Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 3

$A={\mathbb{Z}}^2/⟨(6,9)⟩$. The class of $(2,3)$ has order $3$, so $A$ has torsion and is not free. A unimodular change of basis (Smith normal form, using $\gcd (6,9)=3$) gives $A\cong \mathbb{Z}\oplus \mathbb{Z}/3\mathbb{Z}$.

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The subspace opens are $A\cap U$; quotient opens are those whose inverse image is open; a homeomorphism is a continuous bijection with continuous inverse. Here $q(x,y)=(|x|,y)$ identifies exactly the required pairs and induces a homeomorphism $Y\to [0,1]\times [-1,1]$. An affine rescaling maps that rectangle homeomorphically onto $S$.

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Integrate $(1+z^p{)}^{-1}$ around the wedge $0\le \arg z\le 2\pi /p$. The radial integrals differ by the factor $e^{2\pi i/p}$, the arc vanishes, and the wedge contains the simple pole $e^{i\pi /p}$ with residue $-[p{e}^{-i\pi /p}{]}^{-1}$. Solving the resulting identity gives $I=(\pi /p)\csc (\pi /p)$.

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For fixed boundary values, integration by parts gives $L_u-{\partial }_x{L}_{u_x}-{\partial }_y{L}_{u_y}=0$. For $L=u^2+u_x{u}_y$, this is $2u-2u_{xy}=0$, or $u_{xy}=u$, with the independently prescribed boundary data $u=xy$ on the circle.

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Integration by parts gives ${\tilde{g}}_{\sigma }\prime (k)=-{\sigma }^{2}k{\tilde{g}}_{\sigma }(k)$ and ${\tilde{g}}_{\sigma }(0)=1$, so ${\tilde{g}}_{\sigma }=e^{-{\sigma }^2{k}^2/2}$. The convolution theorem says $\widetilde{f\ast g}=\tilde{f}\tilde{g}$; multiplying the two Gaussian transforms and inverting yields $g_{\sqrt{{\sigma }_1^2+{\sigma }_2^2}}$.

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$L_a={\epsilon }_{abc}{x}_b{p}_c$, with $[x_a,p_b]=i\hslash {\delta }_{ab}$ and position-position and momentum-momentum commutators zero. Expansion gives $[L_a,L_b]=i\hslash {\epsilon }_{abc}{L}_c$ and $[L_a,p_d]=i\hslash {\epsilon }_{adc}{p}_c$. Thus $[L_1,p_2]=i\hslash p_3\ne 0$, so they are not simultaneously diagonalisable.

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Unsteady Bernoulli gives the surface pressure from $p+\rho {\partial }_t\varphi +\rho |\nabla \varphi {|}^2/2=C(t)$. The velocity-squared contribution integrates to zero net force, while the unsteady term gives $F=-m_a\dot{V}$ with added mass $m_a=\frac{1}{2}(4\pi \rho a^3/3)=2\pi \rho a^3/3$. Hence $(m+m_a)a=G$, so $m^{\ast }=m+2\pi \rho a^3/3$: accelerating the bubble also accelerates surrounding fluid.

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The communicating classes are $\{1,2,3,6\}$ and $\{4,5\}$; only the latter is closed. Let $h_i=P_i(T_2<\infty )$, with $h_2=1$ and $h_4=h_5=0$. The first-step equations give $h_3=1/2$, $h_6=(h_1+h_3)/2$, and $h_1=1/3+h_3/3+h_6/6$, hence $h_1=13/22$.

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$V^{\ast }$ is the space of linear functionals and the dual basis satisfies $v_i^{\ast }(v_j)={\delta }_{ij}$. Nondegeneracy and equal finite dimensions make $u\mapsto ⟨u,\cdot ⟩$ an isomorphism; evaluation similarly identifies $V$ with $V^{\ast \ast }$. Riesz representation applied to $v\mapsto ⟨Tv,u⟩$ gives the unique adjoint. If $T$ is diagonalizable, declare an eigenbasis orthonormal to make it self-adjoint. Conversely, for self-adjoint $T$ and invariant $W$, $⟨Tv,w⟩=⟨v,Tw⟩=0$ shows $W^{\perp }$ invariant.

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Generators define a surjection $R^k\to M$, and conversely images of the standard basis generate any quotient. Writing $\varphi (m_i)={\sum }_j{a}_{ij}{m}_j$ with $a_{ij}\in I$ and applying the adjugate to $XI-A$ gives a monic annihilating polynomial whose lower coefficients lie in $I$. Taking $\varphi =1$ when $IM=M$ yields $xM=0$ with $x-1\in I$; for $M=\mathbb{Z}/3$ and $I=(2)$ take $x=3$. The Jacobson radical criterion follows by placing a nonunit $1-rs$ in a maximal ideal. It yields Nakayama’s lemma. Applying the determinant trick to preimages of a finite generating set constructs a polynomial right inverse to any surjective endomorphism, proving injectivity.

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A norm is positive definite, homogeneous, and subadditive; equivalence means mutual bounds by positive constants. Finite-dimensional norm equivalence gives corresponding constants for operator norms, whose $k$th roots tend to one, so $\rho (A)$ is norm-independent. If $\Vert A\Vert <1$, Picard iteration gives $(I-A{)}^{-1}={\sum }_{k\ge 0}{A}^k$. If $\rho (A)<1$, choose $\rho (A)<\alpha <1$ and define $\Vert v{\Vert }_{\alpha }={\sup }_{m\ge 0}{\alpha }^{-m}\Vert A^{m}v\Vert$; it is finite, equivalent to the original norm, and $\Vert Av{\Vert }_{\alpha }\le \alpha \Vert v{\Vert }_{\alpha }$.

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A base covers the space and refines intersections. On $X^k$, products of opens form a base. The weighted metric satisfies the metric axioms and has exactly this topology; convergence is coordinatewise. For $X^{\mathbb{N}}$ the metric $d(f,g)={\sum }_{n\ge 1}{2}^{-n}\min \{1,\rho (f(n),g(n))\}$ induces the stated product topology: finitely many coordinates control a basic neighbourhood and the tail is uniformly small. Thus sequences converge exactly coordinatewise.

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Expanding the derivative limit along real and imaginary increments gives $u_x=v_y$ and $u_y=-v_x$. Since $\mathrm{Re}(z{e}^z)=e^x(x\cos y-y\sin y)$, the required function is $f(z)=z{e}^z$, and $f(0)=0$. Two such functions differ by a holomorphic function with zero real part; the open mapping theorem makes it constant, and the value at zero makes that constant zero.

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Away from $x=y$ the Newton kernel is harmonic, while integrating its normal derivative over a small sphere gives one, proving ${\nabla }^2\hat{G}=\delta$. Green’s second identity is ${\int }_V(u{\nabla }^{2}v-v{\nabla }^{2}u)={\int }_{\partial V}(u{\partial }_{n}v-v{\partial }_{n}u)$. For the half-space, images give $G(x,y)=-(4\pi |x-y|{)}^{-1}+(4\pi |x-y^{\ast }|{)}^{-1}$. Substitution in the boundary formula yields the Poisson kernel $z/[2\pi ((x_1-y_1{)}^2+(x_2-y_2{)}^2+z^2{)}^{3/2}]$; polar integration against the Gaussian gives exactly the stated one-dimensional integral.

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With $P^{\mu }=m{U}^{\mu }$, $U^{\mu }=d{x}^{\mu }/d\tau$, and $F^{\mu \nu }$ the field tensor, separating temporal and spatial components yields the Lorentz force and power equations. The invariants are $E\cdot B$ and $E^2-c^2{B}^2$. For perpendicular fields with $E<cB$, boost with $v=E\times B/B^2$ (here $-E\hat{y}/B$); then $E\prime =0$ and $B\prime =\sqrt{B^2-E^2/c^2}\hat{z}$. Perpendicular motion is circular with $\omega =|q|B\prime /(\gamma m)$ and radius $p/(|q|B\prime )$. Transforming back adds the frame velocity, giving the nonrelativistic $E\times B/B^2$ drift, independent of $m,q$; a parallel velocity would additionally produce a helix.

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Momentum balance gives $\rho u_t=-p_x+\mu u_{yy}$. Steadiness forces $p_x=-G$ and no slip gives $u=Gy(H-y)/(2\mu )$, hence $Q=G{H}^3/(12\mu )$. After removal, separation gives

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Expanding the order condition about $w=1$ yields $\rho (w)={\sigma }_s{\sum }_{l=1}^s{w}^{s-l}(w-1{)}^l/l$ and ${\sigma }_s=(\sum 1/l{)}^{-1}$. Thus BDF2 has $\rho =w^2-4w/3+1/3$, ${\sigma }_2=2/3$, and BDF3 has $\rho =w^3-18w^2/11+9w/11-2/11$, ${\sigma }_3=6/11$. Their first characteristic polynomials satisfy the root condition, so consistency plus Dahlquist equivalence gives convergence. For BDF2 the stability boundary $z=\rho (e^{i\theta })/({\sigma }_2{e}^{2i\theta })$ has nonnegative real part; hence the whole left half-plane is stable.

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Neyman–Pearson rejects for large $g(X)/f(X)$ and is most powerful at its size. For the normal sample, monotone likelihood ratio makes the UMP test reject when $\sqrt{n}\bar{X}>{\Phi }^{-1}(1-\alpha )$; its rejection probability increases in $\mu$, so it also has size $\alpha$ for the composite null $\mu \le 0$. A likelihood-ratio test against a mixture null that has size $\alpha$ under each component bounds every competing test’s mixture power and is therefore UMP for the original two-point null. In the drug problem, the symmetric interval test has equal size $\alpha$ at $\mu =\pm {\mu }_0$ and no larger size farther out; it is the NP test against the equal mixture of those boundary laws, hence is UMP.

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Gradient descent is $x_{t+1}=x_t-\eta \nabla f(x_t)$. Integrating the Hessian along a segment gives the descent lemma. Applying it to the hinted $z$ and convexity yields cocoercivity: $(\nabla f(x)-\nabla f(y){)}^T(x-y)\ge \Vert \nabla f(x)-\nabla f(y){\Vert }^2/\beta$. Apply this to $\varphi =f-\alpha \Vert x{\Vert }^2/2$, whose Hessian lies between $0$ and $(\beta -\alpha )I$, and rearrange to obtain the displayed strengthened inequality. With $y=x^{\ast }$, $\nabla f(x^{\ast })=0$, and $\eta =2/(\alpha +\beta )$, expansion of the squared update gives contraction factor $((\beta -\alpha )/(\beta +\alpha ){)}^2$ per step and the stated bound.

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