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 closed form for regular superfunction expressed as a periodic function sheldonison Long Time Fellow Posts: 645 Threads: 22 Joined: Oct 2008 08/27/2010, 02:09 PM (This post was last modified: 09/03/2010, 01:24 PM by sheldonison.) I edited the title of this post to "closed form for regular superfunction..." Normally, we would look at the regular superfunction (base e) as a limit, $\lim_{n \to \infty} \exp^{[n]} (L+L^{z-n})$ but it has a period of 2Pi*i/L, so perhaps it could also be expressed as an infinite sum of periodic terms, $\sum_{n=0}^{\infty}a_n\times L^{( n*z)}$ $a_0=L$ and I think $a_1=1$. Perhaps there is a closed form limit equation for the other a_n terms in the periodic series? - Sheldon sheldonison Long Time Fellow Posts: 645 Threads: 22 Joined: Oct 2008 08/28/2010, 08:44 PM (This post was last modified: 08/28/2010, 10:46 PM by sheldonison.) (08/27/2010, 02:09 PM)sheldonison Wrote: .... Perhaps there is a closed form limit equation for the other a_n terms in the periodic series? - SheldonOriginally, I intended to solve it mathematically, with limit equations of some sort, but I gave up. Anyway, I brute forced the numerical solutions using complex Fourier analysis, and the results looked good. $a_0=L$ $a_1=1$ $a_2=1/(2*L-2)$ I suspect a_3 would be the reciprical of a polynomial in L, L^2, but I wasn't able to find the pattern... yet. I will continue to pursue both limit equations, and numerical results .... Here are the numerical results. This would provide another way of calculating the regular superfunction. 0.318131505204764 + 1.33723570143069*I 1.00000000000000 - 2.60599763775608 E-46*I -0.151314897155652 - 0.296748836732241*I -0.0369763094090676 + 0.0987305443114970*I 0.0258115979731401 - 0.0173869621265308*I -0.00794441960244236 + 0.000579250181689956*I 0.00197153171916544 + 0.000838273147502224*I -0.000392010935257457 - 0.000393133164925080*I 0.0000581917506305269 + 0.000119532747356117*I -0.00000315362731515909 - 0.0000302507270044311*I -0.00000144282204032780 + 0.00000712739202459367*I 0.000000659214290634412 - 0.00000152248373494640*I -0.000000194922185012021 + 0.000000284379660774925*I 0.0000000534780813335645 - 0.0000000461525820619042*I -0.0000000140401213835816 + 0.00000000762603302713942*I 0.00000000315342989238929 - 0.00000000146720737059747*I -0.000000000591418449135739 + 0.000000000254860555536866*I 1.07938974205158 E-10 - 2.18783515598918 E-11*I -2.47877770137887 E-11 - 1.92092983484722 E-12*I 6.07224941506231 E-12 + 1.28233399551471 E-13*I -1.11638719710692 E-12 + 2.71214581539618 E-13*I 1.27498904804777 E-13 - 5.65381557065319 E-14*I -1.63566889526104 E-14 - 9.67900985098162 E-15*I 7.43073077006837 E-15 + 4.93269953498429 E-15*I -2.43590387189014 E-15 - 5.21055989895944 E-17*I 3.19983418022330 E-16 - 3.06533506041410 E-16*I 2.35111634696870 E-17 + 4.77456396407459 E-17*I -7.31502044718856 E-18 + 1.10974529579819 E-17*I -3.26221285971496 E-18 - 3.99734388033735 E-18*I 1.39071827030212 E-18 - 2.43545980631280 E-19*I -8.14762787710817 E-20 + 3.02121739652423 E-19*I -5.68424637170392 E-20 - 3.30544545268232 E-20*I 1.00134634420191 E-20 - 1.20846493207443 E-20*I 2.44040335072592 E-21 + 3.32815768651263 E-21*I -1.00281770980843 E-21 + 3.16126717352352 E-22*I 7.46727364275524 E-24 - 2.52461237780373 E-22*I 5.59031612947253 E-23 + 1.92941756355972 E-23*I -8.03434017691000 E-24 + 1.15003875138068 E-23*I -2.13560370410329 E-24 - 2.64275348521218 E-24*I 7.66352536099374 E-25 - 3.06816449234638 E-25*I 1.42509561746275 E-26 + 1.97884314138339 E-25*I -4.58748818093499 E-26 - 1.10298477574201 E-26*I 5.73983689586017 E-27 - 9.59735542776205 E-27*I 1.77067703342646 E-27 + 1.97743801995778 E-27*I -5.74269137934548 E-28 + 2.60660071774678 E-28*I -1.76277133832552 E-29 - 1.48518580873011 E-28*I 3.46938198356513 E-29 + 6.74672128662828 E-30*I -4.01277862515764 E-30 + 7.29505202539481 E-30*I -1.34399669008111 E-30 - 1.42300404827861 E-30*I 4.14782554384895 E-31 - 1.97836607707266 E-31*I tommy1729 Ultimate Fellow Posts: 1,384 Threads: 338 Joined: Feb 2009 08/28/2010, 11:21 PM that seems efficient and intresting. in fact i doubt it hasnt been considered before ? sheldonison Long Time Fellow Posts: 645 Threads: 22 Joined: Oct 2008 08/30/2010, 03:09 AM (This post was last modified: 08/30/2010, 03:33 AM by sheldonison.) (08/28/2010, 11:21 PM)tommy1729 Wrote: that seems efficient and intresting. in fact i doubt it hasnt been considered before ?Thanks Tommy! I assume it has been considered, and probably calculated before. I think Kneser developed the complex periodic regular tetration for base e, and probably would've generated the coefficients. But I haven't seen them before. Perhaps Henryk (or someone else) could comment??? I figured out the closed form equation for a couple more terms, and I have an equation that should generate the other terms, but I'm still working it, literally as I write this post! $a_2 = (1/2)/(L - 1)$ $a_3 = (1/6 + a_2)/(L*L - 1)$ $a_4 = (1/24 + (1/2)*a_2*a_2 + (1/2)*a_2 + a_3)/(L*L*L-1)$ What I did is start with the equation: $\text{RegularSuperf}(z) = \sum_{n=0}^{\infty}a_nL^{nz}$ and set it equal to the equation $\text{RegularSuperf}(z) = \exp{(\text{RegularSuperf}(z-1))}$ Continuing, there is a bit of trickery in this step to keep the equations in terms of $L^{nz}$, instead of in terms of $L^{n(z-1)}$. Notice that $L^{n(z-1)}=L^{(nz-n)}=L^{-n}L^{nz}$. $\text{RegularSuperf}(z) = \exp{(\text{RegularSuperf}(z-1))} = \exp{( \sum_{n=0}^{\infty}\exp^{(L^{-n}a_nL^{nz})})}$ This becomes a product, with $a_0=L$ and $a_1=1$ $\text{RegularSuperf}(z) = \prod_{n=0}^{\infty} \exp{(L^{-n}a_nL^{nz})}$ The goal is to get an equation in terms of $L^{nz}$ on both sides of the equation. Then I had a breakthrough, while I was typing this post!!!! The breakthrough is to set $y=L^z$, and rewrite all of the equations in terms of y! This wraps the 2Pi*I/L cyclic Fourier series around the unit circle, as an analytic function in terms of y, which greatly simplifies the equations, and also helps to justify the equations. $\text{RegularSuperf}(z) = \sum_{n=0}^{\infty}a_ny^n = \prod_{n=0}^{\infty} \exp{(L^{-n}a_ny^n)}$ The next step is to expand the individual Tayler series for the $\exp {(L^{-n}a_ny^n)}$, and multiply them all together (which gets a little messy, but remember a0=L and a1=1), and finally equate the terms in $y^n$ on the left hand side equation with those on the right hand side equation, and solve for the individual $a_n$ coefficients. Anyway, the equations match the numerical results. I'll fill in the Tayler series substitution next time; this post is already much more detailed then I thought it was going to be! I figured a lot of this out as I typed this post! - Sheldon Gottfried Ultimate Fellow Posts: 770 Threads: 120 Joined: Aug 2007 08/30/2010, 09:22 AM (08/27/2010, 02:09 PM)sheldonison Wrote: Normally, we would look at the regular superfunction (base e) as a limit, $\lim_{n \to \infty} \exp^{[n]} (L+L^{z-n})$ but it has a period of 2Pi*i/L, so perhaps it could also be expressed as an infinite sum of periodic terms, $\sum_{n=0}^{\infty}a_n\times L^{( n*z)}$ $a_0=L$ and I think $a_1=1$. Perhaps there is a closed form limit equation for the other a_n terms in the periodic series? - Sheldon Hi Sheldon - just to allow me to follow (think I can't involve much) - I don't have a clue from where this is coming, what, for instance, is L at all? I think you've explained it elsewhere before but don't see it at the moment... Would you mind to reexplain in short or to provide the link? Gottfried Gottfried Helms, Kassel tommy1729 Ultimate Fellow Posts: 1,384 Threads: 338 Joined: Feb 2009 08/30/2010, 09:41 AM (08/30/2010, 09:22 AM)Gottfried Wrote: (08/27/2010, 02:09 PM)sheldonison Wrote: Normally, we would look at the regular superfunction (base e) as a limit, $\lim_{n \to \infty} \exp^{[n]} (L+L^{z-n})$ but it has a period of 2Pi*i/L, so perhaps it could also be expressed as an infinite sum of periodic terms, $\sum_{n=0}^{\infty}a_n\times L^{( n*z)}$ $a_0=L$ and I think $a_1=1$. Perhaps there is a closed form limit equation for the other a_n terms in the periodic series? - Sheldon Hi Sheldon - just to allow me to follow (think I can't involve much) - I don't have a clue from where this is coming, what, for instance, is L at all? I think you've explained it elsewhere before but don't see it at the moment... Would you mind to reexplain in short or to provide the link? Gottfried L is the fixpoint of exp. we have been using L for 2 years tommy1729 Ultimate Fellow Posts: 1,384 Threads: 338 Joined: Feb 2009 08/30/2010, 09:46 AM i must say though that i dont completely get how " fourier methods" lead to the closed forms of a_n ... any fourier expert/sheldon want to explain ? further i believe regular superfunction means expanded at upper fixpoint or at least an equivalent of that. sheldonison Long Time Fellow Posts: 645 Threads: 22 Joined: Oct 2008 08/31/2010, 02:35 AM (This post was last modified: 08/31/2010, 09:12 PM by sheldonison.) (08/30/2010, 09:46 AM)tommy1729 Wrote: i must say though that i dont completely get how " fourier methods" lead to the closed forms of a_n ... any fourier expert/sheldon want to explain ? further i believe regular superfunction means expanded at upper fixpoint or at least an equivalent of that.I agree with Tommy when he says "Fourier methods" (complex Fourier analysis of a discreet number of points), won't lead to a closed form, and I didn't use Fourier methods to generate the coefficients for a_4, and a_5. Also, so far, I've only used this method for the regular superfunction for base(e), although it would presumably work for any regular periodic superfunction (using the upper fixed point, for bases

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