# Tetration Forum

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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
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
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
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
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.
hej bo198214

bo198214 Wrote:
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.
Thanks, I understood that ; I will try more carefully:

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
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.
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
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

[attachment=120]
Gottfried Wrote:there is an article of L Euler, which seems to deal with it.

At least I see $e^{\frac{\pi}{2}i}=i$ in the text ...
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 $b^z$.