"겔폰드-슈나이더 정리"의 두 판 사이의 차이

2009년 12월 18일 (금) 16:09 판

(정리) 겔폰드-슈나이더, 1934

\(\alpha \ne 0\),\(\alpha \ne 1\),\(\beta\notin \mathbb{Q}\) 인 복소수 \(\alpha\)와 \(\beta\) 가 대수적수이면, \(\alpha^{\beta} =e^{\beta \log \alpha\) 는 초월수이다.

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 <math>\alpha \ne 0</math>,<math>\alpha \ne 1</math>,<math>\beta\notin \mathbb{Q}</math> 인 복소수 <math>\alpha</math>와 <math>\beta</math> 가 대수적수이면, <math>\alpha^{\beta} =e^{\beta \log \alpha</math> 는 초월수이다.
-'''Comments'''
-* In general, <math>\alpha^{\beta} = \exp\{\beta \log \alpha\}</math> is [http://en.wikipedia.org/wiki/Multivalued_function multivalued], where "log" stands for the [http://en.wikipedia.org/wiki/Complex_logarithm complex logarithm]. This accounts for the phrase "any value of" in the theorem's statement.
-* An equivalent formulation of the theorem is the following: if<math>\alpha</math> and <math>\gamma</math> are nonzero algebraic numbers, and we take any non-zero logarithm of <math>\alpha</math>, then<math>(\log \gamma)/(\log \alpha)</math> is either rational or transcendental.
-* If the restriction that <math>\beta</math> be algebraic is removed, the statement does not remain true in general (choose <math>\alpha=3</math> and <math>\beta=\log 2/\log 3</math>, which is transcendental, then <math>\alpha^{\beta}=2</math> is algebraic). A characterization of the values for <math>\alpha</math> and <math>\beta</math> which yield a transcendental <math>\alpha^{\beta}</math> is not known.
-(wikipedia 의 [http://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem#Statement Gelfond–Schneider theorem 페이지]에서)