Fourier series expansion of f(x)=x

Sooz · #1 $Old$ May 15th, 2008, 11:33 AM

Hi, I'm doing a course on PDEs and we have started looking at Fourier series. I am revising and 2 questions have come up that I am stuck on. One wants the Fourier series expansion of the extension of f(x)=x , -pi<x<pi, and the other the same but with f(x)=|x|.
I can start both but then I get stuck. I know that I need to use complex Fourier series and that you start by finding the Fourier coefficient f^(n) (thats f with a hat, not f to the power n) using integration by parts. This is fine. Its when you have to put everything together at the end and for some reason use f^(0) that I start getting confused. If anyone could help me solve these particualr questions or just give me a general step-by-step guide that would be great. This is what I have so far....(im not very good with the inserting of the symbols so bear with me!)

For f(x)=x, -pi<x<pi
f(x) = 1/(2*pi)*sum(from n= - infinity to infinity) f^(n)*e^(i*n*x) and
f^(n) = integral (between -pi and pi) f(x)*e^(-i*n*x)
I can substitute an x in wherever there is an f(x) and then integrate f^(n) by parts to eventually get f^(n) = (2*pi*((-1)^n)*i)/n
From here I am stuck!
Firstly am I right so far? If so, what do I do next?
Thanks,
Sooz

ThePerfectHacker · #2 $Old$ May 15th, 2008, 12:33 PM

Quote:

Originally Posted by Sooz $View Post$

Hi, I'm doing a course on PDEs and we have started looking at Fourier series. I am revising and 2 questions have come up that I am stuck on. One wants the Fourier series expansion of the extension of f(x)=x , -pi<x<pi

The Fourier Series is given by, $\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n \sin \pi n x+ b_n \cos \pi nx$ .

Where $a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi} x dx = 0$ and $b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} x\cos (\pi nx) dx = 0$ because it is odd function.

And $a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} x \sin \pi n x dx$ .
It remains for you to compute that $a_n$ 's.

Sooz · #3 $Old$ May 15th, 2008, 02:19 PM

Thanks,
I'm just a bit confused as my lecturer said we have to use the complex form of Fourier series. So if I take what you put, is it easy to convert it into the form that I need to use?
Thanks again

Opalg · #4 $Old$ May 15th, 2008, 02:35 PM

Quote:

Originally Posted by Sooz $View Post$

For f(x)=x, -pi<x<pi
f(x) = 1/(2*pi)*sum(from n= - infinity to infinity) f^(n)*e^(i*n*x) and
f^(n) = integral (between -pi and pi) f(x)*e^(-i*n*x)
I can substitute an x in wherever there is an f(x) and then integrate f^(n) by parts to eventually get f^(n) = (2*pi*((-1)^n)*i)/n
From here I am stuck!
Firstly am I right so far? If so, what do I do next?

You're right so far. The formula $\hat{f}(n) = 2\pi(-1)^ni)/n$ is correct, except that it doesn't work when n=0, because you can't divide by 0. That's why you usually have to calculate $\hat{f}(0)$ separately. For the function f(x)=x, you get $\hat{f}(0)=0$ .

All that remains to do is to write down the (complex) Fourier series as $x\sim\frac1{2\pi}\sum_{n=-\infty}^\infty\hat{f}(n)e^{inx}$ . To get round the problem of incorporating the n=0 term into the sum, it's probably easiest to split the sum into two parts, one going from –∞ to –1, and the other one from 1 to ∞. Then

$x\sim\sum_{n=1}^\infty\Bigl(\frac{(-1)^ni}{-n}e^{-inx} + \frac{(-1)^ni}{n}e^{inx}\Bigr) = \sum_{n=1}^\infty\frac{(-1)^ni(e^{inx}-e^{-inx})}n$ .

(This can be written as $\sum_{n=1}^\infty\frac{2(-1)^{n-1}\sin nx}n$ , which is the same as you would get by calculating the real Fourier series. The complex series is not different from the real series, but it's often easier to calculate because integrals with exponentials tend to be more convenient than trigonometrical ones. Also, you only have one integral to deal with, instead of two separate ones for the sines and the cosines.)

For the other function, f(x) = |x|, the method is the same, but you need to split the integral into two intervals, from –π to 0, and from 0 to π. On the first interval, |x|=–x and on the second interval |x|=x. So the formula for the Fourier coefficients becomes $\hat{f}(n) = \int_{-\pi}^0\!(-x)e^{-inx}dx + \int^{\pi}_0\!\!xe^{-inx}dx$ (and again, you'll have to deal with the case n=0 separately).

Sooz · #5 $Old$ May 16th, 2008, 04:02 AM

Thanks so much for your help! I really appreciate it!
Its all making sense now!!!
Sooz

dsprice · #6 $Old$ November 6th, 2009, 06:04 PM

Quote:

Originally Posted by Opalg $View Post$

You're right so far. The formula $\hat{f}(n) = 2\pi(-1)^ni)/n$ is correct...

At the risk of sounding slow, which has never stopped me before, could someone please show step by step how using partial fractions ended up with this equation? I have tried repeatedly to reproduce, but I feel like I'm missing one key, simple, step that would make it all seem clear.

Thanks in advance!

tonio · #7 $Old$ November 7th, 2009, 04:51 AM

Quote:

Originally Posted by dsprice $View Post$

At the risk of sounding slow, which has never stopped me before, could someone please show step by step how using partial fractions ended up with this equation? I have tried repeatedly to reproduce, but I feel like I'm missing one key, simple, step that would make it all seem clear.

Thanks in advance!

$\int\limits_{-\pi}^{\pi}xe^{-inx}\,dx=\left[\frac{ixe^{-inx}}{n}\right]_{-\pi}^{\pi}+\frac{i}{n}\int\limits_{-\pi}^{\pi}e^{-inx}\,dx$ (remember that $\frac{1}{i}=-i$ ) =

$\frac{\pi i}{n}(2\cos n\pi)-\frac{1}{n^2}(e^{-n\pi i}-e^{n\pi i})=\frac{\pi i}{n}(-1)^n\cdot 2+\frac{1}{n^2}\,2\sin n\pi=\frac{2\pi(-1)^ni}{n}$

Tonio

dsprice · #8 $Old$ November 7th, 2009, 08:06 AM

Quote:

Originally Posted by tonio $View Post$

(remember that $\frac{1}{i}=-i$ )

This is precisely the part that I was missing! Thanks!

billym · #9 $Old$ November 7th, 2009, 08:55 AM

Quote:

Originally Posted by ThePerfectHacker $View Post$

The Fourier Series is given by, $\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n \sin \pi n x+ b_n \cos \pi nx$ .

Where $a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi} x dx = 0$ and $b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} x\cos (\pi nx) dx = 0$ because it is odd function.

And $a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} x \sin \pi n x dx$ .
It remains for you to compute that $a_n$ 's.

For this I just did:

$a_0=a_n=0$

$b_n=\frac{2}{\pi}\int_{0}^{\pi} xsin(nx)dx$

$S(x)=\sum_{n=1}^{\infty}b_n sin(nx)$

No good?