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Math Help - Functional problem

  1. #1
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    Functional problem

    Let V=R_n[x] and \alpha_i(p)=p^i(0) be a functional over V( p^i is the number i derivative of p)
    Show that \{\alpha_i\}_\forall\i is a basis of a dual space V^*

    What I first should do is prove that \alpha_i \forall i are linearly independent.
    How should I do it?
    Any ideas?
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  2. #2
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    Quote Originally Posted by wicked View Post
    Let V=R_n[x] and \alpha_i(p)=p^i(0) be a functional over V( p^i is the number i derivative of p)
    Show that \{\alpha_i\}_\forall\i is a basis of a dual space V^*

    What I first should do is prove that \alpha_i \forall i are linearly independent.
    How should I do it?
    Any ideas?
    Hint: notice that \alpha_i(x^k) is k! if i=k, and 0 if i≠k.
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  3. #3
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    Quote Originally Posted by Opalg View Post
    Hint: notice that \alpha_i(x^k) is k! if i=k, and 0 if i≠k.
    \alpha_0,....,\alpha_nare independent iff
    \sum\lambda_i\alpha_i=0

    So if p=a_0+a_1x+....+a_nx^n
    \alpha_0(p)=(a_1+2a_2+...+na_nx^(n-1))(0)=a_0
    .
    .
    .
    \alpha_n(p)=(0+0+...+n!a_nx^0)(0)=n!a_n
    right?
    Then \forall a_1....a_n
    \lambda_0\alpha_0+....+\lambda_n\alpha_n=0

    and therefore it's obvious that \lambda_0=\lambda_1=.....=\lambda_n=0

    And therefore dimV^*=dimV=n+1 => \{\alpha_i\} is basis.
    This is more or less is the answer right?
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