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 Remark on Gottfried's "problem with an infinite product" power tower variation tommy1729 Ultimate Fellow Posts: 1,424 Threads: 346 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

 Messages In This Thread Remark on Gottfried's "problem with an infinite product" power tower variation - by tommy1729 - 07/31/2013, 12:39 AM RE: Remark on Gottfried's "problem with an infinite product" power tower variation - by tommy1729 - 05/02/2014, 11:40 PM RE: Remark on Gottfried's "problem with an infinite product" power tower variation - by tommy1729 - 05/03/2014, 08:30 PM RE: Remark on Gottfried's "problem with an infinite product" power tower variation - by tommy1729 - 05/05/2014, 10:48 PM RE: Remark on Gottfried's "problem with an infinite product" power tower variation - by tommy1729 - 05/06/2014, 09:47 PM

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