When one starts considering tetration then one usually has the choice between chosing
\( x[4](n+1)=x^{x[4]n} \) or \( x[4'](n+1)=(x[4]n)^x \).
Usually people always prefer the first right bracketed law, because the left bracketed law is too simple:
\( x[4']n=x^{x^{n-1}} \), assuming \( x[4']1=x \).
However as I described here it is more appropriate to assume \( x[4']0=x \) getting the even more simple
\( x[4']n=x^{x^n} \).
Though this is a trivial start, it does not remain trivial when considering left pentation:
\( b[5']0=b \)
\( b[5'](n+1)=(b[5']n)[4']b=(b[5']n)^{(b[5']n)^b} \)
For example \( x[5']1=x^{x^x} \), \( x[5']2= \left( {x}^{{x}^{x}} \right) ^{ \left( {x}^{{x}^{x}} \right) ^{x}}=x^{x^x x^{xx^x}}=x^{x^{x+x^{x+1}} \)
So there is no reason to neglect the left hyper operation sequence by being too trivial! But before proceding something about extension of the left tetration to the reals.
There is one quite obvious way for an extension:
We can describe it by \( b[4']n= f^{\circ n}(b) \) where \( f(x)=x^b \). Then we even have the desired \( b[4']0=f^{\circ 0}(b)=b \).
Now \( x^b \) has the fixed point 1 and this is the only analytic fixed point there. So of course we consider regular iteration at this fixed point which indeed yields \( f^{\circ t}(x)=x^{b^t} \) (without proof) so \( b[4']t=b^{b^t} \).
We do similar for left pentation, let \( f(x)=x^{x^b} \) then \( b[5']n=f^{\circ n}(b) \). Now \( f(x) \) has again a fixed point at 1 and we can apply regular iteration there.
Generally:
\( b[k+1']t={f_k}^{\circ t}(b) \) where \( f_k(x)=x[k']b \).
We can show by induction that \( f_k(1)=1 \) for \( k\ge 3 \) and that \( f_k'(1)=1 \) for \( k\ge 4 \). To obtain real values we maybe must assume that \( b\ge 1 \) and \( f_k(x) \) defined only on \( x\ge 1 \) for \( k\ge 4 \). I think it also can be shown that 1 is the only real fixed point of \( f_k \), \( k\ge 4 \).
So the situation here is quite similar to the one of balanced tetration.
\( x[4](n+1)=x^{x[4]n} \) or \( x[4'](n+1)=(x[4]n)^x \).
Usually people always prefer the first right bracketed law, because the left bracketed law is too simple:
\( x[4']n=x^{x^{n-1}} \), assuming \( x[4']1=x \).
However as I described here it is more appropriate to assume \( x[4']0=x \) getting the even more simple
\( x[4']n=x^{x^n} \).
Though this is a trivial start, it does not remain trivial when considering left pentation:
\( b[5']0=b \)
\( b[5'](n+1)=(b[5']n)[4']b=(b[5']n)^{(b[5']n)^b} \)
For example \( x[5']1=x^{x^x} \), \( x[5']2= \left( {x}^{{x}^{x}} \right) ^{ \left( {x}^{{x}^{x}} \right) ^{x}}=x^{x^x x^{xx^x}}=x^{x^{x+x^{x+1}} \)
So there is no reason to neglect the left hyper operation sequence by being too trivial! But before proceding something about extension of the left tetration to the reals.
There is one quite obvious way for an extension:
We can describe it by \( b[4']n= f^{\circ n}(b) \) where \( f(x)=x^b \). Then we even have the desired \( b[4']0=f^{\circ 0}(b)=b \).
Now \( x^b \) has the fixed point 1 and this is the only analytic fixed point there. So of course we consider regular iteration at this fixed point which indeed yields \( f^{\circ t}(x)=x^{b^t} \) (without proof) so \( b[4']t=b^{b^t} \).
We do similar for left pentation, let \( f(x)=x^{x^b} \) then \( b[5']n=f^{\circ n}(b) \). Now \( f(x) \) has again a fixed point at 1 and we can apply regular iteration there.
Generally:
\( b[k+1']t={f_k}^{\circ t}(b) \) where \( f_k(x)=x[k']b \).
We can show by induction that \( f_k(1)=1 \) for \( k\ge 3 \) and that \( f_k'(1)=1 \) for \( k\ge 4 \). To obtain real values we maybe must assume that \( b\ge 1 \) and \( f_k(x) \) defined only on \( x\ge 1 \) for \( k\ge 4 \). I think it also can be shown that 1 is the only real fixed point of \( f_k \), \( k\ge 4 \).
So the situation here is quite similar to the one of balanced tetration.
