3
$\begingroup$

I just wanted to evaluate

$$ \sum_{k=0}^n \cos k\theta $$

and I know that it should give

$$ \cos\left(\frac{n\theta}{2}\right)\frac{\sin\left(\frac{(n+1)\theta}{2}\right)}{\sin(\theta / 2)} $$

I tried to start by writing the sum as

$$ 1 + \cos\theta + \cos 2\theta + \cdots + \cos n\theta $$

and expand each cosine by its series representation. But this soon looked not very helpful so I need some clue about how this partial sum is calculated more efficiently ...

  • 3
    Whenever you already know the answer in terms of a natural number n, the first thought to prove the assertion should always be induction. In this case, it solves it rather simply. Try it.2012-09-06
  • 0
    Wait, you wrote the expanded sum with a 1 at the beginning, but there's no $k=0$ term in the sigma form...2012-09-07
  • 0
    Thanks @BenMillwood, the index k should start from zero of course ...2012-09-07
  • 1
    See also http://math.stackexchange.com/questions/225941/proving-sum-limits-k-0n-coskx-frac12-frac-sin-frac2n12x and the questions linked there.2014-02-21
  • 0
    Write it as the real part of summation of $e^{ik\theta}$2016-03-24

2 Answers 2

10

Use: $$ \sin \frac{\theta}{2} \cdot \cos(k \theta) = \underbrace{\frac{1}{2} \sin\left( \left(k+\frac{1}{2}\right)\theta\right)}_{f_{k+1}} - \underbrace{\frac{1}{2} \sin\left( \left(k-\frac{1}{2}\right)\theta\right)}_{f_{k}} $$ Thus $$ \begin{eqnarray} \sin \frac{\theta}{2} \cdot \sum_{k=1}^n \cos(k \theta) &=& \sum_{k=1}^n \left(f_{k+1} - f_k\right) = f_{n+1}-f_1 \\ &=& \frac{1}{2} \underbrace{\sin\left(\left(n+\frac{1}{2}\right)\theta\right)}_{\sin(\alpha+\beta)}-\frac{1}{2} \underbrace{\sin \frac{\theta}{2}}_{\sin(\alpha-\beta)} = \cos(\alpha) \sin(\beta) \\ &=& \cos\left(\frac{n+1}{2}\theta\right) \sin\left(\frac{n}{2} \theta\right) \end{eqnarray} $$ where $\alpha = \frac{n+1}{2} \theta$ and $\beta = \frac{n}{2} \theta$.

  • 0
    Ah thanks Sasha, I almost suspected that this involves some nifty trigonometric tricks ...2012-09-06
5

Note that $\cos(n\theta) = \Re({e^{in\theta}})$. Thus, our sum can be thought of as $\Re(\sum_{n = 0}^{N}{e^{in\theta}})$. Now, $$\sum_{n = 0}^N{e^{in\theta}} = \frac{e^{i(N+1)\theta}-1}{e^{i\theta} - 1}.$$ Now $$\frac{e^{i(N+1)\theta}-1}{e^{i\theta} - 1} = \frac{e^{i(N+1)\theta/2}}{e^{i\theta/2}}\frac{e^{i(N+1)\theta/2} - e^{-i(N+1)\theta/2}}{e^{i\theta/2} - e^{-i\theta/2}}.$$ Take the real part of the right hand side of the equality and simplify and you will get the result you want. The tricky part is knowing to break up $\frac{e^{i(N+1)\theta}-1}{e^{i\theta} - 1}$ as we did in the above equation. Let me know if you get stuck or don't understand.

  • 0
    That's one way of doing it... why not just multiply by the complex conjugate of the denominator instead?2012-09-07
  • 0
    Also, I'm not sure your geometric formula is quite right. I think your result is a factor of $e^{i\theta}$ short.2012-09-07
  • 0
    And the next line is wrong as well – that numerator is zero! I started off thinking this was good, but now I'm downvoting it for containing an arithmetic error in pretty much every equation.2012-09-07
  • 0
    Hi Ben. I'm sorry for the mistakes. I was really sloppy with the derivation because i wrote this fairly quickly. I hope this is right now.2012-09-07
  • 1
    It still has mistakes and typos. You should have $\mathrm{e}^{i k \theta}$ instead of $\mathrm{e}^{i n \theta}$ in the sum. Moreover the correct evaluation of the sum is: $$\sum_{k=1}^N \exp(i k \theta) = \frac{\exp(i (N+1)\theta) - \exp(i \theta)}{\exp(i \theta)-1}$$2012-09-07
  • 0
    I changed the index notation from $k$ to $n$ and the first term from $0$ to $1$. What I have now should be correct.2012-09-07
  • 0
    Okay, I changed my downvote to an upvote – I like the method in general.2012-09-07