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Statement

Setting (p. 270, Nr. 8). The nodes are the zeros of the Chebyshev polynomial $T_n(\cos\theta)=\cos n\theta$, which the paper also writes as its Jacobi polynomial $J_n(\frac14,\frac14,x)$:

$$x_k=\cos\,(2k+1)\frac{\pi}{2n},\qquad k=0,1,\ldots,n-1,$$

with fundamental functions $l_0(x),\ldots,l_{n-1}(x)$.

Formula (59) (p. 271, stated as the result of Nr. 8). For these nodes, with $x=\cos\theta$ and $0\le\theta\le\pi$,

$$(l_0(x))^2+(l_1(x))^2+\cdots+(l_{n-1}(x))^2 =1+\frac{\cos2\theta+\cos4\theta+\cdots+\cos2(n-1)\theta}{n} =1-\frac1{2n}+\frac1{2n}\,\frac{\sin(2n-1)\theta}{\sin\theta}.$$

Consequences (p. 271), stated as following from (59):

$$(60)\qquad (l_0(x))^2+\cdots+(l_{n-1}(x))^2\le2-\frac1n,\qquad -1\le x\le1,$$ $$(61)\qquad \lim_{n\to\infty}\bigl\{(l_0(x))^2+\cdots+(l_{n-1}(x))^2\bigr\}= \begin{cases}1,&-1<x<1,\\ 2,&x=\pm1.\end{cases}$$

Proof pointer

Pp. 270--271, (52)--(57). The matrix of the values $\varphi_v(\theta_k)$, with $\varphi_0=\sqrt{1/n}$ and $\varphi_v(\theta)=\sqrt{2/n}\cos v\theta$ for $1\le v\le n-1$ at $\theta_k=(2k+1)\pi/(2n)$, is orthogonal. The fundamental functions are therefore an orthogonal transformation of $\varphi_0,\ldots,\varphi_{n-1}$, and the sum of their squares equals $\sum_v\varphi_v(\theta)^2$, which sums to (59).

Read depth

Claims checked: (50), (58)--(61) were read on the page images of the print and the derivation (52)--(57) was followed; (60) and (61) are stated without proof. A second reader checked the statement, hypotheses, label and page against the print; the proof was not independently reviewed.

Dependencies

None in the corpus.

Source. L. Fejér, Bestimmung derjenigen Abszissen eines Intervalles, für welche die Quadratsumme der Grundfunktionen der Lagrangeschen Interpolation im Intervalle ein Möglichst kleines Maximum Besitzt, Ann. Scuola Norm. Sup. Pisa Cl. Sci. (2) 1 (1932), no. 3, 263--276; the edition read is named on the source card.

Bears on

• Problem 1131: (59) gives the integrand of the problem's $I$ in closed form at the Chebyshev nodes; the paper does not integrate it and says nothing about the least value of $I$. Integrating the middle form of (59) against $\sin\theta\,d\theta$, using $\int_0^\pi\cos2v\theta\,\sin\theta\,d\theta=\frac{2}{1-4v^2}$, gives $I=2-\frac{2(n-1)}{n(2n-1)}$ at these nodes, a value made here, not in the paper.