Tetration and imaginary numbers. - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Tetration and imaginary numbers. (/showthread.php?tid=676) Tetration and imaginary numbers. - robo37 - 07/12/2011 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. RE: Tetration and imaginary numbers. - sheldonison - 07/13/2011 (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 RE: Tetration and imaginary numbers. - robo37 - 07/13/2011 (07/13/2011, 01:03 PM)sheldonison Wrote: (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, $^i i \approx 0.424801328697548 + 0.424973603314731i$. The equation I used was $\text{superf}(\text{superf}^{-1}(i)+i)$, where superf is developed from the attracting fixed point for base i. 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 $\text{superf}(\text{superf}^{-1}(i)-i)$? If it is, than the result is $\approx -0.0723270995404099 - 0.323973330391954i$ - Sheldon Wow, thanks for that. It's interesting that the imaginary part is almost as big as the real part with the first resault, but I'm sure that's just coincidence. I'm rather interested with the imaginary and complex plain; I've already found out, with a little help from Google Calculator, that $i root i = 4.81047738$ $i! = 0.498015668 - 0.154949828 i$, $F (i) = 0.379294534 + 0.215939518 i$ and the ith square triangular number is -$0.120450647 - 1.87314977*10[-16i]$, at the moment I'm on the ith partition number, but I'm having difficulty as there seems to be no closed finite function to use. I don't suppose anyone could herlp me out here? $p(i) = \frac1i\cdot\sum_{k=0}^{i-1}\sigma(i-k)\cdot p(k) = ?$ $p(i) = 1+\sum_{k=1}^{\lfloor \frac{1}{2}i \rfloor} p(k,i-k) = ?$