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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 Vandermonde's identity ： ウィキペディア英語版 Vandermonde's identity In combinatorics, Vandermonde's identity, or Vandermonde's convolution, named after Alexandre-Théophile Vandermonde (1772), states that :$=\backslash sum\_^r,\backslash qquad\; m,n,r\backslash in\backslash mathbb\_0,$ for binomial coefficients. This identity was given already in 1303 by the Chinese mathematician Zhu Shijie (Chu Shi-Chieh). See Askey 1975, pp. 59–60 for the history. There is a q-analog to this theorem called the q-Vandermonde identity. == Algebraic proof == In general, the product of two polynomials with degrees ''m'' and ''n'', respectively, is given by :$\backslash biggl(\backslash sum\_^m\; a\_ix^i\backslash biggr)\; \backslash biggl(\backslash sum\_^n\; b\_jx^j\backslash biggr)$ = \sum_^\biggl(\sum_^r a_k b_\biggr) x^r, where we use the convention that ''ai'' = 0 for all integers ''i'' > ''m'' and ''bj'' = 0 for all integers ''j'' > ''n''. By the binomial theorem, :$(1+x)^\; =\; \backslash sum\_^\; x^r.$ Using the binomial theorem also for the exponents ''m'' and ''n'', and then the above formula for the product of polynomials, we obtain :$\backslash begin$ \sum_^ x^r &= (1+x)^\\ &= (1+x)^m (1+x)^n \\ &= \biggl(\sum_^m x^i\biggr) \biggl(\sum_^n x^j\biggr)\\ &=\sum_^\biggl(\sum_^r \biggr) x^r, \end where the above convention for the coefficients of the polynomials agrees with the definition of the binomial coefficients, because both give zero for all ''i'' > ''m'' and ''j'' > ''n'', respectively. By comparing coefficients of ''xr'', Vandermonde's identity follows for all integers ''r'' with 0 ≤ ''r'' ≤ ''m'' + ''n''. For larger integers ''r'', both sides of Vandermonde's identity are zero due to the definition of binomial coefficients. 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「Vandermonde's identity」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.