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Math Help - Lagrange polynomials

  1. #1
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    Smile Lagrange polynomials

    hello
    for i=1,2,...,(n+1)
    let
    ,
    prove that (P_1,P_2,...P_{n+1}) is a basis of \mathbb{R}_{n}[X].
    thanks.

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  2. #2
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    Quote Originally Posted by Raoh View Post
    hello
    for i=1,2,...,(n+1)
    let
    ,
    prove that (P_1,P_2,...P_{n+1}) is a basis of \mathbb{R}_{n}[X].
    thanks.

    obviously we also need to have a_i \neq a_j, for i \neq j. anyway, again, 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. first see that 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
    Last edited by NonCommAlg; December 18th 2009 at 12:51 PM.
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  3. #3
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    Smile

    Quote Originally Posted by NonCommAlg View Post
    obviously we also need to have a_i \neq a_j, for i \neq j. anyway, again, 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

    first see that 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
    if i put x=a_i we'll have P_j(a_i)=0, no ?
    thank you.
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  4. #4
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    Quote Originally Posted by Raoh View Post
    if i put x=a_i we'll have P_j(a_i)=0, no ?
    thank you.
    yes, for a fixed i, we have P_i(a_i)=1 and P_j(a_i)=0, for all j \neq i. by the way, a few words in my previous post were missing. i just edited my post. so you should take a look at it again.
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  5. #5
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    Quote Originally Posted by NonCommAlg View Post
    yes, for a fixed i, we have P_i(a_i)=1 and P_j(a_i)=0, for all j \neq i. by the way, a few words in my previous post were missing. i just edited my post. so you should take a look at it again.
    what if i suppose \sum_{i=1}^{n+1}c_i P_i(X)=0 for some c_i and put X=a_i to get c_i=0 is it okay ?
    thanks.
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  6. #6
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    Quote Originally Posted by Raoh View Post
    what if i suppose \sum_{i=1}^{n+1}c_i P_i(X)=0 for some c_i and put X=a_i to get c_i=0 is it okay ?
    thanks.
    yes, just for the index of your sum you should use j instead of i because when you put x=a_i you're assuming that i is fixed. (do as i did in my solution!)
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  7. #7
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    Quote Originally Posted by NonCommAlg View Post
    yes, just for the index of your sum you should use j instead of i because when you put x=a_i you're assuming that i is fixed. (do as i did in my solution!)
    of course i'm going to do as you did in your solution but at least let me know how did you do it.
    my question,what do you mean by a fixed i,i don't understand your last post,would you please explain it to me.
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