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tiny q: superroots of real numbers x>e
#1
Hi folks -

I've just asked this question in news:sci.math; it is a tiny question and possibly answered anywhere here around ( I didn't follow the superroot-discussion intensely) so maybe we have a link already...

Ok, let's go:

Let's define the n'th iterative root ("srt") via
Code:
f(x,1) = x    f(x,2) = x^x   f(x,3) = x^(x^x)     f(x,k) = ...
as one inverse of f, returning a base if a number and a iteration-count is given, such that, for instance
Code:
srt(y,3) = x  --> f(x,3) = y
and consider the sequence
Code:
srt(3,1) , srt(3,2), srt(3,3),..., srt(3,k),...  (for k=1 ... inf )

Then: what is x in
Code:
x =  lim {k->inf} srt(3,k)

The sequence decreases from 3 down to e^(1/e) + eps but I think, it cannot fall below.

Code:
k       x=srt(3,k)
---------------------
1    3.000000     =srt(3,1)
2    1.825455
4    1.563628
8    1.484080
16    1.457948
32    1.449171
64    1.446164
128    1.445135    =srt(3,128)
...
->inf   -> ??       srt(3,inf)
================================    

compare other limits

inf    1.444668    =e^(1/e)
--------------------------------    
inf    1.442250    =3^(1/3)

On the other hand, it should arrive at 3^(1/3)...

Do I actually overlook something and the sequence can indeed cross e^(1/e)?

<urrks>

Gottfried
Gottfried Helms, Kassel
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#2
Gottfried Wrote:On the other hand, it should arrive at 3^(1/3)...

Do I actually overlook something and the sequence can indeed cross e^(1/e)?

Indeed a very interesting observation, Gottfried.

You only arrive at the expected value if it is , i.e.
only if .

This is because for , where and .

For , for example , is always for each . Suppose otherwise then would , for while .
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#3
Hi Henryk -

It's late, I can't comment/proceed at the moment, let's see tomorrow.
Here are two plots to illustrate the beginning of the trajectory, anyway.

   

   

Nächtle... ;-)

[update] pic changed [/update]
Gottfried


Attached Files Image(s)
   
Gottfried Helms, Kassel
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#4
To be clear: I think its sure that

for

and for I would guess:



this also corresponds to your pictures.
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#5
Yepp, so we have the interesting property, that we have two numbers:
a proper limit (e^(1/e)) for the sequence of srt of increasing order and x^(1/x) as value for "the immediate" evaluation of the infinite expression.

Hmm - surely this should be formulated more smoothly.

Can we then say, that the infinite iterative root for y>e has two values?

... so many questions... Sad

Gottfried
Gottfried Helms, Kassel
Reply
#6
Gottfried Wrote:Can we then say, that the infinite iterative root for y>e has two values?

No, we have two cases and for each case one limit.
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