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x exp(x) musing
#2
(07/01/2009, 03:19 PM)tommy1729 Wrote: let f( x exp(x) ) * exp(x) * (x+1) = f(x)

then F( x exp(x) ) = F(x) + C

inversing ->

G(x+C) = G(x) * exp(G(x))

now notice G(x + n C) =
G(x) * exp(G(x)) * exp(G(x))^exp(G(x)) * ... n times.

thus exp(G(x)) ^^ n = G(x + n C) / G(x + n C - C) ( Form 1 )

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.

Quote:and maybe f(x) can be found by current attempts for tetration.

and G(x) might be found by inversing the ( modified ) carleman matrix of F(x)

It would probably be simpler to find G directly.

Quote:notice also that x exp(x) has a unique real fixed point !!

Moreover, it is at zero, so it doesn't even need shifting.
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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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