Tetration / Powertowers again

Discussion in 'Math Research' started by Gottfried Helms, Jul 16, 2007.

  1. Finding Andrew Robbins' "tetration"-pages recently
    I was inspired to check, whether my previously
    described method to compute the alternating sum
    of increasing powertowers would also allow inter-
    polation for the tetration-operator to fractional
    iterations, in Robbins' notation:

    z(x,y,a) = x^^y.a

    for fractional y (and a=1 for instance).

    This is indeed possible, at least in some region of
    parameters x and y.

    If I use the matrix-logarithm of my Bs-matrix (as
    described in my previous posts) multiply by a fractional
    y and exponentiate, such that

    By = exp(y*log(Bs))

    then I can use to compute

    z(x,y) := z(x,y,1)

    for any real y in a reasonable domain.

    With dim=24 (only providing 24 coefficients for the
    series to be summed) I get for the following examples:

    -- fractional computation applied--- -- original computation applied ---
    z(sqrt(2),1/2) ~ 1.24362162767
    z(sqrt(2),2/2) ~ 1.41421356237 z(sqrt(2),1) ~ 1.41421356237
    z(sqrt(2),3/2) ~ 1.53880541744
    z(sqrt(2),4/2) ~ 1.63252691944 z(sqrt(2),2) ~ 1.63252691944

    For the less good approximable, but still easily manageable case x=2

    -- fractional computation applied--- -- original computation applied ---
    z( 2 ,1/2) ~ 1.45878179055
    z( 2 ,2/2) ~ 2.00000000000 z( 2 ,1) ~ 2.00000000000
    z( 2 ,3/2) ~ 2.74876151102
    z( 2 ,4/2) ~ 3.99999999999 z( 2 ,2) ~ 4.00000000000

    Critical here is the possibility to approximate
    log(Bs) to a good degree - and formally, to describe the
    range of parameters, for which log(Bs) and thus exp(y*log(Bs))
    is also formally defined.

    Reading Andrew Robbins' articles (*1) cursory it seems, that the
    core computations are equivalent, if not identical.
    This matrix method, which computes not only a single
    value z(x,y) for parameters x and y, but always a complete powerseries
    of that value z(x,y) , meaning

    Z(x,y) = [1 , z(x,y) , z(x,y)^2, z(x,y)^3), ....]

    seems to be then the best description for a complete operator
    to perform continuous tetration and its inverses.

    Gottfried Helms

    (*1) Andrew Robbins at http://tetration.itgo.com/

    For an impression:
    protocol - the first 6 terms of the powerseries-vectors for Z(x,y) are given.
    Z(x,y) = [1 , z(x,y) , z(x,y)^2, z(x,y)^3), ....]

    x = sqrt(2)

    Z(sqrt(2),1/2) ~ 1.00000000000 1.24362162767 1.54659475280 1.92337868383 2.39195532940 2.97468738006
    Z(sqrt(2),2/2) ~ 1.00000000000 1.41421356237 2.00000000000 2.82842712475 4.00000000000 5.65685424949
    Z(sqrt(2),3/2) ~ 1.00000000000 1.53880541744 2.36792211274 3.64377137516 5.60705513201 8.62816681301
    Z(sqrt(2),4/2) ~ 1.00000000000 1.63252691944 2.66514414269 4.35091955712 7.10299330132 11.5958277730

    Z(sqrt(2), 1 ) ~ 1.00000000000 1.41421356237 2.00000000000 2.82842712475 4.00000000000 5.65685424949
    Z(sqrt(2), 2 ) ~ 1.00000000000 1.63252691944 2.66514414269 4.35091955712 7.10299330132 11.5958277730

    x = 2

    Z( 2 ,1/2) ~ 1.00000000000 1.45878179055 2.12804354412 3.10435127462 4.52857888176 6.60624045681
    Z( 2 ,2/2) ~ 1.00000000000 2.00000000000 4.00000000000 8.00000000000 16.0000000000 32.0000000000
    Z( 2 ,3/2) ~ 1.00000000000 2.74876151102 7.55569462067 20.7688507896 57.0888799638 156.924683512
    Z( 2 ,4/2) ~ 1.00000000000 3.99999999999 15.9999999846 63.9999975683 255.999858206 1023.99518770

    Z( 2 ,1 ) ~ 1.00000000000 2.00000000000 4.00000000000 8.00000000000 16.0000000000 32.0000000000
    Z( 2 ,2 ) ~ 1.00000000000 4.00000000000 16.0000000000 64.0000000000 256.000000000 1024.00000000

    Gottfried Helms, Jul 16, 2007
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  2. Intuitively one would expect to see the identity

    x(x,y+1) = x^z(x,y)

    That seems to work for your examples, but showing it either numerically
    or analytically would make the argument more powerful.

    Oscar Lanzi III, Jul 21, 2007
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  3. Oscar -

    thanks again for your discussion!

    Am 21.07.2007 16:18 schrieb Oscar Lanzi III:
    Hmm, I've difficulties to understand this, since
    on the left hand I'm missing a "z", did you mean

    z(x,y+1) = x^z(x,y)


    It works exactly like this. One other numerical example:

    x=s=1.2 (I'm using s for the basis in my examples)
    Then for

    V(1)~ * Bs^k (for k=0..9)

    in the 2'nd column of the result-vector I get the entries
    documented in the table-column "matrix",
    corresponding to the direct scalar computation (table-
    column "skalar").

    In table-column "Step 3 matrix" I get the results for

    V(1)~ * (Bs^3)^j for j=0..3

    which all are perfectly compatible.

    VPT(s,9) step 3
    matrix skalar matrix
    B=Bs x^z(x,y) C= Bs^3
    B^0 -> 1 1 C^0 -> 1
    B^1 -> 1.200000000 1.200000000
    B^2 -> 1.244564747 1.244564747
    B^3 -> 1.254718171 1.254718171 C^1 -> 1.254718171
    B^4 -> 1.257043041 1.257043041
    B^5 -> 1.257575982 1.257575982
    B^6 -> 1.257698182 1.257698182 C^2 -> 1.257698182
    B^7 -> 1.257726204 1.257726204
    B^8 -> 1.257732629 1.257732629
    B^9 -> 1.257734103 1.257734103 C^3 -> 1.257734103
    .... ->

    It seems that the Eigenvalues of Bs are the consecutive
    powers of w= 1.257734541 ,where w^w = s = 1.2

    Commenting the analytical side, where I made some
    progress in finding promising heuristics and one
    possible pathway to a final analytical deciphering (see 4).

    1) binary operator vs matrix-operator - notation
    for tetration?

    I've found the collection of notations at A.Robbins
    site and tried to reformulate my results in terms
    of one of this "classical" forms. It doesn't fit
    the notation exactly, so I came to the provisorial

    - with the given form of notation for binary operators
    we use a somehow "anonymous" operator which is thought
    to be an abstract, to be independent of its operands.
    The common use of "+","*","^".or "^^" keeps the
    effect of the binary operator somehow independent of
    its operands.

    - The matrix-approach tells me, that this view of
    things may be not sufficient. For tetration it would be
    better to have a sense instead, that the operator
    itself has a parameter "imprinted", such that in any
    notation of a chosen binary operator-symbol this symbol
    must have a connotation of this "imprinted" parameter,
    which is the parameter s in my case (the base-parameter)
    and may be a constant for the classical binary operations.

    I'm checking, how this concept of an operator
    with "imprint" is also meaningful for the other
    common binary operators, too.
    As I have shown in the other article (*1), the pathway
    to the eta-and zeta/bernoulli-polynomials can
    also be seen as an iteration of an operator (the
    binomial-matrix, which implements x^k -> (x+1)^k
    for all k). But I didn't formulate an explict
    conclusion yet.

    The notation of an "imprinted" binary operator
    seemed then to be a bit artificial and I left
    it because of its only relative benefit or even drawback
    in flexibility and precise expression of what's going
    on, and stick to the simple matrix-formula instead.

    So my notation will be

    V(z)~ = V(x)~ * Bs^y

    using the second column of z for the scalar result
    and which is nearest to one of the notations
    cited by A. Robbins:

    s^^y.x = s^s^s...^s^x (y-fold occurence of s)


    2) Continuous version by Eigensystem-decomposition

    Analytically it appears, that from the finite dimensional
    approximations of Bs with parameters s for the convergent
    cases of the infinite powertower (I will call them
    the "simple cases") one can conclude, that Bs dim->oo has
    a discrete spektrum, and in the simple cases eigenvalues
    between 1..0 or 1..-1 , possibly with a simple geometric
    progression occur.

    In the simple case an eigendecomposition makes sense and

    Bs = Qs * Ds^y * Qs^-1

    I can operate with a continuous parameter y.

    2.a) Inverse operation

    I can even formulate the inverse operation, given x,z, and s,
    searching y

    V(z)~ = V(x)~ Qs * Ds^y * Qs^-1
    (V(z)~*Qs) = (V(x)~*Qs) * Ds^y

    then the parentheses are known row-vectors, which
    may be rewritten as multiplication of the unit-
    rowvector V(1)~ multiplied by known diagonalvectors
    for instance

    V(1)~ Ts = V(1)~ Us * Ds^y
    V(1)~ Ts = V(1)~ Ds^y * Us // interchaning the rightmost

    and then

    Ds^y = (Ts*Us^-1)

    and getting the logarithm of the entries of Ds
    to the base s.

    Just today I found, that the eigenvalues for the
    simple cases seem to be the powerseries of a value
    t, where t is defined by the tetration-expression:

    s = t^t

    I'll check about the cases, where s>e^e^(-1) next
    time, but it is difficult to compute reliable eigen-
    system-decompositions for the current matrices when
    dim>~20 using Pari/GP, even with float-precision
    of 400 or 800.


    3) the infinite alternating sum of powertowers of
    increasing height

    The heuristics about the spectrum of Bs also
    explain the "niceness" of the cases for infinite
    summations of increasing powertowers. It was
    expressed as

    Z~ = V(x)~ * Ms

    where Ms = (I + Bs)^-1

    The spectrum for the simple cases is in the range

    r = [1,,0[ // discrete

    and even the cases , where it ranges r= [1..-1[
    and has a convergence-point for all eigenvalues of
    high index, will be shifted to r' = [2..[0 if the identity-
    matrix is added to Bs, and I+Bs can then be inverted
    still having an analytically accessible set of
    finite eigenvalues.

    The surprising ability of the method to sum power-
    towers of increasing height even in some divergent
    cases is then due to the this discreteness
    of the spectrum of the operator and this shifting,
    which allows then inversion also for a range of
    non-simple cases, as far as eigenvalues r_inf -> 0
    do not occur.


    4) A true triangular operator for a very near related
    operation possibly provides better analytical access
    to the eigensystem-properties

    I just found also another pathway to possibly
    assure these heuristics and base them with
    analytical arguments.

    This will then be due to the fact, that Bs can be
    decomposed in the product of two well known
    matrices, which are triangular and additionally have
    eigenvalues of all 1 if s=exp(1) or generally the power-
    series of log(s).

    Given the known decomposition

    Bs = dV(log(s)) * fS2F * P~


    - dV(log(s)) is a diagonalmatrix containing
    the powers of log(s)

    - fs2F is the factorial similarity-scaling of
    the lower triangular matrix of Stirling-numbers
    2'nd kind (which thus has unitary spectrum)
    using F = diag(0!,1!,2!,...) and f= F^-1

    - P is the lower triangular Pascal-matrix , and
    P~ its transpose.

    one may use a version free of P~

    As = dV(log(s)) * fS2F

    Then this is an operator, which is triangular, has known
    spectrum (the powers of log(s)) but still behaves very
    similar to Bs. It gives, by one iteration,

    V(x)~ * As = V(s^x-1)

    meaning I get the powers of (s^x-1) instead of s^x
    as in the case of using Bs .

    Letting s=e and x=1 I get by powers of As in the
    second column of the result the iterated exponentiation
    for the scalar parameter e:

    (V(1)~ * As ) [1]= e^1-1
    (V(1)~ * As^2 ) [1]= e^(e^1-1)-1
    (V(1)~ * As^3 ) [1]= e^(e^(e^1-1)-1)-1
    or letting s=2

    (V(1)~ * As ) [1]= 2^1-1 =1
    (V(1)~ * As^2 ) [1]= 2^(2^1-1)-1 =1
    (V(1)~ * As^3 ) [1]= 2^(2^(2^1-1)-1)-1 =1

    which is an identical type of result, and also can
    be used to compute alternating infinite increasing
    powertower-sums for some s.

    Possibly I can find an analytical proof, that
    the spectrum of Bs is indeed the powers of
    its parameter s (or some shifting or function
    of it) from its near relation to As.


    Well, I didn't add more new numerical examples here -
    I thought the mentioned analytical aspects are
    the more interesting at this stage of discussion.
    If you would like to see some numerical example,
    I would provide them in my documentation of power-
    tower tables (link was given in previous post)
    and post a notification.

    Gottfried Helms, Jul 23, 2007
  4. I did in fact mean z(y+1,x) on the LHS in my posting. Also for w =
    1.257+ it is w^(1/w) that equals 1.2, not w^w.

    I'm curious about what happens when s drops below 1 and the
    integer-tetration values acquire an oscillatory character. Does this
    lead to the introduction of complex numbers for fractional tetrates?

    Oscar Lanzi III, Jul 24, 2007
  5. Sorry ,

    at some places I was mixing things at the second
    editing of the posting.

    Am 23.07.2007 15:05 schrieb Gottfried Helms:
    this is wrong, this value is in fact the exponential of
    the value, what should have been given here.

    The ordered eigenvalues of Bs are

    [ 1.00000000000 0.229312119617 0.0525840482028 0.0120581594905 ....]

    Let r be the second eigenvalue, r = 0.229312119617; then

    exp(r) = w = 1.257734541

    w^(1/w) = 1.2 = s
    exp(r * exp(-r)) = 1.2 = s

    And the second place is wrong the same way by quickly
    inserting new results in the nearly finished posting:
    It should have been the same way as above,
    first correcting

    s = t^(1/t)

    and then

    t = exp(r)

    where r is the second highest eigenvalue, and the sequence
    of eigenvalues are the powers of r.

    These corrections don't affect the other contents
    of the article.

    Gottfried Helms, Jul 24, 2007
  6. Am 24.07.2007 02:59 schrieb Oscar Lanzi III:
    Again from heuristics I assume the following.

    I observed negative eigenvalues for such cases, and
    also, that the sequence of negative eigenvalues might have
    a convergence point between ]0..-1] for e^-e < s <1.

    In the light of the previous post, it seems, these
    "convergence points" are simply the negative values
    of the geometric series for negative argument r.

    This would fit perfectly the observations, as far as
    I can calculate reasonably reliable eigen-decompositions
    for the parameter s, where e^-e <= s <= e^e^-1.

    s=0.9; Bs = PT_Bs(s); \\
    MEig(VE(Bs,20)) \\ using dimension 20
    Mat(CEV): \\ cached vector of Eigenvalues

    Assume the second highest /absolute/ value indicates r,
    the parameter for the geometric series, then

    %1375 = 0.908699313092
    %1376 = 0.900000000000
    which equals s (accepting as reasonable approximation).


    2) s <1 and towards the lower bound

    Already for s<2/3 I have surprising results, where
    some eigenvalues deviate from the expectation,
    and I didn't find a special critical point yet,
    which apparently lies already in the range

    0.64 < s < 0.666... = 2/3

    and possibly at s=1-log(2)/2 ~ 0.653426409720

    I get sometimes small complex eigenvalues and
    also often one or two exceeding -1 or +1, depending on
    whether I select odd or even dimension. With s=1/2
    this is already very clear, as I have documented
    in the appendix (*1). These exceptional eigenvalues
    are not assignable to the geometric series built
    by the remaining "normal" eigenvalues.
    Trying with Bs^2, where the entries are analytically
    determined (thus *not* by simply squaring the
    finite dimension examples of Bs), it seems a bit
    better, but still numerically questionable.

    It may be due to a numeric instability of the
    eigensystem-method in Pari/Gp and I would be
    interested, how Mathematica or Maple would perform

    3) s exceeding the upper bound

    For s>e^e^-1 the sequence of eigenvalues diverges
    as expected, additionally the sequence is then no more
    a geometric series and also I have eigenvalues in
    the range [1..0[ (finite dimension), generally
    I get the range for the k'th eigenvalues:

    0 < r_k

    The alternating sum of powertowers of increasing height
    is still computable due to this result, since for
    (I + Bs) I get then eigenvalues

    1 < r_k

    and for Ms = (I + Bs)^-1 this is then for the
    eigenvalues of Ms

    1 > r_k

    which explains then the nice and stable computations
    even for the parameters s, which would produce
    divergent cases for the single infinite powertower.



    (*1) last page of
    Gottfried Helms, Jul 25, 2007
  7. A new(?) interesting intermediate result:

    With the taken assumptions concerning the characteristic
    of the spectrum of the matrix Bs, being a geometric series
    related to its parameter s.

    Decompose the given parameter s into the following:

    s = t^(1/t) // for admissible s

    then my assumption concerning the asymptotic spectrum
    for infinite simension is with

    v = log(t)

    about the k'th eigenvalue (the index k=0 for first eigenvalue)

    v_k = v^k

    We know, that the sum of eigenvalues equals the trace of
    a matrix, so under this assumption it should be:

    trace(Bs) = sum {k=0..inf} v_k
    = sum {k=0..inf} v^k
    = (1-v)^-1 (*1)

    The structure of Bs is simple; the sum of its diagonal
    elements are

    trace(Bs) = sum {r=0..inf} log(s)^r/r! * r^r
    = sum {r=0..inf} (log(t)/t )^r* r^r / r! (*2)

    (*1) and (*2) should thus be equal , - and the
    equality can be tested for arbitrary dimension;
    the found equality backs my assumption numerically.

    Is the resulting equality

    sum {r=0..inf} (log(t)/t )^r* r^r / r! = (1-log(t))^-1

    a known one?

    Gottfried Helms, Jul 31, 2007
  8. [snip for brevity]
    Yes. It can be derived directly from one of the identities on the paper "On An
    Application of Lambert's W Function to Infinite Exponentials".

    Specifically, the last of the three identities on page 773, has:

    sum((log(x)/x)^(n-1)*n^(n-1)/n!,n=1..oo)=h(x^(1/x)), when |log(x)/x|<=1/e.

    Multiplying the above by log(x)/x, we get:


    h is a partial inverse of the function f(x)=x^(1/x), therefore when x is in the
    above range,


    Let a_n be the coefficients of A, i.e., a_n=n^(n-1)/n!.

    Then the coefficients b_n of the series for B=A/(1-A), are exactly:


    which are the coefficients of



    sum((log(x)/x)^n*n^n/n!,n=1..oo)=log(x)/(1-log(x)), from which follows:

    I.N. Galidakis, Aug 1, 2007
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