Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))

[Gottfried]

Hej Gotfried,

Congratulations! There are no accidents.

So how do you finally compute the \( k \)th branch of \( h \) or \( W \) with this formula?

Yes, how would You? And particularly, in case

h(e^pi/2) and h(e^-pi/2)?

Waiting impatiently even

Ivars

No chance.... ;-)

Yes, there is the coincidence; but I did not compute the W-function but the function, which gives real bases b (or, to avoid confusion with the parameter-notation in the article:bases s) for exp(u/t) where s=t^(1/t) is the principal branch, given u (or more precisely: given the imaginary part of u), and get real bases s>1 .

I set imag(u) = beta = any real value -pi < beta < pi
compute real(u) according to the above formula thus having u completed,
then compute t = exp(u)
and then compute s = exp(u/t), which is then surely real.

The formula for this is in the file fixpoints.pdf
In this article I've not yet included the extension of the range for beta>pi; but I've done some computations with this and found further fixpoints for the real s in the regions 2*k*pi< beta < 2*(k+1)*pi. However, there is no exact periodicity, the consecutive fixpoints for the same s approach the lower bound 2*k*pi with the index k.

I can find the inverse, to compute a fixpoint t by a given s, only by numerical approximation, since s=f(u) is monotonic, using a binary iterative process.
If I *had* the branch-enabled Lambert-W-function, this would be easier, but Pari/GP doesn't have it and I still have only the python-example from wikipedia for the real-valued region, and have not yet invested in programming it myself.

Gottfried