Find the Fourier series of the function: $f(x)=\begin{cases}x+1 & -1\leq x< 0,\\1-x &0\leq x< 1\end{cases},\;\;f(x+2)=f(x)$ I want to find the Fourier series of the function. For now, I am clueless on how to handle the function, int that it has $f(x+2)=f(x).$
$$f(x)=\begin{cases}x+1 & -1\leq x< 0,\\1-x &0\leq x< 1\end{cases},\;\;f(x+2)=f(x)$$
Question: 
Can anyone explain what this term ($f(x+2)=f(x)$) means in relation to the function? Solution or references will be highly appreciated.

BACKGROUND
First $f(x+2) = f(x)$ is just a way of saying that you got to repeat your function, i.e. it is periodic of period $2$. 
This is needed so that we could derive the fourier series, 
\begin{equation}
 a_0 + \sum\limits_{k=1}^{\infty}
 a_k\cos(k x) + b_k \sin(k x)
\end{equation}
where 
\begin{align}
 a_0 &= \frac{1}{T} \int\limits_{-\frac{T}{2} }^{\frac{T}{2} }
 f(x) \ dx \\
 a_k &= \frac{2}{T} \int\limits_{-\frac{T}{2} }^{\frac{T}{2}}
 f(x) \cos(kx) \ dx \\ 
 b_k &= \frac{2}{T} \int\limits_{-\frac{T}{2} }^{\frac{T}{2} }
 f(x) \sin(kx) \ dx \\
\end{align}
with $T$ being the period of our function, which in our case is $2$, right?
COMPUTING $a_0$
Let's compute $a_0$
\begin{equation}
 a_0 = \frac{1}{T} \int\limits_{-\frac{T}{2} }^{\frac{T}{2} }
 f(x) \ dx
  =
\frac{1}{2} \int\limits_{-1 }^{1 }
 f(x) \ dx
 =
\frac{1}{2} \int\limits_{-1 }^{0}
 (x+1) \ dx 
+\frac{1}{2} \int\limits_{1}^{0}
 (1-x) \ dx 
=
\frac{1}{4}
+
\frac{1}{4}
=
\frac{1}{2}
\end{equation}
COMPUTING $a_k$
Now, to compute $a_k$, we use the formula
\begin{equation}
  a_k = \frac{2}{2} \int\limits_{-1 }^{1}
 f(x) \cos(kx) \ dx 
 =
\int\limits_{-1 }^{0}
(x+1) \cos (kx) \ dx 
+
\int\limits_{0 }^{1}
(1-x) \cos (kx) \ dx 
\end{equation}
that is after starightforward math
\begin{equation}
  a_k = \frac{2}{2} \int\limits_{-1 }^{1}
 f(x) \cos(kx) \ dx 
 =
-2\dfrac{\cos\left(k\right)-1}{k^2}
\end{equation}
COMPUTING $b_k$
Now, to compute $b_k$, we use the formula
\begin{equation}
  b_k = \frac{2}{2} \int\limits_{-1 }^{1}
 f(x) \sin(kx) \ dx 
 =
\int\limits_{-1 }^{0}
(x+1) \sin (kx) \ dx 
+
\int\limits_{0 }^{1}
(1-x) \sin (kx) \ dx 
\end{equation}
that is after staright-forward math
\begin{equation}
  b_k 
  =
\dfrac{\sin\left(k\right)-k}{k^2}
-\dfrac{\sin\left(k\right)-k}{k^2}
=
0
\end{equation}
NOTE on $b_k$
Since $f(x) = f(-x)$ (easy to prove), then $f(x)$ is even, then $b_k = 0$ for all $k$. That's a property btw.
