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    When z is complex and |z|< 1, . \sum_{n=0}^\infty z^n = \frac{1}{1-z} . Use this result to find the sum of the series \sum_{n=0}^\infty a^n \sin n \theta, where theta and a are real and |a| < 1.

    Not really sure where to start with this... I tried putting z^n = r^n exp(in \theta) = r^n (\cos n \theta + i \sin n \theta), then kind of assumed that a is equivalent to r (I'm not sure why ?), to get

    z^n = a^n (\cos n \theta + i \sin n \theta) = a^n \cos n \theta + a^n i \sin n \theta... but I got stuck after that!

    Any tips?
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    Can you prove that \sin n \theta = \dfrac{e^{in\theta} - e^{-in\theta}}{2i} ? Hence what the sum equivalent to?
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    (Original post by GHOSH-5)
    Can you prove that \sin n \theta = \dfrac{e^{in\theta} - e^{-in\theta}}{2i} ? Hence what the sum equivalent to?
    Ah yes, I remember proving that identity before (cba to do it again now :p: )

    So then we get a^n \sin n \theta = \frac{a^n}{2i} (e^{in \theta} - e^{- in \theta})

    But z^n = a^n e^{in \theta}... and... erm... what about a^n e^{-in \theta}? Am I on the right track?
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    (Original post by trm90)
    Ah yes, I remember proving that identity before (cba to do it again now :p: )

    So then we get a^n \sin n \theta = \frac{a^n}{2i} (e^{in \theta} - e^{- in \theta})

    But z^n = a^n e^{in \theta}... and... erm... what about a^n e^{-in \theta}? Am I on the right track?
    Yep, you're on the right track: - split up the sum:

     \dfrac{1}{2i}\displaystyle\sum_{  n=0}^{\infty} a^n(e^{in\theta} -e^{-in\theta}) = \dfrac{1}{2i} \left( \displaystyle\sum_{n=0}^{\infty} a^ne^{in\theta} - \displaystyle\sum_{n=0}^{\infty} a^ne^{-in\theta}\right)

    Can you see how to proceed now?
    Spoiler:
    Show
     a^ne^{in\theta} = (ae^{i\theta})^n and  a^ne^{-in\theta} = (ae^{-i\theta})^n
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    (Original post by GHOSH-5)
    Yep, you're on the right track: - split up the sum:

     \dfrac{1}{2i}\displaystyle\sum_{  n=0}^{\infty} a^n(e^{in\theta} -e^{-in\theta}) = \dfrac{1}{2i} \left( \displaystyle\sum_{n=0}^{\infty} a^ne^{in\theta} - \displaystyle\sum_{n=0}^{\infty} a^ne^{-in\theta}\right)

    Can you see how to proceed now?
    Spoiler:
    Show
     a^ne^{in\theta} = (ae^{i\theta})^n and  a^ne^{-in\theta} = (ae^{-i\theta})^n
    I can sort of see what to do... the first summation term is basically sum of zn which is just 1/(1-z), but I don't really know how what to do with (ae^{-i\theta})^n... is that the same thing as z-n? Or is that the same thing as (z^n)^* (conjugate)? Regardless, not quite sure how I'd evaluate that sum, unless I'm missing something!

    (Thanks by the way!)
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    They are almost the same sum, just the 'z' is different; in the first case, 'z' is ae^{i.theta}, and in the second case, 'z' is ae^{-i.theta}. Both of them sum to 1/(1-z), just 'z' is different...
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    (Original post by GHOSH-5)
    They are almost the same sum, just the 'z' is different; in the first case, 'z' is ae^{i.theta}, and in the second case, 'z' is ae^{-i.theta}. Both of them sum to 1/(1-z), just 'z' is different...
    Ah ok. So would my answer literally look something like:

    \frac{1}{2i} (\frac{1}{1 - z} + \frac{1}{1 - z^*})

    Where z* = a.exp(-i.theta)?

    I feel like I'm overcomplicating things here, though
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    (Original post by trm90)
    Ah ok. So would my answer literally look something like:

    \frac{1}{2i} (\frac{1}{1 - z} + \frac{1}{1 - z^*})

    Where z* = a.exp(-i.theta)?

    I feel like I'm overcomplicating things here, though
    Pretty much, yes. You can simplify your answer though. Leave z and z* in the exponential form, and rewrite everything over one fraction. Then recall that:

     e^{ix} + e^{-ix} = 2\cos x, and the analogous identity for sin(x) we used before.
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    There is another trick here .. (apologies - I've yet to learn LaTex or whatever)

    Let S = the sum you seek
    Let C = sum from n = 0 to inf (a^n cos(ntheta))

    Consider C + iS = sum from n= 0 to inf (a^n . exp(intheta)) which is a GP I can do in one line

    Equate real and imaginary components.
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    I'd go for the C + iS method described in ian.slater's post.
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    (Original post by ian.slater)
    There is another trick here .. (apologies - I've yet to learn LaTex or whatever)

    Let S = the sum you seek
    Let C = sum from n = 0 to inf (a^n cos(ntheta))

    Consider C + iS = sum from n= 0 to inf (a^n . exp(intheta)) which is a GP I can do in one line

    Equate real and imaginary components.
    Nice. :yy:
 
 
 
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