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+--- Thread: tetration bending uniqueness ? (/showthread.php?tid=503)
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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 \( n > 0 \).