# [tetration] Alternating series of powertowers of increasing heights/aconjecture

Discussion in 'Math Research' started by Gottfried Helms, Nov 21, 2007.

1. ### Gottfried HelmsGuest

Define the (integer) tetration (="powertower of integer height") as

Tb(x,h) = b^b^b^...^b^x with h-fold repetion of base-parameter b

for height h=0
Tb(x,0) = x

and for negative height

Tb(x,-h) = lb(lb(...(lb(x))...) with h-fold repetition of lb
where lb(z) is log(z)/log(b)

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With this I conjecture the following identity:

0 = \sum_{h=-inf}^{+inf} (-1)^k Tb(x,h) (1)

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Differently written:

x -b^x + b^b^x - ... + ... (2)
= lb(x) - lb(lb(x)) + ... - ...

or

0 = ... +lb(lb(x)) - lb(x) + x - b^x + b^b^x -... +... (3)

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This would then be another way to compute the extremely divergent series

s = x - b^x + b^b^x -... +... (4)

(which we discussed here some weeks ago) by

s = lb(x) - lb(lb(x)) +... - ... (5)
which can be summed by cesaro or Euler-summation.

x cannot assume an integer power of b here, especially not the
values 1 or 0, since in (5) we would then introduce singularities.

A related discussion of this can be found in a current thread in
the "tetration-forum" at

Gottfried Helms

Gottfried Helms, Nov 21, 2007

2. ### Gottfried HelmsGuest

Am 21.11.2007 12:15 schrieb Gottfried Helms:

Due to a counterexample by Prof. Edgar (see sci.math) I have to
retract this conjecture.

The error may essentially be due to a misconception about the
matrix of Stirling-numbers 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 pascal-matrix, which could be cured by assuming
an non-neglectable 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_{r,r+1}

p_{r,r+1} = binomial(r,r+1)*1/k

then in the sum of all powers of p we get the entries

s_{r,r+1} = binomial(r,r+1)* sum_{k=1..inf} 1/k

By defining

binomial(r,r+1) = r!/(r+1)!(-1)! = 1/(r+1) / (-1)!

assuming zeta(1)/(-1)! = - 1 this leads to the non-neglectable
rational quantities in that subdiagonal of the sum-matrix

s_{r,r+1} = (1/(r+1) / (-1)!) * sum_{k=1..inf} 1/k
= 1/(r+1) * zeta(1)/(-1)!
= - 1/(r+1)

With that correction the infinite series of powers of the
pascal-matrix leads then to a correct matrix, which provides
the coefficients of the (integrals of) the bernoulli-
polynomials, and can be used to express sums of like powers as
expected and described by H.Faulhaber and J.Bernoulli.

---

This suggests then to reconsider the matrix of Stirling-numbers
of 1'st kind with the focus of existence of a similar structure
in there.
Is something known about such an aspect of this matrix?

Gottfried Helms, Kassel

Gottfried Helms, Nov 23, 2007