- #1
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hello everyone
for <br /> i=1,2,...,(n+1)<br /> let P_{i}(X)=\frac{\prod_{1\leq j\leq n+1,j\neq i}(X-a_j)}{\prod_{1\leq j\leq n+1,j\neq i}(a_i-a_j)}
prove that <br /> (P_1,P_2,...P_{n+1})<br /> is basis of <br /> \mathbb{R}_{n}[X]<br />.
i already have an answer but i don't understand some of it.
...
we have B=\{P_1, \cdots , P_{n+1} \} \subset \mathbb{R}_n[x] and \dim_{\mathbb{R}} \mathbb{R}[x]=n+1. so, in order to show that B is a basis for \mathbb{R}_n[x], we only need
to show that B is linearly independent.for any i we have P_i(a_i)=1 and P_j(a_i)=0, for i \neq j. now, to show that B is linearly independent, suppose that \sum_{j=1}^{n+1}c_j P_j(x)=0, for some
c_j \in \mathbb{R}. put x=a_i to get c_i=0. \ \Box
...
what i don't understand,
why should we have P_j(X) and not P_i(X)
why we had to find P_j(a_i) ?
and why we have \sum_{j=1}^{n+1} and not \sum_{i=1}^{n+1} ?
thank you so much.
for <br /> i=1,2,...,(n+1)<br /> let P_{i}(X)=\frac{\prod_{1\leq j\leq n+1,j\neq i}(X-a_j)}{\prod_{1\leq j\leq n+1,j\neq i}(a_i-a_j)}
prove that <br /> (P_1,P_2,...P_{n+1})<br /> is basis of <br /> \mathbb{R}_{n}[X]<br />.
i already have an answer but i don't understand some of it.
...
we have B=\{P_1, \cdots , P_{n+1} \} \subset \mathbb{R}_n[x] and \dim_{\mathbb{R}} \mathbb{R}[x]=n+1. so, in order to show that B is a basis for \mathbb{R}_n[x], we only need
to show that B is linearly independent.for any i we have P_i(a_i)=1 and P_j(a_i)=0, for i \neq j. now, to show that B is linearly independent, suppose that \sum_{j=1}^{n+1}c_j P_j(x)=0, for some
c_j \in \mathbb{R}. put x=a_i to get c_i=0. \ \Box
...
what i don't understand,
why should we have P_j(X) and not P_i(X)
why we had to find P_j(a_i) ?
and why we have \sum_{j=1}^{n+1} and not \sum_{i=1}^{n+1} ?
thank you so much.
