作者clouddeep (fix point)
看板Math
標題Re: [分析] Weierstrass theorem
時間Fri Jan 2 00:18:14 2009
※ 引述《Jer1983 (stanley)》之銘言:
: Let C[a,b] be the collection of real-valued continuous functions.
: For f in C[a,b], if
: ∫^a_b f(x) * x^n dx = 0, n = 0,1,2,...
: Prove f(x) = 0 on [a,b].
: 在證明的過程中
: 很多作者都直接引用 Weierstrass theorem: there exists a sequence of
: polynomials {P_n(x)} converging uniformly to f(x). Hence
: lim ∫_a^b P_n(x) * f(x) dx = ∫ [f(x)]^2 dx = 0
: and we have f(x) = 0.
: 我想問的是...那 x^n for n = 0,1,2... 的作用在哪 ? 好像跟證明無關阿
: 謝謝回覆
polynomial是x^n(n=1,2,3,....)的線性組合,由此關係可以得到:
根據Weierstrass Theorem,有一sequence of polynomial均勻收斂至f(x),
而且∫ [f(x)]^2 dx
= lim∫_a^b P_n(x) * f(x) dx
= the sum of the coefficient times of nth term ∫^a_b f(x) * x^n dx
= 0+0+0+...+0 = 0
例:
如果 P_n(x) = a_n*x^n+a_(n-1)*x^(n-1)+...+a_0
則 ∫ [f(x)]^2 dx = sum of a_k*∫^a_b f(x) * x^k dx from k=n to k=0.
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◆ From: 123.195.33.28
推 Jer1983:the sum of coefficients 那邊可以再解釋一下嗎? 01/02 02:05
※ 編輯: clouddeep 來自: 123.195.33.28 (01/03 23:19)
推 Jer1983:感謝 01/05 19:32