08/29/2007, 06:14 PM
I have to admit I never considered the case \( 0<b<1 \), but I recently discussed it with Gianfranco.
We know that \( {}^{2n}x \) is no more injective if we allow this range.
If we consider one rank below the ordinary exponentiation, we have a similar case there, that \( b^x \) is no more defined for \( -\infty<b<0 \), because \( x^{2n} \) is not injective for \( x<0 \) so \( b^{1/2}=\pm |b|^{1/2} \). This gives also trouble as then \( b^n=b^{\frac{2n}{2}}=\pm b^n \).
On the other hand for tetration \( {}^{\frac{1}{2}} x \) is *not* the inversion of \( {}^2 x \). So the rules may be a bit different here.
So what do you think how tetration looks for \( 0<b<1 \)? And what role plays the range \( e^{-e}<b<1 \)?
We know that \( {}^{2n}x \) is no more injective if we allow this range.
If we consider one rank below the ordinary exponentiation, we have a similar case there, that \( b^x \) is no more defined for \( -\infty<b<0 \), because \( x^{2n} \) is not injective for \( x<0 \) so \( b^{1/2}=\pm |b|^{1/2} \). This gives also trouble as then \( b^n=b^{\frac{2n}{2}}=\pm b^n \).
On the other hand for tetration \( {}^{\frac{1}{2}} x \) is *not* the inversion of \( {}^2 x \). So the rules may be a bit different here.
So what do you think how tetration looks for \( 0<b<1 \)? And what role plays the range \( e^{-e}<b<1 \)?