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Closed form of a summation

  1. Jan 14, 2008 #1
    1. The problem statement, all variables and given/known data

    I am looking for a closed form of the summation:
    sin(x) + sin(3x) + sin(5x) + ... + sin((2n-1*)x)

    2. Relevant equations

    None.

    3. The attempt at a solution

    Through a complete stroke of luck, I believe I have arrived at the correct solution: sin^2(nx)/sin(x)
    I have tested this for many different cases, and I believe it is correct. However, I am having a hard time proving that it is. Can anybody point me in the right direction?
     
  2. jcsd
  3. Jan 14, 2008 #2

    rock.freak667

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    How did you prove that?


    [EDIT: The last term should be sin(2n+1)x not 2n-1
     
  4. Jan 14, 2008 #3

    Hurkyl

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    Gold Member

    Seems to me that mathematical induction would be an obvious thing to try...
     
  5. Jan 14, 2008 #4
    I'm a little rusty on my induction skills. Is this what I need to do?

    1. Show that:
    sin^2(nx)/sin(x) + sin((2(n+1)-1)x) = sin^2((n+1)x)/sin(x)

    2. Show that my formula works for any specific case.
     
  6. Jan 14, 2008 #5

    rock.freak667

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    You need to show that

    [tex]\sum_{n=0} ^N sin(2n-1)x= \frac{sin^2(Nx)}{sinx}[/tex]


    then add the (N+1)th term to each side and show that is can be written as

    [tex]\frac{sin^2((N+1)x)}{sinx}[/tex]
     
    Last edited: Jan 14, 2008
  7. Jan 14, 2008 #6
    I've been trying to get these two sides equal, but I'm not coming up with anything. Which identities should I be using to tackle this problem?
     
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