Past exam of the options II course of the University of Oxford / 2004 (AO2, XMAT 4304)

Options II, Honour School of Mathematics; Friday 25 June 2004, 2.30pm–5.30pm. The complete thirteen-page source was supplied in the disclosed past-papers archive, member “FOI-20210708-1 - Data/4304 (9).pdf”, through the original request. A full editable source transcription is held privately. No exact worked-answer advertisement was found in the bounded availability check.

The permutation inverse is ${\rho }_{g^{-1}}$: applying the two maps successively sends $\omega$ to $(\omega g)g^{-1}=\omega$. With the paper's right-action convention, permutation products also act on the right, andThus $\boxed{\rho (gh)=\rho (g)\rho (h)}$. If function composition is written on the left instead, the same right action is represented by $g\mapsto {\rho }_{g^{-1}}$; keeping these conventions consistent avoids an artificial reversal. The kernel consists precisely of elements fixing every point, so

The rotational tetrahedral group acts transitively on its four vertices. A rotation fixing a vertex must preserve its opposite face and rotate its three vertices cyclically; its stabilizer has order three. The orbit-stabilizer theorem gives $|G|=4\cdot 3=12$. The rotations are the identity, two nontrivial rotations about each of the four vertex-to-opposite-face axes, and a half-turn about each of the three opposite-edge-midpoint axes. Therefore the counts by element order areA rotation fixing all vertices fixes three linearly independent vertex vectors and is the identity. The vertex permutation action is therefore faithful. The listed rotations induce the identity, 3-cycles and double transpositions, all even permutations. Its injective image has order 12, the full order of $A_4$. Hence $\boxed{G\cong A_4}$.

Take $\boxed{K=F[x]/(f)}$ and let $\alpha$ be the residue class of $x$. Division with remainder represents every class by a polynomial of degree less than $n$, so every element is a polynomial in $\alpha$ and $f(\alpha )=0$. A nonzero class represented by $u$ has $\gcd (u,f)=1$ because $f$ is irreducible. Bézout's identity gives $au+bf=1$, making the class of $a$ its inverse. Thus $K$ is a field, containing $F$ through its constant classes.

A splitting field of a nonconstant polynomial is an extension in which it factors into linear factors and which is generated over the original field by its roots. To construct one, choose an irreducible factor and use the preceding quotient construction to adjoin a root. Divide off a linear factor in that extension and repeat on the remaining polynomial. Each stage lowers the remaining degree, so finitely many stages suffice. The field generated by all roots in the resulting field is the required splitting field. This induction works with repeated roots and does not assume separability. The two roman clauses in the source specify properties of this one construction; they are not separate questions.

The roots are $\pm \sqrt{-5}$, so one root already supplies both. The polynomial has no rational root and is irreducible of degree two. Thus its splitting field is $\mathbb{Q}(\sqrt{-5})$, with $\boxed{[K:\mathbb{Q}]=2}$.

Put $\alpha =\sqrt[3]{2}$ and $\zeta =e^{2\pi i/3}$. The three roots are $\alpha ,\alpha \zeta ,\alpha {\zeta }^2$, so the splitting field is $K=\mathbb{Q}(\alpha ,\zeta )$. Eisenstein's criterion at two gives $[\mathbb{Q}(\alpha ):\mathbb{Q}]=3$. This intermediate field is real, while $\zeta$ is not real, and $\zeta$ satisfies $x^2+x+1=0$. Its degree over the intermediate field is therefore exactly two. By the tower law, $\boxed{[K:\mathbb{Q}]=6}$.

Over ${\mathbb{F}}_5$,The quadratic discriminant is $9-16\equiv 3$, while the squares modulo five are $0,1,4$. Thus that factor is irreducible. Adjoining either quadratic root creates a degree-two extension and supplies the other root by the sum-of-roots relation. Consequently the splitting field is ${\mathbb{F}}_{25}$ and $\boxed{[K:{\mathbb{F}}_5]=2}$.

The Legendre symbol isThe distinction between the zero class and a nonzero quadratic residue is essential.

Euler's criterion statesFor $p\nmid a$ the right side is $+1$ or $-1$. The displayed congruence also holds for $p\mid a$, since $p$ is odd and the exponent is positive.

For distinct odd primes,Equivalently the two Legendre symbols agree unless both primes are three modulo four, in which case their signs are opposite. Neither symbol is zero here.

Quadratic reciprocity with the prime three givesThe latter symbol is $+1$ when $p\equiv 1(\mathrm{mod}3)$ and $-1$ when $p\equiv 2(\mathrm{mod}3)$. A prime greater than three occupies one of the classes $1,5,7,11$ modulo twelve. Combining its residues modulo three and four gives signs $+1,-1,-1,+1$ respectively. Hence

For $n\ge 1$, the exponent $2^n$ is positive and even. Since $2^{2j}=4^j\equiv 4(\mathrm{mod}12)$ for every $j\ge 1$, it follows thatHere the natural-number indexing must start at one. If zero is included in that convention, $F_0=3$ is an exception to the printed assertion and must be excluded.

Let $p=F_m$ be prime with $m\ge 1$. The preceding parts give $(3/p)=-1$, so Euler's criterion implies $3^{(p-1)/2}\equiv -1(\mathrm{mod}p)$. The multiplicative order $d$ of three divides $p-1=2^{2^m}$. Every proper divisor of this power of two divides $(p-1)/2$. If $d$ were proper, that half-power would be one, a contradiction. Thereforewhich means precisely that three is a primitive root modulo $p$.

The monotone convergence theorem says that if $0\le f_j\uparrow f$ almost everywhere on a measure space, then $\int f_j\uparrow \int f$, allowing infinity. The dominated convergence theorem says that if $g_j\to g$ almost everywhere and $|g_j|\le H$ for an integrable $H$, then $g$ is integrable and $\int g_j\to \int g$.

For $a\ne 0$, the truncated functions $f_a{1}_{[-j,j]}$ increase to $f_a$. Each has the elementary integral $2\arctan (|a|j)/|a|$. Thus monotone convergence givesFor fixed $x$, eventually $|x|\le n\pi /2$ and $n\sin (x/n)\to x$, so $g_n(x)\to 1/(1+x^2)$. Inside the support the supplied inequality yields $n^2{\sin }^2(x/n)\ge 4x^2/{\pi }^2$. Outside the support $g_n=0$. Consequently the global bound isand this majorant is integrable by the already proved formula with $a=2/\pi$. Dominated convergence now gives $\boxed{\int g_n\to \int (1+x^2{)}^{-1}dx=\pi }$. The majorant is independent of $n$, including on the moving support.

Lebesgue measurability of $g$ means that $g^{-1}(U)$ is a Lebesgue measurable subset of $\mathbb{R}$ for every open subset $U\subseteq \mathbb{R}$. Equivalently one can check all sets $\{x:g(x)>t\}$ with real $t$.

There are two conventions for a step function on the whole real line. Under the bounded-support integration convention it is a finite linear combination of indicators of bounded intervals; under the unrestricted piecewise-constant convention nonzero constant tails are allowed. The conclusions below state both conventions explicitly, since only the first assertion depends on this choice.

Under the bounded-support convention, false in general, true if $f(0)=0$. For the first claim, take $g=0$ and $f\equiv 1$: the composition is one on all of $\mathbb{R}$, and cannot have bounded support. For the second, refine the finitely many defining intervals into disjoint intervals on which $g$ is constant. The composition is constant on each of them and vanishes outside their bounded union because $f(0)=0$. Hence it is another step function.

Under the unrestricted piecewise-constant convention, true in both cases: composition with any function preserves constancy on the same partition, including its unbounded cells. This also applies to the broader measurable-partition convention. Continuity of $f$ is not needed for this particular assertion.

False in both cases. Without the zero-at-zero condition, $g=0$ and $f\equiv 1$ already give an integrable input and a non-integrable composition on $\mathbb{R}$.

Even imposing $f(0)=0$ does not control how fast $f(t)$ approaches zero. Take $g(x)=(1+x^2{)}^{-1}$ and the continuous function $f(t)=\sqrt{|t|}$. The input has integral $\pi$, buthas a divergent integral, because its tails are comparable to $1/|x|$. Thus continuity at zero is weaker than the linear growth estimate that would preserve integrability.

True in both cases. For every open $U$, continuity of $f$ makes $f^{-1}(U)$ open. Thenis Lebesgue measurable by measurability of $g$. The value $f(0)$ has no bearing on this preimage argument.

A space is Hausdorff if distinct points have disjoint open neighbourhoods. In symbols, for $x\ne y$ there are open $U\ni x$ and $V\ni y$ with $U\cap V=\varnothing$.

A space is compact if every open cover has a finite subcover. The finite-intersection formulation says that every family of closed sets with the finite intersection property has nonempty total intersection.

The nested closed sets have the finite intersection property: the intersection of finitely many is the one with the largest index and is nonempty. Compactness gives $\boxed{Y_{\infty }\ne \varnothing }$.

For an open neighbourhood $W$ of the intersection, suppose no $Y_n$ were contained in $W$. Then $Z_n=Y_n\cap (X\setminus W)$ would be nested, nonempty and closed. The same argument gives a point in all $Z_n$, contradicting $Y_{\infty }\subseteq W$. Hence

For connectedness, suppose $Y_{\infty }=A\cup B$ were a separation into two nonempty relatively closed sets. Since $Y_{\infty }$ is compact, $A,B$ are disjoint compact subsets of $X$. The supplied separation fact gives disjoint open neighbourhoods $U_A,U_B$. Their union contains $Y_{\infty }$, so some $Y_n$ lies in that union. But $Y_n$ meets both neighbourhoods, since it contains $A$ and $B$. Its intersections with the two neighbourhoods are then a separation, contradicting its connectedness. Thus the nested intersection is connected.

For the quotient map $\pi$, a subset $U$ of the quotient is open exactly when ${\pi }^{-1}(U)$ is open in $X$. This definition immediately makes $\pi$ continuous. If $g$ is continuous then $g\circ \pi$ is continuous. Conversely, continuity of the composition gives, for every open $V\subseteq Z$, that ${\pi }^{-1}(g^{-1}(V))$ is open. By the defining criterion $g^{-1}(V)$ is open, so $g$ is continuous.

A map $h$ constant on each equivalence class defines a unique map $\bar{h}([x])=h(x)$. It is well-defined by the assumed class constancy, and its composition with $\pi$ is $h$. The preceding criterion proves that $\bar{h}$ is continuous.

For the integer lattice quotient, defineThis continuous map is onto the product of the two unit circles, and its fibres are exactly the lattice equivalence classes. It therefore induces a continuous bijection $\bar{F}:{\mathbb{R}}^2/{\mathbb{Z}}^2\to {\mathbb{T}}^2$. Every class has a representative in $[0,1{]}^2$, so the quotient is the continuous image of that compact square and is compact. The target is Hausdorff as a subspace of ${\mathbb{C}}^2$. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism, provingFinally, if $U\subseteq {\mathbb{R}}^2$ is open, then ${\pi }^{-1}(\pi (U))={\bigcup }_{m\in {\mathbb{Z}}^2}(U+m)$ is open, being a union of translates. The quotient criterion makes $\pi (U)$ open. Thus the quotient map is an open map.

Write the variables as $(u,v)\in {\mathbb{R}}^n\times {\mathbb{R}}^k$, with $a=(u_0,v_0)$. The hypothesis makes $D_{u}f(a)$ invertible. The implicit function theorem supplies neighbourhoods $U,V$ of $u_0,v_0$ and a unique smooth map $\varphi :V\to U$ such thatIts derivative is $D\varphi (v_0)=-(D_{u}f(a){)}^{-1}{D}_{v}f(a)$. Thus the derivative of this parametrization has image exactly $\ker Df(a)$.

Since $h$ has a constrained maximum at $a$, differentiating $v\mapsto h(\varphi (v),v)$ gives zero in every direction. Therefore $Dh(a)$ vanishes on $\ker Df(a)$. In finite-dimensional linear algebra, $(\ker Df(a){)}^{\perp }$ is the row space of $Df(a)$. ConsequentlyThe full-rank hypothesis is what permits this Lagrange multiplier conclusion; it is not a consequence of constrained maximality alone at a singular constraint.

Positive definiteness gives a positive smallest eigenvalue $c$ with $x^{t}Ax\ge c|x{|}^2$. Hence the set $x^{t}Ax=y^{t}Ay=1$ is compact, and $h$ attains a maximum there. Its two constraint gradients are $(2Ax,0)$ and $(0,2Ay)$; they are linearly independent because $x,y$ are nonzero and $A$ is invertible.

Put $s=x^{t}Ay$. At any constrained extremum the multiplier equations areMultiplication by $x^t$ and $y^t$ gives $\lambda =\mu =s^2$. If $s=0$ the objective is zero. Otherwise $Ay=sAx$, so $y=sx$ and normalization gives $s^2=1$. Thus every maximum is at most one, and the value one is attained at $x=y$. This proves $\boxed{(x^{t}Ay{)}^2\le 1}$ on the normalized constraints.

For nonzero arbitrary vectors apply that result to $x/\sqrt{x^{t}Ax}$ and $y/\sqrt{y^{t}Ay}$. If either vector is zero the conclusion is immediate. ThereforeEquality for nonzero vectors holds precisely when they are linearly dependent. This is the Cauchy-Schwarz inequality for the inner product induced by $A$.

For a variation $y+\epsilon \eta$ with $\eta (0)=\eta (1)=0$, stationarity givesThe boundary term vanishes. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equation $d{F}_{y^{\prime }}/dx-F_y=0$.

Differentiating the proposed first integral givesThus it is constant along an extremal when $F_x=0$, proving the Beltrami identity.

For the positive branch relevant to the endpoint data, $F=\sqrt{1+y^{\prime 2}}/y$. The identity becomesThe Euler-Lagrange equation itself gives $y{y}^{\prime \prime }=-(1+y^{\prime 2})$, so $(y^2{)}^{\prime \prime }=-2$. Integrating, or solving the first integral, yields a circle $y^2+(x-c{)}^2=R^2$. The equal endpoint values force $c=1/2$ and $R^2=5/4$. Hence the unique positive smooth extremal isIt stays positive on the entire interval and satisfies the second-order equation, including its turning point. Merely differentiating the first integral after dividing by $y^{\prime }$ would need extra care there. Geometrically this is an arc of a hyperbolic geodesic in the upper half-plane model.

At fractional position $s$ along a straight rigid rod, the velocity is $(1-s)u+sv$. Uniform mass per unit length givesThe string constraints are $x_1^2+y_1^2+(-l+z_1{)}^2=l^2$ and $x_2^2+y_2^2+(-2l+z_2{)}^2=4l^2$. Their first-order terms give $z_1=z_2=0$. Their second-order terms then giveThe rod constraint is $(l+x_2-x_1{)}^2+(y_2-y_1{)}^2+(-l+z_2-z_1{)}^2=2l^2$. Its first-order part is $2l(x_2-x_1)=0$, hence $x_2=x_1$ to first order.

The centre of mass rises by $(z_1+z_2)/2$. Substituting the quadratic string expansions and the linear relation between the horizontal displacements givesVertical velocities contribute only at fourth order to the energy. Substituting $u=({\dot{x}}_1,{\dot{y}}_1,0)$ and $v=({\dot{x}}_1,{\dot{y}}_2,0)$ into the rod energy givesThe decoupled $x_1$ equation is ${\ddot{x}}_1+3g{x}_1/(4l)=0$. For the other two coordinates put $\kappa =3g/(2l)$; their equations areA normal-mode ansatz gives $3{\omega }^4-6\kappa {\omega }^2+2{\kappa }^2=0$, whose roots are ${\omega }^2=g(3\pm \sqrt{3})/(2l)$. The corresponding displacement ratios are $y_2/y_1=1\mp \sqrt{3}$. Therefore the three periods are

Set $a=\sqrt{m\omega /(2\hslash )}x+\sqrt{\hslash /(2m\omega )}d/dx$ and let $a^{†}$ have the opposite derivative sign. On the bound-state domain, $[a,a^{†}]=1$ and $H=\hslash \omega (a^{†}a+1/2)$. Thus every energy eigenvalue has $E/\hslash \omega -1/2=\nu \ge 0$.

The lowering operator maps an eigenstate of $a^{†}a$ of eigenvalue $\nu$ to one of eigenvalue $\nu -1$, with squared norm multiplied by $\nu$. Repeated lowering cannot pass below zero, so it must terminate at zero. Hence $\nu$ is a nonnegative integer. The equation $a{\psi }_0=0$ gives a Gaussian ground state, and repeated raising gives a nonzero square-integrable state for every integer $n\ge 0$. ThereforeThese are also the Gaussian-times-Hermite-polynomial solutions obtained by the suggested series method.

Put $\alpha =m\omega /\hslash$. Normalization gives ${\psi }_0=(\alpha /\pi {)}^{1/4}{e}^{-\alpha x^2/2}$, soThe expectation of $x$ itself is zero by parity; that cancellation does not apply to its absolute value.

Orient the body boundary $C$ counterclockwise. Its outward normal vector satisfies $\mathbf{n}ds=(dy,-dx)$, so the pressure force on the body has complex componentFor steady incompressible flow and irrotational flow, Bernoulli's principle gives $p=p_0-\rho |w^{\prime }{|}^2/2$. The constant-pressure integral vanishes. Write $q=w^{\prime }=u-iv$. The impermeability condition makes velocity tangent to $C$, implying $qdz=\bar{q}d\bar{z}$. Thus $|q{|}^{2}d\bar{z}=q^{2}dz$ along the boundary, andThis proves the Blasius force theorem, with force measured per unit length perpendicular to the flow plane.

For a circle of radius $a$, Milne-Thomson's circle theorem replaces an incident analytic potential $f(z)$ by $f(z)+\overline{f(a^2/\bar{z})}$, making the stream function constant on the circle. Its reflected singularities must lie inside the circle; here the physical source lies outside. With $c=Q/(2\pi )$ and $d=a^2/b$, this yields, up to additive constants,The interior image source and sink have equal and opposite strengths, so there is no extra net source inside the body. No independent vortex term is added. The potential may have logarithmic branches, but the velocity and pressure are single-valued.

Only $d$ and $0$ lie inside $C$. The residues of $(w^{\prime }{)}^2$ areTheir sum is $-2c^2{a}^2/[b(b^2-a^2)]$. The residue theorem and the preceding force formula therefore giveThe force points towards the external source. Reflection symmetry across the horizontal axis independently checks the vanishing vertical component.

The velocity is $\mathbf{u}=\nabla \varphi$ because the flow is irrotational. Incompressible flow requires $\nabla \cdot \mathbf{u}=0$, hence the velocity potential satisfies Laplace's equation:The exact kinematic boundary condition at the free surface is ${\eta }_t+{\varphi }_x{\eta }_x={\varphi }_y$. With constant atmospheric pressure, the dynamic condition from Bernoulli's principle is ${\varphi }_t+\frac{1}{2}|\nabla \varphi {|}^2+g\eta =0$, after choosing the time-dependent potential gauge. For small amplitudes, the products of perturbations are second order. Evaluating the remaining first-order terms at the undisturbed level $y=0$ givesSubstitution of $\varphi =f(y)\sin (kx-\omega t)$ into Laplace's equation gives $f^{\prime \prime }-k^{2}f=0$. For $k\ne 0$, decay as $y\to -\infty$ selects $f(y)=B{e}^{|k|y}$. The two boundary conditions for $\eta =A\cos (kx-\omega t)$ yieldFor a nonzero wave amplitude these are compatible precisely whenWith the conventional $k>0$ and positive frequency, the dispersion relation is $\omega (k)=\sqrt{gk}$. The exponential decay is the infinite-depth boundary condition; retaining a growing exponential would violate it.

The moment-generating function of a random variable $X$ is $M_X(t)=\mathbb{E}[e^{tX}]$ wherever this expectation is finite. Existence on an open interval containing zero is the condition needed for the usual uniqueness and convergence theorems. For the exponential distribution with rate $\lambda >0$,For $t\ge \lambda$ the integral diverges. Differentiating at zero gives mean $1/\lambda$ and variance $1/{\lambda }^2$.

The independent identically distributed central limit theorem states that for variables of common finite mean $\mu$ and finite positive variance ${\sigma }^2$,For the exponential distribution, $\mu =1/\lambda$ and $\sigma =1/\lambda$. Thus the event in question is exactlyThe limiting normal distribution has a continuous distribution function at every real $x$, so convergence in distribution givesHere $\Phi (x)=(2\pi {)}^{-1/2}{\int }_{-\infty }^x{e}^{-u^2/2}du$, including the strict-inequality convention because the limiting distribution is continuous.

For sufficiently large $n$, $p_n=\lambda /n$ lies in $(0,1]$. The waiting time has the geometric distribution on $1,2,\dots$, so summing its geometric series giveswhenever $(1-\lambda /n)e^{t/n}<1$. Fix $t<\lambda$. The expansions $e^{t/n}=1+t/n+O(n^{-2})$ show that the numerator is $\lambda /n+O(n^{-2})$ and the denominator is $(\lambda -t)/n+O(n^{-2})$. ThereforeThe convergence theorem for moment-generating functions says that pointwise convergence of finite moment-generating functions on a fixed open interval about zero to the moment-generating function of a probability distribution implies convergence in distribution to that distribution. For example take $|t|<\lambda /2$; all large $n$ satisfy the convergence condition throughout that interval. The limit is the moment-generating function from part (a), proving

Reading positive entries as directed edges, the communicating class decomposition isFor the first class, paths $1\to 5\to 8\to 9\to 1$ establish mutual reachability, and every transition from one of these states stays in this set. States 2 and 3 communicate, but state 2 can enter the first class and never return. States 4,6,7 communicate through $4\to 6\to 7\to 4$, but state 6 can enter state 2 with no route back. These paths also exclude any larger communicating class. In a finite Markov chain, the closed class is recurrent and the two open classes are transient.

Assume $K\ge 1$ is an integer, $0<p<1$, and put $q=1-p$. There is a discrepancy between the printed state definition and the formula requested later. We first follow the stated cumulative count, which retains previous good items when a defective item appears during phase A. The transition probabilities arewith all others zero. The balance equations imply $q{\pi }_i=q{\pi }_{i-1}$ for $1\le i<K$ and $(p/r){\pi }_K=q{\pi }_{K-1}$. ThusFor comparison, if clearance requires $K$ consecutive good items, each defect in phase A resets the state to zero. Replace its transitions by $P_{i,0}=p$, $P_{i,i+1}=q$, combining contributions if destinations coincide. Phase B retains its two transitions. For this consecutive-clearance inspection model, balance yields ${\pi }_i=q^i{\pi }_0$ and $(p/r){\pi }_K=q^K{\pi }_0$. With $D=1+(r-1)q^K$, normalization givesThis second model is the one giving the printed formula in part (iii). It is a different transition rule, rather than an algebraic simplification of the cumulative model.

For an irreducible positive recurrent Markov chain, the Markov chain ergodic theorem gives a unique stationary distribution $\pi$, with almost-sure long-run occupation fraction ${\pi }_j$ in state $j$. More generally, time averages of an integrable state reward converge to its stationary mean. If the chain is also aperiodic, then ${\mathbb{P}}_i(X_n=j)\to {\pi }_j$ for every initial state $i$; the aperiodicity assumption is needed for this last distributional conclusion.

For either transition rule in part (i), $0<p<1$ permits progression through the phase-A states and a return from phase B to zero. Both chains are finite and irreducible, hence positive recurrent. State $K$ has a positive self-loop, so both are aperiodic. Thereforeusing respectively the cumulative or consecutive stationary distribution given in part (i), for every starting distribution.

The excluded endpoint cases are simple: at $p=0$, phase B is eventually reached and then absorbing, so the limiting law is the point mass at $K$. At $p=1$, the cumulative rule traps the chain at its current phase-A count (or eventually at zero if started in phase B), so the limit can depend on the initial count; under the consecutive rule zero is eventually absorbing. Irreducibility and the preceding stationary formulas must not be asserted unchanged at these endpoints.