Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 4

Legendre formula giveswhose summands are zero or one. If the sum is at least $k$, some nonzero summand has index $j\ge k$, whence $p^k\le p^j\le 2n$.

Grouping the von Mangoldt function by prime givesEvery prime-power contribution to $\log \left(\genfrac{}{}{0ex}{}{2n}{n}\right)$ occurs in $\psi (2n)$ by the first part. Finally $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)={\prod }_{j=1}^n(n+j)/j\ge 2^n$, so $\psi (2n)\ge n\log 2$.

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The Brouwer fixed-point theorem says that every continuous map from a closed two-dimensional disk to itself has a fixed point.

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The equivalent no-retraction theorem says there is no continuous retraction of the closed disk onto its boundary circle. A fixed-point-free map gives a retraction by projecting from $f(x)$ through $x$ to the boundary; conversely a retraction composed with the antipodal boundary map gives a fixed-point-free self-map.

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On the simplex define $f(x)=Ax/(1^{T}Ax)$. Strict positivity makes the denominator positive and maps the simplex continuously into its interior. By Brouwer fixed-point theorem, $f(x)=x$ for some $x$. Thus $Ax=\lambda x$, where $\lambda =1^{T}Ax>0$, and because $Ax$ is strictly positive, so is $x$.

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For Rabin cryptosystem, choose primes $p,q\equiv 3(\mathrm{mod}4)$, publish $N=pq$, and encrypt $x$ as $x^2\mathrm{mod}N$. The factors let the receiver take square roots modulo $p$ and $q$ and combine them using the Chinese remainder theorem; redundancy identifies the intended one of four roots.

For RSA cryptosystem, choose $N=pq$, select $e$ coprime to $\phi (N)$ and $d$ with $ed\equiv 1(\mathrm{mod}\phi (N))$. Encrypt $x$ as $x^e$ and decrypt by raising to $d$ modulo $N$. Rabin inversion is provably equivalent to factoring, while ordinary RSA lacks that reduction; Rabin's disadvantage is its fourfold decryption ambiguity.

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The pumping lemma for regular languages says that sufficiently long accepted words split as $xyz$ with $|xy|\le P$, $|y|>0$, and $x{y}^{k}z$ accepted for all $k\ge 0$. Applying it to $0^P{1}^{P}01$ pumps only zeros, so language (i) is not regular.

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This language is not regular. Intersecting with the regular language $0^{\ast }110^{\ast }$ isolates strings whose two zero blocks must have equal length; the pumping lemma then gives a contradiction.

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A word containing neither $01$ nor $10$ cannot change symbol. The language is therefore $0^{\ast }\cup 1^{\ast }$, hence is regular.

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This is Poisson regression with independent counts and log mean $\log {\mu }_d=1.5624+0.1658d$. The residual deviance $9.90$ on $12$ degrees of freedom has $p=0.625$, and the plotted curve tracks the data, so there is no evidence of poor fit.

The fitted doubling time is $\log 2/0.1658=4.16$ days. A 95% interval for the growth coefficient is $0.1658\pm 1.96(0.0166)=(0.133,0.198)$, corresponding to doubling times $(3.50,5.21)$ days. Seven days, whose rate is about $0.099$, is unsupported.

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The terms are logistic birth and crowding, $rn-r{n}^2/K$, linear predation mortality $-\delta n$, and diffusion $D{n}_{xx}$. The coast is reflecting, so $n_x(0)=0$; lethal habitat beyond $L$ gives the absorbing condition $n(L)=0$.

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Linearizing at extinction gives $n_t=(r-\delta )n+D{n}_{xx}$. The principal eigenfunction for Neumann data at $0$ and Dirichlet data at $L$ is $\cos (\pi x/(2L))$, with growth rateA population is viable exactly when this is positive, requiring $r>\delta$ and $L>(\pi /2)\sqrt{D/(r-\delta )}$. Thus intrinsic growth must beat predation and the refuge must exceed the diffusion-dependent critical size.

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A Riemann P-symbol specifies a second-order Fuchsian equation with regular singular points $a,b,c$ and the two local exponents shown beneath each point. The Fuchs relation is $\alpha +\alpha \prime +\beta +\beta \prime +\gamma +\gamma \prime =1$.

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Set $z\prime =(1-z)/2$, sending $1,-1,\infty$ to $0,1,\infty$. Then rescale the dependent variable byThis shifts the exponent pairs to $(0,m)$ at zero and $(0,-m)$ at one, while leaving the exponents at infinity unchanged.

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The hypergeometric exponent differences give $1-c=m$, $c-a-b=-m$, and exponents $a,b$ at infinity. Hence one may takeso a solution is ${}_2{F}_1(n,1-n;1-m;(1-z)/2)$, with the prefactor from part (i) when returning to the original dependent variable.

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A Lagrange top is an axially symmetric rigid body with one point fixed, centre of mass a distance $l$ along its symmetry axis, principal moments $I_1=I_2,I_3$, mass $M$, and Euler angles $\theta ,\phi ,\psi$ in gravity $g$.

The cyclic coordinates give $p_{\psi }=I_3(\dot{\psi }+\dot{\phi }\cos \theta )$ and $p_{\phi }=I_1\dot{\phi }{\sin }^2\theta +p_{\psi }\cos \theta$; the energy $E$ is the third integral. Eliminating $\dot{\phi },\dot{\psi }$ from $E$ and setting $u=\cos \theta$, with $\dot{u}=-\dot{\theta }\sin \theta$, yields ${\dot{u}}^2=f(u)$ after multiplication by $1-u^2$; all denominators cancel and $f$ is cubic.

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For $I=\frac{1}{2}\sum m_i|x_i{|}^2$, differentiation givesbecause the Newtonian potential is homogeneous of degree $-1$. Bounded positions and velocities imply $\overline{\ddot{I}}=0$, so the virial theorem gives $\overline{T}=-\overline{V}/2$.

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Observed stellar velocities estimate $T$, while visible mass and size predict $V$. Galaxies and clusters often have far more kinetic energy than luminous matter can bind under the virial theorem. The additional gravitational potential is evidence for unseen dark matter.

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Writing $\omega =e^{2\pi i/N}$, $|{\chi }_k⟩=N^{-1/2}{\sum }_j{\omega }^{-jk}|j⟩$. Therefore $S|{\chi }_k⟩={\omega }^k|{\chi }_k⟩$.

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If $|vangle={\sum }_k{c}_k|{\chi }_{k}angle$, then QFT measurement has probabilities $|c_k{|}^2$. Applying $S$ multiplies $c_k$ by the phase ${\omega }^k$, so the probabilities are unchanged.

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The offset state is $S^{x_0}$ applied to the offset-zero periodic state. Part (b) therefore makes its QFT output distribution independent of $x_0$. Explicitly, only outcomes divisible by $A=N/r$ occur, each with probability $1/r$; $x_0$ contributes only a phase.

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$N$ is a Fermat pseudoprime to base $b$ when it is composite, coprime to $b$, and $b^{N-1}\equiv 1(\mathrm{mod}N)$. Here $N\mid b^k-1$ gives coprimality and $k\mid N-1$, so $b^{N-1}=(b^k{)}^{(N-1)/k}\equiv 1(\mathrm{mod}N)$.

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Both $n$ and $m$ divide $b^{2p}-1$, hence so does $N=nm$ after checking $\gcd (n,m)=1$ under the hypothesis. Also $n\equiv p(\mathrm{mod}b-1)$ and the elementary factorization gives $N\equiv 1(\mathrm{mod}2p)$; part (a) with $k=2p$ applies. As odd primes $p$ grow, $N$ grows, yielding infinitely many examples.

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By the Chinese remainder theorem, $b^{n-1}\equiv 1(\mathrm{mod}n)$ iff it holds modulo every $p_j$. Since $n-1=p_j{r}_j-1\equiv r_j-1(\mathrm{mod}{p}_j-1)$ and the order of $b$ modulo $p_j$ divides $p_j-1$, this is equivalent to ${\mathrm{ord}}_{p_j}(b)\mid r_j-1$ for every $j$.

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For each prime $k\ge 5$, choose a prime divisor $p$ of $2^k-1$ and a prime divisor $q\ne 3$ of $2^k+1$. Their orders are $k$ and $2k$, respectively, so $p\equiv 1(\mathrm{mod}k)$, $q\equiv 1(\mathrm{mod}2k)$, and the criterion in part (c) shows $pq$ is a base-two pseudoprime. The supplied gcd fact makes examples from distinct prime $k$ distinct, so there are infinitely many.

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Multiplying the matrices inductively gives columns $(p_n,q_n{)}^T$ and $(p_{n-1},q_{n-1}{)}^T$. Their determinant is $(-1{)}^{n-1}$, and writing $x=(p_n{x}_{n+1}+p_{n-1})/(q_n{x}_{n+1}+q_{n-1})$ givesIf $A{y}^2+By+C=0$, then for rationals $p/q$ near $y$, factorization against the conjugate root gives $|p/q-y|\ge M/q^2$; rationals away from $y$ are handled by reducing $M$. Finally $q_{n+1}\ge a_{n+1}{q}_n$, so the upper bound is at most $1/(a_{n+1}{q}_n^2)$. Unbounded partial quotients contradict the fixed lower bound for a quadratic irrational.

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Differentiating the ridge objective gives $(X^{T}X+\lambda I)\hat{\beta }=X^{T}Y$, hence $\hat{\mu }=X(X^{T}X+\lambda I{)}^{-1}{X}^{T}Y=H_{\lambda }Y$.

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Since $Y-HY=(I-H)\mu +(I-H)\epsilon$, the cross term has mean zero and $H$ is symmetric. Thus the expectation is $\Vert (I-H)\mu {\Vert }^2+{\sigma }^2\mathrm{tr}(I-H{)}^2$, which expands to the stated expression.

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Independence gives $\mathbb{E}\Vert Y^{\ast }-HY{\Vert }^2=\Vert (I-H)\mu {\Vert }^2+n{\sigma }^2+{\sigma }^2\mathrm{tr}{H}^2$. Comparing with part (b) shows the difference is $2{\sigma }^2\mathrm{tr}H$, proving unbiasedness.

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Here $H=hI$, $h=(1+\lambda {)}^{-1}$, and prediction risk isIt decreases then increases, with $h^{\ast }=\Vert \mu {\Vert }^2/(\Vert \mu {\Vert }^2+n{\sigma }^2)$, equivalently ${\lambda }^{\ast }=n{\sigma }^2/\Vert \mu {\Vert }^2$. Its minimum is $n{\sigma }^2+n{\sigma }^2\Vert \mu {\Vert }^2/(\Vert \mu {\Vert }^2+n{\sigma }^2)$; the endpoint limits are $2n{\sigma }^2$ and $n{\sigma }^2+\Vert \mu {\Vert }^2$.

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This bistable reaction-diffusion equation models an invading population with an Allee effect: densities below $r$ decline, while those above it grow toward carrying capacity one.

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The homogeneous equilibria are $0,r,1$. Since $f\prime (0)=-r<0$, $f\prime (1)=r-1<0$, and $f\prime (r)=r(1-r)>0$, the first and last are linearly stable and the middle one unstable.

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Multiplying $Dn\prime \prime +f(n)=0$ by $n\prime$ and using the endpoint conditions gives the logistic frontIt has $n_0(0)=1/2$ and direct differentiation verifies the equation for $r=1/2$.

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With $\xi =x-ct$, the equation is $Du\prime \prime +cu\prime +f(u)=0$. Since $f_r(u)-f_{1/2}(u)=u(1-u)(1/2-r)$ and $n_0\prime =-n_0(1-n_0)/\sqrt{2D}$, the same profile works with

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The speed is positive for $r<1/2$, zero at $1/2$, and negative for $r>1/2$. The state with the larger basin-weighted reaction potential invades the other; raising the Allee threshold favours extinction.

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The Poisson bracket is $\{f,g\}={\sum }_i(f_{q_i}{g}_{p_i}-f_{p_i}{g}_{q_i})$. Canonical variables satisfy $\{Q_i,Q_j\}=\{P_i,P_j\}=0$ and $\{Q_i,P_j\}={\delta }_{ij}$.

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To first order the conditions are ${\partial }_{p_j}\delta q_i-{\partial }_{p_i}\delta q_j=0$, ${\partial }_{q_j}\delta p_i-{\partial }_{q_i}\delta p_j=0$, and ${\partial }_{q_j}\delta q_i+{\partial }_{p_i}\delta p_j=0$. The choices $\delta q_i=\epsilon F_{p_i}$ and $\delta p_i=-\epsilon F_{q_i}$ satisfy them by equality of mixed partials.

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Taylor expansion gives $\delta H=H_q\cdot \delta q+H_p\cdot \delta p=-\epsilon \{F,H\}$. Hamilton's equation gives $\dot{F}=\{F,H\}$, so $F$ is conserved exactly when the generated infinitesimal canonical transformation leaves $H$ invariant.

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For a constant translation vector $a$, take $F=a\cdot {\sum }_i{p}_i$. Then $\delta r_i=\epsilon a$ and $\delta p_i=0$. The potential depends only on position differences and the kinetic energy only on momenta, so this is a symmetry and total momentum is conserved.

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Axiom of foundation says every nonempty set $A$ contains $a$ with $a\cap A=\varnothing$. Define $T_0=x$, $T_{n+1}=\bigcup T_n$, and $\mathrm{TC}(x)={\bigcup }_{n<\omega }{T}_n$ (including $x$ if that convention is desired). Replacement and union form this set, and it is transitive; induction shows every transitive set containing $x$ contains it.

The principle of membership induction says that if $(\forall y\in xP(y))\Rightarrow P(x)$ for every $x$, then $P$ holds for every set. Otherwise Foundation applied to the set of counterexamples in a suitable transitive closure gives a minimal counterexample.

Apply induction to $P(x):F(x)=x$. If it holds for all $y\in x$, then extensionality and preservation plus surjectivity give $z\in F(x)$ iff $z=F(y)$ for some $y\in x$, iff $z\in x$. Hence $F(x)=x$.

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Hall marriage theorem states that a bipartite graph with parts $A,B$ has a matching saturating $A$ iff $|N(S)|\ge |S|$ for every $S\subseteq A$. Starting with a maximum matching, an unmatched vertex and its alternating reachable set would violate Hall unless an augmenting path exists; flipping along that path increases the matching, proving sufficiency.

Every vertex cover meets each edge of a matching, so $\gamma \le \beta$. Endpoints of a maximal matching cover every edge, so $\beta \le 2\gamma$. A disjoint union of $k$ triangles has $(\gamma ,\beta )=(k,2k)$. For $T_3(30)$, $\gamma =15$ and $\beta =30-10=20$. In bipartite graphs the alternating-path proof constructs a cover of size equal to a maximum matching, giving König theorem.

The chromatic index is the minimum number of matchings partitioning the edges. Label vertices of $K_{2^r}$ by the vector space over ${\mathbb{F}}_2$; for each nonzero $a$, pair $x$ with $x+a$. These $2^r-1$ perfect matchings partition all edges, while degree gives the matching lower bound, so $\chi \prime =n-1$.

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An action by field automorphisms is a homomorphism $Go\mathrm{Aut}(L)$. Elements fixed by every group element are closed under addition, multiplication, additive inverses, and inversion of nonzero elements, so $L^G$ is a subfield.

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The orbit polynomial ${\prod }_{\beta \in G\alpha }(X-\beta )$ has coefficients fixed by $G$, lies in $K[X]$, and has distinct roots. Thus $\alpha$ is separable and its degree is the orbit size $|G:G_{\alpha }|$, which divides $|G|$. If $L=K(\alpha )$ and the action is faithful, only the identity fixes $\alpha$, so the degree is $|G|$.

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The relations ${\sigma }^n={\tau }^2=1$ and $\tau \sigma \tau ={\sigma }^{-1}$ give the dihedral group; its $2n$ substitutions $z\mapsto {\zeta }_n^{j}z$ and $z\mapsto {\zeta }_n^j/z$ are distinct. The element $w=z^n+z^{-n}$ is fixed. Since $z$ satisfies $X^{2n}-w{X}^n+1=0$, the extension has degree at most $2n$, while the automorphism group gives degree at least $2n$. Hence $L^G=\mathbb{C}(w)$.

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$S^1=\{z\in \mathbb{C}:|z|=1\}$. A continuous finite-dimensional representation is unitary after averaging an inner product over Haar measure, and commuting unitary matrices diagonalize simultaneously. Its irreducibles are therefore $z\mapsto z^n$, $n\in \mathbb{Z}$. Their characters obey ${\int }_{S^1}{z}^n\overline{z^m}dz={\delta }_{nm}$.

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$SU(2)$ consists of unitary $2\times 2$ matrices of determinant one. Every element is unitarily diagonalizable with eigenvalues $e^{\pm i\theta }$, and two are conjugate exactly when they have the same trace. Thus $t(g)=\mathrm{tr}(g)/2=\cos \theta$ bijects conjugacy classes with $[-1,1]$.

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For $-1<x<1$, the centralizer of the diagonal representative is the maximal torus $T$. Hence the orbit is $SU(2)/T$. Identifying $SU(2)$ with $S^3$ makes this quotient the Hopf quotient $S^3/S^1\cong S^2$.

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If $V$ is the defining representation, ${\chi }_V=2t$. The character of ${\mathrm{Sym}}^{3}V$ is $(2t{)}^3-2(2t)=8t^3-4t$. Therefore $8t^3$ is the character of ${\mathrm{Sym}}^{3}V\oplus V\oplus V$.

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${\mathcal{O}}_K=\mathbb{Z}[\sqrt{30}]$ and $d_K=120$, so every class has an ideal of norm at most $\sqrt{30}$. Factoring the primes $2,3,5$ and checking principal norms shows that the ramified prime ideals represent one nontrivial class of order two; hence $\mathrm{Cl}(K)\cong \mathbb{Z}/2$. The continued fraction of $\sqrt{30}$ gives the least unit greater than one as $11+2\sqrt{30}$, of norm $1$.

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The two-sign map $K^{\ast }\to \{\pm 1{\}}^2$ is surjective, with kernel the totally positive elements, so their index is four. Narrow equivalence is reflexive, symmetric, and transitive because totally positive elements form a group; ideal multiplication makes the classes an abelian group.

The map from the narrow class group to the ordinary class group is onto. Its kernel records sign patterns of principal generators modulo signs realized by units. A norm-$-1$ fundamental unit realizes the two mixed signs, making the kernel trivial; otherwise only equal signs occur and the kernel has order two. Thus the narrow class number is $h_K$ in the first case and $2h_K$ otherwise. Here it is $4$.

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For a finite triangulation, the Euler characteristic is $\chi (X)=V-E+F$, independent of the triangulation.

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Choose a triangulation of the sphere having every branch point as a vertex. Its lift has twice every edge and face, and twice every nonbranch vertex, but only one vertex above each of the $n$ branch points. HenceSince $\chi ({\Sigma }_g)=2-2g$, this gives $n=2g+2$, the degree-two Riemann-Hurwitz formula.

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For fixed $y$, the map $x\mapsto ⟨Tx,y⟩$ is a bounded functional. The Riesz representation theorem gives a unique vector $T^{\ast }y$ representing it. Linearity, uniqueness, and $\Vert T^{\ast }y\Vert \le \Vert T\Vert \Vert y\Vert$ make $T^{\ast }$ the unique bounded adjoint operator.

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A countable orthonormal basis is a frame with $A=B=1$. For any frame, a vector orthogonal to every $e_i$ makes the lower frame bound force its norm to vanish. Therefore the span is dense.

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The upper frame bound gives $\Vert Ux{\Vert }_{{\ell }^2}\le \sqrt{B}\Vert x\Vert$. For $c\in {\ell }^2$, the adjoint is the norm-convergent synthesis seriesbecause its inner product with $x$ is ${\sum }_i{c}_i⟨e_i,x⟩$.

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Given $\epsilon >0$, choose $N$ with ${\sum }_{i>N}|⟨a,e_i⟩{|}^2<{\epsilon }^2$. Every cube point has uniformly bounded tail by this sum, while its first $N$ coordinates lie in a compact finite-dimensional box. A finite net for that box is therefore an $\epsilon$-net for the cube. The coordinate inequalities define a closed set, so the cube is complete and totally bounded, hence compact.

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Strong convergence means $\Vert f_j-f{\Vert }_p\to 0$; weak convergence means $\int f_{j}h\to \int fh$ for every $h\in L^q$. By duality choose $h$ of norm one with $\int fh$ arbitrarily close to $\Vert f{\Vert }_p$. Holder inequality and passage to the limit give $\Vert f{\Vert }_p\le \mathrm{liminf}\Vert f_j{\Vert }_p$.

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Because $1<p<\infty$, $L^p$ is reflexive, so a bounded sequence has a weakly convergent subsequence $g_{j_k}\rightharpoonup g$. Thenand weak convergence handles $\int g_{j_k}h$. Thus the displayed limit holds (with the evident OCR correction $\int gh$ on the right).

Strong convergence of the tests is essential: in $L^2(0,2\pi )$ take the same orthonormal sequence for $g_j$ and $h_j$. Both converge weakly to zero, but $\int g_j{h}_j=1$.

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Choose a nonconstant rational function $t$ on $C$. Functions in $L(D)$ are integral over a finite-dimensional bounded-pole space over $k(t)$; equivalently, evaluation of sufficiently many principal parts embeds $L(D)$ into a finite-dimensional vector space. Hence $L(D)$ is finite dimensional.

If $E=D+\mathrm{div}(h)$, multiplication by $h$ gives the isomorphism $L(E)\to L(D)$. A canonical divisor is the divisor of a nonzero rational differential. Riemann-Roch theorem says $\ell (D)-\ell (K-D)=\mathrm{deg}D+1-g$. Taking $D=0$ gives $1-g=1+g-\mathrm{deg}K$, hence $\mathrm{deg}K=2g-2$.

For a smooth plane curve of degree $d$, the adjunction formula gives $K_C\sim (d-3)H{|}_C$. A cubic has $d=3$, so $K_C\sim 0$.

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The geodesic curvature is the tangential component of $\alpha \prime \prime$, namely $k_g=⟨\alpha \prime \prime ,N\times \alpha \prime ⟩$ up to orientation; a curve is geodesic when $k_g=0$. Gauss-Bonnet theorem for a compact region says $\int K+{\int }_{\partial }{k}_g+\sum \text{exterior angles}=2\pi \chi$.

For $K=0$, a flat cylinder has infinitely many parallel closed geodesics.

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For $K>0$ the number can again be infinite: remove two points away from an infinite family of great circles on the round sphere. The result is a cylinder of positive curvature containing those intersecting simple closed geodesics.