x exp(x) musing

[tommy1729]

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.

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.

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