Thread Rating:
  • 0 Vote(s) - 0 Average
  • 1
  • 2
  • 3
  • 4
  • 5
Tetration and imaginary numbers.
#1
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.
Reply
#2
(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 (), which can be used to develop a superfunction for base i. When I used the attracting fixed point the result I got was, . The equation I used was , 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 ? If it is, than the result is
- Sheldon

Reply
#3
(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 (), which can be used to develop a superfunction for base i. When I used the attracting fixed point the result I got was, . The equation I used was , 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 ? If it is, than the result is
- 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 , and the ith square triangular number is -, 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?


Reply


Possibly Related Threads...
Thread Author Replies Views Last Post
  Spiral Numbers tommy1729 9 7,400 03/01/2016, 10:15 PM
Last Post: tommy1729
  Fractionally dimensioned numbers marraco 3 3,293 03/01/2016, 09:45 PM
Last Post: tommy1729
  A new set of numbers is necessary to extend tetration to real exponents. marraco 7 8,392 03/19/2015, 10:45 PM
Last Post: marraco
  Tommy's conjecture : every positive integer is the sum of at most 8 pentatope numbers tommy1729 0 1,968 08/17/2014, 09:01 PM
Last Post: tommy1729
  Number theoretic formula for hyper operators (-oo, 2] at prime numbers JmsNxn 2 3,740 07/17/2012, 02:12 AM
Last Post: JmsNxn
  A notation for really big numbers Tai Ferret 4 6,868 02/14/2012, 10:48 PM
Last Post: Tai Ferret
  The imaginary tetration unit? ssroot of -1 JmsNxn 2 5,071 07/15/2011, 05:12 PM
Last Post: JmsNxn
  Infinite tetration of the imaginary unit GFR 40 50,739 06/26/2011, 08:06 AM
Last Post: bo198214
  Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) Gottfried 91 84,618 03/03/2011, 03:16 PM
Last Post: Gottfried
  Iteration-exercises: article on Bell-numbers Gottfried 0 2,062 05/31/2008, 10:32 AM
Last Post: Gottfried



Users browsing this thread: 1 Guest(s)