Periodic function (fourier expansion) help

[mooeypoo]

Helloo..

 

I have a function:

 

(have no clue how to write a { 'group' for an equation in LaTeX.. did my best with what I could find..):

[math]

f(\theta) = sin^2(\theta ) \text{ when } 0 \le \theta \le \pi

[/math]

[math]

f(\theta) = sin^2(\theta ) \text{ when } \pi < \theta \le 2\pi

[/math]

 

with values of the function given by periodicity for other domains of theta. Obtain the Fourier series expansion for f(\theta).

 

Okay, so we've only done this once in class on the board, and it was with a completely different equation - one without 'conditions' for that matter, just periodicity (a saw-tooth function).

 

I am having trouble doing this one.

 

Okay, the periodicity on this function is 2pi. So here's what I started with:

 

[math]

a_{n} = \frac{2}{2\pi} \int_{0}^{2\pi} F(\theta ') cos(n\theta) d\theta '

[/math]

[math]

b_{n} = \frac{2}{2\pi} \int_{0}^{2\pi} F(\theta ') sin(n\theta) d\theta '

[/math]

 

And so

[math]

a_{n} = 1/\pi \int_{0}^{2\pi} sin^2(\theta ) cos(n\theta) d\theta '

[/math]

[math]

b_{n} = 1/\pi \int_{0}^{2\pi} sin^2(\theta ) sin(n\theta) d\theta '

[/math]

 

I am not sure how to go from here. I had an idea that I need to check what happens to my cos() and sin() - so, the sin will equal 0 every even n (n=2,4,6,8...) because that would make the theta in it in the domain of 2pi, but that doesn't happen to all values of theta, so that doesn't help me much.

 

How do I expand this!? Anyone has any *useful* resources? I found some websites talking about Fourier Expansion but they don't quite explain, they give out a formula and state that this is the expansion - and my case the function is conditional... so.. I have no clue what to do.

 

Help!

 

Thanks

 

~moo

[D H]

I have a function:

[math]

f(\theta) = sin^2(\theta ) \text{ when } 0 \le \theta \le \pi

[/math]

[math]

f(\theta) = sin^2(\theta ) \text{ when } \pi < \theta \le 2\pi

[/math]

Hi, mooey. Could you write the definition of the function again? You gave the same function for 0 to pi and pi and 2*pi, and that doesn't quite make sense. Why not just say [math]f(\theta)=\sin^2 \theta[/math] and be done with it?

 

BTW, to write a series of equations so they line up nicely:

 

[math]

f(\theta) = \Bigl{\lbrace}

\aligned

\phantom{-}\sin^2 \theta &\quad \text{when $\theta \in [ 0,\pi ]$} \\

-\sin^2 \theta &\quad \text{when $\theta \in ( \pi,2\pi ]$}

\endaligned

[/math]

 

[math]

f(\theta) = \Bigl{\lbrace}

\aligned

\phantom{-}\sin^2 \theta &\quad \text{when $\theta \in [ 0,\pi ]$} \\

-\sin^2 \theta &\quad \text{when $\theta \in ( \pi,2\pi ]$}

\endaligned

[/math]

 

 

 

Note: "Real" LaTeX, as opposed to the scaled-down version available here, gives a lot more capability.

[mooeypoo]

oh, shoot! sorry

 

[math]

f(\theta) = \Bigl{\lbrace}

\aligned

\sin^2 \theta &\quad \text{when $\theta \in [ 0,\pi ]$} \\

0 \phantom{\sin^2} &\quad \text{when $\theta \in ( \pi,2\pi ]$}

\endaligned

[/math]

 

(ha! fixed alignment thanks D H for the sample

 

Sorry 'bout that, the function is 0 in the second condition, and then periodical over other domains.

[D H]

OK then. The integrals are to be taken over one full period, here, zero to 2*pi. However, the function in question is identically zero over the latter half of the interval. The obvious (and correct) thing to do is to just take the integral from 0 to pi (but obviously keep the original scale factor):

 

[math]

\aligned

a_n &= \frac 1 {\pi} \int_0^{\pi}\sin^2\theta \cos n\theta \,d\theta \\

b_n &= \frac 1 {\pi} \int_0^{\pi}\sin^2\theta \sin n\theta \,d\theta

\endaligned

[/math]

 

Note: There is no [math]\theta^{\prime}[/math] stuff in the integrals. I believe that is the source of your confusion. All of the parameters in the integral are the dummy variable [math]\theta[/math]. The a_n and b_n are just numbers derived from the function f. They are not functions, at least not in this case.

 

What you need to do is to use some trig substitutions to evaluate those integrals.