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some questions about sexp
#1
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
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#2
anyone ?
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#3
(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 ) has the following powerseries coefficients at 0:
Code:
0, 1, 1/4, 1/48, 0, 1/3840, -7/92160, 1/645120,...
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#4
(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 ) 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 is actually a singularity of some kind of () and so there is no Taylor expansion there.
moderators note: corrected latex expression

What about (here for base )

does not change sign for all

?
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#5
(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 ) 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).
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#6
(07/01/2010, 10:06 PM)tommy1729 Wrote: i meant for bases > sqrt(e).

What's special about ?
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#7
(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 ) 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 is actually a singularity of some kind of () and so there is no Taylor expansion there.

Yayaya, but these are called "asymptotic Taylor expansions". The Taylor expansions of regular at converge to the given coefficients for though it is not analytic at 0.

But this is sufficient for our case, as for close enough to 0 this one derivative must turn negative.
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#8
(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 ?

Wah Mike, you know that he meant !
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#9
(07/02/2010, 07:17 AM)bo198214 Wrote: Wah Mike, you know that he meant !

Actually, I didn't. I saw "sqrt" and I thought that meant, well, square root, not self-root! Smile I thought maybe he had discovered something new about base .
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#10
(07/02/2010, 07:55 AM)mike3 Wrote:
(07/02/2010, 07:17 AM)bo198214 Wrote: Wah Mike, you know that he meant !

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

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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