Infinite powering by $i$ Find the value of:
$i^{i^{i^{i^{i^{i^{....\infty}}}}}}$
Simply infinite powering by i's and the limiting value.
Thank you for the help.
 A: Another way is to take natural logarithms: 
$$i^{i^{i^{i^{i^{i^{\dots \infty}}}}}}=y$$
$$\ln y= \ln (i)^y$$
$$y\ln i=\ln y$$
$$\ln i=\dfrac{i \pi}{2}$$
$$\dfrac{y.i\pi}{2}= \ln y$$
$$e^{\frac{iy\pi}{2}}=y$$
A: Hint:let $ x = {i^{i^{i^{.^{.^{\infty}}}}}}$
hence $x = {i^{x}}$
$\ln x = x\ln(i)$
$\frac{\ln x}{x} = \ln(0+i)$
A: Here is the Maple version of the graph from sos440...

A: Let us denote $x=i^{i^{i^{i^\cdots}}}$. Then we have $$i^x=x.$$ It looks like the solution is $x= \frac{2i}{\pi} W(-i\pi/2)$ with $W$ Lambert's $W$ function. Now, $W$ is multivalued. You have to figure out which of the different branches $x$ converges to (and if it converges at all). Numerically, you find (using the principal branch of the logarithm to define the exponentiation) that $x= 0.438283 + 0.360592 i$ which corresponds to the principal branch.
Knowing that you should be able to prove the result by some kind of fixed point theorem.
A: Here is a numerical result supporting Fabian's argument.
Here, the complex logarithm
$$ z^{w} := \exp (w \operatorname{Log} z) $$
is defined via the principal value $\mathrm{Log}$ of the logarithm, defined on $\Bbb{C} \setminus (-\infty, 0]$.

A: In clisp:
(loop for i upfrom 1 
      and prev = 0 then x 
      and x = #c(0L0 1L0) then (expt #c(0 1) x) 
      while (< long-float-epsilon (abs (- x prev))) 
      finally (return (values x i)))
#C(0.43828293672703211162L0 0.36059247187138548596L0) ;
393

so, in fewer than 400 iterations you get 20 correct decimal digits.
(The convergence is quadratic).
