Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 2

The dual vector space is $V^{\ast }={\mathrm{Hom}}_{\mathbb{R}}(V,\mathbb{R})$, the vector space of all linear functionals on $V$. For a vector subspace $U\le V$, its annihilator of a vector subspace is

Given a basis $x_1,\dots ,x_n$, define $x_i^{\ast }$ by $x_i^{\ast }(x_j)={\delta }_{ij}$ and extend linearly. If ${\sum }_i{a}_i{x}_i^{\ast }=0$, evaluation at $x_j$ gives $a_j=0$, so the $x_i^{\ast }$ are linearly independent. Every $\varphi \in V^{\ast }$ satisfiesso they also span $V^{\ast }$ and therefore form its dual basis.

Write $\varphi =a{x}_1^{\ast }+b{x}_2^{\ast }+c{x}_3^{\ast }+d{x}_4^{\ast }$. The condition $\varphi \in U^0$ isGaussian elimination gives $b=-2c-3d$ and $a=c+2d$. Hence

Let $S=\{1,x,x^2,\dots \}$. Since a principal ideal domain is an integral domain and $x\ne 0$, $S$ is a multiplicative subset avoiding zero. The stated equivalence relation and operations are precisely those of the localization of a ring $S^{-1}R$.

For completeness, reflexivity and symmetry are immediate, while transitivity follows by multiplying the two cross-multiplication identities and cancelling a nonzero power of $x$. If $(r,x^n)\sim (s,x^m)$ and $(r^{\prime },x^{n^{\prime }})\sim (s^{\prime },x^{m^{\prime }})$, cross multiplication shows that the proposed sums and products are equivalent. Thus the operations do not depend on representatives. The usual fraction calculations then prove associativity, commutativity and distributivity, with identities $[(0,1)]$ and $[(1,1)]$. Therefore $R^{\prime }$ is a well-defined commutative ring.

Let $J$ be an ideal of $R^{\prime }=S^{-1}R$ and contract it toThis is an ideal of $R$, so $I=(d)$ because $R$ is a principal ideal domain. Every $x^n/1$ is a unit in a ring $R^{\prime }$. Consequently, if $r/x^n\in J$, then $r/1=(x^n/1)(r/x^n)\in J$, so $r\in (d)$ and $r/x^n\in (d/1)$. Conversely $d/1\in J$, hence every multiple of it belongs to $J$. Thusso every ideal of $R^{\prime }$ is principal. The localization remains an integral domain; hence $R^{\prime }$ is a principal ideal domain.

The L1 norm properties of absolute homogeneity and the triangle inequality follow from the corresponding properties of the absolute value and the integral. If $\Vert f{\Vert }_1=0$, the nonnegative continuous function $|f|$ has zero integral. Were $|f(x_0)|>0$, continuity would make it bounded below by a positive number on a nontrivial interval, contradicting the zero integral. Thus $f=0$, proving positive definiteness.

Put $M=\Vert f{\Vert }_{\infty }$ and choose the continuous cutoff functionThen $g_n(0)=g_n(1)=0$, $\Vert g_n{\Vert }_{\infty }\le M$, and $f-g_n$ is supported in the two endpoint intervals of total length $2/n$. Therefore

Now suppose ${\int }_0^{1}fg=0$ for every $g\in \mathcal{S}$. Apply this to the above $g_n$. By the uniform norm bound,Hence ${\int }_0^1{f}^2=0$, and continuity again gives $f=0$.

A metric space metric $d:X\times X\to [0,\infty )$ satisfies positivity, $d(x,y)=0$ exactly when $x=y$, symmetry, and the triangle inequality. A subset $U\subseteq X$ is an open set when every $x\in U$ has some $\epsilon >0$ with the open ball $B_d(x,\epsilon )\subseteq U$.

Both $\varnothing$ and $X$ are open. An arbitrary union of open sets is open because a point lies in one member supplying a ball. A finite intersection is open because the minimum of the finitely many available radii supplies a ball. Thus the metric-open sets satisfy the topology axioms.

For $C[0,1]$, the Cauchy-Schwarz inequality gives $d_1(f,g)\le d_2(f,g)$, so the $d_2$ topology is at least as fine as the $d_1$ topology. To see strictness, define the continuous triangular spikesThenThus $f_n$ converges to zero in the L1 norm but not in the L2 norm. The two metrics induce different topologies.

Along a curve $s\mapsto (x(s),y(s))$, the chain rule givesTogether with the second-order partial differential equation, these form a linear system for $(u_{xx},u_{xy},u_{yy})$ whose coefficient matrix isA characteristic curve is precisely one along which this system is singular, equivalently where the principal symbol vanishes on a normal covector. Taking the determinant yields

For a time-independent current, the Maxwell equations reduce to magnetostatics:Write $\mathbf{B}=\nabla \times \mathbf{A}$ using a vector potential and choose the Coulomb gauge $\nabla \cdot \mathbf{A}=0$. The vector identity for the curl of a curl gives the Poisson equationThe free-space Green function therefore givesTaking the curl with respect to $\mathbf{r}$, moving it under the integral, and using $\nabla (1/|\mathbf{r}-{\mathbf{r}}^{\prime }|)=-(\mathbf{r}-{\mathbf{r}}^{\prime })/|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3$ gives the Biot-Savart law

Write $\mathbf{g}=-g{\mathbf{e}}_z$. The vertical component of the balance gives hydrostatic pressure,Its horizontal gradient is ${\nabla }_{h}p=\rho g{\nabla }_{h}h$. The horizontal balance is consequentlyThus $\mathbf{u}\cdot {\nabla }_{h}h=0$: the velocity is tangent to every level set of $h$, so those contours are streamlines of the geostrophic flow.

Across a vertical section joining the contours $h_0$ and $h_0+\Delta h$, an element of horizontal distance normal to the contours is $dn=dh/|{\nabla }_{h}h|$, while $|\mathbf{u}|=(g/f)|{\nabla }_{h}h|$. The volumetric flow rate is thereforeFor dimensional analysis, $[g]=L{T}^{-2}$, $[f]=T^{-1}$ and $[h_0]=[\Delta h]=L$, so the result has dimension $L^3{T}^{-1}$, exactly that of a volume flux.

A simple hypothesis specifies a single probability distribution. For a test with rejection region $R$, its size of a statistical test is the rejection probability under the simple null hypothesis, while its statistical power against the simple alternative is the rejection probability under that alternative. The Neyman-Pearson lemma states that among tests of size at most $\alpha$, a test rejecting for the largest values of the likelihood ratio $f_1/f_0$ has greatest power, with boundary randomization if needed.

Herewhich decreases with $|x|$. The best test therefore rejects for $|X|\le c$. Under $H_0$, ${\mathbb{P}}_0(|X|\le c)=c$, so size $0.05$ requires $c=0.05=1/20$. Its power isThus reject $H_0$ exactly when $|X|\le 0.05$.

A function $f:{\mathbb{R}}^n\to \mathbb{R}$ is a convex function whenfor every $x_0,x_1$ and $t\in [0,1]$.

Fix $b_0,b_1\in \mathbb{R}$, $t\in [0,1]$, and $\epsilon >0$. Since the value function is finite, choose $x_i$ withFor the convex combination $x_t=(1-t)x_0+t{x}_1$, convexity of $g$ givesso $x_t$ is feasible for the intermediate right-hand side. Convexity of $f$ then givesLetting $\epsilon \downarrow 0$ provesso $\varphi$ is convex.

View $X$ as the matrix of a linear map $T:F^m\to F^n$ of matrix rank $r$. Choose vectors $u_1,\dots ,u_r$ whose images form a basis of $\mathrm{im}T$, extend them by a basis of $\ker T$ to a basis of $F^m$, and extend $T{u}_1,\dots ,T{u}_r$ to a basis of $F^n$. In these two bases the matrix of $T$ is the rank normal formThe two changes of basis give invertible $P,Q$ with the displayed matrix equal to $Q^{-1}XP$.

For the block upper triangular matrix $A=\left(\begin{array}{cc}B& D\\ 0& C\end{array}\right)$, every nonzero term in the Leibniz formula for determinants sends the rows belonging to $C$ into the columns belonging to $C$; bijectivity then sends the remaining rows into the $B$ columns. The permutation sum consequently factors into the determinant sums for $B$ and $C$, proving

Finally suppose $L(X)=0$, so $AX=XB$: the map $X:{\mathbb{C}}^m\to {\mathbb{C}}^n$ is an intertwining operator. If $v$ lies in the generalized eigenspace of $B$ for $\lambda$, then for some $k$,Because $A$ and $B$ have no common eigenvalue, $A-\lambda I$ is invertible, hence $Xv=0$. The generalized eigenspaces of $B$ span ${\mathbb{C}}^m$, so $X=0$. Thus the Sylvester equation operator $L$ is injective.

First, every nonzero nonunit in a principal ideal domain factors into irreducible elements. Otherwise repeatedly choose a proper nonunit factor $a_{j+1}$ of $a_j$; then the principal idealsform a strictly increasing chain, contradicting the ascending chain condition, since every principal ideal domain is a Noetherian ring.

Next every irreducible $p$ is a prime element. If $p\mid ab$ but $p\nmid a$, then $\gcd (p,a)=1$. The Bezout identity gives $up+va=1$, so multiplying by $b$ shows $p\mid b$. Thus irreducibles are prime.

Existence of an irreducible factorization and primality give uniqueness: an irreducible on one side divides the product on the other, hence is associate to one factor; cancel and continue. Therefore every principal ideal domain is a unique factorization domain.

Any unit in a ring of $R$ must also be a unit of $\mathbb{Q}[X]$, hence a nonzero constant $q\in \mathbb{Q}$. Membership of both $q$ and $q^{-1}$ in $R$ requires $q,q^{-1}\in \mathbb{Z}$, so

Let $f\in R$ be irreducible. If $\mathrm{deg}f=0$, then $f\in \mathbb{Z}$. A nonzero integer that is neither a unit nor, up to sign, a prime number factors nontrivially in $R$, so necessarily $f=\pm p$ for a prime $p$.

Now suppose $\mathrm{deg}f\ge 1$ and write $m=f(0)\in \mathbb{Z}$. If $m=0$, thenand both factors lie in $R$ and are nonunits. If $|m|>1$, choose a prime $p\mid m$; then $f=p(f/p)$ is again a factorization into two nonunits of $R$, since $(f/p)(0)=m/p\in \mathbb{Z}$. Both cases contradict irreducibility. Hence

Suppose $X=f_1\cdots f_s$ were a product of irreducible elements of $R$. Since degree of a polynomial is additive under products, exactly one factor has positive degree and all the others are constant. By the preceding classification, the positive-degree factor has constant term $\pm 1$, while every constant irreducible is $\pm p$ for a prime $p$. Their product therefore has a nonzero constant term, contradicting $X(0)=0$.

Thus $X$ is not a product of irreducibles, so this integral domain is not an atomic domain and in particular is not a unique factorization domain.

A map $f:A\to \mathbb{R}$ is Lipschitz continuous with Lipschitz constant $L$ whenfor all $y,z\in A$.

Fix $y_0\in A$. The Lipschitz inequality and the triangle inequality implywhile choosing $y=y_0$ gives a finite upper bound. Hence the infimum defining $F(x)$ is a real number.

If $x\in A$, then $f(x)\le f(y)+Ld(x,y)$ for every $y\in A$, while $y=x$ gives the reverse inequality after taking the infimum. Thus $F(x)=f(x)$. For arbitrary $x,z\in X$,Taking infima gives $F(x)\le F(z)+Ld(x,z)$; interchanging $x,z$ givesThis is the McShane extension theorem: $F$ extends $f$ without increasing its Lipschitz constant.

Two norms $N_1,N_2$ are equivalent norms when constants $c,C>0$ satisfyfor every $v$.

For $x,y\in {\mathbb{R}}^n$, the reverse triangle inequality givesby the Cauchy-Schwarz inequality. Hence $g$ is Lipschitz continuous, and therefore continuous.

On the compact Euclidean unit sphere, $g$ is positive and continuous, so the extreme value theorem gives $0<m\le g\le M<\infty$. Homogeneity then yieldsThus every norm on a finite-dimensional vector space is equivalent to the Euclidean norm; comparing two such bounds proves any two norms on $V$ are Lipschitz equivalent.

Let $P_n$ be the finite-dimensional vector space of real polynomials of degree at most $n$. For fixed $a>0$,is a norm: if it vanishes, the polynomial vanishes on an interval and hence is the zero polynomial. Alsois a norm on $P_n$. By finite-dimensional equivalence of norms, there is $C>0$, depending only on $n$ and $a$, such that $M(p)\le CN(p)$. ThereforeThe extreme value theorem supplies $y\in [0,a]$ at which the last supremum is attained, and hence

On an annulus $r<|z-z_0|<R$, the Laurent theorem represents an analytic function uniquely asFor any positively oriented simple closed contour $C$ in the annulus winding once around $z_0$, the coefficients in the Laurent series areThe formula follows from the Cauchy integral formula applied term by term, and it is independent of the chosen contour by the Cauchy integral theorem.

Using the Taylor series of the exponential and series division,Thus the first three nonzero Laurent terms areThe singularities occur at $z=k\pi i$. The nearest nonzero ones are $\pm \pi i$, so this Laurent series is valid on $0<|z|<\pi$.

The function $f$ has a simple pole of residue $1/2$ at every $z=k\pi i$, because $(e^{2z}-1{)}^{\prime }=2$ there. At $z=0$, the term $1/(2z)$ in $g$ cancels the principal part of $f$. For $1\le k\le m$, the summandhas residue $1/2$ at each of $z=\pm k\pi i$, so it cancels the corresponding pole of $f$. Therefore all possible singularities of $F$ in $|z|<(m+1)\pi$ are removable singularities: they are $0$ and $\pm k\pi i$ for $1\le k\le m$. After filling them in, $F$ is analytic throughout that disc.

Write the function from the preceding part as $F_m$. Its expansion at zero isHence the residue theorem givesChoose $R=R_m=(m+\frac{1}{2})\pi$. This circle stays a distance at least $\pi /2$ from every pole of $f$, so $f$ is bounded uniformly on it. MoreoverThus ${\max }_{C_{R_m}}|F_m|=O(\log m)$, and the ML inequality yieldsLetting $m\to \infty$ solves the Basel problem:

The characteristic polynomial of $A$ isWhen $|\mathrm{tr}A|>2$, its discriminant is positive, so $A$ has two distinct real eigenvalues $\lambda ,{\lambda }^{-1}$ and two independent real eigenvectors. Let $P$ have those eigenvectors as columns. Rescaling one column makes $\det P=1$ without changing the diagonalization. Hence $P\in S{L}_2(\mathbb{R})$ andThis is the normal form of a hyperbolic element of PSL2(R).

Write $B=\mathrm{diag}(\lambda ,{\lambda }^{-1})$ and $q={\lambda }^2>0$, so its Möbius transformation is $z\mapsto qz$, with $q\ne 1$. For $z=u+iv$, the Hyperbolic distance in the Poincare half-plane satisfiesAt $z=i$, the final parenthesis is $1$. Since $\cosh$ is strictly increasing on nonnegative arguments, every $z$ off the imaginary axis has

A fixed point of $A$ satisfiesIts discriminant iswhere $ad-bc=1$ was used. Thus the two roots are nonreal complex conjugates, and exactly one lies in the upper half-plane. Consequently every elliptic element of PSL2(R) fixes a point of $\mathbb{H}$.

Take the parabolic element of PSL2(R)It fixes no point in $\mathbb{H}$. A Geodesic in the Poincare half-plane model is determined by its unordered pair of ideal endpoints: either two real numbers or one real number together with $\infty$. Translation adds one to every finite endpoint, so no such pair is invariant. Therefore this $A$ preserves neither a point nor a hyperbolic line in $\mathbb{H}$.

For a variation $y+\epsilon \eta$ with $\eta ={\eta }^{\prime }=0$ at both endpoints, the first variation isTwo applications of integration by parts remove derivatives from $\eta$. The boundary terms vanish, and the fundamental lemma of the calculus of variations gives the higher-order Euler-Lagrange equation

For the stated integrand,so the equation reduces toThe general solution of this linear ordinary differential equation isDecay and finiteness of the integral force $A=B=0$; $y(0)=1$ gives $C=1$, and $y^{\prime }(0)=2$ gives $D=3$. Hence the unique stationary function is

The function is an even function, so every sine Fourier coefficient vanishes. Direct integration givesThusfor the $2\pi$-periodic extension.

The homogeneous solutions $(C_1+C_{2}x)e^{-x}$ are not $2\pi$-periodic unless $C_1=C_2=0$, so there is a unique periodic solution. For a forcing term $a_n\cos (nx)$, a trial solution $A_n\cos (nx)+B_n\sin (nx)$ givesUsing the Fourier series from part (a), for which only odd $n$ occur and $a_n=-4/(\pi n^2)$, yieldsThe coefficients decay fast enough to differentiate the series twice term by term, verifying the equation and periodicity.

In atomic units the Time-independent Schrodinger equation isThe displayed equation is the Radial Schrodinger equation for the hydrogen atom. Its radial derivative terms come from the Laplacian in spherical coordinates, the Coulomb term is the potential $-1/r$, and separation into spherical harmonics uses the orbital angular momentum eigenvalue $l(l+1)$, producing the centrifugal term $-l(l+1)R/r^2$.

For $E=-1/(2n^2)$, substitute $R=r^{\alpha }{e}^{-r/n}$. Dividing the equation by $R$ and comparing powers of $r$ givesThe solution regular at the origin has $\alpha =l$, and when $l=n-1$ both equations agree. This is the circular Coulomb bound state. ThusThe radial probability measure is $|R(r){|}^2{r}^{2}dr$. With $a=2/n$ and the Gamma integral, the mean radius of a circular Coulomb bound state is

At fixed principal quantum number $n$, the allowed orbital angular momentum quantum numbers are $l=0,\dots ,n-1$, and each has magnetic quantum number $m=-l,\dots ,l$. Hence the orbital quantum degeneracy isIncluding electron spin would double this to $2n^2$, but spin is absent from the stated wavefunctions.

Write the jump from side $-$ to side $+$ as $[\mathbf{F}]={\mathbf{F}}_{+}-{\mathbf{F}}_{-}$. A thin rectangular loop in the static Faraday's law and a thin pillbox in Gauss's law for magnetism giveThus the tangential electric field and normal magnetic field are continuous in the plane's rest frame.

For the boost $\mathbf{v}=v\mathbf{n}$, the Lorentz transformation of electromagnetic fields givesThereforeMoreover $[\mathbf{E}]$ is purely normal, so $[{\mathbf{B}}_t^{\prime }]=\gamma [{\mathbf{B}}_t]$. Consequently the transformed tangential boundary condition isUsing the magnetic-field jump across a surface current, $\mathbf{n}\times [\mathbf{B}]={\mu }_0\mathbf{K}$, this can equivalently be writtenThe nonzero jump is consistent with Faraday's law because the magnetic-field discontinuity itself moves through Albert's frame.