Tetration base 1 is a constant function equal to 1, for x>-1, and has a discontinuity at -1. Or at least that's the solution if \( \\[15pt]
{^{x}1=\lim_{b \to 1} ^xb} \).
![[Image: 3ZQkL3L.jpg?1]](http://i.imgur.com/3ZQkL3L.jpg?1)
The oscillatory behavior of bases 0<b<1 suggest that base 0 looks like this on the limit for b=0:
![[Image: ToRZAgn.jpg?1]](http://i.imgur.com/ToRZAgn.jpg?1)
We don't know if tetration has changes of state as the base converges to zero, but \( \\[15pt]
{^{x}0} \) can only be equal to 0 or 1, so the doubt is how many discontinuities it has on a given interval.
Base -1:
\( \\[15pt]
{-1^{-1}=-1} \), so, once \( \\[15pt]
{^{x_0}-1=-1} \), then the function must be periodically equal to -1: \( \\[15pt]
{^{x_0+n}-1=-1} \)
This is a possible solution:
![[Image: Gs1tpK1.jpg?1]](http://i.imgur.com/Gs1tpK1.jpg?1)
Is not the only solution; it may be displaced to the right by any real value d (0<d<1), and still will be a solution valid on the interval -2+d<x.
I wonder if there is a family of continuous solutions, and this one is the evolvent.
{^{x}1=\lim_{b \to 1} ^xb} \).
The oscillatory behavior of bases 0<b<1 suggest that base 0 looks like this on the limit for b=0:
We don't know if tetration has changes of state as the base converges to zero, but \( \\[15pt]
{^{x}0} \) can only be equal to 0 or 1, so the doubt is how many discontinuities it has on a given interval.
Base -1:
\( \\[15pt]
{-1^{-1}=-1} \), so, once \( \\[15pt]
{^{x_0}-1=-1} \), then the function must be periodically equal to -1: \( \\[15pt]
{^{x_0+n}-1=-1} \)
This is a possible solution:
Is not the only solution; it may be displaced to the right by any real value d (0<d<1), and still will be a solution valid on the interval -2+d<x.
I wonder if there is a family of continuous solutions, and this one is the evolvent.
I have the result, but I do not yet know how to get it.