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 regular iteration of sqrt(2)^x (was: eta as branchpoint of tetrational) JmsNxn Long Time Fellow Posts: 739 Threads: 104 Joined: Dec 2010 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. bo198214 Administrator Posts: 1,412 Threads: 91 Joined: Aug 2007 06/09/2011, 07:16 PM (This post was last modified: 06/09/2011, 07:16 PM by bo198214.) (06/09/2011, 06:20 PM)JmsNxn Wrote: 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: Here is some introduction into the topic. sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 06/09/2011, 07:49 PM (This post was last modified: 06/09/2011, 09:17 PM by sheldonison.) (06/09/2011, 06:20 PM)JmsNxn Wrote: @ 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: .... Therefore how do we generate $\exp_{\sqrt{2}}^{\circ \sigma}(z)\,\,;\,\,\R(z) \in (2, 4)$? Do we create a middle super function?Hey James, $\exp_{\sqrt{2}}^{\circ \sigma}(z)=\text{Usexp_{\sqrt{2}}(\text{Uslog_{\sqrt{2}}(z)+\sigma)$ The short answer, is that these functions are imaginary periodic. Usexp(z) has a period of approximately 19.236i. USexp is real valued at the real axis going from 4+delta to infinity. But at exactly half that period, the Usexp(z) function is also real valued from -infinity to infinity, gently making a transition from 4-delta to 2+delta. Lsexp(z) has a period of about 17.143i. And at exactly half that period, the Lsexp(z) function is real valued from -infinity to infinity, also gently making a transition from 4-delta to 2+delta. In going from 4-delta to 2+delta, these two functions can be lined up, so that they are nearly identical, but they differ by a tiny amount! So, for $z \in (2, 4)\,\; \Im(\text{Uslog_{\sqrt{2}}(z))\approx9.62i$. Hope that helps. - Sheldon Here are some more links (to go on wiki page?) http://math.eretrandre.org/tetrationforu...96#pid3296 And another really good thread. http://math.eretrandre.org/tetrationforu...534#pid534 JmsNxn Long Time Fellow Posts: 739 Threads: 104 Joined: Dec 2010 06/09/2011, 11:04 PM Thanks sheldon that was really helpful. Is there any code yet generating these two functions? And my second question still stands, though; why exactly does $\text{Usexp}_{\sqrt{2}}(0) = 5.767053...$, is this point arbitrary? If it was shifted to 8 it wouldn't affect its status as a super function of root 2 would it? It's just a horizontal shift right? sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 06/09/2011, 11:45 PM (This post was last modified: 06/09/2011, 11:52 PM by sheldonison.) (06/09/2011, 11:04 PM)JmsNxn Wrote: Thanks sheldon that was really helpful. Is there any code yet generating these two functions? And my second question still stands, though; why exactly does $\text{Usexp}_{\sqrt{2}}(0) = 5.767053...$, is this point arbitrary? If it was shifted to 8 it wouldn't affect its status as a super function of root 2 would it? It's just a horizontal shift right? Horizontal shift -- yes and no. It is what comes out of the limit equation, that generates Usexp(0), limit as n->infinity. I'll need to dig that equation out. But even if you do a horizontal shift, it gets cancelled out since Uslog is the inverse of Usexp. I do have the lower level primitives, "superf(z)" and a "isuperf(z)" functions in kneser.gp. Type init(sqrt(2)), and those two functions are available. For bases

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