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Math Help - CauchySchwarz Inequality Help :D

  1. #1
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    CauchySchwarz Inequality Help :D

    ||x||=4, ||y||=5,<x,y>=8.
    What is the cauchyshwarz inequality? <This was relatively straight forward.. Not too sure about the rest however..
    What is the norm of 4x-4y?
    What is the cosine angle between x and 4x-4y?
    Find λ, such that 2x-y and λx-y are orthogonal?

    Many thanks
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi,
    Quote Originally Posted by brd_7 View Post
    ||x||=4, ||y||=5,<x,y>=8. (...)
    What is the norm of 4x-4y?
    \|4x-4y\|=4\|x-y\|=4\sqrt{\langle x-y,x-y\rangle}=4\sqrt{\langle x,x\rangle-2\langle x,y\rangle+\langle y,y\rangle}=\ldots
    What is the cosine angle between x and 4x-4y?
    Remember that \langle u,v\rangle = \|u\|\|v\|\cos(u,v).
    Find λ, such that 2x-y and λx-y are orthogonal?
    In other words, solve \langle 2x-y,\lambda x-y\rangle=0 for \lambda.
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  3. #3
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    I got 28 for the norm of 4x-4y.. I hope thats right.. I then gained th result of 66.4 for the theta angle in part b.. Thanks for the help.

    I think im going to need a litrle bit more help with the last part of the question...
    Last edited by brd_7; November 15th 2008 at 07:15 AM.
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by brd_7 View Post
    I got 28 for the norm of 4x-4y.. I hope thats right.. I then gained th result of 66.4 for the theta angle in part b.. Thanks for the help.
    As the angle you've found is correct, I think you mistyped your answer to the first question : \|4x-4y\|=20.
    I think im going to need a litrle bit more help with the last part of the question...
    The dot product is a bilinear function : \begin{cases}\langle u,  v+w\rangle =  \langle u, v\rangle + \langle u, w\rangle & (1)\\\langle u+ v, w\rangle=\langle u,w\rangle +\langle v,w\rangle &(2)\\<br />
\langle \alpha u,v\rangle=\langle u, \alpha v\rangle = \alpha\langle u, v\rangle & (3)\end{cases} so the LHS of \langle 2x-y,\lambda x-y\rangle=0 can be expanded.

    \begin{aligned}<br />
\langle 2x-y,\lambda x-y\rangle &=0\\<br />
 \langle 2x-y,\lambda x\rangle+\langle 2x-y,-y\rangle &=0 \text{\,\,\,\,using (1)}\\<br />
\lambda \langle 2x-y, x\rangle-\langle 2x-y,y\rangle &=0 \text{\,\,\,\,using (3)}\\<br />
\end{aligned}

    This can be further expanded using (2) and (3) to make appear \|x\|^2,\,\|y\|^2 and \langle x ,y\rangle...
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  5. #5
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    Yup, that is what i got sorry

    And this is what i got so far, for the next bit..

    λ[2<x,x> - <x,y>]- [2<x,y> - <y,y>]

    With λ = -3/8.. eventually

    Thanks again for all the help!
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  6. #6
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by brd_7 View Post
    With λ = -3/8.. eventually
    That's it !
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