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

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

  1. 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)


    With this I conjecture the following identity:

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


    Differently written:

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


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


    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


    Comments are much appreciated.

    Gottfried Helms
    Gottfried Helms, Nov 21, 2007
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  2. 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
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