problem in Functions and Periodicity What is the period of  $f(x+\frac{1}{2}) +f(x-\frac{1}{2})=f(x)$ ?
 I tried substituting $x=x+\frac{1}{2}$ and $x=x-\frac{1}{2}$ but that didn't get me anywhere. According to the standard procedures , we are not allowed to directly substitute $x$ as an integer.So any answers and boundary conditions(if any) would be helpful.
 A: If you have the condition of
$$f(x)=f(x+1/2)+f(x-1/2)$$
and let's suppose that the function has a period T ($f(x+nT)=f(x),\forall n $ integer). Let's make the changes of variable $x\to x-1/2$, finding
$$f(x-1/2)=f(x)+f(x-1)\\
f(x)=f(x+1/2)+f(x-1/2)$$
Summing both expressions
$$f(x-1)=-f(x+1/2)$$
similar expression can be found for $f(x+1)=-f(x-1/2)$. Now, let's substitute the expression for $f(x-1/2)$ in the first of the two equations written before, finding
$$f(x-1/2)=f(x)-f(x-1/2)\to f(x-1/2)=f(x)/2$$And again, similar expression can be obtained for the $f(x+1/2)=f(x)/2$. Hence, we find that 
$$f(x+1/2)=f(x-1/2)=f(x)/2$$
Up to here, it would seem that the period of the function is $1$. Let's check this fact by going back to the equations we have computed before, knowing that if the function has a period $1$ it has to fulfill that $f(x-1)=f(x)$. Taking the expression 
$$f(x-1)=-f(x+1/2)=-f(x)/2$$
The only function with a period $1$ and that fulfills this equation is the trivial $f(x)=0$. 
Finally, as we have that $f(x+1)=f(x-1)=-f(x)/2$, you could think that maybe the function has a period of $2$. With a similar reasoning, you can demonstrate that $2$ is a period of the function only if $f(x)=0$.
I am not sure how to systematically demonstrate you can find that the only solution is the trivial function for any value of the period $T$. However, I have checked that if you find some symmetry of the function $f(x+T/2)=f(x-T/2)$ (which would suggests a period $T$), it is not true for $f(x+T)\neq f(x)$. I hope it would be useful for you!
