07/13/2011, 01:03 PM
(This post was last modified: 07/13/2011, 02:51 PM by sheldonison.)
(07/12/2011, 03:22 PM)robo37 Wrote: Thanks for the help I got with my last question, now here's something else.
i^i = 0.207879576..., which is interesting, so I wounder if there is any way to find out what i^^i is? Furthermore, what is i sroot i, i itteratedroot i, and the ith exponential factorial? Thanks.
There is an attracting fixed point (\( \approx 0.438282936727032 + 0.360592471871385i \)), which can be used to develop a superfunction for base i. When I used the attracting fixed point the result I got was, \( ^i i \approx 0.500129061733810 + 0.324266941212720i \). The equation I used was \( \text{superf}(\text{superf}^{-1}(1)+i) \), where superf is developed from the attracting fixed point for base i. edit, I made a correction here
I forget how to figure out the nth sroot.... so you'll have to report back the results for your other questions. Is the "ith sroot" equation perhaps \( \text{superf}(\text{superf}^{-1}(1)-i) \)? If it is, than the result is \( ^{-i} i \approx -1.13983245176083 + 0.702048300301002i \)
- Sheldon
