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Math Help - Cauchy-Schwarz Inequality

  1. #1
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    Cauchy-Schwarz Inequality

    I am supposed to prove that for all real numbers a,b, \theta that
    (acos\theta+bsin\theta)^2\leq a^2+b^2
    I've only gotten up to this point which is where I am not sure as to how to appply the cauchy schwarz inequality,

    (acos\theta+bsin\theta)^2=a^2+b^2-(asin\theta+bcos\theta)^2

    Am I on the right track?
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  2. #2
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    <acosx+bsinx,acosx+bsinx>=<acosx,acosx+bsinx>+<bsi  nx,acosx+bsinx>

    =<acosx,acosx>+2<acosx,bsinx>+<bsinx,bsinx>=.....

    Also, I am assuming inner product is defined as such:

    \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)g(x)dx because if this is the case, everything works out just fine.
    Last edited by dwsmith; June 26th 2010 at 12:47 PM.
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  3. #3
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    It does not specify the inner product. I have posted all the information that was given to me.
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  4. #4
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    Quote Originally Posted by Jukodan View Post
    It does not specify the inner product. I have posted all the information that was given to me.
    Try finishing the problem with that inner product in mind and you will have your problem completed.
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  5. #5
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    Quote Originally Posted by Jukodan View Post
    I am supposed to prove that for all real numbers a,b, \theta that
    (acos\theta+bsin\theta)^2\leq a^2+b^2
    \left( {a\sin (t) - b\cos (t)} \right)^2  = a^2 \sin ^2 (t) - 2ab\sin (t)\cos (t) + b^2 \cos (t) Note the minus.
    So we get \left( {a\cos (t) + b\sin (t)} \right)^2  = a^2  + b^2  - \left( {a\sin (t) - b\cos (t)} \right)^2
    Notice that  a^2  + b^2  - \left( {a\sin (t) - b\cos (t)} \right)^2 \le a^2+b^2
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  6. #6
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    <acosx+bsinx,acosx+bsinx>=<acosx,acosx+bsinx>+<bsi  nx,acosx+bsinx>

    =<acosx,acosx>+2<acosx,bsinx>+<bsinx,bsinx>=.....

    Spoiler:
    a^2<cosx,cosx>+2ab<cosx,sinx>+b^2<sinx,sinx>

    \frac{1}{\pi}\int_{-\pi}^{\pi}(a^2cos^2x)dx=a^2

    \frac{1}{\pi}\int_{-\pi}^{\pi}(2ab*cosx*sinx)dx=0

    \frac{1}{\pi}\int_{-\pi}^{\pi}(b^2sin^2x)dx=b^2

    a^2<cosx,cosx>+2ab<cosx,sinx>+b^2<sinx,sinx>

    a^2+b^2

    \Rightarrow (acosx+bsinx)^2\leq a^2+b^2
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  7. #7
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    Ah-ha!
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  8. #8
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    Quote Originally Posted by Jukodan View Post
    I am supposed to prove that for all real numbers a,b, \theta that
    (acos\theta+bsin\theta)^2\leq a^2+b^2
    I've only gotten up to this point which is where I am not sure as to how to appply the cauchy schwarz inequality,

    (acos\theta+bsin\theta)^2=a^2+b^2-(asin\theta+bcos\theta)^2

    Am I on the right track?
    The CS inequality says that x_1y_1+x_2y_2\leqslant (x_1^2+x_2^2)^{1/2}(y_1^2+y_2^2)^{1/2}. Apply that with x_1=a, x_2=b, y_1=\cos\theta, y_2 = \sin\theta. Then square both sides.
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