Proving a function is periodic! I am having trouble assimilating periodic function. Let me tell you, I have had a semester of fourier analysis already but reviewing the first chapter got me confused on a trivial equation. 
A function is periodic if it satisfies: $x(t) = x(t+T_0)$, where smallest $T_0$ is the fundamental period. 
Lets take, for instance, $\sin(x)$. I can intuitively see that period is $2\pi$ BUT $\sin(0)=\sin(\pi)$ as well. By that I mean, the value of $\sin(x)$ is the same at $x=0$ and $x=pi$, then shouldn't $T_0$ be $\pi$ because $T_0$ is fundamental for smallest value of $T_0$ which i just showed is $1$ and NOT $2$.
please help me on this trivial issue.  
 A: For periodicity with period $p$, you want $\sin(x+p) = \sin x$ for all $x$ (not just particular values of $x$).
Expand: $\sin(x+p) = \sin x \cos p + \cos x \sin p$.
Equate coefficients:  $\cos p = 1$ and $\sin p = 0$.
The only values of $p$ that satisfy both conditions are of the form $p = 2k\pi, k \in \mathbb{Z}$. The fundamental period occurs when $k=1$, i.e. $2\pi$.
(If you take $p = \pi$, for example, you'll find $\cos \pi = -1$, which is a violation of the required conditions).
A: I think that where you're confused is in the fact that the period is the least number  T so that for  all x , we have $$ f(x+T)=f(x)$$. For some x (like, in your example of sinx, $sin(0)=sin(\pi)$, it may be the case that there is some $T' < T$ so that$ f(x+T')=f(x)$, but you must have a $T$ that works for all x.
A: You should look carefully at what your definitions mean exactly. A function $f$ is periodic on a set $D$ with period $T$ if and only if $f(x+T) = f(x)$ for any $x \in D$. If you choose some $T$, and there is just one $x$ that doesn't satisfy that equation, then that $T$ you choose is not even a period, not to say a fundamental period. A fundamental period is subsequently defined as the minimum positive period if it exists. Note that the constant function has arbitrarily small periods and hence no fundamental period.
