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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
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(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,...
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07/01/2010, 11:05 AM
(This post was last modified: 07/02/2010, 07:02 AM by bo198214.)
(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 \( x > 0 \)
?
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(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).
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(07/01/2010, 10:06 PM)tommy1729 Wrote: i meant for bases > sqrt(e).
What's special about \( \sqrt{e} \)?
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07/02/2010, 07:14 AM
(This post was last modified: 07/02/2010, 07:15 AM by bo198214.)
(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.
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(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 \( \sqrt[e]{e} \)!
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(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 \( \sqrt{e} \).
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(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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