Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 4

The Jordan normal form theorem says that every complex square matrix is similar to a direct sum of Jordan blocks, uniquely up to their order. For an eigenvalue $\lambda$, if its blocks have sizes $n_1,\dots ,n_r$, then its algebraic multiplicity and geometric multiplicity areand its contribution to the minimal polynomial is $(t-\lambda {)}^{{\max }_j{n}_j}$. Thuswhere $s_{\lambda }$ is the largest $\lambda$-block size.

For the given matrix,Moreover,HenceThe eigenvalue $-1$ therefore has one Jordan block of size two, so

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A subset $A\subseteq X$ is connected if it cannot be written as $A=U\cup V$, where $U,V$ are disjoint, nonempty sets open in the subspace topology on $A$.

If $f(X)$ were disconnected as $U\cup V$, continuity would make $f^{-1}(U)$ and $f^{-1}(V)$ a disconnection of $X$. Thus the continuous image of a connected space is connected.

For $Y=\{0,1\}$ with the discrete topology, a nonconstant continuous $h:X\to Y$ gives the disconnectionConversely, a disconnection $X=U\cup V$ defines a continuous nonconstant function by assigning $0$ on $U$ and $1$ on $V$. Hence $X$ is connected exactly when every such $h$ is constant.

Finally suppose $C$ is connected but $\mathrm{Cl}(C)=U\cup V$ is a disconnection. The intersections $C\cap U$ and $C\cap V$ cannot both be nonempty, so, after interchanging $U,V$, we have $C\subseteq U$. For any $v\in V$, the relatively open neighbourhood $V$ of $v$ in $\mathrm{Cl}(C)$ is disjoint from $C$, contradicting $v\in \mathrm{Cl}(C)$. Therefore the closure of a connected set is connected.

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A function $f:U\to \mathbb{C}$ is holomorphic when it is complex differentiable at every point of the open connected set $U$. Morera's theorem states that a continuous function on a domain is holomorphic if its integral around every triangle whose interior lies in the domain is zero.

The integrand $e^{tz}/(1+t^2)$ is continuous jointly in $(t,z)$ on $[0,1]\times \mathbb{C}$, so the displayed integral defines a continuous function. For any triangle $T$, Fubini's theorem and the Cauchy integral theorem giveMorera's theorem therefore proves that $f$ is entire.

The function $1/z$ is holomorphic on $\mathbb{C}\setminus \{0\}$ but has no antiderivative there, because an antiderivative would integrate to zero around every closed curve whereasThis is the standard period obstruction to a holomorphic antiderivative.

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The time-independent Schrodinger equation isIntegrating it over $(a-\epsilon ,a+\epsilon )$ and letting $\epsilon \to 0$ shows that ${\psi }^{\prime }$ has no jump because $V$ is finite. A jump in $\psi$ would create a delta term in ${\psi }^{\prime }$ and hence a derivative of a delta in ${\psi }^{\prime \prime }$, so $\psi$ is also continuous.

For a bound state $-V_0<E<0$, defineThen the proposed even wavefunction solves the equation away from the interfaces, andContinuity at $x=a$ gives $B\cos (ka)=A{e}^{-\eta a}$, while continuity of the derivative gives $-Bk\sin (ka)=-A\eta e^{-\eta a}$. Dividing yields the second required relationThus the lowest even state of the finite square well is the first-quadrant intersection, with $0<ka<\pi /2$, of the circle $k^2+{\eta }^2=2m{V}_0/{\hslash }^2$ and the curve $\eta =k\tan (ka)$.

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Faraday's law isChoose the positive circulation to be anticlockwise as viewed from $+z$. Since ${\Phi }_B=\pi B{r}^2$,

If $\dot{r}>0$, the current is clockwise and its field points in the $-z$ direction, reducing the field inside the loop. Since $d\mathbf{F}=Id\mathbf{l}\times \mathbf{B}$, the force due to the imposed field is radially inward. If $\dot{r}<0$, the current is anticlockwise, its field points in the $+z$ direction and increases the field inside, while the force is radially outward. In both cases the magnetic force opposes the radial motion, in accordance with Lenz's law.

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On each interval of width $h=1/N$, Taylor expansion about its midpoint shows that the local midpoint-rule error is $O(h^3)$. Summing over $N$ intervals gives the composite midpoint rule error

To remove the endpoint singularity, set $x=s^2$. Thenwhose transformed integrand is analytic. Applying the composite midpoint rule in $s$ with $s_n=(n+\frac{1}{2})/N$ givesThis has the required form whenThe quadratic substitution for a square-root endpoint singularity therefore restores second-order convergence.

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With states ordered as $(A,B,C)$ and rows representing the current state, the Markov chain has transition matrix

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Starting from the airport, after one step the distribution is $(0,1/2,1/2)$. Multiplying once more by $P$ gives

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Let $p_n=\mathbb{P}(X_n=A)$. Whenever the taxi is away from the airport it returns with probability $3/4$, whereas it must leave whenever it is at the airport. HenceSubtracting the fixed point $3/7$ givesThereforefor every $n\ge 0$, and in particular for the requested $n\ge 1$.

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Two matrices are equivalent when $A^{\prime }=PAQ$ for invertible matrices $P$ and $Q$ of the appropriate sizes. A change of basis in the codomain multiplies a matrix on the left, while a change of basis in the domain multiplies it on the right. Hence two matrices of the same linear map in two pairs of bases are equivalent.

Conversely, start with the map $\alpha :F^n\to F^m$ having matrix $A$ in the standard bases. Given $A^{\prime }=PAQ$, choose domain and codomain bases whose change-of-coordinate matrices produce $Q$ and $P$; then $A^{\prime }$ represents the same map in those bases. This proves the equivalence.

The column rank of $A$ is the dimension of the span of its columns, and its row rank is the dimension of the span of its rows. If $A$ represents $\alpha :F^n\to F^m$, its column rank is $\mathrm{rank}\alpha$. The transpose represents the dual mapso the row rank is $\mathrm{rank}{\alpha }^{\ast }$.

If $\mathrm{rank}\alpha =r$, thenwhose dimension is $m-r$: extend a basis of $\mathrm{im}\alpha$ to one of $F^m$ and use the dual basis. Rank-nullity now gives $\mathrm{rank}{\alpha }^{\ast }=r$. This proves the equality of row rank and column rank.

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For $I\subseteq [m]$, the independent vectors $\{v_i:i\in I\}$ lie in the coordinate subspace supported onwhich has dimension $|S_I|$. Hence $|I|\le |S_I|$, and the stated form of Hall marriage theorem gives an injection $f$ with $f(i)\in \mathrm{supp}(v_i)$.

Let $V$ be the $n\times m$ matrix whose $i$th column is $v_i$. Its column rank is $m$. By equality of row and column rank, it has $m$ linearly independent rows. Let $B$ be their indices. The $m\times m$ submatrix $V_B$ is invertible, so its columns are independent. Those columns are exactly the nonzero coordinates of $B{v}_1,\dots ,B{v}_m$, and therefore these truncated vectors are linearly independent.

Expanding $\det V_B$ gives a permutation $f:[m]\to B$ for whichThus $f(i)\in \mathrm{supp}(v_i)$. Finally, order the coordinates with $B$ first. The matrix whose columns are the $v_i$ together with the $e_j$ for $j\notin B$ is block triangular, with diagonal blocks $V_B$ and an identity matrix. Its determinant is nonzero. Consequentlyis a basis of ${\mathbb{C}}^n$. This is the simultaneous basis exchange from a nonzero minor.

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An $R$-module is free if it has a basis: a subset $B$ such that every element has a unique expression as a finite $R$-linear combination of elements of $B$.

Choose a maximal ideal $\mathfrak{m}$ of the nonzero ring $R$. If $R^n\cong R^m$, quotienting by $\mathfrak{m}$ givesThese are vector spaces over the field $R/\mathfrak{m}$, so equality of dimension gives $n=m$. This proves invariant basis number for a commutative ring.

A direct summand of a free module need not be free. Take $R=\mathbb{Z}/6\mathbb{Z}$. Its ideals $P=(2)$ and $Q=(3)$ satisfyso $P\oplus Q\cong R$. But $P$ has three elements, whereas a finite-rank free $R$-module has $6^r$ elements; hence $P$ is not free.

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Let $P$ be free with basis $(e_i{)}_{i\in I}$. Given a surjection $f:M\to N$ and a map $g:P\to N$, choose $m_i\in M$ satisfying $f(m_i)=g(e_i)$. There is a unique linear map $h:P\to M$ with $h(e_i)=m_i$, and on every basis vectorThus $f\circ h=g$, proving that free modules are projective.

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We prove by induction on $n$ that every submodule $N\subseteq R^n$ is free. The case $n=0$ is immediate. Let $\pi :R^n\to R$ be projection onto the first coordinate. Since $R$ is a principal ideal domain,for some $d$. If $d=0$, then $N\subseteq R^{n-1}$ and induction applies. Otherwise choose $x\in N$ with $\pi (x)=d$. For every $y\in N$, write $\pi (y)=rd$; then $y-rx\in \ker (\pi {|}_N)$. Also $Rx\cap \ker (\pi {|}_N)=0$ because $R$ is a domain and $d\ne 0$. ThereforeThe kernel is a submodule of $R^{n-1}$ and is free by induction, while $Rx\cong R$. Hence $N$ is free. This is the submodule theorem for free modules over a principal ideal domain.

If $P$ is finitely generated and projective, choose a surjection $f:R^n\to P$. Projectivity supplies $h:P\to R^n$ with $f\circ h={\mathrm{id}}_P$, soThus $P\cong h(P)$ is a submodule of the finitely generated free module $R^n$, and the result just proved shows that $P$ is free.

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Differentiability at $x\in {\mathbb{R}}^2$ means that there is a linear map $Df{|}_x:{\mathbb{R}}^2\to \mathbb{R}$ such thatThe partial derivatives are $D_{i}f(x)=Df{|}_x(e_i)$ when the derivative exists; independently, they may be defined by the corresponding one-variable difference quotients.

Suppose the partial derivatives exist near $x=(x_1,x_2)$ and are continuous at $x$. Apply the one-dimensional mean value theorem along the two coordinate segments from $x$ to $x+h$. It gives intermediate points ${\xi }_h,{\eta }_h\to x$ such thatContinuity of the partials makes the remainder after subtractingequal to $o(\Vert h\Vert )$. Hence $f$ is differentiable and this is its derivative.

For the given function, writing $r=\sqrt{x^2+y^2}$, at $r>0$ we haveAt the origin both partial derivatives are zero, since $f(h,0)/h$ and $f(0,h)/h$ tend to zero. Neither partial is continuous there: along its corresponding coordinate axis the cosine term oscillates without a limit. Neverthelessso $f$ is differentiable at the origin with derivative zero. This illustrates that existence of partial derivatives does not imply their continuity.

The final assertion is false. DefineAgain $|g|\le r^2$, so $g$ is differentiable at the origin and it is smooth elsewhere. But along the $x$-axis,is unbounded near zero, and similarly $D_{2}g(0,y)$ is unbounded. Thus both partial derivatives can be unbounded in every neighbourhood of a point even when the function is differentiable everywhere.

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The surface is obtained by revolving the graph $z=f(x)$ about the $x$-axis. It has a central cylindrical section of radius $b$, two smooth transition collars, and unit spherical caps centred at $(\pm 3,0,0)$; its projection onto the $(x,z)$-plane is the region $|z|\le f(x)$.

For a surface of revolution, the Gaussian curvature of a surface of revolution isThus the cylindrical region has $K=0$. On each transition collar, $K<0$ where $f^{\prime \prime }>0$, vanishes at the unique inflection circle, and is positive where $f^{\prime \prime }<0$. The spherical caps have $K>0$.

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The area element isConsequentlyAt $x=2-a$, smooth matching to the cylinder gives $f^{\prime }=0$. At $x=2+a$, the spherical formula givesThe total Gaussian curvature of a surface-of-revolution strip is therefore

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The two curves produced by $y=0$ are meridians, and meridians of a surface of revolution are geodesics. A boundary circle $x=x_0$ is a geodesic precisely when $f^{\prime }(x_0)=0$, as follows either from the geodesic equations or from its geodesic curvatureThe circle at $x=2-a$ is geodesic because it joins the cylinder smoothly. At the other boundary,which vanishes exactly when $a=1$. Hence the cut pieces are geodesic polygons only for

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For $\mathrm{Re}s>0$,This is the Laplace transform time-shift rule in its simplest form.

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Fourier transformation givesFor $x>0$, close the inverse-transform contour in the upper half-plane and take the residue at $k=im$; for $x<0$, close it in the lower half-plane. This gives the one-dimensional modified Helmholtz Green functionThe same formula works for complex $m$ with $\mathrm{Re}m>0$: it decays at both ends, is continuous at zero, and its derivative has the jump $G^{\prime }(0+)-G^{\prime }(0-)=-1$ required by the delta source.

Convolution therefore yields

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Let $U(s,x)={\mathcal{L}}_t\{u(t,x)\}$. The initial data turn the wave equation intoUsing the Green function from part (ii) with $m=s$ givesSincewe obtain the D'Alembert formula with initial velocity

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WriteVariation of $\varphi$ gives $\nabla \cdot \mathbf{u}=0$. Variation of $\alpha$ givesand variation of $\beta$ givesUsing incompressibility, the middle equation becomesThese are the three required Euler-Lagrange equations.

Let $D_t={\partial }_t+\mathbf{u}\cdot \nabla$. Since $D_t\alpha =D_t\beta =0$, differentiating $u_i={\partial }_i\varphi +\beta {\partial }_i\alpha$ givesNowbecause $\mathbf{u}\cdot \nabla \alpha =-{\alpha }_t$. HencewhereThis is the Clebsch-potential variational derivation of incompressible Euler flow.

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For $a=0$ and $\kappa >0$, the heat kernel givesAs $t\downarrow 0$, $K_t$ is a Gaussian of total mass one whose width is of order $\sqrt{\kappa t}$; it becomes a narrow spike at zero and converges to the delta distribution. Thus the convolution tends to $u_0(x)$.

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When $\kappa =0$, the characteristics satisfy $x-at=\text{constant}$ and $u$ is constant along them. Therefore

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Set $v(t,z)=u(t,z+at)$. Then $v_t=\kappa v_{zz}$ and $v(0,z)=u_0(z)$. Applying the heat kernel and returning to $x$ gives the advection-diffusion heat-kernel solution

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For $\psi =A{e}^{-B{x}^2}$,Substitution into the Schrodinger equation and comparison of the constant and $x^2$ coefficients gives

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Put $B=\xi \tan (\varphi +\alpha t)$. Matching the constant and ${\tan }^2$ coefficients in the Riccati equation requiresOne convenient choice is thereforeso

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Since $A^{\prime }/A=-i\hslash B=-\frac{i}{2}\tan (\varphi -it)$ andintegration giveswith the branch chosen continuously from the initial value. The constant $A_0$ fixes normalization.

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Normalizability requires $\mathrm{Re}B>0$. Sincethe given Gaussian integral givesAlso ${\psi }_x=-2Bx\psi$, so integration by parts yieldsThereforeThese are the second moments of a complex Gaussian wave packet.

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Taking the vertical component of the curl of the momentum equation givesThe continuity equation gives ${\eta }_t+h_0\nabla \cdot \mathbf{u}=0$, and henceSince initially $\mathbf{u}=0$ and $\eta ={\eta }_0$,Taking the divergence of the momentum equation and usinggivesDifferentiate continuity in time and substitute the last two identities to obtain

Let the Rossby deformation radius beThe even, decaying steady solution ofwith continuous value and derivative at $x=\pm a$ isThe steady momentum balance is a geostrophic balance:Thus the flow is parallel to the two edges of the raised strip, in opposite $y$-directions on the two sides; for $f>0$, it points toward $+y$ on the left and $-y$ on the right. Its magnitude is concentrated within a few deformation radii of the edges.

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LetDifferentiating the Gaussian log-likelihood, equivalently minimizing the residual sum of squares, gives the ordinary least squares estimatorsassuming $S_{xx}>0$.

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The centered model has residual sum of squaresIts slope normal equation gives

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The intercept normal equation isBecause the centered predictors sum to zero,Meanwhile $\hat{\alpha }=\bar{Y}-\hat{\beta }\bar{x}$, so the two intercept estimates are unequal in general. The parameters themselves satisfy ${\alpha }^{\prime }=\alpha +\beta \bar{x}$.

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Averaging the centered model givesThereforeWriting $z_{0.975}={\Phi }^{-1}(0.975)$, an exact 95% confidence interval is

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Use the residual variance estimatorUnder the Gaussian linear model, $\bar{Y}$ is independent of $s^2$ andConsequently the Student t confidence interval for a centered regression intercept follows fromIt isThe displayed pivotal quantity lies between its 2.5% and 97.5% quantiles with probability $0.95$, which proves the stated coverage.

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Newton's method for minimization usesFor a quantitative local bound, suppose on a convex neighbourhood containing the iterates thatwith $m>0$, and let $x^{\ast }$ be the minimizer. The integral form of the gradient and the Hessian Lipschitz bound give the quadratic convergence bound for Newton's methodIf $q=M\Vert x_0-x^{\ast }\Vert /(2m)<1$, induction yieldsSince $f(x)-f(x^{\ast })\le L\Vert x-x^{\ast }{\Vert }^2/2$,