Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am currently prepping for uni having been a few years out of the studying loop (programming as it happens). Anyway, I've been re-reading my A-level notes/exercises and looking through OpenCourseWare stuff and I noticed this assertation in my notes:

Given $a \cos\theta \pm b \sin\theta = c$ then:

$a \sin\theta \pm b\cos\theta = r\sin(\theta\pm\alpha)+c$

$a \cos\theta \pm b\sin\theta = r\cos(\theta\pm\alpha)+c$

Given $r \in \mathbb{N}$, $0 \leq \theta < 2\pi$.

Now, I've heard of fourier series which have a very similar form to these equestions.

So, my question is, is there a relation between the two?

Please bear in mind I know I'm stepping off a cliff and into "unknown unknowns" territory if they are - I have absolutely no idea what harmonic analysis is and I don't (yet) understand fourier series fully, although I grasp roughly how they work. I'm just interested to know if they are related and of course any further information / direction / reading I should follow up. I ask because this was one of those "odd topics" at A-level that we derived in class from a geometric argument then I've never seen again.

Thanks.

share|improve this question
    
I'd only comment that you have to read up on de Moivre's formula $(\cos(t)+i\sin(t))^n=\cos(nt)+i\sin(nt)$ and Euler's formula $\exp(it)=\cos(t)+i\sin(t)$. –  J. M. Sep 2 '10 at 21:59
    
The assertion is unclear. What are those supposed to mean? –  Aryabhata Sep 2 '10 at 22:27
    
You want to know if there is a relation between Fourier series and what, exactly? –  Mariano Suárez-Alvarez Sep 2 '10 at 22:49

1 Answer 1

up vote 5 down vote accepted

I can't really make sense of those equations, but it looks something like the statement that for any $ a, b \in \mathbb{R} $, then for $ r = \sqrt{a^2 + b^2 }$ and $ \phi = \arctan(b/a) $ we have

$ a \sin(\theta) + b \cos(\theta) = r \sin(\theta + \phi).$

If you visualize x and y axes with $ cos(\theta) $ along the x-axis and $ sin(\theta) $ along the y-axis, then this says we can identify any sinusoidal, i.e. any function like $ r \sin(\theta + \phi) $, by its cartesian coordinates $(a,b)$ in this plane, or its polar coordinate $(r, \phi )$. $r$ is called the magnitude of the sinusodial, and $ \phi $ the phase.

This is directly related to Fourier Analysis. The main idea in Fourier Analysis is to decompose a function into its sinusodial components. For instance, if $f$ is a real-valued function on the interval $ [0, 2\pi] $ that is suitably regular, it turns out you can write $f$ uniquely as an infinite sum of sine and cosines, called the Fourier Series:

$ f(x) = a_0 + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx)). $

The numbers $ a_n $ and $ b_n $ are called the Fourier coefficients of your function.

I think this representation of Fourier series is sometimes difficult to grasp. But using the relationship above, we see we could have also wrote:

$ f(x) = a_0 + \sum_{n=1}^{\infty} r_n \sin(nx + \phi_n) $

In this form I think you can see more clearly what's going on: For each $n$, there is only one sinusoidal component of $f$ of frequency $n$; $r_n = \sqrt{a_n^2 +b_n^2}$ tells you its magnitude and $\phi_n$ tells you its phase.

share|improve this answer
    
You need to enclose latex inside dollar signs like this: $x^n + y^n = z^n$ which looks like $x^n + y^n = z^n$. –  Aryabhata Sep 3 '10 at 2:08
    
@Moron. Thanks for the tip. Should be fixed now. –  Greg O. Sep 3 '10 at 2:51
    
That makes sense. The constraints for $r$ and $\varphi$ don't exist in my notes but the basis of the work was to do the manipulations I assume in preparation for exactly what you've detailed above. Thanks for the answer, +1. –  Ninefingers Sep 3 '10 at 19:30

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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