Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 3

A permutation representation is a group homomorphism $\rho :G\to \mathrm{Sym}(\Omega )$, equivalently a group action on a set $\Omega$. The left regular action takes $\Omega =G$ and $\rho (g)(h)=gh$. Each such map is a bijection, and $\rho (g_1{g}_2)=\rho (g_1)\rho (g_2)$. If $\rho (g)$ is the identity, evaluating at the identity element gives $g=1$. Thus the representation is faithful. Numbering the elements of $G$ identifies $\mathrm{Sym}(G)$ with the symmetric group $S_n$, proving Cayley theorem:For a nonabelian simple group, compose this faithful representation with the sign of a permutation homomorphism $S_n\to \{1,-1\}$. Its kernel is a normal subgroup of $G$, hence either trivial or all of $G$. A trivial kernel would embed $G$ in a group of order two, contradicting noncommutativity. Therefore every represented element is even, and the same representation embeds $G$ in the alternating group $A_n$.

Finally let $S_3$ act on the two cosets of its normal subgroup $A_3$. An even permutation fixes both cosets, whereas an odd permutation exchanges them. This gives $S_3\to S_2$ with kernel $A_3$; explicitly, send even permutations to the identity and odd permutations to $(12)$.

The sequence has uniform convergence on $X$ if there is a function $f:X\to \mathbb{R}$ such thatThe same $N$ must work for every $x$. Equivalently, the supremum of the error over $X$ tends to zero. In pointwise convergence, $N$ is allowed to depend on $x$.

For each fixed $0<x<1$, $x^n\to 0$, so the pointwise limit is $f(x)=1/(1-x)$. Howeverfor every fixed $n$. Hence the error has infinite supremum on $(0,1)$ and the convergence is not uniform. This does not contradict uniform convergence on any smaller closed interval $[0,r]$, $r<1$, where the error is at most $r^n/(1-r)$.

For $0<x<1$, $|x^k/k^2|\le 1/k^2$, and the numerical series ${\sum }_{k\ge 1}{k}^{-2}$ converges. The Weierstrass M-test therefore gives uniform convergence to $f(x)={\sum }_{k\ge 1}{x}^k/k^2$. More explicitly,The bound is independent of $x$, including points arbitrarily close to one.

For each fixed real $x$, $x/n\to 0$, so the pointwise limit is zero. But ${\sup }_{x\in \mathbb{R}}|x/n|=\infty$ for every $n$. Alternatively, $x=n$ gives error one whatever $n$ is. Thus the convergence is not uniform on $\mathbb{R}$. It is uniform on each bounded interval, where $|x/n|\le M/n$.

The pointwise limit is zero, including at $x=0$. Differentiation givesso the maximum on $[0,\infty )$ is attained at $x=1/n$ and equals $1/(en)$. Thereforeand the convergence is uniform, even though the domain is unbounded.

The convention in this question counts a point whenever every open neighbourhood meets $A$, without excluding the point itself. These are adherent points, whose set is the closure $\overline{A}$; they differ from limit points under the convention that requires a different point of $A$ in each neighbourhood.

If $A$ is a closed set and $x\notin A$, the open set $X\setminus A$ is a neighbourhood of $x$ disjoint from $A$. Thus every adherent point lies in $A$. Conversely, if every adherent point lies in $A$, each $x\notin A$ has an open neighbourhood disjoint from $A$. Their union is $X\setminus A$, which is open. Hence $A$ is closed exactly when it contains all the points specified in the question.

The interior of a set and closure areThe neighbourhood description of the closure follows because failure to be adherent is equivalent to lying in an open set disjoint from $A$.

For closure of a connected set, suppose $\overline{A}=U\bigsqcup V$ were a separation into two nonempty relatively open sets. Their intersections with the connected subset $A$ would separate $A$ unless one intersection were empty. Say $A\subseteq U$. Choose $v\in V$ and an open set $O\subseteq X$ with $v\in O$ and $O\cap \overline{A}\subseteq V$. Since $v\in \overline{A}$, the set $O$ meets $A$, contradicting $A\subseteq U$. Thus the closure of a connected set is connected. The empty-set case holds directly.

Use Fourier inversion with the stated Fourier transform convention:For $x<0$, close the contour integral in the lower half-plane, where $|e^{ikx}|=e^{-x\mathrm{Im}k}$ decays. Jordan lemma eliminates the large semicircle. The contour is clockwise, so the residue theorem supplies $-2\pi i$ times the residue at $k=-ip$. That simple pole hasConsequentlyThe minus sign uses both the clockwise orientation and the residue $-i{e}^{px}$; it is consistent with the full odd function $\mathrm{sgn}(x)e^{-p|x|}$ away from zero.

For a finite triangulation of a surface of the sphere, Euler formula for a sphere iswhere $V$, $E$ and $F$ count vertices, edges and triangular faces. Since each face has three edges and each edge borders two faces, also $3F=2E$.

Let $p$ and $h$ count the pentagonal and hexagonal faces. Counting incidences of edges with vertices and faces givesThe Euler formula for a sphere also applies to this polygonal decomposition: triangulating a polygon by adding one diagonal increases both $E$ and $F$ by one, preserving $V-E+F$. Substitution givesThus there are exactly 12 pentagons, independently of the number of hexagons.

Lift the angular coordinate continuously to a real coordinate and write $\Delta z=z_B-z_A$ and $\Delta {\varphi }_m={\varphi }_B-{\varphi }_A+2\pi m$ for a prescribed winding number $m\in \mathbb{Z}$. Assume $z_A<z_B$ for the calculation; reversing the endpoints changes no length. The arc length functional on the circular cylinder isThe Euler-Lagrange equation says $R^2{\varphi }^{\prime }/\sqrt{1+R^2{\varphi }^{\prime 2}}$ is constant. This expression is strictly increasing in ${\varphi }^{\prime }$, so ${\varphi }^{\prime }$ is constant. The candidate is the helixIt is a global minimum within the chosen winding class: the integrand is strictly convex, and Jensen inequality bounds the functional below by $(z_B-z_A)\sqrt{1+R^2(\Delta {\varphi }_m/\Delta z{)}^2}$, with equality only for constant slope. Equivalently, unrolling the cylinder makes the minimizing path a straight segment.

The signed pitch of a helix, measured for an increase of $2\pi$ in angle, isIts geometric magnitude is $|P_m|$. If only the physical endpoints are fixed, choose the integer $m$ minimizing $|\Delta {\varphi }_m|$; opposite points may give two equally short helices. If the lifted endpoint angles are prescribed, use that winding class. A zero angular increment gives a straight generator, the limiting infinite-pitch case.

Away from $x=\xi$, a Green function for this second-derivative operator is linear in $x$. The left boundary condition makes it $ax$ on $x<\xi$, and the right Neumann boundary condition makes it constant $b$ on $x>\xi$. Continuity gives $b=a\xi$. Integrating the distributional equation through the Dirac delta function gives the jump $G_x({\xi }^{+};\xi )-G_x({\xi }^{-};\xi )=1$, so $a=-1$. ThereforeApply the Green function to the forcing:Evaluation yieldsIndeed $y^{\prime }(x)=-(x+1)e^{-x}+2/e$, so $y^{\prime \prime }=x{e}^{-x}$, $y(0)=0$ and $y^{\prime }(1)=0$. A difference of two solutions is linear with these homogeneous boundary conditions, hence zero; the boundary value problem has a unique solution.

For a particle of mass $m$, the probability density and one-dimensional probability current areThe Schrodinger equation and its complex conjugate, for real $V$, giveThe potential terms cancel precisely because $V$ is real. Thus ${\rho }_t+j_x=0$, the continuity equation for probability. Integrating over an interval says that its probability changes by the incoming current minus the outgoing current.

The global time-dependent phase cancels from the probability current. Put $\chi =e^{ikx}+R{e}^{-ikx}$. Direct multiplication givesThe second term is real, because the expression in parentheses is purely imaginary. HenceThe first plane wave travels to the right with incident current $\hslash k/m$; the second travels to the left with reflected current $-(\hslash k/m)|R{|}^2$. Their wave interference changes the probability density but cancels from the net current. The current vanishes for $|R|=1$, is positive for $|R|<1$, and is negative for $|R|>1$. These plane waves are scattering states, not individually normalizable bound states.

Let $T_0=0$ and let $T_{r+1}$ be the first visit to $i$ strictly after $T_r$, with $T_{r+1}=\infty$ if there is none. Put $q={\mathbb{P}}_i(T_1<\infty )$. The Strong Markov property at each finite first return time gives, by induction,The number of visits, including time zero, is $N={\sum }_{r\ge 0}{1}_{\{T_r<\infty \}}={\sum }_{n\ge 0}{1}_{\{X_n=i\}}$. By Tonelli theorem for nonnegative terms,If $q<1$, this geometric series is $1/(1-q)$, and $1-q$ is the probability of never returning. If $q=1$, every term of the final series is one and both sides are infinite. Thuswith the prescribed infinite-value convention. This covers both transient states and recurrent states.

The real and imaginary parts of the operator are uniquelyThey are self-adjoint operators, and $\alpha ={\alpha }_1+i{\alpha }_2$. Conversely such a decomposition has ${\alpha }^{\ast }={\alpha }_1-i{\alpha }_2$. Expanding both products givesConsequently $\alpha$ is a normal operator exactly when its self-adjoint real and imaginary parts commute.

We prove the finite-dimensional spectral theorem for normal operators by induction on dimension. A complex linear operator has an eigenvector $v$ with eigenvalue $\lambda$, by the fundamental theorem of algebra. For a normal operator $T$, the operator $T-\lambda I$ is also normal, andfor every $w$. Applying this to $v$ shows $T^{\ast }v=\overline{\lambda }v$.

Normalize $v$. Its orthogonal complement $W=v^{\perp }$ is invariant under both $T$ and $T^{\ast }$: the adjoint operator identities pair $Tw$ with $v$ through $T^{\ast }v$, and $T^{\ast }w$ with $v$ through $Tv$. The restricted operator on $W$ remains normal, and its adjoint is the restriction of $T^{\ast }$. Induction supplies an orthonormal eigenbasis of $W$, which together with $v$ gives one for $V$.

Conversely, in an orthonormal basis of eigenvectors, $T$ is diagonal and $T^{\ast }$ is the diagonal matrix of the conjugate eigenvalues. The two diagonal operators commute. Thus normality is equivalent to an orthonormal eigenbasis. The zero-dimensional case is immediate.

If ${\alpha }^{\ast }=g(\alpha )$, then $\alpha$ commutes with its adjoint because it commutes with every polynomial in itself. Hence $\alpha$ is a normal operator.

Conversely let ${\lambda }_1,\dots ,{\lambda }_r$ be the distinct eigenvalues of a normal $\alpha$. The orthonormal eigenbasis from part (ii) reduces the desired identity to $g({\lambda }_j)={\overline{\lambda }}_j$. Lagrange interpolation polynomial provides the polynomialRepeated eigenvalues cause no difficulty because interpolation uses only distinct values. If $V=\{0\}$, use $g=0$. This proves both directions of the third equivalence.

An algebraic integer is a complex number satisfying a monic polynomial with integer coefficients. Let $f\in \mathbb{Q}[x]$ be the monic minimal polynomial of $\alpha$. It is irreducible over $\mathbb{Q}$. If $H\in \mathbb{Z}[x]$ is a monic polynomial vanishing at $\alpha$, then $f$ divides $H$ over $\mathbb{Q}$. Gauss lemma for polynomials implies that monic rational factors of a monic integer polynomial have integer coefficients. Thus $f\in \mathbb{Z}[x]$, and its irreducibility over $\mathbb{Q}$ implies irreducibility in $\mathbb{Z}[x]$.

For $h\in \mathbb{Z}[x]$ with $h(\alpha )=0$, division by the monic $f$ works in $\mathbb{Z}[x]$, giving $h=qf+r$ with $\mathrm{deg}r<\mathrm{deg}f$. Since $r(\alpha )=0$, minimality forces $r=0$. Conversely each multiple of $f$ vanishes at $\alpha$. ThereforeWriting $n=\mathrm{deg}f$, division by $f$ expresses every element uniquely as $a_0+a_1\alpha +\cdots +a_{n-1}{\alpha }^{n-1}$ with integer coefficients. A relation among these powers would be a nonzero vanishing polynomial of degree less than $n$. Hence $1,\alpha ,\dots ,{\alpha }^{n-1}$ form a basis of the free module over $\mathbb{Z}$ $\mathbb{Z}[\alpha ]$.

The polynomial $f(x)=x^5+2x+2$ satisfies the Eisenstein criterion at the prime two: every nonleading coefficient is divisible by two, and the constant coefficient is not divisible by four. Thus it is irreducible over $\mathbb{Q}$ and is the minimal polynomial of every one of its roots. Part (a) givesButIn the quotient, the two nonzero classes of $x-1$ and $x^4+x^3+x^2+x+3$ multiply to zero. They are nonzero because their nonzero representative degrees are less than five. The quotient is not an integral domain, so $(5)$ is not a prime ideal.

For ring homomorphisms preserving the identity, an image $\beta$ of $\alpha$ in the Gaussian integers would satisfy $f(\beta )=0$. The minimal polynomial of $\beta$ over $\mathbb{Q}$ would then be $f$, of degree five. This is impossible because $\beta \in \mathbb{Q}(i)$ has degree at most two. Thus there is no unital homomorphism $\mathbb{Z}[\alpha ]\to \mathbb{Z}[i]$. If homomorphisms are instead allowed not to preserve the identity, the zero map exists; every nonzero homomorphism into this integral domain is unital, because the image of one is a nonzero idempotent and hence one.

The multiplicative field norm on this ring isIf an element is a unit, its positive integer norm has a multiplicative inverse in the positive integers, hence is one. The only integer solutions of $a^2+5b^2=1$ are $(a,b)=(1,0),(-1,0)$. Therefore the units are exactly $1$ and $-1$.

The standard failure of unique factorization isThere is no element of norm two or three: $b\ne 0$ gives norm at least five, while $b=0$ gives a square. An alleged nontrivial factorization of $2$ (norm four), $3$ (norm nine), or $1\pm \sqrt{-5}$ (norm six) would require a factor of norm two or three. Thus all four displayed factors are irreducible elements. No factor in the first factorization is associated to a factor in the second, since associates have the same norm and the norms are respectively four, nine and six. Hence $\mathbb{Z}[\sqrt{-5}]$ is not a unique factorization domain.

For a metric space $(X,d)$, uniform continuity meansThe choice of $\delta$ is independent of the points. A Lipschitz continuous function satisfies $|f(x)-f(y)|\le Ld(x,y)$ for one finite constant $L\ge 0$. Choosing $\delta =\epsilon /(L+1)$ proves Lipschitz continuity implies uniform continuity.

Let $(x_n)$ be a Cauchy sequence. For a given $\epsilon >0$, uniform continuity supplies $\delta >0$, and the Cauchy property supplies $N$ with $d(x_m,x_n)<\delta$ for $m,n\ge N$. Then $|f(x_m)-f(x_n)|<\epsilon$, so $(f(x_n))$ is Cauchy in $\mathbb{R}$. By completeness of the real numbers, $(f(x_n))$ converges, even if the original sequence has no limit in $X$.

Suppose the continuous $f$ were not uniformly continuous. There would be an ${\epsilon }_0>0$ and pairs $x_n,y_n\in X$ with $d(x_n,y_n)<1/n$ but $|f(x_n)-f(y_n)|\ge {\epsilon }_0$. Sequential compactness gives a subsequence $x_{n_j}\to x\in X$. The triangle inequality gives $y_{n_j}\to x$ as well. Continuity at $x$ then gives $f(x_{n_j})-f(y_{n_j})\to 0$, a contradiction. Thus $f$ is uniformly continuous.

It need not be Lipschitz continuous. On the sequentially compact interval $[0,1]$, the continuous function $f(x)=\sqrt{x}$ hasNo finite Lipschitz constant exists.

For each $x\in X$, density of $Y$ allows a sequence $y_n\in Y$ with $d(y_n,x)<1/n$. If $f_1,f_2$ are two continuous extensions, continuity givesThus a continuous extension is unique if it exists.

Now assume $g$ is uniformly continuous. Each chosen $y_n\to x$ is a Cauchy sequence, so part (a), applied to $Y$, shows that $g(y_n)$ converges in $\mathbb{R}$. DefineThis is independent of the choice: for another $z_n\to x$ in $Y$, $d(y_n,z_n)\to 0$, and uniform continuity gives $g(y_n)-g(z_n)\to 0$. At $x\in Y$, comparison with the constant sequence $x$ proves $f(x)=g(x)$.

To prove uniform continuity of the extension, choose $\delta >0$ so that $|g(y)-g(z)|<\epsilon /2$ whenever $y,z\in Y$ and $d(y,z)<\delta$. If $d(x,x^{\prime })<\delta /3$, choose sequences $y_n\to x$ and $z_n\to x^{\prime }$ in $Y$. For large $n$, both approximation distances are less than $\delta /3$, so $d(y_n,z_n)<\delta$. Passing to the limit gives $|f(x)-f(x^{\prime })|\le \epsilon /2<\epsilon$. The same $\delta /3$ works for all $x,x^{\prime }$. Therefore the extension exists and is uniformly continuous. No completeness of $X$ is needed; completeness of the target $\mathbb{R}$ is the relevant requirement.

Let $T$ be a closed triangle contained, together with a neighbourhood, in the domain of a holomorphic function $f$. We prove Cauchy theorem for a triangle without assuming continuity of $f^{\prime }$.

Join the edge midpoints to subdivide $T$ into four triangles, each with half its diameter and perimeter. Their positively oriented contour integrals sum to the integral around $T$, because the internal edges cancel. At least one subtriangle has integral of modulus at least one quarter of the original. Repeat that choice to obtain nested closed triangles $T_n$ satisfyingTheir intersection is a single point $z_{\ast }$, by compactness and shrinking diameter. complex differentiability at a point at $z_{\ast }$ givesThe first two terms have polynomial primitives, hence integrate to zero around each triangle. For sufficiently large $n$, $|r|<\epsilon$ on $T_n$, so the complex line integral estimate givesCombining the bounds gives $|{\int }_{\partial T}f(z)dz|\le \epsilon d_0{P}_0$. Since $\epsilon$ is arbitrary,The orientation can be reversed with no change to the conclusion.

For a closed piecewise continuously differentiable curve avoiding $a$, its winding number isEquivalently, for a parametrization $\gamma :[u,v]\to \mathbb{C}$, it is $(2\pi i{)}^{-1}{\int }_u^v{\gamma }^{\prime }(t)/(\gamma (t)-a)dt$, summed over the smooth pieces. A continuous choice of the argument of $\gamma (t)-a$ along the parameter interval changes by an integer multiple of $2\pi$ between the endpoints; that integer is the winding number.

For every closed piecewise smooth curve $\gamma$ in $U$, the winding number of $f\circ \gamma$ about zero is the integerIf $f=g_k^k$ for a holomorphic $k$th root, $g_k$ is nowhere zero, andThus $m_{\gamma }$ is divisible by every $k$ for which a root exists. An infinite subset of the positive integers is unbounded, so a fixed integer divisible by all those $k$ must be zero. Hence all periods of the holomorphic function $f^{\prime }/f$ vanish.

Fix $z_0\in U$. Since a domain is connected and open, its points can be joined by piecewise smooth paths. Vanishing periods makeindependent of path. It is a holomorphic primitive with $H^{\prime }=f^{\prime }/f$. Differentiating gives $(f{e}^{-H}{)}^{\prime }=0$, so $f{e}^{-H}=f(z_0)$. Choose a complex number $c$ with $e^c=f(z_0)$. ThenThis constructs a global holomorphic logarithm without requiring $U$ to be simply connected.

Cross-ratio conventions differ by permutations. Use the following ordering, which gives the signs requested in part (c):Values involving infinity are defined by limits. This is the cross-ratio with second point mapped to zero: the Möbius transformationsends $z_1,z_2,z_4$ to $1,0,\infty$ and sends $z_3$ to $r$.

For $M(z)=(az+b)/(cz+d)$, $ad-bc\ne 0$, the identityshows that all factors cancel in the cross-ratio. This proves Möbius invariance of the cross-ratio, with the cases involving infinity following by limits. Algebraic subtraction also givesThe latter expression is another common ordering for the cross-ratio; keeping the convention explicit prevents an incorrect reciprocal or sign.

With $T$ from part (a), the first, second and fourth points map to $1,0,\infty$. The inverse image of the extended real line under a Möbius transformation is a generalized circle, meaning a circle or a line together with infinity. Thus, if $r=T(z_3)$ is real, all four original points lie on that generalized circle.

Conversely, if the four points lie on one generalized circle, its image under $T$ is another generalized circle containing $0,1,\infty$. A generalized circle containing infinity is a line, and the line through zero and one is the real line. Therefore $T(z_3)=r$ is real. Hence the cross-ratio is real exactly when the four points are concyclic or collinear.

Write $z_j=c+R{e}^{i{\theta }_j}$ with ${\theta }_1<{\theta }_2<{\theta }_3<{\theta }_4<{\theta }_1+2\pi$. In the identitythe phase factors cancel in the cross-ratio. In our convention the two numerator sine factors are negative, while the denominator factors have opposite signs. Thus $r<0$, and the exchanged cross-ratio $1-r$ is greater than one. Consequently $|r|+1=1-r=|1-r|$.

Taking absolute values of the two formulas in part (a) givesMultiplying $|r|+1=|1-r|$ by the common positive denominator proves Ptolemy theorem:The product of the diagonals equals the sum of the products of opposite sides.

Use the Fourier transform $\tilde{f}(k)={\int }_{\mathbb{R}}f(x)e^{-ikx}dx$. For a rapidly decaying smooth function, integration by parts gives $\widetilde{f^{\prime }}=ik\tilde{f}$, and differentiation under the integral gives $\widetilde{xf}=id\tilde{f}/dk$. Both are justified here by Gaussian decay. Since $f^{\prime }=-2a^{2}xf$,The Gaussian integral supplied in the hint gives $\tilde{f}(0)=\sqrt{\pi }/a$. Solving this first-order equation yieldsThe calculation states the transform normalization and the two differentiation properties explicitly.

Taking the spatial Fourier transform of the diffusion equation gives ${\tilde{\theta }}_t=-D{k}^2\tilde{\theta }$ and $\tilde{\theta }(k,0)=\tilde{g}(k)$. Hence $\tilde{\theta }=e^{-D{k}^{2}t}\tilde{g}$. Fourier inversion and the Gaussian transform from part (a), or direct evaluation of the inverse Gaussian integral, give the heat kernelSubstituting the integral for $\tilde{g}$ and interchanging absolutely integrable terms produces the convolution solutionFor the stated nonnegative compactly supported data, understood to be nonzero on a set of positive measure, the kernel is strictly positive everywhere. Therefore $\theta (x,t)>0$ for every real $x$ and every $t>0$. Property P holds: diffusion has infinite propagation speed. The distant value can be extremely small, but is not zero. For continuous initial data, any nontrivial nonnegative $g$ automatically has the required positive-measure set.