Re: Finding the Formula...

On Jul 17, 8:21 pm, Gottfried Helms <he...@xxxxxxxxxxxxx> wrote:
Am 17.07.2008 21:47 schrieb mike3:





On Jul 17, 2:06 am, Gottfried Helms <he...@xxxxxxxxxxxxx> wrote:
Mike -

 I've not consequently read into your tables, but something
 looks similar to what I've calculated recently, and where
 I found an iterative algorithm to compute the numbers.

<snip>

This is interesting. I might have a look at it, however what I
was trying to do was figure out a _formula_ for the nth coefficient
of the natural dxp-tetration series (or even better that with
changeable k so other bases can be used), as opposed to just an
algorithm for generating the coefficients. For example, I noticed
how you said it took many many iterations with huge matrices to
obtain a coefficient...

So then why I asked about the h-step mappings with n inputs --
what that means, and what the general expression for those closed-
form formulae (with respect to h) is for any given n... Because it
seems that will provide a formula, at least for natural-base
dxp-tetration (and so also for e^(1/e)-base normal tetration).

Yepp; I don't have a better formula for this yet too - the search
for such a formula just led to that algorithm.

---------

Just a remark: I said, the computable consumtion in this
algorithm is huge; I should better have said, it is increasing
with the index of the coefficients. For the first, say 4
coefficients it is small, then up to 8 one may manage this
with 8 by 28 matrices (guessed) with the order of matrix-
size about n by  binomial(n,2) for the n'th coefficient.
I've also another algorithm (by symbolic eigensystem-analysis)
but this is also using increasing size of the matrices.

Hmm - I'd be curious whether a principally simpler formula
can be found, since the complexity of the coefficients
just *is* as high as the diagonalization exhibits.

---------

I think, one needs another framework than that of powerseries
if a simpler set of coefficients for a series is sought.
For instance, just recently Henryk Trappmann presented
a binomial-expansion on the series of values of integer
iterates of the function itself, so say

  U°0(x) + U°1(x) + U°2(x) + ... = <something>

then

  a0(x) =  1 U°0(x)
  a1(x) = -1 U°0(x) + 1 U°1(x)
  a2(x) =  1 U°0(x) - 2 U°1(x) + 1 U°2(x)
 ...
and

 U°h(x) =   bi(h,0)a0(x) + bi(h,1)a1(x) + bi(h,2)a2(x) + ...

           (bi(x,y) is binomial-coefficient)
where if h is fractional this is an infinite series,

I checked this method with some fractional h and found,
that I needed 200 terms of this series to get accuracy
to say, 12 digits.
Interestingly, when I took for instance h=0.5 then this
converged slowly, if I took h=1.5 and then the log of the
result (mimicking the decrease of h by 1) then the accu-
racy increased, and up to h=23.5 I got then accuracy
up to nearly 30 digits.

---

Well - this is still not answering your question; if there
were a better function for the computation of the coefficients
of the powerseries-representation around, I would like to see
it, too.


Me too, this really sucks.

There was another approach I just thought of, however.

If my "hunch" is right and the polynomials are of degree n, where
n is the order of the derivative we want to take then one needs only
n derivative points and polynomial interpolation will give the
correct polynomial.

Faa di Bruno's formula provides a way to compute the nth derivative
of as high an (integer) iteration as we want recursively. Could it be
useful?

.

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