07/03/2009, 07:34 PM

(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.

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.

1) interating exp ( 'anything' ) ^^ (2^n) seems serious overkill and overclocked speed !!

so i dont think that can be correct.

2) simply finding G directly ? do you know a fixed point for G perhaps ?

why do you think so ?

3) no we dont need shifting , 0 * exp(0) = 0 , i know that of course , in fact , thats partially why i mentioned it.

regards

tommy1729