09/22/2007, 07:36 AM

GFR Wrote:Thanks, Gottfried.

I also shall read again your postings. My problem is that, as a ... bloody engineer, I am (... was?) more interested to what happens, in a more real environment, for b < eta. But, now, I think that what happens at b > eta should indeed be of the same "nature" (I mean "mathematical nature") of what hapens elsewhere. The ... "function" must be ONE.

Best wishes.

Gianfranco

Gianfranco,

I append one post of mine of the newsgroup sci.math here. There was nearly exactly the same problem posed.

Kind regards -

Gottfried

news://news.t-online.de:119/fd2bf9$q...online.com

> > So as promising as this idea may seem, defining

> > x^^(1/n) as a tetraroot will never give a function that will

> > be continous. This is why Gottried Helms's (and others')

> > functions disagree with the tetraroot -- x^^(1/n) must

> > be less than the nth tetraroot for sufficiently large n in

> > order for the function to be continous at zero.

> >

Hmm, I don't have a final answer for this yet. But with infinite

series, especially in the complex domain, we have some couriosities

everywhere, for instance the non-trivial zeros of the zeta-function.

Also the function (1+1/x)^x x->inf, is a tricky one in this regard.

Here we find, that the quantitative difference of the two approximations

of 1/x->zero in the sum and x->inf in the exponent gives a surprising result.

In a current thread in the tetration forum I posed the question

of s^t = t for real s, complex t, and a correspondent focused

this to the question of what happens, if s->inf.

One can rewrite this as

lim{s->inf} s = t^(1/t)

then

lim{s->inf} 1/s = (1/t)^(1/t)

and then

lim{s'->0} s' = t'^t' (1)

and search for a limit in complex t', possibly sharpening

by the condition, that imag(s')=0, not only in the limit.

That there are solutions for finite s, thus nonzero s' seems

to be obvious by numerical approximation, although I don't have a

authoritative reference for this.

If one resolves into t' = a + bi and log(t') = p + qi and expand

(1) then we arrive at something like

(a+bi)^(a+bi) -> 0

(a+bi)^a * (a+bi)^bi -> 0

(a+bi)^a * exp( bi * log(a+bi)) -> 0

(a+bi)^a * exp( bi * (p + qi)) -> 0

(a+bi)^a * exp( p*bi - q*b) -> 0

(a+bi)^a * exp(-qb)*exp( p*bi ) -> 0

where the last exp(p*bi) is just a rotation, and thus irrelevant for

the convergence to zero.

So it must also

(a+bi)^a * exp(-qb) -> 0

and the exp()-term cannot equal zero, so the first term only

must be zero, if not q or b diverge to infinity.

Now factor a out to have

a^a*(1 + b/a i)^a -> 0

and this reminds me to the above (1 + 1/x)^x - formula.

I'm unable to proceed from here at the moment, but this seems

to focus the core point of the problem.

Possibly we arrive here at the famous indeterminacy, what is

0^0? It depends on the source of the term: whether the exponent

or the base is -from the context of this formula- a limit-expression.

On the other hand, there may be some application of L'Hospital

needed here, I don't see it at the moment.

Hmm.

Gottfried

Gottfried Helms, Kassel