By the way, if none of what I'm saying is making sense, fear not! I will be providing colorful charts to illustrate my points. Some of what I've described can already be seen in the chart I originally posted in another thread:
http://math.eretrandre.org/tetrationforu...440#pid440
[attachment=73]
Notice that the red region and the blue region have some overlap, the red region being the fourth iterated logarithm of the blue region. Notice also that the two regions are nothing alike, and hence the derivatives evaluated on either "branch" are nothing like the derivatives evaluated on the other branch. This is in stark contrast to the basic logarithm, where the branches may have different values, but the derivatives at any point are the same regardless of the branch used to evaluate the derivatives.
Notice also that the top of this chart (the top of the green and yellow regions) is on the line with imaginary part pi*i. If you reflect everything you see there above this line, taking complex conjugates of all points, you get the analytic continuation. At the line with imaginary part 2*pi*i, the slog evaluates to exactly the same values it would on the real line. Here we see the cyclic symmetry, not unlike the symmetry of a sine-like function evaluated using the imaginary part as the variable. This of course is due to the branches of the natural logarithm, or the fact that exponentiating values differing by a multiple of 2*pi*i will give the same result.
To get from the origin to 2*pi*i, we could take the ith iterate of 0, which gets us to the blue cross in the chart, near 0.2+0.85i. Then we'd take the -1.5 partial iterate, i.e., halway between a natural logarithm and a double natural logarithm, which gets us to the intersection of yellow and green dark lines, near 0.05+1.62i. Then we'd take the -ith iterate, to get to -0.366+pi*i. Then we'd take another -ith iterate to get to 0.05+(2*pi-1.62)*i. Then we'd take a 1.5 partial iterate, more than an exponentiation but not quite a double exponentiation, to get to 0.2+(2*pi-0.85)*i. Finally, we'd take an ith iteration to get to 0+2*pi*i.
Adding it all up, we get back to 0 iterations from 0, yet we're at 2*pi, not 0, and we didn't even loop around a singularity.
http://math.eretrandre.org/tetrationforu...440#pid440
[attachment=73]
Notice that the red region and the blue region have some overlap, the red region being the fourth iterated logarithm of the blue region. Notice also that the two regions are nothing alike, and hence the derivatives evaluated on either "branch" are nothing like the derivatives evaluated on the other branch. This is in stark contrast to the basic logarithm, where the branches may have different values, but the derivatives at any point are the same regardless of the branch used to evaluate the derivatives.
Notice also that the top of this chart (the top of the green and yellow regions) is on the line with imaginary part pi*i. If you reflect everything you see there above this line, taking complex conjugates of all points, you get the analytic continuation. At the line with imaginary part 2*pi*i, the slog evaluates to exactly the same values it would on the real line. Here we see the cyclic symmetry, not unlike the symmetry of a sine-like function evaluated using the imaginary part as the variable. This of course is due to the branches of the natural logarithm, or the fact that exponentiating values differing by a multiple of 2*pi*i will give the same result.
To get from the origin to 2*pi*i, we could take the ith iterate of 0, which gets us to the blue cross in the chart, near 0.2+0.85i. Then we'd take the -1.5 partial iterate, i.e., halway between a natural logarithm and a double natural logarithm, which gets us to the intersection of yellow and green dark lines, near 0.05+1.62i. Then we'd take the -ith iterate, to get to -0.366+pi*i. Then we'd take another -ith iterate to get to 0.05+(2*pi-1.62)*i. Then we'd take a 1.5 partial iterate, more than an exponentiation but not quite a double exponentiation, to get to 0.2+(2*pi-0.85)*i. Finally, we'd take an ith iteration to get to 0+2*pi*i.
Adding it all up, we get back to 0 iterations from 0, yet we're at 2*pi, not 0, and we didn't even loop around a singularity.
~ Jay Daniel Fox
