let $\quad-1=x\_0 < x\_1 <\ ...\ < x\_n<1\quad$ be a set of abscissas  

and $\quad(y\_0, y\_1,\ ...\,y\_n)\quad$ a sequence of the corresponding ordinates.



> 
> **Question:** 
> 
> 
> what can be said about the existence and calculation of a pole-free rational function $R(x)$,  
> 
>  with the following properties?: 
> 
> 
> $\quad R(x\_i) = y\_i,\quad\quad\quad 0 \le i\le n$ 
> 
> 
> $\quad \frac{d^îR}{dx^î}(1) = \frac{d^iR}{dx^i}(-1),\quad \forall i\in\mathbb{N}\_0, \left(\frac{d^0R}{dx^0}(x) := R(x)\right)$
> 
> 
> 


Background of my question is [Barycentric Rational Interpolation](https://www.in.tu-clausthal.de/fileadmin/homes/techreports/ifi0606floater.pdf) is an infinitely often differentable analytic alternative to Splines and NURBS and I wonder, if rational interpolation could also provide infinitely often differentiable analytic closed curves through a set of points e.g. in the plane.



$C^\infty$ Periodic Pole-free Rational Interpolation
No such function exists, except possibly when $R(x)$ is a constant. Your condition on the derivatives implies periodicity (the periodic extensions of $R(x)$ from $[-1,1]$ to the whole real line is analytic and agrees with $R(x)$ on $[-1,1]$ and must be equal to $R(x)$ by uniqueness of analytic continuation). No non-constant rational function is periodic (if $R(x)$ is rational, you can find a positive integer $n$ such that $x^n R(1/x)$ is analytic at $x=0$, and that is not true for any periodic analytic function other than a constant).


