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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
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(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
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(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.
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(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
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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)
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(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
Gottfried Helms, Kassel
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(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
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(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
