Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 4

A Hermitian form is conjugate-linear in its first argument, linear in its second, and satisfies $\varphi (w,v)=\overline{\varphi (v,w)}$. For $H=\varphi -\psi$, the hypothesis says $H(v,v)=0$. The complex polarization identitytherefore gives $H(v,w)=0$ for all $v,w$.

The displayed matrix is Hermitian. By Sylvester's criterion, its leading principal minors areHence the form is

The closure $\overline{Z}$ is the intersection of all closed subsets containing $Z$; equivalently, every neighbourhood of a point of $\overline{Z}$ meets $Z$. The subspace is dense when $\overline{Z}=X$. A space is Hausdorff when distinct points have disjoint neighbourhoods.

For Hausdorff $Y$, the diagonal ${\Delta }_Y$ is closed in $Y\times Y$. Thusis closed. Since it contains the dense set $Z$, it equals $X$.

The conclusion fails without Hausdorffness. Let $X=Y=\{0,1\}$ have the Sierpiński space topology $\{\varnothing ,\{1\},X\}$ and let $Z=\{1\}$, which is dense. The identity map and the constant map with value $1$ are continuous and agree on $Z$, but differ at $0$.

If such an $f$ existed, boundedness would make its isolated singularity at zero removable. Let $F$ be the holomorphic extension to the disc. Continuity gives $1\le |F(0)|\le 2$. Equality at either endpoint contradicts the minimum-modulus principle applied to $F$ or the maximum modulus principle, since $F$ is nonconstant. Hence $F(0)\in A$. Surjectivity of $f$ supplies some nonzero $z$ with $f(z)=F(0)$, contradicting injectivity of the extension. Therefore

The first term is radial kinetic energy from the Laplacian, the second is the attractive Coulomb potential, and the third is the centrifugal energy from the angular-momentum eigenvalue $L^2=l(l+1){\hslash }^2$. Single-valued spherical harmonics require

Substitute $g=r^a{e}^{-br}$ into the radial Schrodinger equation and equate the coefficients of $r^{-2}$, $r^{-1}$, and the constant term. Regularity selectsThe corresponding bound-state energy is

The higher state has principal quantum number $l+2$ and the lower one $l+1$. Energy conservation for the emitted photon gives

In vacuum, curl Faraday's law and use the Ampère-Maxwell equation:Since $\nabla \cdot \mathbf{B}=0$, the left side is $-{\nabla }^2\mathbf{B}$. Hence

The divergence equation requires $B_3=0$, and the wave equation requiresFor propagation in the positive $z$ direction, Faraday's law givesThus $\mathbf{E}$, $\mathbf{B}$, and the propagation direction are mutually perpendicular.

The Poynting vector isIt is the electromagnetic energy flux in the propagation direction; its time average is $c(B_1^2+B_2^2)/(2{\mu }_0)$.

Let ${\ell }_i$ be the degree-$n-1$ cardinal polynomial with ${\ell }_i(c_j)={\delta }_{ij}$. Exactness applied to the nonnegative polynomial ${\ell }_i^2$ givesbecause $w>0$ and ${\ell }_i$ is nonzero. Thus every Gaussian quadrature weight is positive.

Symmetry gives nodes $\pm c$ with equal weights $b$. Exactness for $1$ and $x^2$ requiresThereforeOdd moments vanish automatically, so the rule is exact through degree three.

Let $q={\mathbb{P}}_i(T_i<\infty )$. By the Strong Markov property, after every return to $i$ the chance of another return is again $q$. HenceThus $V_i$ is geometric with success parameter $1-q$; when $q=1$, it is infinite almost surely.

In a transient irreducible chain, every state $j$ is transient, so the expected number of visits satisfiesThe terms are nonnegative, and every convergent series has terms tending to zero. Therefore

The rank is $r(\alpha )=\dim \mathrm{im}\alpha$ and the nullity is $n(\alpha )=\dim \ker \alpha$. Extending a basis of the kernel to a basis of $V$, the images of the added vectors form a basis of the image, proving the rank-nullity theorem

Since $\mathrm{im}(\alpha +\beta )\subseteq \mathrm{im}\alpha +\mathrm{im}\beta$, the upper sum bound follows. Applying it to $\alpha =(\alpha +\beta )-\beta$ and interchanging $\alpha ,\beta$ gives the lower bound. For products, the image lies in $\mathrm{im}\alpha$ and is the image under $\alpha$ of $\mathrm{im}\beta$, giving the upper bound; rank-nullity on that restriction gives

All four bounds are sharp. In a fixed basis, nested diagonal projections attain the product upper bound; placing $\mathrm{im}\beta$ with the smallest possible intersection with $\ker \alpha$ attains its lower bound. Taking $\beta =-\alpha$ on a subspace of dimension $\min (r(\alpha ),r(\beta ))$ attains the sum lower bound, while choosing the two images in general position and avoiding cancellation attains the sum upper bound.

Both upper bounds need not be attainable simultaneously over every field. Over ${\mathbb{F}}_2$ with $n=1$ and both ranks one, necessarily $\alpha =\beta =I$. Then $r(\alpha \beta )=1$ but $r(\alpha +\beta )=0$, below its upper bound one.

A polynomial over a unique factorization domain is primitive when its coefficients have no common nonunit factor. If a prime divided every coefficient of a product of two primitive polynomials, reduction modulo that prime would turn the product of two nonzero polynomials over an integral domain into zero, impossible. This proves Gauss lemma for polynomials.

If positive-degree $f,g\in R[X]$ became non-coprime in $F[X]$, they would share a positive-degree primitive factor after clearing content. Gauss's lemma then makes that factor divide both in $R[X]$, contrary to coprimality. Hence they remain coprime over the fraction field.

Regard $f,g$ as polynomials in $Y$ over $\mathbb{C}(X)$. Their coprimality and the preceding argument let Euclid's algorithm producewith coefficients in $\mathbb{C}(X)$. Clearing denominators gives a nonzero $h(X)\in (f,g)$, so $I\cap \mathbb{C}[X]\ne 0$. Interchanging $X,Y$ gives a nonzero $k(Y)\in I$.

In the quotient, $h(X)=k(Y)=0$. Reducing powers by these two univariate relations shows that the finitely many monomialsspan. Therefore $\mathbb{C}[X,Y]/I$ is finite-dimensional.

A space is connected when it has no separation into two nonempty disjoint open sets; the connected subspaces of $\mathbb{R}$ are exactly the intervals. A space is path connected if every two points are joined by a continuous path. A separation would pull back along a path to a separation of $[0,1]$, so path connectedness implies connectedness.

Each of the $n$ interval components of $I_n$ must map into one of the $m$ components of $I_m$. This assignment is constant on every connected component of the uniform function space. Conversely, for a fixed assignment, straight-line interpolation within each target interval gives paths between maps. Hence

Take the topologist's sine curveIt is closed and bounded, hence compact. The oscillating graph is connected and accumulates on every point of the vertical segment, so its closure $S$ is connected. No path joins the graph to the vertical segment, as continuity of the first coordinate and the unbounded oscillation of $\sin (1/x)$ would fail at the first time the path reaches $x=0$. Thus $S$ is not path connected.

Constant maps identify a copy of $S$ inside $C([0,1],S)$. If the function space were path connected, evaluating a path of functions at $0$ would give a path in $S$ between any two constant values. Since $S$ is not path connected,

The constant maps form a connected subspace homeomorphic to $S$. Every $f\in C([0,1],S)$ is joined to the constant map $f(0)$ byUniform continuity of $f$ makes $s\mapsto H(s)$ continuous in the uniform metric. Thus the function space is a union of connected sets, each meeting the connected set of constant maps, and so

The Poincare disc model has metricRotations about the origin are hyperbolic isometries, so a Euclidean circle centred there is a hyperbolic circle. Its radius is the radial length

The hyperbolic law of sines isSet $\gamma =\pi /2$ in the cosine laws for sides and angles and eliminate $\cos \alpha$ and $\cos \beta$. Straight algebra gives the right-triangle identities

Gauss-Bonnet theorem gives the area of a curvature-$-1$ triangle as $\pi -(\alpha +\beta +\gamma )$. Split the regular polygon into $2n$ congruent right triangles. If $\theta$ is a polygon half-angle, the right-triangle identity and part (a) giveThe polygon area is $(n-2)\pi -2n\theta$. Rewriting $\theta =\pi /2-{\cot }^{-1}(\tan \theta )$ yields

A regular hyperbolic hexagon has angle tending to $2\pi /3$ as its radius tends to zero and to zero as its vertices approach the ideal boundary. Continuity therefore gives a radius at which every angle is $\pi /2$.

Two congruent right-angled hexagons glued along three alternating sides form a hyperbolic pair of pants with geodesic boundary. Glue $2g-2$ such pairs of pants along boundary curves in the usual pants decomposition. The right angles match smoothly and produce a closed hyperbolic surface of genus $g$ for every $g\ge 2$.

Repeated integration by parts givessoMultiplication by $e^{at}$ shifts the transform:

The Bromwich inversion integral closes to the left for $t>0$. Equivalently, partial fractions at the two double poles giveTaking residues therefore yields

The transform factors as $s^{-2}(s+2{)}^{-2}$, whose inverse factors are $t$ and $t{e}^{-2t}$. The Laplace convolution theorem givesand direct integration producesconfirming part (b).

The first three initial values vanish and the third derivative at zero equals one. Transforming the differential equation givesso $F(s)=1/[s^2(s+2{)}^2]$. Hence

For variations $y+\epsilon \eta$ with $\eta ={\eta }^{\prime }=0$ at both endpoints, integrating the ${\eta }^{\prime }$ term once and the ${\eta }^{\prime \prime }$ term twice givesThe fundamental lemma yields the higher-order Euler-Lagrange equation

DifferentiateThe product rule cancels all terms exceptOn an extremal with no explicit $x$ dependence both terms vanish, proving the stated conserved quantity.

The Euler-Lagrange equation isPut $A=\sqrt{1+\sqrt{2}}$, $B=\sqrt{\sqrt{2}-1}$, and $L=\pi /2$. The conditions at zero reduce the general solution toDefineThe remaining boundary conditions givewhich determines the unique extremal explicitly.

Integration by parts, using decay of $m$, gives the Fourier transform of a derivative

The standard contour transformholds for $\mathrm{Re}\alpha >0$, with the appropriate upper or lower pole chosen according to the sign of $k$. Therefore

Fourier transformation in $x$ changes Laplace's equation intoImposing the two boundary values givesThe value at $k=0$ is understood by continuity.

Part (b) givesSince $(e^{-aq}-e^{-3aq})/\sinh (aq)=2e^{-2aq}$,Inverting with part (b) yields

The Time-dependent Schrodinger equation is $i\hslash {\partial }_t\psi =\hat{H}\psi$. Differentiate $⟨\hat{A}⟩=⟨\psi |\hat{A}|\psi ⟩$, use the equation and its adjoint, and include explicit time dependence of the operator. This gives Ehrenfest theorem

Substitution of the stated phase intocancels the potential and scalar phase terms, leaving

With $X=x+g{t}^2/2$ and $T=t$, the chain rule gives ${\Phi }_t={\Theta }_T+gt{\Theta }_X$ and ${\Phi }_x={\Theta }_X$. The drift terms cancel, soApplying Ehrenfest's theorem with $[X,P]=i\hslash$ and $[P,P^2]=0$ gives

The phase has unit modulus and $X=x+g{t}^2/2$, henceThe free-particle equations imply $⟨X⟩=a+vt$. ThereforeThe wavepacket centre obeys the classical constant-acceleration trajectory, an instance of the correspondence principle.

Irrotationality gives ${\mathbf{u}}_i=\nabla {\varphi }_i$ and ${\nabla }^2{\varphi }_i=0$. At $z=\zeta$, the exact kinematic conditions areand pressure continuity gives $p_1=p_2$. Bernoulli equation in each fluid is

Linearize at $z=0$ and write $\zeta =Z{e}^{i(kx-\omega t)}$. Decay at infinity givesThe kinematic conditions give $-i\omega Z=-k{A}_1=k{A}_2$, while pressure continuity givesElimination yieldsStable waves require the lower fluid to be denser, ${\rho }_2>{\rho }_1$; otherwise this is the Rayleigh-Taylor instability.

A statistic is sufficient when the conditional law of the sample given it is parameter-free; it is minimal sufficient when it is a function of every sufficient statistic. The likelihood factors asso the factorization theorem makes $(n_0,S)$ sufficient. The likelihood-ratio criterion shows that the ratio for two samples is parameter-independent exactly when both $n_0$ and $S$ agree. Thus

For known ${\lambda }_0$, the likelihood ratio of $\pi =1/2$ to $\pi =0$ iswhich increases with $n_0$. Reject for $n_0>c$, with randomization at $n_0=c$ chosen so that the null rejection probability is exactly $\alpha$. Since the zero indicators form a Bernoulli family with success probability increasing in $\pi$ and have a monotone likelihood ratio in $n_0$, the Karlin-Rubin theorem makes this test uniformly most powerful for $\pi >0$.