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 tetration bending uniqueness ? tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 08/28/2010, 11:23 PM (This post was last modified: 08/28/2010, 11:34 PM by tommy1729.) uniqueness by the absense of bending points with respect to z in sexp(slog(z) + r) for positive z and r. ( i use tet for 'my' sexp further ) let tet(slog(x)) = x tet(0) = 0 = slog(0) consider tet(slog(z) + r) take derivate with respect to z. tet'(slog(z) + r) x 1/tet'(slog(z)) take derivate with respect to z. tet''(slog(z)+r)/tet'(slog(z))^2 - (tet'(slog(z)+r)) * tet''(slog(z))/tet'(slog(z))^3 hence tet''(slog(z)+r)/tet'(slog(z))^2 = (tet'(slog(z)+r)) * tet''(slog(z))/tet'(slog(z))^3 thus tet''(slog(z)+r) = tet'(slog(z)+r) * tet''(slog(z))/tet'(slog(z)) hence solve for positive z : tet''(z+r) = tet'(z+r) * tet''(z)/tet'(z) make symmetric tet''(z+r)/tet'(z+r) = tet''(z)/tet'(z) notice that if a z exists , another one must exist. thus if for some r , a z exists , there exist oo z solutions. take integral on both sides ( this step may be a bit dubious ? ) log(tet'(z+r)) + A = log(tet'(z)) + B hence bending points in sexp(slog(z) + r) correspond to bending points in sexp(z). thus all ( analytic ) sexp(z) with sexp(0) = 0 and positive real to positive real , without bending points are identical !! headscratch ... regards tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 08/28/2010, 11:31 PM as an additional related question : is it true that the q-continuum product (as defined by mike ) never adds bending points to an analytic on Re > 0 function that maps positive reals to positive reals and didnt contain any bending points ? tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 08/29/2010, 06:17 PM (This post was last modified: 08/29/2010, 06:18 PM by tommy1729.) since sexp(0) = 0 in the OP , we can extend : no bending points on sexp follows from no bending points on its half-iterates.* ( * but not necc in reverse ! possibly invalidating this post ) we can find the half-iterate of sexp by using the fixpoint 0. f^[2](x)= sexp(x). generalising this , we end up with no bending points at pent(x). hence the no bending points condition extends to the whole hierarchy : tetration , pentation , sextation , ... tommy1729 tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 08/29/2010, 06:38 PM remark about post number 1 : tet(0) = 0 = slog(0) and 0 is the only positive real fixpoint of tet(x) and slog(x). that is an important detail. regards tommy1729 bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 08/30/2010, 08:56 AM Whats a bending point? tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 08/30/2010, 09:37 AM (08/30/2010, 08:56 AM)bo198214 Wrote: Whats a bending point? f ''(x) = 0 x^3 bends at 0. tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 06/06/2011, 10:56 PM i almost forgot about this post. maybe give it some attention again , assuming i did not make a mistake in the OP. bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 06/07/2011, 07:11 AM (08/28/2010, 11:23 PM)tommy1729 Wrote: tet(0) = 0 = slog(0) As far as I remember tet(0)=1. bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 06/07/2011, 07:47 AM (08/28/2010, 11:23 PM)tommy1729 Wrote: log(tet'(z+r)) + A = log(tet'(z)) + B hence bending points in sexp(slog(z) + r) correspond to bending points in sexp(z).? I dont see how that follows, nor does it seem to be right. For example the fractional iterates of c^x, c>eta have no bending points imho, while the curvature of sexp in (-2,0) seems to be negative for me, while for greater x it appears to be positive. So there must be a bending point somewhere. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 06/07/2011, 10:56 AM Hmm. This makes me wonder about the following conjecture: The "principal" analytic fractional iterates $\exp^t(x)$, $t \ge 0$ of the natural exponential (and perhaps any with $b > \eta$) are uniquely characterized by $\frac{d^n}{dx^n} \exp^t(x) > 0$ for all $x$, all $t \ge 0$ and all $n > 0$. « Next Oldest | Next Newest »

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