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Math Help - Series and complex number

  1. #1
    Senior Member vincisonfire's Avatar
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    Series and complex number

    This series
     1 + \frac{cos(\theta)}{sin(\theta)} + \frac{cos(2\theta)}{sin^2(\theta)} + ... + \frac{cos(n\theta)}{sin^n(\theta)}
    I think that the summation notation would be  \sum_{i=0}^{n} \frac{cos(i\theta)}{sin^i(\theta)} .
    But anyway, I'm having trouble with the top one. I have tried to transform it so that I can use complex numbers to find the answer. I have not succeed. Does someone know the way (or any method) to do this problem?
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  2. #2
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    Quote Originally Posted by vincisonfire View Post
    This series
     1 + \frac{cos(\theta)}{sin(\theta)} + \frac{cos(2\theta)}{sin^2(\theta)} + ... + \frac{cos(n\theta)}{sin^n(\theta)}
    I think that the summation notation would be  \sum_{i=0}^{n} \frac{cos(i\theta)}{sin^i(\theta)} .
    But anyway, I'm having trouble with the top one. I have tried to transform it so that I can use complex numbers to find the answer. I have not succeed. Does someone know the way (or any method) to do this problem?
    What is your goal :S? What is it you're trying to get to?

    PS: Confusing to use i as your index if you're also planning on using complex numbers in the solution!
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  3. #3
    Senior Member vincisonfire's Avatar
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    You're right, bad choice of index.
    I try to find a "nice" general formula for  S_n
    <br />
1 + \frac{cos(\theta)}{sin(\theta)} + \frac{cos(2\theta)}{sin^2(\theta)} + ... + \frac{cos(n\theta)}{sin^n(\theta)} = S_n<br />
    <br />
S_n = <br />
\sum_{k=0}^{n} \frac{cos(k\theta)}{sin^k(\theta)}<br />
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  4. #4
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    Hi

    Your idea seems to be good

    Let <br />
T_n = <br />
\sum_{k=0}^{n} \frac{sin(k\theta)}{sin^k(\theta)}<br />

    Then

    <br />
S_n + i T_n = <br />
\sum_{k=0}^{n} \frac{cos(k\theta)+isin(k\theta)}{sin^k(\theta)}<br />

    <br />
 S_n + i T_n = <br />
 \sum_{k=0}^{n} \frac{e^{ik\theta}}{sin^k(\theta)}<br />

    <br />
  S_n + i T_n = <br />
  \sum_{k=0}^{n} \frac{({e^{i\theta}})^k}{sin^k(\theta)}<br />

    <br />
   S_n + i T_n = <br />
   \sum_{k=0}^{n} \left({\frac{e^{i\theta}}{sin(\theta)}}\right)^k<br />
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  5. #5
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    Quote Originally Posted by running-gag View Post
    Hi

    Your idea seems to be good

    Let <br />
T_n = <br />
\sum_{k=0}^{n} \frac{sin(k\theta)}{sin^k(\theta)}<br />

    Then

    <br />
S_n + i T_n = <br />
\sum_{k=0}^{n} \frac{cos(k\theta)+isin(k\theta)}{sin^k(\theta)}<br />

    <br />
S_n + i T_n = <br />
\sum_{k=0}^{n} \frac{e^{ik\theta}}{sin^k(\theta)}<br />

    <br />
S_n + i T_n = <br />
\sum_{k=0}^{n} \frac{({e^{i\theta}})^k}{sin^k(\theta)}<br />

    <br />
S_n + i T_n = <br />
\sum_{k=0}^{n} \left({\frac{e^{i\theta}}{sin(\theta)}}\right)^k<br />
    Also

    <br />
S_n - i T_n = <br />
\sum_{k=0}^{n} \frac{cos(k\theta)-isin(k\theta)}{sin^k(\theta)}<br />

    <br />
S_n - i T_n = <br />
\sum_{k=0}^{n} \frac{e^{-ik\theta}}{sin^k(\theta)}<br />

    <br />
S_n - i T_n = <br />
\sum_{k=0}^{n} \frac{({e^{-i\theta}})^k}{sin^k(\theta)}<br />

    <br />
S_n - i T_n = <br />
\sum_{k=0}^{n} \left({\frac{e^{-i\theta}}{sin(\theta)}}\right)^k<br />

    Two geometric series!
    Last edited by Jester; January 14th 2009 at 10:31 AM.
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  6. #6
    Senior Member vincisonfire's Avatar
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    Quote Originally Posted by running-gag View Post
    <br />
   S_n + i T_n = <br />
   \sum_{k=0}^{n} \left({\frac{e^{i\theta}}{sin(\theta)}}\right)^k<br />
    I would use  \sum_{k=0}^{n} x^i = \frac{1-x^{n+1}}{1-x} and then take <br />
   S_n = <br />
   Re (\sum_{k=0}^{n} \left({\frac{e^{i\theta}}{sin(\theta)}}\right)^k)<br />
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  7. #7
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    Quote Originally Posted by vincisonfire View Post
    I would use  \sum_{k=0}^{n} x^i = \frac{1-x^{n+1}}{1-x} and then take <br />
   S_n = <br />
   Re (\sum_{k=0}^{n} \left({\frac{e^{i\theta}}{sin(\theta)}}\right)^k)<br />
    Yes
    You can also use the hint given by danny arrigo

    S_n = \frac{(S_n + i T_n) + (S_n - i T_n)}{2}

    I did not perform the job, so I don't know which is the easiest way !
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  8. #8
    Senior Member vincisonfire's Avatar
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    Answer is \frac{sin^{n+2}(x)-cos(x)sin^{n+1}(x)-cos((n+1)x)sin(x)+cos(nx)}{sin^n(x)(\frac{3}{2}-\frac{sin(2x+arctan(\frac{1}{2}))}{2sin(arctan(\fr  ac{1}{2}))})} and both ways of doing it are as long.
    To who it may interest. This is the form with the less "sin" and "cos" in it although it looks as complicated as the raw version.
    I replaced *sin^2(x)-2cos(x)sin(x)+1 by \frac{3}{2}-\frac{sin(2x+arctan(\frac{1}{2}))}{2sin(arctan(\fr  ac{1}{2}))}
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