Riemann zeta function: Difference between revisions
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The zeta function has a [http://mathworld.wolfram.com/SimplePole.html simple pole] at {{nowrap|z {{=}} 1}} which forms a barrier against continuing it with its [[Wikipedia:Euler product|Euler product]] or [[Wikipedia:Dirichlet series|Dirichlet series]] representation. We could subtract off the pole, or multiply by a factor of {{nowrap|(z − 1)}}, but at the expense of losing the character of a Dirichlet series or Euler product. A better method is to multiply by a factor of {{nowrap|(1 − 2^(1 − z))}}, leading to the eta function: | The zeta function has a [http://mathworld.wolfram.com/SimplePole.html simple pole] at {{nowrap|z {{=}} 1}} which forms a barrier against continuing it with its [[Wikipedia:Euler product|Euler product]] or [[Wikipedia:Dirichlet series|Dirichlet series]] representation. We could subtract off the pole, or multiply by a factor of {{nowrap|(z − 1)}}, but at the expense of losing the character of a Dirichlet series or Euler product. A better method is to multiply by a factor of {{nowrap|(1 − 2^(1 − z))}}, leading to the eta function: | ||
<math>\displaystyle{\eta(z) = (1-2^{1-z})\zeta(z) = \sum_{n=1}^\infty (-1)^{n-1} n^{-z} | <math>\displaystyle{\eta(z) = \left(1-2^{1-z}\right)\zeta(z) = \sum_{n=1}^\infty (-1)^{n-1} n^{-z} | ||
= \frac{1}{1^z} - \frac{1}{2^z} + \frac{1}{3^z} - \frac{1}{4^z} + \cdots}</math> | = \frac{1}{1^z} - \frac{1}{2^z} + \frac{1}{3^z} - \frac{1}{4^z} + \cdots}</math> | ||