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 Remark on Gottfried's "problem with an infinite product" power tower variation tommy1729 Ultimate Fellow     Posts: 1,455 Threads: 350 Joined: Feb 2009 07/31/2013, 12:39 AM (This post was last modified: 07/31/2013, 12:42 AM by tommy1729.) Remark on Gottfried's "problem with an infinite product". http://math.eretrandre.org/tetrationforu...hp?tid=785 *** A similar case occurs with power towers. g(x) = x ^ f(x) ^ f^(x) ^ ... An intresting case is ln(x) ^ ln(ln(x)) ^ ln^(x) ^ ... = A(x). Let x >> 10. Define f(x) = arcsinh(x/2) ( YES I USE THIS AGAIN , THE INVERSE OF 2SINH !!! ) Now there exists (small) constants C1,C2,C3,C4 such that f(x) ^ f^(x) ^ f^(x) ^ ... ~ C1 (x + C2)^C3 + C4 Nice. Return of the master tommy1729 tommy1729 Ultimate Fellow     Posts: 1,455 Threads: 350 Joined: Feb 2009 05/02/2014, 11:40 PM (This post was last modified: 05/03/2014, 12:13 PM by tommy1729.) H(exp(x)) = x^(H(x)^a) Such equations seem related. Have they occured here before ? regards tommy1729 *edited* tommy1729 Ultimate Fellow     Posts: 1,455 Threads: 350 Joined: Feb 2009 05/03/2014, 08:30 PM ------------------------------------------------------------------------------ Conjecture B : This thread relates to another : http://math.eretrandre.org/tetrationforu...hp?tid=825 More specifically my g(x) and/or H(x) relates to sheldon's f(x) and/or sheldon's f^[-1](x). Its BTW no coincidence that I gave attention to them again at the same day. In other words I expected this connection for a long time. ------------------------------------------------------------------------------ regards tommy1729 tommy1729 Ultimate Fellow     Posts: 1,455 Threads: 350 Joined: Feb 2009 05/05/2014, 10:48 PM I referred to this thread in http://math.eretrandre.org/tetrationforu...43#pid6943 For simplicity lets replace arc2sinh(x/2) with ln(x). This is a bit informal handwaving, but for the time being it might be good enough. Let f(x) = ln(x) ^ ln(ln(x)) ^ ln^(x) ^ ... As often in math we get the weird result that something is true IFF something else is true. Keeping this in mind : This might be handwaving but lets go : Let eps > 0 ( the famous epsilon ) Assume for sufficiently large x : x^eps < f(x) < x^A for some real A > eps. (the epsilon assumption) This is more powerfull than you might expect ! : We know ln(x^b(x)) / ln(x) = b(x) Based on that : f(x) = ln(x) ^ ln(ln(x)) ^ ln^(x) ^ ... ln(f(x)) / ln(x) = ln( ln(x) ^ ln(ln(x)) ^ ln^(x) ^ ... ) / ln(x) = (ln(ln(x)) * ln(ln(x)) ^ ln^(x) ^ ...) / ln(x) use f(x) = x^b(x) = ln(ln(x)) * ln(x)^b(ln(x)) * (ln(x))^-1 => the best constant fit for b(x) = b(ln(x)) = b => ( existance of b follows from the epsilon assumption ) => b = 1 because ln(ln(x)) * ln(x) / ln(x) = ln(ln(x)) and ln(ln(x)) is smaller than a power of ln(x). By a similar logic we can improve : f(x) = x^b(x) = x^b / ln(x) This gives us : = ln(ln(x)) * ln(x)^b(ln(x)) * (ln(x))^-1 = ln(ln(x)) * f(ln(x)) * (ln(x))^-1 = ln(ln(x)) * ln(x)^b / ln(ln(x)) * (ln(x))^-1 = 1. = b. QED. Hence if x^eps < f(x) < x^A then f(x) = x / (ln(x))^(1+o(1)). WOW. THIS IS THE PNT FOR POWER TOWERS !! However like I said this is a bit handwaving. Because ... " replace arc2sinh(x/2) with ln(x) " ? Is x^eps < f(x) < x^A TRUE ??? Funny thing is also that currently the mellin transform is considered for tetration. The same mellin transform used in the proof for PNT ! So maybe a mellin transform can be used for this power tower problem as well ??? ------------------------------------------------------------------------------ Remark : Its not even clear f(x) is analytic ? I considered Taylor series , four series and dirichlet. And taking derivatives. But if f(x) is not analytic ... That might also be problematic for the mellin transform. On the other hand f(x) is C^oo. So I believe taking derivatives is justified. This brings me to another idea :taking the logarithmic derivative of f(x). That might help decide if the epsilon assumption is true !! ------------------------------------------------------------------------------ regards tommy1729 tommy1729 Ultimate Fellow     Posts: 1,455 Threads: 350 Joined: Feb 2009 05/06/2014, 09:47 PM Lets dig a bit deeper. H(exp(x)) = x^H(x) I use =\$= notation here , =\$= compares two functions and concludes <,=,> for sufficiently large x. What is special about this notation is that we do not switch LHS with RHS. Say H is about x^2 : => exp(2x) =\$= x^(x^2) => exp(2x) =\$= exp(ln(x) x^2) => exp(2x) < exp(ln(x) x^2) Likewise say H is about x^0.5 : => exp(0.5 x) =\$= x^(x^0.5) => exp(0.5 x) =\$= exp(ln(x) x^0.5) => exp(0.5 x) > exp(ln(x) x^0.5) => x^0.5 < H(x) < x^2 ( It should be clear to you WHY =\$= does not switch LHS with RHS now. ) It is easy to generalize this to : x^a < H(x) < x^(1/a) for all 0

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