Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 4

The dual space $V^{\ast }$ is the vector space of linear maps $V\to \mathbb{F}$. If $B=(v_1,\dots ,v_n)$, its dual family satisfies $v_i^{\ast }(v_j)={\delta }_{ij}$. Every functional $f$ obeysso the family spans; evaluation on each $v_j$ proves linear independence. Thus it is a basis without any prior dimension argument.

For $p=a+bt+c{t}^2$,The basis dual to $(f_0,f_1,f_2)$ isas direct substitution gives $f_i(p_j)={\delta }_{ij}$. Hence the three functionals form a basis of $P_2^{\ast }$.

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A contraction satisfies $d(Tx,Ty)\le qd(x,y)$ for some $0\le q<1$. Starting from $x_0$, define $x_{k+1}=T{x}_k$. The geometric bound on successive distances makes $(x_k)$ Cauchy, so completeness gives a limit $x$. Continuity of $T$ gives $Tx=x$. If $Ty=y$, then $d(x,y)\le qd(x,y)$, hence $x=y$. This is the contraction mapping theorem.

If $T^n$ is a contraction, it has a unique fixed point $x$. Since $T^n(Tx)=T(T^{n}x)=Tx$, the point $Tx$ is also fixed by $T^n$, so uniqueness gives $Tx=x$. Every fixed point of $T$ is fixed by $T^n$, so it too must equal $x$. Thus $T$ has exactly one fixed point.

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The local maximum modulus principle says that if $|f|$ has a local maximum at an interior point of a connected domain, then $f$ is constant. On a small circle about the maximum, the mean-value property andforce equality everywhere. Equality in the triangle inequality makes the boundary values identical; Cauchy's formula then makes $f$ constant locally, and the identity theorem makes it constant on the domain.

Write $f=u+iv$. The hypothesis gives $u-v\le 0$. Thereforewith equality at zero. The local maximum modulus principle makes the exponential constant. Differentiation then gives $f^{\prime }=0$, so $f$ is constant; since $f(0)=0$, it is identically zero.

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Separation with Dirichlet walls givesFor $a<b<c$, the ground state $(1,1,1)$ has a nondegenerate energy. The cheapest excitation changes the quantum number in the longest direction, so the first excited state $(1,1,2)$ also has a nondegenerate energy.

If $a<b=c$, the ground-state energy remains nondegenerate, but $(1,2,1)$ and $(1,1,2)$ have equal first-excited energy, giving degeneracy two.

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In vacuum, curl Faraday's law and use $\nabla \cdot E=0$ and the Ampère-Maxwell equation to obtainThe plane wave solves this when $\omega =c|k|$ and $k\cdot E_0=0$. Faraday's law gives

For the stated wave, propagation is along $+x$ and polarization is $(0,1,1)/\sqrt{2}$. ThusThe Poynting vector is electromagnetic energy flux, and its average is the wave intensity.

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Exactness for $1$ and $x$ givesHence

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A two-node Gaussian rule is exact through degree three, the maximal degree $2n-1$. Its nodes are the zeros of the monic quadratic orthogonal to $1,x$ for weight $x$ on $[0,1]$. Writing it as $p=x^2+ux+v$ givesso $u=-6/5$, $v=3/10$. Therefore

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The communicating classes are $\{1,3\}$ and $\{2,4\}$. The first is open because state 3 can enter 4 and cannot return; the second is closed.

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On the closed class, the transition matrix isIts stationary distribution is $(3/11,8/11)$. The class is irreducible and aperiodic because of its self-loops, while the other class is transient and enters it almost surely. Hence

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The symmetric bilinear form associated with a real quadratic form $Q$ is obtained by polarization identity:In coordinates $Q(x)=x^{T}Ax$, replacing $A$ by its symmetric part does not change $Q$, and the formula gives $phi(u,v)=u^{T}Av$; this proves existence. A symmetric form is positive semidefinite when $phi(v,v)\ge 0$ for every $v$, and positive definite when the inequality is strict for every $v\ne 0$.

The diagonalization theorem for real quadratic forms says that some basis puts any symmetric form intowith $r$ further zero coordinates. Sylvester's law of inertia says that $(p,q,r)$ is independent of the diagonalizing basis. To prove this, let $P$ be the span of the positive coordinate vectors and let $N\oplus Z$ be the span of the negative and zero vectors. If $P^{\prime }$ is positive definite for another diagonalization and $\dim P^{\prime }>p$, thenso the two spaces intersect nontrivially. A vector in the intersection would have both positive and nonpositive square, a contradiction. Thus $p^{\prime }\le p$; symmetry gives $p=p^{\prime }$. Applying the same argument to $-\varphi$ gives $q=q^{\prime }$, and then $r=r^{\prime }$.

For the nondegenerate form on $V$, write its inertia as $(p,q,0)$, so $p+q=2n$. The restriction vanishes identically on $E$: polarization gives $phi(u,v)=0$ for $u,v\in E$. Projection of $E$ to the positive coordinate space is injective, since a vector with zero positive projection cannot be isotropic unless it is zero. Hence $k\le p$; projection to the negative space likewise gives $k\le q$. Therefore $k\le \min (p,q)\le n$.

The matrix of $l^2$ is $l{l}^T$. If $l\ne 0$, it has rank one and inertia $(1,0,n-1)$, hence signature one; if $l=0$, its rank and signature are zero. The coefficientwise product $(l^2,s^2)$ has matrix $t{t}^T$, where $t_i=l_i{s}_i$, so it has the same rank-one conclusion when $t\ne 0$ and is zero otherwise.

Finally, diagonalization writes every positive semidefinite form as a sum of squares, say $f={\sum }_a{l}_a^2$ and $g={\sum }_b{s}_b^2$. Bilinearity of coefficientwise multiplication givesa sum of positive semidefinite rank-at-most-one forms. Thus $(f,g)$ is positive semidefinite.

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If $M$ is irreducible and $m\ne 0$, the image $Rm$ is a nonzero submodule, hence $Rm=M$ and the map is onto. Conversely, if every nonzero $m$ is cyclic and $N\le M$ is nonzero, choose $m\in N\setminus \{0\}$. Then $M=Rm\subseteq N$, so $N=M$ and $M$ is irreducible.

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The annihilator $I={\mathrm{Ann}}_R(M)$ is an ideal. For $r\notin I$, choose $m$ with $rm\ne 0$. Irreducibility and part (i) give $s\in R$ such that $srm=m$. The element $(sr-1)$ therefore kills the generator $m$, and hence all of $M$; thus $sr-1\in I$. Every nonzero class in $R/I$ consequently has an inverse, so $R/I$ is a field.

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Make $V$ a $k[x]$-module by $xv=\phi (v)$. The structure theorem over the Euclidean domain $k[x]$ decomposes its torsion module aswhere the $p_{\alpha }$ are monic irreducibles. Each summand is indecomposable: its submodules form a chain, so two nonzero submodules cannot form a direct sum. This is the desired decomposition into invariant indecomposable subspaces.

The characteristic polynomial iswhereas the minimal polynomial is the least common multiple

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The prime ideals of $\mathbb{R}[x]$ are $(0)$ and the maximal ideals generated by the monic irreduciblesFor a nonzero prime $I=(p)$, use the residue classes of $1,x,\dots ,x^{N-1}$, where $N=n\mathrm{deg}p$. Multiplication by $x$ on $\mathbb{R}[x]/(p^n)$ is represented by the companion matrix of the monic polynomial $p^n$. In particular, for $p=x-a$ this may equivalently be written as one size-$n$ Jordan block with eigenvalue $a$. The ideal $(0)$ does not give a finite-dimensional quotient.

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The upper-left block has characteristic and minimal polynomial $(x^2+1{)}^2$, while the lower-right block has characteristic and minimal polynomial $x^2+1$. Hence the corresponding module isIn the power bases from part (ii), an explicit normal form isthe direct sum of the companion matrices of $(x^2+1{)}^2$ and $x^2+1$.

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A map $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$ is differentiable at $x$ if there is a linear map $L$ such thatthen $Df{|}_x=L$. The inverse function theorem says that if $f$ is continuously differentiable near $x$ and $Df{|}_x$ is invertible, then $f$ restricts to a $C^1$ diffeomorphism between neighborhoods of $x$ and $f(x)$.

Since $F(A)=A^{T}A$ is polynomial,Thus $\ker DF{|}_I=T$, the space of skew-symmetric matrices.

DefineThen $D\Phi {|}_I(H)=2H$, so the inverse function theorem supplies open neighborhoods $I\in U$, $0\in V$ on which $\Phi :U\to V$ is a $C^1$ diffeomorphism. Its symmetric part is $A^{T}A-I$ and its skew part is $A-A^T$. Consequentlyand therefore $\Phi (U\cap \mathcal{O})=V\cap T$.

For any $R\in \mathcal{O}$, left multiplication $A\mapsto R^{T}A$ is a homeomorphism preserving $\mathcal{O}$ and carrying $R$ to $I$. Transporting the preceding chart gives a neighborhood of $R$ in $\mathcal{O}$ homeomorphic to an open subset of $T$.

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The disc model isand the upper half-plane model isThe Cayley mapmaps $D$ bijectively to $\mathfrak{h}$. Since $C^{\prime }(z)=2i/(1-z{)}^2$ and $\mathrm{Im}C(z)=(1-|z{|}^2)/|1-z{|}^2$, direct substitution gives $C^{\ast }{g}_{\mathfrak{h}}=g_D$.

Representing $C$ by $K=\left(\begin{array}{cc}i& i\\ -1& 1\end{array}\right)$, the disc isometry corresponding to $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ is represented, up to a nonzero scalar, by

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A hyperbolic triangle is bounded by three hyperbolic geodesic segments or rays. Its vertices may lie in the hyperbolic plane; an ideal vertex is their endpoint on the boundary at infinity. Every ideal angle is zero. Gauss-Bonnet theorem with curvature $-1$ givesso an all-ideal triangle has area $\pi$.

For fixed admissible angles, hyperbolic trigonometry determines all three side lengths from the angles, for exampleThus two such triangles are congruent. An orientation-preserving isometry can send one chosen vertex and oriented tangent to the corresponding data of the other, and the determined side lengths and angles then send the entire triangle to it. Since the orientation-preserving isometry group is $PS{L}_2(\mathbb{R})$, represented by $S{L}_2(\mathbb{R})$, the stated action is transitive.

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The Heaviside factor restricts the integral to $t\ge \alpha$. With $u=t-\alpha$,

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Split the transform into periods and translate each interval:Summing the geometric series yields

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For one period,Part (b), followed by cancellation of $1+e^{-\pi s}$, gives

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The Euler--Lagrange equation isPutting $u=\cos (\varphi /2)$ and simplifying gives

Since $1-\cos \varphi =2(1-u^2)$, $1+\cos \varphi =2u^2$, and ${\dot{u}}^2=(1-u^2){\dot{\varphi }}^2/4$, the transformed functional isIts Euler--Lagrange equation is $\ddot{u}+(g/4a)u=0$, exactly the preceding equation.

For endpoint-vanishing $\eta$, the second variation isWriting $\eta ={\sum }_{n\ge 1}{c}_n{e}_n$ and using orthonormality givesIt is positive definite when $T<\pi /\omega$ and has a negative $e_1$ direction when $T>\pi /\omega$. Therefore

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Spatial Fourier transformation givesHenceInverting and splitting the cosine into exponentials, or applying D'Alembert formula, yieldsThus the initial Gaussian separates into two half-amplitude Gaussian pulses travelling without distortion at speeds $c$ and $-c$.

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Here $L_3=-i\hslash (x_1{\partial }_{x_2}-x_2{\partial }_{x_1})$. The radial factor and $x_3^n$ are annihilated by the angular derivative, whileThe product rule therefore givesReplacing $i$ by $-i$ gives eigenvalue $-m\hslash$ for $(x_1-i{x}_2{)}^m{x}_3^{n}f(r)$.

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Set ${\xi }_j=\sqrt{m\omega /\hslash }{x}_j$. Separation into three one-dimensional oscillators givesThe ground state isThe number of triples of nonnegative integers summing to $N$ iswhich is the degeneracy of $E_N$.

Finally,is a pure level-$N=2$ oscillator state: the constant terms in the two degree-two Hermite contributions cancel. Part (i) gives $L_3\chi =2\hslash \chi$, while $N=2$ gives $H\chi =(7/2)\hslash \omega \chi$.

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In the fluid region $y>\eta (x)$, incompressibility and potential flow giveNo penetration through the exact surface gives

The condition $hk\ll 1$ says the hill slope is small; $|\nabla \varphi |\ll U$ says the disturbance velocity is small compared with the background wind. Dropping the product ${\varphi }_x{\eta }_x$ and Taylor-shifting the boundary from $y=\eta$ to $y=0$ therefore givesThe decaying solution is

Bernoulli's equation isTo first order on the surface,ThusEquivalently, crest minus trough is the negative of this. For $k{U}^2/g\ll 1$, hydrostatic elevation dominates; for $k{U}^2/g\gg 1$, the Bernoulli pressure drop caused by faster crest flow dominates.

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The unrestricted maximum-likelihood estimates are $(\bar{X},\bar{Y})$, while the null fixes both means at zero. HenceUnder the null this is ${\chi }_2^2$, so the size-$\alpha$ generalized likelihood-ratio test rejects exactly when

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Under the null the common-mean estimate is $(m\bar{X}+n\bar{Y})/(m+n)$. Completing squares giveswhereunder the null. The test rejects when

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Put $u=\sqrt{m}\bar{X}$ and $v=\sqrt{n}\bar{Y}$. Test (a) accepts inside the discwhereas $Z$ is the projection of $(u,v)$ onto the unit vectorIf $q_1=[{\Phi }^{-1}(1-\alpha /2){]}^2$, then $q_1$ is the corresponding ${\chi }_1^2$ quantile. Since a ${\chi }_2^2$ variable is stochastically larger than a ${\chi }_1^2$ variable, $q_2>q_1$. Choose a point on that unit-vector line with squared radius strictly between $q_1$ and $q_2$. A whole open neighborhood then makes (b) reject while (a) accepts. The joint normal density of $(\bar{X},\bar{Y})$ is strictly positive everywhere for every true $(\lambda ,\mu )$, so this neighborhood always has positive probability.

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Introduce unrestricted row and column potentials $u_i,v_j$. The dual isPrimal and dual feasible solutions are optimal precisely when complementary slackness holds:Indeed, the primal--dual objective gap is ${\sum }_{ij}{x}_{ij}(c_{ij}-u_i-v_j)$, a sum of nonnegative terms.

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One of the $m+n$ balance equations is redundant, and the remaining $m+n-1$ have full rank. A nondegenerate basic feasible solution therefore has exactly $m+n-1$ positive variables. Equivalently, its positive cells form a spanning tree of the complete bipartite row--column graph.

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The northwest-corner rule constructs an initial basic feasible solution: fill the current cell with the smaller remaining supply and demand, delete the exhausted row or column, and continue.

For a current spanning-tree basis, solve $u_i+v_j=c_{ij}$ on its occupied cells, fixing one potential to zero. The reduced cost of an unoccupied cell isIf all reduced costs are nonnegative, part (a) proves optimality. Otherwise choose a cell with negative reduced cost. Adding its edge to the tree creates a unique even cycle. Mark its cells alternately $+$ and $-$, starting with $+$ at the entering cell, and setAdd $\theta$ on the plus cells and subtract it on the minus cells. Row and column totals are unchanged, the entering cell becomes positive, and a minimizing minus cell leaves the basis. In the nondegenerate case the objective decreases strictly. There are finitely many bases, so no basis repeats and the algorithm terminates at a basis with no negative reduced cost, which is optimal.

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With integer supplies and demands, every northwest-corner allocation is integer. At each pivot, $\theta$ is the minimum of finitely many current allocations on the minus cells, so it remains integer; adding and subtracting it preserves integrality. The algorithm therefore reaches an integer-valued optimal solution.

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