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Math Help - find formulas

  1. #1
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    find formulas

    Find closed formulas for the following two sums:
    1 + cos
    X + cos 2X + . . . + cos nX

    sin
    X + sin 2X + . . . + sinnX

    Hint: Use the sum of the geometric series with the general term eikX.
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  2. #2
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    Quote Originally Posted by ohadch View Post
    Find closed formulas for the following two sums:



    1 + cosX + cos 2X + . . . + cos nX

    sin
    X + sin 2X + . . . + sinnX
    Hint: Use the sum of the geometric series with the general term eikX.
    The given hint is a huge one:


    \sum\limits_{k=0}^ne^{kxi}=\frac{e^{(n+1)xi}-1}{e^{xi}-1}

    Now just use that e^{ir}=\cos r+i\sin r\;\;\forall\,r\in \mathbb{R} , plug in and separate real and imaginary parts.

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    The given hint is a huge one:


    \sum\limits_{k=0}^ne^{kxi}=\frac{e^{(n+1)xi}-1}{e^{xi}-1}

    Now just use that e^{ir}=\cos r+i\sin r\;\;\forall\,r\in \mathbb{R} , plug in and separate real and imaginary parts.


    Tonio
    Hi Tonio,
    Thanks for the help, I did what you said and I got the sum of both series:
    by proving its a geometric series ...

    \sum= \frac{cos[(n+1)x] + isin[(n+1)x] -1}{cosx + isinx -1}

    What do you mean by separating real and imaginary parts, How can I do it here? and will it give me the sum of the original series:
    sinx+sin2x+...+sinnx , even if i changed it to a kind of a complex seris?

    Thanks,
    Ohad.
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  4. #4
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    Quote Originally Posted by ohadch View Post
    Hi Tonio,
    Thanks for the help, I did what you said and I got the sum of both series:
    by proving its a geometric series ...

    \sum= \frac{cos[(n+1)x] + isin[(n+1)x] -1}{cosx + isinx -1}

    What do you mean by separating real and imaginary parts, How can I do it here? and will it give me the sum of the original series:
    sinx+sin2x+...+sinnx , even if i changed it to a kind of a complex seris?

    Thanks,
    Ohad.

    It will give you the sum of BOTH series, cosines and sines:

    \frac{\cos (n+1)x+i\sin (n+1)x-1}{\cos x+i\sin x-1} =\frac{(\cos (n+1)x-1)+i\sin (n+1)x}{(\cos x-1)+i\sin x}\,\frac{(\cos x-1)-i\sin x}{(\cos x-1)-i\sin x}=

    =\frac{\cos (n+1)x(\cos x-1)+\sin (n+1)x\sin x}{(\cos x-1)^2+\sin^2x} -\frac{\cos (n+1)x\sin x-\sin (n+1)x(\cos x-1)}{(\cos x-1)^2+\sin^2x}\,i

    And there you have the right side divided in real and imaginary part. Now use a little trigonometry to deduce that the real part is \frac{\cos nx-\cos (n+1)x}{2(1-\cos x)} ,

    and the imaginary part is \frac{\sin nx - \sin (n+1)x}{2(1-\cos x)}.

    OTOH, in the left side we have the sum \sum\limits_{k=1}^ne^{kxi}=\sum\limits_{k=0}^n(\co  s kx+i\sin kx) , so again separate in real and imaginary parts this sum.

    Tonio
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