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 Growth of superexponential sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 02/27/2013, 06:40 PM (This post was last modified: 02/28/2013, 11:32 AM by sheldonison.) (02/27/2013, 02:19 PM)Balarka Sen Wrote: I made an observation : for very small values of z, it seems likely that as b tends towards infinity, b^^z grows to infinity too, but rather slowly. I mean $\lim_{b \rightarrow \infty} {}^{z} b \rightarrow \infty$ for all $z > 0$Its not a trivial question. And I had a difficult time getting my kneser.gp algorithm to converge for large bases, though it currently works for b>100000. Here is a related question, that may lead to a fruitful investigation path, and possibly a proof. Can you prove that the $\text{slog}_b(e)$ for arbitrarily large base=b approaches arbitrarily close to zero? There is a fairly simple linear approximation one can use for tetration for arbitrary bases, that is continuous, and has a continuous first derivative, and works surprisingly well. The estimation uses a straight line linear estimate between $\log_b(\log_b(e))$ and $\log_b(e)$. For example, for base e, the linear approximation is sexp(-1)=0, sexp(1)=1, with a straight line in between, and $\text{slog}_e(e)=1$. edit: updated approximation equations, and plot If I did my algebra correctly, than using the linear approximation for sexp for arbitrary bases leads to the estimate for bases>=e $\text{slog}_b(e)\approx \frac{1}{1+\log(\log(b))}=k$ and the region from k-1..k has an exponential approximation of $\text{sexp}_b(z)\approx\exp(\frac{z}{k})$. For z>k the approximation switches over to a double exponent. This exponential approximation assumes the linear region includes sexp(0), which is true if base>=e. This approximation gives sexp(0)=1, and sexp(1)=b, which are both exact. Then you you could conjecture that for large enough bases (empirically, b>9), the actual sexp(k)>e. Also, I would conjecture that for b>9 the approximation is less than actual sexp(z) until z=1, where by definition the approximation is exactly correct once again. Anyway, such an slog(e) approximation for large bases goes to zero, but very slowly. For b=googleplex=$10^{10^{100}}$, the approximation is slog(e)=0.0043. For b=10, the approximation is 0.545 and the correct $\text{slog}_{10}(e) \approx0.544$. For b=100, the approximation is 0.396 and the correct value is $\text{slog}_{100}(e) \approx0.374$. For b=100000, the approximation is 0.290, and the correct value is $\text{slog}_{100000}(e) \approx0.250$. Here is a graph of sexp_100000(z). The function is surprisingly well behaved in the region of interest. Here, $k\approx 0.29$, and the linear approximation region would be from -1.71 to -0.71. The actual sexp is in red, and the linear approximation is in green.     - Sheldon « 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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