Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 3

The Smith normal form of an integer $m\times n$ matrix $A$ is a diagonal matrixfor whichwith $U\in G{L}_m(\mathbb{Z})$ and $V\in G{L}_n(\mathbb{Z})$. It exists and is unique.

The structure theorem for finitely generated modules over a principal ideal domain, specialized to $\mathbb{Z}$, says that every finitely generated abelian group has a unique invariant-factor decompositionEquivalently, its finite part is a direct sum of cyclic groups of prime-power order.

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Integer row and column operations giveThus the Smith normal form isThe two relation vectors are the columns of the displayed matrix, so its cokernel is the presented module. Consequently

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By the finite case of the structure theorem, a finite abelian group is a direct sum of cyclic groups of prime-power order. If it is indecomposable, this decomposition can have only one nonzero summand, so the group is cyclic of order $p^n$ for some prime $p$.

Conversely, every nontrivial subgroup of the cyclic group $C_{p^n}$ contains its unique subgroup of order $p$. Hence two nontrivial subgroups cannot have trivial intersection. They therefore cannot be the two summands of an internal direct sum. Thus the indecomposable finite abelian groups are precisely

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Use the global parametrizationIts tangent vectors areandThus $X$ is a regular parametrization and the graph is a smooth surface.

The first fundamental form isChoosing the upward orientation, the Gauss map is

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The Cauchy integral theorem states that if $f$ is holomorphic on a simply connected domain, then its integral around every closed piecewise smooth contour in that domain is zero.

Put $q=b-k$. Completing the square givesAfter the change of variable $z=x-iq/(2a)$, the integral runs along the horizontal line $\mathrm{Im}z=-q/(2a)$. The integrand $e^{-a{z}^2}$ is entire. Apply Cauchy's theorem to a rectangle joining this line to the real axis; the two vertical integrals tend to zero as their real parts tend to $\pm \infty$, because $a>0$. The contour may therefore be shifted to the real line. Hence the Fourier transform of a Gaussian gives

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For a general Lagrangian,Along an Euler-Lagrange trajectory,so energy is conserved when $L$ has no explicit time dependence.

For the given Lagrangian, write $v=|\dot{x}|$. At points where $v\ne 0$,The Euler-Lagrange equation is thereforeIn expanded form its left-hand side isThe momentum has fixed magnitude,so this is constant without needing to solve the equation of motion. The Hamiltonian is

The singular Legendre transform of a degree-one velocity Lagrangian applies here: $p$ determines only the direction of $\dot{x}$, not its magnitude, and the velocity Hessian is not invertible. Thus the usual inverse Legendre transform cannot reconstruct the Lagrangian from this Hamiltonian.

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Set $u(r,\theta )=R(r)\Theta (\theta )$. Periodicity and separation of variables give the angular modes $1$, $\cos n\theta$, and $\sin n\theta$. The radial equation iswhose solutions are $r^{\pm n}$ for $n\ge 1$; the zero mode has solutions $1$ and $\log r$. Thus a general real harmonic function on the annulus is

The boundary data contain only the $n=2$ cosine mode, so writeThe condition at $r=a$ gives $B=-A{a}^4$, and the condition at $r=b$ fixes $A$. The Dirichlet problem on an annulus for one Fourier mode therefore has solution

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The canonical commutation relation is

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For a normalized state, the uncertainty of a Hermitian observable $O$ is

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Positivity of the squared norm of the supplied state gives, for every real $s$,where the stated zero means were used in the last line. This quadratic in $s$ is nonnegative for every real $s$, so its discriminant is nonpositive:This is the quadratic-norm proof of the Heisenberg uncertainty relation, and proves

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The steady Euler equations for an inviscid fluid areTaking the scalar product with $u$ and usinggivesThusThe Bernoulli function $H$ is constant along each streamline: fluid particles in steady flow exchange pressure, kinetic, and potential energy without changing their total mechanical energy density.

Let $y(t)$ be the downward displacement of the free surface from its initial level, and let $U$ be the speed in the tube. Conservation of volume givesThe free surface and the outlet are both at atmospheric pressure, and their vertical separation is $H+h_0-y$. Applying the Bernoulli equation between them, while retaining the small free-surface speed, givesThereforeThe surface reaches the upper tube end when $y=h_0$. Hence the draining time of a uniform tank through a siphon is

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The independent identically distributed sequence $(X_n)$ is itself a Markov chain, since the conditional law of $X_{n+1}$ is its common marginal law and does not depend on the past.

Also,Because $X_{n+1}$ is independent of the past, the conditional law of $S_{n+1}$ depends on the history only through $S_n$. Thus the partial sums $(S_n)$ are a Markov chain.

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The running minima satisfySince $X_{n+1}$ is independent of the past, the conditional law of $L_{n+1}$ depends only on $L_n$. Hence $(L_n)$ is a Markov chain. This is the running minimum of an independent sequence is Markov property.

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The moving sums $(K_n)$ are not necessarily Markov. For a counterexample, let the $X_n$ be independent Bernoulli variables with parameter $1/2$. On the eventwe must have $X_{n-1}=0$ and $X_n=1$, soOn the other hand, $K_n=1$ and $K_{n-1}=2$ force $X_{n-1}=1$ and $X_n=0$, whenceBoth conditioning events have positive probability. Knowledge of $K_n$ alone therefore does not determine the next-step law. The overlapping moving sum need not be Markov.

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Fix $y\in V$. The map $x\mapsto \theta (x,y)$ is a linear functional, so the finite-dimensional Riesz representation theorem gives a unique vector $\beta (y)$ such thatFor scalars $\lambda ,\mu$,Uniqueness givesso $\beta$ is linear.

For $\alpha \in \mathrm{End}(V)$, apply this result toIt produces a unique linear map ${\alpha }^{\ast }$ satisfyingwhich proves existence and uniqueness of the adjoint.

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Suppose first that $\alpha (U)\subseteq U$. If $y\in U^{\perp }$ and $u\in U$, thenThus ${\alpha }^{\ast }(U^{\perp })\subseteq U^{\perp }$.

Conversely, assume that latter inclusion. For $u\in U$ and $y\in U^{\perp }$,Hence $\alpha u\in (U^{\perp }{)}^{\perp }=U$. This proves the adjoint criterion for an invariant orthogonal complement:

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Write $B(x,y)=⟨\alpha x,y⟩$. The hypothesis says $B(x,x)=0$. Expanding at $x+y$ giveswhereas expanding at $x+iy$ givesTherefore $B(x,y)=0$ for every $x,y$, soThis is the complex polarization argument for a vanishing quadratic form.

The conclusion is false over a real inner product space. On ${\mathbb{R}}^2$, the nonzero operatorsatisfies $⟨Jx,x⟩=0$ for every $x$.

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For every $x$,If $\alpha$ is normal, these quantities are equal. Conversely, equality of the norms for all $x$ givesThe operator in parentheses is self-adjoint, so polarization makes it zero. ThereforeThe same equivalence holds over a real inner product space: for a self-adjoint operator $T$, the real polarization identityshows that a vanishing quadratic form forces $T=0$.

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Proceed by induction on $\dim V$. Over $\mathbb{C}$, $\alpha$ has an eigenvector $v$, say $\alpha v=\lambda v$. The normal operator $\alpha -\lambda I$ satisfies the norm equality from part (c), soThus ${\alpha }^{\ast }v=\overline{\lambda }v$. Part (a) now shows that $v^{\perp }$ is invariant under both $\alpha$ and ${\alpha }^{\ast }$. The restriction of $\alpha$ to $v^{\perp }$ is normal. By induction it has an orthonormal eigenbasis, and adjoining the normalized vector $v$ proves the finite-dimensional spectral theorem for normal operators.

Hence

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Suppose $I$ is prime. Ifthen $ab\in I$, so $a\in I$ or $b\in I$. Thus one factor is zero and $R/I$ is an integral domain.

Conversely, if $R/I$ is an integral domain and $ab\in I$, thenso $a+I=0+I$ or $b+I=0+I$. Therefore $a\in I$ or $b\in I$, and $I$ is prime. This proves the prime ideal quotient criterion.

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In a Boolean ring, $(r+1{)}^2=r+1$. Using $r^2=r$ and $1^2=1$ givesso

If $R$ is also a nonzero integral domain, thenforces every $r$ to equal $0$ or $1$. Hence $R\cong {\mathbb{F}}_2$.

If $I$ is prime in a Boolean ring, then $R/I$ is a nonzero Boolean integral domain by part (i), and hence is ${\mathbb{F}}_2$. Since the quotient is a field,This is the prime ideals of a Boolean ring are maximal property.

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Reduce coefficients modulo $I$:This is a surjective ring homomorphism and its kernel consists exactly of polynomials all of whose coefficients lie in $I$, namely $I[X]$. The first isomorphism theorem gives the coefficientwise quotient of a polynomial ringIf $I$ is prime, then $R/I$ is an integral domain, so $(R/I)[X]$ is an integral domain. The quotient criterion therefore shows that $I[X]$ is prime in $R[X]$.

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Take $R=\mathbb{Z}$ and $I=(2)$. The ideal $I$ is maximal, butis not a field, since $X$ is nonzero and not invertible. Therefore

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Let $G={\mathbb{F}}_p^{\times }$, and let $m$ be the least common multiple of the orders of its elements. For every prime power $q^a$ dividing $m$, some element of $G$ has order divisible by $q^a$, and a suitable power of it has order exactly $q^a$. Multiplying these elements over the distinct primes produces, because their orders are coprime, an element of order $m$.

Every element of $G$ is a root of $X^m-1$. A nonzero polynomial of degree $m$ over a field has at most $m$ roots, so $p-1\le m$. On the other hand, every element order divides $p-1$ by Lagrange's theorem, so $m\le p-1$. Hence $m=p-1$, and

The squaring homomorphism has kernel $\{1,-1\}$ because $p$ is odd. Its image $H$ therefore has order $(p-1)/2$ and index two. If $p=3$, then $3=0$ is already a square. Otherwise $2,3\in G$. In the two-element quotient $G/H$, either $2H=H$, or $3H=H$, or both are the nontrivial coset, in which case $6H=H$. Thus one of $2,3,6$ is a square modulo $p$.

Finally,Whichever of $2,3,6$ is a square supplies a root, soThis is the index-two square-class argument for three related residues.

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A function $f:(X,d_X)\to (Y,d_Y)$ is uniformly continuous if for every $\epsilon >0$ there is $\delta >0$ such thatfor all $x,x^{\prime }\in X$.

Suppose $f_n\to f$ uniformly and every $f_n$ is uniformly continuous. Given $\epsilon >0$, choose $N$ such thatfor every $x$. Uniform continuity of $f_N$ supplies $\delta >0$ such that $d_X(x,x^{\prime })<\delta$ impliesThe triangle inequality then gives $d_Y(f(x),f(x^{\prime }))<\epsilon$. Thus the uniform limit theorem for uniformly continuous functions proves that $f$ is uniformly continuous.

Pointwise convergence is insufficient. On $[0,1]$, the uniformly continuous functions $f_n(x)=x^n$ converge pointwise towhich is discontinuous and therefore not uniformly continuous.

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The inequalities implyThus the two metrics induce the same open sets and are equivalent.

The converse fails. On $\mathbb{R}$, letThe metrics are equivalent because $\arctan :\mathbb{R}\to (-\pi /2,\pi /2)$ is a homeomorphism, butso no positive lower comparison constant exists. This is an equivalent metrics need not be bi-Lipschitz equivalent example.

Equivalent metrics on the codomain also need not give the same uniform convergence. Take $X=\mathbb{N}$, $Y=\mathbb{R}$, and use the two metrics above. DefineThenso $f_n\to f$ uniformly for $d_2$, whereasso convergence is not uniform for $d_1$. This is the equivalent codomain metrics need not preserve uniform convergence phenomenon.

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In polar coordinates the metric isAlong a radius, the hyperbolic distance from the origin to Euclidean radius $r$ is thereforeso $r=\tanh (R/2)$. The Riemannian area element isConsequently a hyperbolic disc of radius $R$ has areaThis proves the area of a hyperbolic disc formula from the metric.

For area $\pi /2$ we get $\cosh R=5/4$. Hence $\sinh R=3/4$ and $e^R=2$, so $R=\log 2$. Tangent equal discs have centers at distance $2R$. Since the centers lie successively on the same radial geodesic,If $r_n$ is the Euclidean coordinate of $c_n$, the radial distance formula givesTherefore the radial chain of equal hyperbolic discs has

No such isometry to the stated upper-half-plane configuration exists for $n\ge 3$. The centers $c_0,\dots ,c_n$ lie on one hyperbolic geodesic, so their images under an isometry would also lie on one geodesic. In the upper-half-plane model, geodesics are vertical lines or semicircles orthogonal to the real axis. Neither type can contain three distinct points of the horizontal line $y=1$, while the centers of the distinct discs $D_0^{\prime },\dots ,D_n^{\prime }$ all lie on that line. This is the geodesic obstruction for a horizontal chain of hyperbolic discs.

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For a closed piecewise smooth curve $\gamma$ avoiding $w$, its winding number about $w$ is

The argument principle says that if a positively oriented closed curve $\gamma$ bounds a domain $U$, and a meromorphic function $F$ has no zeros or poles on $\gamma$, thenwhere zeros and poles are counted with multiplicity.

Suppose $F$ and $G$ are holomorphic on a neighbourhood of $\overline{U}$ andFor $0\le t\le 1$, $F+tG$ has no boundary zero, because a zero would imply $|F|=t|G|<|F|$. Its argument-principle count is integer-valued and continuous in $t$, hence constant. Thus Rouché's theorem states that

Now let $w_0=f(z_0)$. Since $|w_0|<1\le |f|$ on the unit circle, Rouché's theorem shows that $f$ and $f-w_0$ have the same number of zeros in the unit disc. The latter has the zero $z_0$, so $f$ has at least one zero there. For any $w$ with $|w|<1$, the same boundary inequality shows that $f-w$ has the same positive number of zeros as $f$. ThereforeThis is the unit-disc image from a boundary modulus lower bound.

Finally, take a sufficiently small positively oriented circle $C$ around zero. The residueis the winding number of $g(C)$ around zero, so $k\in \mathbb{Z}$. The pole is simple, hence its residue is nonzero and $k\ne 0$. Forwe haveThe second term cancels the complete principal part at zero, so the integer residue of a logarithmic derivative proves that $h^{\prime }/h$ has a removable singularity there.

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Separation $y=T(t)X(x)$ and the fixed-end conditions givePutThe initial displacement and zero initial velocity then give the separated solution of the damped string equation

For the triangular initial displacement,Thus the even coefficients vanish and the odd ones alternate in sign.

LetThenOrthogonality of the sine and cosine modes, equivalently the Parseval identity, gives

The sign of its derivative is clearest directly from the equation. Integration by parts, using $y_t=0$ at the fixed endpoints, yieldsThis is the energy dissipation identity for a linearly damped string.

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In electrostatic equilibrium, a nonzero electric field inside a conductor would move its free charges. Hence $E=0$ in the conducting material and the potential is constant throughout each connected conductor. Immediately outside its surface the tangential electric field is zero. A Gaussian pillbox across the surface givesSince $E_{\mathrm{in}}=0$, the electrostatic boundary conditions at a conductor give

For the widely separated shells, let their charges be $Q_1,Q_2$. Their common potential requiresThus the charge sharing between distant connected spheres is

For the charge on connected concentric spherical shells, the potentials at the two radii areEquality forces $Q_1=0$. Henceso all charge lies on the exterior of the outer shell.

For the neutral sphere in the uniform field, the far-field condition gives $\alpha =-E$. Constancy of the potential on $r=R$ gives $\beta =E{R}^3$. ThereforeThe outward normal field at the surface isThe induced charge on a conducting sphere in a uniform electric field is consequentlyFinally,confirming neutrality.

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Take the sphere to move in the positive $z$-direction and use spherical coordinates centred on it. An axisymmetric harmonic potential that decays at infinity has the dipole form $A\cos \theta /r^2$. The no-penetration condition in the laboratory frame isIt fixes $A=-U{a}^3/2$, so the potential flow around a translating sphere isThe laboratory-frame velocity components are