Past exam of the options I course of the University of Oxford / 2012 / P (Numerical Analysis)

A polynomial $p_n\in {\Pi }_n$ is a best polynomial least-squares approximation when it minimizes the weighted error norm over all polynomials of degree at most $n$:Equivalently it minimizes ${\int }_0^1{x}^{-1/2}(f(x)-q(x){)}^{2}dx$. The weight is integrable, so the expression is finite for continuous functions; it defines a genuine norm because a continuous function with zero weighted squared integral must be identically zero.

Take any $q\in {\Pi }_n$. Write $f-q=(f-p_n)+(p_n-q)$. The second term belongs to ${\Pi }_n$, so the assumed orthogonality gives $⟨f-p_n,p_n-q⟩=0$. Expanding the squared norm yields the Pythagorean theorem in this inner product:Taking square roots provesEquality can hold only when $\Vert p_n-q\Vert =0$, hence $p_n=q$, so the best polynomial is unique. Completeness of the ambient space is unnecessary for this orthogonality argument.

Write $p_1(x)=a+bx$. It suffices to impose orthogonality of the error to the basis $1,x$ of ${\Pi }_1$. With the printed weight $x^{-1/2}$ and $f=x^{3/2}$, the normal equations areHere, for example, $⟨f,1⟩={\int }_0^{1}xdx=1/2$ and $⟨f,x⟩={\int }_0^1{x}^{2}dx=1/3$. Solving gives $a=-1/16$, $b=15/16$. ThereforeThe two normal equations make the error orthogonal to every linear polynomial, and part (b) then proves optimality, rather than merely furnishing a candidate fit.

The leading error term can be removed by comparing the two step sizes $h$ and $h/2$:Multiply the second expression by four and subtract the first. The $h^2$ terms cancel, so the first Richardson extrapolation step isRepeat with $R_1(h/2)$ and $R_1(h)$, now cancelling the fourth-order error using the factor $16$. In general, with $R_0(h)=I_h$,Under the stated even-power expansion, successive columns have errors of orders $h^4,h^6,h^8,\dots$. Each step uses the known scaling of the leading error to eliminate its unknown coefficient.

For one interval, the trapezoidal rule givesFor two intervals of width $h=1/2$, the composite trapezoidal rule gives half weights at the endpoints and full weight at the internal node:

Apply the first Richardson extrapolation step to the two computed approximations:The exact integral is $I={\int }_0^1{x}^{4}dx=1/5$, so the actual absolute error isThus the requested strict error bound is verified directly. This first extrapolation of trapezoidal approximations is also the one-panel Simpson's rule value on the three nodes $0,1/2,1$.