation gives the 10th zero as -2623.212352003 as compared with -2623.216046490, for version IV, from my previous post. I am having trouble formalizing why this approximation for the zeros would approximately match the zeros of the version IV approximation, even though it does quite well. This approximation says the zeros are where the real part of the Kneser half iterate at the negative axis goes to zero. $p(z) = \Im(\exp^{0.5}(\pi i + z))$ $z_n = \exp(\pi i + p^{-1} (n-\frac{1}{2})\pi)))\;\;\;$ equation used for zeros of a Weiestrass half iterate approximation The basic problem is how to remove the singularities and the the discontinuity at the negative real axis for the Kneser half iterate. Below, I show the Kneser half iterate, from -10 to 10, with grids every 2 units, singularities at L and L* with cutpoints on a circle of abs(L), extended on the real axis to minus infinity; this is different than the normal way to draw the function with the cutpoints extended so the function is real valued at the real axis. Relatively speaking, this version of the Kneser half iterate is much larger in absolute magnitude at the positive reals, than at the negative reals, which is what the approximations are all based on. Version IV does "Cacuhy type integrals" on this function at carefully chosen radii to minimize the discontinuity. The Weiestrass approximation takes the negative reals of this function as the zeros of an entire function. Meanwhile, I have a new version I am experimenting with that will greatly reduce the discontinuity at the negative real axis, that may allow approximating both the Weiestrass zeros, and the error term for the convergence to the Kneser half iterate... but there will still be a discontinuity cut point at |L|; still working. « Next Oldest | Next Newest »

 Possibly Related Threads… Thread Author Replies Views Last Post Tetration Asymptotic Series Catullus 18 881 07/05/2022, 01:29 AM Last Post: JmsNxn Using a family of asymptotic tetration functions... JmsNxn 15 6,522 08/06/2021, 01:47 AM Last Post: JmsNxn Reducing beta tetration to an asymptotic series, and a pull back JmsNxn 2 1,426 07/22/2021, 03:37 AM Last Post: JmsNxn A Holomorphic Function Asymptotic to Tetration JmsNxn 2 1,787 03/24/2021, 09:58 PM Last Post: JmsNxn An asymptotic expansion for \phi JmsNxn 1 1,489 02/08/2021, 12:25 AM Last Post: JmsNxn Merged fixpoints of 2 iterates ? Asymptotic ? [2019] tommy1729 1 4,032 09/10/2019, 11:28 AM Last Post: sheldonison Another asymptotic development, similar to 2sinh method JmsNxn 0 4,327 07/05/2011, 06:34 PM Last Post: JmsNxn

Users browsing this thread: 2 Guest(s)