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Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))
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.



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

Kind regards -



> > 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)

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.


Gottfried Helms, Kassel

Messages In This Thread
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 09/22/2007, 07:36 AM
RE: Tetration below 1 - by Gottfried - 09/09/2007, 07:04 AM
RE: The Complex Lambert-W - by Gottfried - 09/09/2007, 04:54 PM
RE: The Complex Lambert-W - by andydude - 09/10/2007, 06:58 AM

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