# Series

• Jan 1st 2008, 04:25 PM
slevvio
Series
Show that, for all $\theta \in \mathbb{R}$,

$\frac{cos\theta}{2} + \frac{cos3\theta}{2^2} + \frac{cos5\theta}{2^3} + \ldots = \frac{cos\theta}{1 + 8sin^2\theta}$.

I understand that this has something to do with a geometric series however taking $\frac{a_2}{a_1}$ and $\frac{a_3}{a_2}$ gave me different values... any help with this would be appreciated... thanks a lot and happy new year to all.
• Jan 1st 2008, 04:39 PM
topsquark
Quote:

Originally Posted by slevvio
Show that, for all $\theta \in \mathbb{R}$,

$\frac{cos\theta}{2} + \frac{cos3\theta}{2^2} + \frac{cos5\theta}{2^3} + \ldots = \frac{cos\theta}{1 + 8sin^2\theta}$.

I understand that this has something to do with a geometric series however taking $\frac{a_2}{a_1}$ and $\frac{a_3}{a_2}$ gave me different values... any help with this would be appreciated... thanks a lot and happy new year to all.

Have you looked at a Fourier series for $\frac{cos(\theta)}{1 + 8sin^2(\theta)}$? (Use the sine and cosine form, not the complex exponential.)

-Dan
• Jan 1st 2008, 08:02 PM
ThePerfectHacker
Quote:

Originally Posted by slevvio
Show that, for all $\theta \in \mathbb{R}$,

$\frac{\cos\theta}{2} + \frac{\cos3\theta}{2^2} + \frac{\cos5\theta}{2^3} + \ldots = \frac{\cos\theta}{1 + 8\sin^2\theta}$.

Why is this in the algebra sub-forum??

Note that $\cos z = \frac{1}{2} \left( e^{iz} + e^{-iz} \right)$.

That means the series becomes,
$\sum_{n=0}^{\infty} \frac{\frac{1}{2}\left( e^{i(2n+1)\theta} + e^{-i(2n+1)\theta} \right)}{2^{n+1}} = \sum_{n=0}^{\infty} \frac{e^{i(2n+1)\theta}}{2^{n+2}} + \sum_{n=0}^{\infty}\frac{e^{-i(2n+1)\theta}}{2^{n+2}}$

Now, $\sum_{n=0}^{\infty} \frac{e^{i(2n+1)\theta}}{2^{n+2}} = \frac{e^{i\theta}}{2^2} \sum_{n=0}^{\infty} \frac{e^{2ni\theta}}{2^n} = \frac{e^{i\theta}}{2^2} \sum_{n=0}^{\infty} \left( \frac{e^{2i\theta}}{2} \right)^n = \frac{e^{i\theta}}{2^2} \cdot \frac{2}{2 - e^{2i\theta}}$

Similarly, $\sum_{n=0}^{\infty} \frac{e^{-i(2n+1)\theta}}{2^{n+2}} = \frac{e^{-i\theta}}{2^2}\cdot \frac{2}{2-e^{-2i\theta}}$

Thus series sums to, $\frac{e^{i\theta}}{4 - 2e^{2i\theta}} + \frac{e^{-i\theta}}{4- 2e^{-2i\theta}}$

Now, $\frac{e^{i\theta}}{4 - 2e^{2i\theta}} + \frac{e^{-i\theta}}{4- 2e^{-2i\theta}} = \frac{(4 - 2e^{-2i\theta})e^{i\theta} + (4 - 2e^{2i\theta})e^{-i\theta}}{(4-2e^{2i\theta})(4-2e^{-2i\theta})}$

Which becomes, $\frac{2e^{i\theta} + 2e^{-i\theta}}{2^2(2-e^{2i\theta})(2-e^{-2i\theta})} = \frac{e^{i\theta}+e^{-i\theta}}{2[5 - 2(e^{2i\theta}+e^{-2i\theta})]}$

Use identity above, $\frac{2\cos \theta}{10 - 8\cos 2\theta} = \frac{\cos \theta}{5 - 4\cos 2\theta}$

Double angle, $\frac{\cos \theta}{5 - 4[1 - 2\sin^2 \theta]}$

And we finally (Whew) get, $\frac{\cos \theta}{1 + 8\sin^2 \theta}$. :D