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08/31/2010, 08:04 AM
(This post was last modified: 08/31/2010, 08:32 AM by mike3.)
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(08/31/2010, 07:08 AM)Gottfried Wrote: Hi Sheldon 
I recognize your coefficients.... Gottfried, thanks for your reply. I'm trying to make sense of the coefficients, and would probably need to program the matrix into parigp to verify it.
Earlier, Gottfried wrote:
Quote: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 fixed point of base(e). If you iterate the natural logarithm function hundreds of times (say starting with z=0.5), you get a very good approximation of L. Then the equations describe the regular superfunction for base e, which is complex valued at the real axis, but is analytic and entire.
The rest of what I did basically, it comes down to expressing the regular superfunction as an analytic Fourier series, where all of the terms of the series decay to zero at i*infinity. Such a series is guaranteed to be analytic.
Here is an example of a generalized view of such a Fourier series, not related to the superfunction, with a period of 2Pi.
A full Fourier series would also have the negative coefficients as well, and is often not analytic.
Both of these can be wrapped around the unit circle, with the substitution . In the analytic case, we have an analytic Taylor series. . In the general case, with terms from infinty to +infinty, we have a Laurent series, with singularities inside the unit circle, and an annular ring of convergence. For the full Fourier series, often, the Laurent series only converges on the edge of the unit circle. Hope that helps.
 Sheldon
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08/31/2010, 06:57 PM
(This post was last modified: 08/31/2010, 08:20 PM by sheldonison.)
On to the generic base implementation, for the periodic regular superfunction. Right now, I don't have a working routine for either Mike's suggestion of Bell's polynomial, or Gottfried's matrix routine. So, I'm still doing this "longhand". B is the base. L is the fixed point for base B. I started with my last post, and edited for a generic base.
:: verified if B=sqrt(2), L=4, period = 19.236*I
a_0=L, a_1=1,
substitute:
3) figure out the coefficients for z+1.
4) Apply the definition of the regular superfunction, f(z+1)=B^(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
cancelling, and rearranging terms
I haven't verified these equations, though the last set matched the discreet Fourier analysis for terms a_0 through a_5. In theory, using Mike's Bell expansion, or perhaps Gottfried's matrix, one could write a program that would generate the coefficients for a known interesting base, like sqrt(2). The equations should match the upper super exponential for sqrt(2).
 Sheldon
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08/31/2010, 07:04 PM
(This post was last modified: 09/01/2010, 03:57 PM by Gottfried.)
(08/31/2010, 02:37 PM)sheldonison Wrote: (08/31/2010, 07:08 AM)Gottfried Wrote: Hi Sheldon 
I recognize your coefficients.... Gottfried, thanks for your reply. I'm trying to make sense of the coefficients, and would probably need to program the matrix into parigp to verify it.
Sheldon 
here some Pari/GPstatements to arrive at the coefficients. I just took your last substitution rules in your earlier letter. The "cryptic" functioncalls are in my collection of procedures  "VE" means truncation of vectors/matrices, dFac(1,6) creates a 6x6diagonalmatrix containing the factorials r!^1
Code: \\ extract coefficients at x into columnvector "pc"
\ps 16 \\ use only seriesprecision 16 (coefficients)  if more then the product of the exptaylorseries needs long time...
pc = polcoeffs( exp('a1*x) * exp('a2*x^2) * exp('a3*x^3) * exp('a4*x^4) * exp('a5*x^5)) ~ ;
\\ %box are metacommands to advise PariTTY to send the output into a matrixdisplaywindow
%box >pc dFac(1,5)*Mat(VE(pc,5) ) \\ show the coefficients scaled by factorials in window "pc"
\\ result:
\\ 1
\\ a1
\\ a1^2+2*a2
\\ a1^3+6*a2*a1+6*a3
\\ a1^4+12*a2*a1^2+24*a3*a1+(12*a2^2+24*a4)
\\ now begin to substitute
a1 = 1
pc1=subst(VE(pc,6),'a1,a1)
%box >pc dFac(1,6)*Mat(pc1 ) \\ show the coefficients scaled by factorials
\\ 1
\\ 1
\\ 2*a2+1
\\ 6*a2+(6*a3+1)
\\ 12*a2^2+12*a2+(24*a3+(24*a4+1))
a2 = 1/2/(L1)
pc2=subst(VE(pc1,6),'a2,a2)
%box >pc dFac(1,4)*Mat(VE(pc2,4) ) \\ show only 4 rows
1
1
L/(L1)
((6*a3+1)*L+(6*a3+2))/(L1)
a3 = (1/6 + a2)/(L^21)
pc3=subst(VE(pc2,6),'a3,a3)
%box >pc dFac(1,4)*Mat(VE(pc3,4) )
a4=(1/24 + a3 + a2/2 + a2^2/2)/(L^31)
pc4=subst(VE(pc3,6),'a4, a4)
%box >pc dFac(1,5)*Mat(VE(pc4,5) )
a5=(1/5! + a4+a3/2+a3*a2+a2^2/2+a2/3!)/(L^41)
pc5=subst(VE(pc4,6),'a5, a5)
%box >pc dFac(1,6)*Mat(VE(pc5,6) )
1
\\ 1
\\ L/(L1)
\\ (L^3+2*L^2)/(L^3L^2L+1)
\\ (L^6+5*L^5+6*L^4+6*L^3)/(L^6L^5L^4+L^2+L1)
\\ (L^10+9*L^9+24*L^8+40*L^7+46*L^6+36*L^5+24*L^4)/(L^10L^9L^8+2*L^5L^2L+1)
\\ see the coefficients at powers of L in the numerators equal that matrixentries, for instance last row(revert the order and begin at L^4: [24 36 46 40 24 9 1]
\\ denominators can be factored by (L1),(L^21) and so on
Thanks also for the other explanations. It surely helps to have an entry for my own understanding of the procedure once I've got the whole idea behind...
So far 
Gottfried
Gottfried Helms, Kassel
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08/31/2010, 07:26 PM
(This post was last modified: 08/31/2010, 07:51 PM by mike3.)
Formulas for the Bell polynomials can be found here:
http://en.wikipedia.org/wiki/Bell_polynomials
and here:
http://mathworld.wolfram.com/BellPolynomial.html
There's a method involving taking a matrix determinant, a convolution formula, and a sum over partitions (we are only interested in the "complete" Bell polynomials here).
E.g.
then
(where is like "exponentiating" the sequence with the "diamond" operator)
and
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08/31/2010, 07:34 PM
(This post was last modified: 08/31/2010, 07:39 PM by mike3.)
For the variablebase case,
thus
.
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Now all we need is some way to express the Riemann mapping
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(08/31/2010, 07:34 PM)mike3 Wrote: For the variablebase case,
thus
. Wow! Thanks Mike! Only matrices and statistics, "n choose k" formulas is one of my weak areas, so it might be a few days before I catch up.
I'm kind of intrigued at actually having a closed form for a real valued superfunction, like sqrt(2). I'm guessing that the number of terms required for convergence gets extremely large as z increases. From the terms I computed a couple of days ago, it looks like the terms are decreasing exponentially, which means the series acts like it has a singularity. However, since the regular superfunction is entire, we must have convergence to infinity, so the terms must eventually decrease faster than exponentially.
 Sheldon
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08/31/2010, 08:34 PM
(This post was last modified: 08/31/2010, 08:40 PM by Gottfried.)
Please excuse  this post contains some <aaarrrgggh>s
But it's worth to notice how the bellpolynomials and my for most fellows here: still cryptic  matrixmethod are related. That I didn't see this earlier

In the mathworldwolframlink I find a short description of the Bellpolynomials first kind, fortunately with some example.
This expressed with my matrixformulae is
V(t)~ * dF(1)*S2 * V(x) = exp( (exp(t)1)*x)
or using my standardmatrix for x>exp(x)1 plus one intermediate step
V(t) ~ * fS2F = V(exp(t)1)~
write "y" for "exp(t)1"
V(y)~ * dF(1) * V(x) = F(1)~* dV(y)*V(x)
= F(1)~ * V( y*x)
= exp( y*x)
= exp( (exp(t)1)*x)
Or in one expression:
V(t) ~ * (fS2F * dF(1)) * V(x) = exp( (exp(t)1)*x)
(which I can recognize immediately to be correct because I'm extremely used to that notation)
or the even simpler definition of the vector of Bellpolynomials:
B(x) = S2*V(x)
So now I see at least, how the Bellpolynomials of the *first kind* are related to my matrixlingo.
(So I should rename fS2F into "Bell" perhaps...)

But then follows the Bellpolynomials of *second kind*. And there I'm lost again... No example, no redundancy... as if the reader could not do some error in parsing a complex formula...
Is there possibly meant the iteration of the Bellpolynomials (=matrixpower of S2)?
That would be simple then...
Or, wait  looking at the *subscripted* "x" I think, that they are now coefficients for some arbitrary function developed as a powerseries with reciprocal factorials, ... and then...
Well, this seems to be just the definition of what I found out myself at the very beginning of my fiddling with this subject and called it a "matrixoperator" for some function. With the additional property, that it is (lower) triangular (because of the missing x0 in the formula in mathworld), so for instance all my U_tmatrices for the decremented exponentiation (or "Utetration to base t" in my early speak) were such collections of Bellpolynomials of second kind (but also the schrödermatrices, just all my lowertriangular matrixoperators which I worked with the last years and have the reciprocal factorial scaling <arrggh!>)
So now  if we talk about the Bellpolynomials (first or second kind) and the symbolic representation in terms of the log(L) (or log(t)) then this is just what I solved in my earlier posted link (good to know)! Here it is again: http://go.helmsnet.de/math/tetdocs/APT.htm . And Faa di Bruno/Bellpolynomials and "matrixoperators" is the same and one needs only read about one side of these notations...
Amen 
<cough>
Gottfried
Gottfried Helms, Kassel
