I simply cannot figure out what this means. I read this on an article about the scalar product of $2\pi$ periodic functions. it says that < f,g > goes from $\Bbb R/2\pi \to \Bbb C$ (complex)
Do they really mean divide the real numbers by $2\pi$?
Thanks a lot in advance for the help
PS: I simply can't figure it out using google.
As Jonathan Hebert noted, $\Bbb R / 2\pi$ is probably the set of real numbers modulo $2\pi$. Are you familiar with $\Bbb Z_{n}$ from Abstract Algebra? It's the set of integers modulo the natural number $n$, and the concept here is the same.
Basically, you look at $\Bbb R$ and for each $x \in \Bbb R$, you start forming its equivalence class. You say $y \in \Bbb R$ is equivalent to $x$ if $x - y = 2 \pi t$ for some integer $t$. That means $y = x + 2 \pi t$. So basically, the elements equivalent to $x$ are of the form $x + 2 \pi t$ ($t \in \Bbb Z$).
So for each $x \in \Bbb R$, we have its equivalence class $[x] := \{x + 2 \pi t \mid t \in \Bbb Z \}$. Now look at $\{ [x] \mid x \in \Bbb R \}$, and this set is $\Bbb R / 2 \pi$.
Basically, a function defined on $\Bbb R / 2 \pi$ is equivalent to saying the function is defined on $\Bbb R$ and has period $2 \pi$. (This means if we have a function defined on $\Bbb R / 2 \pi$, we can realize another function from it defined on $\Bbb R$ with period $2\pi$, and if we have a function with period $2 \pi$ on $\Bbb R$, we can realize another function from it defined on $\Bbb R / 2\pi$.)