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andydude Wrote:Have you tried computing or plotting or yet? I would do this but I don't have any code yet for AS(x), and I'm lazy.
Andrew Robbins
Yes, I fiddled with this a bit.
First, it is obvious, that in
Sb(x) = x  b^x + b^b^x  ... + ...
there is some "selfsimilarity", if x itself is a powertower of base b
Sb(0) = s0
Sb(1) = s0
Sb(b) = 1  s0
Sb(b^b) = b  1 + s0
and so on.
For the fixpoints, let t0,t1,t2,... be the fixpoints for base b
t_k = h_k(b)
then
Sb(t) = t  t + t  t +... ... = t*(eta(0)) = t/2
I've added some graphs for different bases for varying x, marking x=t0,t1 and x=0, x=1, x=b
It would be interesting to show graphs of negative heights, aka iterated logarithms...
Gottfried
Gottfried Helms, Kassel
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11/23/2007, 10:47 AM
(This post was last modified: 11/23/2007, 11:23 AM by Gottfried.)
I'm extending a note, which I posted to sci.math.research today.
I hope to find a solution of the problem by a reconsideration of the structure of the matrix of stirlingnumbers 1'st kind, which may contain "Infinitesimals" which become significant if infinite summing of its powers are assumed.
 (text is a bit edited) 
Due to a counterexample by Prof. Edgar (see sci.math) I have to withdraw this conjecture.
The error may essentially be due to a misconception about the matrix of Stirlingnumbers 1'st kind and the infinite series of its powers.

It is perhaps similar to the problem of the infinite series of powers of the pascalmatrix P, which could be cured by assuming a nonneglectable infinitesimal in the first upper subdiagonal
Assuming for an entry of the first upper subdiagonal in row r (r beginning at zero) in the pascal matrix P
which is an infinitesimal quantity and appears as zero in all usual applications of P. But if applied in a operation including infinite series of consecutive powers, then in the matrix Z, containing this sum of all consecutive powers of P we get the entries
By defining
assuming this leads to the nonneglectable rational quantities in that subdiagonal of the summatrix
With that correction the infinite series of powers of the pascalmatrix leads then to a correct matrix,
including the above definitions for the first upper subdiagonal, which provides the coefficients of the (integrals of) the bernoullipolynomials, and can be used to express sums of like powers as expected and described by H.Faulhaber and J.Bernoulli.
(for more details see powerseries of P, page 13 ff )

This suggests then to reconsider the matrix of Stirlingnumbers of 1'st kind with the focus of existence of a similar structure in there.

Does this sound reasonable? It would require a description of the matrix of Stirlingnumbers 1'st kind, which allows such an infinitesimal quantity.
But there is one important remark: this matrix contains the coefficients of the series for logarithm and powers of logarithms. A modification of this matrix would then introduce an additional term in the definition of these series. Something hazardeous...
Gottfried Helms, Kassel
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Joined: Aug 2007
12/26/2007, 07:39 PM
(This post was last modified: 12/27/2007, 11:20 AM by Gottfried.)
I've made a little progress with the problem of deviance of the serial summed alternaing series of powertowers of increasing heights ("Tetraseries"). Since T and Utetration can be mutually converted by shift of their parameter x, I can concentrate on the Utetration here, which has the advantage, that its operator is a triangular matrix, whose integerpowers and eigensystem are easily and exact (within the unavoidable sizetruncation for the actual computation) computable.
Code: Denote the fixed base for
lb(x) = log(1+x)/log(b)
ub(x) = b^x  1
The iterates
ltb(x,h) = ltb(lb(x),h1) ltb(x,0)= x
utb(x,h) = utb(ub(x),h1) utb(x,0)= x
The infinite alternating sums
ASLb(x) = x  ltb(x,1) + ltb(x,2)  ltb(x,3) + ...  ...
ASUb(x) = x  utb(x,1) + utb(x,2)  utb(x,3) + ...  ...
The matrixapproach suggests to compute the ASseries using the geometricseries of the Utetrationmatrices S1b and S2b, which I shall call MLb and MUb here. From here the conjecture was derived, that
Code: ASLb(x) + ASUb(x) = x // matrixcomputation
since
Code: MLb + MUb = I // matrices
However, the computation of ASLb(x) along its partial sums and summation using Cesaro or Eulersummation gives a different result:
Code: ASLb(x) + ASUb(x) = x + db(x) // serialcomputation
The difference of the matrix and serialcomputation may then be expressed by db(x) only.

In my previous posts I already mentioned, that the deviance between the two methods seem to be somehow periodic, wrt x as the variable parameter.
The first useful result was, that I found periodicity for db(x)
Code: db(x) =  db(ltb(x,1)) = db(ltb(x,2)) =  db(ltb(x,3)) ...
for few bases and numerical accessible range for x. The plot showed also a sort of sinuscurve, but where the frequency was somehow distorted.
Today I could produce a very good approximation to a sinuscurve, using fractional tetrates for x.
I computed the k=0..32 fractional Utetrates of x for height 1/16
for base b=sqrt(2), and instead of x I used the index k as xaxis for the plot.
This provided a very good approximation to a sinuscurve for db(x_k)
In the plot I display the two nearlines for ASLb(x_k), ASUb(x_k) and db(x_k) and overlaid a sincurve, whose parameters I set manually by inspection.
The curve for db(x_k) and sin() match very good; I added also a plot for their difference.
If I can manage to make my procedures more handy, I'll check the same for more parameters. The near match of the curves give apparently good hope that this line of investigation may be profitable...
Gottfried
[updated image]
Errorcurve( deviation of db(x) from overlaid sinuscurve)
Gottfried Helms, Kassel
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Joined: Aug 2007
I started rethinking the inconsistency about the tetraseries with increasing negative heights.
First  I don't remember whether I've already linked my compilation concerning this problem. Here is the link tetraseriesproblem
Then I realized, that using a fixpoint we'll have a special case.
Recall the definition
ALU (b,x) = x  log(1+x)/log(b) + log(1+log(1+x)/log(b))/log(b)  ... + ...
which means the infinite alternating sum of towers of negative heights.
If x is a fixpoint of b, where b^t1=t or b=(1+t)^(1/t), then clearly this simplifies to
ALU (b,t) = t  t + t  ... + ... = t * eta(0) = t/2 //by Euleror Cesarosummation. and the conjecture
ASU(b,t) + ALU(b,t)  t =0
holds for this case.
Don't know how to make this helpful for reparature of my conjecture, but perhaps it gives an idea.
Gottfried
@Andrew (I just also reread your note in the other thread): the needed matrix is simple to compute. Just construct the triangular Bellmatrix U (for decremented iterated exponentiation for a base b), and compute
MU = (I + U)^1
and
ML = (I + U^1)^1
Gottfried Helms, Kassel
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06/13/2008, 07:15 AM
(This post was last modified: 06/13/2008, 07:34 AM by Gottfried.)
Assume for the following a fixed base t for the dexp or "U"tetration, so that dxp°h(x):=dxp_t°h(x) and for shortness let me replace dxp by U in the following. Also I change some namings from the previous posts for consistency.
Denote the alternating series of Upowertowers of increasing positive heights
and of increasing negative heights
then my conjecture, based on diagonalization was
which was wrong with a certain systematic error
I have now a description for the error d(x), which fits very well.
First, let's formally write as the twowayinfinite series
Then it is obvious, that as(x) is periodic with 2h, if x is expressed as powertower
where h is integer and r is the fractional remainder of a number y (mod 2)
and define
Then we may discuss as_r = as(x) as
where r = y (mod 2)
We can then rewrite the formula
as
My observation was, that d_r is sinusoidal with r, with a very good fit (I checked also various bases t).
I got now
where "a" indicates the amplitude and "w" a phaseshift (depending on base t).
The following fits the result very well:
so we could rewrite as functional equation for asp
and also can determine all as_r using asn_r for r=0 and r=0.5 (the integer and halfintegeriterate U°0(1) and U°0.5(1)) only.
Note, that the computation of asp_r is exact using the appropriate matrix (I + Ut)^1 of the diagonalizationmethod.
The diagonalizationmethod deviates only for the part asn_r; it gives asn_r  d_r instead of asn_r ; unfortunately, this does not allow to determine d_r correctly with the diagonalizationmethod only (yet).
The benefit of the diagonalizationmethod is here, that its matrix (I+Ut)^1 provides the coefficients for a powerseries for asp_r, which seems to be the analytic continuation for bases t, where the series asp would diverge.
Gottfried Helms, Kassel
Posts: 766
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06/22/2008, 05:25 PM
(This post was last modified: 06/22/2008, 06:51 PM by Gottfried.)
Hi 
few days ago I collected my current results in a msg in sci.math and sci.math reasearch; I think, it fits well here, although I stated most of it in earlier posts already, so this may appear boring. However, since the readers of the newsgroups are not familiar with the matrix/diagonalization concept I explained the problem in terms of sums of the formal powerseries, which may be interesting also here.
Additionally I append here some more notes, which are new, and possibly focus the problem in a more fruitful way.
Because I'm a bit lazy today I don't convert it into latex  just plain text.
I'd like to put it also into the "open problems" section, but I'm a bit unsure how I could shorten the exposition for that thread by appropriate referencing.
Gottfried
Code: I have a new result for the Tetraseries here, which points
to a more fundamental but general effect in summing of this
type of series.
I discuss the Utetration instead of the usual Ttetration here,
because the effect under consideration apparently is the same with
Ttetration, and Utetration is easier to implement using the
matrix/diagonalizationmethod.
I usuallly denote
Ttetration:
Tb (x) = b^x // baseparameter b
Tb°0(x) = x // base b occurs 0 times
Tb°1(x) = b^x // base b occurs 1 times
Tb°h(x) = Tb°(h1)(Tb(x))
= b^b^b^...^b^x // base b occurs h times
Tb°1(x) = log(x)/log(b)
Utetration:
Ut (x) = t^x 1 // baseparameter t
Ut°0(x) = x
Ut°1(x) = t^x 1
Ut°h(x) = Ut°(h1)(Ut(x))
Ut°(1)(x) = log(1+x)/log(t)

For the discussion of the series we assume a fixed baseparameter t
here, so I omit it in the notation of the Utetrationfunction in
the following.
Also I restrict myself to bases t where all of the following series
are conventionally summable using Cesaro/Eulersummation.
The series under discussion are
(Upowertowers of increasing positive heights)
asup(x) = x  (t^x 1) + (t^(t^x  1) 1)  ... +...
= sum {h=0..inf} (1)^h * U°h(x)
(Upowertowers of increasing negative heights)
asun(x) = x  log(1+x)/log(t) + log(1+ log(1+x)/log(t))/log(t) ...+...
= sum {h=0..inf} (1)^h * U°h(x)
(all heights)
asu(x) = asp(x) + asn(x)  x
= sum {h=inf..inf} (1)^h * U°h(x)
The acronyms mean here:
a(lternating) s(ums of) u(tetration with increasing)
p(ositive heights)
n(egative heigths)

Using u=log(t) = 1/2 , t=exp(u)~ 1.648721... the series asup and asun
have bounded terms with alternating signs, so they can be Cesaro or
Eulersummed.
If they are summed this way, evaluated term by term, I call
this "serial summation" in contrast to my matrixapproach.
Values for asup(1) and asun(1), found by serial summation are
asup(1) = 0.596672423492... // serial summation
asun(1) = 0.403327069976... // serial summation
asu(1) = 0.000000506531563910... // serial summation
My earlier conjecture, based on consideration of the matrixmethod,
was, that asu(1) = 0 for each base t, but which was wrong.
Computations give this results:
asup(1) = 0.596672423492... // matrix method
asun(1) = 0.403327576508... // matrix method
asu(1) = 0 // matrix method
where in all checked cases asup(x) appeared as identical for both
methods, and only asun(x) differed (begin of differences of
digits marked by vertical line):
asun(1) = 0.403327  069976... // serial summation
asun(1) = 0.403327  576508... // matrix method
The difference of the two methods occurs systematically, so there is
reason to study this difference systematically as well.
Since most readers here are unfamiliar with the the matrixmethod,
I'll give the examples below in more conventional description using
the explicit powerseries representation of the problem.

Code: But we need some prerequisites.
First note, that if x is seen as Upowertower to base t itself,
then the results in asu(x) are periodic with the integerheight
part of x; so if
x = U°h(1)
then the results for asu(x) occur periodically with k in
x_k = U°(2*k*h)(1) = U°(2*k*floor(h))(x_r)
where x_r is the remaining part of fractional height; so we may
standardize our notation to
x_r = U°r(1)
where r means the fractional part of h and reduce our notations
for asup, asun and asu to
asup_r = asup(x) = asup(U°r(1))
asun_r = asun(x) = asun(U°r(1))
asu_r = asu (x) = asu (U°r(1))
Second: what we also need is the halfiterate U°0.5(1), such that
U°0.5(U°0.5(1)) = t  1
The powerseries for U°1(x) = t^x  1 is simple; it is just the
exponential series reduced by its constant term (use u = log(t))
U°1(x) = ux + (ux)^2/2! + (ux)^3/3! + ...
Using the matrix/diagonalizationmethod one can find the
coefficients a,b,c,... for the U°0.5function as well:
U°0.5(x) = a x + b x^2 + c x^3 + ...
= 0.707107...*x + 0.103553...*x^2 + 0.00534412...*x^3
 0.000124330...*x^4 + 0.0000201543...*x^5 + O(x^6)
If I use this function (actually with 96 terms and higher precision)
then I get
U°0.5(1) = 0.815903...
U°0.5(0.815903...) = 0.648721...
which is very well approximated
U°1(1) = t^1 1 = exp(1/2)  1
= 0.648721...
using 96 sodetermined terms of the powerseries of U°0.5(x).
So we may assume, U°0.5(1)= 0.815903... is determined with
sufficient (and principally with arbitrary) precision.
Now we compute asup_0.5 and the other series by serial summation
asup_0.5 = asup(U°0.5(1)) = asup(0.815903...) = 0.497542... // serial
asun_0.5 = asun(U°0.5(1)) = asun(0.815903...) = 0.318354... // serial
asu_0.5 = asu (U°0.5(1)) = asu (0.815903...) = 0.00000690039... // serial
where
asu_0.5 = asup_0.5 + asun_0.5  x_0.5
= 0.497542... + 0.318354...  0.815903...
while by the matrixmethod we get
asup_0.5 = asup(U°0.5(1)) = asup(0.815903...) = 0.497542... // matrix
asun_0.5 = asun(U°0.5(1)) = asun(0.815903...) = 0.502458... // matrix
asu_0.5 = asu (U°0.5(1)) = asu (0.815903...) = 0 // matrix

Code: The first result is now, that for any r apparently we may describe
the difference between the matrixcomputed results and the serial
results using
d_r = asu_r (//serial)  asu_r (//matrix)
= asu_r (//serial)
computable by
d_r = ampl * sin(2*pi*r + w)
where the amplitude is
ampl = sqrt(d_0^2 + d_0.5^2)
and the constant phaseshift w
w = arg(d_0.5 + d_0*I)
Here we find
d_0 = 0.00000050653156391...
d_0.5 = 0.00000690038760124...
d_0^2+d_0.5^2 = 4.78719232725 E11
ampl = 0.00000691895391461...
w = arg(d_0*I + d_0.5) = 3.06831783019...
= atan(d_0/d_0.5)  Pi = 0.0732748233988...  Pi
so we may as well say, that the error in computing asn(x)=asn_r by the
matrixmethod is the sinusoidal function d_r.
So the matrixmethod must be reconsidered for the case of infinite
series of negative heights.

Code: As I promised in the above, we need not go into details of the
matrixmethod itself; it can be shown, that the coefficients for
the powerseries of asn(x) determined by the matrixmethod and the
following conventional method are the same.
Consider the sequence of powerseries for U°0(x), U°1(x), U°2(x)
which must be alternating summed to give the powerseries for asn(x)
U°0(x) = 0 1 x
U°1(x)= 0 2 x +2/2! x^2 4/3! x^3 +12/4! x^4 48/5! x^5 +...
+U°2(x)= 0 +4 x 12/2! x^2 +64/3! x^3 496/4! x^4 +5072/5! x^5 ...
U°3(x)= 0 8 x +56/2! x^2 672/3! x^3 +11584/4! x^4 262176/5! x^5 +...
+U°4(x)= 0 +16 x 240/2! x^2 +6080/3! x^3 220160/4! x^4 +10442816/5! x^5 ...
U°5(x)= 0 32 x +992/2! x^2 51584/3! x^3 +3825152/4! x^4 371146880/5! x^5 +...
... ... ... ... ... ...

asn(x)= 0 a1 x +a2 x^2 +a3 x^3 +a4 x^4 +a5 x^5 +...
then, when we collect like powers of x, we get divergent sums of
coefficients at each power of x.
However, the second column indicates, that these sums may be
computed by the given analytical continuation of the geometric
series  unfortunately, the composition of the following columns
from geometric series are not obvious.
But if we want to resort to for instance Eulersummation, which gives
regular results if some conditions on the growthrate of the terms
of a infinite sum/a series are given, we may assign values to all a_k.
One of these conditions is, that the growthrate is eventually
geometric, thus the quotient of absolute values of two subsequent
terms must converge to a constant.
I checked this condition and it is satisfied (also backed by
inspection of the general description of terms as given in [1])
Quotients of absolute values of subsequent rowentries for the
leading five columns:
2. 6.00000 16.0000 41.3333 105.667
2. 4.66667 10.5000 23.3548 51.6909
2. 4.28571 9.04762 19.0055 39.8313
2. 4.13333 8.48421 17.3744 35.5409
2. 4.06452 8.23325 16.6588 33.6885
2. 4.03175 8.11453 16.3227 32.8250
2. 4.01575 8.05675 16.1597 32.4078
2. 4.00784 8.02825 16.0795 32.2028
2. 4.00391 8.01409 16.0396 32.1011
2. 4.00196 8.00704 16.0198 32.0505
2. 4.00098 8.00352 16.0099 32.0252
2. 4.00049 8.00176 16.0049 32.0126
2. 4.00024 8.00088 16.0025 32.0063
2. 4.00012 8.00044 16.0012 32.0032
... ... ... ... ...
We see empirically, that the quotients converge to powers of u^1
(where u=1/2 for all computations in this examples)
So the columnwise summation of coefficients using Eulersummation
should give valid results for the final powerseries asn(x)
What I get is, for
... ... ... ... ... ...

asn(x)= a1 x + a2 x^2 +a3 x^3 +a4 x^4 +a5 x^5 +...
the explicite values for coefficients a_k:
asn(x)= 1/3 x +1/15 x^2 +2/405 x^3 0.0010893246 x^4 0.000457736 x^5 +...
// by matrixmethod (= collecting coefficients at like powers of x;
// Eulersums. The coefficients are rational multiples of integer
// polynomials in u.
So, by comparision of the results, we know, that this powerseries
is *false* and needs correction by a component d_r, which follows a
sinuscurve according to the fractional height r of x . Where x is
assumed as Upowertower
x = x_r = U°r(1)
===================================================================
This effect of a sinusoidal component in the determination of
asn(x) when computed by collecting like powers of x of all involved
powerseries seems somehow fundamental to me, and I would like
to find the source of this effect.
May be, it is due to the required *increasing* order of Eulersummation,
where a column c needs order of u^c, and the implicite binomial
transform in Eulersummation with infinite increasing order must
be reflected by special considerations.
Gottfried Helms
[1] http://go.helmsnet.de/math/tetdocs/ContinuousfunctionalIteration.pdf
see page 21
=====================================================================
Code: =====================================================================
I'm adding some more checks here which deal with the special problem
in asn(x) only.
First note, that if x is a fixpoint such that
U°h(x) = x
then the series asn(x) changes to the alternating series
x  x + x  x + ...  ... = x * eta(0) = 1/2 x
and it is interesting, what serial and matrixsummation do, if the
fixpoint is given as parameter.
One fixpoint x0 = 0, since t^0  1 = 0; however, this is not of
interest here
The second fixpoint, which can be computed as limit
xoo = lim{h>inf} U°(h)(1)
is, using u=0.5, t = exp(u)~ 1.648721..., h=400
xoo = ut(1,400) = 2.51286241725233935396547523322...

direct check: (is xoo a fixpoint?)
(t^xoo  1)  xoo = 4.10903766646035512593597628782 E113
log(1+xoo)/u  xoo = 2.33942419508380691068587979906 E113
so we have a very well aproximated fixpoint (I used floatprecision
of 1200 decimal digits)
Serial summation: 
check: is asn(xoo) = xoo/2 ?
asn(xoo) = 1.25643120862616967698273761661
asn(xoo) xoo/2 = 7.45354655251552020322193384272 E114
so indeed, the serial summation behaves as expected.
Matrix: (dim=96) 
direct check: (is xoo a fixpoint?)
%box ESum1(1.2485)*dV(xoo)*Mat(UtI[,2])  xoo*Mat(V(1))
V(xoo)~*UtI[,2]  xoo = 3.78989 E28
Well, I've to check, whether the precision increases with increasing
number of terms.
check: is asn(xoo) = xoo/2 ?
UtMI = I  UtI + UtI^2  UtI^3  ... + ...
= (I + UtI)^1 // geometric series
asn(xoo) = V(xoo)~ * UtMI[,1]
The coefficients in second column of UtMI with powers of xoo form
a divergent series, so I have to apply Eulersummation; but since
Eulersummation may be too weak for this series, I also append
a check with a stronger (however still experimental method PkPow)
%box ESum1(1.32)*dV(xoo)*Mat(UtMI[,2])
%box PkPowSum(1.7,1.1)*dV(xoo)*Mat(UtMI[,2])
asn(xoo) = V(xoo)~* UtMI[,2] = 1.26414... // ESum 1.32
asn(xoo) = V(xoo)~* UtMI[,2] = 1.26487... // PkPow 1.7,1.1
result  xoo/2 = 0.0077142... // ESum 1.32
result  xoo/2 = 0.0084459... // PkPow 1.7,1.1
So, as expected we get the error also if xoo is used.

Well, so I get *some* numbers here. What would be interesting,
is how these numbers can be related to a correctionfactor d_r
of the asnmatrix/asnpowerseries coefficients to give correct
results for any x and base t in asn_t(x).
It is clear, that the powerseries (including the correction component
d_r)
asn(x) = a1 x + a2 x^2 + a3 x^3 + a4 x^4 + a5 x^5 +...
+ d_r
cannot be corrected by a constant a0 = d_r, since d_r is a scaled
sinefunction dependent on x (and also on t). But I've no idea,
how to proceed here.
Gottfried
Gottfried Helms, Kassel
Posts: 766
Threads: 119
Joined: Aug 2007
06/29/2008, 09:41 PM
(This post was last modified: 06/30/2008, 06:21 AM by Gottfried.)
A curious result from study of the tetraseries. (text updated)
I considered the "reverse" of the tetraseries problem.
Instead of asking for the a_lternating s_um of powertowers of increasing p_ositive heights (asp)
Code: asp(x,dxp) = dxp°0(x)  dxp°1(x) + dxp°2(x)  ... + ...
where dxp(x) = exp(x)1 and dxp°h(x) is the h'th integeriterate.
I asked for a function tf(x) where
Code: asp(x,tf) = (e^x1)/2 = tf°0(x)  tf°1(x) + tf°2(x)  ... +...
so I ask: can (e^x  1)/2 be represented by a tetraseries of a function tf(x) and what would that function look like?
Using the matrixoperatorapproach I got the result
Code: tf(x) = x  x^2 + 2/3*x^3  3/4*x^4 + 11/15*x^5  59/72*x^6 + 379/420*x^7  331/320*x^8
+ 1805/1512*x^9  282379/201600*x^10 + 3307019/1995840*x^11  6152789/3110400*x^12
+ 616774003/259459200*x^13  3212381993/1117670400*x^14 + 54372093481/15567552000*x^15
 594671543783/139485265920*x^16 + 58070127447587/11115232128000*x^17
 1209735800444267/188305108992000*x^18 + 26776614379573099/3379030566912000*x^19
 209181772596680209/21341245685760000*x^20 + 1034961114326994557/85151570286182400*x^21
 80852235077445729119/5352384417988608000*x^22
+ 2210690796475549862239/117509166994931712000*x^23
 18624665294361841906483/793412278431252480000*x^24
+ 379264261780067802109819/12926008369442488320000*x^25
 6584114267874407529534167/179240649389602504704000*x^26
+ 5046681464320089079803469/109576837040799744000000*x^27
 326480035696597942691643978259/5646080455772478898176000000*x^28
+ 327920863401689931801359966641/4511103058030460180889600000*x^29
 418419411682443365665393881223739/4573325169175707907522560000000*x^30
+ 15798888070625404329026746075454779/137047310902965380295426048000000*x^31
+ O(x^32)
which can be determined to arbitrary many coefficients by a recursive process on rational numbers.
The floatrepresentation is
Code: tf(x) = 1.00000000000*x  1.00000000000*x^2 + 0.666666666667*x^3  0.750000000000*x^4
+ 0.733333333333*x^5  0.819444444444*x^6 + 0.902380952381*x^7  1.03437500000*x^8
+ 1.19378306878*x^9  1.40068948413*x^10 + 1.65695596841*x^11  1.97813432356*x^12
+ 2.37715218038*x^13  2.87417649515*x^14 + 3.49265533084*x^15  4.26332874559*x^16
+ 5.22437379435*x^17  6.42433870711*x^18 + 7.92434807834*x^19  9.80176020073*x^20
+ 12.1543397362*x^21  15.1058348510*x^22 + 18.8129220299*x^23  23.4741329327*x^24
+ 29.3411740841*x^25  36.7333765543*x^26 + 46.0560972611*x^27  57.8241911808*x^28
+ 72.6919467774*x^29  91.4912883306*x^30 + 115.280540468*x^31 + O(x^32)
so I assume, that this series tf(x) has radius of convergence limited to about x<0.7
Moreover, the iterates of this function seem always to be of a similar form, so the alternating sum of the found coefficients of the iterated functions at like powers of x is divergent for each coefficient (but may be Eulersummed). So this result must be considered in more detail next, since I had inconsistency of the matrixmethod with serial summation either for increasing positive or for increasing negative heights.
However, for x=1/2 or x=1/3 or smaller we can accelerate convergence of asp() by Eulersummation such that I get good (?) approximation to the six'th digit for x=1/3 using the truncated series with 31 terms only.
The process for the generation of these coefficients is a bit tedious yet; so I don't have for instance  the function, whose iterations must be nonalternating summed to get the exp(x)1 value (or (exp(x)1)/2 or some other scalar multiple) which as I guess could have better range of convergence.
I'll post the result, if I got it.
Gottfried
Gottfried Helms, Kassel
Posts: 766
Threads: 119
Joined: Aug 2007
06/30/2008, 12:11 PM
(This post was last modified: 06/30/2008, 12:13 PM by Gottfried.)
An asnwer in the newsgroup sci.math:
Dear Gottfried,
the function tf(x) verifies
exp(tf(x)) + exp(x) = 2*x +2
So
tf(x)=ln( exp(x) +2*x+2)
and series near 0,
x  x^2 + 2/3x^3  3/4 x^4 + 11/15 x^5 ... [I corrected a missing term]
Alain
So I took the long way...
Gottfried
Gottfried Helms, Kassel
Posts: 766
Threads: 119
Joined: Aug 2007
A new result, which I just posted in sci.math; I'll improve the formatting later (a bit lazy...)
Gottfried
Code: Am 30.06.2008 17:00 schrieb alainverghote@gmail.com:
>> >> Gottfried
> >
> > Well, inverse function of tf(x) using lambertW()
> > is Lambert(1/2exp(1/2e^x  1)) +1/2e^x  1
> > series near 0 , x +x^2 +4/3x^3 +29/12x^4 +51/10x^5 ...
> >
> >
Yepp, I got the same series  good!
But now  iterations and especially the sum of iterations
of this functions in the Lambertrepresentation should be
intractable, so I don't assume this can be helpful to get
more insight in the source of the inconsistenyproblem;
remember:
by the formal application of the matrixapproach
asn(x)+asp(x)x = 0 // expected
or
asn(x,f(x)) = x  asp(x,f(x)) // expected
or
asp(x,f°(1)(x)) = x  asp(x,f(x)) // expected
is not true, at least for the function f(x) = exp(x)1
Well  it was a try...

Meanwhile I refined my computationprocess, so I've
now even the function fz(x) with the condition
e^x  1 = fz(x) + fz(fz(x)) + fz(fz(fz(x))) + ..
= sum{h=1..inf} fz°h(x)
I got
fz(x) = 2*(x/4)/1! + 6*(x/4)^2/2! + 10*(x/4)^3/3!  46*(x/4)^4/4!  554*(x/4)^5/5!
 1690*(x/4)^6/6! + 27882*(x/4)^7/7! + 505986*(x/4)^8/8! + 2529590*(x/4)^9/9!
 61918794*(x/4)^10/10!  1726391798*(x/4)^11/11!  14268435022*(x/4)^12/12!
+ 352044609814*(x/4)^13/13! + O(x^14)
where the integer parts of the coefficients are
[2]
[6]
[10]
[46]
[554]
[1690]
[27882]
[505986]
[2529590]
[61918794]
[1726391798]
[14268435022]
[352044609814]
which have to be divided py powers of 4 and by factorials to give the
coefficients of the function.
The float representation of this function is
fz(x) = 0.500000000000*x^1 + 0.187500000000*x^2 + 0.0260416666667*x^3  0.00748697916667*x^4
 0.00450846354167*x^5  0.000573052300347*x^6 + 0.000337655203683*x^7 + 0.000191486449469*x^8
+ 0.0000265917660278*x^9  0.0000162726971838*x^10  0.0000103115449999*x^11
 0.00000177549511523*x^12 + 0.000000842437121499*x^13 + 0.000000632647393830*x^14
+ O(x^15)
Can we give a range for x where this converges ?
The quotients of subsequent coefficients give the following sequence
0.375000000000,0.138888888889,0.287500000000,
0.602173913043,0.127105896510,0.589222316145,
0.567106466538,0.138870223462,0.611944959460,
0.633671534805,0.172185168687,0.474480112208,
0.750972835462,0.212965249501,0.333234112499,
0.928624638217,0.256835516535,0.218432174828,
1.22327362079,0.303250445581,0.126531683586,
1.82969444504,0.353026162019,0.0509774937397,
3.94499797418,0.407780103462,0.0133041813614,13.1183083394,
0.470026310618,0.0698970628404,2.15481130004,
0.543690120454,...

Using 32 coefficients for the function and 60 iterates for the sum
I could approximate e^1 1 relatively well. I got
sum(h=1,60,fz°h(1.0))  ( exp(1)1 ) = 3.24385306514 E13
where the quality of approximation increased when terms of the
function and numbers of iterates are increased.
Fun... :)
Gottfried Helms
Gottfried Helms, Kassel
Posts: 509
Threads: 44
Joined: Aug 2007
This is probably an old result. For the function
I found using Carleman matrices that
is this related to the series above?
