Tetration extension for bases between 1 and eta dantheman163 Junior Fellow Posts: 13 Threads: 3 Joined: Oct 2009 11/05/2009, 03:00 AM (This post was last modified: 11/05/2009, 09:14 PM by dantheman163.) I believe I have found an analytic extension of tetration for bases 1 < b <= e^(1/e). This is based on the assumption (1) The function y=b^^x is a smooth, monotonic concave down function Conjecture: If assumption (1) is true then ${}^x b = \lim_{k\to \infty} (log_{b}^{ok}(x({}^k b- {}^{(k-1)} b)+{}^k b) )$ for $-1 \le x\le 0$ Some properties: This formula converges rapidly for values of b that are closer one. For base eta it converges to b^^x for all x but this is not true for the other bases. Interestingly for b= sqrt(2) and x=1 it seems to be converging to the super square root of 2 I will try to post a proof in the next couple of days I just need some time to type it up. Thanks bo198214 Administrator Posts: 1,630 Threads: 106 Joined: Aug 2007 11/05/2009, 01:44 PM (This post was last modified: 11/05/2009, 01:47 PM by bo198214.) Hey Dan, thank you for this contribution. The formula you mention is the inversion of Lévy's formula (which is generally applicable to functions with f'(p)=1 for a fixed point p of f). See e.g. the tetration methods draft, formula 2.26 (where formula 2.27 is called Lévy's formula). The formula is known to give the regular tetration for $b=e^{1/e}$. However it is new to apply it to $1 f(x+k) > g_k(x+k)$ and $g_k(x+k)\uparrow a$ for $k\to\infty$. We have now $f(x+k)-g_k(x+k)\downarrow 0$ for $k\to\infty$. But I think that does not directly show the convergence $f(x) - \log_b^{\circ k} g_k(x+k)\downarrow 0$. Any ideas? bo198214 Administrator Posts: 1,630 Threads: 106 Joined: Aug 2007 11/07/2009, 05:11 PM (11/07/2009, 09:31 AM)bo198214 Wrote: We have now $f(x+k)-g_k(x+k)\downarrow 0$ for $k\to\infty$. But I think that does not directly show the convergence $f(x) - \log_b^{\circ k} g_k(x+k)\downarrow 0$. Any ideas? Actually I think one can show that $\log_b^{\circ k} g_k(x+k)$ is strictly increasing with $k$ because $\log_b g_k(x+k)$ is concave and hence $g_{k-1}(x+k-1) < \log_b g_k(x+k)$ and thatswhy $\log_b^{\circ k-1} g_{k-1}(x+k-1)<\log_b^{\circ k} g_{k}(x+k)$ also it is bounded and hence must have a limit. But now there is still the question why the limit $f(x)$ indeed satisfies $f(x+1)=b^{f(x)}$? bo198214 Administrator Posts: 1,630 Threads: 106 Joined: Aug 2007 11/07/2009, 08:12 PM (11/07/2009, 05:11 PM)bo198214 Wrote: But now there is still the question why the limit $f(x)$ indeed satisfies $f(x+1)=b^{f(x)}$? Actually it is not the case but we can obtain something very similar. I just read in [1], p. 31, th. 10, that we have the following limit for a function $f(x)=f_1 x+f_2x^2+f_3 x^3 + ...$ with \( 0

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