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x exp(x) musing
#4
(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 !! Smile

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.
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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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