Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 2

A metric space is a complete metric space when every Cauchy sequence converges to a point of that space. It is a totally bounded space when, for every $\epsilon >0$, finitely many open metric balls of radius $\epsilon$ cover the space.

With their usual distances, $\mathbb{R}$ is complete but not totally bounded, and $(0,1)$ is totally bounded but not complete. The Cauchy sequence $1/n$ in $(0,1)$ has its limit outside the space; no finite collection of balls of a fixed radius covers $\mathbb{R}$.

For the requested continuous functions, takeandBoth are continuous and onto. The domain of $f$ is a complete metric space, but its image is not complete because $1/n$ tends to the missing point $0$. The domain of $g$ is a totally bounded space, but its unbounded image is not totally bounded. Uniform continuity, unlike mere continuity, would preserve total boundedness.

Let $R_{ij}$ be the components of an orthogonal matrix relating two Cartesian frames by $v_i^{\prime }=R_{ij}{v}_j$. Repeated indices are summed from $1$ to $3$. The tensor component transformation law for a Cartesian second-rank tensor isFor a proper rotation, $\det R=1$; the transformation law itself also holds for orthogonal changes of Cartesian frame with determinant $-1$.

To find a cubically invariant second-rank tensor, first square each of the prescribed quarter-turns. For example the half-turn about the first axis is $S_x=\mathrm{diag}(1,-1,-1)$. Invariance under $S_{x}A{S}_x^{\mathsf{T}}$ forces $A_{12},A_{13},A_{21},A_{31}$ to vanish. The corresponding half-turns about the other axes eliminate the remaining off-diagonal components, even though symmetry of $A$ was never assumed.

Now write $A=\mathrm{diag}(a,b,c)$. A quarter-turn about the third axis interchanges $a,b$, so $a=b$; a quarter-turn about the first axis gives $b=c$. Conversely a scalar multiple of the identity matrix is invariant under every rotation. Thus the most general answer is

Represent a possibly randomized statistical test by a function $\phi (Y)$ taking values in $[0,1]$, its conditional probability of rejection. Its power function is $\pi (\rho )={\mathbb{E}}_{\rho }\phi (Y)$, and its size of a statistical test is the supremum of this probability over the null hypothesis. At a prescribed significance level $\alpha$, a uniformly most powerful test has at least as much power as every other level-$\alpha$ test at every parameter in the alternative.

Put $S={\sum }_{j=1}^n{Y}_j$. For any fixed alternative $0<{\rho }_1<{\rho }_0$, the likelihood ratio iswhich is strictly increasing in $S$. The Neyman-Pearson lemma therefore selects the upper tail $S>c_{\alpha }$. Its cutoff is determined solely by the null distribution, not by ${\rho }_1$, so the same statistical test is most powerful against every allowed alternative and hence uniformly most powerful.

Under rate $\rho$, $S$ has a gamma distribution with shape $n$ and rate $\rho$. For completeness, convolution starts with $\rho e^{-\rho s}$ and, if the $n$-term density is ${\rho }^n{s}^{n-1}{e}^{-\rho s}/(n-1)!$, convolving once more gives ${\rho }^{n+1}{s}^n{e}^{-\rho s}/n!$. Integrating this density by parts repeatedly gives its upper-tail probability. Consequently reject precisely when $S>c_{\alpha }$, whereand the power function isFor $0<\alpha <1$ the continuous strictly decreasing null tail gives a unique positive cutoff, with exact size $\alpha$ and no boundary randomization.

Use the one-sided product bound for an entire function by consideringThis is an entire function, and $f^2=(u^2-v^2)+2iuv$ gives $|F|=e^{2uv}$. If $uv\le M$, then $|F|\le e^{2M}$ everywhere. The Liouville theorem therefore makes $F$ constant.

Because an exponential never vanishes, differentiating yields $0=F^{\prime }=-2if{f}^{\prime }F$, and hence $(f^2{)}^{\prime }=2f{f}^{\prime }=0$. Thus $f^2$ is constant. If that constant is zero, $f=0$; otherwise $f$ takes values in the two square roots of one nonzero constant. Continuity on the connected space $\mathbb{C}$ prevents switching between those two values. Therefore $f$ is constant.

Let $a_1,a_2,a_3$ denote the columns. The Gram-Schmidt process begins with $\Vert a_1\Vert =2$, soThe second column has $r_{12}=q_1^{\mathsf{T}}{a}_2=4$ and residualThis residual has norm $2$, giving $q_2=\frac{1}{2}(-1,1,-1,1{)}^{\mathsf{T}}$ and $r_{22}=2$.

For the third column the orthogonal projections are $r_{13}=6$ and $r_{23}=4$. Subtracting them givesIts norm is $2$, so $q_3=\frac{1}{2}(1,1,-1,-1{)}^{\mathsf{T}}$ and $r_{33}=2$. Thus the requested thin QR decomposition isThe three displayed columns have pairwise zero inner products and unit norms; direct multiplication gives $QR=A$.

The minimal polynomial divides the annihilating polynomial $t^2(t-1)$, since $A^3-A^2=0$. Every nonconstant monic divisor can occur in dimension four: use Jordan blocks of sizes at most two at eigenvalue $0$ and of size one at eigenvalue $1$, adding scalar blocks as needed. The complete list isFor example the first four occur for $0$, $J_2(0)\oplus 0\oplus 0$, $I$, and $\mathrm{diag}(0,0,1,1)$, respectively; the final one occurs for $J_2(0)\oplus 0\oplus 1$.

Failure to be diagonalizable requires a size-two Jordan block at zero. The condition $A^2\ne 0$ requires at least one block at one, since every allowed zero block squares to zero. A size-two zero block and one scalar block at one leave only one dimension. It can carry zero or one. Thus, up to permutation of blocks, the only possible Jordan normal forms are

Write the holomorphic function as $f=u+iv$, with $u^2+v^2=C$. If $C=0$ then $f=0$. If $C>0$, differentiating the constant modulus with respect to both coordinates, and using the Cauchy-Riemann equations $u_y=-v_x$, $v_y=u_x$, givesThe coefficient matrix has determinant $-(u^2+v^2)=-C\ne 0$, so $u_x=v_x=0$. The Cauchy-Riemann equations then also give $u_y=v_y=0$. Along a straight segment inside the disk, both real components have zero derivative and hence are constant. Thus $f$ is constant throughout the disk.

Use the convention that a sesquilinear form is linear in its first argument and conjugate-linear in its second: for complex scalars $a,b$,No symmetry or positivity is assumed. Put $\chi =\varphi -\psi$. Its diagonal vanishes. Expansion at $v+w$ giveswhile expansion at $v+iw$ givesThese two equations force $\chi (v,w)=\chi (w,v)=0$. This is the polarization argument for a vanishing quadratic form, and proves $\varphi =\psi$ on every pair of vectors.

For the final deduction, the linear map $\alpha$ makes $\psi (v,w)=\varphi (\alpha v,\alpha w)$ another sesquilinear form. Its diagonal agrees with that of $\varphi$, so the result just proved gives

Use the position-space momentum operator $P_j=-i\hslash {\partial }_j$ and set $F=(x+iy)z$. Direct differentiation givesTo compute the squared orbital angular momentum, write $D_x=y{\partial }_z-z{\partial }_y$, $D_y=z{\partial }_x-x{\partial }_z$, $D_z=x{\partial }_y-y{\partial }_x$, so $L_j=-i\hslash D_j$. Expanding the products, including derivatives of their variable coefficients, givesFor example $D_x^2=y^2{\partial }_z^2+z^2{\partial }_y^2-2yz{\partial }_y{\partial }_z-y{\partial }_y-z{\partial }_z$; summing the three cyclic expressions proves the identity.

Here $\Delta F=0$ and $EF=2F$, so $E^{2}F=4F$. ThereforeThis is the angular momentum of a homogeneous harmonic polynomial, with angular quantum numbers $\ell =2,m=1$. The polynomial is a nonzero angular eigenfunction; by itself it is not a normalizable full-space wavefunction.

The contraction mapping theorem says that a self-map $g$ of a nonempty complete metric space satisfying $d(gx,gy)\le kd(x,y)$ for a constant $0\le k<1$ has exactly one fixed point. Apply it to $g=f^N$ to obtain $p$ with $f^{N}p=p$.

No continuity of $f$ itself is needed: composition givesThus $fp$ is also a fixed point of $f^N$, whose uniqueness forces $fp=p$. Conversely any fixed point of $f$ is fixed by $f^N$, so it must equal $p$. The unique fixed point of the contracting iterate is also the unique fixed point of $f$. The usual nonempty-space hypothesis is necessary; an empty space has no fixed point.

The strict distance inequality implies $d(fx,fy)\le d(x,y)$ for all points, so $f$ is Lipschitz continuous. Henceis a continuous function, since $|g(x)-g(y)|\le d(fx,fy)+d(x,y)\le 2d(x,y)$. On the nonempty compact metric space, $g$ attains its minimum at some $p$.

If $fp\ne p$, the strict distance inequality applied to $fp,p$ givesa contradiction. Thus $fp=p$. If $p,q$ were different fixed points, then $d(p,q)=d(fp,fq)<d(p,q)$, again impossible. There is exactly one fixed point. This proves strict distance decrease on a compact metric space without assuming a uniform contraction constant.

Let $K=\overline{O(x)}$. It is nonempty, closed and bounded in finite-dimensional Euclidean space, hence compact by the Heine-Borel theorem. The strict distance inequality again makes $f$ Lipschitz continuous. If $y\in K$, choose a sequence $y_j\in O(x)$ converging to $y$. Then $f{y}_j\in O(x)$, and continuity gives $f{y}_j\to fy$, so $fy\in K$. This proves $f(K)\subseteq K$.

The restriction to $K$ satisfies strict distance decrease on a compact metric space, so part (b) produces a fixed point there. If a second fixed point existed anywhere in ${\mathbb{R}}^n$, the same strict distance inequality applied to the two points would contradict equality of their distance. Therefore the fixed point is unique in all of ${\mathbb{R}}^n$. This is the bounded-orbit fixed point for a strictly distance-decreasing Euclidean map.

Under a Cartesian frame change $B_i^{\prime }=R_{ik}{B}_k$, where $R$ is an orthogonal matrix. Orthogonality gives $B_k^{\prime }{B}_k^{\prime }=B_k{B}_k$ and $R_{ik}{R}_{j\ell }{\delta }_{k\ell }={\delta }_{ij}$. Thereforewhich is the tensor component transformation law. Also $T_{ij}=T_{ji}$ directly, proving that it is a symmetric Cartesian second-rank tensor for every choice of the two scalars.

For $\mathbf{B}\ne 0$, the direction of $\mathbf{B}$ is an eigenvector with eigenvalue $(\alpha +\beta )|\mathbf{B}{|}^2$. Every vector perpendicular to $\mathbf{B}$ is an eigenvector with eigenvalue $\beta |\mathbf{B}{|}^2$, a two-dimensional eigenspace. If $\alpha =0$, these eigenvalues coincide and every nonzero vector is an eigenvector. If $\mathbf{B}=0$, the tensor is zero and again every nonzero vector is an eigenvector, now with eigenvalue zero.

For the position-dependent field with zero divergence, the product rule givesContraction of the two Levi-Civita symbols in the curl and cross product givesConsequentlyThese constants give the magnetic part of the Maxwell stress tensor, with its physical prefactor suppressed. They are forced if the identity is to hold for every divergence-free field: $\mathbf{B}=(y,0,0)$ forces $2\beta =-1$; at the origin $\mathbf{B}=(y,1,0)$ then forces $\alpha =1$.

Apply the divergence theorem separately to each component:Since the whole vector field vanishes on $S$, all components of $T$ vanish there, and hence the vector volume integral is zero. The argument assumes a differentiable field and a boundary for which the divergence theorem applies.

For independent observations from a normal distribution with unknown mean and positive variance, the maximum likelihood estimation of the mean givesIndeed ${\sum }_i(X_i-\mu {)}^2=S_{XX}+n(\bar{X}-\mu {)}^2$, which is minimized at $\mu =\bar{X}$. The likelihood isFor fixed $\mu$, differentiation in ${\sigma }^2$ gives the maximizer $[S_{XX}+n(\bar{X}-\mu {)}^2]/n$. Thus the unrestricted fit has ${\hat{\sigma }}^2=S_{XX}/n$, whereas under the null hypothesis $\mu ={\mu }_0$ the fitted variance is $[S_{XX}+n(\bar{X}-{\mu }_0{)}^2]/n$. Both maximize rather than minimize the likelihood: it tends to zero at either variance endpoint when the residual sum of squares is positive.

The restricted-to-unrestricted likelihood ratio is thereforeThe generalized likelihood-ratio test rejects for small $\Lambda$, equivalently large $|T|$. Under the null, $Z=\sqrt{n}(\bar{X}-{\mu }_0)/\sigma$ is standard normal, $U=S_{XX}/{\sigma }^2$ has a chi-squared distribution with $n-1$ degrees of freedom, and they are independent. To see the independence, apply an orthogonal matrix to the standardized normal sample, taking its first row to be $(1,\dots ,1)/\sqrt{n}$: the transformed coordinates remain independent standard normal variables, and $U$ is the sum of squares of the other $n-1$ coordinates. Hence $T=Z/\sqrt{U/(n-1)}$ has Student's t-distribution with $n-1$ degrees of freedom.

For this two-sided test at $5\%$, the cutoff is the $97.5\%$ percentile of $t_9$, namely $2.26$. The data giveSince $|T|<2.26$, do not reject the manufacturer's mean-hardness claim at the $5\%$ level. This is a failure to reject, not proof that the mean equals $75$.

Define $\mathcal{T}$ to be the family of all unions of members of $\mathcal{B}$, including the empty union. We will verify that this is a topology and that $\mathcal{B}$ is a basis of a topology. The two conditions play different roles: the covering condition supplies the whole space, while the local intersection condition supplies closure under finite intersections.

By the covering condition, $X={\bigcup }_{B\in \mathcal{B}}B$ belongs to $\mathcal{T}$. The empty set belongs to $\mathcal{T}$ as the empty union. An arbitrary union of sets that are themselves unions of members of $\mathcal{B}$ is still a union of members of $\mathcal{B}$. Thus $\mathcal{T}$ contains $X,\varnothing$ and is closed under arbitrary unions, as required for a topology.

Take $U,V\in \mathcal{T}$ and $x\in U\cap V$. Their representations as unions provide $B_1,B_2\in \mathcal{B}$ with $x\in B_1\subseteq U$ and $x\in B_2\subseteq V$. The intersection condition supplies $B\in \mathcal{B}$ with $x\in B\subseteq B_1\cap B_2\subseteq U\cap V$.

It follows that $U\cap V$ is the union of all members of $\mathcal{B}$ contained in it, and hence belongs to $\mathcal{T}$. Induction gives closure under every finite intersection. Together with part (i), this proves that $\mathcal{T}$ is a topology. Each $B\in \mathcal{B}$ belongs to $\mathcal{T}$ as a union with one member, and every open set is a union of such members by construction. Thus $\mathcal{B}$ is a basis for $\mathcal{T}$. Here the printed subset sign is interpreted as inclusion, allowing equality, as in the usual basis criterion.

The lexicographic order is a total order. Each point $p=(a,b)$ lies in the order interval between $(a,b-1)$ and $(a,b+1)$, so the order intervals cover ${\mathbb{R}}^2$.

If $p$ lies in both $⟨x_1,y_1⟩$ and $⟨x_2,y_2⟩$, let $x=\max (x_1,x_2)$ and $y=\min (y_1,y_2)$ in that total order. Then $x<p<y$ andIn particular an order interval contains $p$ and is contained in the intersection. Both conditions of part (a) hold, proving that these intervals form a basis of a topology. The resulting topology is the lexicographic order topology on the real plane.

For every $a\in \mathbb{R}$, the nonempty order intervalis open in the lexicographic order topology on the real plane. The sets $U_a$ are pairwise disjoint.

Suppose $\{B_n:n\in \mathbb{N}\}$ were a countable basis of a topology. For each $a$, the basis of a topology property gives a member containing $(a,0)$ and contained in $U_a$. Let $n(a)$ be the least index of such a member. The same nonempty $B_n$ cannot lie in two disjoint $U_a$, so $a\mapsto n(a)$ is an injection from the uncountable set $\mathbb{R}$ into $\mathbb{N}$, a contradiction. This topology is not second countable.

For distinct interpolation nodes $x_0,\dots ,x_n$, let $p_n$ be the unique polynomial of degree at most $n$ satisfying $p_n(x_i)=f(x_i)$. Existence follows from the Lagrange polynomial expressionuniqueness follows because the difference of two interpolants has $n+1$ distinct zeros but degree at most $n$.

Define the divided difference $f[x_0,\dots ,x_n]$ to be the coefficient of $x^n$ in $p_n$, even when that coefficient is zero. For $n=0$ this gives $f[x_0]=f(x_0)$. Permuting the nodes does not change the interpolation conditions, so it does not change their unique interpolant or this coefficient. The divided difference is symmetric in all the nodes. Equivalently its symmetric expression is

For $n\ge 1$, let $p_L,p_R$ be the interpolation polynomials of degree at most $n-1$ on nodes $x_0,\dots ,x_{n-1}$ and $x_1,\dots ,x_n$, respectively. The polynomialhas degree at most $n$. At $x_0$ it equals $p_L(x_0)$; at $x_n$ it equals $p_R(x_n)$. At each common node both interpolants equal $f$, and the displayed weights sum to one, so $p$ also equals $f$ there. Thus it is the full interpolant by uniqueness.

Its coefficient of $x^n$ is the difference of the leading coefficients of $p_R,p_L$ divided by $x_n-x_0$. By the definition of a divided difference,

Write $D=f[x_0,\dots ,x_n]$ and let $D_i$ be the divided difference with $x_i$ omitted. By symmetry from part (a), reorder the full list so that its first node is $x_i$ and its last is $x_j$. The recurrence from part (b) then has first omitted-node term $D_i$ and second omitted-node term $D_j$, and givesThis proves the requested identity for every pair $i\ne j$ and fixes its sign explicitly; it is one of the omitted-node identities for divided differences.

Continue with $D,D_i$ from part (c). Comparing omission of $x_i$ and $x_n$ givesComparing omission of $x_0$ and $x_n$ gives $D_0=D_n+(x_n-x_0)D$. Eliminating $D$ provesHere $D_n=f[x_0,\dots ,x_{n-1}]$ and $D_0=f[x_1,\dots ,x_n]$, so this is exactly the required omitted-node identities for divided differences. The original nodes need not be ordered, and the formula remains valid even if $\gamma$ lies outside $[0,1]$.

If $U$ is a unipotent matrix, $N=U-I$ is a nilpotent matrix. If $Uv=\lambda v$ for a nonzero eigenvector, then $N^{m}v=(\lambda -1{)}^{m}v=0$ for some $m$, forcing $\lambda =1$.

Conversely, suppose the only eigenvalue is $1$. Over $\mathbb{C}$, the characteristic polynomial is then $(t-1{)}^n$. The Cayley-Hamilton theorem gives $(U-I{)}^n=0$, so $U-I$ is nilpotent. Thus $U$ is unipotent if and only if its only eigenvalue is one.

Choose an invertible change-of-basis matrix $P$ so that $J=PT{P}^{-1}$ is the Jordan normal form of $T$. Write each block aswhere $N_{m_j}$ has ones on its first superdiagonal and all other entries zero. DefineEvery ${\lambda }_j\ne 0$ because $T$ is invertible, so $D_0$ is an invertible diagonal matrix. The matrix $N$ is strictly upper triangular and hence nilpotent. On each block $D_0$ is a scalar multiple of the identity, so it commutes with $N$ on that block and therefore on the whole space. This gives

For $D=P^{-1}{D}_{0}P$, conjugating $U=D^{-1}T$ givesBecause $D_0^{-1}$ commutes with $N$, and $N^n=0$,Thus $U-I$ is similar to a nilpotent matrix, and is itself nilpotent. Therefore $U$ is a unipotent matrix.

The definition gives $T=DU$. The matrix $D=P^{-1}{D}_{0}P$ is invertible and diagonalizable. In the same basis $U$ is $I+D_0^{-1}N$, which commutes with $D_0$ because $N$ does. Conjugating back givesThis is the multiplicative Jordan decomposition in an affine algebraic group, here applied to the general linear group. The explicit construction also handles repeated eigenvalues and nontrivial Jordan blocks.

The residue is the coefficient of $(z-a{)}^{-1}$ in the Laurent series:Equivalently it is $\frac{1}{2\pi i}{\int }_{|z-a|=r}f(z)dz$ for a positively oriented circle small enough to enclose no other singularities.

If the pole has order $k+1$, multiplying the Laurent series by $(z-a{)}^{k+1}$ gives a function $h$ with a holomorphic extension across $a$:Its coefficient of $(z-a{)}^k$ is exactly $c_{-1}$. The Taylor series coefficient formula gives $h^{(k)}(a)=k!c_{-1}$. Continuity of this derivative at $a$ therefore proves

Integrate $F(z)=(1+z^2{)}^{-(k+1)}$ along the interval $[-R,R]$ closed by the positively oriented upper semicircle, with $R>1$. There is just one enclosed pole, at $i$, of order $k+1$. There,Differentiating $k$ times givesBy part (b),On the semicircle $|1+z^2|\ge R^2-1$, so its integral has absolute value at most $\pi R/(R^2-1{)}^{k+1}$, tending to zero. The real integral converges absolutely, so the residue theorem and the limit $R\to \infty$ give

Use an inner product linear in its first argument. The adjoint operator ${\alpha }^{\ast }$ is the unique linear operator such thatIn a finite-dimensional inner product space, existence follows by choosing an orthonormal basis: the matrix of the adjoint is the conjugate transpose of the matrix of $\alpha$. Uniqueness follows from nondegeneracy of the inner product.

If $\alpha (W)\subseteq W$, then for $w\in W$ and $v\in W^{\perp }$,Thus ${\alpha }^{\ast }v\in W^{\perp }$. Conversely, if ${\alpha }^{\ast }(W^{\perp })\subseteq W^{\perp }$, the same equality gives $⟨\alpha w,v⟩=0$ for every $v\in W^{\perp }$, so $\alpha w\in (W^{\perp }{)}^{\perp }=W$. This proves the adjoint criterion for an invariant orthogonal complement.

The equality $(W^{\perp }{)}^{\perp }=W$ uses the finite-dimensional setting appropriate here. In a general Hilbert space it holds for closed subspaces, and an adjoint is assured for bounded operators. Without closedness the converse only implies invariance of the closure. For example in ${\ell }^2$, take $W$ to be the finite-support sequences and $\alpha v=⟨v,e_1⟩(2^{-j}{)}_{j\ge 1}$. Then $W^{\perp }=\{0\}$ is automatically adjoint-invariant, but $\alpha e_1\notin W$. Thus the finite-dimensional or closed-subspace hypothesis cannot be omitted from an infinite-dimensional reading.

A normal operator satisfies ${\alpha }^{\ast }\alpha =\alpha {\alpha }^{\ast }$. Moving an operator across the inner product givesOne of these norms is zero exactly when the other is zero. Therefore