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

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 Tetra-series here, which points `

to a more fundamental -but general- effect in summing of this

type of series.

I discuss the U-tetration instead of the usual T-tetration here,

because the effect under consideration apparently is the same with

T-tetration, and U-tetration is easier to implement using the

matrix-/diagonalization-method.

I usuallly denote

T-tetration:

Tb (x) = b^x // base-parameter b

Tb°0(x) = x // base b occurs 0 times

Tb°1(x) = b^x // base b occurs 1 times

Tb°h(x) = Tb°(h-1)(Tb(x))

= b^b^b^...^b^x // base b occurs h times

Tb°-1(x) = log(x)/log(b)

U-tetration:

Ut (x) = t^x -1 // base-parameter t

Ut°0(x) = x

Ut°1(x) = t^x -1

Ut°h(x) = Ut°(h-1)(Ut(x))

Ut°(-1)(x) = log(1+x)/log(t)

-------------------------------------------

For the discussion of the series we assume a fixed base-parameter t

here, so I omit it in the notation of the U-tetration-function in

the following.

Also I restrict myself to bases t where all of the following series

are conventionally summable using Cesaro/Euler-summation.

The series under discussion are

(U-powertowers of increasing positive heights)

asup(x) = x - (t^x -1) + (t^(t^x - 1) -1) - ... +...

= sum {h=0..inf} (-1)^h * U°h(x)

(U-powertowers 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

Euler-summed.

If they are summed this way, evaluated term by term, I call

this "serial summation" in contrast to my matrix-approach.

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 matrix-method,

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 matrix-method,

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 U-powertower to base t itself,

then the results in asu(x) are periodic with the integer-height

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 half-iterate 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-/diagonalization-method one can find the

coefficients a,b,c,... for the U°0.5-function 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 so-determined 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 matrix-method 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 matrix-computed 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 phase-shift 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 E-11

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

matrix-method is the sinusoidal function d_r.

So the matrix-method 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 `

matrix-method itself; it can be shown, that the coefficients for

the powerseries of asn(x) determined by the matrix-method 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- Euler-summation, 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 row-entries 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 column-wise summation of coefficients using Euler-summation

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 matrix-method (= collecting coefficients at like powers of x;

// Euler-sums. 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 U-powertower

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 Euler-summation,

where a column c needs order of u^-c, and the implicite binomial-

transform in Euler-summation with infinite increasing order must

be reflected by special considerations.

Gottfried Helms

[1] http://go.helms-net.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 matrix-summation 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 E-113

log(1+xoo)/u - xoo = 2.33942419508380691068587979906 E-113

so we have a very well aproximated fixpoint (I used float-precision

of 1200 decimal digits)

Serial summation: ----------------------------------------------------

check: is asn(xoo) = xoo/2 ?

asn(xoo) = 1.25643120862616967698273761661

asn(xoo)- xoo/2 = -7.45354655251552020322193384272 E-114

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 E-28

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 Euler-summation; but since

Euler-summation 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 correction-factor d_r

of the asn-matrix/asn-powerseries 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

sine-function dependent on x (and also on t). But I've no idea,

how to proceed here.

Gottfried

Gottfried Helms, Kassel