Zeta function, product of summations
- To: mathgroup at smc.vnet.net
- Subject: [mg43440] Zeta function, product of summations
- From: "Diana" <diana53xiii at earthlink.remove13.net>
- Date: Wed, 17 Sep 2003 07:58:56 -0400 (EDT)
- Reply-to: "Diana" <diana53xiii at earthlink.remove13.net>
- Sender: owner-wri-mathgroup at wolfram.com
Hi math folks,
I am trying to show that the inverse of the Zeta(s) function is "the
infinite summation as n goes from 1 to infinity" of MobiusMu(n)/(n^s).
The Zeta(s) function is defined as "the infinite summation as n goes from 1
to infinity" of 1/(n^s).
To do this, I am first calculating the product of two generic summations:
{"the infinite summation as n goes from 1 to infinity" of a_n/(n^s)} times
{"the infinite summation as n goes from 1 to infinity" of b_n/(n^s)}.
What I believe is the answer is:
{"the infinite summation as n goes from 1 to infinity" of the summation of
divisors "d" of n of [a_d * (b_(n/d))]/(n^s)}.
Using this formula, I am trying to show that:
{"the infinite summation as n goes from 1 to infinity" of 1/(n^s) of the
summation of divisors "d" of n of MobiusMu(n/d)/(n^s)}
is one.
I am trying to test my formulas on Mathematica, and understand how to
program regular summations. But, I am unsure how to code the summation of
divisors "d" of n with Mathematica.
I hope that my text is understandable, without math fonts.
Can someone help with the proof/Mathematica code?
Thanks in advance,
Diana M.
--
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"God made the integers, all else is the work of man."
L. Kronecker, Jahresber. DMV 2, S. 19.