Complex Fourier Series for f(x) = 2 - x, -2<x<2

[csnsc14320]

Having some trouble with this, any help is appreciated

Homework Statement

Give the complex Fourier series for f(x) = 2 - x, -2&lt;x&lt;2

Homework Equations

f(x) = \sum_{n=-\infty}^\infty C_ne^{\frac{i n \pi x}{l}

C_n=\frac{1}{2l} \int_{-l}^l f(x)e^{\frac{-i n \pi x}{2}} dx

The Attempt at a Solution

f(x) = 2 - x, l = 2

f(x) = \sum_{n=-\infty}^\infty \frac{1}{4} \int_{-2}^2 ((2-x)e^{\frac{-i n \pi x}{2}}dx) e^{\frac{i n \pi x}{2}


now here, I don't know which steps I should take next. Should I take this integral for general n? Or should I break it up into the case for when n=0, n>0, and n<0?

I've kind of tried both to no avail :(

 

[lanedance]

from a quick look here's a ew helpful (hopefully) tips:

the sum over negative & positive n values in the complex exponent, is equivalent to summing only over positive n's with both sin & cos terms, & this one might be easier to evaluate in terms of sin's & cos's

also you should find a proof of the integral of sin(2.pi.n.x/L).sin(2.pi.n.x/L) over L = 0 for integers m not equal p and similar for cosines & mixes, this will allow you to write each coefficient in terms of single integral

now consider your function
g(x) = 2-x

first consider the constant part, 2. Any integral of a sinusoid over an integer times period will be zero, real or complex so only the n= 0 (or constant cos part will contribute)

now -x is an odd function, the cos integrals will be zero, (assuming it is just repeated),so only sin's will contribute...

so you should be able to reduce the integrals to required to ~x.sin(n.pi.x/2)

 

[csnsc14320]

from a quick look here's a ew helpful (hopefully) tips:

the sum over negative & positive n values in the complex exponent, is equivalent to summing only over positive n's with both sin & cos terms, & this one might be easier to evaluate in terms of sin's & cos's

also you should find a proof of the integral of sin(2.pi.n.x/L).sin(2.pi.n.x/L) over L = 0 for integers m not equal p and similar for cosines & mixes, this will allow you to write each coefficient in terms of single integral

now consider your function
g(x) = 2-x

first consider the constant part, 2. Any integral of a sinusoid over an integer times period will be zero, real or complex so only the n= 0 (or constant cos part will contribute)

now -x is an odd function, the cos integrals will be zero, (assuming it is just repeated),so only sin's will contribute...

so you should be able to reduce the integrals to required to ~x.sin(n.pi.x/2)

yeah, i think that was part a), they had us find the Fourier series in sin/cos form, now they want us to re-derive it with complex exponentials

as a side note, i didnt think about splitting up the 2 - x into both 2 and the -x part, that makes sense now and seems quicker for part a :)

 

[lanedance]

not sure if you are allowed, but you could always write your complex exponential as:
e^{i x } = cos(x) + i sin(x)
then use what you know about sin & cos integrals - evaluating the integrals will essentially be the same then...

and the negative n ones will just be complex conjugates of the poistives

 

[csnsc14320]

not sure if you are allowed, but you could always write your complex exponential as:
e^{i x } = cos(x) + i sin(x)
then use what you know about sin & cos integrals - evaluating the integrals will essentially be the same then...

and the negative n ones will just be complex conjugates of the poistives

i think they want it purely in the e^ix form :(

just stuck on whether to worry about positive/negative cases and whether i do or don't how to simplify it to something reasonable

 

[lanedance]

hmm.. by separating 2 & -x should be able to integrate *2 easy,for n = 0

then simple int by parts on the *(-x) for when n is not equal to zero

knowing what is going to happen to the terms based on what you have already done (sin & cos), i would attempt it for a generic n & -n, then show what happens when you sum the result... so express you results as a sum from 0 to n