0
$\begingroup$

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 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.

$\endgroup$
2
  • $\begingroup$ What is $R^i(x)$ in your notation? $\endgroup$ Aug 28, 2016 at 6:46
  • $\begingroup$ @IgorKhavkine it is meant to be the $i$-th derivative of $R(x)$ and $R^0(x):=R(x)$ $\endgroup$ Aug 28, 2016 at 10:51

1 Answer 1

1
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.