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closed form for regular superfunction expressed as a periodic function
#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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