# Fourier Series

• Apr 20th 2013, 05:39 PM
shumaterm
Fourier Series
Hey, I have some questions from my Differential class that I am really confused on. I still don't completely understand Fourier, but I really don't understand this. I don't want the answers, just someone to explain how to go about doing these. Thanks!

Problem 1

Let $f(t)=t^2$ for $0 and suppose $f(t+2)=f(t)$ for all $t \epsilon R-2Z$. By $2Z$, we denote even integers. Define $f(2k)=1$ for $k \epsilon Z$. Calculate the Fourier Series for $f$.

Problem 2

Show, by the result of the last problem, $\sum_{n=1}^\infty \frac{\sin{nt}}{n}=\frac{\pi-t}{2}$. Derive a series which converges to $\pi$ by setting $t$ to a particular value and doing a little algebra. How many terms in this series would you have to sum in order to obtain the digits 3.14 in $\pi$ with certainty?
• Apr 20th 2013, 07:46 PM
chiro
Re: Fourier Series
Hey shumaterm.

For problem 1, calculate a_n and b_n using the standard formula's and show us what you got as an answer.
• Apr 21st 2013, 01:37 PM
shumaterm
Re: Fourier Series
would $a_0$ be 1 and $a_n$ be $\frac{4(\pi n \sin{\pi n} + \cos{\pi n}-1)}{\pi^2n^2}$? I know the general forumla for $t\epsilon(-\pi,\pi)$, but I'm not sure what happens when the interval changes to a different period... Thank you for your help. I really do appreciate it.
• Apr 21st 2013, 01:40 PM
shumaterm
Re: Fourier Series
Oops. also, in problem 1, $f(t)=t$, not $f(t)=t^2$.
• Apr 21st 2013, 04:43 PM
chiro
Re: Fourier Series
Its going to be a lot better if you setup the integral.

Hint: The Integrals will involve sin(pi*n*t)*f(t)dt and cos(pi*n*t)*f(t)dt as well as f(t)dt. (You will have to normalize them as well).
• Apr 22nd 2013, 08:56 AM
shumaterm
Re: Fourier Series
So would $a_0=\int_0^2 f(t)dt$, $a_n=\int_0^2 f(t)cos(\frac{\pi n t}{2})dt$, and $b_n=\int_0^2 f(t)sin(\frac{\pi n t}{2})dt$? I don't know where these really originate from or how to normalize the function, my professor didn't really explain it. Sorry for my ignorance haha.
• Apr 22nd 2013, 06:30 PM
chiro
Re: Fourier Series

Fourier series - Wikipedia, the free encyclopedia

Basically the idea is that you take the original fourier series (which is defined over [-pi,pi)) and you scale them for the general interval.

This gives the integration with respect to cos(2*pi*n*t/tau) and sin(2*pi*n*t/tau) where tau is the length of the interval f(t + tau) = f(t).

To normalize the integrals you need to calculate Integral over Main Branch cos^2(2*pi*n*t/tau)dt and Integral over main branch sin^2(2*pi*n*t/tau)dt and take the square roots of these and divide all your a_n's and b_n's by this square root value.

It has the same meaning as if you had a normal vector v and normalize by calculating v / ||v|| where ||v|| is the length of the vector.

The normalization for a_0 will be sqrt(2) since Integral (0 to 2) 1dt = 2 and taking square root is SQRT(2).

Once you calculate the normalized values, you can then plug them into the fourier series expansion.
• Apr 23rd 2013, 06:59 AM
shumaterm
Re: Fourier Series
Ok. I think I understand the scaling. Normalization is still a little unclear to me. Why do we choose $\cos^2{\frac{2 \pi n t}{\tau}}dt$ to integrate? Is this just finding the " $||v||$"? Which you find by the length? I think it's making more sense.

Why would you find $a_0$ by doing $\int_0^2 1dt$ and not $\int_0^2 t dt$? This is where the first part doesn't make sense to me. In my mind, I would find $\int_0^2 t dt$ and then divide by the length, since the function of $t$ goes from 0 to 2, the length of the vector would be $2\sqrt{2}$. That would make $a_0=\sqrt{2}$?

In the same way, to find $a_n$ I would do $\int_0^2 t \cos{\pi n t}$ which would give me $\frac{2\sin{2 \pi n}}{\pi n}+\frac{\cos{2 \pi n}}{\pi^2 n^2}-\frac{1}{\pi^2 n^2}$, and then I'm not sure. haha
• Apr 23rd 2013, 08:42 PM
chiro
Re: Fourier Series
Yes its basically the idea of finding the length and dividing by it: just like in a normal vector in R^n (or a similar space).

The integral will be t^2 instead of t by the way since f(t) = t^2 not t.
• Apr 24th 2013, 06:05 AM
shumaterm
Re: Fourier Series
Ok, I got some help from the tutoring center yesterday. I think I understand it now. Would you mind to check and see if I'm doing this correctly? Also, I meant for $f(t)=t$, not $t^2$. My apologies.

$a_0=\frac{1}{2}\int_{-2}^2 f(t)dt=\frac{1}{2}\int_{-2}^0 (t+2)dt + \frac{1}{2}\int_0^2 tdt=2$

$a_n=\frac{1}{2}\int_{-2}^2 f(t)cos{\frac{\pi n t}{2}}dt=\frac{1}{2}\int_{-2}^0(t+2)cos{\frac{\pi n t}{2}}dt+\frac{1}{2}\int_0^2 t cos{\frac{\pi n t}{2}}dt$ which after some calculations we find $a_n=0$

$b_n=\frac{2}{\pi n}[(-1)^n-1]$ which was found by the same method as we found $a_n$

Then we get $f = 2+\sum_{n=1}^\infty\frac{2}{\pi n}[(-1)^n-1]sin{\frac{\pi n t}{2}}$

Does this seem somewhat correct?

Thank you so much for your help.
• Apr 24th 2013, 05:52 PM
chiro
Re: Fourier Series
The answer for a0 is not right since you have to normalize it. This means dividing by SQRT(2) which would leave a0 = 2/SQRT(2) = SQRT(2).

I have no idea what you are doing with the limits though: the principle branch is [0,2) but you are calculating limits for [-2,2). If the limits are really [-2,2) as opposed to [0,2) then you have to use different normalization constants as well as different scaling factors within the sine and cosine terms.
• Apr 26th 2013, 06:53 AM
shumaterm
Re: Fourier Series
Gotcha. I went and talked to my professor yesterday. I have it figured out now. Thank you so much for your help!