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 some questions about sexp tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 06/28/2010, 11:18 PM 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 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 06/29/2010, 10:55 PM anyone ? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 07/01/2010, 09:39 AM (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,... mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 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 $x > 0$ ? tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 07/01/2010, 10:06 PM (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). mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 07/01/2010, 10:28 PM (07/01/2010, 10:06 PM)tommy1729 Wrote: i meant for bases > sqrt(e). What's special about $\sqrt{e}$? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 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. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 07/02/2010, 07:17 AM (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 $\sqrt[e]{e}$! mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 07/02/2010, 07:55 AM (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 $\sqrt{e}$. tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 07/02/2010, 12:27 PM (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 « Next Oldest | Next Newest »

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