tetration bending uniqueness ? - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: tetration bending uniqueness ? (/showthread.php?tid=503) Pages: 1 2 tetration bending uniqueness ? - tommy1729 - 08/28/2010 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 RE: tetration bending uniqueness ? - tommy1729 - 08/28/2010 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 ? RE: tetration bending uniqueness ? - tommy1729 - 08/29/2010 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 RE: tetration bending uniqueness ? - tommy1729 - 08/29/2010 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 RE: tetration bending uniqueness ? - bo198214 - 08/30/2010 Whats a bending point? RE: tetration bending uniqueness ? - tommy1729 - 08/30/2010 (08/30/2010, 08:56 AM)bo198214 Wrote: Whats a bending point? f ''(x) = 0 x^3 bends at 0. RE: tetration bending uniqueness ? - tommy1729 - 06/06/2011 i almost forgot about this post. maybe give it some attention again , assuming i did not make a mistake in the OP. RE: tetration bending uniqueness ? - bo198214 - 06/07/2011 (08/28/2010, 11:23 PM)tommy1729 Wrote: tet(0) = 0 = slog(0) As far as I remember tet(0)=1. RE: tetration bending uniqueness ? - bo198214 - 06/07/2011 (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. RE: tetration bending uniqueness ? - mike3 - 06/07/2011 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 $n > 0$.