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im going to ask some questions about sexp , and i mean all proposed solutions of sexp.

(question 1)
does the following hold :

d f^n / d x^n sexp(slog(x) + k) > 0 for all positive integer n and all positive real k ?

(question 2)

since slog(z) (base e) has period 2pi i why doesnt sexp(z) look like a log spiral ?

or does it , like having a branch cut at real x < -2 ?

(question 3)

what happens to limit cycles and n-ary fixpoints ??

sure we can set the fixpoints exp(L) = L at oo i but how about the fixpoints of exp(exp(.. q)) = q and limit cycles of the exp iterations ...

e.g. let e^q1 = q2 , e^q2 = q1 , if we want half-iterates , 1/3 iterates and sqrt(2) iterates to have the same fixpoints , this is a problem , not ?

perhaps we can 'hide' L at +/- oo i ( like kouznetsov ) and 'hide' the other points at - oo ??

(question 4)

do all 'analytic in the neigbourhood of the positive reals' sexp have the fixpoints exp(L) = L at oo i ?

( i know that [L,sexp(slog(L)+o(1))] cannot be in the analytic zone , but maybe [L,sexp(slog(L)+o(1))] isnt part of the analytic zone )

(question 5)

slog(z) base x is not holomorphic in x in a domain containing the interval [a,b] if eta is between a and b.

why is that ? i know that the real fixpoint dissappears but still ...

sorry if those are FAQ or trivial Q.

regards

tommy1729
anyone ?
(06/28/2010, 11:18 PM)tommy1729 Wrote: [ -> ](question 1)
does the following hold :

d f^n / d x^n sexp(slog(x) + k) > 0 for all positive integer n and all positive real k ?

Though tempting for uniqueness this seems not to be the case, the half iterate of exp(x)-1 (which is conjugate to $e^{x/e}$) has the following powerseries coefficients at 0:
Code:
0, 1, 1/4, 1/48, 0, 1/3840, -7/92160, 1/645120,...
(07/01/2010, 09:39 AM)bo198214 Wrote: [ -> ]
(06/28/2010, 11:18 PM)tommy1729 Wrote: [ -> ](question 1)
does the following hold :

d f^n / d x^n sexp(slog(x) + k) > 0 for all positive integer n and all positive real k ?

Though tempting for uniqueness this seems not to be the case, the half iterate of exp(x)-1 (which is conjugate to $e^{x/e}$) has the following powerseries coefficients at 0:
Code:
0, 1, 1/4, 1/48, 0, 1/3840, -7/92160, 1/645120,...

That Taylor series does not converge (I've seen this one before, it's from the parabolic regular iteration, right?), though. So $0$ is actually a singularity of some kind of $\mathrm{dxp}^{1/2}_{e^{1/e}}(x)$ ($\mathrm{dxp}_b(x) = \exp_b(x) - 1$) and so there is no Taylor expansion there.
moderators note: corrected latex expression

What about (here for base $e^{1/e}$)

$\frac{d^n}{dx^n} \mathrm{sexp}_{e^{1/e}}(x)$ does not change sign for all $x > 0$

?
(07/01/2010, 09:39 AM)bo198214 Wrote: [ -> ]
(06/28/2010, 11:18 PM)tommy1729 Wrote: [ -> ](question 1)
does the following hold :

d f^n / d x^n sexp(slog(x) + k) > 0 for all positive integer n and all positive real k ?

Though tempting for uniqueness this seems not to be the case, the half iterate of exp(x)-1 (which is conjugate to $e^{x/e}$) has the following powerseries coefficients at 0:
Code:
0, 1, 1/4, 1/48, 0, 1/3840, -7/92160, 1/645120,...

i meant for bases > sqrt(e).
(07/01/2010, 10:06 PM)tommy1729 Wrote: [ -> ]i meant for bases > sqrt(e).

What's special about $\sqrt{e}$?
(07/01/2010, 11:05 AM)mike3 Wrote: [ -> ]
(07/01/2010, 09:39 AM)bo198214 Wrote: [ -> ]Though tempting for uniqueness this seems not to be the case, the half iterate of exp(x)-1 (which is conjugate to $e^{x/e}$) has the following powerseries coefficients at 0:
Code:
0, 1, 1/4, 1/48, 0, 1/3840, -7/92160, 1/645120,...

That Taylor series does not converge (I've seen this one before, it's from the parabolic regular iteration, right?), though. So $0$ is actually a singularity of some kind of $\mathrm{dxp}^{1/2}_{e^{1/e}}(x)$ ($\mathrm{dxp}_b(x) = \exp_b(x) - 1$) and so there is no Taylor expansion there.

Yayaya, but these are called "asymptotic Taylor expansions". The Taylor expansions of regular $\mathrm{dxp}^{1/2}_{e^{1/e}}$ at $x_0\neq 0$ converge to the given coefficients for $x_0\to 0$ though it is not analytic at 0.

But this is sufficient for our case, as for $x_0$ close enough to 0 this one derivative must turn negative.
(07/01/2010, 10:28 PM)mike3 Wrote: [ -> ]
(07/01/2010, 10:06 PM)tommy1729 Wrote: [ -> ]i meant for bases > sqrt(e).

What's special about $\sqrt{e}$?

Wah Mike, you know that he meant $\sqrt[e]{e}$!
(07/02/2010, 07:17 AM)bo198214 Wrote: [ -> ]Wah Mike, you know that he meant $\sqrt[e]{e}$!

Actually, I didn't. I saw "sqrt" and I thought that meant, well, square root, not self-root! I thought maybe he had discovered something new about base $\sqrt{e}$.
(07/02/2010, 07:55 AM)mike3 Wrote: [ -> ]
(07/02/2010, 07:17 AM)bo198214 Wrote: [ -> ]Wah Mike, you know that he meant $\sqrt[e]{e}$!

Actually, I didn't. I saw "sqrt" and I thought that meant, well, square root, not self-root! I thought maybe he had discovered something new about base $\sqrt{e}$.

i meant sqrt(e) !

because my method works for bases > sqrt(e) , not for all bases > e^(1/e).

mike was correct.

tommy1729
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