Even function I don't think is even My professor provided the following function:

which he says is even with a period of $2\pi$. 
I get what he means, it's supposed to be a sawtooth-like function. But is this really even? In a function like that, every single line portion would be a different line, right? This function is just going to keep going further and further away, naturally, as we move from $(0,0)$. 
Am I misunderstanding something here?
 A: The function is even about the line $x=\frac{\pi}{2}$; i.e. about the line $x=\frac{\pi}{2}$, if the left hand part of the function is the reflection of the right hand part of the function, then the function is even. 
Or, in other words, shift the origin of the co-ordinate system to $(\frac{\pi}{2},0)$ and suitably change the functions according to the shifting of the axes. Then you will find that the function is even i.e. $f(-x)=f(x)$.
A: Let $X$ be a set of real numbers, and say that $X$ is symmetric if $X$ is invariant under reflection in the origin, i.e., if $-x \in X$ if and only if $x \in X$.
Let $X$ be symmetric. A real-valued function defined on $X$ is even if $f(-x) = f(x)$ for all $X$ in $X$, and is odd if $f(-x) = -f(x)$ for all $X$ in $X$. 
As Did notes, your function is defined on $[0, \pi]$, an interval that is not symmetric, and does not have length $2\pi$.
Since your professor mentions an even function with period $2\pi$, it's fairly clear from context that you're meant to consider the even extension of $f$, defined on $[-\pi, \pi]$ by requiring $f(-x) = f(x)$ for $-\pi \leq x < 0$. (There is also an odd extension, defined by requiring $f(-x) = -f(x)$ for $-\pi \leq x < 0$.)
On the guess that you're studying Fourier series, note carefully that a piecewise-$C^{1}$ function $f$ on $[0, \pi]$ has:


*

*A cosine series, the Fourier series of the even extension.

*A sine series, the Fourier series of the odd extension.
The function $f$ itself does not have a Fourier series with respect to the standard trigonometric basis because $f$ (as initially defined, on an interval of length $\pi$) is not $2\pi$-periodic.
