[Math] Interpolation- Barycentric coefficients for nodes that are Chebyshev points of the second kind.

interpolationnumerical methodspolynomials

So I came across the following theorem:

If the interpolation node are Chebyshev points of the second kind given by :
$$ x_k=\cos \left( \frac{j\pi}{n}\right) \qquad ( 0 \leq j \leq n) $$

Then the barycetric coefficients are :

$\beta_j=(-1)^j\theta_j $

Where $\theta_j=\frac{1}{2}$ when $j=0$ or $j=n$ and $\theta_j=1$ in other cases.

But how come? I can't seem to come across the proof of this, and I can't seem to get this coefficients by calculating $\beta_i=1/\prod\limits_{j=0,j\neq i}^{j=n} \left( \cos \left( \frac{i\pi}{n}\right) – \cos \left( \frac{j\pi}{n}\right)\right)$

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The formula for $\beta_i$ is $(X - x_i)/P_n(X) = 1/P_n'(x)$ where $P_n$ is the monic polynomial with roots $x_i = \cos \frac{i\pi}{n}$. The polynomial is $P_n(x) = (1 + \cos (n \arccos x))/2^n$, which can be differentiated and the derivative evaluated at interpolation points.

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[Math] Computation of coefficients of Lagrange polynomials

I guess, the easiest way to get expression for Lagrange coefficients is to use a determinant representation. "Beginner's guide to mapping simplexes affinely", section "Lagrange interpolation", describes a general formula of Lagrange polynomial that interpolates $(a_0;b_0)$, $\dots$, $(a_n;b_n)$ $$ P(x) = (-1) \frac{ \det \begin{pmatrix} 0 & b_0 & b_1 & \cdots & b_n \\ x^n & a_0^n & a_1^n & \cdots & a_n^n \\ x^{n-1} & a_0^{n-1} & a_1^{n-1} & \cdots & a_n^{n-1} \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ 1 & 1 & 1 & \cdots & 1 \\ \end{pmatrix} }{ \det \begin{pmatrix} a_0^n & a_1^n & \cdots & a_n^n \\ a_0^{n-1} & a_1^{n-1} & \cdots & a_n^{n-1} \\ \cdots & \cdots & \cdots & \cdots \\ 1 & 1 & \cdots & 1 \\ \end{pmatrix} }. $$ The standard representation can be obtained from this expression by Laplace expansion along the first row. But we can do Laplace expansion along the first column as well and it will lead us to formula for coefficients at $x^i$. Result should look as follows $$ c_i = (-1)^{n-i} \frac{ \det \begin{pmatrix} b_0 & b_1 & \cdots & b_n \\ a_0^n & a_1^n & \cdots & a_n^n \\ % x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\ \cdots & \cdots & \cdots & \cdots \\ a_0^{i-1} & a_1^{i-1} & \cdots & a_n^{i-1} \\ a_0^{i+1} & a_1^{i+1} & \cdots & a_n^{i+1} \\ \cdots & \cdots & \cdots & \cdots \\ 1 & 1 & \cdots & 1 \\ \end{pmatrix} }{ \det \begin{pmatrix} a_0^n & a_1^n & \cdots & a_n^n \\ a_0^{n-1} & a_1^{n-1} & \cdots & a_n^{n-1} \\ \cdots & \cdots & \cdots & \cdots \\ 1 & 1 & \cdots & 1 \\ \end{pmatrix} }, $$ where $c_i$ is coefficient at $x^i$ in the polynomial. For practical example you may want to check "Workbook on mapping simplexes affinely", section "Lagrange interpolation".

Numerical Methods – Hermite Interpolation of e^x: Analyzing Derivative Behavior

I have found the cause, it seems that the Hermite Interpolation like other Interpolation methods most likely is unstable in the sense that if $\left(\tilde{x}_{i},\tilde{f}(\tilde{x}_{i})\right)$ are the perturbed values of $(x_i,f(x_i))$ say due to round off error or measurement then the resulting interpolating polynomials will wildly deviate from the true Hermite Interpolating polynomial (which of course is unique). Especially when the Lebesgue constant is large. To remedy this increase the working precision in the second line of the code above

x[k_] := N[Cos[(k + 1/2) \[Pi]/(n + 1)],20]

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[Math] Computation of coefficients of Lagrange polynomials

Numerical Methods – Hermite Interpolation of e^x: Analyzing Derivative Behavior

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