Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 4

A Jordan normal form of a complex $n\times n$ matrix $A$ is a similarity transformation $P^{-1}AP=J$ in which $J$ is a direct sum of Jordan blocksFor each eigenvalue, the multiset of block sizes is uniquely determined by $A$; only the order of the blocks is arbitrary. The change-of-basis matrix $P$ is generally not unique.

For the given matrix, expansion of the characteristic polynomial givesThe upper-left $2\times 2$ minor of $A-I$ is $-5$, so its rank is $2$ and the eigenspace for $1$ has dimension $1$. Thus the two occurrences of the eigenvalue $1$ belong to a single size-$2$ Jordan block. More explicitly, the Jordan chain and the remaining eigenvector can be chosen asConsequently, with $P=(v_1{v}_2{v}_3)$, for which $\det P=1$, the answer is

Write $q_n(X)=1+X+\cdots +X^{n-1}$, a monic primitive polynomial. If $n=ab$ with $a,b>1$, grouping terms gives a factorization in the polynomial ring $\mathbb{Z}[X]$:Both factors have positive degree, so $q_n$ is not an irreducible polynomial.

Conversely, if $n=p$ is a prime number, change variables by $X=Y+1$. The binomial theorem givesEvery coefficient except the leading one is divisible by $p$, because $p\mid \left(\genfrac{}{}{0ex}{}{p}{k}\right)$ for $1\le k<p$, whereas the constant coefficient $p$ is not divisible by $p^2$. The Eisenstein criterion makes this polynomial irreducible over $\mathbb{Q}$. The substitution $X\mapsto Y+1$ is an invertible ring homomorphism, so $q_p$ is also irreducible over $\mathbb{Q}$; the Gauss lemma for polynomials transfers this to $\mathbb{Z}[X]$. Hence

Both functionals are finite on $X$: a continuous function on the compact metric space $[0,1]$ is bounded and integrable. Homogeneity follows from $|af|=|a||f|$, and the triangle inequality follows by integrating, or taking the supremum of, $|f+g|\le |f|+|g|$. Each is nonnegative. If $\Vert f{\Vert }_{\infty }=0$, then $f$ vanishes everywhere. If a continuous function $f$ is nonzero at some point, continuity supplies an interval of positive length on which $|f|$ is bounded below by a positive number; hence $\Vert f{\Vert }_1>0$. Thus both are norms.

For the nonnegative $f_n$, direct integration gives the L1 normTherefore $\boxed{f_n⟶0\text{ in }\Vert \cdot {\Vert }_1}$. On the other hand, differentiation places the maximum of $t^n(1-t)$ at $t=n/(n+1)$, giving the supremum normFor $t<1$, exponential decay implies $n{t}^n(1-t)\to 0$, while $f_n(1)=0$. Thus the pointwise convergence is to zero throughout $[0,1]$. Any convergence in the supremum norm would imply uniform convergence, hence the same pointwise limit zero, contradicting the displayed positive limiting norm. Therefore there is no convergence in the supremum norm.

Set $h=f-g$, an entire function. By the Bolzano-Weierstrass theorem, a subsequence of the distinct zeros $z_i$ converges to a point $a$ in the closed unit disc. An analytic function has a convergent Taylor series in a neighbourhood of each point. If the first nonzero coefficient of $h$ at $a$ had index $m$, thenwith $k$ analytic and nonzero in a sufficiently small disc. The zero $a$ would then be isolated, contradicting the accumulating distinct zeros. All Taylor coefficients at $a$ vanish, so $h$ vanishes on a disc.

To justify continuation across the whole plane, let $Z$ be the set of points at which $h$ vanishes on a neighbourhood. This set is nonempty and open. It is also closed: if points of $Z$ tend to $b$, continuity of each derivative gives $h^{(m)}(b)=0$ for every $m$, and the Taylor series at $b$ gives a zero neighbourhood. Since $\mathbb{C}$ is a connected space, $Z=\mathbb{C}$. This proves the identity theorem here and gives $\boxed{f\equiv g}$.

Without boundedness, the distinct zeros need not have an accumulation point in the plane. For example,are distinct entire functions agreeing at all the indicated positive integers.

For the half-range Fourier cosine series on $(0,1)$, the coefficients areTwo integrations by parts, with $k=n\pi$, giveThusThe even, $2$-periodic extension is continuous and piecewise continuously differentiable, so the Fourier series converges to it, including at $x=1$. Equivalently, the summable coefficient bound gives uniform convergence and permits taking the limit from the interior. Evaluating at $x=1$, where $(-1{)}^n\cos (n\pi )=1$, yieldsThis is the Basel problem.

For a time-independent Quantum Hamiltonian, a stationary state is an energy eigenstate. The Schrödinger equation separates its time dependence asIts probability density $|\psi (x,t){|}^2=|\chi (x){|}^2$ is time independent. Linearity of the Schrödinger equation givesPut $A_{jk}={\int }_{-\infty }^{\infty }{\chi }_j^{\ast }\hat{A}{\chi }_{k}dx$. The operator for a time-independent observable is a Hermitian operator, so $A_{21}=\overline{A_{12}}$. Substituting the evolved wavefunction into the expectation value givesand therefore, for a normalized state,The normalization is automatic for orthogonal ${\chi }_1,{\chi }_2$, in particular when $E_1\ne E_2$: self-adjointness of $\hat{H}$ gives $(E_1-E_2)⟨{\chi }_1,{\chi }_2⟩=0$. If the energies coincide, normalization of the individual eigenstates alone does not ensure normalization of their displayed sum; its squared norm is $1+\mathrm{Re}⟨{\chi }_1,{\chi }_2⟩$. For a nonzero sum not having unit norm, the physical expectation value is the displayed numerator divided by this squared norm. With orthonormal energy states, the stated formula holds in both the distinct and degenerate cases.

The differential Gauss law is $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$. On a large ball $B_R$, the product rule and $\nabla \varphi =-\mathbf{E}$ giveThe divergence theorem consequently expresses the electrostatic energy asFor a localized charge distribution with the electric potential chosen to vanish at infinity, $\varphi =O(R^{-1})$ and $\mathbf{E}=O(R^{-2})$, so the surface term is $O(R^{-1})$ and tends to zero. More generally, it suffices to assume that this boundary integral vanishes. Passing to all space givesThe physical decay and potential reference are relevant: $\mathbf{E}\to 0$ by itself does not specify the additive constant in $\varphi$. Adding a constant $C$ changes the charge-based expression by $CQ/2$ when the total charge is $Q$, while leaving the field unchanged. The zero-at-infinity potential is the usual reference implicit in this all-space electrostatic energy formula.

Spherical symmetry and the integral Gauss law give $4\pi r^2{E}_r(r)=Q/{ϵ}_0$ for $r>R$. Together with the stipulated zero interior field,The electric field jumps across the ideal charged surface; its value exactly on that surface is not needed for the volume integral. The electrostatic energy of a uniformly charged spherical shell isAs a check, the shell has constant electric potential $Q/(4\pi {ϵ}_{0}R)$, with zero reference at infinity, so $U=Q\varphi (R)/2$ gives the same result.

With $x=\cos \theta$, the weighted inner product for the Chebyshev polynomials becomesThe identity $2\cos (n\theta )\cos (m\theta )=\cos ((n-m)\theta )+\cos ((n+m)\theta )$ and the vanishing of ${\int }_0^{\pi }\cos (k\theta )d\theta$ for positive integers $k$ give zero when $n\ne m$, $\pi /2$ when $n=m>0$, and $\pi$ when $n=m=0$. ThusFor Gaussian quadrature, choose the $n$ distinct nodes as the zeros of the degree-$n$ orthogonal polynomial for the weight, here $T_n$, and choose the weights as the weighted integrals of their Lagrange interpolation basis functions. If a polynomial $p$ has degree at most $2n-1$, divide it as $p=q{T}_n+r$, where both $q$ and $r$ have degree at most $n-1$. Orthogonality makes the integral of $q{T}_n$ zero, while its quadrature sum also vanishes; interpolation integrates $r$ exactly. This proves degree $2n-1$ exactness. It is optimal because the squared nodal polynomial, of degree $2n$, has zero quadrature sum and strictly positive integral.

For $n=3$, $T_3(x)=4x^3-3x$, so the nodes are $-\sqrt{3}/2,0,\sqrt{3}/2$. Their interpolation basis functions areThe weighted zeroth and second moments are $\pi$ and $\pi /2$, and the first moment is zero. Integrating these three basis functions gives equal weights $\pi /3$, hence the Chebyshev–Gauss quadrature rulewhich is exact for every polynomial of degree at most $5$.

The communicating classes are $T=\{1,2\}$, $B=\{3,4\}$ and $C=\{5,6,7\}$. The class $T$ is open: it can reach $B$ or $C$ and cannot be reached from either. To remain in $T$ for two steps has probability $1/6$, so ${\mathbb{P}}_i(X_{2k}\in T)=(1/6{)}^k$ for $i\in T$. Its states are transient states. All returns within $T$ have even length and a two-step return is possible, so their period of a state in a Markov chain is $2$.

The closed communicating class $B$ alternates deterministically between $3$ and $4$, and therefore consists of positive recurrent states of period $2$. The class $C$ is finite, closed and irreducible, so its states are also positive recurrent states. The self-loop at $7$ gives period $1$ there, and the period is constant on a communicating class, so every state of $C$ has period $1$.

For $C$, solve the stationary distribution equations:Hence $({\pi }_5,{\pi }_6,{\pi }_7)=(1/4,1/4,1/2)$. Finite irreducibility and aperiodicity imply convergence to this stationary distribution from any state of $C$. If $h_i$ is the probability of eventually entering $C$, the Markov property givesso $h_1=2/5$ and $h_2=3/10$. Conditioning on the finite entrance time and then letting the remaining time tend to infinity proves that the limiting probabilities in $C$ are $h_i{\pi }_j$. The probability of entrance after a large cutoff tends to zero, which justifies this conditioning limit.

For $B$, convergence depends on the parity balance on entry to a period-two Markov class. Entry always occurs at $3$. Starting from $1$, the probabilities of first entry at time $2k+1$ and at time $2k+2$ are both $\frac{1}{4}(1/6{)}^k$. Each parity therefore has total probability $3/10$, so both $p_{13}^{(n)}$ and $p_{14}^{(n)}$ tend to $3/10$, despite the recurrent class being periodic. Starting from $2$, the corresponding odd- and even-entry probabilities are $\frac{1}{2}(1/6{)}^k$ and $\frac{1}{12}(1/6{)}^k$, with totals $3/5$ and $1/10$. Consequentlyso neither limit exists. From $3$ or $4$, the probabilities at $3$ and $4$ alternate between zero and one; from $C$ they are identically zero. Finally every probability of being in $T$ tends to zero, from all starting states.

All $49$ requested conclusions are encoded below, with $\ast$ denoting a nonexistent limit:The class classification is $T$: transient, period $2$; $B$: recurrent, period $2$; $C$: recurrent, period $1$.

A Hermitian matrix satisfies $A=A^{†}$, where $†$ means conjugate transpose. If $Av=\lambda v$ with $v\ne 0$, then $v^{†}Av=\lambda \Vert v{\Vert }^2$ is real, since it equals its complex conjugate $v^{†}{A}^{†}v$. Thus every eigenvalue is real. To prove the finite-dimensional spectral theorem, choose a unit eigenvector $v$; one exists because the complex characteristic polynomial has a root. Its orthogonal complement is invariant, since for $w\perp v$,The restriction to that complement is again Hermitian. Induction on dimension supplies an orthonormal basis of eigenvectors, so $A=U\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)U^{†}$ for a unitary matrix $U$.

In this basis, $x^{†}Ax={\sum }_i{\lambda }_i|(U^{†}x{)}_i{|}^2$. Therefore $\boxed{A>0⟺{\lambda }_i>0\text{ for every }i}$. In that case the square root of a matrixis Hermitian, positive definite and satisfies $C^2=A$. For uniqueness, any positive-definite Hermitian $D$ satisfying $D^2=A$ commutes with $A$ and hence preserves each eigenspace of $A$. Its restriction to the $\lambda$-eigenspace is Hermitian, and its positive eigenvalues all solve ${\mu }^2=\lambda$. Thus that restriction is $\sqrt{\lambda }I$, forcing $D=C$. This is the principal square root of a positive semidefinite matrix, here strictly positive definite.

Now let $D=\sqrt{B}$ and $X=C-D$. Expanding without assuming that $C$ and $D$ commute givesThe Hermitian matrix $X$ has $X^2\ge 0$ because $v^{†}{X}^{2}v=\Vert Xv{\Vert }^2$, whereas $A-B>0$. Hence $CX+XC>0$. If $Xv=\mu v$, $v\ne 0$, thenSince $C>0$, every real eigenvalue $\mu$ of $X$ is positive. The positive-eigenvalue criterion now gives the strict finite-dimensional operator monotonicity of the square root:

An R-module $M$ is a free module if it has a basis $(e_i{)}_{i\in I}$: every $m\in M$ has a unique expression as a finite sum ${\sum }_i{r}_i{e}_i$, with $r_i\in R$. Equivalently, it is a direct sum of copies of $R$. Let $q:M\to M/N$ be the quotient map. When $M/N$ is free, choose a basis $({\overline{f}}_j)$ and lifts $f_j\in M$. The universal property of a free module defines an $R$-linear map $s:M/N\to M$ by $s({\sum }_j{r}_j{\overline{f}}_j)={\sum }_j{r}_j{f}_j$. It satisfies $q\circ s=\mathrm{id}$, so every $m\in M$ decomposes uniquely asThis gives a split short exact sequence and $M\cong N\oplus (M/N)$. If $N$ is also free, the union of its basis and the lifted quotient basis is a basis of $M$: spanning follows from the decomposition, and applying $q$ to a putative relation kills all quotient coefficients before independence in $N$ kills the remaining ones. Thus free submodule and free quotient imply a free module.

For the first claim, take $R=\mathbb{Z}$ andHere $N\cong N^{\prime }\cong \mathbb{Z}$ are free modules and $M/N\cong M^{\prime }/N^{\prime }\cong \mathbb{Z}/2\mathbb{Z}$. But $M$ is a torsion-free module, whereas $M^{\prime }$ contains the nonzero torsion element $(0,\overline{1})$. Therefore claim (1) is false.

For the second claim, the same lifting construction works without assuming that $N$ is free. Since both isomorphic quotient modules are free, it yieldsThus claim (2) is true. Equivalently, this uses the fact that free modules are projective.

For a parametrized curve on the surface, the energy of a curve iswhere the first fundamental form coefficients are evaluated at $(u(t),v(t))$. A fixed-endpoint variation of a curve has coordinate variation fields $\eta ,\xi$ vanishing at $a,b$. Differentiating under the integral and integrating their derivative terms by parts givesThe fields can be chosen independently with compact support in $(a,b)$. Stationarity and the fundamental lemma of the calculus of variations therefore force both coefficients to vanish. These Euler-Lagrange equations are the geodesic equations:

The coordinate curves are assumed to be geodesics as geometric curves; use their arc-length parametrization in the geodesic equations. On $u=c$, choose increasing arc length $s$, so $\dot{u}=0$ and $\dot{v}=1/\sqrt{G}$. The transverse equation becomesThus $(F/\sqrt{G}{)}_v=G_u/(2\sqrt{G})=(\sqrt{G}{)}_u$. On $v=d$, similarly $\dot{v}=0$, $\dot{u}=1/\sqrt{E}$, and the transverse equation gives $(F/\sqrt{E}{)}_u=E_v/(2\sqrt{E})=(\sqrt{E}{)}_v$. Hence the geodesic coordinate grid identities are

The angle $\theta$ between the oriented coordinate tangent vectors satisfies $\cos \theta =F/\sqrt{EG}$. The constant-angle assumption therefore gives $F=k\sqrt{E}\sqrt{G}$ for a constant $k$. The first fundamental form is positive definite, so $EG-F^2>0$ and hence $|k|<1$. Substitution into the geodesic coordinate grid identities from part (i) givesEliminating either derivative yields $(1-k^2)(\sqrt{E}{)}_v=0$ and $(1-k^2)(\sqrt{G}{)}_u=0$. Since $E,G>0$ and $1-k^2>0$, this provesIf the angle is understood between unoriented curves, its constancy initially fixes $|F|/\sqrt{EG}$; continuity fixes the sign on each connected chart when this quantity is nonzero, and the zero case already has $F=0$. Thus the same conclusion holds. In fact constant-angle geodesic coordinate grids are locally flat, although only the displayed derivatives are needed here.

Two norms $\Vert \cdot {\Vert }_a,\Vert \cdot {\Vert }_b$ are Lipschitz equivalent norms when constants $c,C>0$ satisfy $c\Vert x{\Vert }_a\le \Vert x{\Vert }_b\le C\Vert x{\Vert }_a$ for every vector $x$. Applying these inequalities to $x_j-x_k$ shows that a sequence is a Cauchy sequence for one norm exactly when it is so for the other; applying them to $x_j-x$ likewise equates convergence. If one norm is complete, a Cauchy sequence for the other converges using the first norm and hence also the second. Interchanging the norms proves completeness is preserved in both directions.

Write $x={\sum }_i{x}_i{e}_i$ and set $C={\sum }_i\Vert e_i\Vert$. If $\Vert x{\Vert }_2=1$, then $|x_i|\le 1$, and the triangle inequality givesScaling a nonzero $x$ gives $\Vert x\Vert \le C\Vert x{\Vert }_2$ for every $x$. The reverse triangle inequality now impliesso $x\mapsto \Vert x\Vert$ is continuous, indeed Lipschitz, in the Euclidean norm.

Suppose there were no positive lower bound on the Euclidean unit sphere. We could choose $x_j$ with $\Vert x_j{\Vert }_2=1$ but $\Vert x_j\Vert <1/j$. The Bolzano-Weierstrass theorem supplies a subsequence converging in the Euclidean norm to $x$. Continuity of both norms gives $\Vert x{\Vert }_2=1$ and $\Vert x\Vert =0$, contradicting definiteness. Therefore some $m>0$ satisfies $\Vert x\Vert \ge m$ on the unit sphere. Scaling again proves the finite-dimensional equivalence of norms:

A basis of a topology is a family $\mathcal{B}$ of open sets such that every open set is a union of members of $\mathcal{B}$. Equivalently, whenever $x\in U$ and $U$ is open, there is $B\in \mathcal{B}$ with $x\in B\subseteq U$. For a metric space, all open balls form a basis. For the product topology on $X\times Y$, the rectangles $U\times V$ with $U,V$ open in their respective factors form a basis; rectangles made from factor bases also suffice.

If $C\subseteq X$ and $D\subseteq Y$ are countable dense subsets, then $C\times D$ is countable. Every nonempty basic rectangle meets it, since $U\cap C$ and $V\cap D$ are nonempty. Every nonempty open set contains such a rectangle, proving that a product of two separable spaces is separable.

Suppose a metric space has a countable dense set $C$. The balls $B(c,r)$ with $c\in C$ and positive rational $r$ form a countable basis. Indeed, given $x\in U$ open, choose $ϵ>0$ with $B(x,ϵ)\subseteq U$, then choose $c\in C$ with $d(c,x)<ϵ/3$ and rational $r$ satisfyingThe triangle inequality gives $x\in B(c,r)\subseteq B(x,ϵ)\subseteq U$. Conversely, from any countable basis select one point of each nonempty member. These points form a countable dense subset, because every nonempty open set contains a basis member. This direction holds for arbitrary topological spaces. Thus a metric space is separable exactly when it is a second-countable space. Intersecting a countable basis with a subspace gives a countable basis for its subspace topology; selecting points as above shows that every subspace of a separable metric space is separable.

For the Sorgenfrey line, $\mathbb{Q}$ meets every nonempty interval $[a,b)$, so it is a countable dense subset. In the Sorgenfrey plane, however, for any $x\in \mathbb{R}$ and $ϵ>0$,Indeed, $(t,-t)$ in that rectangle requires both $t\ge x$ and $t\le x$. Every singleton is therefore open in the subspace topology on $Y$, which is the discrete topology. A dense subset of a discrete space must contain every point. Since $Y$ is in bijection with the uncountable set $\mathbb{R}$, it has no countable dense subset. Hence $X$ is separable and $Y$ is not separable, illustrating why the metric assumption in the inheritance statement matters.

Choose the principal branch of the complex logarithm on $H$, with $-\pi <\arg z<\pi$. ThenThus $-\pi <\mathrm{Re}w<\pi$ and the imaginary part ranges over all real numbers. Conversely, for $w=x+iy$ in the strip, $z=e^{iw}=e^{-y}{e}^{ix}$ is in $H$ and its principal logarithm is $iw$. Therefore $\boxed{w=-i\log z:H⟶S}$ is an analytic bijection, with analytic inverse $z=e^{iw}$. Every other continuous branch on the connected domain is $\log z+2\pi ik$ for some fixed integer $k$, so its image is the translated strip $\boxed{S+2\pi k}$.

For $G$, introduce the Möbius transformation $\zeta =(z-1)/(z+1)$, whose inverse is $z=(1+\zeta )/(1-\zeta )$. Writing $\zeta =a+ib$, its defining circle inequalities transform as follows:andHence $G$ maps bijectively onto the wedge $\pi /2<\arg \zeta <3\pi /4$. The principal branch of the complex logarithm is analytic throughout this wedge. Multiplying its logarithm by $-8i$ expands the interval of real parts to $(4\pi ,6\pi )$, and subtracting $5\pi$ centers it at zero. The required conformal map is thereforeIts real part is $8\arg \zeta -5\pi \in (-\pi ,\pi )$, and its imaginary part $-8\log |\zeta |$ takes every real value. More explicitly, its inverse on $S$ is obtained from $\zeta =\exp (i(W+5\pi )/8)$ and $z=(1+\zeta )/(1-\zeta )$, which verifies both surjectivity and injectivity.

Use the Fourier transform convention $\tilde{F}(\omega )={\int }_{\mathbb{R}}F(x)e^{-i\omega x}dx$. Reindexing the convergent periodization of an integrable function gives $g(\tau +2\pi )=g(\tau )$; absolute integrability also ensures that the defining sum is absolutely convergent almost everywhere in a period. To calculate its Fourier series coefficients, uniform convergence permits passing the limit of the partial sums through the finite-interval integral. Changing variables $x=\tau +2\pi k$ and using $e^{2\pi ink}=1$ givesThe sum of these integrals is absolutely convergent, bounded by ${\int }_{\mathbb{R}}|F|$, so the last rearrangement is justified. This proves the proposed Fourier expansion at the level of coefficients.

There is a genuine missing hypothesis in the pointwise assertion as printed. Take $F(x)=1_{\{0\}}(x)$. It is absolutely integrable with integral zero, and its periodization has only finitely many nonzero terms on $[0,2\pi ]$, so the series converges uniformly there. Yet $g(0)=1$ while $\tilde{F}(n)=0$ for every $n$. Thus the displayed pointwise identity does not follow from the stated assumptions. Additional regularity ensuring Fourier convergence to the prescribed values of $g$, for example a continuous piecewise continuously differentiable periodization, makes the coefficient identity into the intended Poisson summation formula.

The particular function $F(x)=e^{-|x|}$ satisfies that stronger condition. Direct integration yieldsOn $0\le \tau \le 2\pi$, summing the two geometric tails gives $g(\tau )=\cosh (\tau -\pi )/\sinh \pi$, a continuous periodic function which is piecewise smooth. Its Fourier series has absolutely summable coefficients and converges to $g$. At $\tau =0$,Consequently

If the primed frame moves at speed $v$ in the positive $x$ direction relative to the unprimed frame, the one-dimensional Lorentz transformation isFor a trajectory with velocity $q=dx/dt$, differentiating these expressions gives $q^{\prime }=(q-v)/(1-vq/c^2)$. Solving for $q$, when the particle has primed-frame speed $u$, proves the relativistic velocity-addition formulaFor the train problem the sign of the frame velocity reverses: the train frame $S$ moves at $+v$ relative to the station frame $S^{\prime }$. Put $\gamma =(1-v^2/c^2{)}^{-1/2}$. The transformation appropriate to the subsequent parts is therefore