11/12/2007, 08:23 PM
Ivars Wrote:So we have just limit log b log b log b log b ... log b (+-i*2pik) to be concerned with for branches in numerator. Right?
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So we have lim k, n-> infintylog b log b log b ...... log b ( +-i*2pik/(pi/2+- i*2pi/k))
So the simplest starting value is then +-i*2pik/(pi/2+-i*2pik)?
Are these k in numerator and denominator the same or running independently ?
I am not sure what you mean. The method is about repeated applying of \( \log_{k,b} \) and not of \( \log_b \). You can not pull out the \( 2\pi i k \).
