# Divergence of the Lagrange interpolation on the Chebyshev nodes

Faber theorem states that for every $\lbrace x_k^{(n)} \rbrace$ there exists a continuous function $f$ such that $\| f - L_n \|_{\infty} \not\rightarrow 0$, where $L_n$ is an interpolation polynomial on the points $x_0^{(n)}, x_1^{(n)}, \ldots, x_n^{(n)}$.

Can somebody give me an example for a continuous function which does not converge uniformly on the Chebyshev nodes?

I've only found positive results on this topic.

We interpolate $f(x) = \frac{x^2}{\sin(\arccos(x))^2} = \frac{x^2}{1-x^2}$ on the Chebyshev-nodes in the $(-1,1)$ interval.

Lemma 1. [1, 6.4 lemma]. If $x_1, x_2, \ldots, x_{n+1}$ are the roots of the polynomial $\omega_{n+1}(x)$, then the $\ell_i(x)$ Lagrange-polynomials can be written in the following form $$\ell_i(x) = \frac{\omega_{n+1}(x)}{(x-x_i)\cdot \omega'_{n+1}(x_i)}.$$ We also need a trigonometric identity.

Proposition 1. Let $n$ be an arbitrary positive, even integer, then the following holds $$\sum^{n-1}_{k=0} (-1)^k \cot\left(\frac{(2k+1) \pi}{2n}\right) = n.$$ Proof. A similar statement can be found in [2, 141. b.], that is, $$\sum^{n-1}_{k=0} (-1)^k \cot\left(\frac{(2k+1) \pi}{4n}\right) = n.$$ Using that $\cot(2x)=\frac{1}{2}(\cot(x) - \tan(x))$ holds for all $x$, it is sufficient to show that $$\sum^{n-1}_{k=0} (-1)^k \tan\left(\frac{(2k+1) \pi}{4n}\right) = -n,$$ from which follows the statement.

I prove the last inequality based on the solution of [2, 141. b.]. We know [2, 132. c] that $$\tan(n\alpha) = \frac{\binom{n}{1} \tan \alpha - \binom{n}{3} \tan^3 \alpha + \binom{n}{5} \tan^5 \alpha - \ldots}{1 - \binom{n}{2} \tan^2 \alpha + \binom{n}{4} \tan^4 \alpha - \ldots}.$$ If $\alpha$ is one of the following numbers $\frac{\pi}{4n}, \frac{5\pi}{4n}, \ldots, \frac{(4n-3)\pi}{4n}$, then $\tan(n\alpha) = 1$, thus the $\alpha$ numbers give all of the roots of the following polynomial $$1 - \binom{n}{1} x - \binom{n}{2} x^2 + \binom{n}{3}x^3 + \binom{n}{4}x^4 - \binom{n}{5}x^5 - \ldots - \binom{n}{2k-1} x^{2k-1} - \binom{n}{2k} x^{2k}$$ where $n = 2k$ and $n \equiv 2 \pmod{4}$. If $n \equiv 0 \pmod{4}$, then $$1 - \binom{n}{1} x - \binom{n}{2} x^2 + \binom{n}{3}x^3 + \binom{n}{4}x^4 - \binom{n}{5}x^5 - \ldots + \binom{n}{2k-1} x^{2k-1} + \binom{n}{2k} x^{2k}$$ we get the roots of the polynomial above.

Notice that $\tan\left( \frac{(4n-3)\pi}{4n} \right) = -\tan\left( \frac{3\pi}{4n} \right), \tan\left( \frac{(4n-7)\pi}{4n} \right) = -\tan\left( \frac{7\pi}{4n} \right), \ldots, \tan\left( \frac{(2n+1)\pi}{4n} \right) = -\tan\left( \frac{(2n-1)\pi}{4n} \right),$ that is, the sum of the roots is $\sum^{n-1}_{k=0} (-1)^k \cot\left(\frac{(2k+1) \pi}{2n}\right),$ and applying Viète-formula we have $$-\frac{\binom{n}{2k-1}}{\binom{n}{2k}} = -\frac{n}{1} = -n.$$ Proposition 2. Let $L_{n-1}$ be the $n$th interpolating polynomial on the Chebyshev-nodes of $f$, then $L_{n-1}(0) = -\cos(\frac{\pi}{2} \cdot n)$ for all $n \in \mathbb{N}$.

Proof. By definition \begin{equation}L_{n-1}(x) = \sum^{n-1}_{k=0} \ell_k(x) f(x_k), \end{equation} where $x_k = \cos \left( \frac{(2k+1)\pi}{2n} \right), k = 0,1,\ldots,n-1$. By Lemma 1. we can easily calculate the Lagrange-polynomials at $0$. Since $T_n'(x) = \frac{n\cdot \sin(n \cdot \arccos(x))}{\sqrt{1-x^2}}$, we have that $$T_n'(x_k) = \frac{n\cdot\sin\left(\frac{2k+1}{2}\pi \right)}{\sin\left(\frac{2k+1}{2n}\pi \right)}.$$ Furthermore $T_n(0) = \cos(\frac{\pi}{2}n)$, thus $$\ell_k(0) = \frac{\sin\left(\frac{2k+1}{2n}\pi \right) \cos(\frac{\pi}{2}n)}{-n\cdot\sin\left(\frac{2k+1}{2}\pi \right) \cos \left( \frac{(2k+1)\pi}{2n} \right)}.$$ Now, we calculate $f(x_k)$: $$f(x_k) = \frac{\cos \left( \frac{(2k+1)\pi}{2n} \right)^2}{\sin \left( \frac{(2k+1)\pi}{2n} \right)^2}.$$ Since $\sin\left(\frac{2k+1}{2}\pi \right) = (-1)^k$, we get that $$\ell_k(0)f(x_k) = (-1)^{k+1} \frac{\cot \left( \frac{(2k+1)\pi}{2n} \right) }{n}.$$ From which follows that $$L_{n-1}(0) = \frac{-1 \cdot \cos(\frac{\pi}{2}n)}{n} \sum^{n-1}_{k=0} (-1)^{k} \cot \left( \frac{(2k+1)\pi}{2n} \right),$$ however if $n$ is even by Proposition 1. we have $$\sum^{n-1}_{k=0} (-1)^{k} \cot \left( \frac{(2k+1)\pi}{2n} \right) = n,$$ and we obtain what we need.

If $n$ is odd, it is true that $L_{n-1}(0) = f(0) = 0$, since $0$ is a root of the $n$th Chebyshev-polynomial. Since $T_n(0) = \cos(\frac{\pi}{2} \cdot n) = 0$, it is sufficient to take that term of (1) when $x_k = 0$. Then $$f(x_k) \frac{T_n(x)}{x_k T_n'(x_k)} = f(x_k) = 0,$$ since $\frac{T_n(x)}{xT_n'(x)} = 1$ if $x = 0$ and this is a root of $T_n(x)$.

Therefore, we get that $L_{n}(0) \not\rightarrow f(0)$ when $n \rightarrow \infty$.

References.

 J.C. Mason, David C. Handscomb, Chebyshev Polynomials, CRC Press, 2002.

 A. M. Yaglom, I. M. Yaglom Challenging Mathematical Problems With Elementary Solutions, Vol. 2, Courier Corporation, 1987

• But $f$ is not continuous on the segment $[-1,1]$. – Fedor Petrov Jul 12 '15 at 11:01
• Yes, it is not. But $f \in C(-1,1)$ and Chebyshev-nodes also lie in $(-1,1)$. I think you wanted to point out that Faber theorem states that there exists such a function on a closed interval. In this case, the solution above is indeed a weaker result. It is worth mentioning that if we map Chebyshev-nodes to the $[-1/2,1/2]$ closed interval then $f$ converges "very well" (according to Matlab tests) on that points. – Benjamin Jul 28 '15 at 20:02

Marcinkiewicz and Grûnwald (1936) have independently shown that there exists a function continuous on $[-1,1]$ such that its Lagrange interpolants on the Chebyshev nodes diverge at all points of $[-1,1]$. There is an explicit construction of such a function in the book by I.P. Natanson, Constructive Function Theory, Vol. 3, pp. 35-46.
The proof is elementary, but takes a few pages, and it gives more than what you are looking for. I don't know of a reference, with a possibly simpler example, which would only exhibit non uniform convergence.