11/12/2007, 09:56 AM

The best way to illustrate it is through this graph:

where the thin dotted line is tetration base-1, the solid line is tetration base-e, and the thick dotted line is tetration base-infinity. Neither of these are very interesting (because they are straight lines), and base-infinity isn't really solvable, but it doesn't matter because you can take limits. In the limit as the base goes to 1 the curve gets closer and closer to the thin dotted line, whereas in the limit as the base goes to infinity, the curve gets closer and closer to the thick dotted line, so these aren't really functions but asymptotes of tetration! Some of these trends can be seen on one of my graphs on my website. However, I didn't come to this conclusion by looking at graphs; I have proof. So the super-logarithms of the previously mentioned bases would be the reflection of those asymptotes across the y=x line.

Andrew Robbins

where the thin dotted line is tetration base-1, the solid line is tetration base-e, and the thick dotted line is tetration base-infinity. Neither of these are very interesting (because they are straight lines), and base-infinity isn't really solvable, but it doesn't matter because you can take limits. In the limit as the base goes to 1 the curve gets closer and closer to the thin dotted line, whereas in the limit as the base goes to infinity, the curve gets closer and closer to the thick dotted line, so these aren't really functions but asymptotes of tetration! Some of these trends can be seen on one of my graphs on my website. However, I didn't come to this conclusion by looking at graphs; I have proof. So the super-logarithms of the previously mentioned bases would be the reflection of those asymptotes across the y=x line.

Andrew Robbins