Finding Fourier Series for f(x)

[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
[tex]1+ \sum_{k=0}^\infty \frac{2(-1)^k}{(2k+1}\pi} cos((2k+1)x)[/itex].

The sine series, $\sum b_n sin(nx)$ is a little more complicated but the same idea.