I ask because I want to observe how logarithmic semi-operators behave for bases less than or equal to \( e^{1/e} \). I haven't a clue where I might going about getting these. I think it was Sheldon who posted coefficients for me before, but they were base 2, and the graphs didn't look pretty; so I just wonder if more erratic bases will give different results. I know that \( log_{e^{1/e}}(x) \) has a fix point at x = e, so I wonder if that might change anything and might shift the hump that appears with base 2.
If anybody wonders what I'm talking about it's \( 0 \le q \le 1 \) and \( r:f(x) = f^{\alpha r}(x) = f^{[r]}(x) \):
\( x\, \{-q\}\, y = {\small (1-q):}\log_{b}({\small (q-1):}\log_{b}x+ {\small (q-1):}\log_{b}y) \) which behaves as addition
\( x\, \{1-q\}\, y = {\small (q-1):}\log_{b}({\small (1-q):}\log_{b}x + {\small (1-q):}\log_{b}y) \) which behaves as multiplication
\( x\, \{2-q\}\, y = {\small (q-1):}\log_{b}(y [{\small (1-q):}\log_{b}x]) \) which behaves as exponentiation
Thanks for reading this. Any help would be greatly appreciated.
If anybody wonders what I'm talking about it's \( 0 \le q \le 1 \) and \( r:f(x) = f^{\alpha r}(x) = f^{[r]}(x) \):
\( x\, \{-q\}\, y = {\small (1-q):}\log_{b}({\small (q-1):}\log_{b}x+ {\small (q-1):}\log_{b}y) \) which behaves as addition
\( x\, \{1-q\}\, y = {\small (q-1):}\log_{b}({\small (1-q):}\log_{b}x + {\small (1-q):}\log_{b}y) \) which behaves as multiplication
\( x\, \{2-q\}\, y = {\small (q-1):}\log_{b}(y [{\small (1-q):}\log_{b}x]) \) which behaves as exponentiation
Thanks for reading this. Any help would be greatly appreciated.
