(10/10/2010, 08:02 AM)Ansus Wrote: So my advice is to avoid confusion by using continous sum but not indefinite continous sum, but a continous sum with fixed lower limit which is defined unambiguiously.
So, then we have
\( \sum_x f(x) \)
as definitely ambiguous and not well defined if \( f \) is not entire, even with Fourier sums, as powerful as they are, while
\( \sum_{n=n_0}^{x-1} f(n) \)
is not. The direction of the singularities is toward the right of a singularity in \( f \) if \( \Re(n_0) < \Re(S) \), where \( S \) is the singularity point, and to the left otherwise. And if \( \Re(n_0) = \Re(S) \), it goes in both directions, i.e. \( \frac{1}{2}\left(\psi(-x) + \psi(x+1)\right) \).
Thus, the laws
\( \sum_{n=a + \alpha}^{b + \alpha} f(n) = \sum_{n=a}^{b} f(n + \alpha) \)
and
\( F(b+1) - F(a) = \sum_{n=a}^{b} f(n) \),
where \( F(z) = \sum_{n=n_0}^{z-1} f(z) \) for some \( n_0 \), are only guaranteed to hold when all numbers \( a \), \( b \), and \( \alpha \) are integers, or \( f \) is entire.
Also, I made an interesting observation about the function
\( F(x) = \sum_{n=0}^{x-1} e^{-n^2} \),
a sort of sum-analogue of the error function. With the Fourier-based sum, it does not settle to a limit as \( x \rightarrow \infty \), but instead goes to a small 1-cyclic wobble about the limit of the discrete sum. Namely, \( F(1/2) = \frac{1}{2} \), and so
\( \lim_{n \rightarrow \infty} F\left(n + \frac{1}{2}\right) = \frac{1}{2} + \sum_{n=0}^{\infty} e^{-{\left(n + \frac{1}{2}\right)}^2} \ne \sum_{n=0}^{\infty} e^{-n^2} \).
The value 1/2 for the function at 1/2 can also be obtained from Faulhaber's formula when it is organized as a sum over the Bernoulli polynomials. I think the half-integers are the only other places than the integers where that formula converges.
The message here being, of course, that we can't necessarily expect continuum sums and discrete sums to behave in the same way, so the idea we may have to sacrifice some identities may not be a surprise.