11/19/2009, 10:28 PM
I just found a simple way to convert between a Taylor series at 0 and a Dirichlet series.
This requires a fractional differintegral.
For any function \( f(z) \) with \( f(0) = 0 \):
\( f(z) = \sum_{n=1}^{\infty} c_n z^n = \sum_{n=1}^{\infty} \frac{(\ln z)^n}{n!} g(-n) \)
\( g(s) = \sum_{n=1}^{\infty} \frac{c_n}{n^{s}} = \mathbb{D}^{-s}[f \circ \exp](0) \)
Pretty cool, huh?
Using this conversion, the Riemann Zeta function can be expressed as: \( \zeta(s) = \mathbb{D}_z^{-s}\left[\frac{1}{1 - e^z}\right](0) \)
which I believe is somewhat well known due to its connection with Bernoulli polynomials, and 1/(1 - e^z) is well known to be the generating function for Bernoulli polynomials. Still, I've never seen this formula before, it is interesting to see it like this.
This requires a fractional differintegral.
For any function \( f(z) \) with \( f(0) = 0 \):
\( f(z) = \sum_{n=1}^{\infty} c_n z^n = \sum_{n=1}^{\infty} \frac{(\ln z)^n}{n!} g(-n) \)
\( g(s) = \sum_{n=1}^{\infty} \frac{c_n}{n^{s}} = \mathbb{D}^{-s}[f \circ \exp](0) \)
Pretty cool, huh?
Using this conversion, the Riemann Zeta function can be expressed as: \( \zeta(s) = \mathbb{D}_z^{-s}\left[\frac{1}{1 - e^z}\right](0) \)
which I believe is somewhat well known due to its connection with Bernoulli polynomials, and 1/(1 - e^z) is well known to be the generating function for Bernoulli polynomials. Still, I've never seen this formula before, it is interesting to see it like this.
