Math stuff and other things...but mostly math.: Trigonometry and complex numbers (they're related???)

Hopefully you know that they're related. Anyway, some random stuff. The AIME class is pretty good. This week focused on solving difficult equations. The class is mostly strategies for solving problems. I have to admit that it helped, most of the practice problems we were given afterwards I probably wouldn't have been able to do before the class.

School has a couple weeks left. Really not much else to say, nothing to rant about. So I guess I'll go right to the problem.

$\text{Evaluate:} \sum_{n=1}^{\infty} \frac{\cos{(n\theta})}{2^n}$

$\text{Given that } \cos{\theta} = 1/5$

Ooomph. I mentioned complex numbers in the title, and I'm going to incorporate them. When you first look at this, complex numbers are most definitely NOT what come to mind immediately. After all, theres not a single square-root-of-negative-one in there. however, the cos and the n inside of the function should eventually tip you off. I'll show you why first.

First off, rewrite that with $\frac{1}{2^n} = (\frac{1}{2})^n$ . It'll make sense soon.

Now consider the cis function. You should be familiar with that, its the polar representation of complex numbers, and comes up a lot when dealing with them. If you aren't familiar with it, then you probably won't have much luck when looking at complex number problems (With some exceptions, of course. Then again, if you aren't, then I doubt you'll be able to do the harder problems which don't use it).

$r\mathrm{cis }(\theta)=r\cos \theta + ir\sin \theta$

If you're familiar with powers of complex numbers, then you also know that:

$(r\mathrm{cis}(\theta))^n = r^n\mathrm{cis}(n\theta)= r^n\cos{(n\theta)} + ir^n\sin{(n\theta)}$

Now you should see that the real element of this is very similar to our expression, and if r = 1/2, then it IS our expression. This means that we can rewrite our summation as:

$\sum_{n=0}^{\infty}(\mathrm{Re}(\frac{1}{2}\mathrm{cis}{\theta})^n)$

$= \mathrm{Re}(\sum_{n=0}^{\infty}(\frac{1}{2}\mathrm{cis}{\theta})^n)$

The summation is the definition of an infinite geometric series, so we can rewrite it as:

$\frac{1}{1 - \frac{1}{2}\mathrm{cis}{\theta}}$

We know how to express cis, so we just need to find sin of theta to complete it, using the trigonometric identity:

$cos^2{\theta} + sin^2{\theta} = 1$

$(\frac{1}{5})^2 + sin^2{\theta} = 1$

$\frac{1}{25} + sin^2{\theta} = 1$

$sin^2{\theta} = \frac{24}{25}$

$sin{\theta} = \sqrt{\frac{24}{25}} = \frac{\sqrt{24}}{5} = \frac{2\sqrt{6}}{5}$

Now we can find our infinite series:

$\frac{1}{1 - \frac{1}{2}(\frac{1}{5} + \frac{2\sqrt{6}}{5}i)}$

$= \frac{1}{1 - \frac{1}{10} - \frac{2\sqrt{6}}{10}i}$

$= \frac{1}{\frac{9}{10} - \frac{2\sqrt{6}}{10}i}$

$= \frac{1}{\frac{9}{10} - \frac{2\sqrt{6}}{10}i} \cdot \frac{\frac{9}{10} + \frac{2\sqrt{6}}{10}i}{\frac{9}{10} + \frac{2\sqrt{6}}{10}i}$

$= \frac{\frac{9}{10} + \frac{2\sqrt{6}}{10}i}{(\frac{9}{10})^2 + (\frac{2\sqrt{6}}{10})^2}$

$= \frac{\frac{9}{10} + \frac{2\sqrt{6}}{10}i}{\frac{81}{100} + \frac{24}{100}}$

$= \frac{\frac{9}{10} + \frac{2\sqrt{6}}{10}i}{\frac{105}{100}}$

$= \frac{\frac{9}{10} + \frac{2\sqrt{6}}{10}i}{\frac{21}{20}}$

$= \frac{20}{21}(\frac{9}{10} + \frac{2\sqrt{6}}{10}i})$

$= (\frac{20}{21})(\frac{9}{10}) + (\frac{20}{21})(\frac{2\sqrt{6}}{10}i)$

We only want the real element of this, so you should see that there's no point in calculating the unreal part now.

$Re = (\frac{20}{21})(\frac{9}{10}) = \frac{180}{210} = \frac{18}{21} = \frac{6}{7}$

And theres the answer.

I'd like to take a moment to talk about the problem solving aspect of this. I'll admit that I had a hint, it was in the "Complex Numbers" section of my book, so I knew I would be using those to solve it. But I could see somebody looking at this and trying to use trig identities to figure it out. Obviously you can't do this up till infinity.

The main part of solving this consisted of recognizing that it was the real element of a series of complex powers. Once that was figures out, then the rest was simple. However, that first step is usually the hardest part of solving a problem.

We can learn how to solve generic types of problems, and how to apply these skills to slightly different problems. However, solving problems which we've never seen before is usually a lot harder. This was something we discussed in our first AIME class:

"It is one thing to understand a few methods for solving equations -- even odd equations. It is another thing to creatively manipulate equations into forms that we can apply these methods to. When examining a difficult or unusual equation we should think about what kind of equation we have and what similar kinds of equations we know we can solve."

Applying this to the previous problem, recognizing that the expression was similar to powers of cis was key.