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    I am a bit stuck on this question... some help would be appreciated.



    I think I can do a) and b)... but when I get to c I am stuck. Ill post what I have done...

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    (Original post by importunate)
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    Not studied fourier transfroms, so don't know why you're doing it the way you are.

    But, applying the formula from part (a), using IBP, gives the desired result, after using the hint, and multiplying top and bottom by appropriate amounts to get into the form they desire.
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    (Original post by ghostwalker)
    Not studied fourier transfroms, so don't know why you're doing it the way you are.

    But, applying the formula from part (a), using IBP, gives the desired result, after using the hint, and multiplying top and bottom by appropriate amounts to get into the form they desire.
    Sorry I get what you are saying... great. Ill just have to see if it works now... have tried doing it over plus and minus infinity and it was horrible.
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    still cant do it. some ideas would be appreciated . thanks
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    (Original post by importunate)
    still cant do it. some ideas would be appreciated . thanks
    Well f(x) = a-x for 0<= x <= a

    So you want \displaystyle\frac{2}{\sqrt{2 \pi}}\int_0^a (a-x)cos(kx)dx

    Upper limit is a, since f(x) = 0 for x>= a
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    dont you have to include the -x+a section?
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    (Original post by importunate)
    dont you have to include the -x+a section?
    Sorry, don't understand that statement.

    If you're refering to the part of the graph from -a to 0, then it seems to me that the whole point of part (a) was to show that you don't need to consider it for an even function, as is the case here.
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    Thanks... but when I integrate this I get:

    -(coa(ak)-1)*root2/(k^2*rootpi....


    this isnt the answer I am looking for...
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    (Original post by importunate)
    Thanks... but when I integrate this I get:

    -(coa(ak)-1)*root2/(k^2*rootpi....


    this isnt the answer I am looking for...
    That's because you haven't applied the hint and multiplied top and bottom by appropriate amounts yet to get it into the form requested. Looks OK so far....
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    i can get the result but i end up with an erroneous 4 on bottom. ill post my workings.
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    (Original post by importunate)
    ...
    \displaystyle\frac{2a^2}{\sqrt{2  \pi}}\frac{(1-cos(ak))}{a^2k^2}

    \displaystyle=\frac{4a^2}{     \sqrt{2\pi}}\frac{\sin^2(\frac{a  k}{2})}{a^2k^2}

    \displaystyle=\frac{a^2}{     \sqrt{2\pi}}\frac{\sin^2(\frac{a  k}{2})}{a^2k^2/4}

    \displaystyle=\frac{a^2}{     \sqrt{2\pi}}\frac{\sin^2(\frac{a  k}{2})}{(\frac{ak}{2})^2}

    QED.
 
 
 
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