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A complex sum

Posted by Chris on November 20, 2011 – 1:58 pm

Evaluate Sum(n = 0 to ∞, cos(n x)/2^n), where cos(x) = 1/5.

Indirect hint, really an excuse to point out the remarkable proof of a distant cousin to the above problem, see: http://en.wikipedia.org/wiki/Gaussian_integral


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  1. 1. Carc Said:

    7/6?

  2. 2. Chris Said:

    Hi Carc. I’m going to assume you made a “typo” somewhere: it’s 6/7.

    I’ve just skimmed the link I gave. Sadly my memory of the proof was faulty. So, keeping my hint very indirect, see http://en.wikipedia.org/wiki/Methods_of_contour_integration#Example_.28III.29_.E2.80.93_trigonometric_integrals (I used to do that sort of stuff 40 years ago – it’s much easier than it looks).

  3. 3. slavy Said:

    Hi Chris. I think the hint is slightly misleading. For those who are working on the problem: Don’t try to come up with some fancy integral of a real-valued function and solve it as a (part of a) contour integral of a complex function! The only thing you need is actually this link http://en.wikipedia.org/wiki/Euler%27s_formula

  4. 4. Chris Said:

    Hi Slavy. I probably was being too indirect. All I was really trying to avoid saying is, “use Euler’s formula”. I’m dying to know if that’s what Carc did.

  5. 5. Chris Said:

    Letting “Re” be “the real part of”, then cos(nx) = Re(einx)
    So the original sum is Re(Sum(n = 0 to ∞, einx / 2n))
    = Re(Sum(n = 0 to ∞, (eix/2)n))

    But 1 + z + z2 + z3 + … = 1/(1 – z), for |z| < 1.
    So Sum … = 1/(1 – eix/2) = 1/(1 – (cos(x) + i sin(x))/2).
    We are given that cos(x) = 1/5 => sin(x) = ±(2/5)√6
    (I used sin2(x) + cos2(x) = 1, to get that)
    => Sum … = 1/(1 – ((1/5) ± i (2/5)√6)/2)
    => Sum… = 10/(10 -(1 ± 2 i √6)) = 10/(9 -± 2 i √6))

    Using 1/(x + iy) = (x – iy)/((x + iy)(x – iy)) = (x – iy)/(x2 + y2)
    => Sum… = (90 ± 20 i √6)/105. The real part is 90/105 = 6/7.

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