0
$\begingroup$

Use $ 2\cos{n\theta} = z^n + z^{-n} $ to express $\cos\theta + \cos3\theta + \cos5\theta + ... + \cos(2n-1)\theta $ as a geometric progression in terms of $z$. Hence find the sum of this progression in terms of $\theta$. Any tips/help would be appreciated. I have the common ratio as $z^2$ and the first term as $z^{1-2n},$ and I can put these into the original formula, but I can't seem to get the answer I'm looking for.

z= $\cos\theta + i\sin\theta$ where i is the imaginary unit

$\endgroup$
  • $\begingroup$ To get multicharacter exponents, put them in braces. So z^{-n} gives $z^{-n}$, while z^-n gives $z^-n$. Also a backslash before cos will give you the proper font: $\cos$ instead of $cos$ $\endgroup$ – Ross Millikan Oct 9 '13 at 16:07
  • $\begingroup$ It's a bit easier to use $z = e^{i \theta}$, sure that you don't want to use that? Otherwise the solution gets really ugly really fast. $\endgroup$ – DanielV Oct 9 '13 at 16:45
1
$\begingroup$

$\begin{align} answer & = \sum_{i = 1}^{n} {\cos(\theta (2i - 1))} \\ & = \sum_{i = 0}^{n - 1} {\cos(\theta (2i + 1))} \\ & = \sum_{i = 0}^{n - 1} {\frac{z^{2i + 1} + z^{-2i - 1}}{2}} \\ & = \frac z 2 \sum_{i = 0}^{n - 1} {z^{2i} + \frac 1 {2z} \sum_{i = 0}^{n - 1} z^{-2i}} \\ & = \frac z 2 \text{geometric sum in } z^2 + \frac 1 {2z} \text{geometric sum in }z^{-2} \\ & = \frac z 2 \frac {z^{2n} - 1} {z^2 - 1} + \frac 1 {2z} \frac {z^{-2n} - 1} {z^{-2} - 1}\\ & = \frac z 2 \frac {z^{2n} - 1} {z^2 - 1} + \frac 1 {2z} \frac {z^{-2n} - 1} {z^{-2} - 1} \frac {z^2}{z^2} \\ & = \frac z 2 \frac {z^{2n} - 1} {z^2 - 1} + \frac z 2 \frac {z^{-2n} - 1} {1 - z^{2}}\\ & = \frac z 2 \frac {z^{2n} - z^{-2n}} {z^2 - 1} \end{align}$

$\endgroup$
0
$\begingroup$

$$ z^{1-2n} + z^{1-2n + 2} + \cdots + z^{2n-1} = z^{1-2n} \left(1 + z^2 + \cdots + z^{4n-2}\right) = z^{1-2n} \frac{1-z^{4n-2}}{1 - z^2}.$$

$\endgroup$
0
$\begingroup$

Regroup sum into two geometric progressions: $$2\left(\cos\theta + \cos3\theta + \cos5\theta + ... + \cos(2n-1)\theta\right)\\ =z+\dfrac{1}{z}+z^3+\dfrac{1}{z^3}+\ldots+z^{2n-1}+\dfrac{1}{z^{2n-1}}\\= z+z^3+\ldots+z^{2n-1}+\dfrac{1}{z}+\dfrac{1}{z^3}+\ldots+\dfrac{1}{z^{2n-1}}$$

$\endgroup$
  • $\begingroup$ This is helpful, and I can get the formula for the sum of the progression, but I still can't manage to express the sum in terms of theta? $\endgroup$ – Arkel Oct 9 '13 at 17:05
  • $\begingroup$ Remember that $z=e^{i\theta}$ $\endgroup$ – M. Strochyk Oct 9 '13 at 17:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.