Re: Logarithm of repeated exponential

In article <1247669680.848240@athprx04>,
"I.N. Galidakis" <morpheus@xxxxxxxxxxxx> wrote:

master1729 wrote:
World Wide Wade wrote :

In article
<20f2d3ea-1e87-45dc-86b2-a917f89e9370@xxxxxxxxxxxxxxxx
legroups.com>,
mike3 <mike4ty4@xxxxxxxxx> wrote:

Hi.

I noticed this.

log(3) ~ 1.098612288668109691395245237
log(log(3^3)) ~ 1.192660116284808707569579569
log(log(log(3^3^3))) ~ 1.220795907132767865324020633
log(log(log(log(3^3^3^3)))) ~ 1.221729301870251716316203810
(calculated indirectly via identity log(x^y) = y log(x).)
log(log(log(log(log(3^3^3^3^3))))) ~ 1.221729301870251827504003124
(calculated indirectly via identity log(log(x^x^y)) = y log(x) + log
(log(x)).)

It seems to be stabilizing on some weird value, around 1.2217293. What
is this? And we seem to run out of log identities here making it
infeasible to compute further approximations.

Has this been examined before?

Here's a fairly elementary proof that this sequence
converges (super
fast). Let L_n denote the n-fold logarithm, and let
x^{n} denote the
n-fold tower exponential. Your sequence is then a_n =
L_n(3^{n}).

Claim 1; a_n is strictly increasing: L_(n+1)(3^{n+1})
=
L_n(L(3^{n+1})) = L_n(3^{n}*L_1(3)) > L_n(3^{n})
(using L_1(3) > 1 and
the fact that L_n is strictly increasing).

i agree.

and so does Ioannis.



Claim 2: a_(n+1) - a_n < 1/[e^{1}*e^{2}*...*e^{n-1}]
(which shows a_n
converges rapidly). Proof: From above a_(n+1) =
L_n(3^{n}*L_1(3)) <
L_n(2*3^{n}). By the mean value theorem, L_n(2*3^{n})
- L_n(3^{n}) =
3^{n}*L_n'(c) for some c between 3^{n} and 2*3^{n}.
Now L_n'(c) =
1/[L_(n-1)(c)*L_(n-2)(c)*...*L_1(c)*c], and c > 3^{n}
e^{n}. Because
L_k(e^{n}) = e^{n-k}, we're done.

this confuses me ?

I don't understand Wade's argument, either, but that doesn't mean he is
wrong.
Wade is almost never wrong, but see below.

what about Ioannis post then ?

http://mathforum.org/kb/message.jspa?messageID=6783965&tstart=0

you cannot both be correct.

No offense but on first sight ; Im betting on Ioannis :)

I cannot find fault with my argument either, but I wouldn't bet on me. In any
case, there's another bit which I find strange, so let's see if someone can
explain it.

Clearly:

log[3](x) < log(x) (i.e., the log base-3 function lies below the natural log
function).

Now iterate n-1 times:

log^{(n-1)}[3](x) < log^{(n-1)}(x)

The above is of course supported by Maple. For n=4, for example:

f:=x->log[3](x);
g:=x->log(x);
with(plots):
p1:=plot((f@@3)(x),x=1..10,color=red):
p2:=plot((g@@3)(x),x=1..10,color=green):
display({p1,p2});

Image (red is base 3 log iterate):

http://misc.virtualcomposer2000.com/log3e.gif

Hence:

log^{(n-1)}[3](3^^n) < log^{(n-1)}(3^^n), for all n\in N.

But the left term is just 3, for all n\in N. So if the sequence indeed
converges
to a, then, taking limits, we have:

3 <= a.

The sequence is log^{n}(3^^n), leading to 1 < a. Nothing strange there.

We have two operators here: One is the power function 3^x and the other is
log(x). I find it highly suspicious that n-1 iterates of log cancel the
growth
of n iterates of 3^x, which grows faster than the inverse of log, which is
e^x.
Something is very weird, here.

Not really. log^{n+1}(3^^(n+1)) = log^{n}(3^^n*log(3)). So to get to
the next term, you dilate by a fixed amount in the previous term. The
continued interation of logs should imply the effect of that fixed
dilate -> 0 as n -> oo.

What _I think_ is happening, is that all of us are "correct" more or less,
and
this series is probably not convergent "conventionally" (whatever that may
mean,
anyway),

No, as I showed, it converges extremely rapidly in the conventional
sense, as the numerical evidence suggests. In fact log^{n}(x^^n)
converges for any x >= e (with the same rapid convergence).

The estimate a_(n+1) - a_n < 1/[e^{1}*e^{2}*...*e^{n-1}] (from my
previous post, using my notation now) was obtained using the mean
value theorem. Is that causing trouble for you, or is it the
convergence of a_n from this estimate that is bothering you?

but is nevertheless convergent using stronger methods, in the same
way
some series are divergent using conventional methods, but convergent using,
for
example, Euler, Abel or Cessaro summations.

This is as much as I understand.

regards

tommy1729
.

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