Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability (resit) / 4

By the fundamental theorem of calculus,

For the composite trapezoidal rule, the mesh width is $h=1/N$ and the nodes are $x_n=n/N$. The endpoint weights are halves and the interior weights are ones, soThis formula is an approximation to the definite integral, not an exact equality with it for finite $N$.

Let $S_N=N^{-3}{\sum }_{n=1}^N{n}^2$. The expression in part (ii) givesBecause $3x^2$ is continuous on $[0,1]$, the composite trapezoidal rule converges to its definite integral, which is $1$ by part (i). In fact its standard composite trapezoidal error bound is $|T_N-1|\le 1/(2N^2)$, since $f^{\prime \prime }=6$. Therefore

Taking the partial derivatives of the parabolic coordinates transformation, holding the other coordinate fixed each time, givesThus its Jacobian matrix is $J=\left(\begin{array}{cc}t& s\\ -s& t\end{array}\right)$.

The determinant of the Jacobian matrix is $s^2+t^2$. Wherever $(s,t)\ne (0,0)$, the inverse function theorem permits a local inverse of the parabolic coordinates transformation, andReading off its entries givesThese are local formulas on a chosen branch: $(s,t)$ and $(-s,-t)$ have the same image, and at the origin the Jacobian matrix is singular, so this inverse-derivative formula does not apply there.

Apply the multivariable chain rule to a differentiable function $f(s,t)$ on the chosen local branch of the parabolic coordinates:Substituting part (ii) gives

For $y=1+x^2$, the derivatives are $y^{\prime }=2x$ and $y^{\prime \prime }=2$. Its endpoint values are $y(0)=1$ and $y(2)=5$, so it satisfies both the ordinary differential equation and the boundary conditions. Moreover any solution of $y^{\prime \prime }=2$ has the form $x^2+Ax+B$; the boundary conditions force $B=1$ and $A=0$. Thus the unique solution of the boundary value problem is

The endpoint boundary conditions areThe finite difference equation can be written $y_{n+1}-2y_n+y_{n-1}=2h^2$. Set $d_n=y_n-y_{n-1}$. Then $d_{n+1}-d_n=2h^2$, so $d_n=d_1+2h^2(n-1)$. Summing from $1$ to $n$ givesUsing $y_N=5$ and $h=2/N$ yields $N{d}_1=4-h^{2}N(N-1)=4/N$, hence $d_1=4/N^2=h^2$. ThereforeThis also covers $N=1$, where there are no interior equations. Uniqueness follows because the difference of two solutions has zero second finite difference, hence is linear in $n$, and vanishes at both endpoints. This is an exact finite difference solution for constant second derivative: the method is exact at the nodes here because the centered second finite difference differentiates this quadratic polynomial exactly.