08/29/2007, 05:52 PM
Thanks for the proposal, Gianfranco.
I think that most of it makes sense.
What regards the symbols I would suggest:
\( \log[n]_b(x) \) and \( \exp[n]_b(x) \) for the hyper operations of rank n. So that for example \( \text{slog}_b(x)=\log[4]_b(x) \) and \( \log[3]_b(x)=\log_b(x) \) and \( \log[2]_b(x)=\frac{1}{b}x \) and \( \log[1]_b(x)=-b+x \).
But anyway the 4th operation is the bottle neck. If we solved this satisfactory then all the other higher operations probably follow at once. So nobody perhaps will speak particularely about \( \log[5]_b(x) \) but rather about all hyper logarithms. Thatswhy it is important to have the separate writings \( \text{slog}_b(x) \) and the \( \text{sexp}_b(x) \) and \( {}^xb \).
Also the hyperroots are not yet discussed here. Appropriately I have no alternate suggestion
.
I think that most of it makes sense.
What regards the symbols I would suggest:
\( \log[n]_b(x) \) and \( \exp[n]_b(x) \) for the hyper operations of rank n. So that for example \( \text{slog}_b(x)=\log[4]_b(x) \) and \( \log[3]_b(x)=\log_b(x) \) and \( \log[2]_b(x)=\frac{1}{b}x \) and \( \log[1]_b(x)=-b+x \).
But anyway the 4th operation is the bottle neck. If we solved this satisfactory then all the other higher operations probably follow at once. So nobody perhaps will speak particularely about \( \log[5]_b(x) \) but rather about all hyper logarithms. Thatswhy it is important to have the separate writings \( \text{slog}_b(x) \) and the \( \text{sexp}_b(x) \) and \( {}^xb \).
Also the hyperroots are not yet discussed here. Appropriately I have no alternate suggestion
.