Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 2

Each endomorphism has three distinct eigenvalues in a three-dimensional complex vector space. Choose a nonzero eigenvector for each eigenvalue; these vectors are linearly independent and hence form an eigenbasis. Both operators are therefore diagonalizable with diagonal matrix $D=\mathrm{diag}(1,2,3)$.

Their characteristic polynomials areIf a polynomial $p$ annihilates either operator, applying it to each eigenvector gives $p(1)=p(2)=p(3)=0$. Thus its minimal polynomial must contain all three factors. Conversely, their product annihilates the diagonal matrix, soFinally, send an ordered eigenbasis of $S$ to the correspondingly ordered eigenbasis of $T$. The resulting invertible map $C$ satisfies $CS=TC$, and hence $T=CS{C}^{-1}$: the two endomorphisms are conjugate.

A Euclidean domain is an integral domain $R$ with a function $\delta :R\setminus \{0\}\to {\mathbb{Z}}_{\ge 0}$ such that, for every $a\in R$ and nonzero $b\in R$, there are $q,r\in R$ satisfying $a=bq+r$, with either $r=0$ or $\delta (r)<\delta (b)$.

For the integers, take $\delta (b)=|b|$. Given $a$ and $b\ne 0$, put $m=|b|$ and choose the integer $q_0=\lfloor a/m\rfloor$. Then $a=m{q}_0+r$ with $0\le r<m$. Setting $q=q_0$ when $b>0$ and $q=-q_0$ when $b<0$ gives $a=bq+r$. A nonzero remainder has $\delta (r)=r<|b|=\delta (b)$, and the integers are an integral domain. Thus $\mathbb{Z}$ is a Euclidean domain with Euclidean function $|b|$.

Write $v_2(m)$ for the 2-adic valuation of a nonzero integer $m$. Then $d(a,b)=2^{-v_2(a-b)}$ for unequal arguments. It is positive for $a\ne b$, vanishes exactly when $a=b$, and is symmetric because $v_2(-m)=v_2(m)$.

For the triangle inequality, if $2^r$ divides $a-b$ and $2^s$ divides $b-c$, then $2^{\min (r,s)}$ divides their sum $a-c$. ThusCoincident arguments also satisfy these inequalities immediately. Consequently $d$ is a metric, and indeed an ultrametric induced by the 2-adic absolute value.

For the sequence, $x_n-(-1)=2^n$, soThe same sequence diverges in the usual Euclidean metric; the convergence here depends on the 2-adic absolute value.

True. Let $\{V_{\alpha }\}$ be an open cover of $f(S)$ in its subspace topology. Write $V_{\alpha }=f(S)\cap W_{\alpha }$ with $W_{\alpha }$ open in $Y$. Since $f$ is continuous, the sets $S\cap f^{-1}(W_{\alpha })$ are open in $S$ and cover $S$. By compactness, finitely many cover $S$. Their corresponding $V_{\alpha }$ then cover $f(S)$, proving that the continuous image of a compact space is compact.

False. Give $X=\mathbb{R}$ its usual topology, let $Y$ be a one-point space, and take the constant continuous map $f:X\to Y$. The set $K=Y$ is compact, but $f^{-1}(K)=\mathbb{R}$ is not compact. For example, the open cover $\{(-n,n):n\ge 1\}$ has no finite subcover. Thus continuity alone does not preserve compactness under inverse images.

False. The real line with its usual metric is complete and is homeomorphic to $(0,1)$ with its usual metric, throughIts inverse is $h^{-1}(u)=\tan (\pi (u-1/2))$, so both maps are continuous. However, $(1/n{)}_{n\ge 2}$ is a Cauchy sequence in $(0,1)$ and has no limit there: its only possible real limit is zero. Hence $(0,1)$ is not a complete metric space. Completeness is a property of the metric, not just of its topology.

Use the mixed Dirichlet-Neumann modes on an interval. To justify their Fourier series, extend $y$ evenly across $x=1$ to $[0,2]$, then oddly across $x=0$, and repeat with period four. The condition $y(0)=0$ makes the odd extension continuous, and $y^{\prime }(1)=0$ makes the even reflection continuously differentiable. The resulting periodic function has a convergent Fourier sine series. Reflection about $x=1$ eliminates the even sine harmonics, leavingAlternatively these values follow directly from $\sin {\lambda }_{n}0=0$ and ${\lambda }_n\cos {\lambda }_n=0$. Fourier orthogonality givesThe extension argument also supplies completeness, not merely orthogonality, of this Fourier sine basis.

Expand the forcing as $x\cos \pi x=\sum b_n\sin {\lambda }_{n}x$. Product-to-sum and integration by parts giveEach mode of the differential operator has eigenvalue $-({\lambda }_n^2+{\alpha }^2)$. Hence, for real $\alpha$, the solution isThe coefficients are $O(n^{-4})$, so the series and its first two differentiated series converge uniformly. Therefore it solves the equation and the boundary conditions term by term. Uniqueness follows by multiplying the homogeneous equation by its real solution and integrating: the boundary terms vanish and ${\int }_0^1[(y^{\prime }{)}^2+{\alpha }^2{y}^2],dx=0$, forcing $y=0$.

In a steady state, Maxwell's equations imply $\nabla \times \mathbf{E}=0$, since the magnetic field is time independent. The divergence of the steady Ampère equation gives $\nabla \cdot \mathbf{j}=0$. With uniform electrical conductivity, Ohm's law is $\mathbf{j}=\sigma \mathbf{E}$, soFor the usual straight-wire model, the cylindrical side is insulating and the two flat end faces are equipotentials. Put $\mathbf{E}=-\nabla V$. The electric potential solves Laplace's equation, with constant end values and zero normal derivative on the side. The linear axial potential $V(z)=V(0)-\Delta Vz/l$ satisfies all these conditions. It is unique: the difference $h$ of two solutions has zero end values and zero side normal derivative, and integration by parts gives $\int |\nabla h{|}^{2}dV=0$. ThusThis proves uniform current in a straight homogeneous wire with the usual contact conditions. Integrating the electric current density over a cross-section and using Ohm's law then givesThe second expression is the rate of Ohmic heating.

Strictly, steady current and uniform conductivity alone do not force uniformity; the contact conditions are implicit in the wire model. For example, let $\chi$ be a nonconstant transverse Neumann eigenfunction on the disk, satisfying ${\Delta }_{\perp }\chi =-{\lambda }^2\chi$ and zero normal derivative at its edge. Then $V=-E_{0}z+C\chi \cosh (\lambda (z-l/2))$ is harmonic and preserves an insulating cylindrical side, but its nonconstant end potentials produce nonuniform current. Its added axial current has zero cross-sectional mean. This supplies a counterexample if arbitrary contacts are permitted.

A bend generally redistributes the electric current density: the inner side of the bend carries more current than the outer side, as illustrated by $j_{\theta }\propto 1/r$ in an annular conductor. Exact resistance and local Ohmic heating can therefore change. For a slender, gently curved wire the straight-wire formulas remain the leading approximation; the identity $\dot{Q}=I^{2}R$ remains valid for the total steady dissipation.

With coincident origins and aligned axes, the Lorentz transformation isIts inverse replaces $v$ by $-v$. Along the worldline of the origin of $S^{\prime \prime }$, $d{x}^{\prime }/d{t}^{\prime }=u$. Differentiating the inverse Lorentz transformation givesTaking the ratio yields the relativistic velocity-addition formulaFor $|u|,|v|\ll c$, the denominator tends to one, recovering $V\simeq u+v$. Simultaneously ${\gamma }_v\to 1$, so $x^{\prime }\simeq x-vt$ and $t^{\prime }\simeq t$: these are the Galilean transformation and its Newtonian velocity rule.

The free surface is a material surface. For $F(x,y,t)=y-\eta (x,t)$, the material derivative vanishes on $F=0$. Since the velocity is $\nabla \varphi$, this gives the exact kinematic boundary condition for a free-surface graphAt the surface the pressure equals the constant atmospheric pressure. Absorbing that constant into the time-dependent potential gauge in Unsteady Bernoulli equation yields the exact dynamic boundary condition for an inviscid interfaceThere is no surface-tension term in this model.

Write $\varphi =Ux+\phi$ and linearize in the small perturbation $\phi$ and the surface displacement. The base value is $F=U^2/2$, and evaluation at the displaced surface can be replaced by evaluation at zero to first order. ThusIncompressibility requires Laplace's equation ${\phi }_{xx}+{\phi }_{yy}=0$. For $k>0$, the complex perturbation $b{e}^{ky}{e}^{i(kx-\omega t)}$ satisfies this equation and decays with depth. Put $\eta =\mathrm{Re}\{a{e}^{i(kx-\omega t)}\}$ and $\Omega =\omega -kU$. The linear boundary conditions becomeThey are consistent precisely whenConsequently the two laboratory phase velocities of a deep-water gravity wave areIf propagation of a wave packet is intended, its group velocities are instead $d\omega /dk=U\pm \frac{1}{2}\sqrt{g/k}$.

The row player maximizes the payoff and the column player minimizes it. Row three is weakly dominated by row one, so the row player can discard it. On the remaining two rows, column two is weakly dominated by column three, so the column player can discard column two. The reduced zero-sum game has payoff matrixIf the first row is chosen with probability $r$, the expected payoffs against its two columns are $3-r$ and $2+2r$. Equalizing them gives $r=1/3$ and value $8/3$. If the first column is chosen with probability $s$, the payoffs of the two rows are $4-2s$ and $2+s$. Equalizing them gives $s=2/3$.

Restoring the original indexing, the optimal mixed strategies and value of a zero-sum game areTo verify these against every original strategy, the row mixture gives the three column payoffs $(8/3,3,8/3)$, each at least $v$. The column mixture gives the three row payoffs $(8/3,8/3,7/3)$, each at most $v$. Thus each mixed strategy guarantees the stated value, including against the discarded strategies. The strict inequalities for column two and row three also force those probabilities to be zero in any optimal strategy, and the remaining equalities determine the displayed mixtures uniquely.

The integral is linear in its first argument and conjugate-linear in its second, and complex conjugation of the integrand shows $⟨g,f⟩=\overline{⟨f,g⟩}$. Thus it is a Hermitian form.

If $f(z)={\sum }_{j=0}^N{a}_j{z}^j$, Fourier orthogonality givesThis is nonnegative and vanishes only when every coefficient vanishes, that is, only for the zero polynomial. Hence the Hermitian form is positive definite.

For nonnegative integers $m,n$, Fourier orthogonality givesEvery polynomial is a finite linear combination of these monomials, and distinct monomials are linearly independent. ThereforeThis is an algebraic basis: no infinite linear combinations are needed in the polynomial vector space.

For $d$ variables, integrate over the product of $d$ unit circles:For multi-indices $\alpha ,\beta \in {\mathbb{Z}}_{\ge 0}^d$, applying one-variable Fourier orthogonality in each variable givesIf $f={\sum }_{\alpha }{a}_{\alpha }{z}^{\alpha }$, its squared norm is ${\sum }_{\alpha }|a_{\alpha }{|}^2$, a finite sum. The form is therefore a positive definite Hermitian form andThis is the monomial orthonormal basis on the unit torus.

A field, for example $\mathbb{Q}$, is a Noetherian ring: its only ideals are $(0)$ and the whole field, both finitely generated.

For a non-Noetherian example, take $R=\mathbb{Q}[x_1,x_2,\dots ]$, the polynomial ring in countably many variables, each individual polynomial using only finitely many variables. The ascending chain of idealsis strictly increasing. Indeed, if $x_{n+1}$ belonged to $(x_1,\dots ,x_n)$, setting the first $n$ variables to zero would give the impossible equality $x_{n+1}=0$. The chain never stabilizes, contradicting the ascending chain condition. Thus $\mathbb{Q}$ is Noetherian, while $\mathbb{Q}[x_1,x_2,\dots ]$ is not.

Hilbert's basis theorem: if $R$ is a commutative Noetherian ring with identity, then the polynomial ring $R[x]$ is Noetherian. Iteration gives the same assertion for every finite number of variables.

Here is a bounded-degree coefficient proof of Hilbert basis theorem. Let $I$ be any ideal of $R[x]$. For $n\ge 0$, define $J_n\subseteq R$ to be the coefficients of $x^n$ in the polynomials of $I$ of degree at most $n$, including the zero coefficient. Closure under addition and scalar multiplication makes $J_n$ an ideal of $R$. Multiplication of a polynomial by $x$ shows $J_n\subseteq J_{n+1}$. By the ascending chain condition, there is $N$ such that $J_N=J_{N+1}=\cdots$.

Each $J_n$ for $0\le n\le N$ is finitely generated. Choose generators $a_{n,1},\dots ,a_{n,r_n}$ and corresponding polynomials $f_{n,j}\in I$ of degree at most $n$ whose coefficient of $x^n$ is $a_{n,j}$. There are only finitely many chosen polynomials. We prove they generate $I$ by induction on degree.

Take a nonzero $f\in I$ of degree $d$ and leading coefficient $a$. If $d\le N$, express $a={\sum }_j{c}_j{a}_{d,j}$ and subtract ${\sum }_j{c}_j{f}_{d,j}$. If $d>N$, use $a\in J_d=J_N$, express $a={\sum }_j{c}_j{a}_{N,j}$, and subtract ${\sum }_j{c}_j{x}^{d-N}{f}_{N,j}$. In both cases the leading coefficient cancels, leaving a polynomial in $I$ of smaller degree. For degree zero the remainder is zero, and the induction completes the claim.

Thus every ideal of $R[x]$ is finitely generated, proving the Hilbert basis theorem. The proof uses no cancellation of nonzero elements of $R$, so it remains valid when $R$ has zero divisors.

For the spherical circle, use a rotation to put the center at the north pole. Points at spherical distance $r$ have polar angle $r$ and admit the parametrizationIts speed is $|{\gamma }^{\prime }(\theta )|=\sin r$, so the arc length isFor small $r$, the Taylor series gives $C(r)=2\pi r-(\pi /3)r^3+O(r^5)$. Thus its circumference is smaller than the Euclidean circumference $2\pi r$ for the same intrinsic radius. Their ratio is $\sin r/r=1-r^2/6+O(r^4)$ and tends to one. The leading agreement expresses local Euclidean behavior; the cubic deficit reflects the positive Gaussian curvature of the unit sphere.

Define the Möbius transformationThe three prescribed distinct points go to $0,1,\infty$, respectively, and the specified cross-ratio is exactly $T(z_4)$.

To identify the preimage of the extended real line directly, write $T(z)=(az+b)/(cz+d)$. The condition $T(z)=\overline{T(z)}$ becomesAfter division by $2i$, this has the real form $A|z{|}^2+Bz+\overline{B}\overline{z}+C=0$, with $A,C$ real. If $A=0$ it is a line; otherwise completing the square gives a circle. It contains $z_1,z_2,z_3$, including the pole as a point mapped to infinity, so it is nondegenerate. Three distinct noncollinear points determine one circle; if the three are collinear, their unique generalized circle is their line.

Consequently $T(z_4)$ is real exactly when the fourth point is on the same generalized circle as the first three. This proves the real cross-ratio criterion for a generalized circle:

For the first assertion, fix $x_0\in [0,1]$ and $\epsilon >0$. By uniform convergence, choose $N$ such that $|f_N(x)-f(x)|<\epsilon /3$ for every $x$. By continuity of $f_N$ at $x_0$, there is $\delta >0$ such that $|f_N(x)-f_N(x_0)|<\epsilon /3$ whenever $|x-x_0|<\delta$. The triangle inequality then givesThus a uniform limit of continuous functions is continuous, including at the endpoints with their relative neighborhoods.

The converse implication involving integrals is false. Set $f_1=0$ and, for $n\ge 2$, use the shrinking continuous spikes with vanishing integralTheir supports are $[1/n-1/n^2,1/n+1/n^2]$. For each $x>0$ they eventually vanish at $x$, and they vanish at zero for every $n$. Hence $f_n\to 0$ pointwise, with continuous limit. Each triangular area is $1/n^2$, so ${\int }_0^1{f}_n\to 0={\int }_0^{1}0$. Yet $\sup |f_n|=1$, excluding uniform convergence.

The final assertion is also false. Take $f_n(x)=x^n$. Then the pointwise limit isThis limit is discontinuous at one but is Riemann integrable, with integral zero. Each $f_n$ is continuous and ${\int }_0^1{f}_n=1/(n+1)\to 0$. Thus convergence of the integrals to the integral of an integrable pointwise limit does not enforce its continuity.

Use the conformal map of a vertical half-strip by sine followed by a Cayley transform:We verify that this maps the entire half-strip bijectively onto the unit disc. First,so the sine image lies in the upper half-plane. Also $\cos z\ne 0$ in the domain, making the map locally a conformal map. If $\sin z=\sin w$, the identity for a difference of sines gives either $z-w=2\pi n$ or $z+w=(2n+1)\pi$. The real-part bounds force $n=0$ in the first case, while positive imaginary parts exclude the second. Thus sine is injective.

For surjectivity, given any $w$ in the upper half-plane, solveNo root is on the unit circle, since $w=(t-t^{-1})/(2i)$ would then be real. The product of the two roots is $-1$, so exactly one root has $|t|<1$. For this root,which forces $\mathrm{Re}t>0$. Choose its logarithm with argument in $(-\pi /2,\pi /2)$ and put $z=-i\log t$. Then $\mathrm{Im}z=-\log |t|>0$, its real part lies in the required interval, and $\sin z=w$. This proves that sine maps onto the upper half-plane.

Finally, for $\mathrm{Im}w>0$, $|w-i|<|w+i|$, and the Möbius transformation $(w-i)/(w+i)$ is a bijection to the unit disc. Its inverse is $w=i(1+\zeta )/(1-\zeta )$. Both constituent derivatives are nonzero, completing the proof for the displayed conformal map.

The Sturm-Liouville operator satisfies the Lagrange identity for a Sturm-Liouville operatorIndeed the potential terms cancel, and differentiating the right side gives the remaining two derivative terms. Integrate over $[0,1]$. The common Dirichlet boundary conditions make the endpoint term zero, while $L(y_0;{\lambda }_0)=0$. ThereforeThis is the necessary solvability condition imposed by the nonzero homogeneous solution; its proof here does not rely on merely quoting the Fredholm alternative.

For the nonlinear problem, chooseThe orthogonality condition then givesWith $y=ϵy_0+{ϵ}^2{y}_1$, where the correction remains bounded as $ϵ\to 0$, the normalization givesDivide by ${ϵ}^3$ and use $\mu =O(1)$ to obtainThe positive quartic integral determines the leading direction of the nonlinear shift from the linear eigenvalue.

The orbital angular momentum operators are $\mathbf{L}=\mathbf{x}\times \mathbf{p}$, soThe canonical commutation relations are $[x_i,x_j]=[p_i,p_j]=0$ and $[x_i,p_j]=i\hslash {\delta }_{ij}$. On a common domain of smooth wavefunctions, the product commutator identity givesContract this with ${ϵ}_{iab}{ϵ}_{jcd}$. The first term contributes $i\hslash ({\delta }_{ij}\mathbf{x}\cdot \mathbf{p}-x_j{p}_i)$ and the second contributes $i\hslash (x_i{p}_j-{\delta }_{ij}\mathbf{x}\cdot \mathbf{p})$. Their sum isThis verifies the angular momentum commutation relations explicitly. Similarly,

In the position representation $\mathbf{p}=-i\hslash \nabla$. For a radial wavefunction, $\nabla {\Psi }_0={\Psi }_0^{\prime }(r)\mathbf{x}/r$, parallel to $\mathbf{x}$, soWrite ${\Phi }_i=h(r)x_i$ with $h(r)={\Psi }_0^{\prime }(r)/r$. The operator $L_3=-i\hslash (x_1{\partial }_2-x_2{\partial }_1)$ annihilates the radial factor. It gives $L_3{x}_1=i\hslash x_2$, $L_3{x}_2=-i\hslash x_1$, and $L_3{x}_3=0$, henceThus these components, when nonzero, are the specified eigenfunctions with eigenvalues $0,+\hslash ,-\hslash$. This is an example of angular momentum of a radial function times a linear polynomial.

In SI units, the two curl Maxwell equations areUse the vector identity $\nabla \cdot (\mathbf{E}\times \mathbf{B})=\mathbf{B}\cdot (\nabla \times \mathbf{E})-\mathbf{E}\cdot (\nabla \times \mathbf{B})$ to obtainIntegration over the fixed volume and the divergence theorem give Poynting's theorem:The first term is the rate of change of stored electromagnetic energy. The surface integral is the outward energy flux carried by the Poynting vector. The right side is minus the power delivered by the field to matter; for an Ohmic conductor it is minus the Ohmic heating rate. Thus energy lost from the field either flows out or does work on charges.

For the specified transverse plane wave, both divergence Maxwell equations hold because $E_y$ depends only on $x,t$. Faraday's equation requiresHere we take the magnetic wave associated with the electric wave, without an additional static background field. Ampère's vacuum equation then gives $k^2={\mu }_0{ϵ}_0{\omega }^2$, equivalentlyThus all four vacuum Maxwell equations are satisfied. The Poynting vector is $\mathbf{P}=(k{E}_0^2/({\mu }_0\omega )){\cos }^2(kx-\omega t){\mathbf{e}}_x$, and averaging over a period givesFor propagation in the positive $x$ direction, $\omega =ck$ and this is $E_0^2{\mathbf{e}}_x/(2{\mu }_{0}c)$. It represents the mean energy transmitted per unit area per unit time, the wave's time-averaged energy flux per unit area, directed along its propagation.

Let $x_{\ast }=A^{-1}b$ and $e^{(k)}=x^{(k)}-x_{\ast }$. The Richardson iteration givesBy the spectral theorem, a real symmetric positive definite matrix has an orthonormal eigenbasis with positive eigenvalues ${\lambda }_1,\dots ,{\lambda }_n$. The error in the $i$th eigenvector direction is multiplied by $(1-\tau {\lambda }_i{)}^k$. Thus convergence for every initial vector occurs exactly when $|1-\tau {\lambda }_i|<1$ for every $i$, or $0<\tau {\lambda }_i<2$. Since $\rho (A)={\lambda }_{\max }$,Necessity also follows from an initial error in an offending eigenvector direction. At $\tau =0$ it remains unchanged; at $\tau =2/\rho (A)$ the largest-eigenvalue error alternates sign; outside the interval some error does not decay. Hence neither endpoint can be included.

The Neyman-Pearson lemma concerns a simple hypothesis with density $f_0$ against a simple alternative with density $f_1$. Allow a randomized test $0\le \phi \le 1$. Choose a threshold $c\ge 0$ and, if necessary, randomize on the equality set so thatThen ${\phi }_{\ast }$ is a most powerful test among all tests of size at most $\alpha$. To prove it, let $E_0\phi \le \alpha$. Pointwise, $({\phi }_{\ast }-\phi )(f_1-c{f}_0)\ge 0$: above threshold both factors are nonnegative; below threshold both are nonpositive; on the threshold their product vanishes. ThereforeThis proves the lemma, including regions with $f_0=0$, where a positive alternative density must be rejected. It also shows that ties in the likelihood ratio are the only discretionary region when the inequality is strict off the threshold.

For the stated family, $R=|X|$ has the Gamma distribution with shape $\theta$ and unit rate. At $\theta =1$ it has the exponential distribution, and the likelihood ratio for $\theta =2$ against $\theta =1$ isbecause $\Gamma (1)=\Gamma (2)=1$. The most powerful test therefore rejects for large $|X|$. Since ${\mathrm{Pr}}_1(|X|>c)=e^{-c}$, its exact size-$\alpha$ critical value and rejection rule areThe distribution is continuous, so boundary randomization is unnecessary. Under $\theta =2$, integration by parts gives the power function of a statistical testFor every fixed $\theta >1$, the likelihood ratio against the same null is $|x{|}^{\theta -1}/\Gamma (\theta )$, strictly increasing in $|x|$. The size constraint still selects exactly the same critical value $c=-\log \alpha$, independent of the alternative. The Neyman-Pearson lemma therefore makes this single test most powerful against each $\theta >1$ separately. Hence it is uniformly most powerful for the composite alternative $\theta >1$.

Let $G_0,G_1,\dots$ be independent with $\mathrm{Pr}(G_n=j)=q{p}^j$ for $j\ge 0$. The transition law is realized by the geometric-arrival single-service queueThe chain is an irreducible Markov chain: repeated zero arrivals reach zero from every state, and any state can be reached from zero in one step. It is an aperiodic Markov chain because the zero state has self-transition probability $q>0$.