Results 1 to 2 of 2

Math Help - Inner product

  1. #1
    Newbie
    Joined
    Nov 2009
    Posts
    23

    Inner product

    V is the space of polynomials with real coefficients with degree ≤ n.
    The inner product is defined like this: <p,q> = the integral from -1 to 1 of pq(x)
    I need to prove that if: p ≠ 0, <p,x^i> = 0, i < k then: deg p ≥ k

    Thanks for any kind of help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    By contradiction, assume \deg p < k then p belongs to the subspace of the polynomials of degree less or equal to k-1, which is generated by: 1,...,x^{k-1}.
    But p is orthogonal to each of the elements of a set that generates it, hence it must be the \bold{0} polynomial (*)! contradiction!

    <br />
(*) p(x)=a_0\cdot 1 + a_1\cdot x^1+...+a_{k-1}\cdot x^{k-1} now : <p,p>=a_0\cdot <1,p> + ...+a_{k-1}\cdot <x^{k-1},p>=0 hence p=\bold{0}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: October 18th 2011, 05:40 AM
  2. Replies: 6
    Last Post: September 7th 2010, 10:03 PM
  3. multivariable differential for inner product(scalar product)?
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: October 23rd 2009, 06:40 PM
  4. Replies: 4
    Last Post: September 2nd 2009, 05:07 AM
  5. Replies: 1
    Last Post: May 14th 2008, 12:31 PM

Search Tags


/mathhelpforum @mathhelpforum