• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 slog_b(sexp_b(z)) How does it look like ? sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 04/28/2014, 11:33 PM (This post was last modified: 04/28/2014, 11:43 PM by sheldonison.) (04/28/2014, 09:01 PM)tommy1729 Wrote: ------------------------------------------------------------------------- Do you agree that sexp(z) + 2pi i is another branch of sexp(z) ?? ------------------------------------------------------------------------- ....Like if exp(sexp(u)+2 pi i) = exp(sexp(u)) = sexp(u+1) then it seems logical to conclude sexp(z) + 2pi i is another branch of sexp(z). So for example, sexp(1)=e, is there a sense that sexp(0)=log(sexp(1))+2pi i = 1+2pi i? Not if we stay in the upper half of the complex plane, where there are no singularities. There are no degrees of freedom in picking the logarithmic branch for sexp(z)=log(sexp(z+1)), since we analytically continue, smoothly from sexp(z+1) to sexp(z). If sexp(1)=e, then we cannot analytically continue to sexp(0)=1+2pi i unless we circle a logarithmic singularity. So, can we get to a place on the Riemann surface where sexp(0)=1+2pi i? Well, it just so happens that sexp(z) has a logarithmic singularity at z=-2. And if you start at sexp(0)=1, and circle counterclockwise around the singularity, then you pass through the region between sexp(-3) and sexp(-2). In this segment, imag(z)=pi I in the upper half of the complex plane and imag(z)= -pi i from the lower half of the complex plane. So, we can go through this region, and then back to sexp(0). And then we are in a strange branch of sexp(z); which I will call sexp_l(z) for logarithmic. Now, sexp_l(0) = 1 + 2 pi i. sexp_l(1)=e + 2pi i. The real axis of sexp_l is not real valued, but real valued + 2pi i. In the sexp_l branch, the defining equation is sexp_l(z+1)= exp(sexp_l(z)-2 pi i) + 2 pi i. I'm quite sure this is not what Tommy has in mind ... but yeah, it just so happens to be right there on the Riemann surface. But other than that really weird branch of sexp(z), if you avoid the real axis less then or equal to -2, where all the singularities are, then nope, it has no meaning to talk about sexp(0)=1+2pi i. And if you are talking about that wierd branch, one should be crystal clear that you are referring to that part of the function where sexp(z+1)<>exp(sexp(z)). Now, back to the main branch. slog(1+2pi i) ~= 1.9433573+0.66346938i. But this is also completely unrelated to the concept of a branch of sexp(1). - Sheldon « Next Oldest | Next Newest »

 Messages In This Thread slog_b(sexp_b(z)) How does it look like ? - by tommy1729 - 04/23/2014, 09:15 PM RE: slog_b(sexp_b(z)) How does it look like ? - by sheldonison - 04/23/2014, 10:46 PM RE: slog_b(sexp_b(z)) How does it look like ? - by tommy1729 - 04/23/2014, 11:36 PM RE: slog_b(sexp_b(z)) How does it look like ? - by sheldonison - 04/24/2014, 09:50 AM RE: slog_b(sexp_b(z)) How does it look like ? - by tommy1729 - 04/24/2014, 10:30 PM RE: slog_b(sexp_b(z)) How does it look like ? - by sheldonison - 04/25/2014, 04:54 PM RE: slog_b(sexp_b(z)) How does it look like ? - by sheldonison - 04/25/2014, 10:48 PM RE: slog_b(sexp_b(z)) How does it look like ? - by tommy1729 - 04/26/2014, 08:53 AM RE: slog_b(sexp_b(z)) How does it look like ? - by sheldonison - 04/27/2014, 04:52 AM RE: slog_b(sexp_b(z)) How does it look like ? - by tommy1729 - 04/27/2014, 03:27 PM RE: slog_b(sexp_b(z)) How does it look like ? - by sheldonison - 04/28/2014, 03:12 AM RE: slog_b(sexp_b(z)) How does it look like ? - by tommy1729 - 04/24/2014, 10:33 PM RE: slog_b(sexp_b(z)) How does it look like ? - by tommy1729 - 04/24/2014, 10:43 PM RE: slog_b(sexp_b(z)) How does it look like ? - by tommy1729 - 04/27/2014, 03:39 PM RE: slog_b(sexp_b(z)) How does it look like ? - by tommy1729 - 04/28/2014, 09:01 PM RE: slog_b(sexp_b(z)) How does it look like ? - by sheldonison - 04/28/2014, 11:33 PM RE: slog_b(sexp_b(z)) How does it look like ? - by tommy1729 - 04/29/2014, 11:52 AM

Users browsing this thread: 1 Guest(s)