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 Super-logarithm on the imaginary line andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 11/15/2007, 08:40 AM I just found a way to calculate slog on the imaginary axis! It depends very much on Jay's observation that slog is imaginary-periodic. Let $S(x) = \text{slog}_e(i x)$. S is periodic with period $2\pi$, because $S(x + 2\pi) = \text{slog}(i (x + 2\pi)) = \text{slog}(i x + 2i\pi) = \text{slog}(i x) = S(x)$. Since S is periodic, we can use Fourier series to represent it. Let $R(x) = S(-i\ln(x))$, then $R(e^{ix}) = S(x)$. The Taylor series coefficients of R will then be the Fourier series coefficients of S. In terms of the super-logarithm, $R(x) = S(-i\ln(x)) = \text{slog}(-ii\ln(x)) = \text{slog}(\ln(x)) = \text{slog}(x) - 1$. This means the Fourier series coefficients of $\text{slog}(ix)$ are the Taylor series coefficients of $\text{slog}(x) - 1$ which we already know. In other words, $\text{slog}(ix) = \text{slog}(e^{ix}) - 1$, so: $ \text{slog}_e(ix)_4 = -2 + \frac{12}{13}e^{ix} + \frac{16}{65}e^{2ix} - \frac{12}{65}e^{3ix} + \frac{1}{65}e^{4ix}$ The nice thing about this is that it seems to bypass the radius of convergence problem near z=i since its a Fourier series and not a Taylor series. Is this right? I've included a plot with multiple approximations, which seem to converge much faster than doing analytic continuation section-by-section. The top line is the imaginary part, and the bottom line is the real part, and the "y" axis is $\text{slog}(ix)$: PDF version Andrew Robbins « 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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