Tetra-series

[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 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