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+--- Thread: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) (/showthread.php?tid=54)
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - GFR - 09/21/2007
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
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Gottfried - 09/22/2007
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\(q2e\)03$1@news.t-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
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Gottfried - 10/20/2007
I've just uploaded a short discussion of the fixpoint-detail, as discussed in my previous postings. I'm trying to show, that by selection of a parameter beta I can construct a complex fixpoint, and it is made sure, that I get a purely real base b > eta. The graph and the formulae seem to indicate that the full range eta < b < inf can be constructed by a continuous function f(beta) = b for 0<= beta < pi, which would then confirm Gianfrancos and my own earlier formulae.
See
Fixpoints
It's another sketchy manuscript of mine, but... ;-)
Gottfried
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 10/30/2007
Hello,
I am new here- I was wandering about the imaginary unit for some time to be able to understand physics better, but I was quite positively surprised that you have confirmed by rigorous analysis what I just yesterday found out by studying the possible content of i- a definition ( or perhaps one of the definitions of i) -of imaginary unit.
Perhaps there are many infinite values involved as well as this definition arises from Lambert function of a logarithm, and logarithm may take infinitely many values, but still
I keep - i sign):-i = power tower (e^pi/2) = h (e^pi/2)
i = - h(e^pi/2)
so here z = e^pi/2 = 4,81047738097.........
h(z) = - i
Derivation:
h(e^pi/2) = -W( -ln e^pi/2)/ ln e^pi/2 = -W(-pi/2) / pi/2
But W( -pi/2) = i*pi/2 so
h(e^pi/2) = - i*pi/2 / pi/2 = - i .
but - h(e^pi/2) = W(-pi/2) / pi/2 = i*pi/2/pi/2 = i
There must be many more such beauties outside radius of convergence of h(z).
Best regards,
Ivars Fabriciuss
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - bo198214 - 11/02/2007
Hey Ivars, welcome on board!
Ivars Wrote:-i = power tower (e^pi/2) = h (e^pi/2)
But be aware that this is mathematically wrong!
Correct is:
power tower (e^(pi/2)) = oo
If you mean something different for example an infinite valued function or a fixed point of x^x then you have to explicitely state that.
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 11/03/2007
hej bo198214
bo198214 Wrote:Thanks, I understood that ; I will try more carefully:Ivars Wrote:-i = power tower (e^pi/2) = h (e^pi/h2)
But be aware that this is mathematically wrong!
Correct is:
power tower (e^(pi/2)) = oo
If you mean something different for example an infinite valued function or a fixed point of x^x then you have to explicitely state that.
Analytic continuation of h(z) defined as = -W(-ln(z))/ln(z) at point z=e^(pi/2) is - i.
also, in this case:
h(e^pi/2)^i = e^(pi/2)
h(e^pi/2)^-i= e^(-pi/2) as
-i^i=e^(pi/2) = 4,810477.. -i^-i = e^(-pi/2)=0,207879
i^i = e^(-pi/2)= 0,207879.. i^-i = e^(pi/2) = 4,810477
and
i*ln h(e^pi/2) = pi/2
ln h(e^pi/2) = -i pi/2 = pi/2*(h(e^pi/2)
(h(e^pi/2))^2 = -i^2 = e^-ipi = i^6
However, as e^-pi/2 > e^-e,
h(e^-pi/2) = 0,474541.......= (2/pi)*W(pi/2)
So there is no symmetry and i will not be a result of analytic continuation of infinite tetration of anything, while -i is.
Best regards,
Ivars
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - bo198214 - 11/03/2007
Ivars Wrote:However, as e^-pi/2 > e^-e,
h(e^-pi/2) = 0,474541.......= (2/pi)*W(pi/2)
So there is no symmetry and i will not be a result of analytic continuation of infinite tetration of anything, while -i is.
This depends on how you analytically continue h, on which path.
W (and hence h) has a singularity at \( -\frac{1}{e} \), so if you continue on different paths around \( -\frac{1}{e} \), you possibly get different results (similar to the analytic continuation of the logarithm with singularity at 0).
Surely on another path/branch
\( h(e^{-\pi/2})=i \).
Edit: Perhaps its not at all "Surely", was just my conjecture.
Edit2: Now its not a conjecture anymore but regarded as nonsense XD, instead \( h(e^{\pi/2})=i \) on another branch.
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 11/03/2007
Hej Bo,
Would be interesing to see a concrete example where h(z) is analytically continued to get h(z) = i, on any branch, any z. Might contain interesting information, whatever the outcome.
Ivars
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Gottfried - 11/03/2007
Ivar -
there is an article of L Euler, which seems to deal with it. Its index in the Euler-archive is E489, and I found it available at some university-library with a digitizing service. I found the attached snippet interesting, and may be it inspires to have a deeper look into it.
Gottfried
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - bo198214 - 11/03/2007
Gottfried Wrote:there is an article of L Euler, which seems to deal with it.
Wow, Gottfried! Where did you dig out this article?!
However it seems to be beneficial being able to read Latin ...
At least I see \( e^{\frac{\pi}{2}i}=i \) in the text ...
And there we have it already, Ivar.
One branch of the LambertW function gives \( W(-\frac{\pi}{2})=-\frac{\pi}{2}i \), hence for this branch \( h(e^{\frac{\pi}{2}})=i \). (Forget what I said about \( h(e^{-\frac{\pi}{2}}) \), that was a somewhat misled speculation).
We can imagine the branches of \( h(b) \) given by the fixed points of \( b^z \).