
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 for and suppose for all . By , we denote even integers. Define for . Calculate the Fourier Series for .
Problem 2
Show, by the result of the last problem, . Derive a series which converges to by setting 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 with certainty?

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.

Re: Fourier Series
would be 1 and be ? I know the general forumla for , 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.

Re: Fourier Series
Oops. also, in problem 1, , not .

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).

Re: Fourier Series
So would , , and ? 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.

Re: Fourier Series
I suggest you take a look at this page:
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.

Re: Fourier Series
Ok. I think I understand the scaling. Normalization is still a little unclear to me. Why do we choose to integrate? Is this just finding the " "? Which you find by the length? I think it's making more sense.
Why would you find by doing and not ? This is where the first part doesn't make sense to me. In my mind, I would find and then divide by the length, since the function of goes from 0 to 2, the length of the vector would be . That would make ?
In the same way, to find I would do which would give me , and then I'm not sure. haha

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.

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 , not . My apologies.
which after some calculations we find
which was found by the same method as we found
Then we get
Does this seem somewhat correct?
Thank you so much for your help.

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.

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!