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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 Lagrange interpolation formula ： ウィキペディア英語版 Lagrange polynomial In numerical analysis, Lagrange polynomials are used for polynomial interpolation. For a given set of distinct points $x\_j$ and numbers $y\_j$, the Lagrange polynomial is the polynomial of the least degree that at each point $x\_j$ assumes the corresponding value $y\_j$ (i.e. the functions coincide at each point). The interpolating polynomial of the least degree is unique, however, and it is therefore more appropriate to speak of "the Lagrange form" of that unique polynomial rather than "the Lagrange interpolation polynomial", since the same polynomial can be arrived at through multiple methods. Although named after Joseph Louis Lagrange, who published it in 1795, it was first discovered in 1779 by Edward Waring and it is also an easy consequence of a formula published in 1783 by Leonhard Euler.〔 .〕 Lagrange interpolation is susceptible to Runge's phenomenon, and the fact that changing the interpolation points requires recalculating the entire interpolant can make Newton polynomials easier to use. Lagrange polynomials are used in the Newton–Cotes method of numerical integration and in Shamir's secret sharing scheme in cryptography. ==Definition== Given a set of ''k'' + 1 data points :$(x\_0,\; y\_0),\backslash ldots,(x\_j,\; y\_j),\backslash ldots,(x\_k,\; y\_k)$ where no two $x\_j$ are the same, the interpolation polynomial in the Lagrange form is a linear combination :$L(x)\; :=\; \backslash sum\_^\; y\_j\; \backslash ell\_j(x)$ of Lagrange basis polynomials :$\backslash ell\_j(x)\; :=\; \backslash prod\_\}\; \backslash frac\; =\; \backslash frac\; \backslash cdots\; \backslash frac\; \backslash frac\; \backslash cdots\; \backslash frac,$ where $0\backslash le\; j\backslash le\; k$. Note how, given the initial assumption that no two $x\_i$ are the same, $x\_j\; -\; x\_m\; \backslash neq\; 0$, so this expression is always well-defined. The reason pairs $x\_i\; =\; x\_j$ with $y\_i\backslash neq\; y\_j$ are not allowed is that no interpolation function $L$ such that $y\_i\; =\; L(x\_i)$ would exist; a function can only get one value for each argument $x\_i$. On the other hand, if also $y\_i\; =\; y\_j$, then those two points would actually be one single point. For all $i\backslash neq\; j$, $\backslash ell\_j(x)$ includes the term $(x-x\_i)$ in the numerator, so the whole product will be zero at $x=x\_i$: :$\backslash ell\_(x\_i)\; =\; \backslash prod\_\; \backslash frac\; =\; \backslash frac\; \backslash cdots\; \backslash frac\; \backslash cdots\; \backslash frac\; =\; 0.$ On the other hand, :$\backslash ell\_i(x\_i)\; :=\; \backslash prod\_\; \backslash frac\; =\; 1$ In other words, all basis polynomials are zero at $x=x\_i$, except $\backslash ell\_i(x)$, for which it holds that $\backslash ell\_i(x\_i)=1$, because it lacks the $(x-x\_i)$ term. It follows that $y\_i\; \backslash ell\_i(x\_i)=y\_i$, so at each point $x\_i$, $L(x\_i)=y\_i+0+0+\backslash dots\; +0=y\_i$, showing that $L$ interpolates the function exactly. 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「Lagrange polynomial」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.