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Integral inequality

  • Thread starter mohbb
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Homework Statement



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)

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
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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618
Use a different inner product. Define <f,g> to be the integral of f(x)*g(x)*cos(x)*dx.
 

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