proof: Limit of self-super-roots is e^1/e. TPID 6
#1
In reply to
http://math.eretrandre.org/tetrationforu...73#pid4073

First it is easy to see that for :

( is the lower fixed point of )

Hence for we have for all :
(*)

We also know that for , quite fast, particularly for each there is an such that for all :
(**) .

Now we lead proof by contradiction, suppose that
where .

Then there must be a subsequence and such that this subsequence stays always more than apart from :
.
I.e. there is and such that
either or .

By (*) and (**) we have such that for all :
and .
As is monotone increasing for we have also
and .

This particularly means and hence none of the can be the self superroot, in contradiction to our assumption.
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#2
Wow! Very nice! You make it seem so easy. Smile I've been working on that one for while, ever since the xsrtx thread.
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#3
The same method of proof could possibly be used to easily prove that, possibly for all k>4, limit of self-hyper-k-root(x) as x -> infinity = (defined as the largest real x such that , i.e. where the maximum of self-hyper-(k-1)-root function occurs; let's establish this notation); yeah I know, I only substituted the pentation-analogues into the proof and quickly checked.
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#4
(07/10/2010, 05:19 AM)Base-Acid Tetration Wrote: The same method of proof could possibly be used to easily prove that, possibly for all k>4

The thing is: to define the hyper k-self-root you need a hyper (k-1) operation defined on the reals.
And we still have several methods of doing this without equality proofs.
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