# Tetration Forum

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I don't have a powerful enough computer to determine this, but I would like someone to tell me that I'm wrong.

I will let the data speak for itself:

e^(1/e) = 1.444...
Let d = 1/e

Set infinity to be some arbitrarily high number, e.g. 9.99e10000000

Take the following sequences:

...^ (1.4444 + d) ^ (1.4444 + d) ^ (1.4444 + d) ^ (1.4444 + d) ^ (1.4444 + d)

...^ (1.4444 + d^2) ^ (1.4444 + d^2) ^ (1.4444 + d^2) ^ (1.4444 + d^2) ^ (1.4444 + d^2)

...^ (1.4444 + d^3) ^ (1.4444 + d^3) ^ (1.4444 + d^3) ^ (1.4444 + d^3) ^ (1.4444 + d^3)

Continue to increase n toward infinity...
...^ (1.4444 + d^n) ^ (1.4444 + d^n) ^ (1.4444 + d^n) ^ (1.4444 + d^n) ^ (1.4444 + d^n)

Each of these sequences reaches "infinity" after the following iterations:
8, 12, 16, 25, 41, 66, 108, 178, 293, 482, 794 when using (d,d^2,d^3,d^4,d^5,d^6,d^7,d^8,d^9,d^10) respectively.

This looks like a geometric series based close to sqrt(e) = (1.645...).

Perhaps the number of iterations to get to "infinity" approaches sqrt(e)???

Bo, I'm awaiting your superior mathematical intellect.

Ryan
(04/21/2010, 07:19 PM)rsgerard Wrote: [ -> ]e^(1/e) = 1.444...
Let d = 1/e

Set infinity to be some arbitrarily high number, e.g. 9.99e10000000

I can further generalize this conjecture:

if d= 1/c, for any constant > 1

the infinite tetration of e^(1/e) + d, will reach "infinity" after 1/sqrt© iterations. I can post the data if anyone is interested:

For example, when d=1/10 we reach "infinity" after:
12, 34, 104, 325, 1024 iterations for d=(1/10,1/100,1/10^3,1/10^4)

This series grows at sqrt(10) for each iteration approximately.

Ryan
(04/21/2010, 07:48 PM)rsgerard Wrote: [ -> ]
(04/21/2010, 07:19 PM)rsgerard Wrote: [ -> ]e^(1/e) = 1.444...
Let d = 1/e

Set infinity to be some arbitrarily high number, e.g. 9.99e10000000

I can further generalize this conjecture:

if d= 1/c, for any constant > 1

the infinite tetration of e^(1/e) + d, will reach "infinity" after 1/sqrt© iterations. I can post the data if anyone is interested:

For example, when d=1/10 we reach "infinity" after:
12, 34, 104, 325, 1024 iterations for d=(1/10,1/100,1/10^3,1/10^4)

This series grows at sqrt(10) for each iteration approximately.

Ryan

Hm, so what you are saying is that
$\lim_{y\to\infty} \frac{\operatorname{slog}_{\eta+1/c^{n+1}}(y)}{\operatorname{slog}_{\eta+1/c^n}(y)}\to \sqrt{c}$
Or at least
$\lim_{n\to\infty}\lim_{y\to\infty} \frac{\operatorname{slog}_{\eta+1/c^{n+1}}(y)}{\operatorname{slog}_{\eta+1/c^n}(y)}\to \sqrt{c}$
where $\eta=e^{1/e}$ and $\operatorname{slog}_b$ is the inverse function of $f(z)=\exp_b^{\circ z}(1)$

Sounds really interesting, however I have no idea how to tackle.
(04/22/2010, 12:41 PM)bo198214 Wrote: [ -> ]
(04/21/2010, 07:48 PM)rsgerard Wrote: [ -> ]
(04/21/2010, 07:19 PM)rsgerard Wrote: [ -> ]e^(1/e) = 1.444...
Let d = 1/e

Set infinity to be some arbitrarily high number, e.g. 9.99e10000000

I can further generalize this conjecture:

if d= 1/c, for any constant > 1

the infinite tetration of e^(1/e) + d, will reach "infinity" after 1/sqrt© iterations. I can post the data if anyone is interested:

For example, when d=1/10 we reach "infinity" after:
12, 34, 104, 325, 1024 iterations for d=(1/10,1/100,1/10^3,1/10^4)

This series grows at sqrt(10) for each iteration approximately.

Ryan

Hm, so what you are saying is that
$\lim_{y\to\infty} \frac{\operatorname{slog}_{\eta+1/c^{n+1}}(y)}{\operatorname{slog}_{\eta+1/c^n}(y)}\to \sqrt{c}$
Or at least
$\lim_{n\to\infty}\lim_{y\to\infty} \frac{\operatorname{slog}_{\eta+1/c^{n+1}}(y)}{\operatorname{slog}_{\eta+1/c^n}(y)}\to \sqrt{c}$
where $\eta=e^{1/e}$ and $\operatorname{slog}_b$ is the inverse function of $f(z)=\exp_b^{\circ z}(1)$

Sounds really interesting, however I have no idea how to tackle.

i noticed that too , very long ago.

perhaps the count till 'oo' is the confusing part.

what if we replace d with -d and count until we reach 'e' (instead of 'oo')

then would the limit also give sqrt© ?

if so , i think we are close to a proof.

or at least arrive at showing these limits depend on earlier conjectured limits ( such as the limit by gottfried )

regards

tommy1729
i couldnt help noticing that sqrt(e) occurs here , just as it does in my " use sinh " thread.

could there be a link ??

it seems sqrt(e) is the number 3 constant after e^1/e.

1) e^1/e
2) fixpoint e^x
3) sqrt(e)
(04/22/2010, 12:41 PM)bo198214 Wrote: [ -> ](...)
Sounds really interesting, however I have no idea how to tackle.

Hmm, I do not really see a good possibility to tackle this. Just want to note that one can rewrite this

$\lim_{n\to \infty} b\^\^^n < c\^\^^m < b\^\^^{(n+1)}$

where $b=\eta+1/\exp(k)$ and $c=\eta+1/ \exp(1+k)$ and then

$m \to n*1.64... = n*\exp (0.5)$ with some k
(02/28/2011, 02:43 PM)Gottfried Wrote: [ -> ]
(04/22/2010, 12:41 PM)bo198214 Wrote: [ -> ](...)
Sounds really interesting, however I have no idea how to tackle.

Hmm, I do not really see a good possibility to tackle this. Just want to note that one can rewrite this

$\lim_{n\to \infty} b\^\^^n < c\^\^^m < b\^\^^{(n+1)}$

where $b=\eta+1/\exp(k)$ and $c=\eta+1/ \exp(1+k)$ and then

$m \to n*1.64... = n*\exp (0.5)$ with some k

Hmm. This is quite an old post.

I remember thinking I know how Gottfried arrived at this.

But I seem to have forgotten now.

Maybe I should have posted my ideas back then.

Could you plz explain Gottfried ?

regards

tommy1729
(04/22/2010, 12:41 PM)bo198214 Wrote: [ -> ]Hm, so what you are saying is that
$\lim_{y\to\infty} \frac{\operatorname{slog}_{\eta+1/c^{n+1}}(y)}{\operatorname{slog}_{\eta+1/c^n}(y)}\to \sqrt{c}$

Maybe I am wrong but I seem to disagree with that.

Does this not contradict the base change ?

Since the base change dictates that $\lim_{y\to\infty} slog_a(y) - slog_b(y) = Constant$

regards

tommy1729