Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 3

Write $R=\mathbb{Z}[\sqrt{-13}]$ and $I=(2,1+\sqrt{-13})$. The mapis a surjective ring homomorphism: in ${\mathbb{F}}_2$, the image of $\sqrt{-13}$ is $1$ and $1^2=-13=1$. Its kernel is exactly $I$. Indeed, both generators lie in the kernel; conversely, if $a+b$ is even, thenand $a-b$ is even. Hence $R/I\cong {\mathbb{F}}_2$ and the index of a subgroup is $[R:I]=2$.

If $I=(\alpha )$ were a principal ideal, multiplication by $\alpha =a+b\sqrt{-13}$ would identify its additive lattice with a sublattice of indexThus principality would require $a^2+13b^2=2$. This Diophantine equation has no integer solution: $b=0$ would give $a^2=2$, while $b\ne 0$ makes the left side at least $13$. Therefore

A function $f:A\to \mathbb{R}$ is uniformly continuous if, for every $\epsilon >0$, there is a $\delta >0$ such thatThe same $\delta$ must work at every pair of points in the domain.

The function $x\mapsto x^2$ is not uniformly continuous on $\mathbb{R}$. Take $x_n=n$ and $y_n=n+1/n$. Then $|x_n-y_n|\to 0$, whereasThis contradicts the sequential criterion for uniform continuity.

The square-root function is uniformly continuous on $[0,\infty )$. For $x,y\ge 0$,so $|\sqrt{x}-\sqrt{y}|\le \sqrt{|x-y|}$. Given $\epsilon >0$, choosing $\delta ={\epsilon }^2$ proves uniform continuity.

The function $x\mapsto \cos (1/x)$ is uniformly continuous on $[1,\infty )$. Its derivative satisfiesso the mean value theorem givesThus the function is Lipschitz, which implies uniform continuity.

A compact space is one for which every open cover has a finite subcover. A metric space is sequentially compact when every sequence in it has a convergent subsequence whose limit belongs to the space.

First, compactness makes $X$ totally bounded: for every positive integer $k$, the open cover by balls of radius $1/k$ has a finite subcover. Given a sequence, repeatedly choose a ball of radius $1/k$ containing infinitely many terms of the subsequence retained at the preceding stage. Taking a diagonal subsequence $(x_{n_k})$, any two terms with indices at least $k$ lie in one ball of radius $1/k$, so their distance is below $2/k$. The diagonal subsequence is therefore a Cauchy sequence.

A compact metric space is complete. To see this without assuming sequential compactness, let $(y_n)$ be Cauchy and put $F_n=\overline{\{y_m:m\ge n\}}$. These nonempty closed sets are nested and have the finite intersection property; compactness gives a point $y\in {\bigcap }_n{F}_n$. Given $\epsilon >0$, a sufficiently late tail has diameter below $\epsilon /2$, and because $y$ lies in its closure, every point of that tail lies within $\epsilon$ of $y$. Thus $y_n\to y$.

The Cauchy subsequence constructed above consequently converges in $X$. Hence every compact metric space is sequentially compact.

The Möbius transformationmaps the unit disc conformally onto the upper half-plane. On the upper semicircle of its boundary, $w$ approaches the positive real axis; on the lower semicircle it approaches the negative real axis. If $0<\arg w<\pi$, the harmonic functionhas the required boundary limits.

For $z=x+iy$,Since the imaginary part is positive inside the disc,Therefore the solution of the Dirichlet problem isThe exceptional boundary endpoints are precisely the jump points of the prescribed data.

On the unit sphere, the spherical excess formula gives the area of a spherical triangle asFor a sphere of radius $R$, the right side is multiplied by $R^2$.

Join an interior point of a convex spherical $n$-gon to its vertices by geodesics. The resulting $n$ spherical triangles contain all polygon angles and angles summing to $2\pi$ at the interior point. Adding their spherical excesses givesFor a regular polygon with interior angle $\alpha$, positivity of area requires $n\alpha >(n-2)\pi$, while strict convexity requires $\alpha <\pi$. Both limiting values can be approached by regular polygons, so

Sincefor $x>0$, the second-derivative criterion shows that $f(x)=x^a/a$ is strictly convex. Put $b=a/(a-1)$, so $1/a+1/b=1$. For dual variable $p>0$, the stationary equation in the Legendre transform isIt is the unique global maximizer of $px-f(x)$, and henceApplying the same calculation with the conjugate exponent $b$ givesUnder the Legendre-Fenchel transform on all real dual variables, the same supremum additionally gives $f^{\ast }(p)=0$ for $p\le 0$.

Now take $a=r$, $b=s$, and apply the Fenchel–Young inequality $f(x)+f^{\ast }(y)\ge xy$. This yields Young's inequality for productsfor positive $x,y$ and conjugate exponents $r,s$.

Let $\omega =e^{2\pi i/N}$. Using the conventionthe discrete Fourier transform of the supplied sequence follows from the binomial theorem:The orthogonality of roots of unity givesSince $m,k$ both lie between zero and $N-1$, it follows that

For square-integrable wavefunctions with the stated decay, multiplication by the real coordinate $x$ is Hermitian becauseFor $p_x=-i\hslash {\partial }_x$, integration by parts givesand the boundary term vanishes. The same arguments apply to $y$ and $p_y$.

If $F$ and $G$ are Hermitian on a common suitable domain, then $(FG{)}^{†}=GF$, soBecause $x$ commutes with $p_y$ and $y$ with $p_x$,is Hermitian. Using $p_{x}x=x{p}_x-i\hslash$ and its $y$ analogue,so $D$ is Hermitian as well.

Finally set $R=x{\partial }_y-y{\partial }_x$ and $S=x{\partial }_x+y{\partial }_y$. Direct evaluation on a smooth test function gives $[R,S]=0$: rotations commute with radial dilations. The scalar term in $D$ also commutes with everything, and therefore

States $i,j\in S$ communicate when each is accessible from the other: there exist nonnegative integers $m,n$ such that $P^m(i,j)>0$ and $P^n(j,i)>0$. A communicating class is an equivalence class for this relation.

The period of a state in a Markov chain $a$ isA state is aperiodic when this greatest common divisor is one.

Suppose $a$ and $b$ communicate. Choose $r,s$ with $P^r(a,b)>0$ and $P^s(b,a)>0$. Then $r+s$ is a possible return time to $a$, so $d(a)$ divides $r+s$. If $n$ is any possible return time to $b$, concatenating the paths gives a return to $a$ of length $r+n+s$. Thus $d(a)$ also divides $r+n+s$, and subtraction shows that $d(a)$ divides every such $n$. Therefore $d(a)\mid d(b)$. Interchanging $a$ and $b$ gives the reverse divisibility, soHence period is constant on a communicating class.

The symmetric bilinear form associated with a quadratic form is obtained by polarization:To diagonalize it, argue by induction on $\dim V$. If $\varphi (v,v)=0$ for every $v$, polarization gives $\varphi =0$, so every basis works. Otherwise choose $e$ with $\varphi (e,e)\ne 0$. Thenand diagonalize the restriction to $e^{\perp }$ inductively. Thus some basis gives a diagonal matrix with positive, negative, and zero entries. By Sylvester's law of inertia, their counts $n_{+},n_{-},n_0$ are basis-independent. In the convention relevant here, the signature of a quadratic form is

Suppose $R$ lies in the radical of a bilinear form, so $\varphi (r,v)=0$ for all $r\in R,v\in V$. Then $q(r)=0$ andHence $q^{\prime }(v+R)=q(v)$ is a well-defined quadratic form on the quotient vector space $V/R$. In an adapted diagonal basis, quotienting by $R$ merely removes zero diagonal directions, so $n_{+}$ and $n_{-}$, and therefore the signature, are unchanged.

Now let $W=\mathrm{span}(e,f)$. Its Gram matrix iswhich has determinant $-1$. Thus $W$ is nondegenerate and has one positive and one negative square: it is a hyperbolic plane (quadratic form) and has signature zero. Nondegeneracy gives the orthogonal direct sumSignature is additive under orthogonal direct sums, so

The Eisenstein integers are $\mathbb{Z}[\omega ]$, where ${\omega }^2+\omega +1=0$. Their multiplicative algebraic norm isGiven $\alpha ,\beta \ne 0$, choose an Eisenstein integer $q$ nearest to the complex number $\alpha /\beta$. The hexagonal lattice has covering radius $1/\sqrt{3}$, soWith $r=\alpha -q\beta$, multiplicativity givesThus division with remainder decreases the nonnegative integer norm, proving that $\boxed{\mathbb{Z}[\omega ]\text{ is a Euclidean domain}}$.

Every Euclidean domain is a principal ideal domain: in a nonzero ideal choose an element of least positive Euclidean value and divide every other element by it; minimality forces every remainder to vanish. Every principal ideal domain is a unique factorization domain: the ascending-chain condition supplies factorizations into irreducibles, and Euclid lemma follows because every irreducible generates a prime ideal. Therefore

Factor the equation in the Eisenstein integers:Let $\pi =1-\omega$. Its norm is $3$, andso $3$ is a unit times ${\pi }^2$. Suppose $x\equiv 2(\mathrm{mod}3)$, say $x=3k+2$. Since $\omega =1-\pi$,The first term is divisible by ${\pi }^2$ and the second has discrete valuation exactly one, so $v_{\pi }(x+\omega )=1$. Conjugation gives $v_{\pi }(x+{\omega }^2)=1$ as well. ConsequentlyBut $v_{\pi }(y^3)=3v_{\pi }(y)$ is divisible by three, a contradiction. Hence

A local form of the Picard-Lindelof theorem is as follows. Let $F(t,y)$ be continuous on a rectangle about $(t_0,y_0)$ and uniformly Lipschitz in $y$ there. Then some interval about $t_0$ admits a unique continuously differentiable solution ofThe solution is the unique fixed point of the associated integral operator. If $F$ is continuous on $[a,b]\times {\mathbb{R}}^n$ and globally Lipschitz in $y$ with one constant, the solution exists uniquely throughout $[a,b]$.

Since $1\le t\le b$ and $c\ge 0$,Taking suprema givesThus the weighted norm and the usual uniform norm are equivalent norms.

For $f,g\in X$, the Lipschitz condition givesChoose $c$ so that $c+1>Mb$. Then $\varphi$ is a contraction mapping. The space $X$ is complete in the sup norm, and therefore also in the equivalent weighted norm.

For initial value $y_0$, defineIt has the same contraction constant. The Banach fixed-point theorem gives a unique fixed point $f$, which satisfies the integral equation. The fundamental theorem of calculus then givesConversely every solution satisfies that integral equation, proving uniqueness on $[1,b]$. This weighted-norm argument is the Bielecki norm method.

For a closed piecewise smooth path $\gamma$ avoiding $w$, its winding number isA meromorphic function has a zero of order $m>0$ at $z_0$ whenwith $g$ holomorphic; it has a pole of order $m$ at $w_0$ when $(z-w_0{)}^{m}f(z)$ extends holomorphically and nonvanishingly there.

The argument principle states that, if $f$ has no zero or pole on $\gamma$,For a positively oriented simple contour this is the number of enclosed zeros minus poles, counted with multiplicity. It follows from the residue theorem because the residue of $f^{\prime }/f$ at a zero of order $m$ is $m$, while at a pole of order $m$ it is $-m$.

LetOn the imaginary axis,Thus $p(iy)$ always lies in the open right half-plane, has no zero there, and its continuous argument has net change zero as the imaginary-axis part of a large right-half-disc contour is traversed. On the right semicircle, $p(z)/z^4\to 1$ uniformly, so the argument change tends to that of $z^4$, namely $4\pi$. The argument principle therefore givesroots in the open right half-plane, counted with multiplicity.

A geodesic triangulation of a closed smooth surface decomposes it into finitely many triangular discs whose edges are geodesic arcs and whose intersections are common faces. Its Euler number isThe Gauss-Bonnet theorem for a closed surface statesFor any triangulation of a closed surface, each triangular face has three edges and every edge belongs to two faces, so $3F=2E$. The sum of all vertex valencies is $2E$. Hence the average valency isThe torus has Euler characteristic $\chi =V-E+F=0$. Using $F=2E/3$ gives $V-E/3=0$, so $E=3V$ and

For the sphere, $\chi =2$, and the same incidence identity givesTherefore

Subdividing one face in the stated way adds one vertex, three edges, and two net faces. Starting from any spherical triangulation and repeating this operation makes $V\to \infty$ while preserving the sphere. Since every resulting average isthese averages approach six from below. Thus spherical triangulations can have average valency arbitrarily close to six.

If $K<0$ everywhere, compactness makes the integral negative, so $\chi (S)<0$. For every triangulation, $3F=2E$ again yields the general identitySince $V\ge 1$ and $\chi (S)<0$,This supplies bounds depending only on the surface.

In the sense of distribution theory, $H^{\prime }(t)=\delta (t)$. Since $\sin (\alpha t)$ vanishes at zero,and therefore

Let $G_{\mathrm{ret}}(\mathbf{x},t;\mathbf{y})$ solveThe spatial Fourier transform turns this into the preceding oscillator equation with $\alpha =ck$, soInverting with the supplied radial integral gives the three-dimensional retarded Green function

For zero initial displacement and initial velocity $f$, convolution with this kernel gives the Kirchhoff formulaThus the value at $(\mathbf{x},t)$ depends only on the initial data on the sphere of radius $ct$, not on its interior. This is the Huygens principle and describes propagation at the finite wave speed $c$.

Inside the infinite square well, the stationary Schrodinger equation for unit mass isIts normalized eigenstates and energies areThe general normalized time-dependent solution is

Immediately after removing the barrier, the normalized wavefunction isIts Fourier sine series coefficients in the full-well basis areThey areThe Born rule therefore gives

For a boost of speed $v$ in the $x$ direction, withthe Lorentz transformation of electromagnetic fields isExpanding $E_y^{\prime }{B}_y^{\prime }+E_z^{\prime }{B}_z^{\prime }$, the mixed terms cancel and the remaining transverse terms acquire the factor ${\gamma }^2(1-v^2/c^2)=1$. Together with the unchanged longitudinal product, this proves the Lorentz invariantAn independent invariant isIf only $E_y=E$ and $B_y=B$ are nonzero, thenand direct subtraction gives

Now put $cB=\lambda E$ and $\beta =v/c$. Invariance of the dot product gives ${\mathbf{E}}^{\prime }\cdot {\mathbf{B}}^{\prime }=\lambda E^2/c$, while the displayed components giveHenceAt $\beta =0$ this equals $\mathrm{sgn}\lambda$, and as $|\beta |\uparrow 1$ it tends continuously to zero with the same sign. Thus boosts realize every $0\le \theta <\pi /2$ when $\lambda >0$, and every $\pi /2<\theta \le \pi$ when $\lambda <0$.

Let the perturbed interface be $z=\eta (x,y,t)$. For inviscid disturbances about rest, the initially irrotational velocities remain potential flows,in the upper layer $0<z<h$ and lower layer $-h<z<0$. The rigid walls impose no-penetration boundary conditionand zero normal derivative at $x,y=0,2h$. Linearizing the kinematic and pressure-continuity conditions at $z=0$ gives

The horizontal Neumann eigenfunctions arewhere $m,n$ are nonnegative integers, not both zero. For time dependence $e^{-i\omega t}$, the vertical factors satisfying the rigid lids areThe kinematic condition gives their interface valuesSubstitution into the dynamic condition yields the interfacial gravity-wave dispersion relationFor ${\rho }_2>{\rho }_1$, the lowest nontrivial wavenumber is $k=\pi /(2h)$, with the two degenerate modes $(m,n)=(1,0)$ and $(0,1)$. If ${\rho }_1\ll {\rho }_2$, the relation reduces to the finite-depth free-surface result ${\omega }^2\simeq gk\tanh (kh)$. If ${\rho }_1>{\rho }_2$, then ${\omega }^2<0$ and disturbances grow exponentially: this is the Rayleigh-Taylor instability.

Bilinearity and symmetry follow from the corresponding properties of the integral. Since $w(x)>0$,If this integral is zero, the nonnegative continuous integrand must vanish everywhere, so $f=0$. Therefore the displayed formula is a real inner product on $C[a,b]$.