How Do You Find the Fourier Series for a Piecewise Function?

[Juggler123]

I need to find the Fourier series for the function f;

0 if -\pi \prec x \leq -\frac{\pi}{2}

1+x if -\frac{\pi}{2} \prec x \prec \frac{\pi}{2}

0 if \frac{\pi}{2} \leq x \leq \pi

I've never done a Fourier series computation before so I don't really know if any of what I'm doing is correct.

I've got than a(0)=1, a(n)=\frac{2sin\frac{n\pi}{2}}{n\pi} and
b(n)=\frac{2sin\frac{n\pi}{2}}{n^{2}\pi} - \frac{cos\frac{n\pi}{2}}{n}

I know the formula for a Fourier series but none of the examples I've seen are in the form of the a(n) and b(n) that I've got so I don't know where to go next.
Could anyone help please? Thankyou.

 

[HallsofIvy]

You are going to use the fact that sin(n\pi/2)= 0 if n is even, aren't you? And that sin(n\pi/2)= 1 if n is "congruent to 1 mod 4" (i.e. n= 4k+ 1 for some integer k) while sin(n\pi/2)= -1 if n is "contruent to 3 mod 4" (i.e. n= 4k+ 3 for some k. That is, a_2= 0, a_3= -1/3\pi, a_4= 0, a_5= 1/5\pi, etc.

Like wise cos(n\pi/2)= 0 if n is odd, cos(n\pi/2)= 1 if n is "congruent to 2 mod 4" (divisible by 2 but not by 4), and cos(n\pi/2)= -1 if n is "congruent to 0 mod 4" (divisible by 4). Notice that for all n, one of the (2sin(n\pi)/2)/n\pi or cos(n\pi/2)/n\pi is 0 so you are not actually "subtracting".

 

[Juggler123]

I can see the sequences that a(n) and b(n) make, but I can't write down a formula for them. I'm using trial and error really but I can't find any functions dependent on n that will describe the sequences a(n) and b(n) exactly. Is there a formula for caculating these type of functions?

 

[HallsofIvy]

You said that a(n)= \frac{2sin\frac{n\pi}{2}}{n\pi}

Since sin(n\pi/2)= 0 for n even, for a(n) to be non-zero, n must be of the form 2k+1 for some integer, k. But then sin(n\pi/2)= sin((2k+1)\pi/2)= sin(k\pi+ \pi/2)= 1 if k is even and -1 if k is odd. That is, sin((2k+1)\pi/2)= (-1)^k.

So a(n)= 0 if n is even and, if n is odd, n= 2k+1, a(n)= a(2k+1)= 2(-1)^k/(2k+1)\pi. You could write the Fourier cosine series as
1+ \sum_{k=0}^\infty \frac{2(-1)^k}{(2k+1}\pi} cos((2k+1)x)[/itex].<br /> <br /> The sine series, \sum b_n sin(nx) is a little more complicated but the same idea.