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 Growth of superexponential Balarka Sen Junior Fellow Posts: 25 Threads: 7 Joined: Feb 2013 02/27/2013, 07:24 PM (02/27/2013, 06:40 PM)sheldonison Wrote: If I did my algebra correctly, than using the linear approximation for sexp for arbitrary bases leads to the estimate for $\text{slog}_b(e)=\frac{1}{1+\log(\log(b))}$, where with some algebra you could also come up with an approximation for the slope sexp_b(z)=e, and show that the slope gets arbitrarily large for superexponentially large bases, and then conjecture that for large enough bases, the approximation for slog(e) is an overestimate for the analytic slog, and that the slope for sexp(z) is increasing in this region.... That's an interesting approach! I was searching for a sequence diverges to infinity but grows slower than b^^z for z in the interval [0, 1], however. Finding such a sequence would be enough to prove the divergence of the limit we are concerned about. I will give it a try with the you suggested. Thank you very much for your reply, it's much appreciated. Balarka . « Next Oldest | Next Newest »

 Messages In This Thread Growth of superexponential - by Balarka Sen - 02/26/2013, 11:19 AM RE: Growth of superexponential - by tommy1729 - 02/26/2013, 10:00 PM RE: Growth of superexponential - by Balarka Sen - 02/27/2013, 02:19 PM RE: Growth of superexponential - by sheldonison - 02/27/2013, 06:40 PM RE: Growth of superexponential - by Balarka Sen - 02/27/2013, 07:24 PM RE: Growth of superexponential - by tommy1729 - 03/01/2013, 12:11 AM RE: Growth of superexponential - by tommy1729 - 03/06/2013, 11:51 PM RE: Growth of superexponential - by tommy1729 - 03/06/2013, 11:55 PM

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