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Super-logarithm on the imaginary line

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Super-logarithm on the imaginary line
andydude Offline
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#4
11/15/2007, 05:52 PM
Right, since the real part of all points in the "backbone" of the slog are less than the logarithm of the radius of convergence, the exponential of them is within the radius of convergence (of the series expansion about z=0).

Andrew Robbins
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RE: Super-logarithm on the imaginary line - by andydude - 11/15/2007, 05:52 PM

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