06/09/2011, 06:20 PM
@ Sheldon's root 2 postings:
Hmm, this is all very interesting. I myself am interested in \( \text{sexp}_{\sqrt{2}}(z) \); since I think this may be the natural base semi-operators work in. Therefore I have a few questions:
I understand \( \text{Lsexp}_{\sqrt{2}}(\text{Lsexp}_{\sqrt{2}}^{-1}(z) + \sigma) = \exp_{\sqrt{2}}^{\circ \sigma}(z)\,\,;\,\,\R(z) < 2 \)
and
\( \text{Usexp}_{\sqrt{2}}(\text{Usexp}_{\sqrt{2}}^{-1}(z) + \sigma) = \exp_{\sqrt{2}}^{\circ \sigma}(z)\,\,;\,\,\R(z) > 4 \)
Therefore how do we generate \( \exp_{\sqrt{2}}^{\circ \sigma}(z)\,\,;\,\,\R(z) \in (2, 4) \)?
Do we create a middle super function?
Secondly, why doesn't \( \text{Usexp}_{\sqrt{2}}(0) = 8 \)? Is there a reason it isn't like this? because if it was centered at 8 it would give the beautiful result:
\( \text{Usexp}_{\sqrt{2}}(\sigma) = 4\,\,\bigtriangleup_{\sigma}^{\small{\sqrt{2}}}\,\,4\,\,:\,\,\R(\sigma) \le 2 \)
It would also be consistent with the cheta function, where it's centered at 2 times the fix point.
Hmm, this is all very interesting. I myself am interested in \( \text{sexp}_{\sqrt{2}}(z) \); since I think this may be the natural base semi-operators work in. Therefore I have a few questions:
I understand \( \text{Lsexp}_{\sqrt{2}}(\text{Lsexp}_{\sqrt{2}}^{-1}(z) + \sigma) = \exp_{\sqrt{2}}^{\circ \sigma}(z)\,\,;\,\,\R(z) < 2 \)
and
\( \text{Usexp}_{\sqrt{2}}(\text{Usexp}_{\sqrt{2}}^{-1}(z) + \sigma) = \exp_{\sqrt{2}}^{\circ \sigma}(z)\,\,;\,\,\R(z) > 4 \)
Therefore how do we generate \( \exp_{\sqrt{2}}^{\circ \sigma}(z)\,\,;\,\,\R(z) \in (2, 4) \)?
Do we create a middle super function?
Secondly, why doesn't \( \text{Usexp}_{\sqrt{2}}(0) = 8 \)? Is there a reason it isn't like this? because if it was centered at 8 it would give the beautiful result:
\( \text{Usexp}_{\sqrt{2}}(\sigma) = 4\,\,\bigtriangleup_{\sigma}^{\small{\sqrt{2}}}\,\,4\,\,:\,\,\R(\sigma) \le 2 \)
It would also be consistent with the cheta function, where it's centered at 2 times the fix point.