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 base holomorphic tetration bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 11/13/2009, 12:16 PM Give me the weekend and I will make a complex contour plot of some regular tetrapower, or I ask Dmitrii, perhaps he is faster in make such plot. The singularities should be quite visible in such a plot. bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 11/16/2009, 12:30 AM (This post was last modified: 11/16/2009, 12:33 AM by bo198214.) (11/13/2009, 12:16 PM)bo198214 Wrote: Give me the weekend and I will make a complex contour plot of some regular tetrapower Here are finally the pictures (damn did that take time, as a side effect I discovered a bug in the python mpmath.lambertw function). I show the conformal map of z[4]0.5 inside the upper half of the Shell-Thron region via regular iteration, i.e. $z[4]0.5 = \exp_b^{\circ 0.5}(1)$.         For the accuracy I checked the difference $\exp_b^{\circ 1/2}(\exp_b^{\circ 1/2}(1))- b$. It is always under $10^{-8}$. I computed the regular iteration as a mixture of the powerseries development at the lower fixed point, together with (if 1 is not inside the convergence radius of this powerseries) the equality $\log_b^{\circ n} \circ \exp_b^{\circ t} \circ \exp_b^{\circ n} = \exp_b^{\circ t}$, i.e. the exponentials move the argument towards the fixed point until it is inside its convergence radius then the powerseries can be applied and the same number of logarithms have to finish the computation. I assumed the convergence radius to be at least the distance of the fixed point from the image of the Shell-Thron region. bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 11/16/2009, 12:55 AM (This post was last modified: 11/16/2009, 12:57 AM by bo198214.) If we choose the original region closer to the Shell-Thron region the regular tetra-power gets chaotic:         This seems to indicate that the function has singularities on the Shell-Thron boundary. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 11/16/2009, 01:10 AM (This post was last modified: 11/16/2009, 01:10 AM by mike3.) What about the vicinity of the area $b = e^{1/e}$? And what formula are you using for this, the sum one or the limit one? Which requires more numerical precision, esp. near the edge? bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 11/16/2009, 09:01 AM (This post was last modified: 11/16/2009, 09:03 AM by bo198214.) (11/16/2009, 01:10 AM)mike3 Wrote: And what formula are you using for this, the sum one or the limit one? Which requires more numerical precision, esp. near the edge? Mike, I took the extra time to describe it in my previous post: (11/16/2009, 12:30 AM)bo198214 Wrote: I computed the regular iteration as a mixture of the powerseries development at the lower fixed point, together with (if 1 is not inside the convergence radius of this powerseries) the equality $\log_b^{\circ n} \circ \exp_b^{\circ t} \circ \exp_b^{\circ n} = \exp_b^{\circ t}$, i.e. the exponentials move the argument towards the fixed point until it is inside its convergence radius then the powerseries can be applied and the same number of logarithms have to finish the computation. I assumed the convergence radius to be at least the distance of the fixed point from the image of the Shell-Thron region. Quote:What about the vicinity of the area $b = e^{1/e}$? In which form/region you want it to be shown? Different sectors, or decreasing distance to the boundary? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 11/16/2009, 11:39 AM (11/16/2009, 09:01 AM)bo198214 Wrote: Mike, I took the extra time to describe it in my previous post: (11/16/2009, 12:30 AM)bo198214 Wrote: I computed the regular iteration as a mixture of the powerseries development at the lower fixed point, together with (if 1 is not inside the convergence radius of this powerseries) the equality $\log_b^{\circ n} \circ \exp_b^{\circ t} \circ \exp_b^{\circ n} = \exp_b^{\circ t}$, i.e. the exponentials move the argument towards the fixed point until it is inside its convergence radius then the powerseries can be applied and the same number of logarithms have to finish the computation. I assumed the convergence radius to be at least the distance of the fixed point from the image of the Shell-Thron region. Ah, I see now. How much numerical precision was required, though? Quote:What about the vicinity of the area $b = e^{1/e}$? In which form/region you want it to be shown? Different sectors, or decreasing distance to the boundary? Region? I suppose a small circle around e^{1/e}, excluding the wedge that lies outside the STR? bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 11/16/2009, 12:35 PM (11/16/2009, 11:39 AM)mike3 Wrote: Ah, I see now. How much numerical precision was required, though? Nearly nothing. I computed with 100 bits precision and 10 terms of the powerseries. This gives already error under $10^{-8}$, at least for the shown regions. Quote:Region? I suppose a small circle around e^{1/e}, excluding the wedge that lies outside the STR? Yes I meant how to approach the STR boundary. Because if you go really close then you dont see much in the chaos (as it is shown in my previous pictures). I will do it today evening (MET). bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 11/16/2009, 06:29 PM So here you get the wished pictures, with my invention of a sector plot. It shows that $b=e^{1/e}$ behaves quite regularly, while a randomly choosen other base on the STR boundary does not.                 mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 11/16/2009, 08:08 PM (This post was last modified: 11/16/2009, 08:09 PM by mike3.) Interesting. So perhaps $b = e^{1/e}$ isn't a singularity after all, though there are other points on the boundary that are, which would agree with the results from the Taylor experiment. bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 11/17/2009, 10:20 AM (This post was last modified: 11/17/2009, 10:22 AM by bo198214.) (11/16/2009, 08:08 PM)mike3 Wrote: So perhaps $b = e^{1/e}$ isn't a singularity after all, though there are other points on the boundary that are, which would agree with the results from the Taylor experiment. Unfortunately from the picture you can not conclude that it is no singularity. Isolated singularities have this behaviour to take on nearly every value in each neighborhood, branchpoints however can be quite wellbehaved, e.g. the pictures for sqrt are:         My guess is that the "proper" singularities (i.e. those with chaotic behaviour taking nearly every point in each neighborhood) are dense on the TSR boundary. So that no continuable path leads outside the TSR. « Next Oldest | Next Newest »

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