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 Simple Question about e^(1/e) rsgerard Junior Fellow Posts: 11 Threads: 5 Joined: May 2008 04/20/2010, 10:07 PM My question is related to the infinite tetration regarding e^(1/e). Perhaps it would not be called a true tetration since each term is slightly different. Code:#This is arbitrary but should be high to represent infinity. #j=0 iterations=1000 for i in range(1,iterations):     p=e^(1/e)+1/i^(X)     j = (p)^j What is the smallest value of X??? In plain English: e^(1/e) = 1.444 roughly If X were 1: ......^(1.4444 + 1/4) ^ (1.4444 + 1/3) ^ (1.4444 + 1/2) ^ (1.4444 + 1/1) If X were 3: ......^(1.4444 + 1/64) ^ (1.4444 + 1/27) ^ (1.4444 + 1/8.0) ^ (1.4444 + 1/1) Based on my program 1 < X < 3 In fact, X seems to be close to 2.374 which is a constant that I haven't seen before. When X is extremely large, the sequence never seems to get past e (although I could be wrong). Any advise from the genius' here would be greatly appreciated. Ryan Forgive the lack of formatting. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/21/2010, 09:38 AM (04/20/2010, 10:07 PM)rsgerard Wrote: What is the smallest value of X??? You mean: What is the infimum of x such that your sequence still converges? Interesting question, I never heard about that, and honestly I can not imagine that there exist results about it. One would have to start showing that it converges for x=3 and that it diverges for x=1. Also there is slight difference, the loop that you give computes: ^(1.444 + 1/4^x) ^ (1.444 + 1/3^x) ^ (1.444 + 1/2^x) because you start with j=0. If you would start with j=1 what would produce the sequence you give: ^(1.444 + 1/4^x) ^ (1.444 + 1/3^x) ^ (1.444 + 1/2^x) ^ (1.444 + 1) but probably a different value of x. rsgerard Junior Fellow Posts: 11 Threads: 5 Joined: May 2008 04/21/2010, 06:43 PM Bo, I appreciate your look into this question. I am working on some interesting conjectures related to infinity using iteration theory. Good catch on me starting with j=1. Thanks very much. Ryan (04/21/2010, 09:38 AM)bo198214 Wrote: (04/20/2010, 10:07 PM)rsgerard Wrote: What is the smallest value of X??? You mean: What is the infimum of x such that your sequence still converges? Interesting question, I never heard about that, and honestly I can not imagine that there exist results about it. One would have to start showing that it converges for x=3 and that it diverges for x=1. Also there is slight difference, the loop that you give computes: ^(1.444 + 1/4^x) ^ (1.444 + 1/3^x) ^ (1.444 + 1/2^x) because you start with j=0. If you would start with j=1 what would produce the sequence you give: ^(1.444 + 1/4^x) ^ (1.444 + 1/3^x) ^ (1.444 + 1/2^x) ^ (1.444 + 1) but probably a different value of x. tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 04/22/2010, 02:09 PM (04/20/2010, 10:07 PM)rsgerard Wrote: My question is related to the infinite tetration regarding e^(1/e). Perhaps it would not be called a true tetration since each term is slightly different. Code:#This is arbitrary but should be high to represent infinity. #j=0 iterations=1000 for i in range(1,iterations):     p=e^(1/e)+1/i^(X)     j = (p)^j What is the smallest value of X??? In plain English: e^(1/e) = 1.444 roughly If X were 1: ......^(1.4444 + 1/4) ^ (1.4444 + 1/3) ^ (1.4444 + 1/2) ^ (1.4444 + 1/1) If X were 3: ......^(1.4444 + 1/64) ^ (1.4444 + 1/27) ^ (1.4444 + 1/8.0) ^ (1.4444 + 1/1) Based on my program 1 < X < 3 In fact, X seems to be close to 2.374 which is a constant that I haven't seen before. When X is extremely large, the sequence never seems to get past e (although I could be wrong). Any advise from the genius' here would be greatly appreciated. Ryan Forgive the lack of formatting. im not sure i got it correct. but it seems to remind me of my ' tommy-zeta function '. see http://www.research.att.com/~njas/sequences/A102575 regards tommy1729 andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 04/28/2010, 08:35 PM (04/20/2010, 10:07 PM)rsgerard Wrote: Any advise from the genius' here would be greatly appreciated. I don't believe in genius. I believe in patience, it will get you farther than genius alone. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/26/2010, 04:55 PM By some incidence I just found the article: Barrow, D. F. (1936). Infinite exponentials. Amer. math. Monthly, 43, 150–160. He states there in Theorem 7: The infinite exponential $E_{i=0}^\infty (e^{1/e}+\epsilon_i)$, where $\epsilon_i$ are all positive or 0, (a) will converge if $\lim_{n\to\infty} \epsilon_n n^2 < \frac{e^{1/e}}{2e}$ (b) will diverge if $\lim_{n\to\infty} \epsilon_n n^2 > \frac{e^{1/e}}{2e}$. However the numbering is kinda reverse to yours. $E_{i=0}^n a_i$ means $a_0\^a_1\dots \^a_n$. I.e. the tower will be enlarged on top (infinity on top), while your tower gets enlarged on the bottom (infinity at bottom). But anyway its good to know that people dealt already with such questions. And perhaps reading this article helps you with your problem. rsgerard Junior Fellow Posts: 11 Threads: 5 Joined: May 2008 05/26/2010, 05:21 PM Thanks very much for your reference. I'm still doing a lot of research on this. This is extremely helpful, I'll follow up in the future. Ryan (05/26/2010, 04:55 PM)bo198214 Wrote: By some incidence I just found the article: Barrow, D. F. (1936). Infinite exponentials. Amer. math. Monthly, 43, 150–160. He states there in Theorem 7: The infinite exponential $E_{i=0}^\infty (e^{1/e}+\epsilon_i)$, where $\epsilon_i$ are all positive or 0, (a) will converge if $\lim_{n\to\infty} \epsilon_n n^2 < \frac{e^{1/e}}{2e}$ (b) will diverge if $\lim_{n\to\infty} \epsilon_n n^2 > \frac{e^{1/e}}{2e}$. However the numbering is kinda reverse to yours. $E_{i=0}^n a_i$ means $a_0\^a_1\dots \^a_n$. I.e. the tower will be enlarged on top (infinity on top), while your tower gets enlarged on the bottom (infinity at bottom). But anyway its good to know that people dealt already with such questions. And perhaps reading this article helps you with your problem. tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 05/26/2010, 06:04 PM (05/26/2010, 04:55 PM)bo198214 Wrote: By some incidence I just found the article: Barrow, D. F. (1936). Infinite exponentials. Amer. math. Monthly, 43, 150–160. He states there in Theorem 7: The infinite exponential $E_{i=0}^\infty (e^{1/e}+\epsilon_i)$, where $\epsilon_i$ are all positive or 0, (a) will converge if $\lim_{n\to\infty} \epsilon_n n^2 < \frac{e^{1/e}}{2e}$ (b) will diverge if $\lim_{n\to\infty} \epsilon_n n^2 > \frac{e^{1/e}}{2e}$. and i assume it diverges if > or < is replaced by =. since we dont have an upper bound. « Next Oldest | Next Newest »

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