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 tiny q: superroots of real numbers x>e Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 02/02/2009, 06:54 PM Hi folks - I've just asked this question in news:sci.math; it is a tiny question and possibly answered anywhere here around ( I didn't follow the superroot-discussion intensely) so maybe we have a link already... Ok, let's go: Let's define the n'th iterative root ("srt") via Code:f(x,1) = x    f(x,2) = x^x   f(x,3) = x^(x^x)     f(x,k) = ...as one inverse of f, returning a base if a number and a iteration-count is given, such that, for instance Code:srt(y,3) = x  --> f(x,3) = yand consider the sequence Code:srt(3,1) , srt(3,2), srt(3,3),..., srt(3,k),...  (for k=1 ... inf ) Then: what is x in Code:x =  lim {k->inf} srt(3,k) The sequence decreases from 3 down to e^(1/e) + eps but I think, it cannot fall below. Code:k       x=srt(3,k) --------------------- 1    3.000000     =srt(3,1) 2    1.825455 4    1.563628 8    1.484080 16    1.457948 32    1.449171 64    1.446164 128    1.445135    =srt(3,128) ... ->inf   -> ??       srt(3,inf) ================================     compare other limits inf    1.444668    =e^(1/e) --------------------------------     inf    1.442250    =3^(1/3) On the other hand, it should arrive at 3^(1/3)... Do I actually overlook something and the sequence can indeed cross e^(1/e)? Gottfried Gottfried Helms, Kassel bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 02/02/2009, 10:21 PM (This post was last modified: 02/02/2009, 10:31 PM by bo198214.) Gottfried Wrote:On the other hand, it should arrive at 3^(1/3)... Do I actually overlook something and the sequence can indeed cross e^(1/e)? Indeed a very interesting observation, Gottfried. You only arrive at the expected value if it is $, i.e. $\lim_{k\to\infty} \text{srt}(x,k)=\sqrt[x]{x}$ only if $1\le x\le e$. This is because $1\le {^\infty}b\le e$ for $1\le b\le e^{1/e}$, where $x={^\infty}b$ and $b=\sqrt[x]{x}$. For $x>e$, for example $x=3$, is always $\text{srt}(x,k) > e^{1/e}$ for each $k$. Suppose otherwise $\text{srt}(x,k)\le e^{1/e}=:y$ then would $x\le {^k}y$, for $y\le e^{1/e}$ while ${^k}y\le e$. Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 02/03/2009, 12:15 AM (This post was last modified: 02/03/2009, 09:52 AM by Gottfried.) Hi Henryk - It's late, I can't comment/proceed at the moment, let's see tomorrow. Here are two plots to illustrate the beginning of the trajectory, anyway.         Nächtle... ;-) [update] pic changed [/update] Gottfried Attached Files Image(s)     Gottfried Helms, Kassel bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 02/03/2009, 10:01 AM To be clear: I think its sure that $\lim_{k\to\infty}\text{srt}(x,k) = \sqrt[x]{x}$ for $1\le x\le e$ and for $x>e$ I would guess: $\lim_{k\to\infty}\text{srt}(x,k) = e^{1/e}$ this also corresponds to your pictures. Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 02/03/2009, 11:36 AM Yepp, so we have the interesting property, that we have two numbers: a proper limit (e^(1/e)) for the sequence of srt of increasing order and x^(1/x) as value for "the immediate" evaluation of the infinite expression. Hmm - surely this should be formulated more smoothly. Can we then say, that the infinite iterative root for y>e has two values? ... so many questions... Gottfried Gottfried Helms, Kassel bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 02/03/2009, 12:46 PM Gottfried Wrote:Can we then say, that the infinite iterative root for y>e has two values? No, we have two cases and for each case one limit. « Next Oldest | Next Newest »

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