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From: chiaki@kit.ac.jp (Tsukamoto Chiaki)
Newsgroups: fj.sci.math
Subject: Re: 不等式Σ_{n=1}^m |1/n^z|≦Σ_{n=1}^m |1/n^Re(z)|の証明
Date: Mon, 31 Jan 2011 11:31:56 GMT
Organization: Kyoto Institute of Technology
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工繊大の塚本と申します.

In article <3bdc4e67-b09a-4d3a-92db-553594285cdb@d16g2000yqd.googlegroups.com>
KyokoYoshida <kyokoyoshida123@gmail.com> writes:
> m,nを自然数,zを複素数とする時,次の不等式が成立する事を示せ。
> 
> Σ_{n=1}^m |1/n^z|≦Σ_{n=1}^m |1/n^Re(z)|
> 
> はどのようにして示せますでしょうか。

 |n^z|
  = |\exp((\log n) z)|
  = |\exp((\log n) Re(z) + i (\log n) Im(z))|
  = |\exp((\log n) Re(z)) \exp(i (\log n) Im(z))|
  = |\exp((\log n) Re(z))| |\exp(i (\log n) Im(z))|
  = |n^{Re(z)}| |\cos((\log n) Im(z)) + i \sin((\log n) Im(z))|
  = n^{Re(z)}

# \exp(i \theta) が絶対値 1 の複素数を表すことは宜しいか.

 \sum_{n=1}^m |1/n^z| = \sum_{n=1}^m 1/n^{Re(z)} となります.
-- 
塚本千秋@数理・自然部門.基盤科学系.京都工芸繊維大学
Tsukamoto, C. : chiaki@kit.ac.jp