Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 1

For $A=(a_{ij})\in M_n(\mathbb{C})$, the determinant isForany nonzero term in the Leibniz formula for determinants must match every row in the lower block to a column in the $N$ block. The remaining upper rows must then match the $M$ columns. The sum consequently factors into the determinant sums for the two diagonal blocks:

Move the first $n-k$ columns ofpast its last $k$ columns. This requires $k(n-k)$ column interchanges and produces the block diagonal matrixEach interchange reverses the determinant's sign, so part (a) gives

A unit-speed curve on a smooth embedded surface is a geodesic exactly when its acceleration is normal to the surface, equivalently when its tangential acceleration vanishes.

Every unit-speed geodesic on the cylinder through $(1,0,0)$ has the formIndeed, unrolling the cylinder to its universal cover ${\mathbb{R}}^2$ turns these curves into straight lines. Directly,which is parallel to the cylinder's radial normal vector, verifying the geodesic characterization. Such a geodesic is closed exactly when $b=0$; its image is then the horizontal circle $z=0$.

Yes. In polar coordinates $(r,\theta )$ on ${\mathbb{R}}^2\setminus \{0\}$, use the Riemannian metricThe coordinate $u=\log r$ identifies this surface isometrically with the flat cylinder $\mathbb{R}\times S^1$. Through each point, the circle $u=\text{constant}$ is a closed geodesic, and every other geodesic has nonzero linear motion in $u$ and is not closed. Thus every point lies on a unique closed geodesic.

The Taylor series of the complex sine isTherefore, for $z\ne 0$,and the right side defines a holomorphic function at $z=0$ with value $1$. Thus the apparent singularity is a removable singularity.

The resulting power series isThe ratio test shows convergence for every $z\in \mathbb{C}$, so its radius of convergence is

Forthe Euler-Lagrange equation isHere this becomesBecause the forcing is resonant with the complementary solution, a particular integral is $(x/2)\cos x$. HenceandThe Neumann boundary conditions giveThus

Let the columns of $A$ be $a_1,a_2,a_3$. The Gram-Schmidt process givesNext,soFor the final column,andThus $q_3=\frac{1}{2}(-1,-1,1,-1{)}^T$ and $r_{33}=2$. The reduced QR decomposition isThe columns of $Q$ are orthonormal, and direct multiplication gives $QR=A$.

The Rao-Blackwell theorem says that if $U$ is an estimator with finite second moment and $T$ is a sufficient statistic, thenhas the same expectation as $U$ and satisfieswith equality only when $U$ is already a function of $T$ almost surely.

Here $\hat{\theta }=1_{\{X_1=1\}}$ is unbiased becauseConditional on $T=t$, every weak composition $(x_1,\dots ,x_n)$ of $t$ has the same probability $p^n(1-p{)}^t$. There aresuch compositions. For $t\ge 1$, those with $x_1=1$ correspond to weak compositions of $t-1$ into $n-1$ parts, of which there areThe Rao-Blackwell estimator is thereforeFor $n\ge 2$ and $0<p<1$, $\hat{\theta }$ is not determined by $T$; for example, conditional on $T=1$, the sole failure can occur in any coordinate. The variance inequality is consequently strict:

Starting from $x_0\in {\mathbb{R}}^n$, gradient descent repeatedly computes the gradient and updatesstopping when the gradient norm, step, or objective decrease is sufficiently small.

The Hessian bounds say that $f$ is $\alpha$-strongly convex and has $\beta$-smooth gradient. With $\eta =1/\beta$,Thus the iteration count isconvergence becomes slower linearly with the condition number $\kappa =\beta /\alpha$.

Forthe Hessian matrix is $\mathrm{diag}(2,200,20000)$, soTakeThenwhose Hessian is $2I$ and whose condition number is $1$.

The adjoint operator ${\alpha }^{\ast }$ is defined bySince $V$ is a nonzero finite-dimensional complex vector space, $\alpha$ has an eigenvector $v$ with eigenvalue $\lambda$. Normality givesfor every $x$, by expanding both sides and using $\alpha {\alpha }^{\ast }={\alpha }^{\ast }\alpha$. Taking $x=v$ yields

For the common eigenvector $v$, its orthogonal complement $v^{\perp }$ is invariant under both $\alpha$ and ${\alpha }^{\ast }$. Indeed, if $w\perp v$, thenand the analogous calculation applies to ${\alpha }^{\ast }w$. The restriction of $\alpha$ to $v^{\perp }$ is therefore again normal. Induction on $\dim V$, beginning with the normalized vector $v/\Vert v\Vert$, produces an orthonormal eigenbasis. This proves the finite-dimensional spectral theorem for normal operators.

If $Av=\lambda v$ for a nonzero real vector $v$, then skew-symmetry givesThusso a real skew-symmetric matrix has no nonzero real eigenvalue.

Over $\mathbb{C}$, the real skew-symmetric matrix $A$ is normal because $A^{\ast }=-A$. Part (a) gives an orthonormal complex eigenbasis, and the nonzero eigenvalues are purely imaginary pairs $\pm i\lambda$.

For each $\lambda >0$, choose a unit real vector $x$ in the kernel of $A^2+{\lambda }^{2}I$ and putThen $x\perp y$, $\Vert y\Vert =1$, andThus $A$ has matrixon the orthonormal basis $(x,y)$. Distinct such invariant planes are orthogonal; complete them by an orthonormal basis of $\ker A$. Taking these basis vectors as the columns of an orthogonal matrix $R$ gives

A Euclidean domain is an integral domain $R$ equipped with a function $\delta :R\setminus \{0\}\to \mathbb{N}$ such that for $a,b\in R$, $b\ne 0$, there are $q,r\in R$ withFor the Gaussian integers $\mathbb{Z}[i]$, take $\delta (z)=N(z)=|z{|}^2$. Choosing a Gaussian integer nearest to $a/b$ makes the remainder norm smaller than $N(b)$.

The units are precisely the elements of norm one:Unique factorization in this Euclidean domain givesThe displayed factors have prime norms $2$ or $5$, so they are irreducible; factors appearing together are nonassociate.

Now suppose $x^2+4=y^3$. Necessarily $y>0$. First let $y$ be odd. Then $x$ is odd, and $x+2i$ and $x-2i$ are coprime in $\mathbb{Z}[i]$: a common Gaussian prime would divide $4i$, while their product has odd norm. Henceup to a unit, which can be absorbed into the cube. Comparing imaginary parts givesChecking $b\mid 2$ yields only $b=-2$, $a=\pm 1$, and therefore

If $y$ is even, congruence modulo $8$ gives $x=2X$ with $X$ odd and $y=2Y$, whereEach of $X\pm i$ contains exactly one factor $1+i$, so the coprime quotients are cubes up to units. ThusComparing the imaginary part after the four possible units reduces toEach integer factor must have absolute value one. Substitution then gives $a^2+b^2=1$, so $Y=1$ and $X=\pm 1$. HenceAll four pairs satisfy the equation, so the complete answer is

This is always true. Given $\epsilon >0$, choose $N_A,N_B$ such thaton $A$ for $n\ge N_A$ and on $B$ for $n\ge N_B$. For $n\ge \max (N_A,N_B)$ the same inequality holds at every point of $X=A\cup B$. Hence $f_n\to f$ uniformly on $X$.

This may be false when the limiting functions are unbounded. On $X=\mathbb{R}$, letBoth sequences converge uniformly to their stated limits, butis unbounded for every $n$. Thus the products do not converge uniformly.

For every $(x,y)\in [1,2{]}^2$,Moreover,The bound is independent of $(x,y)$ and tends to zero, so

This is always true. Uniform convergence makes $(f_n)$ a uniformly Cauchy sequence: for every $\epsilon >0$, there is $N$ such thatfor every $x\in A$ and $m,n\ge N$. Letting $x\to x_0$ through points of $A\setminus \{x_0\}$ and using the assumed limits givesThus $(y_n)$ is a Cauchy sequence. Since $Y$ is a complete metric space, $(y_n)$ converges in $Y$.

This is always true. Since $f$ is bounded and $f_n\to f$ uniformly, there are $M$ and $N$ such thatfor every $x\in X$ and $n\ge N$. The continuous function $g$ is uniformly continuous on the compact interval $[-M-1,M+1]$. Therefore, given $\epsilon >0$, a sufficiently small uniform bound on $|f_n-f|$ impliesHence $g\circ f_n\to g\circ f$ uniformly.

Every orientation-preserving isometry of the Poincare half-plane model has the formwith matrices identified up to multiplication by $-I$. Thus $G\cong PSL(2,\mathbb{R})$.

For $z=x+iy\in \mathbb{H}$, the affine map $(z-x)/y$ sends $z$ to $i$, proving transitivity on points. Real translations and positive dilations act transitively on finite boundary points, while inversion $z\mapsto -1/z$ exchanges $0$ and $\infty$; hence the action on $\mathbb{R}\cup \{\infty \}$ is transitive. A hyperbolic line is determined by its unordered pair of boundary endpoints, and a real Möbius transformation can send any such pair to $\{0,\infty \}$. Therefore $G$ is also transitive on hyperbolic lines.

By point transitivity, it suffices to take $P=i$. The conditionis equivalent to $d=a$, $c=-b$, and the determinant condition becomes $a^2+b^2=1$. Thus the stabilizer consists ofmodulo $\pm I$. Writing $a=\cos \phi$, $b=\sin \phi$ identifies it with rotations of the tangent plane, and the doubled matrix angle removes the $\pm I$ quotient. Hence

An isometry sends the geodesic through $P,Q$ to the imaginary axis and sends the two points to $ir$ and $i/r$ for some $r>0$. The transformationbelongs to $PSL(2,\mathbb{R})$ and exchanges $ir$ with $i/r$. Conjugating it back gives an orientation-preserving isometry exchanging $P$ and $Q$.

For uniqueness, the quotient of two such isometries fixes both $P$ and $Q$. An orientation-preserving hyperbolic isometry fixing two distinct interior points fixes their connecting geodesic and both tangent directions there, hence is the identity. Therefore the exchanging isometry is unique; geometrically it is the hyperbolic half-turn about the midpoint of the segment $PQ$.

Use an orientation-preserving isometry to send $P$ to $i$ and $\ell$ to the imaginary axis, whose boundary points are $0,\infty$. A geodesic $m$ through $i$ meeting it at angle $\theta$ has boundary pointsIndeed $uv=-1$, so the semicircle with endpoints $u,v$ passes through $i$, and its tangent there makes angle $\theta$ with the vertical.

The cross-ratio is invariant under the normalizing Möbius transformation. With the suitable ordering $(u,v,\infty ,0)$ and the convention used here,

The hypothesis makes $\Omega$ star-shaped with centre $z_0$. For small $h$ with the triangle having vertices $z_0,w,w+h$ contained in $\Omega$, the vanishing triangular integral givesParametrizing the final segment,by continuity. Thus $g$ is holomorphic andThis is the triangle-integral construction behind Morera's theorem.

The set $D$ is star-shaped with centre $z_0$: if $z\in D$, every point of $[z_0,z]$ has its segment to $z_0$ contained in the same segment and therefore avoids the origin. Applying part (a) and the stated triangular-integral fact to $1/z$ gives a holomorphic functionChoose either square root $c$ of $z_0$ and defineThen $h$ is holomorphic. Moreover,and its value at $z_0$ is one. Hence

Because $\cos 0=1$, continuity gives a disc $U$ about zero on which $\cos z\ne 0$. The function $-\tan z=(\log \cos z{)}^{\prime }$ has a holomorphic primitive $G$ on a sufficiently small disc, normalized by $G(0)=0$. Thenis holomorphic and satisfies $f(0)=1$ and $f(z{)}^2=\cos z$ on $U$.

No such square root exists throughout $|z|<2$. The point $z=\pi /2$ lies in that disc and is a simple zero of $\cos z$, since $-\sin (\pi /2)=-1$. Every zero of the square of a holomorphic function has even order of a zero of a holomorphic function, contradicting this simple zero.

Take positive $y$ vertically downward. For a short string element, the transverse tension resultant is $T{y}_{xx}dx$, while gravity and linear drag contribute $\mu gdx$ and $-2k\mu y_{t}dx$. Newton's second law givesSince $T=\mu c^2$, the damped wave equation isChoosing upward displacement reverses the sign of the gravity term.

With gravity omitted, the initial data contain only the first fixed-end normal mode, so writeThenBecause $k=\pi c/l$, this oscillator is critically damped, andTherefore

The first mode remains the solution from part (b). The applied force excites only the third mode, sowhere, using $\pi c/l=k$,A particular solution isAdding the decaying complementary solution and imposing the initial conditions gives

As $t\to \infty$, all transients decay but the driven third-mode oscillation remains. The string approaches a periodic steady-state response, rather than coming to rest.

With gravity omitted, its mechanical energy isWriting $s=\alpha /(34k^2)$ and letting $t\to \infty$ givesThe limiting energy is periodic because the external force continually supplies the energy dissipated by drag.

For the one-dimensional quantum harmonic oscillator,This is the Time-dependent Schrodinger equation with harmonic potential.

Fordirect differentiation givesandSubstitution into the Schrodinger equation and equality of the coefficients of $x^2,x,1$ give the necessary and sufficient Riccati equation system

Put $u=a+i\omega t$. ThenThe $G$ equation from part (b) is satisfied whensoWith this value, the $F$ equation follows from ${\tanh }^{2}u=1-{\mathrm{sech}}^{2}u$: both its constant and ${\mathrm{sech}}^{2}u$ coefficients agree. Thus the stated $F,G$ satisfy both shape equations; the remaining scalar equation determines the normalization factor $N(t)$.

The probability density isCompleting the square shows that, when $\mathrm{Re}F>0$,where

Writing $\theta =\omega t$ andthe supplied identities giveUsing $A=m\omega /(2\hslash )$ in part (d),The peak executes simple harmonic motion about the origin with angular frequency $\omega$ and amplitude $\sqrt{\hslash /(m\omega )}/\sinh a$.

At each fixed time, $|\psi (x,t){|}^2$ is a bell-shaped Gaussian centred at $x=h(t)$, with width determined by $\mathrm{Re}F(t)>0$. Completing the square as in part (d) makes the density symmetric about $h(t)$. Therefore its position expectation value isThe sketch is a Gaussian peak whose centre oscillates between the two turning positions found in part (e), while its width varies periodically through $\mathrm{Re}F(t)$.

For a coaxial Gaussian cylinder of radius $r$ and length $L$, symmetry makes the electric field radial and constant on the curved surface. Gauss's law givessoSince $\mathbf{E}=-\nabla \varphi$, a potential relative to an arbitrary reference radius $r_0$ isAn infinite line charge has no finite convention $\varphi (\infty )=0$.