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We are given a function $f:[-1,1]\rightarrow \mathbb{R}$ and the Tschebyscheff nodes, $x_j= $ cos$(\frac{2j+1}{2n}\pi), \,j=0, 1, ....n-1.$ We require that lim$_{n\to \infty} {\left\lVert f- p_n\right\rVert}_{C[-1,1]}=0,\,\,p_n$ being the interpolation polynomial related to the Tschebyscheff nodes. I need to state the conditions on $f$ for the requirement to be fulfilled. We know that the Weierstrass approximation theorem requires only continuity of $f$ to find a polynomial of degree $\leq n$ in order to achieve an approximation with a given error $\epsilon. $ What does the additional condition on Tschebyscheff nodes change in respect to the requirements on $f$ ? Is the continuity condition on $f$ sufficient ?

user249018
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1 Answers1

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Remember that the interpolation error at $x$ when interpolating at points $x_0, \dotsc, x_n$ in $[a, b]$ is given for some $\zeta \in [a, b]$ (depending on $x_0, \dotsc, x_n$ and $x$) by:

$\begin{align*} f(x) - p_n(x) &= \frac{1}{(n + 1)!} f^{(n+1)}(\zeta) \prod_{0 \le j \le n}(x - x_j) \end{align*}$

(this assumes $f$ has enough derivatives). The Chebyshev nodes minimize the maximal value of the product.

vonbrand
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  • Thanks. But the Weierstrass theorem assumes only continuity for $f$ to be approximated as good as one may want. What makes the difference ? – user249018 Jan 30 '21 at 14:36