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Math Help - Please help check my solutions for vector space questions

  1. #1
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    Please help check my solutions for vector space questions

    Please help correct my work

    Please help check my solutions for vector space questions-screen-shot-2012-09-21-12.24.21-pm.png

    Please help check my solutions for vector space questions-screen-shot-2012-09-21-12.24.35-pm.png
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  2. #2
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    Re: Please help check my solutions for vector space questions

    From looking at the images you posted it seems that you are proving:
    |V\cdot U|\le\|V\|\|U\|

    The standard proof follows from the fact \frac{V\cdot U}{\|V\|\|U\|}=\cos(\theta) where \theta is the angle between the two non-zero vectors V~\&~U.

    Then note that 0\le |\cos(\theta)|\le 1~.
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  3. #3
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    Re: Please help check my solutions for vector space questions

    first of all, you have \|U\| and \|V\| wrong.

    \|U\| = \sqrt{U \cdot U} = \sqrt{\int_0^{\frac{\pi}{2}} \sin^2x\ dx}

     = \sqrt{\int_0^{\frac{\pi}{2}}(1 - \cos^2x)\ dx}

     = \sqrt{\frac{\pi}{2} - \int_0^{\frac{\pi}{2}} \cos^2x\ dx}

    = \sqrt{\frac{\pi}{2} - \int_0^{\frac{\pi}{2}} \frac{1 + \cos 2x}{2}\ dx}

     = \sqrt{\frac{\pi}{2} - \frac{\pi}{4} - \frac{1}{4}\int_0^{\frac{\pi}{2}} \cos 2x\ 2dx}

     = \sqrt{\frac{\pi}{4} - \frac{1}{4}(\sin (\pi) - \sin (0))}

     = \frac{\sqrt{\pi}}{2}

    while:

    \|V\| = \sqrt{V \cdot V} = \sqrt{\int_0^{\frac{\pi}{2}} \cos^2x\ dx}

     = \sqrt{\int_0^{\frac{\pi}{2}} \frac{1+\cos 2x}{2}\ dx}

     = \sqrt{\frac{\pi}{4} + \frac{1}{4}(\sin(\pi) - \sin(0))}

     = \frac{\sqrt{\pi}}{2}, as well.

    thus:

    \|U\|\|V\| = \frac{\pi}{4}.

    on the other side of the inequality we have:

    |U \cdot V| = \left|\int_0^{\frac{\pi}{2}} \sin x \cos x\ dx \right|

     = \left| \frac{\sin^2(\frac{\pi}{2})}{2} - \frac{\sin^2(0)}{2} \right|  = \left|\frac{1}{2} \right| = \frac{1}{2}

    since  1 < \frac{\pi}{2} , we indeed have for these particular U,V:

    |U \cdot V| \leq \|U\|\|V\|.

    *********

    for your second problem, you are also calculating the norms of U and V incorrectly. use the definition of the inner product you are given!
    Last edited by Deveno; September 21st 2012 at 01:37 PM.
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  4. #4
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    Re: Please help check my solutions for vector space questions

    Thank you so much!! Deveno! I missed the obvious point

    Could you please re-check my second answer again?
    Please help check my solutions for vector space questions-screen-shot-2012-09-21-10.02.07-pm.png
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  5. #5
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    Re: Please help check my solutions for vector space questions

    still incorrect. for starters, you have an extra 2-n term in \|U\| and \|V\|.

    secondly, you don't seem to get that the inner product of two vectors should be a SCALAR, not an "expression".

    so for U, you should have:

    \|U\| = \sqrt{\sum_{n=1}^{\infty} \frac{1}{2^n}} = 1

    and for V, you should have:

    \|V\| = \sqrt{\sum_{n=1}^{\infty} \frac{1}{8^n}} = \frac{1}{\sqrt{7}}

    while for the dot product, you should have:

    |U \cdot V| = \left| \sum_{n=1}^{\infty} \frac{1}{4^n} \right| = \frac{1}{3}
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