Results 1 to 5 of 5

Math Help - Orthogonal functions

  1. #1
    Member billym's Avatar
    Joined
    Feb 2008
    Posts
    183

    Orthogonal functions

    I have two functions and I want to show that they are orthogonal over a given interval. Do I just multiply them together and show that the integral over the interval is zero?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member girdav's Avatar
    Joined
    Jul 2009
    From
    Rouen, France
    Posts
    678
    Thanks
    32
    I think it's the definition of orthogonal function, isn't it?
    What are the two functions?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,419
    Thanks
    1856
    Quote Originally Posted by billym View Post
    I have two functions and I want to show that they are orthogonal over a given interval. Do I just multiply them together and show that the integral over the interval is zero?
    If your definition of inner product is <f, g>= \int_a^b f(x)g(x)dx. But there are other ways of defining the inner product in function spaces, usually involving a "weighting function", so the inner product is <f(x),g(x)>= \int_a^b f(x)g(x)\mu(x)dx. For example, the Leguerre polynomials are orthogonal using the inner product defined by <f, g>= \int_0^\infty f(x)g(x)e^{-x}dx. In that case the " e^-{x} is necessary because the interval of definition is infinite. Bessel functions are orthogonal using the inner product defined by <f, g>= \int_0^1 x f(x)g(x)dx. Here, the "x" factor is necessary because the Bessel functions may have a pole of order 1 at 0.

    In general, two vectors in an innerproduct space are orthgonal if and only if their inner product is 0. Check to see what inner product you should be using.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member billym's Avatar
    Joined
    Feb 2008
    Posts
    183
    Mine are

    f(x)=\frac{3x^2-1}{2} and g(x)=\frac{5x^3-3x}{2} on x \in [-1,1]

    Is \int_a^b f(x)g(x)dx the inner product I should use? (I hope so because I already stapled this thing together).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
    Posts
    397
    actually it turn out quite nice:


    h(x) = f(x) \cdot g(x) = \frac{{x\left( {3x^2  - 1} \right)\left( {5x^2  - 3} \right)}}<br />
{4}

    now you can easily see that h(x) is an odd function since h(x) = -h(-x)
    and we all know that the integral of an odd function over a symmetrical interval is 0.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: August 15th 2011, 05:32 AM
  2. Orthogonal basic && basic for the orthogonal complement
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 5th 2009, 07:33 AM
  3. Orthogonal Functions
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 12th 2009, 05:51 AM
  4. Prove these functions are orthogonal
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 20th 2009, 06:59 PM
  5. how do u find the functions orthogonal to this
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 1st 2009, 01:20 AM

Search Tags


/mathhelpforum @mathhelpforum