2
$\begingroup$

for the complex function $f_{(2)}(z)=z^z$,where in the complex plane does the inverse $z_{(2)}(f)$ not exist, same for inverse of functions $f_{(3)}(z)=z^{z^z}$ being $z_{(3)}(f)$ and so on

for instance $z_{(2)}(e^{-\pi /2})=i$ since $i^i=e^{-\pi /2}$

what is $z_{(2)}(i)$

what is $z_{(3)}(e^{-\pi /2})$

what is $z_{(2)}(1+i)$

$\endgroup$
  • 1
    $\begingroup$ Don't we have the Lambert-W-function which under the assumption of "principal values" of the logarithm gives an answer for $z^z =a $ to find out $z$ -even if $a$ and/or $z$ is complex? So I seem not really to understand what your problem is here? Also I have done a little essay on "super-roots" (a term that I don't really like) of higher orders, getting powerseries-solutions for them, for instance what you call $z_{(3)}()$ Perhaps you find here in MSE something when searching for "superroots", or in wikipedia with the same term. For instance math.stackexchange.com/q/3314712 $\endgroup$ – Gottfried Helms Oct 12 at 7:34
1
$\begingroup$

at "$z_{(2)}(i)$" and "$z_{(2)}(1+i)$" : (Pari/GP)

\\ input to Pari/GP         \\   output from Pari/GP
\\ -------------------------\\-----------------------------
t=exp(LambertW(log(I)))     \\ %666 = 1.36062 + 1.11944*I
t^t                         \\ %667 =           1.00000*I

t=exp(LambertW(log(1+I)))   \\ %668 = 1.39402 + 0.577732*I
t^t                         \\ %669 = 1.00000 + 1.00000*I                

For $t = z_{(3)}(x)$ I used my series-representation. The value $x=\exp(-\pi/2)$ seems too much out of radius of convergence, so I could not yet obtain a result. But for $x=\exp(\pi/2) \approx 4.81048$ I could obtain $t \approx 1.77053$ and $t^{t^t} \approx 4.81048 $ with error of about $-0.0000000045...$. Here I had to use a procedure for summing divergent series (because the found powerseries at this argument diverges strongly), which is an adaption of the Noerlund-summation with a manually adapted "order" for the summation, but likely the better known Borel-summation should have been possible too.

$\endgroup$
  • $\begingroup$ thank you ... are there closed form expressions for these $\endgroup$ – phdmba7of12 Oct 13 at 14:38
  • 1
    $\begingroup$ @phdmba7of12 - no as far as I know, no more "closed" than the LambertW is "closed" form... $\endgroup$ – Gottfried Helms Oct 13 at 17:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.