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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク removable singularity ： ウィキペディア英語版 removable singularity In complex analysis, a removable singularity of a holomorphic function is a point at which the function is undefined, but it is possible to define the function at that point in such a way that the function is regular in a neighbourhood of that point. For instance, the (unnormalized) sinc function :$\backslash text(z)\; =\; \backslash frac$ has a singularity at ''z'' = 0. This singularity can be removed by defining ''f''(0) := 1, which is the limit of ''f'' as ''z'' tends to 0. The resulting function is holomorphic. In this case the problem was caused by ''f'' being given an indeterminate form. Taking a power series expansion for $\backslash frac$ shows that :$\backslash text(z)\; =\; \backslash frac\backslash left(\backslash sum\_^\; \backslash frac\; \backslash right)\; =\; \backslash sum\_^\; \backslash frac\; =\; 1\; -\; \backslash frac\; +\; \backslash frac\; -\; \backslash frac\; +\; \backslash cdots.$ Formally, if $U\; \backslash subset\; \backslash mathbb\; C$ is an open subset of the complex plane $\backslash mathbb\; C$, $a\; \backslash in\; U$ a point of $U$, and $f:\; U\backslash setminus\; \backslash \; \backslash rightarrow\; \backslash mathbb\; C$ is a holomorphic function, then $a$ is called a removable singularity for $f$ if there exists a holomorphic function $g:\; U\; \backslash rightarrow\; \backslash mathbb\; C$ which coincides with $f$ on $U\backslash setminus\; \backslash$. We say $f$ is holomorphically extendable over $U$ if such a $g$ exists. == Riemann's theorem == Riemann's theorem on removable singularities states when a singularity is removable: Theorem. Let $D\; \backslash subset\; C$ be an open subset of the complex plane, $a\; \backslash in\; D$ a point of $D$ and $f$ a holomorphic function defined on the set $D\; \backslash setminus\; \backslash$. The following are equivalent: # $f$ is holomorphically extendable over $a$. # $f$ is continuously extendable over $a$. # There exists a neighborhood of $a$ on which $f$ is bounded. # $\backslash lim\_(z\; -\; a)\; f(z)\; =\; 0$. The implications 1 ⇒ 2 ⇒ 3 ⇒ 4 are trivial. To prove 4 ⇒ 1, we first recall that the holomorphy of a function at $a$ is equivalent to it being analytic at $a$ (proof), i.e. having a power series representation. Define :$$ h(z) = \begin (z - a)^2 f(z) & z \ne a ,\\ 0 & z = a . \end Clearly, ''h'' is holomorphic on ''D'' \ , and there exists :$h\text{'}(a)=\backslash lim\_\backslash frac=\backslash lim\_(z\; -\; a)\; f(z)=0$ by 4, hence ''h'' is holomorphic on ''D'' and has a Taylor series about ''a'': :$h(z)\; =\; c\_0\; +\; c\_1(z-a)\; +\; c\_2\; (z\; -\; a)^2\; +\; c\_3\; (z\; -\; a)^3\; +\; \backslash cdots\; \backslash ,\; .$ We have ''c''0 = ''h''(''a'') = 0 and ''c''1 = ''h''(''a'') = 0; therefore :$h(z)\; =\; c\_2\; (z\; -\; a)^2\; +\; c\_3\; (z\; -\; a)^3\; +\; \backslash cdots\; \backslash ,\; .$ Hence, where z≠a, we have: :$f(z)\; =\; \backslash frac\; =\; c\_2\; +\; c\_3\; (z\; -\; a)\; +\; \backslash cdots\; \backslash ,\; .$ However, :$g(z)\; =\; c\_2\; +\; c\_3\; (z\; -\; a)\; +\; \backslash cdots\; \backslash ,\; .$ is holomorphic on ''D'', thus an extension of ''f''. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「removable singularity」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.