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+--- Thread: Taylor-Dirichlet conversion (/showthread.php?tid=392)
Taylor-Dirichlet conversion - andydude - 11/19/2009
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.
RE: Taylor-Dirichlet conversion - andydude - 11/20/2009
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 \( \lim_{n \to -1} \frac{z^n\zeta(-n)}{n!} = -\frac{1}{z} \)