Re: Why is it so difficult?

Am 10.08.2007 09:56 schrieb mike3:
Hi.

Why is it so difficult, anyway, to extend tetration to real numbers?
For example, why is it easier to define 2 raised to the one half
power, but hard to define 2 tetrated to the one half tower?

It seems to lose much of its mystery, if we extend our notion
of binary operators to such of powerseries, or more precisely,
sets of powerseries, collected into infinite matrices, which
finally can be seen as "operators" in matrix-form.

Then tetration for integral height can be computed by powers
of that operator, and of fractional or real height by fractional
of real powers of that matrix. Fractional or real powers (even
complex powers) of such a matrix can be obained, if an eigensystem-
decomposition is possible (and a usable approximation can be
given).
So, if in the matrix-equation

V(x)~ * Bs = V(s^x)~ (* see footnote)

(where the matrix Bs is the tetration-operator for a certain
parameter s) and its iterates

V(s^x)~ * Bs = V(s^s^x) ~
...
expressed as

V(x) ~ * Bs^y = V(s^s^s^...^s^x)~ // s occurs y-times

an eigensystem decomposition of Bs is available, such that

Bs = Q * D * Q^-1

then we may extend tetration to arbitrary operands y by

V(x)~ * Q * D^y * Q^-1 = V({s,x}^^y)

where {s,x}^^y means the expression

s^s^s^..^s^x with y-fold occurence of s

as well as its continuous extension, since a power of
a diagonal matrix is just the diagonal matrix of the powers
of its entries.

Then also we have simple arithmetic in the position of y:

V(x)~ * Bs^y = V(x)~ * (Bs^(y/m))^m

and the like.

If this view of an binary operator as a matrix-product on
powerseries-vectors can be re-extended to the other common
binary operators +,*,^ analoguusly, then it seems to me, that
the special difficulty of tetration may be the observation,
that the parameter s is "enclosed" in the matrix Bs - not only
in its eigenvalues D but also in its eigenvector-matrices, so
that for iterates the scalar decomposition of the involved
computations the parameter s cannot easily be extracted or
factored out. For the other operators the operator-matrix is
possibly constant or something more easy, but I did not have
time to make a proposal for all three operators.
Well, for the + - operator one may replace Bs by the pascal-
matrix P, then we have

P * V(x) = V(x+1)
P^y * V(x) = V(x+y)

and continuous and even complex powers of P are easily obtainable
using the extremely simple matrix-logarithm of P, writing then

P^y * V(x) = exp( log(P) * y) * V(x) = V(x+y) // y any complex number

or by the occurence of a simple toeplitz-matrix, if P is taken
to powers.


Gottfried

(*) for definitions of the matrices see my previous postings
or http://go.helms-net.de/math/binomial_new/10_4_Powertower.pdf
This article needs some brush up, I hope, I'll have it ready
these days

--
---

Gottfried Helms, Kassel
.

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