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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 Trigonometric moment problem ： ウィキペディア英語版 Trigonometric moment problem In mathematics, the trigonometric moment problem is formulated as follows: given a finite sequence , does there exist a positive Borel measure ''μ'' on the interval (2''π'' ) such that :$\backslash alpha\_k\; =\; \backslash frac\backslash int\_0\; ^\; e^\backslash ,d\; \backslash mu(t).$ In other words, an affirmative answer to the problems means that are the first ''n'' + 1 ''Fourier coefficients'' of some positive Borel measure ''μ'' on (2''π'' ). == Characterization == The trigonometric moment problem is solvable, that is, is a sequence of Fourier coefficients, if and only if the (''n'' + 1) × (''n'' + 1) Toeplitz matrix :$$ A = \left(\begin \alpha_0 & \alpha_1 & \cdots & \alpha_n \\ \bar & \alpha_0 & \cdots & \alpha_ \\ \vdots & \vdots & \ddots & \vdots \\ \bar & \bar\right) is positive semidefinite. The "only if" part of the claims can be verified by a direct calculation. We sketch an argument for the converse. The positive semidefinite matrix ''A'' defines a sesquilinear product on C''n'' + 1, resulting in a Hilbert space :$(\backslash mathcal,\; \backslash langle\; \backslash ;,\backslash ;\; \backslash rangle)$ of dimensional at most ''n'' + 1, a typical element of which is an equivalence class denoted by (). The Toeplitz structure of ''A'' means that a "truncated" shift is a partial isometry on $\backslash mathcal$. More specifically, let be the standard basis of C''n'' + 1. Let $\backslash mathcal$ be the subspace generated by and $\backslash mathcal$ be the subspace generated by . Define an operator :$V:\; \backslash mathcal\; \backslash rightarrow\; \backslash mathcal$ by :$V()\; =\; ()\; \backslash quad\; \backslash mbox\; \backslash quad\; k\; =\; 0\; \backslash ldots\; n-1.$ Since :$\backslash langle\; V(),\; V()\; \backslash rangle\; =\; \backslash langle\; (),\; ()\; \backslash rangle\; =\; A\_\; =\; A\_\; =\; \backslash langle\; (),\; ()\; \backslash rangle,$ ''V'' can be extended to a partial isometry acting on all of $\backslash mathcal$. Take a minimal unitary extension ''U'' of ''V'', on a possibly larger space (this always exists). According to the spectral theorem, there exists a Borel measure ''m'' on the unit circle T such that for all integer ''k'' :$\backslash langle\; (U^$ *)^k (e_ ), (e_ ) \rangle = \int_ dm . For ''k'' = 0,...,''n'', the left hand side is :$$ \langle (U^ *)^k (e_ ), (e_ ) \rangle = \langle (V^ *)^k (e_ ), (e_ ) \rangle = \langle (), (e_ ) \rangle = A_ = \bar. So :$$ \int_ dm = \int_^k dm = \alpha_k. Finally, parametrize the unit circle T by ''eit'' on (2''π'' ) gives :$\backslash frac\; \backslash int\_0\; ^\; e^\; d\backslash mu(t)\; =\; \backslash alpha\_k$ for some suitable measure ''μ. 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「Trigonometric moment problem」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.