Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 4

If $\dim V=n$, then $\dim \mathrm{End}(V)=n^2$. The $n^2+1$ endomorphisms $I,T,\dots ,T^{n^2}$ are therefore linearly dependent, so some nonzero polynomial $p$ satisfies $p(T)=0$. The minimal polynomial is the monic polynomial of least degree annihilating $T$. Uniqueness follows because the difference of two such monic polynomials would be a nonzero annihilator of smaller degree.

A Jordan block $J_r(\lambda )$ has minimal polynomial $(x-\lambda {)}^r$, and the minimal polynomial of a direct sum is the least common multiple of those of its blocks. Thus an eight-dimensional example is

Let $T$ consist of all torsion elements of a module in $M$. If $rm=0$ and $sn=0$ with $r,s\ne 0$, then $rs(m+n)=0$. Since $R$ is an integral domain, $rs\ne 0$. Also $r(am)=0$ for every scalar $a\in R$. Thus $T$ is the torsion submodule.

If $r(m+T)=0$ in $M/T$ with $r\ne 0$, then $rm\in T$, so some $s\ne 0$ satisfies $srm=0$. The domain property again gives $sr\ne 0$, hence $m\in T$. Therefore $M/T$ is a torsion-free module.

Finally, any submodule $N$ with the stated properties satisfies $N\subseteq T$. Conversely, if $m\in T$, its coset in $M/N$ is torsion and therefore zero, so $m\in N$. Hence $\boxed{N=T}$ is the unique choice.

Uniform convergence means that for every $\epsilon >0$ there is $N$ such that $|s_n(x)-s(x)|<\epsilon$ for every $x\in \mathbb{R}$ and every $n\ge N$. Equivalently, ${\sup }_x|s_n(x)-s(x)|\to 0$. The nonzero constant functions $s_n=1/n$ converge uniformly to zero. In contrast, the continuous moving bumps $s_n(x)=\max (0,1-|x-n|)$ converge pointwise to zero but have supremum one for every $n$.

On the compact interval,Thus uniform convergence permits the stated interchange with integration.

Choose $\boxed{s(x)=1/(1+|x|)}$. It is continuous, nonnegative and tends to zero at infinity, but its improper integral diverges logarithmically. This even choice also makes the prescribed cutoff continuous at both $n$ and $-n$: beyond either endpoint it remains at $s(n)$ for another unit of distance and then decreases as $s(n)/(|x|-n{)}^2$.

Inside $[-n,n]$, $s_n=s$. Outside, both functions lie between zero and $1/(n+1)$, so ${\sup }_x|s_n-s|\le 1/(n+1)\to 0$. Nevertheless each cutoff has finite integral:Here the value of the minimum at $t=0$ is its continuous limiting value one. This example shows that uniform convergence on an unbounded domain does not by itself ensure convergence of the integrals over that domain.

Rouché's theorem states that if $f,g$ are holomorphic on and inside a simple closed contour and $|g|<|f|$ on it, then $f$ and $f+g$ have the same number of zeros inside, counted with multiplicity. On $|z|=1$, the power series for sine givesApply Rouché's theorem with $f=2z$ and $g=-\sin z$. There is one zero inside and none on the boundary. Since $z=0$ is a zero and $(2z-\sin z{)}^{\prime }{|}_{z=0}=1$, it is simple. Exactly one complex number qualifies: $z=0$.

At a regular constrained stationary point, $\nabla g\ne 0$ and the directional derivative of $f$ vanishes along the tangent to $g=0$. Therefore $\nabla f=\lambda \nabla g$ for a Lagrange multiplier $\lambda$. Solve this equation together with the constraint, then compare candidate values and any exceptional nonregular points. Proportionality of the two gradients givesIf $\nabla g=0$ at an exceptional point, this determinant vanishes trivially, although the regular multiplier method need not identify every such point.

For the given ellipse, maximize $r^2=x^2+y^2$. The constraint is $r^2+xy=4$, and $xy\ge -r^2/2$, so $4\ge r^2/2$. Equality requires $x=-y$ and the constraint then gives $(x,y)=(2,-2)$ or $(-2,2)$. ThusThe constraint gradient vanishes only at the origin, which is not on this ellipse, so the regularity condition holds everywhere relevant.

Put $D=1+i\hslash t/m$ and write the Gaussian wave packet as $\psi ={\pi }^{-1/4}{D}^{-1/2}{e}^{-x^2/(2D)}$. Differentiation givesHence $\boxed{i\hslash {\psi }_t=-({\hslash }^2/2m){\psi }_{xx}}$, the free Time-dependent Schrödinger equation.

With $\tau =\hslash t/m$, its probability density isThe stated Gaussian integral shows its integral is one. Symmetry gives $⟨x⟩=0$; differentiating that Gaussian integral with respect to its coefficient gives $⟨x^2⟩=(1+{\tau }^2)/2$. Thus the position standard deviation isIt is not a stationary state: its probability density changes with time as the packet spreads. A stationary state would have only an overall time-dependent phase and a fixed position density.

With consistent flux and loop orientations, Faraday's law is $V=-d\Phi /dt$. The magnetic force on a current-carrying wire is $d\mathbf{F}=Id\mathbf{\ell }\times \mathbf{B}$.

Let $z(t)$ be the top edge height and $v=-\dot{z}$ the downward speed. While $0<z<l$, only the top horizontal edge lies in the field, and the linked magnetic flux is $Bwz$. Therefore the induced electromotive force has magnitude $Bwv$, and Ohm's law gives current magnitude $I=Bwv/R$. Its force opposes the downward motion and has upward magnitude $IBw=B^2{w}^{2}v/R$. Forces on the vertical sides cancel horizontally. HenceFor a release from rest, $v=v_{\mathrm{terminal}}(1-e^{-B^2{w}^{2}t/(mR)})$. The terminal speed is the formal limit while partial overlap is maintained; after the entire loop leaves the field this magnetic braking term vanishes.

For distinct Gaussian quadrature nodes, let ${\ell }_i$ be the Lagrange interpolation polynomial with ${\ell }_i(x_j)={\delta }_{ij}$. Since ${\ell }_i^2$ has degree $2n-2$, exactness givesThe last inequality follows from $w>0$ on the interval and the nonzero polynomial ${\ell }_i$ being nonzero on a subinterval. It does not require prior knowledge of the signs of the weights.

For $n=2$, the nodes are the roots of the supplied orthogonal polynomial $p_2$:Indeed exactness for $1$ gives $b_1+b_2={\int }_{-1}^1(1+x^2)dx=8/3$, and exactness for $x$ gives equality of the weights.

Use coordinates $(i,j)\in \{0,\dots ,7{\}}^2$, with the starting corner $(0,0)$. The bishop stays on the 32 squares where $i+j$ is an even number. Its moves form a random walk on a graph, with an edge between distinct squares whenever $|i-i^{\prime }|=|j-j^{\prime }|$. If $d(i,j)$ is the number of reachable squares, the transition probability to each neighbor is $1/d(i,j)$.

The graph is connected: from $(i,j)$ move along its antidiagonal to $((i+j)/2,(i+j)/2)$, then along the main diagonal to the corner. The midpoint coordinates are integers because $i+j$ is an even number, and this square always lies on the board. Thus this is the full communicating state space of the bishop random walk.

The invariant distribution is proportional to degree. Its normalizing constant can be counted by summing the ordered pairs along each diagonal. The diagonals of one orientation have lengths $8,6,6,4,4,2,2$, contributing $144$; the other have lengths $1,3,5,7,7,5,3,1$, contributing $136$. ThereforeFor any neighboring states $s,t$, $\pi (s)P(s,t)=1/280=\pi (t)P(t,s)$, proving the detailed balance equations and hence that the chain is a reversible Markov chain.

The corner has seven neighbors. Kac's lemma gives its expected first positive return time:

Two real symmetric bilinear forms are congruent if an invertible linear map takes one to the other; in matrices, $B=S^{\mathsf{T}}AS$ for some invertible $S$. The spectral theorem for real symmetric matrices gives an orthogonal eigenbasis. Rescaling each basis vector with nonzero eigenvalue then reduces any such form toSylvester's law of inertia asserts that $(p,q,r)$ is independent of the congruence reduction. To prove it, $p$ is the greatest dimension of a subspace on which the form is positive definite: the positive coordinate subspace attains it, whereas any subspace of dimension greater than $p$ meets the nonpositive coordinate subspace of dimension $n-p$ and therefore cannot be positive definite. Applying the same argument to the negative of the form determines $q$. Then $r=n-p-q$ is determined too.

The rank of a quadratic form is $p+q$, and its signature of a quadratic form is $p-q$ under the signed-difference convention. They determine $p,q,r$ and hence the congruence class. Counting all pairs $p,q\ge 0$ with $p+q\le n$ givesFor $n\ge 1$, exactly two classes have every level set bounded: the positive-definite and negative-definite classes. Their nonempty nonzero levels are ellipsoids, and their zero level is the origin. A degenerate form has an entire null line in its zero level. An indefinite nondegenerate form also has a nonzero isotropic vector, for example the sum of one positive and one negative normalized basis vector, whose scalar multiples make the zero level unbounded.

For the displayed three-dimensional form,Its leading principal minors are $3,14,32$, all positive. Sylvester's criterion makes it positive definite, so the surface is a bounded ellipsoid. Directly, its quadratic expression iswhich also proves positivity and boundedness of the unit level.

Prove the submodule theorem for free modules over a principal ideal domain by induction on the rank of $R^n$. For a submodule $N$, project onto its first coordinate. The image is an ideal $aR$. If $a=0$, then $N$ is a submodule of $R^{n-1}$ and induction applies. Otherwise choose $v\in N$ projecting to $a$. Every $w\in N$ has first coordinate $ra$ and hence $w-rv$ lies in the projection kernel. The domain property shows $Rv$ meets that kernel only at zero. Thusand the kernel is a free module by induction. This proves that $N$ is free, of rank at most $n$.

To prove free modules are projective, choose a basis $(e_{\lambda })$ of a free module $P$. For the given surjection $f$, choose $m_{\lambda }\in M$ with $f(m_{\lambda })=g(e_{\lambda })$ and extend $e_{\lambda }\mapsto m_{\lambda }$ linearly. Finite support of basis expansions makes the resulting module homomorphism well defined, and $f\circ h=g$.

Finally, take a finite generating set of a projective module $P$ to obtain a surjection $\pi :R^n\to P$. Projectivity applied to ${\mathrm{id}}_P$ gives a section $s:P\to R^n$ with $\pi s={\mathrm{id}}_P$. Thus $s$ is injective and identifies $P$ with a submodule of $R^n$. The proved submodule theorem for free modules over a principal ideal domain shows that every finitely generated projective module over a principal ideal domain is free.

Parametrize the graph by $X(u,v)=(u,v,f(u,v))$. The first fundamental form has coefficients $E=1+f_u^2$, $F=f_u{f}_v$, $G=1+f_v^2$. With unit normal $(-f_u,-f_v,1)/\sqrt{1+|\nabla f{|}^2}$, the second fundamental form has coefficients $e=f_{uu}/\sqrt{1+|\nabla f{|}^2}$, $f_{\mathrm{II}}=f_{uv}/\sqrt{1+|\nabla f{|}^2}$, and $g=f_{vv}/\sqrt{1+|\nabla f{|}^2}$. Consequently the Gaussian curvature of a graph surface is

A compact embedded surface has a highest point in the vertical direction. At that point its tangent plane is horizontal, so it is locally a graph with a local maximum of $f$. Its Hessian matrix is negative semidefinite, giving $\det D^{2}f\ge 0$ and hence $K\ge 0$. Thus the Gaussian curvature cannot be everywhere negative.

An embedded ring torus, with major radius $R$ larger than minor radius $r>0$, is a compact surface without boundary. Its standard parametrization is $((R+r\cos \theta )\cos \varphi ,(R+r\cos \theta )\sin \varphi ,r\sin \theta )$, and the Gaussian curvature of a ring torus isSince the denominator is positive, curvature is positive on the outer side and negative on the inner side, so this example changes sign.

Write $L(f)={\sup }_{x\ne y}|f(x)-f(y)|/d(x,y)$ and $B(f)={\sup }_z|f(z)|$. The quantity in the question is the bounded Lipschitz norm $\Vert f{\Vert }_{\mathrm{BL}}=L(f)+B(f)$. Both terms are nonnegative and absolutely homogeneous, and each satisfies the triangle inequality. Therefore their sum does too; finite values are preserved by addition and scalar multiplication, making the domain a vector space. If the sum is zero, $B(f)=0$ forces $f=0$. Thus it is a norm.

For the sequence bounded by one, both $L(f_i)\le 1$ and $B(f_i)\le 1$. Given real $x$, choose a nearby rational $q$. ThenFirst make $|x-q|$ small, then use convergence at $q$ to make the last term small. The values at every real $x$ form a Cauchy sequence, so $f_i$ converges pointwise to a real function $f$.

To preserve the sum bound rather than merely bound its two terms separately, use every fixed triple $x\ne y,z$:Pass to the pointwise limit and take the independent suprema over $(x,y)$ and $z$. This gives $\boxed{\Vert f{\Vert }_{\mathrm{BL}}\le 1}$, the pointwise closure of the bounded Lipschitz unit ball.

To obtain diagonal compactness for bounded Lipschitz functions, enumerate the rationals for an arbitrary sequence in this unit ball. Values at the first rational lie in $[-1,1]$, so the Bolzano-Weierstrass theorem supplies a convergent subsequence there. Repeatedly extract nested subsequences for successive rationals. The diagonal subsequence argument gives one subsequence converging at every rational. Applying the preceding argument proves there is a pointwise convergent subsequence with limit of bounded Lipschitz norm at most one.

If an open set meets $\overline{A}$, it meets $A$. Thus if two open sets meet the closure of an irreducible set, they both meet $A$, their intersection meets $A$ by irreducibility, and therefore it also meets $\overline{A}$. This proves irreducibility of the closure. A singleton is irreducible, so its closure is an irreducible closed subset.

In a Hausdorff space, two distinct points have disjoint open neighborhoods. An irreducible set cannot contain both, so every nonempty irreducible closed subset is a singleton, which is closed and has a unique generating point. Hence every Hausdorff space is a sober topological space. An infinite set with the cofinite topology is itself irreducible, because any two nonempty open sets meet, but every singleton is closed. Its whole space is therefore not the closure of a singleton, proving that it is not sober.

For the proposed sobrification, let $\hat{X}$ consist of nonempty irreducible closed subsets. The basic operations areThe last identity uses irreducibility of each member of $\hat{X}$. Hence these sets already form a topology. The map $U\mapsto \hat{U}$ is injective: $\overline{\{x\}}$ meets an open $U$ exactly when $x\in U$, so equality of the hats forces equality of the original opens. It is surjective onto the topology just constructed by definition.

Now let $C$ be a nonempty irreducible closed subset of $\hat{X}$. Its complement is $\hat{W}$ for a unique open $W\subseteq X$. Put $F=X\setminus W$. ThenIf opens $U,V$ meet $F$, choose points in those intersections. Their singleton closures lie in $C$, so $\hat{U},\hat{V}$ meet $C$. Irreducibility of $C$ forces $\widehat{U\cap V}$ to meet $C$, and hence $U\cap V$ meets $F$. Thus $F$ is nonempty, closed and irreducible, so $F$ is itself a point of $\hat{X}$.

The closure of this point in $\hat{X}$ is exactly $\{A:A\subseteq F\}$: inclusion makes every open meeting $A$ meet $F$, whereas a point of $A\setminus F$ is separated by the open $X\setminus F$. Hence $C=\overline{\{F\}}$. Any other point $G$ with the same closure satisfies $G\subseteq F$ and $F\subseteq G$, forcing $F=G$. This proves the constructed space is sober.

Use the Fourier transform convention $\hat{u}(k,y)={\int }_{\mathbb{R}}u(x,y)e^{-ikx}dx$ and inverse factor $1/(2\pi )$. Laplace equation gives ${\hat{u}}_{yy}-k^2\hat{u}=0$. Selecting its decaying Fourier branch gives $\hat{u}(k,y)={\hat{u}}_0(k)e^{-|k|y}$, the standard bounded solution for bounded continuous boundary data. The inverse transform of the multiplier is the Poisson kernel for the upper half-plane:The convolution theorem therefore gives the Poisson integralFor suitably regular boundary data, this kernel is an approximate identity as $y\downarrow 0$, recovering $u_0$.

For $a>0$, contour integration gives ${\hat{u}}_0(k)=(\pi /a)e^{-a|k|}$. Multiplication by $e^{-y|k|}$ and inversion yieldThe stated form assumes $a>0$; if $a<0$, replace it by $|a|$ because the boundary function only depends on $a^2$. The case $a=0$ has singular boundary data and is outside this regular calculation.

The transform argument requires a growth condition. A zero-boundary harmonic function with pointwise vertical decay shows that the printed conditions alone do not ensure uniqueness:vanishes at $y=0$ and tends to zero as $y\to \infty$ for every fixed $x$. It can be added without changing those conditions, but its growth in $x$ excludes it from the bounded solution class. The displayed Poisson integral is the standard bounded solution for regular bounded boundary data; pointwise vertical decay by itself cannot justify a uniqueness claim.

Away from $x=\xi$, the Dirichlet Green function solves the homogeneous equation. The endpoint conditions select $A{y}_1(x)$ on the left and $B{y}_2(x)$ on the right. Continuity and the unit derivative jump needed for a Dirac delta function giveWith Wronskian $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2\ne 0$, solve to obtain $A=y_2(\xi )/W(\xi )$ and $B=y_1(\xi )/W(\xi )$. ThusThe first derivative term creates no extra delta because $G$ is continuous. The value at $x=\xi$ is the common limiting value.

For the specified constant-coefficient operator, takeTherefore its explicit Green function isFor the forcing step and the homogeneous endpoint conditions inherited from this Green-function problem, $y(x)=2{\int }_{x_0}^{1}G(x,\xi )d\xi$. When $x<x_0$, every integration point lies on the first branch, soWithout those endpoint conditions, an arbitrary homogeneous solution could also be added.

Choose the photon direction in the $x^{\prime }$–$y^{\prime }$ plane. Its four-momentum in the stellar rest frame isLet $\beta =v/c$, $\gamma =(1-{\beta }^2{)}^{-1/2}$. The Lorentz transformation to $S$ givesUsing $p^0=h\nu /c$ and $p^x=(h\nu /c)\cos \theta$ yields the relativistic Doppler effect and relativistic aberration:

For the oppositely directed transverse photons, their rest-frame spatial momenta cancel and their total energy is $E=2h{\nu }^{\prime }$. Boosting this combined four-vector gives laboratory energy $\gamma E$ and momentumThe star loses rest mass $E/c^2$. At unchanged speed its momentum therefore decreases by $\gamma (E/c^2)v$, exactly the photon momentum. Its laboratory energy loss is likewise $\gamma E$. There is no rest-frame recoil from the symmetric emission, and constant speed is consistent with four-momentum conservation because the stellar mass decreases.

For small irrotational surface gravity waves, write the velocity field as $\nabla \varphi$. Incompressibility gives ${\nabla }^2\varphi =0$. The rigid walls have ${\varphi }_x=0$ at $x=0,a$, ${\varphi }_y=0$ at $y=0,b$, and ${\varphi }_z=0$ at $z=-h$. At $z=0$, the linear kinematic boundary condition and pressure condition areSeparate variables for a rectangular standing surface-gravity mode:The horizontal cosine factors and vertical hyperbolic cosine satisfy all rigid-wall conditions. Put $\eta =A(t)\cos (m\pi x/a)\cos (n\pi y/b)$ and $\lambda =k\tanh kh$. The free-surface conditions give $\dot{A}=\lambda B$ and $\dot{B}=-gA$, henceBoth integers are nonnegative and not both zero. The zero pair would be a spatially uniform displacement, excluded by fixed fluid volume; a spatially constant potential is only a gauge mode.

The applied pressure changes the dynamic condition to ${\varphi }_t+g\eta =-p/\rho$. For its specified spatial mode, the two amplitude equations becomeWhen $\Omega \ne \omega$, the corresponding particular oscillation isFree oscillations with amplitudes $C\cos \omega t+D\sin \omega t$ may be added. If the mode starts from rest at $t=0$, its total amplitude instead isNear resonance, the response grows too large for the linearization. At exact resonance the rest-start limit is $A=(\epsilon h/2)(\omega t\cos \omega t-\sin \omega t)$, with secular growth. Dissipation or nonlinear effects would be required to limit it.

A sufficient statistic is a function of the sample whose conditional sample distribution, given that statistic, does not depend on the parameter. The factorization criterion for sufficiency states, for a dominated family, that $T$ is sufficient exactly when the joint density has form $g_{\theta }(T(x))h(x)$, where $h$ is independent of $\theta$.

Let $m=X_{(1)}={\min }_i{X}_i$ and $M=X_{(n)}={\max }_i{X}_i$. The joint uniform density isso $\boxed{T=(m,M)}$ is sufficient for $(a,b)$ by the factorization criterion for sufficiency. The uniform order statistics have expectationsFor example, the standardized minimum has survival probability $(1-t{)}^n$ on $[0,1]$, whose integral is $1/(n+1)$; the maximum follows by reflection. Solving these expectation equations gives the unbiased estimatorThe assumption $n\ge 2$ makes the denominator positive.

Interpret a beneficial trade as weakly improving both participants and strictly improving at least one; otherwise the identity trade would always remain available. Put $g_j=\nabla U_j(y_j)$. For a globally strictly increasing concave function, every component of its gradient is strictly positive: concavity makes a coordinate derivative nonincreasing along that coordinate, and monotonicity makes it nonnegative. If it vanished at a finite point, it would vanish farther along the coordinate, contradicting strict increase on the whole real line.

To construct a pairwise improvement from nonparallel utility gradients, suppose two positive gradients $g,h$ are not proportional. Strict Cauchy-Schwarz inequality givesChoose $t$ between these bounds and put $\delta =g-th$. Then $g\cdot \delta >0$ and $h\cdot \delta <0$. For sufficiently small $\epsilon >0$, the feasible transfer $y_i\mapsto y_i+\epsilon \delta$, $y_j\mapsto y_j-\epsilon \delta$ strictly improves both utilities by differentiability. This contradicts the absence of beneficial pairwise trades.

Hence all gradients are proportional. Choose $v=g_1$; positivity makes each proportionality scalar positive:Now take $\boxed{a_j=1/{\lambda }_j}$. The concave supporting-tangent inequality gives, for any other allocation $(z_j)$ with the same total endowment,Thus the final allocations globally maximize the social planner's weighted objective. They need not be unique when utilities are not strictly concave, but the required positive supporting weights exist.