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Infinite sum of sines

  1. Apr 8, 2015 #1
    1. The problem statement, all variables and given/known data
    I have this exercise: Calculate
    ##\sum\limits_{k=0}^\infty t^{k}sin{(kx)}##
    Where x and t are real and t is between 0 and 1.
    2. Relevant equations
    ?
    3. The attempt at a solution
    The ratio test says that this sum does have a limit, and tk obviously converges, as t is between 0 and 1. However, how am I supposed to solve it for the sines? A summatory inside a summatory? I don't think so, this is an exam question and we haven't done anything similar. I've been stuck with this for a day now...
     
    Last edited: Apr 8, 2015
  2. jcsd
  3. Apr 8, 2015 #2

    Orodruin

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    How about trying to rewrite the sine in some intelligent fashion?
     
  4. Apr 8, 2015 #3
    Damn, how did I not think of that?
    So the answer is
    ##\frac{2i}{2i-te^{ix}}-\frac{2i}{2i-te^{-ix}}##?
     
  5. Apr 8, 2015 #4

    Matternot

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    I get:

    ##\frac{tsin(x)}{1-2t cos(x)+t^2}##

    Your answer may be the same.

    What was your method?
     
    Last edited: Apr 8, 2015
  6. Apr 8, 2015 #5

    Matternot

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    ##\frac{tsin(x)}{1-2tcos(x)+t^2}##
     
    Last edited: Apr 8, 2015
  7. Apr 8, 2015 #6

    LCKurtz

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    That isn't what I get. Show your work, and, since the answer must be real, put it in real form.

    [Edit, added]I didn't see Stephen's post; I agree with his answer.
     
  8. Apr 8, 2015 #7
    Crap! I'm being stupid today! my previous expression should have been
    ##\frac{1}{2i}(\frac{1}{1-te^{ix}}-\frac{1}{1-te^{-ix}})##
    Which, if simplified, gives the same answer you got. Thanks, guys!
     
  9. Apr 9, 2015 #8

    Matternot

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    It seems like your method is unnecessarily complicated. Did you write sin as:

    ##\frac{e^{ix}-e^{-ix}}{2i}##

    or as

    ##Im(e^{ix})##?
     
  10. Apr 9, 2015 #9
    I used
    ##\sin{(x)}=\frac{e^{ix}-e^{-ix}}{2i}##

    How do you do it with

    ##Im(e^{ix})##?
     
  11. Apr 9, 2015 #10

    Orodruin

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    In exactly the same way, you just move the Im outside of the sum since ##t## is real. The difference is that you get only one term.
     
  12. Apr 9, 2015 #11

    Matternot

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    It's considered easier to use

    ##Im(e^{ix})##

    Try using that method and see which you prefer.
     
  13. Apr 9, 2015 #12
    Okay, I've just tried it and it's definitely easier and there's less things that you can do wrong. I wouldn't have thought of that myself, as ##Im(z)## doesn't feel like a "proper" function to me, but I see it can be useful if you know what you're doing. Thanks for the help, guys!
     
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