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Small angle forula problem

  1. Mar 2, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]S_N(x)= \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{\sin ((2 n-1)x)}{2 n-1}[/tex]

    By considering a suitable small angle formula show that the value of the sum at this point is

    [tex]S_N \Big( \frac{\pi}{2 N} \Big)=\frac{2}{\pi} \int_0^{\pi} \frac{\sin (\mu)}{\mu} \; d{\mu}[/tex]


    2. Relevant equations

    i have no idea how to get the suitable small angle formula working with this problem


    3. The attempt at a solution

    I have shown that

    [tex]S_N(x) [/tex]

    can be written as

    [tex]S_N(x)=\frac{2}{\pi} \int_0^{x} \frac{\sin (2 N t)}{\sin (t) } \; d{t}[/tex]

    my guess for suitable small angle formula is

    [tex]\sin (x) \approx x [/tex] when x is small


    Thank you for any help
     
    Last edited: Mar 2, 2009
  2. jcsd
  3. Mar 3, 2009 #2
    anyone got any ideas? :P
     
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