Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 2

Differentiability at $(a,b)$ means that there is a linear map $L:{\mathbb{R}}^2\to \mathbb{R}$ such thatNecessarily $L(h,k)=f_x(a,b)h+f_y(a,b)k$ when those partial derivatives exist. Under the stated assumptions, apply the one-variable mean value theorem separately to the two coordinate increments:for some intermediate $\theta ,\eta \in (0,1)$, omitting a term if its increment is zero. Existence of the partial derivatives in a neighborhood supplies the differentiability along each segment required for that theorem. Their continuity at $(a,b)$ makes the difference from $f_x(a,b)h+f_y(a,b)k$ bounded by $(|h|+|k|)\epsilon (h,k)$, where $\epsilon (h,k)\to 0$. Since $|h|+|k|\le \sqrt{2}\sqrt{h^2+k^2}$, this is the required little-o remainder and proves Fréchet differentiability.

For the particular function, $f(h,0)=f(0,k)=0$, so the defining difference quotients give $f_x(0,0)=f_y(0,0)=0$. However $f(t,t)=1/2$ for every $t\ne 0$. It is not continuous at the origin, and therefore not differentiable there. This illustrates why existence of partial derivatives alone is insufficient.

Let $M_{ij}={\int }_S{x}_i{x}_{j}dS$. A rotation matrix $Q$ maps the unit sphere and its surface measure to themselves. Changing variables therefore gives $Q_{ip}{Q}_{jq}{M}_{pq}=M_{ij}$, so $M$ is an isotropic second-rank tensor. Reflection symmetry also directly makes its off-diagonal entries zero, while interchange of coordinates makes its three diagonal entries equal. Its trace ishence $M_{ij}=(4\pi /3){\delta }_{ij}$. Oddness gives ${\int }_S{x}_{i}dS=0$. Expanding the defining product for $T$ now givesThe action on a vector is $Tv=\lambda v+\mu y(y\cdot v)$. For $y\ne 0$, its eigenvalues and eigenspaces are thereforeThe latter has multiplicity two. If $y=0$, the single eigenvalue $4\pi /3$ has multiplicity three and every vector is an eigenvector apart from the excluded zero vector.

Assume the known variances are positive and write $\bar{x}=n^{-1}{\sum }_i{x}_i$. The likelihood times the normal prior distribution is proportional toCollecting the quadratic and linear terms in $\theta$ completes the square with precision $v^{-1}=n/{\sigma }^2+1/{\tau }^2$ and mean $m=v(n\bar{x}/{\sigma }^2+\mu /{\tau }^2)$. Thus normal-normal conjugacy givesFor quadratic loss, the posterior risk of reporting $a$ is $\mathbb{E}[(a-\theta {)}^2\mid x]=v+(a-m{)}^2$, uniquely minimized at $a=m$. For absolute-error loss, differentiating the posterior risk gives $2F_{\theta \mid x}(a)-1$, so its minimizer is a posterior median. The normal posterior is symmetric about $m$ and has a strictly increasing distribution function, hence that median is uniquely $m$. Therefore both optimal point estimates equal $m$, the posterior mean.

For the cofinite topology ${\tau }_1$, the empty set and $\mathbb{N}$ are included. A nonempty union of cofinite sets is cofinite because its complement is contained in the finite complement of any one member. A finite intersection is cofinite because its complement is a finite union of finite sets; intersections involving the empty set are empty. This proves all topology axioms.

For the tail topology on the natural numbers ${\tau }_2$, $\mathbb{N}=I_1$. Any nonempty collection of tails has a smallest starting index, so its union is that tail; a finite intersection of tails is the tail with largest starting index. Empty sets and empty-family conventions supply the remaining cases. Thus ${\tau }_2$ is also a topology.

We classify continuous maps between cofinite and tail topologies. Continuity is equivalent toIf $f$ has unbounded image, each $E_k$ is nonempty, hence cofinite. For every threshold $k$, eventually $f(n)\ge k$, which is exactly $f(n)\to \infty$. If the image is bounded, it is a nonempty finite subset of $\mathbb{N}$ and has a maximum $N$. Then $E_N$ is nonempty and cofinite, but $f(n)\le N$ always, so $f(n)=N$ except at finitely many indices. These exhaust the two possibilities and prove necessity. Their sufficiency is checked separately below.

If $f(n)\to \infty$, then for each $k$ the complement of $f^{-1}(I_k)$ contains only finitely many indices. Thus every nonempty open tail has a cofinite preimage, and the empty set has empty preimage. Hence $f$ is continuous from the cofinite space to the tail-topology space.

Suppose $f(n)\le N$ for all $n$ and $f(n)=N$ eventually. If $k>N$, then $f^{-1}(I_k)=\varnothing$. If $k\le N$, the preimage contains all sufficiently large indices and is cofinite. Every open set therefore has open preimage, proving $f$ is continuous. Together with the introductory argument this proves the claimed if-and-only-if classification.

Eliminate the first entries of rows two, three and four using multipliers $a^3,a^2,a$. All remaining subdiagonal entries are then already zero. The unit-lower-triangular LU decomposition isMultiplication verifies $LU=A$. Since $\gamma \ne 0$, all pivots are nonzero. Forward substitution in $Ly=b$ givesBackward substitution in $Ux=y$ gives $x_4=0$, $x_3=-a^2$, $x_2=0$, and $x_1=\gamma +a^4=1$. ThusDirect substitution yields $Ax=(\gamma ,0,0,a\gamma {)}^T$ as a check.

Associate to $c=(c_1,\dots ,c_n{)}^T$ the polynomial $p(t)={\sum }_{j=1}^n{c}_j{t}^{j-1}$. The $i$th component of $Ac$ is exactly $p(a_i)$. If $Ac=0$, this polynomial of degree at most $n-1$ has $n$ distinct roots. Repeated use of the factor theorem would make ${\prod }_{i=1}^n(t-a_i)$ divide any nonzero such polynomial, contradicting its degree. Hence $p$ is the zero polynomial and all $c_j=0$. The converse is immediate, proving $\ker A=\{0\}$.

A square matrix has full column rank if and only if its kernel is zero, by rank-nullity theorem. Its row rank equals its column rank. Therefore this Vandermonde matrix has row rank $n$, and its rows $v_1,\dots ,v_n$ are a basis and span ${\mathbb{R}}^n$.

Take the unit circle counterclockwise and $n\ge 0$. The integrand is a Laurent polynomial with its only possible singularity at zero. Its expansion isThe $z^{-1}$ term is the term $j=n$, so the residue is $(-1{)}^n(\genfrac{}{}{0ex}{}{2n}{n})$. The residue theorem gives

Parametrize the same contour by $z=e^{it}$, $0\le t\le 2\pi$. Then $dz/z=idt$ and $z-z^{-1}=2i\sin t$. Part a becomesCanceling the nonzero common factors gives

The positive-definite symmetric form $b$ is a real inner product. Choose a $b$-orthonormal basis by the Gram-Schmidt process. The stated identity means that the matrix $K$ of $\psi$ satisfies $K^T=-K$. If $d=\dim U$, thenAn invertible $K$ has nonzero determinant, so $d$ is even.

To cover a possibly singular map, first show $\ker \psi =(\mathrm{im}\psi {)}^{\perp }$. If $x\in \ker \psi$, then $b(x,\psi y)=-b(\psi x,y)=0$ for every $y$, giving one inclusion. Conversely orthogonality to the image gives $b(\psi x,y)=0$ for all $y$, so nondegeneracy of $b$ gives $\psi x=0$. Positivity then implies $\mathrm{im}\psi \cap \ker \psi =\{0\}$.

The image is invariant under $\psi$, and its restricted map $\psi {|}_{\mathrm{im}\psi }$ is injective by that trivial intersection. In finite dimension it is consequently invertible. The restricted inner product is positive definite and the same skew-adjoint identity holds, so the first argument applied to this restriction makes $\dim \mathrm{im}\psi$ even. Thusincluding rank zero. This is the even rank of a skew-adjoint linear map.

A Hermitian operator satisfies $⟨u,Av⟩=⟨Au,v⟩$ for all admissible states in its domain; in finite dimension this is $A^{\ast }=A$. For an unbounded physical observable one specifies a self-adjoint realization, rather than only a formal differential expression.

Here $p=-i\hslash d/dx$ and $H=-{\hslash }^2{d}^2/(2md{x}^2)+V(x)$. For states with sufficient regularity and decay, integration by parts givesThe multiplication term cancels because $V$ is real. Thus the Quantum Hamiltonian is Hermitian on the stated admissible domain. Equivalent boundary conditions must remove this boundary term if the system is on a bounded interval.

The Schrödinger equation $i\hslash \dot{\psi }=H\psi$ and its conjugate imply, for a time-independent observable $A$ and normalized state,This follows by differentiating both factors in $⟨\psi ,A\psi ⟩$ and moving $H$ using Hermiticity. The canonical commutator $[x,p]=i\hslash$ givesConsequently the two Ehrenfest theorem identities areThey require the displayed expectations and domain operations to exist; real-valuedness of a potential alone is not a proof about every possible singular operator domain.

First $N(A)$ is finite: by Cauchy-Schwarz, $\Vert Ax{\Vert }_2\le ({\sum }_{ij}{a}_{ij}^2{)}^{1/2}\Vert x{\Vert }_2$. It is nonnegative and absolutely homogeneous. If $N(A)=0$, every $A{e}_j=0$, so every column is zero and $A=0$; the converse is immediate. The triangle inequality for the Euclidean norm gives $N(A+B)\le N(A)+N(B)$ after taking suprema. Thus $N$ is a norm.

For any vector $x$, rescaling gives $\Vert Ax{\Vert }_2\le N(A)\Vert x{\Vert }_2$. Applying this twice givesFurthermore $|a_{ij}|\le \Vert A{e}_j{\Vert }_2\le N(A)$, proving the requested strict entry bound when $N(A)<\epsilon$.

For a polynomial in matrix entries, expand each monomial at $A+H$. The terms involving exactly one entry of $H$ give a linear map $Df(A)[H]$, with coefficients polynomial in the entries of $A$. Every remaining term has at least two entries of $H$ and is $O(N(H{)}^2)$ for $N(H)\le 1$, because $|h_{ij}|\le N(H)$. This proves differentiability. Its derivative is continuous as an operator-valued function: each coefficient is continuous, and for $N(H)\le 1$ every $h_{ij}$ is bounded by one, so the operator-norm difference is bounded by the sum of the coefficient differences. Coordinate continuity follows from the same entry bound. Hence every such polynomial function is continuously differentiable.

In the Leibniz expansion of $\det (I+H)$, the identity permutation supplies $1+{\sum }_i{h}_{ii}$ plus terms of degree at least two. Any nonidentity permutation moves at least two indices, so each of its terms has at least two $H$ entries. ConsequentlyIf $A$ is invertible, multiplicativity of the determinant and the preceding expansion giveThe adjugate identity gives $\mathrm{adj}A=\det (A)A^{-1}$. Thus $d^{\prime }(A)[H]=\mathrm{tr}((\mathrm{adj}A)H)$ for invertible $A$.

For singular $A$, take invertible $A_r\to A$, using the permitted density result. Both $d^{\prime }$ and the adjugate entries are continuous polynomial expressions, so taking limits for every fixed $H$ proves the derivative of the determinant formula on all matrices:Here the standard adjugate is the transpose of the cofactor array, as required by $(\mathrm{adj}A)A=(\det A)I$. In components this derivative is ${\sum }_{ij}{\mathrm{cof}}_{ij}(A)h_{ij}$, also valid at singular matrices.

For an orthogonal change of Cartesian coordinates $x_i^{\prime }=Q_{ij}{x}_j$, an ordinary rank-$n$ tensor transforms byAn isotropic tensor is invariant under such rotations. Orthogonality gives $Q_{ip}{Q}_{jq}{\delta }_{pq}={\delta }_{ij}$, so transforming any of the three products of two Kronecker deltas leaves it unchanged. Their linear combination with arbitrary scalar coefficients is therefore isotropic.

For completeness, these three products exhaust the rank-four isotropic tensors in three dimensions. Half-turns about coordinate axes force a nonzero component to contain each coordinate label an even number of times. Axis interchanges then leave just the values $X=c_{1111}$, $\alpha =c_{1122}$, $\beta =c_{1212}$ and $\gamma =c_{1221}$. Rotating the first two axes by $\pi /4$ gives $X=\frac{1}{2}X+\frac{1}{2}(\alpha +\beta +\gamma )$, so $X=\alpha +\beta +\gamma$. These component values are exactly those of the three-delta-product expression. Thus isotropy of the elastic medium gives this form for its elasticity tensor.

Contracting with the symmetric strain givesThese are the Lamé parameters. Taking the trace determines the scalar part of the strain:Since ${\sum }_{ij}{e}_{ij}^2=3p^2+{\sum }_{ij}{d}_{ij}^2$, the stored strain energy density isA nonzero traceless test strain proves necessity of $\mu \ge 0$, and a pure scalar strain proves necessity of $3\lambda +2\mu \ge 0$. Conversely these two conditions make both terms nonnegative for every strain. Thus nonnegative isotropic elastic strain energy is equivalent to

Model the two samples as independent multinomial counts with fixed row totals $N_A=N_B=500$ and three category probabilities. The null hypothesis is $p_{Aj}=p_{Bj}=p_j$ for every category $j$. Up to combinatorial factors the likelihood is ${\prod }_{ij}{p}_{ij}^{O_{ij}}$. Without the null restriction its maximizers are $O_{ij}/N_i$; under the null, maximizing ${\sum }_j(O_{Aj}+O_{Bj})\log p_j$ subject to ${\sum }_j{p}_j=1$ givesThe observed column totals are $243,281,476$, giving fitted expected counts $(121.5,140.5,238)$ in each row.

To derive the null distribution, put $D=O_A-O_B$. Under the null its mean is zero and covariance is $1000(\mathrm{diag}p-p{p}^T)$. The multivariate central limit theorem gives, asymptotically, the standardized contrast $Z_j=D_j/\sqrt{1000p_j}$ a centered normal distribution with covarianceThis is the orthogonal projection onto a two-dimensional plane. In an orthonormal basis of that plane, the two nonzero coordinates are independent standard normal variables, so ${\sum }_j{Z}_j^2$ tends to ${\chi }_2^2$. The pooled estimates are consistent; substituting them for $p_j$ leaves this limit unchanged by Slutsky's theorem. Since $O_{Aj}-E_{Aj}=D_j/2$ and $O_{Bj}-E_{Bj}=-D_j/2$, the resulting statistic is precisely the Pearson chi-squared test of homogeneity statisticThis supplies the two degrees of freedom rather than treating all six cells as independent. Every fitted expected count is large, supporting the approximation.

For these data,It exceeds the two-degree-of-freedom 95th percentile $5.99$, but not the 99th percentile $9.21$. Hence reject equal score distributions at the 5% level, but not at the 1% level. The asymptotic p-value is $\mathrm{Pr}({\chi }_2^2\ge 7.56906)=e^{-7.56906/2}\simeq 0.02272$.

Define $F:[0,1]\to \mathbb{C}$ by $F(s)=s{e}^{2\pi i/s}$ for $s>0$ and $F(0)=0$. It is continuous away from zero, and $|F(s)-F(0)|=s\to 0$, so it is continuous at zero too. Its image is exactly $X\cup \{0\}$, using $s=1/t$. The interval is compact, so its continuous image is compact.

For any two image points $F(s_0),F(s_1)$, the explicit pathis continuous and joins them. Thus the reciprocal spiral with its endpoint is also path-connected. Infinite winding causes no discontinuity at the endpoint because its radius tends to zero.

The set $Y$ is path-connected as the continuous image of $[1,\infty )$, hence connected. Its closure is $Y$ together with the unit circle. Indeed $g(k+\theta /(2\pi ))\to e^{i\theta }$ as $k\to \infty$, and any convergent sequence of spiral points either has bounded parameters, giving a point of $Y$, or has parameters tending to infinity along a subsequence, giving radius one. The closure of a connected set is connected: a separation of its closure would induce a separation of the dense original set, with neither open part missing that set. The closed disk is connected and meets this closure in the unit circle; their union $Y\cup \overline{D}$ is therefore connected.

To show it is not path-connected, suppose a path $\alpha :[0,1]\to Y\cup \overline{D}$ starts in $Y$ and ends in the disk. Let $u_{\ast }$ be its first contact with the disk. The disk is closed, so this first contact exists, and $u_{\ast }>0$. Before it, the path belongs to $Y$ and has a continuous uniquely determined parameterContinuity at contact forces $|\alpha (u_{\ast })|=1$, and hence $t(u)\to \infty$ as $u\uparrow u_{\ast }$. The intermediate value theorem supplies times approaching $u_{\ast }$ at which $t(u)$ is an integer, and other times at which it is an integer plus $1/2$. These times must approach contact because $t$ is bounded on every compact interval strictly before it. Along the first sequence $\alpha (u)\to 1$; along the second it tends to $-1$. This contradicts continuity at $u_{\ast }$. Thus no path joins the spiral to the disk, proving failure of path-connectedness for the spiral accumulating on a circle.

Exactness on the polynomial basis $1,x,x^2$ requiresSolving givesThus the approximation is the unit-spacing centered second difference, $\mu (f)=f(-1)-2f(0)+f(1)$.

The error $e(f)=f^{\prime \prime }(0)-f(-1)+2f(0)-f(1)$ is linear and annihilates all degree-at-most-two polynomials. The Peano kernel theorem with derivative order three givesFor $t\ne 0$, the second derivative of the truncated quadratic at $x=0$ is $1_{\{t<0\}}$. Its values at $x=-1,0,1$ then giveThe value at the single point $t=0$ is irrelevant to the integral. This is the centered second-derivative Peano kernel. Its absolute integral isConsequentlyIts changing sign explains why cubic polynomials also have zero error: their third derivative is constant and ${\int }_{-1}^{1}K=0$. No fourth derivative is needed for the claimed bound.

By the rank-nullity theorem, $\mathrm{nullity}A=n-\mathrm{rank}A\ge n-m>0$. Thus a first dependent column prefix exists. Minimality of $k$ makes the first $k-1$ columns independent, so $\mathrm{rank}{A}_{k-1}=k-1$. Adding one column cannot reduce this rank, while nonzero nullity of $A_k$ implies rank at most $k-1$. HenceFor $k=1$ the earlier prefix is the zero-dimensional empty matrix, and the same argument applies.

Choose any nonzero vector $b$ in the one-dimensional kernel of $A_k$. It gives ${\sum }_{j=1}^k{a}_{ij}{b}_j=0$ for every row. Because no column is zero, $b$ must have at least two nonzero entries: a relation supported at a single index would make that column zero.

Suppose, contrary to the claim, that $A_k\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_k)b=0$. Its kernel is spanned by $b$, so there is a scalar $c$ with ${\lambda }_j{b}_j=c{b}_j$ for every $j$. At the two nonzero entries this gives two equal weights ${\lambda }_j=c$, contradicting their distinctness. Thus the weighted sum is nonzero in at least one row. This proves that distinct diagonal weights break a first column dependence. Some entries of $b$ may be zero; only two nonzero entries are needed.

Assume for contradiction that $n>m$. Apply part b to the matrix of the given relation coefficients, whose columns are nonzero. For its first dependent prefix choose $b_1,\dots ,b_k$ withMultiply the $j$th given vector relation by $b_j$ and sum. Interchanging the finite sums givesThe $w_i$ are a basis, so all $z_i$ must be zero, contradicting part b. Therefore $n\le m$.

Choose a fixed reference basis and let $V,W$ have columns $v_i,w_i$. Both matrices are invertible. The vectors $v_i+\lambda w_i$ are dependent exactly when $\det (V+\lambda W)=0$. This is a polynomial in $\lambda$ of degree $m$, whose leading coefficient is $\det W\ne 0$. A nonzero degree-$m$ polynomial has at most $m$ distinct real roots. Thus there are at most $m$ exceptional parameters. This is the determinant argument for the matrix pencil $V+\lambda W$.

Use $\hat{f}(\lambda )=(2\pi {)}^{-1/2}{\int }_{\mathbb{R}}{e}^{-i\lambda x}f(x)dx$. The transform statement requires sufficient decay and regularity for the transform and its derivatives; functions in the Schwartz space suffice, and all the functions constructed below belong to that space. Integration by parts twice and differentiation under the integral giveTransforming the differential equation therefore givesThe differential equation by itself does not ensure that an ordinary Fourier transform exists: $f=e^{x^2/2}$ solves it with $\mu =1$ but grows too rapidly. Thus the usual transformability assumptions are implicit in this part.

Set $f=p(x)e^{-x^2/2}$. Differentiating givesThus the polynomial must satisfy $p^{\prime \prime }-2x{p}^{\prime }-(\mu +1)p=0$. If it has degree $n$ and nonzero leading coefficient, comparison of highest powers gives $\mu =-(2n+1)$. For this value the equation becomes the Hermite differential equationWrite $p={\sum }_{j=0}^n{a}_j{x}^j$, with coefficients beyond $n$ zero. Comparing powers yieldsThe equation at $j=n-1$ forces $a_{n-1}=0$, and downward recurrence removes all terms of the opposite parity and determines all same-parity terms from $a_n$. Taking $a_n=1$ produces the explicit polynomialIt satisfies every coefficient equation, so existence as well as uniqueness up to a nonzero scalar is proved. HenceThe monic $p_n$ is $2^{-n}$ times the usual physicists' Hermite polynomial.

Multiplication by $x$ transforms to $id/d\lambda$. Write $p_n(x)={\sum }_{j=0}^n{a}_j{x}^j$ and use $\hat{g}=cg$ for the Gaussian $g=e^{-x^2/2}$. ThenSuccessive differentiation of the Gaussian gives a polynomial of degree $j$ with leading term $(-1{)}^j{\lambda }^j$ multiplying that same Gaussian. Thus $q_n$ has leading coefficient $c{a}_n(-i{)}^n\ne 0$. This proves $q_n$ is a polynomial of degree exactly $n$, rather than merely at most $n$.

By part a the transformed function solves the same equation with ${\mu }_n$, and by part c its polynomial factor has degree $n$. The uniqueness proved in part b therefore gives $q_n=c_n{p}_n$. Comparing their leading coefficients givesFor the unitary convention used above, the Gaussian integral gives $c=1$, so $c_n=(-i{)}^n$. With the unnormalized phase-$e^{-i\lambda x}$ convention, $c=\sqrt{2\pi }$ instead. This proves that Hermite functions are Fourier eigenfunctions, with the normalization-dependent scalar fully specified.

A Hermitian two-by-two matrix has the formAddition and real scalar multiplication preserve this condition, and the two diagonal entries plus the real and imaginary parts of $z$ give four independent real coordinates. Thus $S$ is a real four-dimensional vector space. The trace product is real because $\overline{\mathrm{tr}(AB)}=\mathrm{tr}(BA)=\mathrm{tr}(AB)$; this also proves symmetry of $b$.

Directly, $\mathrm{tr}(A^2)={\alpha }^2+{\delta }^2+2|z{|}^2$ and $(\mathrm{tr}A{)}^2={\alpha }^2+2\alpha \delta +{\delta }^2$. HenceSince $\mathrm{tr}I=2$ and $\mathrm{tr}(AI)=\mathrm{tr}A$,

The displayed matrices are the three Pauli matrices, in the order ${\sigma }_3,{\sigma }_1,{\sigma }_2$. They satisfy $\mathrm{tr}{A}_i=0$ and $\mathrm{tr}(A_i{A}_j)=2{\delta }_{ij}$. Every Hermitian matrix has a unique expression $aI+{\sum }_i{v}_i{A}_i$ with all four coefficients real. Therefore they form a basis together with $I$.

The preceding trace identities give $b(I,I)=-1$, $b(I,A_i)=0$, and $b(A_i,A_j)={\delta }_{ij}$. The Gram matrix is thusConsequently rank $=4$ and inertia signature $(3,1)$. If signature is defined as the number of positive squares minus the number of negative squares, its value is $2$. This is the Hermitian matrix representation of Minkowski four-vectors, with the negative-determinant sign convention.