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On the stability of unevenly spaced samples for interpolation and quadrature

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Unevenly spaced samples from a periodic function are common in signal processing and can often be viewed as a perturbed equally spaced grid. In this paper, the question of how the uneven distribution of the samples impacts the quality of interpolation and quadrature is analyzed. Starting with equally spaced nodes on \([-\pi ,\pi )\) with grid spacing h, suppose the unevenly spaced nodes are obtained by perturbing each uniform node by an arbitrary amount \(\le \alpha h\), where \(0 \le \alpha < 1/2\) is a fixed constant. A discrete version of the Kadec-1/4 theorem is proved, which states that the nonuniform discrete Fourier transform associated with perturbed nodes has a bounded condition number independent of h, for any \(\alpha <1/4\). Then, it is shown that unevenly spaced quadrature rules converge for all continuous functions and interpolants converge uniformly for all differentiable functions whose derivative has bounded variation when \(0\le \alpha <1/4\). Though, quadrature rules at perturbed nodes can have negative weights for any \(\alpha >0\), a bound on the absolute sum of the quadrature weights is provided, which shows that perturbed equally spaced grids with small \(\alpha \) can be used without numerical woes. While the proof techniques work primarily when \(0 \le \alpha < 1/4\), it is shown that a small amount of oversampling extends our results to the case when \(1/4\le \alpha <1/2\).

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  1. For two functions \(g_1(N)\) and \(g_2(N)\), one writes \(g_1(N) = \varOmega (g_2(N))\) if there is a constant \(C>0\) that is independent of N such that \(g_1(N)\ge Cg_2(N)\) for all N.

  2. Note that \(\sum _{j =1}^{N}1/(j+\alpha )> \sum _{j=1}^N 1/(j+1) >\log (N) - 1/2\).

  3. If \(\alpha \ge (1-2\alpha _0)/2\), then one can pick some \(\alpha _1\) such that \(\alpha< \alpha _1 < \alpha _0\) and \(\alpha < (1-2\alpha _1)/2\). Since Eq. (5.2) holds if \(\alpha _0\) is replaced by \(\alpha _1\), it also holds for \(\alpha _0\).

  4. Note that we do not necessarily have \(\left| x_k - {\tilde{x}}_k\right| \le \alpha h\) for all k. Instead, we have \(\left| x_k - {\tilde{x}}_k\right| \le 2\alpha h\).

  5. Otherwise, swap the roles of \(k_1\) and \(k_2\).

  6. Note that j may not be between \(k_1\) and \(k_2\). However, the claim follows regardless of how j compares to \(k_1\) and \(k_2\).


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This work is partially supported by the National Science Foundation grants DMS-1818757, DMS-1952757 and DMS-2045646. We are thankful for many conversations with Anthony Austin, Nick Trefethen, and Kuan Xu regarding unevenly spaced trigonometric interpolation over several years. In private communication, Heather Wilber gave an initial proof of the discrete Kadec-1/4 theorem in December 2017, which we adapted for our purposes. It was Laurent Demanet who brought our attention to the conditioning of NUDFT matrices and we also benefited from Alex Barnett’s wisdom on the subject. The research direction became more focused after brain storming sessions during Cornell’s math REU program in 2021 and we thank Aparna Gupte, Yunan Yang, and Liu Zhang for discussions regarding the MZ inequalities and quadrature rules.

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This work is partially supported by the National Science Foundation Grants DMS-1818757, DMS-1952757 and DMS-2045646.


Appendix A: A quadrature weight at perturbed nodes is negative

To show that the perturbed nodes in Eq. (4.1) have an associated quadrature weight that is negative when N is sufficiently large, we need a few trigonometric inequalities that are technical to derive.

Let \(p, q, r > 0\) be such that \(0< q-p-2r< q-p+2r < \pi \). The following trigonometric inequality holds:

$$\begin{aligned} \begin{aligned} \sin (p-r)\sin (q+r)&= \frac{\cos (q-p+2r)-\cos (p+q)}{2} \\&< \frac{\cos (q-p-2r)-\cos (p+q)}{2} = \sin (p+r)\sin (q-r). \end{aligned}\nonumber \\ \end{aligned}$$

The next lemma bounds the values of \(\ell _0\), the trigonometric Lagrange polynomial for \({\tilde{x}}_0\) associated with \(\{{\tilde{x}}_j\}_{j=-N}^N\) (see Eq. (4.3)), at an equally spaced node. This helps us bound the weight \({\tilde{w}}_0\) in Theorem 4.1.

Lemma A.1

Suppose N is even and \({\tilde{x}}_{-N},\ldots ,{\tilde{x}}_N\) are the perturbed nodes given in Eq. (4.1). We have

$$\begin{aligned} \left| \ell _0(kh)\right| = -\ell _0(kh) \ge \frac{\sin \left( \alpha h/2\right) }{\sin \left( (\left| k\right| + \alpha )h/2\right) }, \qquad -N \le k \le N,\quad k\ne 0, \end{aligned}$$

where \(h = 2\pi /(2N+1)\).


Since \(\ell _0(-x) = \ell _0(x)\) for all x, it suffices to show that Eq. (A.2) holds for \(k > 0\). By a simple counting argument, one can verify the equation \(\ell _0(kh) = \prod _{j=-N, j \ne 0}^N (\sin ((kh - {\tilde{x}}_j)/2) / \sin (-{\tilde{x}}_j/2)) \le 0\). Let \(s(x) = \sin (\left| xh/2\right| )\). Then, the numerator of \(\left| \ell _0(kh)\right| \) can be written as

$$\begin{aligned}&\prod _{\begin{array}{c} j = -N, \\ j \ne 0 \end{array}}^N \sin \left( \left| \frac{kh-{\tilde{x}}_{j}}{2}\right| \right) \\&\quad = \prod _{j=-N}^{-N+k-1} \sin \left( \left| \frac{kh-{\tilde{x}}_{j}}{2}\right| \right) \prod _{j=-N+k}^{-1} \sin \left( \left| \frac{kh-{\tilde{x}}_{j}}{2}\right| \right) \prod _{j=1}^{k} \sin \left( \left| \frac{kh-{\tilde{x}}_{j}}{2}\right| \right) \\&\qquad \prod _{j=k+1}^{N} \sin \left( \left| \frac{kh-{\tilde{x}}_{j}}{2}\right| \right) \\&\quad = \prod _{j=N-k+1}^N s(j + (-1)^{j+k}\alpha ) \prod _{j=k+1}^N s(j + (-1)^{j+k}\alpha ) \prod _{j=0}^{k-1} s(j + (-1)^{j+k+1}\alpha )\\&\qquad \prod _{j=1}^{N-k} s(j + (-1)^{j+k}\alpha ) \\&\quad = \prod _{j=1}^N s(j + (-1)^{j+k}\alpha ) \prod _{j=k+1}^N s(j + (-1)^{j+k}\alpha ) \prod _{j=0}^{k-1} s(j + (-1)^{j+k+1}\alpha ). \end{aligned}$$

Similarly, the denominator of \(\left| \ell _0(kh)\right| \) can be written as

$$\begin{aligned} \prod _{\begin{array}{c} j = -N, \\ j \ne 0 \end{array}}^N \sin \left( \left| \frac{{\tilde{x}}_j}{2}\right| \right) = \prod _{j=1}^N \sin ^2 \left( \left| \frac{{\tilde{x}}_j}{2}\right| \right) = \prod _{j=1}^N (s(j + (-1)^j\alpha ))^2. \end{aligned}$$

Using Eq. (A.1), we find the for every odd \(1 \le j \le N\), we have

$$\begin{aligned} s(j + \alpha )s((j+1) - \alpha ) \ge s(j - \alpha )s((j+1) + \alpha ). \end{aligned}$$

Hence, for every odd \(m_1\) and even \(m_2\) such that \(1\le m_1<m_2\le N\), Eq. (A.3) gives us

$$\begin{aligned} \prod _{j=m_1}^{m_2} s(j + (-1)^{j+1}\alpha ) \ge \prod _{j=m_1}^{m_2} s(j + (-1)^{j}\alpha ). \end{aligned}$$

We now consider the cases when k is even and odd separately.

Case I: \({\textbf{k}}\) is even. If k is even, then we have

$$\begin{aligned} -\ell _0(kh)&= \left| \ell _0(kh)\right| = \left( \prod _{j=1}^{k-2} \frac{s(j + (-1)^{j+1} \alpha )}{s(j + (-1)^{j} \alpha )}\right) \frac{s(k-1 + \alpha )}{s(k-1 - \alpha )} \frac{s(\alpha )}{s(k+\alpha )}\\&\ge \frac{s(\alpha )}{s(k+\alpha )} = \frac{\sin (\alpha h/2)}{\sin ((k+\alpha )h/2)}, \end{aligned}$$

where we used Eq. (A.4) with \(m_1 = 1\) and \(m_2 = k-2\).

Case II: \({\textbf{k}}\) is odd. If k is odd, then we have the following inequality on \(-\ell _0(kh)\):

$$\begin{aligned} -\ell _0(kh)&= \left| \ell _0(kh)\right| \ge \left( \prod _{j=k+2}^{N} \frac{s(j + (-1)^{j+1} \alpha )}{s(j + (-1)^{j} \alpha )}\right) \frac{s(k+1 - \alpha )}{s(k+1 + \alpha )} \frac{s(\alpha )}{s(k-\alpha )} \\&\ge \frac{s(k+1 - \alpha )}{s(k+1 + \alpha )} \frac{s(\alpha )}{s(k-\alpha )} \ge \frac{s(\alpha )}{s(k+\alpha )} = \frac{\sin (\alpha h/2)}{\sin ((k+\alpha )h/2)}, \end{aligned}$$

where the sequence of three inequalities are obtained from Eq. (A.4) by setting (1) \(m_1 = 1\) and \(m_2 = N\), (2) \(m_1 = k+2\) and \(m_2 = N\), and (3) \(m_1 = k\) and \(m_2 = k+1\), respectively. The proof is complete. \(\square \)

Appendix B: Only quadrature rules at ample perturbed nodes can have negative weights

For any \(0<\alpha <1/2\), we define \(N^{{\textrm{neg}}}_\alpha \) to be the smallest integer such that for \(N = N^{{\textrm{neg}}}_\alpha \) there exists a set of \(\alpha \)-perturbed quadrature nodes \(\{{\tilde{x}}_j\}_{j=-N}^{N}\) so that the associated exact quadrature rule on \({\mathscr {T}}_N\) has a negative weight. By Theorem 4.1, we know that \(N^{{\textrm{neg}}}_\alpha \) is finite for every \(\alpha \). Here, we derive an implicit lower bound Eq. (B.3) on \(N^{{\textrm{neg}}}_\alpha \) and provide a closed formula in Eq. (B.5), which proves \(\log ( N^{{\textrm{neg}}}_\alpha ) = \varTheta (\alpha ^{-1})\) as \(\alpha \rightarrow 0\). This is a rather technical result.

First, we define an equivalence relation on the indices. Let \(-N \le j, k \le N\) and define \(d(j,k) = \min \{\left| j-k\right| ,2N+1-\left| j-k\right| \}\). For a fixed j, define an equivalence class on \(\{-N, \ldots , j-1, j+1, \ldots , N\}\) by \(k_1 \sim _j k_2\) if and only if \(d(j,k_1) = d(j,k_2)\) and denote the equivalence class by \(S^j_{d(j,k_1)}\). Note that each of \(S^j_1, \ldots , S^j_N\) contains exactly 2 elements.

Lemma B.1

Let \(0\le \alpha < 1/2\), \(h = 2\pi /(2N+1)\), and \(\{{\tilde{x}}_k\}_{k=-N}^{N}\) be a set of \(\alpha \)-perturbed nodes. For fixed \(-N\le j\le N\) and for any \(-N \le k \le N\) with \(k \ne j\), we have

$$\begin{aligned} \left| \frac{\prod _{m=-N,m \ne j,k}^{N} \sin \left( \frac{x_{k} - {\tilde{x}}_m}{2}\right) }{\prod _{m=-N,m\ne j}^{N} \sin \left( \frac{{\tilde{x}}_j - {\tilde{x}}_m}{2}\right) }\right| \le \frac{1}{\sin \left( \frac{d(j,k)h}{2}\right) } \prod _{m=1}^N \frac{\sin ^2\left( \frac{mh + \alpha h}{2}\right) }{\sin \left( \frac{mh - 2\alpha h}{2}\right) \sin \left( \frac{mh}{2}\right) }, \end{aligned}$$

where \(x_k = {\tilde{x}}_j + (k-j)h\).Footnote 4


Let \(d = d(j,k)\) and \(j'\) be the element in \(S^k_{d}\) that is not equal to j. Then, we can write the numerator and denominator of Eq. (B.1) as

$$\begin{aligned} A&= \prod _{\begin{array}{c} m=-N,\\ m \ne j,k \end{array}}^{N} \sin \left( \frac{x_{k} - {\tilde{x}}_m}{2}\right) = \sin \left( \frac{x_k - {\tilde{x}}_{j'}}{2}\right) \prod _{\begin{array}{c} m=1, m \ne d, \\ S^k_m = \{k_1, k_2\} \end{array}}^{N} \left[ \sin \left( \frac{x_{k} - {\tilde{x}}_{k_1}}{2}\right) \sin \left( \frac{x_{k} - {\tilde{x}}_{k_2}}{2}\right) \right] ,\\ B&= \prod _{\begin{array}{c} m=-N,\\ m\ne j \end{array}}^{N} \sin \left( \frac{{\tilde{x}}_j - {\tilde{x}}_m}{2}\right) = \prod _{\begin{array}{c} m=1,\\ S^j_m = \{k_1, k_2\} \end{array}}^{N} \left[ \sin \left( \frac{{\tilde{x}}_j - {\tilde{x}}_{k_1}}{2}\right) \sin \left( \frac{{\tilde{x}}_j - {\tilde{x}}_{k_2}}{2}\right) \right] . \end{aligned}$$

Suppose \(S_m^k = \{k_1,k_2\}\), where \(1 \le m \le N\). Then, the distance between the two nodes satisfies \(d_{\mathbb {T}}(e^{ix_k}, e^{i{\tilde{x}}_{k_1}}) + d_{\mathbb {T}}(e^{ix_k}, e^{i{\tilde{x}}_{k_2}}) \le 2mh + 2\alpha h\) and \(d_{\mathbb {T}}(e^{ix_k}, e^{i{\tilde{x}}_{k_1}}), d_{\mathbb {T}}(e^{ix_k}, e^{i{\tilde{x}}_{k_2}}) \le mh+2\alpha h\), where \(d_{\mathbb {T}}(e^{ix}, e^{iy}) = \min \{\left| x-y\right| , 2\pi - \left| x-y\right| \}\) is the “distance" between \(e^{ix}\) and \(e^{iy}\) on the unit circle for \(x,y\in [-\pi ,\pi )\). Therefore, by elementary calculus, we have

$$\begin{aligned} \left| \sin \left( \frac{x_{k} - {\tilde{x}}_{k_1}}{2}\right) \sin \left( \frac{x_{k} - {\tilde{x}}_{k_2}}{2}\right) \right| \le \sin ^2 \left( \frac{mh+\alpha h}{2}\right) , \end{aligned}$$

where the lefthand side of the inequality above is maximized when \(d_{\mathbb {T}}(e^{ix_k}, e^{i{\tilde{x}}_{k_1}}) = d_{\mathbb {T}}(e^{ix_k}, e^{i{\tilde{x}}_{k_2}}) = mh+\alpha h\). Similarly, suppose \(S_m^j = \{k_1,k_2\}\). We have \(d_{\mathbb {T}}(e^{i{\tilde{x}}_j}, e^{i{\tilde{x}}_{k_1}}) + d_{\mathbb {T}}(e^{i{\tilde{x}}_j}, e^{i{\tilde{x}}_{k_2}}) \ge 2mh-2\alpha h\) and \(d_{\mathbb {T}}(e^{i{\tilde{x}}_j}, e^{i{\tilde{x}}_{k_1}}), d_{\mathbb {T}}(e^{i{\tilde{x}}_j}, e^{i{\tilde{x}}_{k_2}}) \ge mh-2\alpha h\). Hence,

$$\begin{aligned} \left| \sin \left( \frac{{\tilde{x}}_j - {\tilde{x}}_{k_1}}{2}\right) \sin \left( \frac{{\tilde{x}}_j - {\tilde{x}}_{k_2}}{2}\right) \right| \ge \sin \left( \frac{mh - 2\alpha h}{2}\right) \sin \left( \frac{mh}{2}\right) , \end{aligned}$$

where the lefthand side of the inequality above is minimized when \(d_{\mathbb {T}}(e^{i{\tilde{x}}_j}, e^{i{\tilde{x}}_{k_i}}) = mh\) and \(d_{\mathbb {T}}(e^{i{\tilde{x}}_j}, e^{i{\tilde{x}}_{k_{3-i}}}) = mh - 2\alpha h\) for \(i = 1\) or 2. This gives us

$$\begin{aligned} \left| \frac{A}{B}\right| \le \frac{\sin \left( \frac{dh + 2\alpha h}{2}\right) \prod _{m=1, m \ne d}^{N} \sin ^2\left( \frac{mh+\alpha h}{2}\right) }{\prod _{m=1}^N \sin \left( \frac{mh - 2\alpha h}{2}\right) \sin \left( \frac{mh}{2}\right) } \le \frac{1}{\sin \left( \frac{dh}{2}\right) } \prod _{m=1}^N \frac{\sin ^2\left( \frac{mh+\alpha h}{2}\right) }{\sin \left( \frac{mh - 2\alpha h}{2}\right) \sin \left( \frac{mh}{2}\right) }, \end{aligned}$$

as desired. \(\square \)

It is worth observing that A/B in the proof of Lemma B.1 is almost the Lagrange polynomial at \(x_j\). This connection is made precise in the next lemma.

Lemma B.2

Using the same notation as Lemma B.1, if \(\ell _j\) is the jth trigonometric Lagrange basis polynomial for \({\tilde{x}}_j\) associated with \(\{{\tilde{x}}_k\}_{k=-N}^{N}\), then

$$\begin{aligned} \left| \ell _j(x_{k_1}) + \ell _j(x_{k_2})\right| \le \frac{\pi \alpha }{d} \prod _{m=1}^N \frac{(m + \alpha )^2}{(m - 2\alpha )m}, \end{aligned}$$

where \(S^j_d = \{k_1, k_2\}\).


Let \(d = d(j,k)\) and suppose that \(1 \le d \le N\). For \(i = 1, 2\), we have

$$\begin{aligned} \ell _j(x_{k_i})&= \frac{\prod _{m=-N,m\ne j}^{N} \sin \left( \frac{x_{k_i} - {\tilde{x}}_m}{2}\right) }{\prod _{m=-N,m\ne j}^{N} \sin \left( \frac{{\tilde{x}}_j - {\tilde{x}}_m}{2}\right) } = \frac{\sin \left( \frac{x_{k_i} - {\tilde{x}}_{k_i}}{2}\right) \prod _{m=-N,m\ne j, k_i}^{N} \sin \left( \frac{x_{k_i} - {\tilde{x}}_m}{2}\right) }{\prod _{m=-N,m\ne j}^{N} \sin \left( \frac{{\tilde{x}}_j - {\tilde{x}}_m}{2}\right) }. \end{aligned}$$

By Lemma B.1, we find that

$$\begin{aligned} \left| \frac{\prod _{m=-N,m\ne j, k_i}^{N} \sin \left( \frac{x_{k_i} - {\tilde{x}}_m}{2}\right) }{\prod _{m=-N,m\ne j}^{N} \sin \left( \frac{{\tilde{x}}_j - {\tilde{x}}_m}{2}\right) }\right| \le \frac{\pi }{dh} \prod _{m=1}^N \frac{(mh + \alpha h)^2}{(mh - 2\alpha h)(mh)} = \frac{\pi }{dh} \prod _{m=1}^N \frac{(m + \alpha )^2}{(m - 2\alpha )m}, \end{aligned}$$

where we need the fact that \((2/\pi )x \le \sin x\) for all \(0 \le x \le \pi /2\) and the fact that \(\sin y / \sin x \le y / x\) for all \(0< x \le y < \pi \). We now prove Eq. (B.2) by considering the cases \(\ell _j(x_{k_1})\ell _j(x_{k_2}) \le 0\) and \(\ell _j(x_{k_1})\ell _j(x_{k_2}) > 0\) separately.

Case 1: \({{\varvec{\ell }}}_{\textbf{j}}(\textbf{x}_{\textbf{k}_{\textbf{1}}}){{\varvec{\ell }}}_{\textbf{j}}(\textbf{x}_{\textbf{k}_{\textbf{2}}}) \le \textbf{0}\). Since \(d_{\mathbb {T}}(e^{ix_{k_i}}, e^{i{\tilde{x}}_{k_i}}) \le 2 \alpha h\) for \(i = 1, 2\), we have

$$\begin{aligned}&\left| \ell _j(x_{k_1}) + \ell _j(x_{k_2})\right| \le \max \{\left| \ell _j(x_{k_1})\right| , \left| \ell _j(x_{k_2})\right| \} \\&\le \max _{i = 1, 2}\left[ \frac{d_{\mathbb {T}}(e^{ix_{k_i}}, e^{i{\tilde{x}}_{k_i}})}{2}\right] \frac{\pi }{dh} \prod _{m=1}^N \frac{(m + \alpha )^2}{(m - 2\alpha )m} \le \frac{\pi \alpha }{d} \prod _{m=1}^N \frac{(m + \alpha )^2}{(m - 2\alpha )m}, \end{aligned}$$

where we used the fact that \(\sin (x) \le x\) for all \(0 \le x \le \pi /2\).

Case 2: \({\varvec{\ell }}_{\textbf{j}}(\textbf{x}_{\textbf{k}_{\textbf{1}}}){{\varvec{\ell }}}_{\textbf{j}}(\textbf{x}_{\textbf{k}_{\textbf{2}}}) > \textbf{0}\). We claim that

$$\begin{aligned} \frac{\prod _{m=-N,m\ne j, k_1}^{N} \sin \left( \frac{x_{k_1} - {\tilde{x}}_m}{2}\right) }{\prod _{m=-N,m\ne j}^{N} \sin \left( \frac{{\tilde{x}}_j - {\tilde{x}}_m}{2}\right) } \frac{\prod _{m=-N,m\ne j, k_2}^{N} \sin \left( \frac{x_{k_2} - {\tilde{x}}_m}{2}\right) }{\prod _{m=-N,m\ne j}^{N} \sin \left( \frac{{\tilde{x}}_j - {\tilde{x}}_m}{2}\right) } < 0. \end{aligned}$$

The claim follows because if we assume, without loss of generality,Footnote 5 that \(k_1 < k_2\), then there are an even number of integers m, not including j, such that \(k_1< m < k_2\).Footnote 6 When \(m \ne k_1, k_2\), the signs of \(\sin ((x_{k_1}-{\tilde{x}}_m)/2)\) and \(\sin ((x_{k_2}-{\tilde{x}}_m)/2)\) are different if and only if \(k_1< m < k_2\). The claim follows from the fact that \(\sin ((x_{k_2}-{\tilde{x}}_{k_1})/2) \sin ((x_{k_1}-{\tilde{x}}_{k_2})/2) < 0\) because the remaining terms multiply to a positive number. Hence, we have \(\sin ((x_{k_1} - {\tilde{x}}_{k_1})/2) \sin ((x_{k_2} - {\tilde{x}}_{k_2})/2) < 0\) so that \((x_{k_1} - {\tilde{x}}_{k_1})(x_{k_2} - {\tilde{x}}_{k_2}) < 0\). Let \(\delta := {\tilde{x}}_j - jh\). Then, \(x_{k_i} - {\tilde{x}}_{k_i} < \alpha h + \delta \) if \(x_{k_i} - {\tilde{x}}_{k_i} > 0\) and \({\tilde{x}}_{k_i} - x_{k_i} < \alpha h - \delta \) if \(x_{k_i} - {\tilde{x}}_{k_i} < 0\). Hence, \((x_{k_1} - {\tilde{x}}_{k_1})(x_{k_2} - {\tilde{x}}_{k_2}) < 0\) implies \(d_{\mathbb {T}}(e^{ix_{k_1}}, e^{i{\tilde{x}}_{k_1}}) + d_{\mathbb {T}}(e^{ix_{k_2}}, e^{i{\tilde{x}}_{k_2}}) \le 2\alpha h\). By elementary calculus, we find that

$$\begin{aligned} \left| \sin \left( \frac{x_{k_1} - {\tilde{x}}_{k_1}}{2}\right) \right| + \left| \sin \left( \frac{x_{k_2} - {\tilde{x}}_{k_2}}{2}\right) \right| \le 2\sin \left( \frac{\alpha h}{2}\right) \le \alpha h. \end{aligned}$$

Putting this together, we obtain

$$\begin{aligned} \left| \ell _j(x_{k_1}) + \ell _j(x_{k_2})\right|&= \left| \ell _j(x_{k_1})\right| + \left| \ell _j(x_{k_2})\right| \\&\le \left( \left| \sin \left( \frac{x_{k_1} - {\tilde{x}}_{k_1}}{2}\right) \right| + \left| \sin \left( \frac{x_{k_2} - {\tilde{x}}_{k_2}}{2}\right) \right| \right) \frac{\pi }{dh} \prod _{m=1}^N \frac{(m + \alpha )^2}{(m - 2\alpha )m} \\&\le \frac{\pi \alpha }{d} \prod _{m=1}^N \frac{(m + \alpha )^2}{(m - 2\alpha )m}, \end{aligned}$$

as desired. \(\square \)

We are now ready to prove a lower bound on \(N^{{\textrm{neg}}}_\alpha \).

Theorem B.1

We have

$$\begin{aligned} g(N^{{\textrm{neg}}}_\alpha ) > \frac{1}{\pi \alpha }, \qquad g(N):= \prod _{m=1}^{N} \left[ \frac{(m + \alpha )^2}{(m - 2\alpha )m}\right] \left[ \sum _{d=1}^{N} \frac{1}{d}\right] \end{aligned}$$

for \(0<\alpha <1/2\).


We define \(g: {\mathbb {N}}\rightarrow {\mathbb {R}}\) as in Eq. (B.3). It suffices to show that \(g(N) \le 1/(\pi \alpha )\) implies that a quadrature rule at \(\alpha \)-perturbed nodes of degree N contains no negative weight. Let \(\{{\tilde{x}}_j\}_{j=-N}^{N}\) be a set of \(\alpha \)-perturbed nodes. Let \({\tilde{w}}_j\) be the quadrature weight associated with \({\tilde{x}}_j\) and let \(\ell _j\) be the corresponding trigonometric Lagrange basis polynomial. Let \(\{x_k\}_{k=-N}^{N}\), which may depend on j, be defined as in Lemma B.1. Then, we have \({\tilde{w}}_j = (2\pi /(2N+1))\sum _{k=-N}^{N} \ell _j(x_k)\). By Lemma B.2, we have

$$\begin{aligned}&\sum _{k=-N}^{N} \ell _j(x_k) = \ell _j(x_j) + \sum _{k=-N, k \ne j}^{N} \ell _j(x_k) \ge 1 - \sum _{d=1, S^j_d=\{k_1,k_2\}}^{N} \left| \ell _j(x_{k_1}) + \ell _j(x_{k_2})\right| \\&\ge 1 - \sum _{d=1}^{N} \left[ \frac{\pi \alpha }{d} \prod _{m=1}^N \frac{(m + \alpha )^2}{(m - 2\alpha )m}\right] = 1-\pi \alpha \prod _{m=1}^N \left[ \frac{(m + \alpha )^2}{(m - 2\alpha )m}\right] \left[ \sum _{d=1}^N \frac{1}{d}\right] \ge 0. \end{aligned}$$

This proves all quadrature weights are non-negative. \(\square \)

For small \(\alpha \), we can make the statement in Theorem B.1 more explicit.

Corollary B.1

For \(0<\alpha <0.15\) and a real number \(L > 0\) such that

$$\begin{aligned} (\alpha + L)e^{4L} \le \frac{\varGamma (1+\alpha )^2}{\pi \varGamma (1-2\alpha )}, \end{aligned}$$

where \(\varGamma \) is the gamma function, we find that \(N^{{\textrm{neg}}}_\alpha \) satisfies the following inequality:

$$\begin{aligned} \log (N^{{\textrm{neg}}}_\alpha +1+\alpha ) + \frac{1}{2N^{{\textrm{neg}}}_\alpha -1} > (1-\gamma ) + \frac{L}{\alpha }, \end{aligned}$$

where \(\gamma \approx 0.57722\) is the Euler–Mascheroni constant [16].


We aim to show that Eq. (B.4) implies Eq. (B.5). First, we use Gautschi’s inequality [57] to obtain

$$\begin{aligned}{} & {} \prod _{m=1}^N \frac{(m + \alpha )^2}{(m - 2\alpha )m} = \frac{\varGamma (1 - 2\alpha ) \varGamma (N + 1 + \alpha )^2}{\varGamma (N + 1 - 2\alpha )\varGamma (N + 1)\varGamma (1 + \alpha )^2}\nonumber \\{} & {} \le \frac{\varGamma (1 - 2\alpha )}{\varGamma (1 + \alpha )^2} (N + 1 + \alpha )^{4\alpha }. \end{aligned}$$

Assume that \(\log (N+1+\alpha ) + 1/(2N-1) \le (1-\gamma ) + L/\alpha \), for an integer \(N>0\). We have \(\log (N+1+\alpha ) \le L/\alpha \). Hence, \(N+1+\alpha \le e^{L/\alpha }\). By Eq. (B.6), we find that

$$\begin{aligned} g(N)\le & {} \frac{\varGamma (1 - 2\alpha )}{\varGamma (1 + \alpha )^2} (N + 1 + \alpha )^{4\alpha } \left( \log (N) + \gamma + \frac{1}{2N - 1}\right) \\{} & {} \le \frac{\varGamma (1 - 2\alpha )}{\varGamma (1 + \alpha )^2} e^{4L} \left( 1 + \frac{L}{\alpha }\right) \\\le & {} \frac{1}{\pi \alpha }, \end{aligned}$$

where we used the fact that \(\sum _{d=1}^N d^{-1} \le \log (N) + \gamma + 1/(2N-1) \le 1+L/\alpha \) [16] and the last inequality follows from Eq. (B.4). Since g(N) is an increasing function of N, we know from Eq. (B.3) that \(N < N^{{\textrm{neg}}}_\alpha \). Moreover, this holds for all \(N > 1\) that satisfies \(\log (N+1+\alpha ) + 1/(2N-1) \le (1-\gamma ) + L/\alpha \). When \(\alpha < 0.15\), we see that \(N^{{\textrm{neg}}}_\alpha \ge 2\) by Eq. (B.3) and this proves Eq. (B.5). \(\square \)

Corollary B.1 implies that \(\log (N^{{\textrm{neg}}}_\alpha ) = \varOmega (\alpha ^{-1})\) and by Theorem 4.1, we conclude that \(\log ( N^{{\textrm{neg}}}_\alpha ) = \varTheta (\alpha ^{-1})\) as \(\alpha \rightarrow 0\).

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Yu, A., Townsend, A. On the stability of unevenly spaced samples for interpolation and quadrature. Bit Numer Math 63, 23 (2023).

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