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 Extension of tetration to other branches mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 10/25/2009, 10:56 PM (This post was last modified: 10/25/2009, 11:01 PM by mike3.) (10/25/2009, 09:22 PM)bo198214 Wrote: (10/25/2009, 08:47 PM)mike3 Wrote: So an infinite path might be one way of imagining it, or the limit of infinitely many finite paths. Ya, but there are no infinite paths. Only little before I discussed that in this post. If you want to come to a branch value you need a path between two points (or a closed path). And a path between two points or a closed path can not be infinite, can it? So then it would be better to imagine it as a limiting value approached by an infinitely long sequence of results of of finite paths. Every value of $\tet_b(z)$, can be indexed by a finite sequence $[s_0, s_1, ..., s_k]$ of integers relative to some principal branch, which I call a "branch code". This is what gets plugged into the limit formula I gave. This sequence means that to get to some non-principal value for a given $z$, you start at the principal branch, wind $s_0$ times (positive = counterclockwise, negative = clockwise) around $z = -2$, then wind $s_1$ times around $z = -3$, then wind $s_2$ times around $z = -4$, then wind $s_3$ times around $z = -5$, and so on until you have wound $s_k$ times around $z = -2 - k$, and then finally come back to (not winding around any more singularities!) the value $z$ that you wanted to evaluate at. However, we can also assign a sort of value to an infinitely long branch code $[s_0, s_1, s_2, s_3, s_4, ...]$ by considering the $limit$ of the branches obtained with the finite partial branch codes $[s_0]$, $[s_0, s_1]$, $[s_0, s_1, s_2]$, $[s_0, s_1, s_2, s_3]$, $[s_0, s_1, s_2, s_3, s_4]$, and so on. Note that this is not a value of $\mathrm{tet}_b(z)$ as an analytic multivalued function because, as you say, no finite path (hence none that ends at the point "z" we are evaluating at) gets you there. That such limit values exist makes me wonder what the structure of the Riemann surface must look like. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 10/26/2009, 04:28 AM (10/25/2009, 10:56 PM)mike3 Wrote: That such limit values exist makes me wonder what the structure of the Riemann surface must look like. *nods* there seem to be a lot of accumulation points in the branches of one point. Very unlike log which has none. So do you make a picture? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 10/26/2009, 04:41 AM (10/26/2009, 04:28 AM)bo198214 Wrote: (10/25/2009, 10:56 PM)mike3 Wrote: That such limit values exist makes me wonder what the structure of the Riemann surface must look like. *nods* there seem to be a lot of accumulation points in the branches of one point. Very unlike log which has none. So do you make a picture? As mentioned, I do not have software that is capable of graphing a multilayered 3d sheet graph. Would just a graph on the real axis be good? i.e. many branches plotted on the same real-axis graph? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 10/26/2009, 05:07 PM (10/26/2009, 04:41 AM)mike3 Wrote: Would just a graph on the real axis be good? i.e. many branches plotted on the same real-axis graph? Give it a try. I mean those multilayered pictures are anyway not really helpful as they only show imaginary or real part or the modulus. You wouldnt see accumulation points on those pictures. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 10/27/2009, 08:59 PM (10/26/2009, 05:07 PM)bo198214 Wrote: Give it a try. I mean those multilayered pictures are anyway not really helpful as they only show imaginary or real part or the modulus. You wouldnt see accumulation points on those pictures. Because the points (actually "sheets" as they're not just isolated points) are complex numbers, and so it needs to accumulate in both? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 10/28/2009, 07:42 AM (10/27/2009, 08:59 PM)mike3 Wrote: Because the points (actually "sheets" as they're not just isolated points) are complex numbers, and so it needs to accumulate in both? You mean in the real and imaginary part? Of course. The sequence of branches of the real axis (drawn in the complex plane) should converge to a point. Perhaps every point of the complex is then the limit point of some branch sequence. « Next Oldest | Next Newest »

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