Results 1 to 2 of 2

Thread: Inner product

  1. #1
    Nov 2009

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

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

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Oct 18th 2011, 04:40 AM
  2. Replies: 6
    Last Post: Sep 7th 2010, 09:03 PM
  3. multivariable differential for inner product(scalar product)?
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Oct 23rd 2009, 05:40 PM
  4. Replies: 4
    Last Post: Sep 2nd 2009, 04:07 AM
  5. Replies: 1
    Last Post: May 14th 2008, 11:31 AM

Search Tags

/mathhelpforum @mathhelpforum