Lagmental Vicfred

Chebyshev Nodes Suppress Endpoint Oscillation by Vicfred

tags: mathematics, numerical analysis, root finding, interpolation

Placing interpolation nodes at cosine points reduces the largest nodal-product magnitude. The example is deliberately concrete; it is a test of the statement, not a substitute for it.

Start locally

Numerical analysis replaces an exact object \(x^\ast\) by approximations \(x_n\) with controlled error. Stability asks how rounding or data perturbations affect the answer.

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

There are two layers here: the object \(\mathsf D\) and the law \(\mathsf C\). Writing them separately makes the direction of \(\Longrightarrow\) visible and keeps an accidental converse from slipping in.

$$ \prod_{k=0}^{n}(x-x_k)=2^{-n}T_{n+1}(x) $$

Compute before generalising

Now evaluate one representative case. The result should agree with the structural law above, but it is obtained without assuming the conclusion.

$$ T_n(\cos\theta)=\cos(n\theta),\qquad\|T_n\|_{\infty,[-1,1]}=1 $$

The global view

The formula is reusable precisely because it says which pieces are structural and which belong only to the worked example.

$$ \begin{aligned} \mathsf{D}\;&:\quad x_k=\cos\frac{(2k+1)\pi}{2(n+1)},\qquad k=0,\ldots,n,\\[5pt] \mathsf{C}\;&:\quad \prod_{k=0}^{n}(x-x_k)=2^{-n}T_{n+1}(x). \end{aligned} $$

Edge conditions

Fast local convergence is not a global guarantee. Newton's method can diverge, interpolation can oscillate, and a small residual can coexist with a large forward error.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] \prod_{k=0}^{n}(x-x_k)=2^{-n}T_{n+1}(x) \end{gathered}} $$

The notation is dense, but it is doing honest work: every delimiter records scope and every index records dependence. Removing one should require a mathematical reason.

This article was posted on Tue 16 March 2021. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.