Find a rational function $R(x)$ such that:


$1)$ For $i\in\{1,\dots,g\}$, $x\_{i}=x\_{i-1}+g$ with $x\_0=0$. 


$2)$ For $i\in\{0,\dots,g-1\}$ $R(x\_i)=R(x\_i+1)=\dots=R(x\_i+g-1)=i+1$.


$3)$ $R(x\_g)=g+1$.


So $R(x)$ is defined at $g^2+1$ points. 


These points and the value taken by these points have an arithmetic progression structure and I know that Lagrange interpolation gives $O(g^2)$ polynomial.


However is there a degree $O(g)$ rational function $R(x)$ that takes advantages of the structure in the conditions?



Minimum degree rational function interpolation
The answer is no. I assume that the degree of a rational function is the maximum of degrees of its numerator and denominator (in the irreducible form).


Consider any interval $(x\_i+j,x\_i+j+1)$ with $0\leq i\leq g-1$ and $0\leq j\leq g-2$. If the denominator of $R(x)$ has no roots on this interval, then $R'(x)$ should have a root on it. Let $d$ be the degree of $R(x)$. Then there are at most $d$ roots of the denominator, and at most $2d-2$ roots of the numerator of $R'(x)$, since this numerator has degree at most $2d-2$. Thus $g(g-1)\leq d+(2d-2)$, so $d\geq \frac{g(g-1)+2}3$.


