Past exam of the approximation theory and finite element methods course of the University of Oxford / 2009 / A / 1 / d

Use the Chebyshev polynomial identities $T_0=1$, $T_1=x$, and $T_{j+1}=2x{T}_j-T_{j-1}$. Induction shows that $T_j$ has degree $j$ with leading coefficient $2^{j-1}$ for $j\ge 1$. The trigonometric definition also givesThereforeThe leading coefficient is $2^{1-n}{n}^{-1}\cdot n{2}^{n-1}=1$, and the degree is $n+1$. The endpoints are roots through $x^2-1$; the other $n-1$ points $\cos (s\pi /n)$ are roots through $T_n^{\prime }$. They are distinct, so this monic polynomial is exactly ${\pi }_n$.

Differentiating the product ${\pi }_n(x)=(x-x_s){\pi }_{n,s}(x)$ at $x_s$ gives ${\pi }_n^{\prime }(x_s)={\pi }_{n,s}(x_s)=w_s^{-1}$. For an interior node, differentiate the trigonometric form using $d/dx=-(\sin \theta {)}^{-1}d/d\theta$:At ${\theta }_s=s\pi /n$, this is $2^{1-n}n(-1{)}^s$. At either endpoint, taking the limit makes the two terms equal, doubling the value: ${\pi }_n^{\prime }(x_s)=2^{2-n}n(-1{)}^s$ for $s=0,n$.

After removing the common factor $2^{n-1}/n$, the barycentric interpolation formula therefore usesThese Chebyshev-Lobatto node weights differ in magnitude by only a factor of two.