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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 Malgrange preparation theorem ： ウィキペディア英語版 Malgrange preparation theorem In mathematics, the Malgrange preparation theorem is an analogue of the Weierstrass preparation theorem for smooth functions. It was conjectured by René Thom and proved by . ==Statement of Malgrange preparation theorem== Suppose that ''f''(''t'',''x'') is a smooth complex function of ''t''∈R and ''x''∈R''n'' near the origin, and let ''k'' be the smallest integer such that :$f(0,0)=0,\; (0,0)=0,\; \backslash dots,\; \}(0,0)=0,\; \}(0,0)\backslash ne0.$ Then one form of the preparation theorem states that near the origin ''f'' can be written as the product of a smooth function ''c'' that is nonzero at the origin and a smooth function that as a function of ''t'' is a polynomial of degree ''k''. In other words, :$f(t,x)\; =\; c(t,x)\backslash left(t^k+a\_(x)t^+\backslash cdots+a\_0(x)\; \backslash right)$ where the functions ''c'' and ''a'' are smooth and ''c'' is nonzero at the origin. A second form of the theorem, occasionally called the Mather division theorem, is a sort of "division with remainder" theorem: it says that if ''f'' and ''k'' satisfy the conditions above and ''g'' is a smooth function near the origin, then we can write :$g=qf+r\backslash$ where ''q'' and ''r'' are smooth, and as a function of ''t'', ''r'' is a polynomial of degree less than ''k''. This means that :$r(x)=\backslash sum\_t^jr\_j(x)$ for some smooth functions ''r''''j''(''x''). The two forms of the theorem easily imply each other: the first form is the special case of the "division with remainder" form where ''g'' is ''t''''k'', and the division with remainder form follows from the first form of the theorem as we may assume that ''f'' as a function of ''t'' is a polynomial of degree ''k''. If the functions ''f'' and ''g'' are real, then the functions ''c'', ''a'', ''q'', and ''r'' can also be taken to be real. In the case of the Weierstrass preparation theorem these functions are uniquely determined by ''f'' and ''g'', but uniqueness no longer holds for the Malgrange preparation theorem. 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「Malgrange preparation theorem」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.