Ivars Wrote:PLEASE post at least one numerical example how to use this limit formula e.f for k=-2, h(e^(pi/2)).
I am not familiar (entirely my fault) with the notations You use - what is n, how to take the limit, what is z, what is b (I assume b=e^(pi/2) in my case.
Yes the notation, mainly you have to know what \( f^{\circ n}(x) \) means, it is simply the \( n \) times repetition of \( f \) applied to \( x \), for example \( f^{\circ 3}(x)=f(f(f(x))) \).
In our example is then \( b=e^{\frac{\pi}{2}} \). Because \( k=-2 \) we compute \( h_{-2}(b)=\overline{h_{2-1}(b) \) which means the complex conjugate of \( h_1(b) \). To compute this we consider \( \log_{1,b}(z) = \frac{\log(z)+2\pi i}{\pi/2}=\frac{2}{\pi}\log(z)+4i \) and then compute \( h_1(b)=\lim_{n\to\infty} \log_{1,b}(z)^{\circ n}(-1) \). This can of course only approximately computed. Have a look at some sample values:
\( \log_{1,b}(-1)=6*I \)
\( \log_{1,b}^{\circ 2}(-1)=\log_{1,b}(\log_{1,b}(-1))=1.140669505+5.000000000*I \)
\( \log_{1,b}^{\circ 10}(-1)=1.021323305+4.868353812*I \)
\( \log_{1,b}^{\circ 100}(-1)=1.021323316+4.868353806*I \)
But you can see that the value converges, \( h_1(b)\approx 1.021323+4.8683538*I \) hence \( h_{-2}(b)\approx 1.021323-4.8683538*I \).
This is indeed the same value, you get when computing
\( \text{LambertW}(-2,-\pi/2)/(-\pi/2) \) with Maple.