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+--- Thread: Exact and Unique solution for base e^(1/e) (/showthread.php?tid=8)
RE: Exact and Unique solution for base e^(1/e) - bo198214 - 08/02/2009
(08/01/2009, 09:04 PM)jaydfox Wrote: The cheta + base-change approach?Yes I was referring to that.
Quote:Or the "accelerated" method of finding the power series for Andrew's approach? (The latter is more a computational shortcut, but it gives/uses insight into the structure in the complex plane.)
Maybe this can also be mentioned, however the paper should not focus too much on computation. (Though computed graphs are highly appreciated, I particularly remember the beautiful graphs of the regular slog you made. I hope you can revive it for the paper as lots of my own sage scripts got lost in the crash.)
RE: Exact and Unique solution for base e^(1/e) - tommy1729 - 08/02/2009
so tetration base e^1/e is linked to iterations of e^x - 1.
and more general base b^1/b iteration is linked to iterations of b^x - 1.
what if b < e ?
we get again the multiple fixpoint problem.
but i wonder if one of those fixpoints is " better " than the other , for instance if one is equal to the newton method for bases < eta.
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i might be crazy but im thinking about linking iterations : tetration b^1/b , b^x - 1 to a THIRD function that we can iterate.
this in an attempt to attack b > e.
regards
tommy1729
RE: Exact and Unique solution for base e^(1/e) - bo198214 - 08/03/2009
(08/02/2009, 10:34 PM)tommy1729 Wrote: so tetration base e^1/e is linked to iterations of e^x - 1.
Yes, the link is a linear transformation \( \tau \).
If you set \( \tau(x)=e(x+1) \), and \( f(x)=e^{x/e} \) and \( g(x)=e^x-1 \) you have the relation:
\( g = \tau^{-1} \circ f \circ \tau \).
Iff \( f \) has a fixed point \( p \) then \( \tau^{-1} \circ f\circ \tau \) has the fixed point \( \tau^{-1}(p) \). The fixed points are merely mapped linearly. Also the regular iteration of the conjugate is equal to the conjugate of the regular iteration. So nothing new really occurs.
Quote:and more general base b^1/b iteration is linked to iterations of b^x - 1.
Yes just set \( \tau(x)=b(x+1) \).
Quote:what if b < e ?
we get again the multiple fixpoint problem.
yes, the number of fixed points of \( b^{x/b} \) and of \( b^x-1 \) is the same, by linear conjugation nothing new occurs.