(04/20/2010, 10:40 AM)bo198214 Wrote: \( \sigma(z)=\sum_{n=0}^{\infty}\sum_{k=-\infty}^{\infty} \eta_n \rho^n_k e^{(\kappa n + 2\pi i k) z}=\sum_{n=0}^{\infty}\sum_{k=-\infty}^{\infty} \sigma_{n,k} e^{(\kappa n + 2\pi i k) z} \)
So we have a double exponential series instead of a single series, but nevertheless you again can apply your exponential summation. Though I in the moment have not the time to carry it out myself (so either you do it or I do it later).
Eh. I'm not sure if this method is going to work. Take what happens when \( n = 0 \). Then we have coefficients multiplying \( e^{2\pi i k z} \) for integer \( k \). But this does not continuum-sum under the given method: we get \( \frac{e^{2\pi i k z} - 1}{e^{2\pi i k} - 1} \), but the denominator \( e^{2\pi i k} - 1 \) is 0 when \( k \in \mathbf{Z} \). This gives a division by zero. There is no joy trying to use a limit (of, say, \( e^{az} \) as \( a \) approaches \( 2\pi i k \)) -- this singularity explodes to infinity.
Yet it seems we can continuum-sum the tetrational regardless, by using the periodic-approximation method.) What's going on here?
(And as an aside, what did you think of the graph of the tetration of a complex base outside the Shell-Thron Region?)