Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 2

The zero element is torsion. If $m,n\in T$, choose nonzero $r,s\in R$ with $rm=0$ and $sn=0$. Since $R$ is an integral domain, $rs\ne 0$, andThus $m+n\in T$. If $a\in R$, then $r(am)=a(rm)=0$, so $am\in T$. Hence $T$ is closed under addition and scalar multiplication and is an submodule of $M$.

If $m\in T_1$, then $rm=0$ for some nonzero $r$, andThus $\varphi (T_1)\subseteq T_2$. The formulais consequently independent of the representative and defines an $R$-module homomorphism on the quotient module.

If $\varphi$ is injective and $\Phi (m+T_1)=T_2$, then $\varphi (m)$ is torsion, so $r\varphi (m)=0$ for some nonzero $r$. Injectivity gives $rm=0$, hence $m\in T_1$. Therefore

The converse fails. For $R=\mathbb{Z}$, takeThen $\varphi$ is not injective, but quotienting $M_1$ by its torsion submodule leaves $\mathbb{Z}$, and $\Phi :\mathbb{Z}\to \mathbb{Z}$ is the identity.

Because $f$ is a bijection and $e$ is a metric,is nonnegative, symmetric, obeys the triangle inequality, and vanishes exactly when $x=y$. Hence it is a metric. Its open sets are precisely the inverse images under $f$ of $e$-open sets in $N$. Since $f:(M,d)\to (N,e)$ is a homeomorphism, these are exactly the $d$-open sets. Thus $d^{\prime }$ and $d$ are equivalent metrics.

Leta homeomorphism $\mathbb{R}\to (0,1)$, and defineThis metric induces the standard topology on $\mathbb{R}$. However, the sequence $x_n=n$ is $d^{\prime }$-Cauchy because $h(n)\to 1$. If it converged to $x\in \mathbb{R}$, then continuity of $h$ would give $h(x)=1$, impossible. Therefore $(\mathbb{R},d^{\prime })$ is not complete.

The method of characteristics givesAlong a characteristic,Thus the characteristics are circles $x^2+y^2=C$, traversed counterclockwise, and $u$ is constant on each circle. Their intersections with the $x$-axis have $x^2=C$, where the boundary value is $f(C)$. HenceThe characteristic sketch is the family of concentric circles centred at the origin.

The uniform charge density isSpherical symmetry and Gauss's law give

Take the first centre as the origin, so the second centre is at $\mathbf{d}$. In the overlap, each point lies inside both spheres. Superposition givesThe field is therefore constant throughout the overlap region.

The divergence isThus the flow is incompressible.

A particle released at the origin at time $s$ satisfiesIntegration givesIts distance from the origin is therefore

At a fixed observation time $t$, varying the release time $0<s<t$ givesHence the dye streakline is the parabolic arc

LetIgnoring the immaterial choice of density values at the sample points, the likelihood function isBetween consecutive order statistics, $N(\theta )$ is constant while $(1+1/\theta {)}^{N(\theta )}$ decreases with $\theta$. A maximum can therefore be moved to the left endpoint of one of these intervals. With an equivalent version using $1_{\{x\le \theta \}}$, the maximum is attained at a sample order statistic. Thus the maximum-likelihood estimator coincides with one of $X_1,\dots ,X_n$.

The density is an equal mixture of the uniform distributions on $[0,1]$ and $[0,\theta ]$. HencesoThus $\tilde{\theta }$ is unbiased.

Also,The central limit theorem givesReplacing $\theta$ in the asymptotic standard error by the consistent estimator $\tilde{\theta }$ gives the Wald confidence intervalIt may be intersected with the parameter space $(0,1)$.

The Lagrange sufficiency theorem says that for a convex differentiable objective and convex differentiable inequality constraints, any feasible point satisfying the Karush-Kuhn-Tucker conditions with nonnegative multipliers is a global minimizer.

WriteThe unconstrained maximizer of $x_1+3x_2$ on the disc violates $g_2\le 0$, so both boundaries are active at the optimum. Their intersections areand $(3,4)$ gives the smaller objective.

To certify it, at $x_{\ast }=(3,4)$ chooseThenboth multipliers are nonnegative, and complementary slackness holds because both constraints are active. The sufficiency theorem therefore gives

Let $f_1,f_2\in V^{\ast }$ satisfy $f_1(v)f_2(v)=0$ for every vector $v$. If $f_1=0$ there is nothing to prove. Otherwise choose $u$ with $f_1(u)\ne 0$. For every $v\in \ker f_1$ and every real number $t$,so the hypothesis gives $f_2(u+tv)=0$. Taking two values of $t$ shows that $f_2(u)=f_2(v)=0$. Sincewe obtain $f_2=0$. Thus one of the two linear functionals is zero.

Sylvester's law of inertia says that, in a suitable basis, every real quadratic form isThe rank is $r=p+m$, and the signature is $\sigma =p-m$; both are independent of the chosen basis.

If $q=f_1{f}_2$, its polar form has image contained in $\mathrm{span}\{f_1,f_2\}$, so $r\le 2$. If $r=2$, the two functionals must be independent and, after taking their sum and difference, $q$ is a difference of two squares. Hence $\sigma =0$. If $r=1$, then $|\sigma |=1$. In every case,Conversely, this inequality leaves onlyup to a linear change of coordinates. These factor respectively as $0$, $(\pm x_1)x_1$, and $(x_1-x_2)(x_1+x_2)$. Therefore

Finally suppose that $q$ takes both positive and negative values. Sylvester's law supplies normalized basis vectorswith $q(e_i)=1$, $q(f_j)=-1$, and the $z_l$ spanning the radical. The $p+m+k$ vectorsare linearly independent and all satisfy $q(v)=0$. They form the required basis of isotropic vectors.

A Sylow subgroup of a finite group $G$ is a subgroup whose order is the largest power of a prime $p$ dividing $|G|$. The Sylow theorems say that such subgroups exist, that every $p$-subgroup lies in one, that all Sylow $p$-subgroups are conjugate, and that their number $n_p$ divides $|G|/p^a$ and obeys $n_p\equiv 1(\mathrm{mod}p)$.

For the last congruence, let one Sylow $p$-subgroup $P$ act by conjugation on the set of all Sylow $p$-subgroups. Every orbit other than a fixed point has size divisible by $p$. If $Q$ is fixed, then $P,Q\le N_G(Q)$. They are Sylow subgroups of this normalizer, so they are conjugate within $N_G(Q)$; because every element of $N_G(Q)$ fixes $Q$ under conjugation, this forces $P=Q$. Thus there is exactly one fixed point and $n_p\equiv 1(\mathrm{mod}p)$.

Now let $H<A_n$ have index $m$ with $1<m<n$. The coset action gives a homomorphismIts kernel is a normal subgroup of the simple group $A_n$. It cannot be all of $A_n$, since the action is transitive and nontrivial, so it is trivial. This would embed $A_n$ into $S_m$, contrary to $|A_n|=n!/2>m!=|S_m|$. Hence no such subgroup exists.

Suppose finally that a group $G$ of order $90$ were simple. The Sylow count satisfiesand simplicity excludes $n_5=1$, so $n_5=6$. Conjugation on these six subgroups gives a nontrivial homomorphism $G\to S_6$, which simplicity makes injective. Its image lies in $A_6$, because the composite with the sign homomorphism $S_6\to \{\pm 1\}$ must be trivial. It would therefore be a subgroup of $A_6$ of index $360/90=4$, contradicting the result just proved. Thus

The implicit function theorem states that if $F(x_0,y_0)=0$ and the partial derivative $D_{y}F(x_0,y_0)$ is an invertible linear map, then near $(x_0,y_0)$ the zero set of $F$ is uniquely the graph $y=g(x)$ of a continuously differentiable function.

If $f$ is a differentiable bijection with differentiable inverse $g$, the chain rule applied to $g\circ f$ and $f\circ g$ givesThus $D{f}_x$ is an isomorphism with inverse $D{g}_{f(x)}$.

If a continuously differentiable map $F:{\mathbb{R}}^n\to {\mathbb{R}}^n$ has invertible derivative everywhere, the inverse function theorem makes it a local diffeomorphism. In particular it is an open map, so its image is open. Its image need not be closed: $x\mapsto \arctan x$ has nonzero derivative everywhere and image $(-\pi /2,\pi /2)$.

For the given map of elementary symmetric polynomials,and direct evaluation of the determinant givesHence the critical set isIts complement is the Zariski-open set on which the three coordinates are pairwise distinct. Each point has one of the six possible strict coordinate orderings, and each ordering defines a nonempty convex open region. A continuous path cannot change an ordering without crossing $C$. Therefore ${\mathbb{R}}^3\setminus C$ has exactlyconnected components.

Withthe coordinate tangent vectors satisfyThus the first fundamental form isGeometrically, $S$ is the half of a helicoid with positive radius, winding upwards indefinitely.

Consider the upper half of the catenoid, parametrized as the surface of revolutionIt has the same first fundamental form, so $(u,v)\mapsto Y(u,v)$ defines a local isometry. By the invariance of Gaussian curvature under local isometry, the curvatures agree; directly, the conformal metric above gives

For an oriented regular surface with unit normal $N$, the Gauss map sends each point $p$ to $N(p)\in S^2$. The outward normal to the catenoid isIts image is the open southern hemisphere, apart from its limiting equator and south pole. The Gauss map is one-to-one after taking $v$ modulo $2\pi$ and reverses orientation. Since the Jacobian of the Gauss map is the Gaussian curvature, its signed spherical area is the total Gaussian curvature:Equivalently, $KdA=-{\mathrm{sech}}^{2}ududv$, whose integral over $u>0$ and $0\le v<2\pi$ is $-2\pi$.

The helicoid has no angular identification: its height coordinate is $v$. ConsequentlyThe catenoid has total curvature $-2\pi$. A global Riemannian isometry preserves both the Gaussian curvature and the area element, hence preserves total curvature. ThereforeThe same obstruction is visible in the parameter map: increasing $v$ by $2\pi$ gives a different point of the helicoid but the same point of the catenoid.

Close the contour in the upper half-plane. The degree assumption and Jordan lemma make the integral over the large semicircle tend to zero, while the absence of real zeros avoids indentations of the contour. The residue theorem therefore givesRepeated roots are handled by the usual higher-order residue formula.

Apply the formula toThe upper-half-plane poles areand their residues are $e^{i{z}_k}/(4z_k^2)$. Writing $c=1/\sqrt{2}$, their sum isSince the cosine part of $x{e}^{ix}/(1+x^4)$ is odd, its integral vanishes, while the sine part is even. Henceso

For fixed endpoint values, the first variation of the functionalvanishes exactly when the Euler-Lagrange equationholds. If $f$ has no explicit $x$ dependence, differentiation along an extremal gives the Beltrami identityFor a further constraint $J[z]=J_0$, introduce a constant Lagrange multiplier $\lambda$ and apply the Euler-Lagrange equation to $I+\lambda J$; $\lambda$ is then chosen so that the constraint holds.

Enforce the fixed arc length with a multiplier. Apart from an irrelevant common factor, the augmented integrand isThe Beltrami identity givesAfter absorbing constants into $z_0$ and a positive scale $B$, integration yields the catenaryThe endpoint conditions and prescribed length giveThe second equation determines $B>0$ implicitly because $L>2a$, and the first then determines $z_0$.

For the lower semicircleFor a general potential, the Beltrami first integral isSubstitution of the semicircle therefore requiresThus, up to an arbitrary additive constant and a multiplicative strength,The additive constant is absorbed by the length multiplier. The semicircle has length $\pi a$, as required.

Substitution of $y=e^{-x}$ givesFor a second solution $y=ax+b$, the ordinary differential equation reduces to $a\lambda -b=0$. Taking $a=1$ gives the linearly independent solution

The solution $x+\lambda$ satisfies the left boundary condition, while $e^{-x}$ satisfies decay at infinity. The proposed expression for $x<\xi$ is therefore the correct left homogeneous solution. For $x>\xi$, write $G=C{e}^{-x}$. Continuity at $x=\xi$ givesHence the Green function isIndeed, its derivative has the required unit jump,which produces the Dirac delta distribution in $L[G]$.

For $\lambda =2$, the Green representation of the boundary value problem isSplitting the integral at $\xi =x$ givesThereforeIt satisfies $y(0)=2y^{\prime }(0)$ and tends to zero as $x\to \infty$.

Inside the infinite square well, the time-independent Schrodinger equation has Dirichlet boundary conditions at $0$ and $a$. Its normalized energy eigenstates and energy eigenvalues are

Expand the normalized initial wavefunction in the orthonormal energy basis:The Time-dependent Schrodinger equation then gives the unitary time evolution

Forthe phase of the $n$th stationary state isReflection in the midpoint givessoThe minus sign is a global quantum phase and therefore does not change the physical state. The relative even--odd phases first acquire this reflection pattern at $T$. Applying it twice givesIf $f$ vanishes on the right half of the well, then $\psi (x,T)$ vanishes on the left half. The Born rule consequently gives zero probability of finding the particle in $0\le x\le a/2$ at time $T$.

The specified initial state is normalized. The requested energy is $E_2$, and its probability amplitude isThus the Born rule givesUnitary evolution changes each energy coefficient only by a phase, so this probability remains $1/2$ at both $T$ and $2T$. If an energy measurement returns $E_2$, the state collapses to the nondegenerate eigenstate ${\chi }_2$; a subsequent energy measurement therefore returns $E_2$ with probability

The Maxwell equation $\nabla \times \mathbf{E}=-\partial \mathbf{B}/\partial t$, integrated over a fixed spanning surface $S$, gives by Stokes theoremThus the electromotive force and magnetic flux obey Faraday's law

By cylindrical symmetry the field is azimuthal and constant on a circle of radius $r$. The Ampère's law for a steady current givessoIn the plane $x=0$ with $y>0$, it points in the $-x$ direction.

At time $t$, let the centre of the loop have coordinate $Y=d+vt$. With normal $+{\mathbf{e}}_x$, the magnetic flux isFaraday's law and Ohm's law therefore give a current of magnitudeThe original flux points into the plane and decreases in magnitude as the loop recedes. By Lenz's law, the induced current reinforces the into-plane field: it is clockwise when viewed from the $+x$ side.

Translation parallel to the wire leaves every point of the loop at the same distance from the wire, so its magnetic flux is constant. Equivalently, the line integrals of the motional field $\mathbf{u}\times \mathbf{B}$ cancel on opposite sides. Hence

Writing $f,f^{\prime },f^{\prime \prime }$ at $y_n$, the implicit stage has the expansionThus one step isThe exact Taylor expansion has third-order termThe coefficients agree through order two for every real $a$, while the $f^{\prime \prime }{f}^2$ coefficient prevents order three for any $a$. Hence the order of a Runge-Kutta method is(The linear special case has one extra matched term when $a=1/3$, but the order for general nonlinear equations remains two.)

Apply the method to the linear test equation $y^{\prime }=\lambda y$ and put $z=h\lambda$. Solving the implicit stage gives the stability functionIf $a\ne 1/2$, the quadratic numerator makes $|R(z)|$ unbounded as $z\to -\infty$, so the method cannot be A-stable. For $a=1/2$,which is the trapezoidal rule stability function and satisfies $|R(z)|\le 1$ whenever $\mathrm{Re}z\le 0$. Therefore

The chain is an irreducible birth-death process. The detailed balance equations areso ${\pi }_{i+1}={\pi }_i/2$. Normalization gives the stationary distributionAn irreducible countable-state Markov chain that possesses a stationary probability distribution is positive recurrent. Hence $X$ is positive recurrent.