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Thread: Functional problem

  1. #1
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    Functional problem

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

    What I first should do is prove that $\displaystyle \alpha_i \forall i$ 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 $\displaystyle V=R_n[x]$ and $\displaystyle \alpha_i(p)=p^i(0)$ be a functional over $\displaystyle V$( $\displaystyle p^i$ is the number i derivative of p)
    Show that $\displaystyle \{\alpha_i\}_\forall\i$ is a basis of a dual space $\displaystyle V^*$

    What I first should do is prove that $\displaystyle \alpha_i \forall i$ are linearly independent.
    How should I do it?
    Any ideas?
    Hint: notice that $\displaystyle \alpha_i(x^k)$ is $\displaystyle k!$ 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 $\displaystyle \alpha_i(x^k)$ is $\displaystyle k!$ if i=k, and 0 if i≠k.
    $\displaystyle \alpha_0,....,\alpha_n$are independent iff
    $\displaystyle \sum\lambda_i\alpha_i=0 $

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

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

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