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 Crazy conjecture connecting the sqrt(e) and tetrations! rsgerard Junior Fellow Posts: 11 Threads: 5 Joined: May 2008 04/21/2010, 07:19 PM 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 rsgerard Junior Fellow Posts: 11 Threads: 5 Joined: May 2008 04/21/2010, 07:48 PM (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 bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 04/22/2010, 12:41 PM (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. tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 04/22/2010, 02:43 PM (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 tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 06/24/2010, 07:56 PM 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) Gottfried Ultimate Fellow Posts: 787 Threads: 121 Joined: Aug 2007 02/28/2011, 02:43 PM (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 tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 03/26/2014, 12:40 AM (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 tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 03/27/2014, 11:20 PM (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 « Next Oldest | Next Newest »

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