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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 )

and we have a solution for tetration.

now that was very briefly the idea , i know alot relates to it.

another related idea is to take the integral ;

because sexp'(x) = sexp(x) * sexp(x-1) * sexp(x-2) * ... * C2.

so sexp(x) relates to integral G(x). ( " Form 2 " not a formula yet actually :p )

now we have 2 strongly related ideas.

and keep in mind that our base needs to be bigger than eta.

for real bases > eta ; Form 1 or Form 2 might apply to one of the solutions for tetration ... ( or both sometimes !? since Derivate = prod )

i know , i know , ive only given ideas and equations , not solutions or proofs.

but still , i think it is worth consideration.

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)

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

i used 'x' instead of 'z' because mainly we want R > eta -> R

btw G(x) should be strictly increasing on the reals.

i hope that idea was not retarded regards

tommy1729
(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.
(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
(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.
Thank you Ben, I couldnt have explained it better (07/05/2009, 07:12 PM)bo198214 Wrote: [ -> ]Thank you Ben, I couldnt have explained it better yes thank you Ben.

despite my mistakes , i think the basic idea still has potential ...

not ?

back to the drawing board for me regards

tommy1729