Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 4

The dual map reverses the direction of the linear map:Here the dual spaces consist of complex-linear functionals. Composition with $\alpha$ is linear both as a functional on $V$ and as an operation on $W^{\ast }$, so this does define a linear map between the dual spaces.

Let $(v_1,\dots ,v_n)$ and $(w_1,\dots ,w_m)$ be the chosen bases, and let $(v^1,\dots ,v^n)$ and $(w^1,\dots ,w^m)$ be their dual bases. If $A$ represents $\alpha$, then $\alpha (v_i)={\sum }_j{A}_{ji}{w}_j$. ThereforeThe coefficient in row $i$, column $j$ of the matrix of ${\alpha }^{\ast }$ is thus $A_{ji}$, giving the matrix transpose $A^{\mathsf{T}}$. This calculation introduces no complex conjugation: the algebraic dual map uses complex-linear functionals, whereas a Hermitian transpose belongs to the inner-product adjoint construction.

The Fundamental theorem of finitely generated abelian groups expresses every finite abelian group of this order as a product of cyclic groups of prime-power order. Their exponents correspond to the partitions of an integer $4$:These are pairwise nonisomorphic: the largest order of a group element distinguishes all except the two groups with largest element order $p^2$, and those have respectively $p^2$ and $p^3$ elements annihilated by multiplication by $p$. The classification exhausts the possibilities, so there are five isomorphism classes.

The sequence has uniform convergence to $f$ on $(0,1)$ precisely whenThe same $N$ must work throughout the interval; equivalently ${\sup }_{0<x<1}|f_n(x)-f(x)|\to 0$.

For each fixed $x>0$, continuity of the logarithm gives the pointwise convergence $f_n(x)\to \log x$. But at $x_n=1/n$, which belongs to $(0,1)$ for $n>1$,Consequently the uniform error cannot tend to zero. In fact $f_n(x)-\log x=\log (1+1/(nx))$ is unbounded as $x\downarrow 0$ for every fixed $n$. The pointwise limit is $\log x$, and convergence is not uniform.

For each fixed $x$, $x/n\to 0$, so the pointwise limit is $1$. Moreover $1-\cos u=2{\sin }^2(u/2)\le u^2/2$ givesThus the limit is $1$, with uniform convergence on $(0,1)$.

The argument principle states that if $h$ is meromorphic on a neighbourhood of a positively oriented simple closed contour and its interior, with neither zeros nor poles on the contour, thenwhere $N$ and $P$ count its interior zeros and poles with multiplicity. This integer is also the winding number of $h(\gamma )$ about zero.

Fix $z_0\in U$. Injectivity prevents $f$ from being constant on a neighbourhood of $z_0$, so its Taylor expansion has the formChoose a small closed disk $D\subset U$, centred at $z_0$, such that $h$ and $mh+(z-z_0)h^{\prime }$ are nonzero throughout $D$. Then $f-f(z_0)$ has its sole zero at $z_0$ in $D$, of multiplicity $m$, while $f^{\prime }$ has no zero in $D\setminus \{z_0\}$. On its boundary let $\delta =\min |f-f(z_0)|>0$.

Choose $0<|w|<\delta$. For every $0\le t\le 1$, the holomorphic function $f-f(z_0)-tw$ is nonzero on $\partial D$. By the argument principle, its number of zeros is an integer given by a contour integral varying continuously with $t$, hence remains $m$. At $t=1$ none of these zeros is $z_0$, so their derivatives are nonzero and they are simple zeros. Thus there are $m$ distinct points taking the value $f(z_0)+w$. Injectivity forces $m=1$, proving $f^{\prime }(z_0)\ne 0$ everywhere in $U$.

Take the wave speed $c>0$ and introduce characteristic coordinates $\xi =x+ct$, $\eta =x-ct$. The chain rule givesThe wave equation therefore says $y_{\xi \eta }=0$. Integrating first with respect to $\eta$ gives $y_{\xi }=F(\xi )$, and integrating with respect to $\xi$ gives $y=f(x+ct)+g(x-ct)$ with twice differentiable $f,g$. Conversely substitution shows that every such sum solves the wave equation. The two terms are travelling waves of unchanged shape, moving respectively to the left and right at speed $c$.

The zero initial displacement gives $g(x)=-f(x)$, and the initial velocity gives $2c{f}^{\prime }(x)=\psi (x)$. Hence the D'Alembert formula with initial velocity isFor a classical twice differentiable solution one may take $\psi \in C^1$. Differentiation gives $y_t(x,0)=\psi (x)$ and $y(x,0)=0$ directly. The coordinate calculation also works for any nonzero signed $c$, with physical speed $|c|$. If the printed constant is allowed to be $c=0$, the equation instead gives $y=A(x)+tB(x)$ and these initial data give $y=t\psi (x)$; the displayed travelling-wave representation requires $c\ne 0$.

For the wavefunction of a particle of mass $m$, the probability density and probability current areFor a real potential $V$, the Time-dependent Schrödinger equation and its complex conjugate readMultiplying by ${\Psi }^{\ast }$ and $\Psi$ respectively yieldsThe potential terms cancel because $V$ is real. This is the continuity equation, expressing local conservation of probability.

For the given superposition the common time phase cancels in ${\Psi }^{\ast }{\Psi }_x$, andThe expression in the second parentheses is purely imaginary, so the second term is real and contributes nothing to the probability current. ThusIt is the incoming probability current minus the reflected probability current.

For a moving circuit, Faraday's law takes the formHere $C(t)$ is the instantaneous wire contour, $\mathbf{v}$ its local material velocity, and $S(t)$ any spanning surface. The contour orientation and surface normal vector obey the right-hand convention. The integral on the right is the magnetic flux, differentiated with both the field and circuit position allowed to vary. The electric field term is the electric force per unit electric charge around the wire, and $\mathbf{v}\times \mathbf{B}$ supplies the motional electromotive force. Their sum is the electromotive force driving charge around the moving circuit. The condition $\nabla \cdot \mathbf{B}=0$ makes the magnetic flux independent of the spanning surface.

Choose the initial loop normal vector to be $+{\mathbf{e}}_y$, so rotation about $+{\mathbf{e}}_z$ gives $\mathbf{n}(t)=(-\sin \Omega t,\cos \Omega t,0)$. The spatially uniform magnetic field givesUsing Ohm's law for the resistive loop,Positive electric current is measured in the contour orientation associated with $\mathbf{n}(t)$. Reversing that convention reverses the signed answer. When $\Omega =\omega$, the magnetic flux is constant and the induced electric current vanishes. This uses the usual thin resistive-circuit approximation with self-inductance neglected.

Let $L(f)=f^{\prime }(0)-\mu (f)$. Direct evaluation on $1,x,x^2$ gives $L=0$, so the Peano kernel theorem givesThis representation also follows by applying $L$ to the Taylor expansion of $f$ with integral remainder. The derivative at zero of the truncated quadratic is zero for $t\ge 0$, and its sampled values giveThe Peano kernel is nonnegative, andHence $|L(f)|\le \frac{1}{3}\Vert f^{(3)}{\Vert }_{\infty }$. For $f(x)=x^3/6$, the third derivative is $1$, while $f^{\prime }(0)=0$ and $\mu (f)=-1/3$. Thus equality holds, proving the least constant is $c=1/3$, attained by $f(x)=x^3/6$.

For a time-homogeneous Markov chain with transition matrix $P$, two states are communicating states when each can be reached from the other with positive probability in finitely many steps:The convention $P^0=I$ includes communication of a state with itself.

Let $T_i^{+}=\inf \{n\ge 1:X_n=i\}$ be the first positive return time. A recurrent state is one for whichThe subscript means that the Markov chain starts at $i$. The positive-time condition distinguishes a genuine return from the automatic visit at time zero.

Communication is reflexive because $(P^0{)}_{ii}=1$, and symmetric by its definition. If $i\leftrightarrow j$ and $j\leftrightarrow k$, choose positive-probability paths $i\to j$ in $m$ steps and $j\to k$ in $n$ steps. The Chapman-Kolmogorov equation givesCombining the reverse paths gives a positive-probability path $k\to i$ as well. Thus communication is transitive and is an equivalence relation.

To transfer recurrence, first establish the recurrence criterion by return probabilities. Put $r_i={\mathbb{P}}_i(T_i^{+}<\infty )$ and let $N_i={\sum }_{n\ge 0}{1}_{\{X_n=i\}}$ under ${\mathbb{P}}_i$. After each return the Strong Markov property restarts the chain at $i$, so ${\mathbb{P}}_i(N_i\ge k+1)=r_i^k$. ThereforeFor $r_i<1$ this equals $(1-r_i{)}^{-1}$; for $r_i=1$ it is infinite. The expectation identity follows by summing the nonnegative indicators, so remains valid when the expectation is infinite. Consequently $i$ is a recurrent state exactly when its return-probability series diverges.

If $i\leftrightarrow j$, choose $a,b$ with $(P^a{)}_{ji}>0$ and $(P^b{)}_{ij}>0$. For every $n\ge 0$,Summing over $n$, recurrence of $i$ forces the return-probability series for $j$ to diverge. The proved criterion shows $j$ is recurrent, so recurrence is a property of the communicating class.

The Cayley-Hamilton theorem states that every square complex matrix satisfies its own characteristic polynomial: if ${\chi }_A(t)=\det (tI-A)=t^n+c_{n-1}{t}^{n-1}+\cdots +c_0$, thenFor a proof, expand the adjugate matrix as $\mathrm{adj}(tI-A)={\sum }_{k=0}^{n-1}{B}_k{t}^k$. The adjugate identity givesComparing coefficients yields $-A{B}_0=c_{0}I$, $B_{k-1}-A{B}_k=c_{k}I$ for $1\le k\le n-1$, and $B_{n-1}=I$. Multiply the middle identities on the left by $A^k$, include the first identity, and add $A^n{B}_{n-1}=A^n$. The terms involving the $B_k$ cancel consecutively:Their right sides sum to $c_{0}I+{\sum }_{k=1}^{n-1}{c}_k{A}^k+A^n$. This proves the theorem without any substitution of noncommuting matrices into a scalar polynomial identity.

For a $2\times 2$ matrix, the characteristic polynomial is $t^2-\mathrm{tr}(A)t+\det A$. The given matrix trace and determinant therefore make it $t^2+1$. By the Cayley-Hamilton theorem, $A^2=-I$, and hence $A^4=I$. Since $A^2\ne I$, the order of a group element $A$ cannot be $1$ or $2$. If $A^3=I$, then $A=A^4(A^3{)}^{-1}=I$, also impossible. Thus the order is $4$.

First, the orbit-stabilizer theorem applied to a face gives $|G|=20\cdot 3=60$: its stabilizer subgroup consists of the identity and the two rotations through $120^{\circ }$ about the axis through the centres of that face and its opposite. A rotation preserving a face must preserve its outward normal and cyclically permute its three vertices, so there are no further stabilizing rotations.

The conjugacy classes are as follows:

There are $30/2=15$ edge axes, $20/2=10$ face axes, and $12/2=6$ vertex axes. They give respectively $15$, $2\cdot 10=20$, and $4\cdot 6=24$ distinct nonidentity rotations, already accounting for all $59$ nonidentity elements.

To justify the class assertions, conjugation by a rotation carries the rotation axis to its image and preserves the rotation angle about the transported oriented axis. The transitive group action on edges puts all half-turns in one conjugacy class. The transitive group action on faces puts the $120^{\circ }$ rotations defined using outward face normals in one conjugacy class; opposite faces encode opposite signs about the same unoriented axis. The same reasoning with outward vertex directions gives a class of twelve $72^{\circ }$ rotations and a class of twelve $144^{\circ }$ rotations, including their inverses. These last classes cannot merge: the matrix trace of a three-dimensional rotation is $1+2\cos \theta$, which differs for $\theta =2\pi /5$ and $4\pi /5$. Thus the list is complete.

The face action gives $|G|=60$, as calculated above. Every normal subgroup contains the identity and is a union of whole conjugacy classes. Thus its possible order is $1$ plus a subset sum of $15,20,12,12$. The distinct resulting numbers areBy Lagrange's theorem for finite groups, a subgroup order divides $60$. Only $1$ and $60$ in this list do so. Consequently the normal subgroups are $\{1\}$ and $G$, and the rotational symmetry group of a regular icosahedron is a simple group.

In the Poincaré half-plane model, $\mathbb{H}=\{x+iy:y>0\}$ has Riemannian metricIts orientation-preserving isometries are the Möbius transformations $z\mapsto (az+b)/(cz+d)$ with real coefficients and $ad-bc=1$, with a simultaneous sign change representing the same map. The full isometry group also contains their compositions with $z\mapsto -\overline{z}$, which reverse orientation. The hyperbolic lines are vertical Euclidean lines and upper semicircles centred on the real axis; each meets the boundary orthogonally.

A hyperbolic line with finite endpoints $p<q$ is mapped to the imaginary axis byIts coefficient determinant is $q-p>0$, so rescaling the coefficients makes this an orientation-preserving isometry. The endpoints map to $0$ and $\infty$, which characterize the imaginary-axis hyperbolic line. For a vertical line with endpoints $p,\infty$, use $T(z)=z-p$. If $T_1,T_2$ send the two prescribed lines to the imaginary axis, $T_2^{-1}{T}_1$ sends the first to the second. This proves transitivity on hyperbolic lines.

For two ultraparallel hyperbolic lines, choose their common perpendicular and map it to the imaginary axis. A hyperbolic line perpendicular to that axis is a semicircle centred at zero: orthogonality at $iy$ forces the circle centre to have real coordinate zero. Thus the reflecting lines become semicircles of distinct radii $r_1,r_2>0$. The hyperbolic reflection in the radius-$r$ semicircle is $S_r(z)=r^2/\overline{z}$; it fixes that semicircle pointwise and reverses orientation. Their composition isIts $k$th power is $z\mapsto {\lambda }^{k}z$, and no positive power is the identity since $\lambda >0$ and $\lambda \ne 1$. Conjugation by the chosen isometry preserves order of a group element, so the original composition has infinite order. Reversing the order of the reflections replaces $\lambda$ by ${\lambda }^{-1}$ and gives the same conclusion.

The contraction mapping theorem says that if $(X,d)$ is a nonempty complete metric space and $T:X\to X$ satisfies $d(Tx,Ty)\le qd(x,y)$ for a constant $0\le q<1$, then $T$ has exactly one fixed point.

For the proof, start at $x_0\in X$ and set $x_n=T^n{x}_0$. Iterating the contraction bound gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$. The triangle inequality then gives, for $m>n$,Thus $(x_n)$ is a Cauchy sequence and has a limit $x_{\ast }$ by completeness. Since the contraction is continuous, $T{x}_{\ast }=\lim T{x}_n=\lim x_{n+1}=x_{\ast }$. If $y_{\ast }$ is another fixed point, then $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing distance zero. This proves existence and uniqueness.

For the integral map, the fundamental theorem of calculus shows that $Tf$ is continuous whenever $f$ is continuous. In the supremum norm,Hence $\Vert Tf-Tg{\Vert }_{\infty }\le \frac{3}{2}\max (a^2,b^2)\Vert f-g{\Vert }_{\infty }$. We can chooseThe space $C[a,b]$ with this supremum norm is complete: a uniformly Cauchy sequence has pointwise real limits, the Cauchy bound passes to those limits to give uniform convergence, and a uniform limit of continuous functions is continuous. The contraction mapping theorem therefore gives a unique continuous $y$ with $y(x)=1+{\int }_0^{x}3ty(t)dt$. The fundamental theorem of calculus makes this $y$ differentiable, with $y^{\prime }=3xy$ and $y(0)=1$. Conversely every solution of that initial-value problem obeys the integral equation, so uniqueness transfers to the ordinary differential equation. Explicitly $y(x)=e^{3x^2/2}$ on $[-1/2,1/2]$, as direct differentiation confirms.

The closure $\mathrm{cl}(A)$ is the intersection of all closed subsets containing $A$; equivalently it consists of the points whose every open neighbourhood meets $A$.

Suppose $f$ is continuous and $x\in \mathrm{cl}(A)$. For any open neighbourhood $V$ of $f(x)$, continuity makes $f^{-1}(V)$ an open neighbourhood of $x$, so it meets $A$. Hence $V$ meets $f(A)$, showing $f(x)\in \mathrm{cl}(f(A))$. This proves $f(\mathrm{cl}(A))\subseteq \mathrm{cl}(f(A))$.

Conversely suppose this inclusion holds for every subset $A$. For a closed subset $C\subseteq Y$, let $A=f^{-1}(C)$. Then $f(A)\subseteq C$ and therefore $\mathrm{cl}(f(A))\subseteq C$. By the assumed inclusion, $f(\mathrm{cl}(A))\subseteq C$, so $\mathrm{cl}(A)\subseteq f^{-1}(C)=A$. Since $A\subseteq \mathrm{cl}(A)$ always, $A$ is closed. Thus every closed set has closed preimage, equivalently every open set has open preimage. This is continuity, proving the claimed equivalence.

For each integer $n\ge 1$, the open balls $\{B(x,1/n):x\in M\}$ cover the compact metric space $M$. Compactness supplies a finite subcover with centres forming a finite set $S_n$. Define $S={\bigcup }_{n\ge 1}{S}_n$, a countable union of finite sets and therefore countable.

For any $x\in M$ and $\epsilon >0$, choose $n$ with $1/n<\epsilon$. One of the covering balls contains $x$, so there is $s\in S_n\subseteq S$ with $d(x,s)<1/n<\epsilon$. Every open neighbourhood of $x$ therefore meets $S$, giving $x\in \mathrm{cl}(S)$. Thus $\mathrm{cl}(S)=M$, and $S$ is a countable dense subset. If $M$ is empty, simply take $S=\varnothing$.

For an integer $n\ge 0$ and $\mathrm{Re}s>0$, the definition of the Laplace transform gives $F_0(s)={\int }_0^{\infty }{e}^{-st}dt=1/s$. For $n\ge 1$, integration by parts givesThe endpoint term vanishes because exponential decay dominates every fixed power of $t$. Induction therefore gives $F_n(s)=n!/s^{n+1}$. Multiplication by $e^{at}$ replaces $e^{-st}$ by $e^{-(s-a)t}$ in the defining integral, soThe second formula holds for $\mathrm{Re}(s-a)>0$ and is the Laplace transform shift theorem.

By the Bromwich inversion formula, for $t>0$ and $\gamma >0$,Close the upward vertical segment by a left semicircle of radius $R$ centred at $\gamma$. On that arc $\mathrm{Re}s\le \gamma$, so $|e^{st}|\le e^{\gamma t}$, while $|s^2(s+1{)}^2|$ is bounded below by a constant times $R^4$ for large $R$. Its length is $\pi R$, hence the arc integral tends to zero as $O(R^{-3})$. The contour is positively oriented and encloses the double poles at $0$ and $-1$.

Their residues areandThe residue theorem therefore yieldsThe value at zero follows by continuity and is $f(0)=0$.

Part (i) gives $\mathcal{L}(t)=1/s^2$ and $\mathcal{L}(t{e}^{-t})=1/(s+1{)}^2$. The convolution theorem for Laplace transforms therefore givesRepeated integration by parts yieldsSubstitution gives $f(t)=t-2+(t+2)e^{-t}$, agreeing with the contour integration result.

Write $F=\mathcal{L}f$. Using the Laplace transform of a derivative and the initial data givesThe transformed equation is $(s^4+2s^3+s^2)F=1$, so $F=1/[s^2(s+1{)}^2]$. The preceding inverse Laplace transform therefore givesFor a direct check, $f^{\prime }=1-(t+1)e^{-t}$, $f^{\prime \prime }=t{e}^{-t}$, $f^{(3)}=(1-t)e^{-t}$, and $f^{(4)}=(t-2)e^{-t}$. These satisfy both the ordinary differential equation and all four initial conditions.

For fixed endpoints, the Euler-Lagrange equation is $F_y-\frac{d}{dx}{F}_{y^{\prime }}=0$. Since $F$ has no explicit dependence on $x$, differentiating givesThus $F-y^{\prime }{F}_{y^{\prime }}=C$, the Beltrami identity.

With $r^{\prime }=dr/d\theta$, the arc length element and the prescribed speed give the travel-time functionalFor the circular road $r=a$, this is $T_{\mathrm{circle}}=\pi /(2ka)$.

For the integrand $F=\sqrt{r^{\prime 2}+r^2}/r^2$, the Beltrami identity simplifies toThe candidate $r=a(\cos \theta +\sin \theta )$ has $r^{\prime 2}+r^2=2a^2$ and the required endpoints. To prove it really minimizes the travel time, rather than merely satisfies a first integral, put $u=1/r$. ThenThis is the Euclidean arc length integrand for the inverted curve $(X,Y)=(u\cos \theta ,u\sin \theta )$, whose endpoints are $(1/a,0)$ and $(0,1/a)$. Every connecting curve has length at least their Euclidean distance $\sqrt{2}/a$, by the triangle inequality. Equality is attained by the straight line segment $X+Y=1/a$. In polar form it gives $u(\cos \theta +\sin \theta )=1/a$, hence the proposed $r$. It stays strictly positive throughout the interval. ConsequentlyThis travel-time minimization with speed proportional to squared radius is globally solved by the inversion: its minimum is strictly less than $\pi /(2ka)$.

Let the photon momenta be ${\mathbf{p}}_1,{\mathbf{p}}_2$, with $|{\mathbf{p}}_i|=E_i/c$. Four-momentum conservation gives total energy $E_1+E_2$ and total momentum ${\mathbf{p}}_1+{\mathbf{p}}_2$ for the pion. Its mass shell relation therefore givesSince $0\le \theta \le \pi$, the sine is nonnegative, and taking the positive square root yields