# Tetration Forum

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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
(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.
$a_0=L$
$a_1=1$
$a_2=1/(2*L-2)$
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

that seems efficient and intresting.

in fact i doubt it hasnt been considered before ?
(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!
$a_2 = (1/2)/(L - 1)$
$a_3 = (1/6 + a_2)/(L*L - 1)$
$a_4 = (1/24 + (1/2)*a_2*a_2 + (1/2)*a_2 + a_3)/(L*L*L-1)$

$\text{RegularSuperf}(z) = \sum_{n=0}^{\infty}a_nL^{nz}$
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

(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
(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
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.
(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.
$\text{RegularSuperf}(z) = \sum_{n=0}^{\infty}a_nL^{nz}$

2) do the substitution, $y=L^z$, 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.
$\text{RegularSuperf}(z) = \sum_{n=0}^{\infty}a_ny^n$

3) figure out the coefficients for z+1. $y_{z+1}=L^{z+1}=L*L^z=L*y$
$\text{RegularSuperf}(z+1)=
\sum_{n=0}^{\infty}a_nL^ny^n$

4) Apply the definition of the regular superfunction, f(z+1)=exp(f(z))
$\text{RegularSuperf}(z+1)=\exp({\text{RegularSuperf}(z)})$
$\text{RegularSuperf}(z+1) = \sum_{n=0}^{\infty}a_nL^ny^n =
\exp({\sum_{n=0}^{\infty}a_ny^n})$

5) express this as a product. Then expand the Taylor series for each term in the product, and substitute in $a_0=L$ $a_1=1$.

$\sum_{n=0}^{\infty}a_nL^ny^n=
\prod_{n=0}^{\infty}\exp({a_ny^n})$

$\sum_{n=0}^{\infty}a_nL^ny^n=
\prod_{n=0}^{\infty}( {
\sum_{p=0}^{\infty}({(a_ny^n)^p/p!})} ) =$

$
\\(L)\times
\\(1 + y + y^2/2! + y^3/3! + y^4/4! + y^5/5! ....)\times
\\(1+a_2y^2+a_2^2y^4/2!+a_2^3y^6/3!+a_2^4y^8/4! ...)\times
\\(1+a_3y^3+a_3^2y^6/2!+a_3^3y^9/3!+a_3^4y^{12}/4! ...)\times
\\(1+a_4y^4+a_4^2y^8/2!+a_4^3y^{12}/3!+ ...)\times
\\(1+a_5y^5+a_5^2y^{10}/2!+a_5^3y^{15}/3! + ...)\times ...$

6) equate terms with the same y coefficient, and figure out the pattern (I haven't yet)
$a_2L^2y^2=Ly^2/2! + La_2y^2$
$a_3L^3y^3=Ly^3/3! + La_3y^3 + La_2y^3$
$a_4L^4y^4=Ly^4/4! + La_4y^4 + La_3y^4 + La_2y^4/2+ La_2^2y^4/2$
$a_5L^5y^5=Ly^5/5! + La_5y^5 + La_4y^5 + La_3y^5/2 +
La_3a_2y^5+La_2^2y^5/2 + La_2y^5/6$

cancelling, and rearranging terms
$a_2L-a_2=1/2$
$a_3L^2-a_3=1/6 + a_2$
$a_4L^3-a_4=1/24 + a_3 + a_2/2+ a_2^2/2$
$a_5L^4-a_5=1/120 + a_4 + a_3/2 + a_3a_2+a_2^2/2 + a_2/6$

$a_2=0.5/(L-1)$
$a_3=(1/6 + a_2)/(L^2-1)$
$a_4=(1/24 + a_3 + a_2/2+ a_2^2/2)/(L^3-1)$
$a_5= (1/120 + a_4 + a_3/2 + a_3a_2+a_2^2/2 + a_2/6)/(L^4-1)$

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

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

$\prod_{n=0}^{\infty} \exp(a_n y^n) = \exp\left(\sum_{n=0}^{\infty} a_n y^n\right) = \exp(a_0) \exp\left(\sum_{n=1}^{\infty} a_n y^n\right)$.

This then becomes

$\exp(a_0) \exp\left(\sum_{n=1}^{\infty} a_n y^n\right) = \exp(a_0) \sum_{n=1}^{\infty} \frac{\sum_{k=1}^n B_{n,k}(1! a_1, ..., (n-k+1)! a_{n-k+1})}{n!} y^n = \exp(a_0) \sum_{n=1}^{\infty} \frac{B_n(1! a_1, ..., n! a_n)}{n!} y^n$.

Thus the equations to solve are

$a_n L^n = \exp(a_0) \frac{B_n(1! a_1, ..., n! a_n)}{n!}$.

Since $a_0 = L$ and $a_1 = 1$, this is

$a_n L^n = L \frac{B_n(1, 2! a_2, ..., n! a_n)}{n!}$.

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

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

for details.
(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 $a_5$, 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]
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