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closed form for regular superfunction expressed as a periodic function
#1
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,

but it has a period of 2Pi*i/L, so perhaps it could also be expressed as an infinite sum of periodic terms,

and I think
.
Perhaps there is a closed form limit equation for the other a_n terms in the periodic series?
- Sheldon
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#2
(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?
- Sheldon
Originally, 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.



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


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#3
that seems efficient and intresting.

in fact i doubt it hasnt been considered before ?
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#4
(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!




What I did is start with the equation:

and set it equal to the equation


Continuing, there is a bit of trickery in this step to keep the equations in terms of , instead of in terms of . Notice that .


This becomes a product, with and


The goal is to get an equation in terms of on both sides of the equation. Then I had a breakthrough, while I was typing this post!!!! The breakthrough is to set , 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.



The next step is to expand the individual Tayler series for the , and multiply them all together (which gets a little messy, but remember a0=L and a1=1), and finally equate the terms in on the left hand side equation with those on the right hand side equation, and solve for the individual 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



Reply
#5
(08/27/2010, 02:09 PM)sheldonison Wrote: Normally, we would look at the regular superfunction (base e) as a limit,

but it has a period of 2Pi*i/L, so perhaps it could also be expressed as an infinite sum of periodic terms,

and I think
.
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
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#6
(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,

but it has a period of 2Pi*i/L, so perhaps it could also be expressed as an infinite sum of periodic terms,

and I think
.
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 Big Grin
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#7
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.
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#8
(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<eta).

I'm going to start over, and add a little more detail as well. What I'm doing here is a lot like what is done for intuitive tetration, but here the equations are exact. Here are the steps. Hopefully, these steps will be a little clearer.

1) realize that the regular superfunction is periodic, and express that superfunction as the infinite sum of a periodic/Fourier series. For base(e), the fixed point is "L", and the period is 2Pi*i/L.


2) do the substitution, , and rewrite the equations in terms of "y". This is because all of the periodic terms decay to the fixed point, at +I*infinity, and at -real infinity. The equation in y has the same coefficients a_0...a_n, as the equation in z.


3) figure out the coefficients for z+1.


4) Apply the definition of the regular superfunction, f(z+1)=exp(f(z))



5) express this as a product. Then expand the Taylor series for each term in the product, and substitute in .





6) equate terms with the same y coefficient, and figure out the pattern (I haven't yet)




cancelling, and rearranging terms










And that would give you the four non-trivial terms I have closed form solutions for, although I haven't verified , and wouldn't be surprised if it has a typo (edit: a_5 is also verified correct).
- Sheldon

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#9
The product of exp is the exponential of a formal power series. This can be expressed using the Bell polynomials:

.

This then becomes

.

Thus the equations to solve are

.

Since and , this is

.

This is derived from Faà di Bruno's formula, see

http://en.wikipedia.org/wiki/Fa%C3%A0_di...7s_formula

for details.
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#10
(08/31/2010, 02:35 AM)sheldonison Wrote: And that would give you the four non-trivial terms I have closed form solutions for, although I haven't verified , and wouldn't be surprised if it has a typo.
- Sheldon

Hi Sheldon -

I recognize your coefficients. I came to the same coefficients, when I developed the eigensystem for the decremented exponentiation (dxp) which I called U-tetration. They occur as coefficients in the powerseries-expansion for the dxp (when including the iteration-height parameter h - which makes it also the superfunction for dxp).
In

http://go.helms-net.de/math/tetdocs/APT.htm

I've described the procedure to solve for that coefficients based on the eigensystem/schröder-function-decomposition and gave some example coefficients-matrices.

Here is another notation for the decomposition of your a_k coefficients in matrix-notation: [update] I updated the powers of L to simplify the matrix-columns - two errors corrected [/update]
Code:
a0 = 1/0! /1              * 1   * [ 1                                                ]
a1 = 1/1! /1              * L   * [ 1                                                ]
a2=  1/2! /(L-1)          * L^2 * [ 2 + 1*L                                          ]
a3=  1/3! /(L-1)(L^2-1)   * L^3 * [ 6 + 6*L + 5*L^2 + 1*L^3                          ]
a4=  1/4! /(L-1)...       * L^4 * [24 +36*L +46*L^2 +40*L^3 +  24*L^4 + 9*L^5 + 1*L^6]
...
where I treat the coefficients in the brackets as matrix A:
Code:
A =
1   .  .  .   .  .  .  ...
1   .  .  .   .  .  .  ...
2   1  .  .   .  .  .  ...
6   6  5  1   .  .  .  ...
24 36 46 40  24  9  1  ...
... ... ...

Now the rows are known to me and match exactly the last columns of the A-coefficients-matrices in section "Coefficients of the powerseries for fractional iterates"
There last column in A3 is [ 2 1 ], of A4 is [6 6 5 1], of A5 is [24 36 46 40 24 9 1 ] of A6 is [120 240 390 480 514 416 301 160 54 14 1] from where I am confident, that a5 in your case is explicitely

Code:
a5 = 1/5! / (L-1)/(L^2-1)/(L^3-1)/(L^4-1)
        * L^5 * [120 + 240*L + 390*L^2 + 480*L^3 + 514*L^4
                   + 416*L^5 +301*L^6 +160*L^7 + 54*L^8 +  14*L^9 +1*L^10 ]

However, I do not know the further exact relation of this to your approach, for instance I'm using general bases and also the log of the fixpoint (I called it "u", u=log(L) in this case) and its (fractional) h'th powers depending on the height h.
I think I'll have to go through it step by step to find where and how your and my concepts match and where/how they differ in detail to make it possibly helpful for your considerations.

Gottfried


[update] Mike's and this msg seem to have crossed. Possibly the reference to the Faa di Bruno-formula is the more relevant/conclusive one for your problem [/update]
Gottfried Helms, Kassel
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