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 HELP NEEDED: Exponential Factorial and Tetrations rsgerard Junior Fellow Posts: 11 Threads: 5 Joined: May 2008 11/12/2009, 10:42 PM I have been studying exponential factorials and have been looking for the equivalent tetration. For example: 5^4^3^2^1= 5.9 e16 10^9^8^7^6^5^4^3^2^1 = 10 e363879 I believe as n goes to infinity the exponential factorial can be written as a tetration: n^(n-1)^(n-2)...^2^1 = (n/alpha)^(n/alpha)^(n/alpha)^(n/alpha)...repeated n times where alpha is Feigenbaum constant 2.5029... I have been testing this on this site: http://www.ttmath.org/online_calculator It does seem very very close for up to 25^24^23^...^2^1 can be written as an equivalent tetration. I would anyones input to see if they can help me determine it can be written as a tetration more rigorously. Thanks very much Ryan Gerard dantheman163 Junior Fellow Posts: 13 Threads: 3 Joined: Oct 2009 11/12/2009, 11:19 PM (11/12/2009, 10:42 PM)rsgerard Wrote: I have been studying exponential factorials and have been looking for the equivalent tetration. For example: 5^4^3^2^1= 5.9 e16 10^9^8^7^6^5^4^3^2^1 = 10 e363879 Well what you are saying is not really the exponential factorial. What you wrote above is the same as $x^{(x-1)!}$ The value of $5^{4^{3^{2^1}}}$ is acutaly more like 6.206e+183230 As for (11/12/2009, 10:42 PM)rsgerard Wrote: n^(n-1)^(n-2)...^2^1 = (n/alpha)^(n/alpha)^(n/alpha)^(n/alpha)...repeated n times I would have to look into this more to see how close to $x^{(x-1)!}$ it is nuninho1980 Fellow Posts: 95 Threads: 6 Joined: Apr 2009 11/12/2009, 11:19 PM (This post was last modified: 11/12/2009, 11:21 PM by nuninho1980.) aah! it's normal. sorry. I removed. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 11/12/2009, 11:46 PM Honestly, I have never heard of the Feigenbaum constant before. Is 2.5029 the value for exponential functions? or all functions? andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 11/12/2009, 11:51 PM (This post was last modified: 11/12/2009, 11:53 PM by andydude.) (11/12/2009, 11:19 PM)dantheman163 Wrote: Well what you are saying is not really the exponential factorial. Correct. 5^4^3^2^1 = 5^(4^(3^(2^1))) = $5^{4^{3^{2^1}}} \ne 5^{(4\cdot 3 \cdot 2 \cdot 1)}$ = (((5^4)^3)^2)^1. I think the problem here is that ttmath.org evaluates (a^b^c) incorrectly. It should use right-associative (^), but it seems to use left-associative (^), which is wrong. rsgerard Junior Fellow Posts: 11 Threads: 5 Joined: May 2008 11/13/2009, 02:27 AM (11/12/2009, 11:51 PM)andydude Wrote: (11/12/2009, 11:19 PM)dantheman163 Wrote: Well what you are saying is not really the exponential factorial. Correct. 5^4^3^2^1 = 5^(4^(3^(2^1))) = $5^{4^{3^{2^1}}} \ne 5^{(4\cdot 3 \cdot 2 \cdot 1)}$ = (((5^4)^3)^2)^1. I think the problem here is that ttmath.org evaluates (a^b^c) incorrectly. It should use right-associative (^), but it seems to use left-associative (^), which is wrong. Thanks for pointing out that this site is using left-association to evaluate. I guess I'm still curious to see if this constant really applies to these functions. I'll do a little more research myself. Thanks so much everyone. « Next Oldest | Next Newest »

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