Consider a $d$-dimensional convex rational polytope $P\subset\mathbb{Q}^d\subset\mathbb{R}^d$. Then, it's a standard fact that in general the function counting the number of lattice points inside the multiples $t\cdot P$ of $P$ is a quasi-polynomial instead of a polynomial, as in the integral case. This is the so-called Ehrhart quasi-polynomial of $P$.


This means that if $L(t,P)=\#\{x\in\mathbb{Z}^d:x\in t\cdot P\}$, then there exists a finite number of polynomial functions $f\_1,\dots, f\_D$, all of degree $\dim P$, such that $L(t,P)=f\_i(t)$ whenever $t\equiv i\mod D$ (n.b. $D$ is the period of the polytope and if the polytope is integral we have $D=1$ and recover the usual Ehrhart polynomial of $P$).


What I want to know is how much can these polynomial functions differ from one another. That is, are there known bounds to the order of $f\_i-f\_j$ for all $i$ and $j$? How sharp are these bounds?


The case I'm dealing with is for $\dim P=2$, so if there are sharper results for this case than in general that'd be great.



How different can the constituents of an Ehrhart quasi-polynomial be?
Let $d=\dim(P)$. First, since $L(t,P)$ is non-decreasing in $t$, for any positive integer $n$ we have 
$$f\_i((n-1)D+i) \leq f\_j((n-1)D+j) \leq f\_i(nD+i) \leq f\_j(nD+j)$$
whenever $i \leq j$. Thus it is easy to see that the coefficients of $t^d$ in $f\_i,f\_j$ must be the same ([in fact](https://en.wikipedia.org/wiki/Ehrhart_polynomial#Interpretation_of_coefficients) this coefficient is a normalized volume of $P$). Thus $f\_i(t)-f\_j(t)$ is $O(t^{d-1})$.


It is easy to see that $\Omega(t^{d-1})$ can be achieved. For example, if $P$ is the $d$-cube with side-length $1/2$ in the positive quadrant with a vertex at the origin, then 
$$f\_1(t)=\left(\frac{1}{2}\right)^d (t+1)^d$$ and 
$$f\_2(t)=\left(\frac{1}{2}\right)^d (t+2)^d$$
so the difference is at least $\frac{d}{2^d}t^{d-1}$. Probably one can improve this constant.


