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 Taylor-Dirichlet conversion andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 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. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 11/20/2009, 10:22 AM (This post was last modified: 11/20/2009, 10:30 AM by andydude.) Hmm. If the above is true, then $\sum_{n=0}^{\infty} \frac{z^n}{n!} \zeta(-n) = \frac{1}{1 - e^z}$ (which is false) but $\sum_{n=0}^{\infty} \frac{z^n}{n!} \zeta(-n) = \frac{e^z}{1 - e^z} + \frac{1}{z} = \frac{1 + e^z(z-1)}{z(1 - e^z)}$ is the actual result. So I must have made a slight mistake... I'll have to think on this. However, a similar form gives: $\sum_{n=-1}^{\infty} \frac{z^n}{n!} \zeta(-n) = \frac{e^z}{1 - e^z}$ because $\lim_{n \to -1} \frac{z^n\zeta(-n)}{n!} = -\frac{1}{z}$ « Next Oldest | Next Newest »

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