# -- tetration: news on series of increasing heights of powertowers(tetra-series)

Discussion in 'Math Research' started by Gottfried Helms, Jun 20, 2008.

1. ### Gottfried HelmsGuest

I have a new result for the Tetra-series here, which points
to a more fundamental -but general- effect in summing this
type of series.

I discuss the U-tetration instead of the usual T-tetration here,
because the effect under consideration is apparently 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) = 0.5, t=exp(1/2) 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 to be identical for both
methods, and only asun(x) differed:

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.

----------------------------------------------------------
But we need some prerequisites.

First note, that if x is seen as U-powertower to base t itself,
then the results are periodic with the integer-height part of x;
so if

x = U¡h(1)

then the results occur periodically with k in

x_k = 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

U¡1(1) = t^1 -1 = exp(1/2) - 1 = 0.648721...

with very good approximation using, for instance, 96 so-determined
terms of the powerseries for U¡0.5(x).

So we may assume, U¡0.5(1)= 0.815903... is determined with
sufficient (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...) = 0.318361... // matrix
asu_0.5 = asu (U¡0.5(1)) = asu (0.815903...) = 0 // matrix

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

The first result is now, that -apparently- for any r 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 = arctan(d_0.5/d_0)

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.

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

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 the 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

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:

... ... ... ... ... ...
-------------------------------------------------------------------------------
asn(x)= -a1 x + a2 x^2 -a3 x^3 +a4 x^4 -a5 x^5 +...

with explicite values for coefficients

asn(x) // matrix-method (= collecting coefficients at like powers of x, Euler-sums)
= 1/3 x +1/15 x^2 +2/405 x^3 -0.0010893246 x^4 -0.000457736 x^5 +...

So, by comparision, we know, that this result is false and needs
correction by a component, 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

[
Moderator's Note
The non-ASCII character in, for example, Tb!0 is a degree-sign.
But it is best not to use any non-ASCII characters here.
]

Gottfried Helms, Jun 20, 2008

2. ### Gottfried HelmsGuest

Sorry, I have to append some corrections (for unimportant
statements) Also I exchange the unreadable degree-sign
by double-caret here

Am 20.06.2008 13:43 schrieb Gottfried Helms:
(...)
precisely the result is periodically for asu(x)
only; for asup(x) and asun(x) the beginning of
the series is simply shifted, so we have for
integer h

asup(U^^(2h)(1)) = asup(1) - (U^^0(1)-U^^1(1)+...+U^^(h-1)(1))
asun(U^^(2h)(1)) = asun(1) - (U^^1(1)- ...+U^^h(1))

But this does not affect the following argumentation, since
the focus is on asu(x) only.

Gottfried Helms

Gottfried Helms, Jun 20, 2008

3. ### Gottfried HelmsGuest

Sorry, I have to append some corrections (for unimportant
statements) Also I exchange the unreadable degree-sign
by double-caret here

Am 20.06.2008 13:43 schrieb Gottfried Helms:
(...)
precisely the result is periodically for asu(x)
only; for asup(x) and asun(x) the beginning of
the series is simply shifted, so we have for
integer h

asup(U^^(2h)(1)) = asup(1) - (U^^0(1)-U^^1(1)+...+U^^(h-1)(1))
asun(U^^(2h)(1)) = asun(1) - (U^^1(1)- ...+U^^h(1))

But this does not affect the following argumentation, since
the focus is on asu(x) only.

Gottfried Helms

Gottfried Helms, Jun 20, 2008