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Fix $n \geq 2$ an integer and let $x_{k+1} = k/n , k = 0, 1,\cdots, n-1.$ Let now $L_k$ be the Lagrange polynomial given by $$L_k(x) = \prod_{j =1, j \neq k}^{n} \frac{x-x_j}{x_k - x_j}.$$ Define the matrix $A = [a_{i,j}]$ whose entries are given by

$$a_{i,j} := \int_{0}^{x_i}L_j(x)\, dx.$$ I'm interested in finding the infinity norm of $A$ $$\|A\|_\infty := \max_{ 1\leq i\leq n} \sum_{j=1}^n |a_{i,j}|.$$ My attempt: It seems (I used Mathematica for $n = 2, ..., 8$) that the maximum is reached in the last line of the matrix where we have $a_{n,j} \geq 0.$ Thus, we have $$\|A\|_\infty = \sum_{j=1}^n |a_{n,j}| = \sum_{j=1}^n\int_{x_1}^{x_n}L_j(x).$$ It's clear that $\sum_{j=1}^nL_j(x) =1.$ Then, we have $$\|A\|_\infty = n-1.$$ Thank you for any hint.

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  • $\begingroup$ I am curious where this question is coming from. Why do you suspect that $\left\| A \right\|_{\infty} = n-1$? I am basically trying to answer a similar question but did not find that equality. $\endgroup$
    – Daniel
    May 18 '21 at 12:38
  • $\begingroup$ Also, shouldn't it be $\sum_{j=1}^{n} \left| a_{n,j} \right| = \sum_{j=1}^{n} \left| \int_{x_1}^{x_n} L_j(x)~dx \right|$ ? Then you can't use that $\sum_{j=1}^{n} L_j(x) = 1$ any more, can you? $\endgroup$
    – Daniel
    May 18 '21 at 12:41
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If you let

\begin{equation} \Lambda_n := \max_{0 \leq x \leq 1} \sum_{j=1}^{n} \left| l_j(x) \right| \end{equation}

denote the Lebesgue constant, then it is easy to show that

\begin{equation} \left\| A \right\|_{\infty} \leq \Lambda_n. \end{equation}

While there are few direct formulas for the Lebesgue constant, there are a few estimates on how it grows with $n$. For equidistant points, in particular, you have

\begin{equation} \Lambda_n = \frac{2^n}{e n \log(n)} \ \text{as} \ n \to \infty. \end{equation}

I believe this bound is extremely pessimistic, but I was not able to significantly improve it so far. Hence my interest in the source for your hypothesis that $\left\| A \right\|_{\infty} = n - 1$.

Addendum: I don't think the hypothesis holds for equidistant nodes. If I set $n=15$, I get $\left\| A \right\|_{\infty} \approx 20.3$. From there, the norm very quickly grows.

Eisinberg, A.; Fedele, G.; Franzè, G., Lebesgue constant for Lagrange interpolation on equidistant nodes, Anal. Theory Appl. 20, No. 4, 323-331 (2004). ZBL1069.41002.

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