Results 1 to 2 of 2

Thread: Sum of series by method of differences

  1. #1
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338

    Sum of series by method of differences

    If $\displaystyle f(r)\equiv\cos 2r\theta$, find $\displaystyle f(r)-f(r+1)$. Use yur result to find the sum of the first n terms of the series
    $\displaystyle \sin 3\theta+\sin 5\theta+\sin 7\theta+...$

    I've got $\displaystyle f(r)-f(r+1)\equiv\cos 2r\theta-\cos 2(r+1)\theta$
    $\displaystyle \equiv\cos 2r\theta-\cos (2r\theta+2\theta)$
    $\displaystyle \equiv\cos 2r\theta-\cos 2r\theta\cos 2\theta+\sin 2r\theta\sin 2\theta$
    $\displaystyle \equiv\cos 2r\theta(1-\cos 2\theta)+\sin 2r\theta\sin 2\theta$
    $\displaystyle \equiv2\cos 2r\theta\sin^2\theta+sin 2r\theta\sin 2\theta$
    $\displaystyle \equiv\sin\theta(\cos 2r\theta\sin\theta+\sin 2r\theta\cos\theta)+\cos 2r\theta\sin\theta$
    $\displaystyle \equiv\sin\theta(\sin(2r+1)\theta)+\cos 2r\theta\sin\theta$
    Now I don't know how to continue, or how this relates to the series.
    Thanks for any help!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Jul 2010
    From
    Vancouver
    Posts
    432
    Thanks
    17
    Your calculations are almost all good, except that you make a mistake in the last 2 lines. It is actually

    $\displaystyle \displaystylef(r)-f(r+1)\equiv\cos 2r\theta-\cos 2(r+1)\theta$
    $\displaystyle \displaystyle\equiv\cos 2r\theta-\cos (2r\theta+2\theta)$
    $\displaystyle \displaystyle\equiv\cos 2r\theta-\cos 2r\theta\cos 2\theta+\sin 2r\theta\sin 2\theta$
    $\displaystyle \displaystyle\equiv\cos 2r\theta(1-\cos 2\theta)+\sin 2r\theta\sin 2\theta$
    $\displaystyle \displaystyle\equiv2\cos 2r\theta\sin^2\theta+sin 2r\theta\sin 2\theta$
    $\displaystyle \displaystyle\equiv2\cos 2r\theta\sin^2\theta+2 \sin 2r\theta\sin \theta\cos\theta$
    $\displaystyle \displaystyle\equiv 2\sin\theta(\cos 2r\theta\sin\theta+ \sin 2r\theta\cos\theta)$
    $\displaystyle \displaystyle\equiv 2\sin\theta(\sin((2r+1)\theta))$


    This helps by getting that

    $\displaystyle \displaystyle \sin((2r+1)\theta) = \frac{f(r)-f(r+1)}{2\sin\theta}$

    So now, the sum on top is

    [Math]\displaystyle \sin 3\theta+\sin 5\theta+\sin 7\theta+... = \frac{1}{2\sin\theta}\left(\sum_{r = 1}^n (f(r)-f(r+1)\right)[/tex]

    Which is a telescoping sum which sums to

    $\displaystyle \displaystyle \frac{f(1)-f(n+1)}{2\sin\theta} = \frac{\cos(3\theta)-cos((2n+1)\theta)}{2\sin\theta}$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Method of Differences
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 20th 2011, 10:16 PM
  2. method of differences
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Sep 3rd 2009, 07:27 AM
  3. [SOLVED] Method of differences problems
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Feb 11th 2009, 06:50 AM
  4. Summation: Method of differences.
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: May 20th 2008, 04:06 AM
  5. Method of Differences
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Jan 11th 2008, 11:28 PM

Search tags for this page

Click on a term to search for related topics.

/mathhelpforum @mathhelpforum