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 Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 12/30/2007, 01:09 PM One application of h(1/(n^n)) = 1/n as n-> infinity I will try to apply my infinite tetration formula h(1/(n^n)) = 1/n to see what happens: So as n-> infinity (e^n)(n!)/((n^n)(sqrt n))=sqrt 2pi (Clawson) 1/(n^n) = (sgrt(2pi*n)) / ((e^n)*n!) h(1/(n^n)) = h( sgrt(2pi*n)/(e^n)*n!) = 1/n as n-> infinity which is? 0? If I knew how h works on composite arguments, may be it would be possible to study this deaper. Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 12/30/2007, 03:37 PM (This post was last modified: 12/30/2007, 03:38 PM by Ivars.) The initial spider graph of Gottfried looks very much like Caley transform, except that it is has 2 conjugate angled maps of slightly elongated circles, and the lines that should be inside Unit circle in Cayleys transform have moved to the negative part of real axis. http://mathworld.wolfram.com/CayleyTransform.html What transform would produce such a map from real axis as Gottfrieds spider? By the way, the result h(e^pi/2) = +-i also is a result of transform (e^pi/2)^ ((i-1/i+1)) and (e^pi/2)^((-i+1)/(-i-1)) Here the angle by which reals are turned is +-pi/2. As the values go above e^pi/2, angles increase but later decrease again. So the transform is definitely more complex than just: e^Real^((i-Real)/(i+Real) which is true in case Real=e^pi/2. Gottfried Ultimate Fellow Posts: 753 Threads: 114 Joined: Aug 2007 12/30/2007, 05:33 PM (This post was last modified: 12/30/2007, 05:34 PM by Gottfried.) Ivars Wrote:The initial spider graph of Gottfried looks very much like Caley transform, except that it is has 2 conjugate angled maps of slightly elongated circles, and the lines that should be inside Unit circle in Cayleys transform have moved to the negative part of real axis. http://mathworld.wolfram.com/CayleyTransform.html What transform would produce such a map from real axis as Gottfrieds spider? By the way, the result h(e^pi/2) = +-i also is a result of transform (e^pi/2)^ ((i-1/i+1)) and (e^pi/2)^((-i+1)/(-i-1)) Here the angle by which reals are turned is +-pi/2. As the values go above e^pi/2, angles increase but later decrease again. So the transform is definitely more complex than just: e^Real^((i-Real)/(i+Real) which is true in case Real=e^pi/2. Hi Ivars - please have a look at Henryk's curve; I think it is much better. http://math.eretrandre.org/tetrationforu...931#pid931 (sorry, I'm fully engaged with the tetra-series-problem, which drives me crazy (but makes me learn a lot), such that I've currently no room for other discussion ... :-( ) Gottfried Gottfried Helms, Kassel andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 01/05/2008, 08:00 PM That wasn't Henryk's, it was my curve. Its ok, we all need time. Andrew Robbins Gottfried Ultimate Fellow Posts: 753 Threads: 114 Joined: Aug 2007 01/06/2008, 02:55 AM andydude Wrote:That wasn't Henryk's, it was my curve. Its ok, we all need time. Andrew Robbins urrks - yes, you're right, sorry. i had something of Henryk's in my mind, don't know what actually. Gottfried Gottfried Helms, Kassel Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 02/07/2008, 10:49 AM (This post was last modified: 02/08/2008, 10:38 AM by Ivars.) I have plotted all 8 "spirals " of the form 4 : ( +-t^+-1/t), 4: (+- 1/t)^+-(t) in polar coordinates and there are certainly interesting forms , crossing points, and regions. I do not know how to get image in here, but it is in the atachment. Attached Files Image(s)         andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 04/14/2008, 01:01 AM I just thought I'd add something I found recently. A cardioid is a shape that is very similar to this, and I thought I'd try and approximate it. It turns out its not a cardioid, but its very close, and can be approximated with the parametric equations: $x(t) = a \cos(t) (1+\cos(ct))$ $y(t) = a \sin(t) (1+\cos(ct))$ where a = 1.36 and c = 1.17 Andrew Robbins Attached Files Image(s)     Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 04/14/2008, 10:35 AM andydude Wrote:I just thought I'd add something I found recently. A cardioid is a shape that is very similar to this, and I thought I'd try and approximate it. It turns out its not a cardioid, but its very close, and can be approximated with the parametric equations: $x(t) = a \cos(t) (1+\cos(ct))$ $y(t) = a \sin(t) (1+\cos(ct))$ where a = 1.36 and c = 1.17 Andrew Robbins Perhaps this can be also used? It has similar shape and parametrization, but different place for coefficients, and is more general then cardioid. Limacon? Ivars andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 04/25/2008, 11:25 PM I found something interesting with Sage recently, and I'd thought I'd share. It seems that Sage implements complex exponentiation differently, or incorrectly, I'm not quite sure, but this plot is different than the same one produced with Mathematica, so I'm wondering if we need to start worrying about how Sage implements (^) and if this might cause problems... Running the Sage code: Code:contour_plot(imag((x+y*i)**(1/(x+y*i))), (x, -1, 3), (y, -2, 2),     fill=False, contours=0, plot_points=1000)produces:     which, as you can see, is equivalent to the one made in Mathematica, except that the side with negative real part is the same as the side with positive real part. Is there any reason why this should be? Andrew Robbins Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 04/28/2008, 08:11 AM Its interesting as well. Its like an exact mirror image. With +-I being one of the minimums/maximums, balance point. I wonder should it be so, without any mathematical thought whatsoever. But Gottfried achieved assymetric picture in PARI as well, so that does make this last one result questionable? Ivars « Next Oldest | Next Newest »

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