Past exam of the approximation of functions course of the University of Oxford / 2011 / 1

The Lagrange interpolation polynomial isThe interpolation nodes are distinct, so these weights are well defined. Applying the same Lagrange interpolation to the constant polynomial one gives $1={\sum }_r{w}_r{\pi }_{n,r}(x)$. Dividing this identity and the formula for $p$ by ${\pi }_n(x)$ gives the barycentric interpolation formulaThis displayed quotient is used away from the interpolation nodes; at $x=x_r$ its removable singularity is evaluated by setting $p(x_r)=u_r$. Multiplying every weight by the same nonzero constant leaves the quotient unchanged.

Chebyshev points are generally more reliable at high degree. The Chebyshev-Lobatto nodes, or the affine images of the roots of a Chebyshev polynomial, cluster near the interval endpoints and control the amplification measured by the Lebesgue constant of interpolation. Equally spaced interpolation nodes can produce large endpoint oscillations and the Runge phenomenon. This is a comparison of numerical robustness and general approximation behaviour, not a claim that one set minimizes the error for every individual function.

Assume $h>0$. The functional is the third forward difference: each forward difference lowers the polynomial degree by one, so it annihilates every polynomial of degree at most two. Directly, $X[1]=-1+3-3+1=0$, $X[x]=3h-6h+3h=0$, and $X[x^2]=3h^2-12h^2+9h^2=0$.

The Peano kernel theorem now gives $X[u]={\int }_0^{3h}K(t)u^{(3)}(t)dt$, where the Peano kernel of the third forward difference isHere $s_{+}=\max (s,0)$ is the positive-part truncated power function. The normalized quadratic cardinal B-spline isIts three polynomial pieces agree with the displayed kernel after rescaling, giving $\boxed{K(t)=h^2{B}_2(t/h)}$. Its knots are $0,h,2h,3h$ in the physical variable, and $K$ is positive throughout $(0,3h)$.

The kernel mass can be evaluated without integrating each piece: apply its representation to $u(x)=x^3/6$, whose third derivative is one. This gives ${\int }_0^{3h}K(t)dt=X[x^3/6]=h^3$. The weighted mean value theorem for integrals with a nonnegative weight therefore yieldsContinuity of $u^{(3)}$ and strict positivity of the weight in the interior place an attained weighted-average value in that open interval, including the constant-derivative case.