03/12/2008, 03:38 PM
Concerning:
[b -> -00] lim (e[5]b) = - 1.85...
[b -> -1.85..] lim (penta-ln b) = - oo
The problem, for me, was again produced by the TeX animals. After the fifth slash, I fall asleep and I try to compensate using the copy-and-glue technology. Unfortunately, it was late and I was very tired. I just wanted to show that the tails of the sln and sexp (base e) plots cross themselves in a point "sigma", which has real, negative but not integer coordinates. That's all. But, this is the past.
@Henryk -
I agree on the Andydude overall scheme, concerning the odd/even hyperations. I just wished to draw the attention to the need of finding the right numerical values. It is also clear (... for me, I hope not to be wrong) that in the odd (3, 5, 7, ..) ranks y = b[2n+1]x have horizontal asymptotes (for x -> -oo), say for y = - k, and, therefore, their corresponding hyperlogs have vertical asymptotes for x -> - k (with y -> -oo). It is also clear that, for even ranks (2, 4, 6, ..) the contrary is true (horizontal asymptotes for the hyperlogs and vertical asymptotes for the hyperops). However, the problem remains of finding (exactly) such numbers. I think that we must completely solve the "tetration" business, before going further.
Probably, you are also right, even formula [x -> -2n+1]lim(b[2n]x) = - oo may be wrong.
@Ivars -
The role of the selfroot is essential. Nevertheless, it is important as a solution of the y = b[s]y functional equation. But, unfortunately (so to say) it is not its unique "functional root". Another solution can be found by considering y = b[s](b[s]y), or [/b]s-log(y) = y = b[s]y.
E.g. for rank 3 (exponentiation), we should consider (supposing base b) an expression such: [/b]log x = x = b^x. I. e. also other intersection points between the exp and the log, which will give other "branches" of the functional roots, are relevant. And, the same should be valid for other hyperops ranks.
GFR
bo198214 Wrote:Actually, believe or not, what I wanted to say was:GFR Wrote:\( \lim_{b\rightarrow-\infty}(e \begin{tabular}{|c|}\hline 5 \\\hline\end{tabular} b) = -1.85.. \), and:Wah, the second one is wrong! As Andydude wrote, the odd hyperexponentials are bounded below. The even hyper exponentials b[2n]x are not bounded below but only defined on -2n+2=-(2n-3)-1<x.
\( \lim_{b\rightarrow-1.85..}(e \begin{tabular}{|c|}\hline 5 \\\hline\end{tabular} b) = -\infty \)
We have the general relation
\( \lim_{x\to -2n+2} \) b[2n]x = \( -\infty \).
in this case:
\( \lim_{x\to -4} \) b[6]x = \( -\infty \)
[b -> -00] lim (e[5]b) = - 1.85...
[b -> -1.85..] lim (penta-ln b) = - oo
The problem, for me, was again produced by the TeX animals. After the fifth slash, I fall asleep and I try to compensate using the copy-and-glue technology. Unfortunately, it was late and I was very tired. I just wanted to show that the tails of the sln and sexp (base e) plots cross themselves in a point "sigma", which has real, negative but not integer coordinates. That's all. But, this is the past.
@Henryk -
I agree on the Andydude overall scheme, concerning the odd/even hyperations. I just wished to draw the attention to the need of finding the right numerical values. It is also clear (... for me, I hope not to be wrong) that in the odd (3, 5, 7, ..) ranks y = b[2n+1]x have horizontal asymptotes (for x -> -oo), say for y = - k, and, therefore, their corresponding hyperlogs have vertical asymptotes for x -> - k (with y -> -oo). It is also clear that, for even ranks (2, 4, 6, ..) the contrary is true (horizontal asymptotes for the hyperlogs and vertical asymptotes for the hyperops). However, the problem remains of finding (exactly) such numbers. I think that we must completely solve the "tetration" business, before going further.
Probably, you are also right, even formula [x -> -2n+1]lim(b[2n]x) = - oo may be wrong.
@Ivars -
The role of the selfroot is essential. Nevertheless, it is important as a solution of the y = b[s]y functional equation. But, unfortunately (so to say) it is not its unique "functional root". Another solution can be found by considering y = b[s](b[s]y), or [/b]s-log(y) = y = b[s]y.
E.g. for rank 3 (exponentiation), we should consider (supposing base b) an expression such: [/b]log x = x = b^x. I. e. also other intersection points between the exp and the log, which will give other "branches" of the functional roots, are relevant. And, the same should be valid for other hyperops ranks.
GFR