# Tetration Forum

Full Version: tiny q: superroots of real numbers x>e
You're currently viewing a stripped down version of our content. View the full version with proper formatting.
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 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.
$\lim_{k\to\infty} \text{srt}(x,k)=\sqrt[x]{x}$ only if $1\le x\le e$.

This is because $1\le {^\infty}b\le e$ for $1\le b\le e^{1/e}$, where $x={^\infty}b$ and $b=\sqrt[x]{x}$.

For $x>e$, for example $x=3$, is always $\text{srt}(x,k) > e^{1/e}$ for each $k$. Suppose otherwise $\text{srt}(x,k)\le e^{1/e}=:y$ then would $x\le {^k}y$, for $y\le e^{1/e}$ while ${^k}y\le e$.
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.

[attachment=436]

[attachment=438]

Nächtle... ;-)

[update] pic changed [/update]
Gottfried
To be clear: I think its sure that

$\lim_{k\to\infty}\text{srt}(x,k) = \sqrt[x]{x}$ for $1\le x\le e$

and for $x>e$ I would guess:

$\lim_{k\to\infty}\text{srt}(x,k) = e^{1/e}$

this also corresponds to your pictures.
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...

Gottfried
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.