Why Does the Fourier Series of |sin(x)| Treat n=1 Differently?

[Kqwert]

Homework Statement

Hello,

i am trying to do find the Fourier series of abs(sin(x)), but have some problems. As the function is even, bn = 0. I have calculated a0, and I am now working on calculating an. However, when looking at the solution manual, they have set up one calculation for n > 1 (i.e. a2, a3, a4, a5... and so on) and one for n = 1 (i.e. a1). Why?

Homework Equations

The Attempt at a Solution

 

[stevendaryl]

Well, you need to compute $\int cos(nx) sin(x) dx$. You can use a trig identity to rewrite that in terms of $sin((n+1) x)$ and $sin((n-1)x)$. The case $n=1$ is special, because the second term is zero.

 

[Kqwert]

Thank you!

 

[Ray Vickson]

Homework Statement

Hello,

i am trying to do find the Fourier series of abs(sin(x)), but have some problems. As the function is even, bn = 0. I have calculated a0, and I am now working on calculating an. However, when looking at the solution manual, they have set up one calculation for n > 1 (i.e. a2, a3, a4, a5... and so on) and one for n = 1 (i.e. a1). Why?

Homework Equations

The Attempt at a Solution

What interval are you using? $[-\pi,\pi]?$ $[0, \pi]?$ $[0, 2\pi]?$