Past exam of the numerical analysis course of the University of Oxford / 2015 / 1

For $a<b$, the one-panel trapezoidal rule isIt integrates the linear interpolation between the two endpoint samples. Indeed the affine functionhas definite integral $T(f)$. The two quadrature nodes are $a,b$ and both quadrature weights are $(b-a)/2$.

It suffices to assume $f\in C^2[a,b]$. The polynomial interpolation error for the endpoint linear interpolant gives, for every interior $x$, some ${\xi }_x\in (a,b)$ withLet $M={\max }_{[a,b]}|f^{\prime \prime }|$, which exists by continuity of the second derivative and the extreme value theorem. Since $(x-a)(b-x)\ge 0$,Integrating this pointwise inequality and using the triangle inequality for an integral yieldsThusThis derivation does not require choosing ${\xi }_x$ as a continuous function of $x$; the pointwise upper bound is sufficient.

The constant is sharp. Take the quadratic polynomial $f(x)=(x-a{)}^2$ and write $L=b-a$. Its exact integral is $L^3/3$, its trapezoidal rule value is $L^3/2$, and $\max |f^{\prime \prime }|=2$. ConsequentlyEquality holds, so no smaller universal constant than $1/12$ works under these assumptions. This is the sharp one-panel trapezoidal error bound.

The trapezoidal rule is exact for every affine function $f(x)=\alpha +\beta x$, equivalently every polynomial of polynomial degree at most one: the linear interpolant is the function itself. These functions have $f^{\prime \prime }=0$, so the preceding upper bound is also zero. There is no contradiction: zero error is precisely what the bound predicts.

The degree of precision of a quadrature rule is one, since the quadratic polynomial in part (i) is not integrated exactly. This does not exclude exactness for particular higher-degree or nonpolynomial functions. For example $f(x)=(x-m{)}^3$, $m=(a+b)/2$, has zero integral and opposite endpoint values, so the trapezoidal rule also gives zero. Its second derivative is generally nonzero, making the upper bound positive; an upper bound need not be attained. Signed quadrature error can cancel.

On a partition of an interval $a=x_0<x_1<\cdots <x_N=b$, apply the trapezoidal rule separately to each panel and add:For uniform spacing $h=(b-a)/N$, this composite trapezoidal rule becomesEach interior sample occurs in two panels; their half-quadrature weights combine to $h$. The endpoint quadrature weights are $h/2$.

Use the nonpolynomial function $f(x)=1+\sin x$ on $[-1,1]$. Its exact integral isbecause the sine is an odd function.

For every uniform partition of an interval with $N\ge 1$ panels, the quadrature nodes satisfy $x_{N-j}=-x_j$. Their composite trapezoidal rule quadrature weights agree, so the two sampled sine values cancel. If there is a node at zero its sine value is already zero. The constant term is integrated exactly, givingThis is symmetric trapezoidal exactness for odd functions, applied after adding a constant function so that the resulting definite integral is nonzero. It is exactness for this particular integrand, rather than a claim that every nonpolynomial function is integrated exactly.

Write $E(h)$ for the supplied approximation. The asymptotic expansion iswhere $O$ is Big O notation. At the refined parameter $h/3$,Multiply this identity by nine and subtract the first one. The unknown leading truncation error coefficient cancels:Thus the Richardson extrapolation estimate isThe two supplied values giveThe correction to the finer approximation is $(3-2)/8=1/8$. Neglecting the higher-order remainder would also give $c=9/8$.

The conclusion is asymptotic: the leading error order improves from two to at least three when both parameter values are in the range of the stated asymptotic expansion. With no bound on the remainder constant, the two finite observations alone do not guarantee that $25/8$ is closer to $Y$. Indeed, for any proposed value $L$, the polynomial modelsatisfies $E(1)=2$, $E(1/3)=3$ and the required asymptotic expansion with true limit $L$. This is why finite observations do not certify Richardson extrapolation accuracy: asymptotic cancellation provides an improved-order estimator, while a numerical error certificate needs quantitative remainder information.