# Wave Function in specific range

1. Oct 22, 2011

### ZedCar

I obtained the following from a book.

Question is:

Periodic Sawtooth described by the following;

f(x) = x/2∏ for 0<x<2∏
f(x+2∏) = f(x) for -∞<x<+∞

The solution is:

If x = 0
y = 0

If x = 2∏
y = 2∏/2∏ = 1

If x = 4∏
y = f(2∏+2∏) = 2∏ = 1

Can anyone explain to me why when x = 4∏ y = 1 ? I'm not clear on that bit.

I would just think if you're putting x = 4∏ into the 2nd equation in the question you would get y = f(4∏+4∏) = f(8∏)

I know a full rotation is 2 ∏, so I can see how 8∏ would be the same as 2∏, but then how did they go from 2∏ to 1 in the part above which I have emboldened?

Thank you

2. Oct 22, 2011

### LCKurtz

You were given f(x) = x/2∏ for 0<x<2∏. Your function is not defined at x = 0 or at x = 2∏. So you don't have f(0) = 0 and f(2∏) = 1. The value of a function like that at the two end points is irrelevant in calculating things like the Fourier Series so they are usually left undefined. If you insist on defining f(0)= 0, then you have no choice but to also choose f(2∏) = 0 if you want the function to be periodic.