Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 4

A quadratic form on a finite-dimensional real vector space $V$ is a function $q(v)=B(v,v)$ for some symmetric bilinear form $B$. A positive-definite quadratic form satisfies $q(v)>0$ for every $v\ne 0$.

Completing squares givesSet $u=x+y$, $v=y+z$, $w=z$. The corresponding ordered basis in the original coordinates isand the diagonal form is $u^2+v^2+2w^2$. All diagonal coefficients are positive, so the form is positive definite.

Use the presentation $D_6=⟨r,s:r^3=s^2=1,srs=r^{-1}⟩$. Its subgroup $⟨r⟩$ is the unique subgroup of order $3$, hence is a characteristic subgroup. An automorphism must send $r$ to $r$ or $r^{-1}$ and $s$ to one of the three reflections $r^{j}s$. Thus there are at most six automorphisms.

The center of a group of $D_6$ is trivial, so conjugation embeds $D_6$ as a group of six inner automorphisms. These exhaust the possibilities. Hence every automorphism of $D_6$ is inner.

Let $D_8=⟨r,s:r^4=s^2=1,srs=r^{-1}⟩$, the symmetry group of a square. The mappreserves the relations and is an automorphism. Conjugation by a rotation sends $s$ to $r^{2j}s$, while conjugation by a reflection also inverts $r$. Therefore no conjugation fixes $r$ and sends $s$ to $rs$. This is an outer automorphism of the nonabelian group $D_8$.

The Bolzano-Weierstrass theorem in $\mathbb{R}$ says that every bounded real sequence has a convergent subsequence. For a bounded sequence in ${\mathbb{R}}^n$, successively extract subsequences on which the first, second, and so on through the $n$th coordinate converge. The final diagonal subsequence converges in Euclidean space, proving the theorem in ${\mathbb{R}}^n$.

Put $K=D\setminus {\bigcup }_{z\in S}{B}_{\epsilon }(z)$. Suppose the asserted $\delta$ did not exist. Then there would be $x_k\in D$ and $y_k\in K$ with $\Vert x_k-y_k\Vert <1/k$ but $|f(x_k)-f(y_k)|\ge \epsilon$. Since $K$ is closed and bounded, Bolzano-Weierstrass gives a subsequence $y_k\to y\in K$, and then $x_k\to y$. The point $y$ is not in $S$, because otherwise $y\in B_{\epsilon }(y)$, so $f$ has continuity at $y$. Both $f(x_k)$ and $f(y_k)$ tend to $f(y)$, a contradiction. Hence the required uniform $\delta >0$ exists.

The Cauchy integral formula and its differentiated form areIf $f$ is entire and $|f|\le M$, the Cauchy estimate gives $|f^{\prime }(a)|\le M/\rho$. Letting $\rho \to \infty$ yields $f^{\prime }(a)=0$ for every $a$. Thus $f$ is constant, which is Liouville theorem.

The inequality implies $u(u-1)\ge v^2\ge 0$, so $u\le 0$ or $u\ge 1$. Since $u$ is continuous and $\mathbb{C}$ is connected, its image lies wholly in one of these intervals. If $u\ge 1$, then $g$ has no zero and $|1/g|\le 1$; Liouville theorem makes $1/g$, hence $g$, constant. If $u\le 0$, then $|g-1|\ge 1$, so the same argument applied to $1/(g-1)$ proves that $g$ is constant.

Separation of variables with $u=X(x)Y(y)$ and the three homogeneous boundary conditions gives $X_n=\sin (n\pi x)$ and $Y_n=\sinh (n\pi y)$. The boundary value at $y=1$ contains only the third Fourier sine series mode. ThereforeThis harmonic function has all four prescribed boundary values.

The Born rule and probability current in one dimension giveMultiply the Time-dependent Schrodinger equation by ${\Psi }^{\ast }$, subtract its complex conjugate multiplied by $\Psi$, and rearrange to obtain the continuity equation ${\rho }_t+j_x=0$.

For the stated superposition, the interference terms in $j$ are real and cancel from its imaginary part. Henceindependent of $x$ and $t$.

Taking the divergence of the Ampère-Maxwell equation and using the Gauss's law $\nabla \cdot E=\rho /{\epsilon }_0$ giveswhich is charge conservation from Maxwell equations ${\rho }_t+\nabla \cdot J=0$.

By Ohm's law, $J=\sigma E$, so $\nabla \cdot J=\sigma \rho /{\epsilon }_0$. Thereforewith charge-relaxation rate $\sigma /{\epsilon }_0$ and time ${\epsilon }_0/\sigma$.

Gaussian elimination without pivoting givesThus $\det A=3(\lambda -20)$ and the solution is unique exactly when $\lambda \ne 20$.

For $\lambda =20$, forward substitution in a triangular system in $Ly=b$ gives $y=(1,1,3,\mu -10{)}^T$. Consistency of $Ux=y$ requires $\boxed{\mu =10}$. backward substitution in a triangular system then gives

A distribution $\pi$ is an invariant distribution whenfor every $j$, equivalently $\pi P=\pi$.

Detailed balance means ${\pi }_i{p}_{ij}={\pi }_j{p}_{ji}$ for all $i,j$. Summing over $i$ givesso detailed balance implies invariance.

Let $D_i$ be the degree of a vertex and putIf $i$ and $j$ are adjacent, then ${\pi }_i{p}_{ij}=1/{\sum }_k{D}_k={\pi }_j{p}_{ji}$; if they are not adjacent, both sides vanish. Thus the walk is in detailed balance, and its invariant distribution is proportional to vertex degree.

A self-adjoint operator $T$ satisfies $⟨Tx,y⟩=⟨x,Ty⟩$. If $Tx=\lambda x$, then $\lambda ⟨x,x⟩=⟨Tx,x⟩=⟨x,Tx⟩=\overline{\lambda }⟨x,x⟩$, so $\lambda \in \mathbb{R}$. An eigenvector exists over $\mathbb{C}$; its orthogonal complement is $T$-invariant. Induction on dimension produces an orthonormal basis of eigenvectors, proving the finite-dimensional spectral theorem.

A self-adjoint $T$ is positive definite exactly when every eigenvalue is positive, since $⟨Tx,x⟩={\sum }_i{\lambda }_i|x_i{|}^2$ in an eigenbasis. The quadratic forms of two positive-definite operators add, so their sum is positive definite.

The same eigenbasis shows that $\mathrm{spec}T\subset [a,b]$ exactly when $T-\lambda I$ is positive definite for every $\lambda <a$ and negative definite for every $\lambda >b$. Finally, if $v$ is a unit eigenvector of $\alpha +\beta$ with eigenvalue $\gamma$, the Rayleigh quotient bounds giveThus every eigenvalue of $\alpha +\beta$ lies in $[a+c,b+d]$.

The structure theorem for finitely generated modules over a principal ideal domain says that a finitely generated module over a Euclidean domain $R$ iswith the rank and nonunit invariant factors of a linear operator unique up to associates.

Make $F^n$ an $F[X]$-module by $Xv=Av$. It is finitely generated and torsion, so the structure theorem gives cyclic summands $F[X]/(d_i)$ with $d_1\mid \cdots \mid d_k$. In the power basis of each summand, multiplication by $X$ has the companion matrix of $d_i$. Their block diagonal sum is the rational canonical form, unique because the invariant factors are unique.

The characteristic polynomial is $(t-\frac{1}{2}{)}^3$. For $N=A-\frac{1}{2}I$, one finds $N\ne 0$ but $N^2=0$, so the minimal polynomial is $(t-\frac{1}{2}{)}^2$. Hence the invariant factors are $t-\frac{1}{2}$ and $(t-\frac{1}{2}{)}^2$, and the rational canonical form is

A complete metric space is one in which every Cauchy sequence converges. On $I=(0,1]$ defineThe map $x\mapsto 1/x$ is an isometry from $(I,d)$ onto the closed subset $[1,\infty )$ of $\mathbb{R}$, so $d$ is a metric and is complete. That map and its inverse are continuous in the ordinary Euclidean metrics, so it is also a homeomorphism. Therefore $d$ induces exactly the usual subspace topology on $I$.

If $Y$ is closed and $(y_n)$ is Cauchy in $Y$, completeness of $X$ gives $y_n\to y\in X$, and closedness gives $y\in Y$. Conversely, if $Y$ is complete and $y$ lies in its closure, choose $y_n\in Y$ with $d(y_n,y)<1/n$. This sequence is Cauchy, so it converges in $Y$; uniqueness of limits makes its limit $y$. Thus $Y$ is complete exactly when it is closed in $X$.

Choose $x\in X$ and iterate $x_{n+1}=f(x_n)$. The contraction mapping estimatemakes $(x_n)$ Cauchy by a geometric series. Its limit $x_0$ satisfies $f(x_0)=x_0$ because $f$ is Lipschitz. Two fixed points would satisfy $d(x_0,y_0)\le \lambda d(x_0,y_0)$, so they coincide. This proves the Banach fixed-point theorem.

Now suppose the final assertion failed. There would be $m_k,n_k\to \infty$ with $f(a_{m_k})=a_{n_k}$. Since $a_n\to a$ and $f$ is continuous, the two sides tend respectively to $f(a)$ and $a$, so $f(a)=a$. Uniqueness would give $a=x_0$, a contradiction. Hence the required $\ell$ and $m$ exist.

Since $U_m\cap U_n=U_{\gcd (m,n)}$ when the greatest common divisor is at least $2$, and is empty otherwise, finite intersections of the $U_n$ are again among the allowed unions. Arbitrary unions are built into the definition; the empty union is empty, and $X={\bigcup }_{n\in X}{U}_n$. Thus these sets form a topology.

The space is not Hausdorff: every neighborhood of $4$ contains its divisor $2$, and every neighborhood of $2$ contains $2$, so $2$ and $4$ cannot have disjoint neighborhoods.

The cover $X={\bigcup }_{n\in X}{U}_n$ has no finite subcover, because every $U_n$ is finite. Hence $X$ is not compact.

For the stated pasting lemma for closed subspaces, if $C\subset Z$ is closed, thenis a union of two closed subsets of $Y$; therefore $f$ is continuous.

For $m,n\in X$, let $q=mn$. On $[0,\frac{1}{2}]$ define a path equal to $m$ before the endpoint and equal to $q$ at the endpoint; it is continuous because every open set containing $q$ also contains its divisor $m$. On $[\frac{1}{2},1]$ define it to equal $q$ at the first endpoint and $n$ thereafter. This is likewise continuous. The two closed-interval paths agree at $1/2$, so the pasting lemma joins them into a path from $m$ to $n$. Thus $X$ is path-connected.

The Laplace transform isUsing the supplied Gaussian integral after $t=x^2$ gives

Taking the Laplace transform in $t$ givesThe bounded resolvent solution supplied in the hint can be writtenPart (a) inverts the kernel, yielding the heat kernel

Differentiate the parametrized surface. The coefficients of the first fundamental form simplify toIf $c={\sigma }_u\times {\sigma }_v$, then $|c{|}^2=E$. MoreoverThus the second-fundamental-form coefficients satisfy $M=1/\sqrt{E}$ and $N=0$. The Gaussian curvature is consequently

The Euler-Lagrange equations are $\ddot{x}+{\omega }^{2}x=0$ and $\ddot{y}+{\omega }^{2}y=0$. They implyWritinggives the conserved value

The equations areHencewhich is not generally zero. Conservation requires $\beta /\alpha =1$. Then the potential is $\alpha (x^2+y^2{)}^2/4$, so the action has rotational symmetry. By Noether's theorem, $J$ is the conserved angular momentum, up to the sign convention used in its definition.

Remove a small disc about $x$ and apply Green second identity to $u(x_0)$ and $G_0(x;x_0)$. Since ${\nabla }_{x_0}^2{G}_0=\delta (x_0-x)$ and ${\nabla }^{2}u=-\rho$, the shrinking inner boundary contributes $u(x)$ and gives

For the unit disc, the method of images gives the Dirichlet Green functionwhich vanishes on the boundary by the hinted identity. The boundary representation with $u(1,{\theta }_0)=\delta ({\theta }_0-\alpha )$ evaluates the normal derivative of $G_D$ at ${\theta }_0=\alpha$, giving the Poisson kernel

Let $\varphi$ be the velocity potential. The linearized deep-water equations are ${\nabla }^2\varphi =0$ for $z<0$, ${\eta }_t={\varphi }_z$ and ${\varphi }_t+g\eta =0$ at $z=0$, with zero normal velocity at the four side walls. The normal modes arewhere $m,n\ge 0$ are not both zero. The free-surface conditions give the dispersion relation for a deep-water gravity wave:

The initial displacement is the $(m,n)=(3,4)$ mode, so $k=5\pi /a$. Release from rest selects a cosine in time:Velocity decays as $e^{kz}$, so it falls to $1/e$ of its surface value at depth $a/(5\pi )$.

With the uniform prior, the binomial likelihood gives the Beta distribution posterior $\mathrm{Beta}(s+1,n-s+1)$. Under squared-error loss, the Bayes estimator is the posterior mean:

The production manager's prior gives posterior density proportional to $p^{s+1}(1-p{)}^{n-s}$. Minimizing $\mathbb{E}[(1-p)(\hat{p}-p{)}^2\mid X=s]$ gives the weighted posterior meanusing the supplied beta function. Cross multiplication shows ${\hat{p}}_P>{\hat{p}}_G$ exactly when $n>2s$, so it is greater unless $s\ge n/2$.

A cut is a partition $(S,T)$ of the vertices with the source in $S$ and sink in $T$; its capacity is the sum of capacities of directed edges from $S$ to $T$. The max-flow min-cut theorem says that maximum flow value equals minimum cut capacity. With integral capacities, there is an integral max-flow theorem attaining the maximum.

Construct a bipartite graph with one vertex for each row and column and an edge $r_i\to c_j$ whenever $A_{ij}=1$. Add source-to-row and column-to-sink edges of capacity $1$, and give row-to-column edges infinite capacity. An integral flow is precisely a set of independent $1$s, so its maximum value is the largest such set.

A finite-capacity cut places some rows on the sink side and some columns on the source side; these lines cover every $1$, since an uncovered row-to-column edge would cross the cut with infinite capacity. Its capacity is the number of selected lines. Conversely every line cover defines such a cut. Max-flow min-cut therefore proves the maximum number of independent $1$s equals the minimum number of covering lines, the matrix form of König's theorem for bipartite matching.