if $0\le b<1$, then $a↑↑b = a^b$

if $b\ge1$, then $a↑↑b = a^{a↑↑(b-1)}$

if $b<0$, then $a↑↑b = \log_a(a↑↑(b+1))$

so for example,

$2↑↑\pi = 2^{2^{2^{2^{0.1415926...}}}} = 21.5963561$

How can I extend this to complex numbers?


enter image description here


closed as unclear what you're asking by user8795, Shailesh, Arnaldo, JonMark Perry, mlc May 19 '17 at 3:29

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ I suppose you could do it that way, but does it have nice continuity/analyticity characteristics? $\endgroup$ – Oscar Lanzi May 19 '17 at 0:31
  • $\begingroup$ I just want to ask how to extend this, if you could tell me what you are unsure what is being asked that would be appreciated. $\endgroup$ – Theoretical May 19 '17 at 15:24
  • $\begingroup$ Are these edits acceptable? $\endgroup$ – Theoretical May 19 '17 at 16:29
  • $\begingroup$ Presuming your question is focused on $a↑↑z$ where $a$ is real and $z$ is complex. I think this is interesting and will try applying this to an old question of mine concerning the nature of $i↑↑n$. $\endgroup$ – Simply Beautiful Art May 19 '17 at 23:19
  • $\begingroup$ Here's the graph I made of $i↑↑x$ for $-2<x<9$. $\endgroup$ – Simply Beautiful Art May 19 '17 at 23:52