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

Define the Lagrange cardinal polynomialsDistinct nodes make every denominator nonzero. At $x_s$ these polynomials satisfy ${\ell }_r(x_s)={\delta }_{rs}$, so the Lagrange interpolation polynomial isIt has degree at most $n$ and takes the prescribed data values. If two such polynomials existed, their difference would have $n+1$ distinct roots but degree at most $n$, forcing that difference to be zero. This also proves uniqueness.

With $w_s=1/{\pi }_{n,s}(x_s)$, the Lagrange interpolation polynomial can be written, away from its nodes,Apply the same formula to the constant data $u_s=1$. Uniqueness of polynomial interpolation gives ${\sum }_s{\ell }_s=1$, and henceDivide the two identities to obtain the barycentric interpolation formulaAt $x=x_r$, use the continuous limiting value $u_r$. Multiplying every weight by the same nonzero constant cancels between numerator and denominator, so only relative weights are needed in computation.

For uniformly spaced nodes,There are $n-s$ negative factors, coming from $j>s$. Thus the weights fixed by part (b) satisfy this formula exactly. The printed sign $(-1{)}^s$ agrees for even $n$; for odd $n$ it differs by the common factor $-1$. Rescaling all weights by $(-1{)}^n$ gives the printed convention without changing the barycentric interpolation formula.

For even $n$, the required magnitude isStirling's formula givesThus central weights are exponentially larger in magnitude than endpoint weights. Their alternating signs can cause severe cancellation in polynomial interpolation at equispaced nodes; they reflect the unfavorable high-degree interpolation geometry associated with Runge's phenomenon. A barycentric evaluation formula improves the implementation but does not remove the intrinsic sensitivity or guarantee convergence for every smooth function. The comparison with the bounded relative weights at Chebyshev-Lobatto nodes is decisive.

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.