Past exam of the options I course of the University of Oxford / 2012 (AO1, 4303)

Source: Mathematics Part A release, member FOI-20210708-1 - Data/4303 (1).pdf, dated 20 June 2012. All nineteen questions and all fifty-nine printed source leaf parts are answered below.

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The degree of a field extension is the dimension of the larger field, regarded as a vector space over the smaller field:The scalar multiplication is the multiplication of $K$ restricted to scalars in $F$. A finite extension has finite dimension; otherwise the degree is infinite.

The notation $K=F(\delta )$ means that $K$ is the smallest subfield containing both $F$ and $\delta$, namely the simple field extension generated by $\delta$. Its elements can be expressed as rational functions in $\delta$ with coefficients in $F$ and nonzero denominator.

Write $d={\delta }^2\in F$. The element $\delta$ is a root of $X^2-d$, so its minimal polynomial over $F$ has degree at most two. Since $\delta \notin F$, that degree cannot be one. The degree of the simple field extension equals the degree of the minimal polynomial, givingIn particular $1,\delta$ form an $F$-basis, so every element is uniquely $a+b\delta$. This argument also works in characteristic two; separability is not needed for this part.

Choose $\alpha \in K\setminus F$. Since $[K:F]=2$, the linearly independent elements $1,\alpha$ form a basis of $K$ over $F$. Thus there are $a,b\in F$ with ${\alpha }^2=a+b\alpha$. The assumption on the characteristic of a field allows division by two. SetCompleting the square givesAlso $\delta \notin F$, since otherwise $\alpha =\delta +b/2$ would lie in $F$. Because $\alpha \in F(\delta )$ and $\delta \in F(\alpha )$, these generated subfields agree. The basis representation shows $F(\alpha )=K$, henceThe square-root generator is obtained by translating an arbitrary generator to remove the linear term of its quadratic relation.

A left group action of $G$ on a set $S$ is a map $G\times S\to S$, written $(g,s)\mapsto g\cdot s$, satisfyingfor every $s\in S$ and $g,h\in G$, where $e$ is the identity element of $G$. In particular each $g$ acts by a bijection, whose inverse is the action of $g^{-1}$.

Given a group action, define $\rho (g)(j)=g\cdot j$. Each $\rho (g)$ is a permutation, since $\rho (g^{-1})$ is its inverse. The action law gives, for every $j$,Thus $\boxed{\rho :G\to S_n\text{ is a group homomorphism}}$.

Conversely, given a group homomorphism $\rho :G\to S_n$, put $g\cdot j=\rho (g)(j)$. Since a homomorphism sends the identity to the identity permutation, $e\cdot j=j$. AlsoThese are exactly the two group action axioms, so the construction defines the required action.

The three nonzero vectors in ${\mathbb{F}}_2^2$ are $e_1$, $e_2$ and $e_1+e_2$. An invertible linear map permutes this three-element set, so part (b) gives a group homomorphismIf a matrix acts trivially on all three vectors, it fixes both basis vectors $e_1,e_2$, so it is the identity matrix. Thus the kernel of a group homomorphism is trivial and $\rho$ is injective.

To count the domain, its first column may be any of the three nonzero vectors. The second column must be outside the two-element span of the first, leaving two choices. HenceAn injection between these finite sets of equal size is surjective, proving

For a positive integer $n$, Euler's totient function counts the integers in a complete residue system that are coprime to $n$:For $n>1$ this is equally the count with $0<k<n$. The convention gives $\phi (1)=1$.

Let $R_n=\{k\in \mathbb{Z}:0<k<n,\gcd (k,n)=1\}$. The map $k\mapsto n-k$ sends $R_n$ to itself becauseApplying it twice returns $k$, so it is a bijection. Reindexing the finite sum by this bijection givesThere is no need to assume that every residue belongs to a pair of distinct elements; the reindexing also handles the fixed point when $n=2$.

Write $S={\sum }_{k\in R_n}k$. Adding the two equal sums from part (b), and using $|R_n|=\phi (n)$, givesThereforeEquivalently, the average of the positive reduced residues is $n/2$, a consequence of their symmetry under reflection about that midpoint.

The positive integers below $30$ coprime to $30$ areThere are eight, so $\phi (30)=8$. Their sum can be grouped into four pairs summing to $30$:This agrees with the formula.

A sigma-algebra $\mathcal{F}$ on $\Omega$ contains $\Omega$, is closed under complements relative to $\Omega$, and is closed under countable unions. Consequently it contains the empty set and is closed under countable intersections. A measure space is a triple $(\Omega ,\mathcal{F},\mu )$ with a map $\mu :\mathcal{F}\to [0,\infty ]$ satisfying $\mu (\varnothing )=0$ and countable additivity on pairwise disjoint measurable sets.

For the increasing sequence, set $C_1=A_1$ and $C_n=A_n\setminus A_{n-1}$ for $n\ge 2$. These measurable sets are disjoint, their union is $A$, and $A_N={\bigcup }_{n=1}^N{C}_n$. Thus countable additivity and finite additivity giveThe identity permits the value infinity; it uses no subtraction of infinite measures. This is continuity from below of a measure.

For a decreasing sequence, the analogous assertion is not always true. On $\Omega =\mathbb{N}$ with the counting measure, take $B_n=\{n,n+1,\dots \}$. Then ${\bigcap }_n{B}_n=\varnothing$, so $\mu (B)=0$, whereas $\mu (B_n)=\infty$ for every $n$ and their limit is infinity.

A sufficient extra condition for continuity from above of a measure is $\mu (B_1)<\infty$. Indeed $B_1\setminus B_n$ increases to $B_1\setminus B$, and the increasing-sequence result givesMore generally it suffices that one of the $B_n$ has finite measure. The missing finiteness assumption is exactly what the counterexample violates.

On any bounded parameter interval $|t|\le T$, the derivative of the integrand has the integrable majorantThe dominated convergence theorem therefore justifies differentiation under the integral sign at every real $t$, givingUse integration by parts in this absolutely convergent integral, with $x{e}^{-x^2}dx=-\frac{1}{2}d(e^{-x^2})$. Both boundary terms vanish, since the sine is zero at $x=0$ and the Gaussian tends to zero at infinity. HenceIt follows that $f^{\prime }=-4t^{3}f$, or equivalently $(e^{t^4}f(t){)}^{\prime }=0$. This identity holds on all of $\mathbb{R}$, including zero. With the supplied Gaussian integral value,The fourth power comes from the printed phase $2x{t}^2$; the parameter has been differentiated with that phase throughout.

The map is $K$-Lipschitz continuous whenThe same finite Lipschitz constant $K$ must work for all pairs of points.

The map is uniformly continuous when, for every $ϵ>0$, there is a $\delta >0$ such thatThe choice of $\delta$ may depend on $ϵ$ but must be independent of the locations of $x$ and $x^{\prime }$.

Given $ϵ>0$, choose $\delta =ϵ/K>0$. If $d_X(x,x^{\prime })<\delta$, the Lipschitz condition yieldsThis single $\delta$ works for every pair, proving the map is uniformly continuous:

The distance to a set is finite and nonnegative because $A$ is nonempty. For $x,x^{\prime }\in X$ and every $a\in A$, the triangle inequality givesTaking the infimum over $a$ yields $D_A(x)\le d(x,x^{\prime })+D_A(x^{\prime })$. Interchanging $x$ and $x^{\prime }$ gives the opposite inequality. Together these implyThus the distance to a set is $1$-Lipschitz continuous. No closedness, compactness or attainment of the infimum was used.

For fixed $x\in X$, the function $a\mapsto d(x,a)$ on $A$ is continuous, indeed $1$-Lipschitz continuous by the reverse triangle inequality. A continuous real-valued function on a nonempty compact space attains its minimum by the extreme value theorem. Hence some $a_0\in A$ satisfiesThe metric is symmetric, so this is also $d(a_0,x)$. Compactness supplies attainment of the infimum.

A basis of a topology $\mathcal{B}$ is a family of open subsets such that every open set is a union of members of $\mathcal{B}$. Equivalently, for every $U\in \mathcal{T}$ and $x\in U$, some $B\in \mathcal{B}$ satisfiesThe empty open set is represented by the empty union.

Define $\mathcal{T}$ to be all unions of members of $\mathcal{B}$, including the empty union. Then $\varnothing \in \mathcal{T}$ and the covering assumption gives $X\in \mathcal{T}$. Arbitrary unions of such sets are again unions of basis members.

For two members $U={\bigcup }_i{B}_i$ and $V={\bigcup }_j{B}_j^{\prime }$, distribute intersection over union:Each intersection is a union of members of $\mathcal{B}$ by the second assumption, so $U\cap V\in \mathcal{T}$. Repeating gives closure under all finite intersections. These are precisely the topology axioms. Every $B\in \mathcal{B}$ belongs to $\mathcal{T}$, and every set in $\mathcal{T}$ is a union of such members by construction. Thus

The product topology on $X\times Y$ has as a basis all open rectanglesThus a subset $W\subseteq X\times Y$ is open precisely when, around every $(x,y)\in W$, there are open $U\ni x$ and $V\ni y$ with $U\times V\subseteq W$. These rectangles cover the product, and their pairwise intersections are rectangles because $(U\times V)\cap (U^{\prime }\times V^{\prime })=(U\cap U^{\prime })\times (V\cap V^{\prime })$. Hence they define a topology by part (b).

This is equivalently the coarsest topology making the two coordinate projection maps continuous: the preimages of factor-open sets are the open strips $U\times Y$ and $X\times V$, whose finite intersections are the basis rectangles.

Let $\{B_i:i\in I\}$ and $\{C_j:j\in J\}$ be countable bases of the two factors. The familyis countable, since the Cartesian product of two countable sets is countable. It is a basis of a topology for the product topology: given a product-open neighbourhood of $(x,y)$, choose a contained open rectangle $U\times V$, then basis members with $x\in B_i\subseteq U$ and $y\in C_j\subseteq V$. Their product contains $(x,y)$ and is contained in the original neighbourhood. ThereforeIn standard terminology, a finite product of second-countable spaces is second-countable.

The function is continuously differentiable when all its first partial derivatives exist on $U$ and are continuous there. Equivalently, $h$ has a Fréchet derivative at every point and the map $x\mapsto Dh(x)$ is continuous. For a scalar-valued function,On the open domain, continuous first partial derivatives ensure that this linear expression is the total derivative.

For differentiable maps with compatible domains, the multivariable chain rule isIn Jacobian matrix notation this is matrix multiplication. In particular, if both maps are continuously differentiable, the composition is continuously differentiable. Componentwise,

Let $L(x,y,z)=(x-y,y-z,z-x)$. This is a linear map, hence smooth; composing it with the given $C^1$ function proves that $g=f\circ L$ is continuously differentiable by the chain rule. Write $f_1,f_2,f_3$ for the first partial derivatives, evaluated at $L(p)$. ThenAdding these identities givesGeometrically, $L(p+t(1,1,1))=L(p)$, so $g$ is constant along every line in the diagonal direction; its derivative in that direction must vanish.

An embedded $k$-dimensional submanifold $M\subseteq {\mathbb{R}}^n$ is locally a coordinate $k$-plane: for every $p\in M$, there are open neighbourhoods $U\ni p$, $V\subseteq {\mathbb{R}}^n$ and a diffeomorphism $\Phi :U\to V$ withFor the present $C^1$ setting, $\Phi$ and its inverse are continuously differentiable; in the smooth convention both are smooth.

A tangent vector to $M$ at $p$ is a velocity ${\gamma }^{\prime }(0)$ of a continuously differentiable curve $\gamma :(-ϵ,ϵ)\to M$ with $\gamma (0)=p$. The tangent space is the collection of all these vectors:In the local coordinates above, it is $D\Phi (p{)}^{-1}({\mathbb{R}}^k\times \{0{\}}^{n-k})$, so it is a $k$-dimensional linear subspace of ${\mathbb{R}}^n$.

The regular level set theorem says that for a $C^1$ map $g:U\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^r$, if $Dg$ has rank $r$ on $g^{-1}(c)$, then $g^{-1}(c)$ is an $(n-r)$-dimensional submanifold andHere $r=1$ and the given rank condition makes the zero set a two-dimensional submanifold. It is nonempty because $g(0)=f(0)=0$. Part (c) givesConsequentlyThere is also a direct curve proof: $g(p+t(1,1,1))=g(p)=0$, so the entire line through $p$ in this direction lies in $M$. Its velocity at zero is exactly the required tangent vector.

For variations vanishing at the endpoints, the Euler-Lagrange equations areDifferentiate the stated energy expression using the chain rule. The acceleration terms cancel, leavingAlong a stationary trajectory the two brackets vanish. Thus $dH/dt=-F_t$, and in the absence of explicit time dependence $\boxed{H\text{ is constant}}$.

In plane polar coordinates, the squared speed is ${\dot{r}}^2+r^2{\dot{\theta }}^2$, so the kinetic energy is $T=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)$. An inverse-square central gravitational force has potential energy proportional to $1/r$. With the sign convention printed in this paper, write $V=km/r$, givingFor attraction to a star of mass $M_{\star }$, this convention has $k=-G{M}_{\star }<0$. Keeping $k$ real and signed is essential; no sign is changed in differentiating the printed functional.

Since $F_r=mr{\dot{\theta }}^2+km/r^2$, $F_{\dot{r}}=m\dot{r}$, $F_{\theta }=0$ and $F_{\dot{\theta }}=m{r}^2\dot{\theta }$, the equations of motion areThe second equation gives the conserved angular momentum $J=m{r}^2\dot{\theta }$; equivalently $r\ddot{\theta }+2\dot{r}\dot{\theta }=0$. The autonomous-functional identity gives a second conserved quantity,This is the total mechanical energy; for attractive gravity the potential contribution is negative because $k<0$.

Away from a collision, let $r=q_1-q_2$ and $D=|r|>0$. The Lagrangian isDifferentiating $D^2=r\cdot r$ with respect to $x_1$ and $x_2$ gives $2D{\partial }_{x_1}D=2(x_1-x_2)$ and $2D{\partial }_{x_2}D=-2(x_1-x_2)$. ThereforeThe other coordinates give ${\nabla }_{q_1}D=r/D$ and ${\nabla }_{q_2}D=-r/D$. Substituting these into the Euler-Lagrange equations yields equal and opposite central forces:Adding the equations shows that the derivative of the total linear momentum is zero, so $P=m_1{\dot{q}}_1+m_2{\dot{q}}_2$ is constant. For the total angular momentum, the terms ${\dot{q}}_i\times {\dot{q}}_i$ vanish, givingThus both stated vectors are conserved quantities.

The total energy is also constant. Define $E=\frac{1}{2}{m}_1|{\dot{q}}_1{|}^2+\frac{1}{2}{m}_2|{\dot{q}}_2{|}^2+V(D)$. Since $\dot{D}=r\cdot ({\dot{q}}_1-{\dot{q}}_2)/D$,Hence $\boxed{E=T+V(D)\text{ is conserved}}$. The explicit cancellation establishes energy conservation for an arbitrary smooth separation-dependent potential, throughout any collision-free trajectory.

The interior time-independent Schrödinger equation for a free particle isConfinement by the rectangular infinite well imposes Dirichlet boundary conditions: $\psi =0$ at $x=0,b$ and $y=0,\sqrt{5}b$. The wave function is square-integrable and may be normalized to unit probability on the rectangle.

Separation of variables gives products of the one-dimensional well states,The Hamiltonian is the sum of the independent $x$ and $y$ kinetic-energy operators, so the one-dimensional energies add. Put $C={\pi }^2{\hslash }^2/(2m{b}^2)$. ThenFor $n=1$, increasing $\ell$ gives $6C/5,9C/5,14C/5,21C/5$. The smallest energy with $n\ge 2$ is also $21C/5$, attained at $(2,1)$; all $n\ge 3$ give larger energies. Thus the ground and first three distinct excited energy levels areThe largest listed energy therefore has energy-level degeneracy $\boxed{2}$, since the two product eigenfunctions are linearly independent and no other pair solves $5n^2+{\ell }^2=21$.

For a steady flow the local time derivative in the Euler equations vanishes. The vector identityreduces to a gradient when the vorticity $\nabla \times u$ is zero. With constant density and gravity $-gk$, the momentum equation becomesThus on each connected flow region,This is Bernoulli's theorem for a steady irrotational flow: irrotationality makes the same constant apply throughout a connected region, rather than only along an individual streamline.

This is a Rankine vortex. Write $r=\sqrt{x^2+y^2}$. Inside the core the velocity is $u=(\Gamma y/a^2,-\Gamma x/a^2,0)$, soThis is the core value obtained from the inner formula; outside, the given irrotationality makes the vorticity zero. For nonzero $\Gamma$, the sharp boundary $r=a$ has different one-sided velocity gradients, so a single classical vorticity is not defined exactly on that edge; the inner trace is the boxed value and the outer trace is zero.

Inside the core, direct differentiation gives $(u\cdot \nabla )u=-{\Gamma }^2(x,y,0)/a^4$. The steady Euler equations therefore give $p_r=\rho {\Gamma }^{2}r/a^4$ and $p_z=-\rho g$. Integrating,Outside, $|u{|}^2={\Gamma }^2/r^2$, so part (a) givesMatching pressure at the core edge $r=a$ yields $A=B-\rho {\Gamma }^2/a^2$. Hence the complete pressure field, up to one additive constant $B$, isBoth the velocity and pressure are continuous at $r=a$. The inside expression has been derived from the momentum equation because the core is rotational.