## 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

November 20th, 2011 at 2:23 pm

7/6?

November 20th, 2011 at 2:43 pm

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).

November 22nd, 2011 at 1:57 am

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

November 22nd, 2011 at 6:47 am

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.

November 26th, 2011 at 7:47 pm

Letting “Re” be “the real part of”, then cos(nx) = Re(e

^{inx})So the original sum is Re(Sum(n = 0 to ∞, e

^{inx}/ 2^{n}))= Re(Sum(n = 0 to ∞, (e

^{ix}/2)^{n}))But 1 + z + z

^{2}+ z^{3}+ … = 1/(1 – z), for |z| < 1.So Sum … = 1/(1 – e

^{ix}/2) = 1/(1 – (cos(x) + i sin(x))/2).We are given that cos(x) = 1/5 => sin(x) = ±(2/5)√6

(I used sin

^{2}(x) + cos^{2}(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)/(x

^{2}+ y^{2})=> Sum… = (90 ± 20 i √6)/105. The real part is 90/105 = 6/7.