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Integral inequality

  1. Mar 28, 2010 #1
    1. The problem statement, all variables and given/known data

    Hi,

    I must show that ([tex]\int cos(x)f(x)dx[/tex])^2 <= 2 [tex]\int cos(x)f(x)^2dx[/tex]
    (the integrals are from -pi/2 to pi/2)

    3. The attempt at a solution

    I know that I should use cauchey-schwarz inequality to solve this where <f,g> = [tex]\int f(x)g(x)dx[/tex] In this case i just set g(x) = cos x
    Therefore i get
    ([tex]\int cos(x)f(x)dx[/tex])^2 <= [tex]\int f(x)^2dx[/tex] [tex]\int cos^2(x)dx[/tex] I calculated then integral of cos^2(x) which is 1/2(x + sin(2x)/2) since cos^2(x) = (1 + cos(2x))/2

    However this leaves me with pi/2 to get:
    ([tex]\int cos(x)f(x)dx[/tex])^2 <= pi/2 [tex]\int f(x)^2dx[/tex]

    Howcome I am not getting the same answer as the one I should be proving, why does the question have a cos in the integral?

    Thank you
     
  2. jcsd
  3. Mar 28, 2010 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Use a different inner product. Define <f,g> to be the integral of f(x)*g(x)*cos(x)*dx.
     
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