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 Regular slog for base sqrt(2) - Using z=2 andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 11/25/2007, 01:15 AM Two comments. @Henryk The infinitely iterated exponential with branch indecies is probably the best way: $H_k(x) = {}^{\infty}(x)_k = \frac{W_k(-\log(x))}{-\log(x)} = e^{-W_k(-\log(x))}$ with the value substituted for x as in $H_k(\sqrt{2})$. Using the Lambert W function branches, this gives $H_{-1}(\sqrt{2}) = 4$ and $H_{0}(\sqrt{2}) = 2$ which are exactly those that you would expect. @Jay Why is it 18-periodic? I understand it would be periodic in the imaginary direction just like base-e, but why 18? Andrew Robbins « Next Oldest | Next Newest »

 Messages In This Thread Regular slog for base sqrt(2) - Using z=2 - by jaydfox - 11/15/2007, 10:20 AM RE: Regular slog for base sqrt(2) - Using z=2 - by jaydfox - 11/15/2007, 10:37 AM RE: Regular slog for base sqrt(2) - Using z=2 - by jaydfox - 11/15/2007, 10:53 AM RE: Regular slog for base sqrt(2) - Using z=2 - by jaydfox - 11/15/2007, 11:06 AM RE: Regular slog for base sqrt(2) - Using z=2 - by jaydfox - 11/15/2007, 11:39 AM RE: Regular slog for base sqrt(2) - Using z=2 - by jaydfox - 11/20/2007, 05:15 AM RE: Regular slog for base sqrt(2) - Using z=2 - by jaydfox - 11/20/2007, 05:20 AM RE: Regular slog for base sqrt(2) - Using z=2 - by bo198214 - 11/23/2007, 09:27 AM RE: Regular slog for base sqrt(2) - Using z=2 - by andydude - 11/25/2007, 01:15 AM RE: Regular slog for base sqrt(2) - Using z=2 - by jaydfox - 11/26/2007, 07:09 PM RE: Regular slog for base sqrt(2) - Using z=2 - by jaydfox - 11/25/2007, 03:18 AM RE: Regular slog for base sqrt(2) - Using z=2 - by andydude - 11/27/2007, 07:10 PM RE: Regular slog for base sqrt(2) - Using z=2 - by Gottfried - 03/10/2010, 12:47 PM RE: Regular slog for base sqrt(2) - Using z=2 - by Ivars - 11/25/2007, 09:05 AM

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