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 x exp(x) musing tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 07/01/2009, 03:19 PM 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 BenStandeven Junior Fellow Posts: 27 Threads: 3 Joined: Apr 2009 07/03/2009, 03:40 PM (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. tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 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 BenStandeven Junior Fellow Posts: 27 Threads: 3 Joined: Apr 2009 07/04/2009, 11:25 PM (This post was last modified: 07/04/2009, 11:53 PM by BenStandeven.) (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. bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 07/05/2009, 07:12 PM Thank you Ben, I couldnt have explained it better tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 07/05/2009, 07:33 PM (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 « Next Oldest | Next Newest »

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