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 x exp(x) musing BenStandeven Junior Fellow  Posts: 27 Threads: 3 Joined: Apr 2009 07/04/2009, 11:25 PM (This post was last modified: 07/04/2009, 11:53 PM by BenStandeven.) (07/03/2009, 07:34 PM)tommy1729 Wrote: (07/03/2009, 03:40 PM)BenStandeven Wrote: Actually, I think you get exp(G(x)) ^^ 2^n = exp(G(x + n C)) = G(x + n C)/G(x + n C - C), where the operation is "symmetric" tetration (referring to the association method). I don't see any obvious way to get the usual "top down" tetration from this. 1) interating exp ( 'anything' ) ^^ (2^n) seems serious overkill and overclocked speed !! so i dont think that can be correct. Here's how it is derived: exp(G(x)) ^^ 2^n = [exp(G(x)) ^^ 2^(n-1)] ^ [exp(G(x)) ^^ 2^(n-1)] (by definition); then by induction we get exp(G(x)) ^^ 2^n = [exp(G(x + (n-1)C))] ^ [exp(G(x + (n-1)C))] = exp(G(x + (n-1)C)*exp(G(x + (n-1)C))) = exp(G(x + n C)). For the base case, we note that exp(G(x)) ^^ 2^0 = exp(G(x)) ^^ 1 = exp(G(x)) = exp(G(x + 0 C)). Note that symmetric tetration grows more slowly than normal tetration; the " ^^ 2^n" actually only grows a bit faster than a " ^^ n+1" for top-down tetration. For example, "H^^8" is [(H^H)^(H^H)] ^ [(H^H)^(H^H)] = H^(H^[H^(H+1) + H + 1]), which only has four layers of H's. Quote:2) simply finding G directly ? do you know a fixed point for G perhaps ? why do you think so ? We can use regular iteration at the fixed point of x exp(x) to find G. « Next Oldest | Next Newest »

 Messages In This Thread x exp(x) musing - by tommy1729 - 07/01/2009, 03:19 PM RE: x exp(x) musing - by BenStandeven - 07/03/2009, 03:40 PM RE: x exp(x) musing - by tommy1729 - 07/03/2009, 07:34 PM RE: x exp(x) musing - by BenStandeven - 07/04/2009, 11:25 PM RE: x exp(x) musing - by bo198214 - 07/05/2009, 07:12 PM RE: x exp(x) musing - by tommy1729 - 07/05/2009, 07:33 PM

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