Chủ đề này có 3 trả lời

[tientthegioi]

let $P(x)$ be a polynomical  satisfying the conditions
i/ $degP(x)\leq 2n$
ii/$ \forall k\in Z$  and $ k\in [-n;n] $ satisfy the inequality $P(k) \leq1 $
prove that $x \in [-n;n] , we have inequality $  P(x) \leq2^{2n}$

Bài viết đã được chỉnh sửa nội dung bởi dark templar: 06-04-2013 - 10:19

[tientthegioi]

let $P(x)$ be a polynomical  satisfying the conditions
i/ $degP(x)\leq 2n$
ii/$ \forall k\in Z$  and $ k\in [-n;n] $ satisfy the inequality/P(k)/\leq1 $
prove that $\in x[-n;n] , we have inequality $  /P(x)/ \leq2^{2n}$

let http://dientuvietnam...mimetex.cgi?P(x) be a polynomical satisfying the conditions
i/
ii/satisfy the inequality
prove that , we have inequality QUOTE]

[QUANVU]

let $P(x)$ be a polynomical  satisfying the conditions
i/ $degP(x)\leq 2n$
ii/$ \forall k\in Z$  and $ k\in [-n;n] $ satisfy the inequality/P(k)/\leq1 $
prove that $\in x[-n;n] , we have inequality $  /P(x)/ \leq2^{2n}$

P(x) là đa thức thỏa mãn:
1)degP 2n
2)|P(k)| 1 khi k=-n,-n+1,...,n
Chứng minh |P(x)| 4^n x [-n;n].

[Hr MiSu]

Đặt A={-n,-n+1,...,0,...n}, dễ thấy |A|= 2n+1

Áp dụng công thức nội suy Lagrange cho 2n+1 điểm -n, ... ,0,1,...,n ta được:

$\left | f(x) \right |=\left | \sum_{i\in A}^{ }\frac{\prod_{j\neq i\in A}^{ }(x-j)}{\prod_{j\neq i\in A}^{ }(j-i)} \right |$

Để ý $\prod_{j\neq i\in A}^{ }(x-j)\leq (2n)!, \prod_{j\neq i\in A}^{ }(j-i)=(n-i)!(n+i)!,$

Thế nên $\left | f(x) \right |\leqslant \sum_{i=-n}^{n}\frac{(2n)!}{(n-i)!(n+i)!}=\sum_{i=0}^{n}\binom{2n}{i}=2^{2n}$, đpcm