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Fix an interval $[a,b]$. For which integers $n>1$, does there exist $n+1$ distinct points $\{x_0,x_1,...,x_n\}$ in $[a,b]$ such that for every continuous function $f:[a,b] \to (0,\infty)$, the unique interpolating polynomial $p_n(x)$ of $f$ at the nodes $\{x_0,x_1,...,x_n\}$ satisfy $p_n(x)\ge 0,\forall x\in [a,b]$ ?

compare with Does every positive continuous function have a non-negative interpolating polynomial of every degree?

In the present question, we do not want to let the nodes vary with the function.

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  • $\begingroup$ This question is really about polynomials only--there is no need to mention any (continuous) functions. $\endgroup$ – Wlod AA Oct 28 '18 at 6:36
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Let $n \ge 2$. Given any points $x_0 < x_1 < \dots < x_n$, there is a quadratic function positive at all those points, but negative somewhere in $[x_0,x_n]$.

Indeed, let $c \in [x_0,x_n]$ be any point other than those $n+1$ points. There is a quadratic $\phi(x) = -1+m(x-c)^2$ that is positive at those points. Simply take $m$ large enough.

The function $f$ to be approximated has values $f(x_j) = \phi(x_j)$, linear between, and constant on $(-\infty,x_0]$ and on $[x_n,\infty)$. So $f$ is continuous on $(-\infty,\infty)$ and $f(x) > 0$ for all $x \in (-\infty,\infty)$.

The interpolating polynomial $p_n$ with $p_n(x_j) = f(x_j)$ for $j=0,1,\dots, n$ is actually $\phi$ itself, so $p_n(c) = -1$.

For $n=1$, take $x_0,x_1$ the two endpoints. Your interpolating polynomial is degree $1$, the graph is a straight line, so if it is positive at the endpoints, then it is also positive at all points between.

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    $\begingroup$ and for $n=2$ ... ? $\endgroup$ – user521337 Oct 27 '18 at 0:49
  • $\begingroup$ For $n=2$, you need nodes $x_0,x_1,x_2$ ... also, is your $f(x)$ positive only at the nodes ? If it so, then you've got it wrong, I want $f:[a,b]\to (0,\infty)$, that means $f(x)>0,\forall x\in [a,b]$ $\endgroup$ – user521337 Oct 27 '18 at 16:23
  • $\begingroup$ OK, $n+1$ nodes means the approximation has degree $n$ Fixing. $\endgroup$ – Gerald Edgar Oct 27 '18 at 17:34

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