Past exam of the options I course of the University of Oxford / 2008 (XMAT 4303, AO1)

Options I, Honour School of Mathematics Part A, Trinity Term; Wednesday 18 June 2008, 9.30am–12.30pm. The full nine-page source, including its intact cover and all nineteen questions, is preserved in a private editable transcription. It was supplied through the past-papers disclosure, in original Part A bundle pages 71–79. No exact worked-answer advertisement was found in the bounded availability check.

Fix ${\omega }_1$. For any $u,v\ne {\omega }_1$, apply the given two-point transitivity property to $({\omega }_1,u)$ and $({\omega }_1,v)$. The resulting element fixes ${\omega }_1$ and sends $u$ to $v$, so the stabilizer is transitive on the complement.

The whole group is transitive on $\Omega$: extend any desired first-coordinate pair by choosing a different second point on each side. By the orbit-stabilizer theorem, $|G|=n|{\mathrm{Stab}}_G({\omega }_1)|$. Applying the same theorem inside the stabilizer to its orbit of size $n-1$ shows that $n-1$ divides the stabilizer order. HenceIn the action on ordered pairs there are exactly two orbits. One is the diagonal, transitive because the action on $\Omega$ is transitive. The other comprises distinct-coordinate pairs and is transitive by hypothesis. Equality of the two coordinates is preserved by every group element, so the two sets cannot merge into one orbit. Faithfulness of the action is not required.

Eisenstein criterion says that a primitive integer polynomial is irreducible over $\mathbb{Q}$ if some prime divides all coefficients except the leading one, does not divide the leading coefficient, and its square does not divide the constant coefficient. The passage from primitive integer factorization to rational factorization is Gauss lemma for polynomials.

Translate the variable:This is Eisenstein at two. Translation is an automorphism of $\mathbb{Q}[x]$, so $x^8+1$ itself is irreducible.

Set $\zeta =e^{i\pi /8}$ and $c=\cos (\pi /8)$. Since $\zeta$ is a root of that irreducible degree-eight polynomial, $[\mathbb{Q}(\zeta ):\mathbb{Q}]=8$. Also $c=(\zeta +{\zeta }^{-1})/2$, so $\mathbb{Q}(c)\subseteq \mathbb{Q}(\zeta )$ and $\zeta$ solves $T^2-2cT+1=0$. The extension degree is at most two. It is not one, since every element of $\mathbb{Q}(c)$ is real and $\zeta$ is not. The tower law for field extensions now gives

The quadratic reciprocity law states that distinct odd primes $a,b$ satisfyFirst suppose $p\ne q$. If the odd prime $r$ equalled one of them, divisibility of $p-q$ by $r$ would force the other to equal $r$ too, a contradiction. Thus reciprocity applies to $r,p$ and $r,q$. The modular congruence modulo $4r$ gives both $p\equiv q(\mathrm{mod}r)$ and $p\equiv q(\mathrm{mod}4)$. Therefore $(p/r)=(q/r)$, and the reciprocity signs are identical. Rearranging the two reciprocity identities proves $\boxed{(r/p)=(r/q)}$. If $p=q$, the equality is immediate, including a zero Legendre symbol when $p=r$.

For general odd $n>1$, factor $n={\prod }_j{r}_j^{e_j}$. If $p=q$ there is nothing to prove. Otherwise neither $p$ nor $q$ divides $n$, since any such divisor, together with $4n\mid p-q$, would force $p=q$. Each $4r_j$ divides $p-q$, so the first result and multiplicativity of the Legendre symbol giveFor $n=1$, both symbols equal one.

A null set has Lebesgue outer measure zero: for every $\epsilon >0$ it can be covered by countably many open intervals with total length below $\epsilon$. If $E_j$ are null, cover $E_j$ by intervals of total length below $\epsilon 2^{-j}$, for $j\ge 1$. The union of all these covers is countable and covers ${\bigcup }_j{E}_j$ with total length below $\epsilon$. As $\epsilon$ is arbitrary, the countable union is null.

Let $A_n=\{x:f(x)\ge 1/n\}$. To avoid presupposing finite measure of $A_n$, first truncate it: $A_{n,k}=A_n\cap [-k,k]$ is measurable and has finite measure. Its indicator divided by $n$ is integrable and is bounded above by $f$. The allowed monotonicity property givesThus every $A_{n,k}$ is null. Their countable union $A_n$ is null by part (a), and so is $E={\bigcup }_n{A}_n$. Every positive value of $f$ is at least $1/n$ for some integer $n$, hence $\{f>0\}=E$. Consequently $f=0$ outside a null set.

The integrand is positive and continuous on the interval in question. Substitute $t=\log (1+x)$ to obtainIf $\alpha \ge 1$, the exponential factor is at most one and the comparison integral of $t^{-2}$ converges. If $\alpha <1$, exponential growth dominates $t^2$, so the integrand eventually exceeds one and the integral diverges. ThusThe logarithmic square is what makes the boundary value $\alpha =1$ integrable.

The function is measurable and $0\le {\cos }^2(1/x)\le 1$. The interval has finite measure, so its integral is at most one. Hence it is integrable, despite its oscillations near zero.

As $x\to \infty$, ${\cos }^2(1/x)\to 1$. It is therefore at least $1/2$ on some infinite tail. The integral of that lower bound is infinite, so it is not integrable on this interval.

A map of topological spaces is continuous when the preimage $f^{-1}(U)$ is open in $X$ for every open set $U$ of $Y$. Equivalently, the preimage of every closed subset of $Y$ is closed in $X$.

For every $\epsilon >0$ there must be a $\delta >0$ such thatThe choice of $\delta$ may depend on $x_0$ and $\epsilon$. This is the epsilon-delta definition of continuity.

A basis for a topology is a family of open sets such that every open set is a union of members of the family. Equivalently, for every point in an open set there is a basis member containing that point and contained in the open set.

For a basis $\mathcal{B}$ on $Y$, continuity is equivalent to $f^{-1}(B)$ being open for every $B\in \mathcal{B}$. Necessity follows because basis sets are open. Conversely, taking preimages of a union of basis sets gives a union of open preimages, so every open set has an open preimage.

Each interval $(a,b)$ is open in the usual metric: if $x\in (a,b)$, the ball with radius $\min (x-a,b-x)/2$ is contained in it. Conversely, if $U$ is open and $x\in U$, the metric definition supplies $r>0$ with $(x-r,x+r)\subseteq U$. Thus every open set is a union of open intervals, which proves they form a basis.

If $f$ is continuous, the sets in question are preimages of the open rays $(-\infty ,\lambda )$ and $(\lambda ,\infty )$, so they are open. Conversely, suppose all those sets are open. For every $a<b$,is open as the intersection of two open sets. Part (i) and the basis criterion of part (b) then prove $f$ is continuous.

A subset is connected when it has no separation into two disjoint nonempty relatively open sets. It is path-connected when every pair of its points can be joined by a continuous path from $[0,1]$ lying in the subset.

Path-connectedness implies connectedness; the reverse fails in general, for example for the topologist's sine curve. They are equivalent for open subsets of every Euclidean space ${\mathbb{R}}^n$. More generally this holds for open subsets of locally path-connected spaces.

If $f(A)$ were separated by two disjoint nonempty relatively open sets, their preimages under the restriction $f{|}_A$ would be disjoint nonempty relatively open sets covering $A$. Continuity makes them open in $A$, contradicting its connectedness. Therefore $f(A)$ is connected.

The circle is path-connected, using circular arcs, and hence connected. Its continuous image is connected by part (ii). A connected subset of the real line is an interval: if it contained values on opposite sides of a missing value $c$, intersection with $(-\infty ,c)$ and $(c,\infty )$ would separate it. Thus $f(C)$ is an interval. It is also compact, so it is a closed bounded interval, possibly reduced to one point.

For $f=(f_1,\dots ,f_k)$, the Jacobian matrix is the matrix of its derivative in standard bases, with entries $(Df{)}_{ij}=\partial f_i/\partial x_j$. Here

The inverse function theorem states: if $f$ is continuously differentiable on an open subset of ${\mathbb{R}}^n$ and $Df(a)$ is invertible, then some open neighbourhood of $a$ is mapped bijectively onto an open neighbourhood of $f(a)$, with continuously differentiable inverse. Its derivative at $f(a)$ is $(Df(a){)}^{-1}$.

The present map is smooth, and at $a=(\pi /2,\pi )$,Also $f(a)=(0,0)$, so the theorem applies locally and givesThe inverse is defined near the image point $(0,0)$ and takes values near $a$; no assertion of a global inverse is required.

With fixed endpoint values, a smooth extremal satisfies the Euler-Lagrange equation $F_y-(d/dx)F_{y^{\prime }}=0$. If $F$ has no explicit $x$ dependence, differentiation givesThus $F-y^{\prime }{F}_{y^{\prime }}$ is constant, the Beltrami identity.

For this integrand, $F_y=2y$ and $F_{y^{\prime }}=1+2y^{\prime }$, so the equation is $y^{\prime \prime }=y$. ConsequentlyThe two constants are fixed uniquely by the prescribed values at distinct endpoints. Equivalently, writing these values as $y_1,y_2$,Its first integral is $y^2-y^{\prime 2}$, which is constant. Using only that first integral without the Euler-Lagrange equation could admit spurious constant functions, so the differential equation was used directly.

For any nonzero endpoint-vanishing perturbation $\eta$, the linear terms in $J[y+\eta ]-J[y]$ vanish by the Euler-Lagrange equation and integration by parts; also $\int {\eta }^{\prime }=0$. The remainder isHence the extremal is the unique strict global minimum, and cannot be a maximum.

A coordinate absent from the Lagrangian is cyclic. Its Euler-Lagrange equation givesThus the corresponding generalized momentum $\boxed{p_1=\partial L/\partial {\dot{q}}_1}$ is conserved.

The energy function is $E={\sum }_a{\dot{q}}_a\partial L/\partial {\dot{q}}_a-L$. Along a solution,Therefore time independence gives the conserved quantity

Put $A(r)=r^2/(r-2m{)}^2$ and $B(r)=r^3/(r-2m)$, so $L=A{\dot{r}}^2+B{\dot{\phi }}^2$. Direct differentiation gives $A^{\prime }=-4mr/(r-2m{)}^3$ and $B^{\prime }=2r^2(r-3m)/(r-2m{)}^2$. The Euler-Lagrange equations are $2A\ddot{r}+A^{\prime }{\dot{r}}^2-B^{\prime }{\dot{\phi }}^2=0$ and $(d/dt)(2B\dot{\phi })=0$, equivalentlyThe cyclic angular coordinate and time independence give the two conserved quantitiesThe expression for $E$ equals $L$ here because $L$ is homogeneous quadratic in the velocities.

For a constant radius and nonzero angular motion, the radial equation requires $(r_{\ast }-3m){\dot{\phi }}^2=0$, so $\boxed{r_{\ast }=3m}$. This lies outside $2m$, and the angular equation then gives $\phi =\Omega t+{\phi }_0$, with any nonzero constant $\Omega$. This is the photon sphere radius in the stated coordinates. If both velocities are zero, the displayed variational equations also admit a stationary configuration at any radius; the moving circular photon solution is the nonzero-speed case just found.

Gauss law states that for outward-oriented boundary $S=\partial V$,For a continuously differentiable field and continuous density, the divergence theorem turns the left side into ${\int }_V\nabla \cdot \mathbf{E}$. Thus the differential law implies the integral law. Conversely, if the integral law holds for every suitable volume, the integral of $\nabla \cdot \mathbf{E}-\rho /{ϵ}_0$ vanishes on every small ball. A nonzero continuous value would keep one strict sign in some small ball, a contradiction. Hence $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$. Weaker regularity gives the corresponding almost-everywhere or distributional formulation.

For the radial field, the flux on a radius-$r$ sphere is $4\pi r^{2}E(r)$, while its enclosed charge is $4\pi {\int }_0^r{s}^2\rho (s)ds$. ThereforeThis uses the stated density as the full charge distribution, with no additional point charge at the origin. Since it vanishes beyond $a$, the total charge isand for $r>a$ the field is $\boxed{\mathbf{E}=Q{\mathbf{e}}_r/(4\pi {ϵ}_0{r}^2)}$.

Use the stream function convention $u={\psi }_y$, $v=-{\psi }_x$. It gives $u=-\sin x\sin y$, $v=-\cos x\cos y$. Thenso the flow is incompressible.

The vorticity points normally to the plane and has scalar componentIt is nonzero in general, so the flow is rotational.

For this steady, inviscid, constant-density flow without a body force, Bernoulli's theorem says thatIts constant may vary from one streamline to another because the flow has vorticity.

The convective acceleration isSteady Euler equations therefore give $p_x=-\rho \sin x\cos x$ and $p_y=\rho \sin y\cos y$. Integrating these compatible equations yieldswith arbitrary additive constant $p_0$. As a check, $p/\rho +(u^2+v^2)/2=p_0/\rho +1/2-{\psi }^2$, indeed constant on each streamline. This is a Steady Euler flow with vorticity proportional to its stream function.

Condition (1b) is the kinematic free-surface boundary condition: fluid particles on the surface remain there, since the material derivative of $y-\eta (x,t)$ vanishes. Condition (1c) is the dynamic free-surface boundary condition, obtained from the unsteady Bernoulli equation with constant atmospheric pressure and a choice of potential gauge. Condition (1d) is impermeability: no normal flow through the horizontal bed.

Write the surface displacement and the perturbation velocity potential as small quantities about rest, and drop their quadratic products. Evaluating the remaining surface terms at the unperturbed level $y=0$ givesThe velocity components are $u={\varphi }_x$, $v={\varphi }_y$. The shift of the evaluation point from zero to $\eta$ contributes a product of perturbations, as do $u{\eta }_x$ and the kinetic-energy term, so they disappear at this order.

For positive wavenumber $k$, the phase velocity is $c_p=\omega /k$, the speed of a fixed phase. The group velocity is $c_g=d\omega /dk$, the leading speed of a narrow wave packet. Differentiating the supplied dispersion relation givesFor fixed $k>0$, the shallow-depth limit uses $\tanh s/s\to 1$:

The inverse transformation is $x=2u+v$, $y=v$, with absolute Jacobian determinant two. The original positivity constraints become $v>0$ and $2u+v>0$. ThusIntegrating out $v$ givesThis is the centred Laplace distribution with scale one.

The Markov property says that conditional on the present state, the next-state law does not depend on the earlier history. For this time-homogeneous chain, $\mathbb{P}(X_{n+1}=j\mid X_0=i_0,\dots ,X_n=i)=p_{ij}$ whenever the conditioning event has positive probability.

States communicate when each is accessible from the other by a finite positive-probability path; the equivalence classes are the communicating classes. A class is closed if no one-step transition leaves it, and open otherwise. Probability nonnegativity adds the implicit restriction $p\ge 0$ to the printed upper bound.

At $p=0$, every state is absorbing, so the classes are the singletons, all closed. For $0<p\le 1/2$, every neighbouring pair communicates, so the whole state space is one closed class, making the chain irreducible. There are no open classes in either case.

For $m=3$ and $p>0$, detailed balance along the path gives $2p{\pi }_0=p{\pi }_1$, ${\pi }_1={\pi }_2$, and $p{\pi }_2=2p{\pi }_3$. After normalization,For $0<p<1/2$, the self-loops have positive probability; the finite irreducible chain is aperiodic, so ${\mathbb{P}}_i(X_n=j)\to {\pi }_j$ for all fixed starting states and targets. For $p=1/2$, every step changes parity. The chain has period two, and for each fixed $i,j$ one parity of times has probability zero while ${\pi }_j>0$. Thus the stated convergence from a fixed starting state holds exactly when $0<p<1/2$. This is the stationary law of a reflecting nearest-neighbor walk on a finite path.

For a two-way contingency table, Pearson's statistic isHere $O_{i+},O_{+j}$ are the observed margins and $N$ their grand total. Under the independence null, its usual large-sample reference distribution has $(r-1)(c-1)$ degrees of freedom for an $r$-by-$c$ table.

Under the null, the common fitted CC frequency is its pooled proportion $133/3874$. Multiplying by the fixed number of cases givesThis estimates the common type frequency from both groups, rather than using the case frequency alone.

With both column totals fixed, each column is a three-category multinomial sample. Under the null the two columns share one probability vector, which has two free parameters because its three probabilities sum to one. Under the alternative each has its own vector, giving four free parameters. Their difference is two, the reference degrees of freedom for the association test.

Use two degrees of freedom from part (iii). Since $8.42>5.99$, the statistic exceeds the supplied five-percent threshold. Thus reject no association at the five-percent level: the observed genotype frequencies provide evidence of a difference between cases and controls. The approximate p-value is $e^{-8.42/2}\approx 0.0148$ under the ${\chi }_2^2$ reference law.

The recessive alternative distinguishes CC from the other two types while assigning CT and TT the same disease-risk category. Pooling those two types therefore removes a distinction that this alternative does not use. The resulting counts are $83,1856$ for cases and $50,1885$ for controls, arranged as CC versus non-CC.

The resulting two-by-two table has $\boxed{(2-1)(2-1)=1}$ reference degree of freedom. Since $8.41>3.84$, reject no association at the five-percent level using the supplied chi-squared test. The observed CC proportion is higher among cases, $83/1939$, than among controls, $50/1935$, so the direction agrees with the proposed increased-risk category.

The normal equations for linear least squares are $X^{T}X\hat{\theta }=X^{T}Y$. Under full column rank, $\hat{\theta }=(X^{T}X{)}^{-1}{X}^{T}Y$. Since $\mathrm{Cov}(Y)={\sigma }^{2}I$,Here $\bar{x}=0$ and $\bar{z}=2$, so the design matrix isIts columns are pairwise orthogonal and have squared lengths $6,28,4$. Consequently $X^{T}X=\mathrm{diag}(6,28,4)$, givingBoth estimators are linear functions of the Jointly Gaussian observation vector, so they are Jointly Gaussian. Zero covariance then implies independence. This final implication uses the normal error model, whereas the variance calculation itself needs only its second moments.