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I am trying to establish the following partial sum equivalence: $$\sum_{n=1}^{N}\cos (n\pi u) = \frac{1}{2}\left[\frac{\sin(N+\frac{1}{2})\pi u}{\sin(\pi u)/2} - 1\right]$$

with a hint that the series $$x + x^2 + x^3 + \ldots x^m= \frac{1-x^{m+1}}{1-x}$$

My approach is to represent $$\cos(n\pi u) = \frac{1}{2}\left[e^{in\pi u} + e^{-in\pi u}\right]$$ and then letting $x = e^{i\pi u}$ I can reduce this to $$\cos(n\pi u) = \frac{1}{2}\left[x + x^2 + \ldots x^{N} + x^{-1} + x^{-2} + \ldots x^{-N}\right]$$ Now, I notice that the first and last terms sum to $$x + \frac{1}{x^N} = \frac{x^N + 1}{x^N}$$ likewise, the second and the second-to-last sum to $$x^2 + \frac{1}{x^{N-1}} = \frac{x^N + 1}{x^{N-1}}$$ so that the first sum is $x$ multiplied by the second. Now let $$a = \frac{1 + x^N}{x^N}$$ and we reduce this series to $$\cos(n\pi u) = \frac{1}{2}\left[a\left(1 + x + x^2 + \ldots x^{N-1}\right)\right]$$ which is equivalent to $$a\left[\frac{x}{1-x}\right]$$ Then, using the value of $a$, I get $$\cos(n\pi u) = \frac{1}{2}\left[\frac{x^2}{1-x}\right]$$

I am, however, not sure if I have proceeded in the right direction. Thanks

Victoria
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  • It isn't $\cos(\pi u)$ that is equal to that sum of powers of $x$, it is the sum of the cosines, i;e; the expression you are trying to evaluate. What I would do next is add $1$ to the sum, we get two geometric series. Sum them with the finite geometric series formula. At the end (peek at the formula on the right) you will be subtracting that $1$ you just added. – André Nicolas Oct 10 '15 at 04:02
  • See the method I applied in my answer here : http://math.stackexchange.com/questions/986835/finding-the-sum-of-sin0-circ-sin1-circ-sin2-circ-cdots-sin/986860#986860 – Spencer Oct 10 '15 at 05:15
  • consider $$\bigg(\sum_{n=-N}^{N}e^{i\pi nu}\bigg) -1$$ and use the geometric sum formula – Gappy Hilmore Oct 10 '15 at 06:04
  • I appreciate this hint very much, I was able to solve it using this, and liberal doses of Wolfram Alpha for trig identities. Thanks for the idea! – Victoria Oct 11 '15 at 01:32

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