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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 )


tet(slog(x)) = x

tet(0) = 0 = slog(0)


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


tet''(slog(z)+r)/tet'(slog(z))^2 = (tet'(slog(z)+r)) * tet''(slog(z))/tet'(slog(z))^3


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).


all ( analytic ) sexp(z) with sexp(0) = 0 and positive real to positive real ,

without bending points are identical !!

headscratch ...

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 ?
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 , ...

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.


Whats a bending point?
(08/30/2010, 08:56 AM)bo198214 Wrote: [ -> ]Whats a bending point?

f ''(x) = 0

x^3 bends at 0.
i almost forgot about this post.

maybe give it some attention again , assuming i did not make a mistake in the OP.
(08/28/2010, 11:23 PM)tommy1729 Wrote: [ -> ]tet(0) = 0 = slog(0)

As far as I remember tet(0)=1.
(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.
[Image: attachment.php?aid=796]

Hmm. This makes me wonder about the following conjecture:

The "principal" analytic fractional iterates , of the natural exponential (and perhaps any with ) are uniquely characterized by

for all , all and all .
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