RamaWolf
- 95
- 2
(1) Let s be a complex number like s = a + b i, then we define \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}
Our aim:
to compute ζ(\frac{1}{2}+14.1347 i) with the help of the programming language Aribas
(2) Web Links
Aribas: http://www.mathematik.uni-muenchen.de/~forster/sw/aribas.html
Dirichlet eta: http://en.wikipedia.org/wiki/Dirichlet_eta_function
Euler transformation of alterning series: http://en.wikipedia.org/wiki/Euler_transform
(3) Books: Knopp: Theorie und Anwendung der unendlichen Reihen; Henrici: Applied
and Computational Complex Analsis; Derbyshire: Prime Obsession
(4) Basics: The Riemann zeta in (1) is defined only for s.Re > 1 (i.e. the real part of s);
we use the 'alternating zeta' or Dirichlet eta defined as
\eta(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s}
defined for s.Re > 0
The Riemann zeta and the Dirichlet eta are related through:
ζ(s) := η(s) / (1-2^{1-s})
With the Euler transformation of the Dirichlet eta, the Riemann zeta looks like:
\zeta(s)=\frac{1}{1-2^{1-s}}<br /> \sum_{n=0}^\infty \frac {1}{2^{n+1}}<br /> \sum_{k=0}^n (-1)^k {n \choose k} (k+1)^{-s}
To be continued with the programming details
Our aim:
to compute ζ(\frac{1}{2}+14.1347 i) with the help of the programming language Aribas
(2) Web Links
Aribas: http://www.mathematik.uni-muenchen.de/~forster/sw/aribas.html
Dirichlet eta: http://en.wikipedia.org/wiki/Dirichlet_eta_function
Euler transformation of alterning series: http://en.wikipedia.org/wiki/Euler_transform
(3) Books: Knopp: Theorie und Anwendung der unendlichen Reihen; Henrici: Applied
and Computational Complex Analsis; Derbyshire: Prime Obsession
(4) Basics: The Riemann zeta in (1) is defined only for s.Re > 1 (i.e. the real part of s);
we use the 'alternating zeta' or Dirichlet eta defined as
\eta(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s}
defined for s.Re > 0
The Riemann zeta and the Dirichlet eta are related through:
ζ(s) := η(s) / (1-2^{1-s})
With the Euler transformation of the Dirichlet eta, the Riemann zeta looks like:
\zeta(s)=\frac{1}{1-2^{1-s}}<br /> \sum_{n=0}^\infty \frac {1}{2^{n+1}}<br /> \sum_{k=0}^n (-1)^k {n \choose k} (k+1)^{-s}
To be continued with the programming details