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 Super-logarithm on the imaginary line jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 11/15/2007, 09:08 AM Actually, if you think about it, we're still limited by the same radius of convergence, because it's the same series. HOWEVER, because we're plugging e^z into the series, rather than z, we're limited by abs(e^z) < 1.37445, not abs(z)<1.37445. This essentially means that for all complex values with real part less than 0.31813 (real part of primary fixed point), the series will converge. So if you plug e^(iz) into the series, and use only real z, then (because the imaginary part is 0, and hence the real part of iz is 0) you will always be inside the radius of convergence. Effectively, you're calculating points on the unit circle with the original series. ~ Jay Daniel Fox « Next Oldest | Next Newest »

 Messages In This Thread Super-logarithm on the imaginary line - by andydude - 11/15/2007, 08:40 AM RE: Super-logarithm on the imaginary line - by jaydfox - 11/15/2007, 09:08 AM RE: Super-logarithm on the imaginary line - by jaydfox - 11/15/2007, 09:25 AM RE: Super-logarithm on the imaginary line - by andydude - 11/15/2007, 05:52 PM

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