08/15/2009, 10:46 PM

Now I will show the case of n=5.

I find log(log(log(-1))), and for simplicity, I will keep using the principal branch. This gives me a value of 0.664571922248822 + 0.941029487312600*I. I will show two graphs.

The first shows the straight line with endpoints 0 and log(log(log(-1))), in the color of blue. The image of the first exponentiation of this line is shown in a sort of cyan color, and it has the same endpoint as the straight line from the case of n=4, incidentally.

The second exponentiation gets to pi*i, and is show in a sort of yellow-green hue. Finally, the third exponentiation gets to -1, and is shown in red.

This second graph is simply zoomed out, so you can see the full curve of the third exponentiation.

I find log(log(log(-1))), and for simplicity, I will keep using the principal branch. This gives me a value of 0.664571922248822 + 0.941029487312600*I. I will show two graphs.

The first shows the straight line with endpoints 0 and log(log(log(-1))), in the color of blue. The image of the first exponentiation of this line is shown in a sort of cyan color, and it has the same endpoint as the straight line from the case of n=4, incidentally.

The second exponentiation gets to pi*i, and is show in a sort of yellow-green hue. Finally, the third exponentiation gets to -1, and is shown in red.

This second graph is simply zoomed out, so you can see the full curve of the third exponentiation.

~ Jay Daniel Fox