(10/05/2010, 11:40 AM)mike3 Wrote: I mean it's not something weird like \( \psi(z + 1) + e^e \sin(2\pi z) + e^{\sin(4\pi z)} \) or something contrived and wobbly and all that (yes, I know these aren't rigorous terms.). Instead, it's just:
\( \sum_z \frac{1}{z + 1} = \psi(-z) \).
And note also that
\( \psi(-z)=\psi(z + 1)+\pi \cot(\pi z) \)
so the both solutions only differ by a periodic function (which can be filtered out on each continous range separately in the process of finding the natural value by requiring monotonous derivatives of higher order for example).

