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By knowing that a discrete distribution of points go asymptotically to the density:

$\displaystyle p(x)= \frac{1}{\pi \sqrt{1-x²}}$ in $[-1, 1]$

I am able to conclude that interpolating at those points give nice results.

But my textbook says that

$x_j=\cos(j\pi/N), \quad j=0,1,...,N$

satisfy the asymptotic density above. Although my intuition says it does (that $\sqrt{1-x²}$ makes it very evident), I want to justify it properly. How can I do this?

Aloizio Macedo
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1 Answers1

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Consider the points $z_j = (\cos (\pi j/N), \sin (\pi j/N))$, $j=0,\dots,N$. They are uniformly distributed over the upper half of the unit circle, which implies that the number of points in a subarc of length $l$ is asymptotic to $Nl/\pi$.

The Chebyshev nodes are obtained by orthogonally projecting this uniform distribution onto the $x$-axis. Consider an interval $[a,b]\subset [-1,1]$. The number of Chebyshev nodes in this interval is the number of uniformly distributed points on the part of upper half-circle lying above this interval. That part of half-circle has equation $y=\sqrt{1-x^2}$, $a\le x\le b$, and therefore its length is $$\int_{a}^b \sqrt{1+(y')^2}\,dx = \int_a^b \frac{1}{\sqrt{1-x^2}}\,dx$$ Hence, the number of Chebyshev nodes in $[a,b]$ is asymptotic to $$N\cdot \frac{1}{\pi} \int_a^b \frac{1}{\sqrt{1-x^2}}\,dx$$ i.e., the nodes have asymptotic density $\frac{1}{\pi} \frac{1}{\sqrt{1-x^2}}$.

  • why is it the case that if the number of nodes is the one you mentioned (which I understood where you got from) that the nodes have that particular density. I do not understand how you can draw that conclusion (of course, in common terms, it makes perfect sense, but rigorously, I cannot prove it, or at least sketch a more "applied" proof) – Probability-Stats-Optimisation Oct 13 '20 at 12:41