Past exam of the options I course of the University of Oxford / 2004 (AO1, XMAT 4303)

Original solutions to the paper released at the source request. The complete editable transcription is retained in the private examination archive.

A group action is a map $G\times \Omega \to \Omega$, $(g,\omega )\mapsto g\omega$, satisfying $1\omega =\omega$ and $(gh)\omega =g(h\omega )$. Its group orbit and stabilizer areThe stabilizer contains the identity. If $g,h$ fix $\omega$, then $gh$ fixes it; and $g\omega =\omega$ implies $g^{-1}\omega =\omega$ after applying $g^{-1}$. Thus it is a subgroup.

The map from left cosets $g{G}_{\omega }$ to group-orbit points $g\omega$ is well-defined, surjective and injective: $g\omega =h\omega$ is equivalent to $h^{-1}g\in G_{\omega }$. Therefore $|\mathrm{Orb}(\omega )|=[G:G_{\omega }]=|G|/|G_{\omega }|$ for finite $G$.

For conjugation in $S_3$, the group orbits are exactly the conjugacy classes, determined by cycle type:Conjugation sends $(i_1\cdots i_k)$ to $(g(i_1)\cdots g(i_k))$, so these displayed sets are invariant and each is one group orbit. Their sizes $1,3,2$ also agree with centralizer orders $6,2,3$.

The inclusion $F\subseteq K$ makes $K$ a vector space over $F$ by its own addition and multiplication by elements of $F$. The extension is finite when this vector space has a finite basis. Its degree of a field extension isBeing a finite field extension concerns vector-space dimension, not the cardinality of the underlying field.

Since $K\subseteq L$ and $L/F$ is finite, an $F$-linearly independent set in $K$ remains independent in $L$, so $K/F$ is finite. A finite $F$-spanning set of $L$ also spans it over $K$, so $L/K$ is finite.

Take an $F$-basis $a_1,\dots ,a_m$ of $K$ and a $K$-basis $b_1,\dots ,b_n$ of $L$. Expanding coefficients twice shows that all products $a_i{b}_j$ span $L$ over $F$. If ${\sum }_{i,j}{c}_{ij}{a}_i{b}_j=0$ with $c_{ij}\in F$, independence of the $b_j$ forces ${\sum }_i{c}_{ij}{a}_i=0$ for each $j$, and then independence of the $a_i$ forces every coefficient zero. Thus these products form an $F$-basis, proving

The polynomial $x^2-2$ is an irreducible polynomial over $\mathbb{Q}$, so $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$. This intermediate field is real, so it does not contain $i$. Hence $x^2+1$ has no root in it and adjoining $i$ has degree two. The tower law givesA basis is $1,\sqrt{2},i,i\sqrt{2}$. Square-bracket adjoining notation in this algebraic example gives the same field as round-bracket adjoining notation.

By the fundamental theorem of arithmetic, every integer greater than one has a prime factorization, and that factorization is unique up to reordering the factors. Equivalently, it has a unique finite prime-exponent representation $\boxed{n=\prod _p{p}^{v_p(n)}}$, with nonnegative integer exponents. The unit integer has the empty product.

Write $a={\prod }_p{p}^{r_p}$ and $b={\prod }_p{p}^{s_p}$. Divisibility is exactly comparison of prime exponents. Thus the greatest common divisor uses $\min (r_p,s_p)$ and the least common multiple uses $\max (r_p,s_p)$. Since their sum is $r_p+s_p$ for each prime,

Write $a=6r+1$ and $b=6s+1$. Then $ab=6(6rs+r+s)+1$, so $\boxed{ab\equiv 1(\mathrm{mod}6)}$. Repetition gives the same result for any finite product of numbers congruent to one.

Suppose only finitely many primes are congruent to five, and list them as $p_1,\dots ,p_k$. Consider $N=6p_1\cdots p_k-1>1$. It is divisible by neither two nor three, so every prime divisor is congruent to one or five modulo six. If all its prime factors were congruent to one, their product would be congruent to one by part (i), whereas $N\equiv 5$. Hence some prime divisor $q$ has $q\equiv 5(\mathrm{mod}6)$.

But no listed $p_i$ divides $N$, since $N\equiv -1(\mathrm{mod}{p}_i)$. This gives a new such prime, a contradiction. Therefore infinitely many primes are congruent to five modulo six. The argument does not require the specially constructed integer itself to be prime.

For $x>0$ the geometric ratio $e^{-2x}$ has modulus less than one, andEvery summand is nonnegative. The monotone convergence theorem therefore permits integration of the sum:The integral is finite because its integrand is bounded above by $e^{-x}$, whose integral is one. The displayed paired sums are the even partial sums of the alternating harmonic series; its odd partial sums differ by a term tending to zero. Thus the required equality holds with the usual series ordering. One can additionally evaluate it by $t=e^{-x}$:Grouping into nonnegative pairs supplies the justified interchange; absolute convergence of the ungrouped integrated alternating series is not needed and would be false.

The indicated measurable function is nonnegative. Tonelli's theorem applies without first assuming integrability and givesThus $h$ is integrable. Fubini's theorem now allows the reverse order. For $x>1$,ConsequentlyAt $x=1$ the quotient has the removable limit $b-a$; that endpoint does not create an additional divergence.

A metric space is a set $X$ with a function $d:X\times X\to [0,\infty )$ satisfying $d(x,y)=0⟺x=y$, symmetry and the triangle inequality. A subset $U$ is open if every $x\in U$ has some open open ball $B(x,r)$ contained in $U$.

For finitely many open sets $U_1,\dots ,U_k$ and $x$ in their intersection, choose radii $r_j>0$ with $B(x,r_j)\subseteq U_j$. The positive minimum radius gives a ball contained in the intersection. Thus every finite intersection is open; the empty intersection is $X$, also open.

For distinct points put $\delta =d(x_1,x_2)>0$. The balls $B(x_1,\delta /3)$ and $B(x_2,\delta /3)$ are disjoint: a common point would make $d(x_1,x_2)<2\delta /3$ by the triangle inequality. This proves the Hausdorff condition.

A singleton $\{x\}$ is closed because each $y\ne x$ has a ball of radius $d(x,y)/2$ avoiding it, making its complement open. Finite unions of closed sets are closed, as their complements are finite intersections of open sets. Hence every finite subset is closed, including the empty subset.

The product topology is generated by the basis of a topology of open rectangles $U\times V$, with $U$ open in $X$ and $V$ open in $Y$. For a fixed $y$,Both possibilities are open; checking this basis of a topology proves that $i_y$ is continuous. The analogous map $j_x:Y\to X\times Y$, $j_x(y)=(x,y)$, is continuous by the same argument.

If either factor is empty the product is empty and connected. Otherwise let $h:X\times Y\to \{0,1\}$ be continuous. For each $y$, $h\circ i_y$ is constant since $X$ is connected. Thus $h(x,y)=h(x_0,y)$ for any fixed $x_0\in X$. The right-hand side is $h\circ j_{x_0}$ and is constant in $y$ since $Y$ is connected. Therefore $h$ is constant on the whole product. The given characterization proves $X\times Y$ is connected.

The total derivative is the unique linear map $D{f}_a:{\mathbb{R}}^m\to {\mathbb{R}}^n$ satisfyingThis describes the best first-order linear approximation without choosing coordinates. The chain rule isThe evaluation point $f(a)$ in the outer derivative matters, particularly when composing a quadratic matrix map with the exponential.

Expand at a matrix increment $H$:The last term is $O(\Vert H{\Vert }^2)$ in a matrix norm, so it contributes no derivative. HenceThis is a linear operator on matrix increments, not a scalar entrywise derivative.

The requested base point is the identity matrix $I$, not the zero matrix. Since $I$ commutes with every $H$,The matrix power series justifies the quadratic remainder near zero. ConsequentlyAt zero the derivative would instead be $H$; replacing the printed base point would lose the scalar factor.

Here $\exp (I)=eI$. Applying part (a) at that matrix gives $D{s}_{eI}(K)=e(K^t+K)$, while part (b) gives $D{\exp }_I(H)=eH$. ThereforeBoth the outer evaluation point and the inner derivative contribute a factor of $e$.

Multiply the equation for $y_i$ by $y_j$ and the equation for $y_j$ by $y_i$, then subtract. The $q$ terms cancel and the derivative terms combine into a Wronskian boundary derivative:Integrating over the interval eliminates the boundary term because both eigenfunctions vanish at both endpoints. Since their eigenvalues differ,This is orthogonality in a weighted inner product in the regular Sturm-Liouville setting; the endpoint values must justify the integration by parts.

If $\lambda =0$, only a linear function can solve the equation, and zero endpoint values make it zero. If $\lambda <0$, the corresponding hyperbolic-sine solution also cannot vanish at both endpoints unless it is zero. For $\lambda =k^2>0$, the first endpoint gives $y=C\sin kx$, and the second requires $k=n\pi$ for a positive integer $n$. Thus $\boxed{{\lambda }_{\min }={\pi }^2}$.

The admissible trial function $y=\sin \pi x$ has quotientTherefore the minimum, or infimum before its attainment is established, is $\boxed{\le {\pi }^2}$ as requested. The sharp Poincaré inequality supplies the reverse inequality on the finite-energy zero-trace class, so the exact minimum is also ${\pi }^2$ and is attained by this trial function. The one-sided assertion in the source does not require a full variational existence argument.

The inertia tensor is $I_{ij}=\int (r^2{\delta }_{ij}-x_i{x}_j)dm$. The mixed moments vanish by reflection in $x$ or $y$. Every diagonal integrand is even in $z$, so its integral over the upper hemisphere is half its integral over the corresponding full ball of mass $2M$. A uniform ball's coordinate moments are each $(2M)a^2/5$, obtained by spherical symmetry and $⟨r^2⟩=3a^2/5$. Hence the hemisphere has each coordinate second moment $M{a}^2/5$ andThe tensor parallel axis theorem gives $I_O=I_C+M(|d{|}^{2}I-d{d}^t)$ with $d=(0,0,3a/8)$. ThereforeTranslation along the symmetry axis changes the two transverse moments but leaves the axial one unchanged.

The Time-dependent Schrödinger equation isA normalized quantum state has ${\int }_{{\mathbb{R}}^3}|\psi {|}^2{d}^{3}x=1$; the Hamiltonian operator acts on wave functions satisfying the appropriate domain and boundary conditions.

For hard-wall confinement, the stationary spatial modes and energies areTheir stationary time dependence is $e^{-i{E}_{n}t/\hslash }$. The correct triple-angle identity is $\sin 3\theta =3\sin \theta -4{\sin }^3\theta$; the printed hint has its right side reversed in sign. Thus the specified initial state expands asThe coefficients have squared moduli summing to one. The first requested energy is $E_1$, so its probability is $\boxed{9/10}$. The other requested energy is $2E_1$, which is not $n^2{E}_1$ for any positive integer $n$, so its probability is $\boxed{0}$. The occupied excited energy is instead $E_3=9E_1$, with probability $1/10$.

For a material volume, conservation of mass and the Reynolds transport theorem give ${\rho }_t+\nabla \cdot (\rho u)=0$. Constant material density, or equivalently incompressibility with the stated constant initial density, reduces this to $\nabla \cdot u=0$.

In an fluid undergoing inviscid flow the stress is the isotropic pressure stress $-pI$. Momentum conservation for a material volume then givesTransport and the divergence theorem localize this toFor a steady flow, $u_t=0$. For an irrotational flow the vorticity is zero, so the supplied identity gives $(u\cdot \nabla )u=\nabla (|u{|}^2/2)$. Uniform downward gravity is $g=-\nabla (gz)$, and constant density permits pulling $1/\rho$ inside the pressure gradient. ThereforeThe Bernoulli invariant is constant throughout each connected fluid region. Without irrotationality the corresponding steady statement generally gives a constant only along each streamline.

With the complex velocity convention $w^{\prime }(z)=u-iv$, differentiation gives $w^{\prime }=-i\Gamma /(2\pi z)$. Thus $u=-\Gamma y/(2\pi r^2)$ and $v=\Gamma x/(2\pi r^2)$. Resolving into polar components gives $\boxed{u_r=0,u_{\theta }=\Gamma /(2\pi r)}$.

Superposition for the opposite-signed pair givesup to a constant and branch choices. A vortex moves with the regular velocity generated by the other vortex, excluding its own singular field. At $z=ia$, the other contribution is $i\Gamma /[2\pi (2ia)]=\Gamma /(4\pi a)$; at $z=-ia$ it is the same real number. Hence the pair keeps its separation and translates parallel to the real axis withfor $a>0$.

For a vortex at $z_1=x_1+i{y}_1$ in the first quadrant, reflect it across either wall with opposite strength, then across both walls with its original strength. Thus the images are ${\bar{z}}_1$ and $-{\bar{z}}_1$ of strength $-\Gamma$, and $-z_1$ of strength $+\Gamma$. The potential isOn either real or imaginary axis, $z^2$ is real and the numerator and denominator of $(z^2-z_1^2)/(z^2-{\bar{z}}_1^2)$ are conjugates. Its modulus is one, making the stream function constant on both walls and enforcing no normal flow. A regular entire background flow could be added only with additional far-field data; the displayed image solution is the one generated by the specified vortex and its walls.

A marginal density is obtained by integrating out the other variable: $\boxed{f_X(x)={\int }_{\mathbb{R}}{f}_{X,Y}(x,y)dy}$. Here independence gives $f_{X,Y}=e^{-(x^2+y^2)/2}/(2\pi )$.

The polar transformation has absolute Jacobian determinant $r$. Consequentlywith density zero outside this range. Integrating over the angle gives $\boxed{f_R(r)=r{e}^{-r^2/2}}$, the Rayleigh distribution density. Integrating over radius, using ${\int }_0^{\infty }r{e}^{-r^2/2}dr=1$, gives $\boxed{f_{\Theta }(\theta )=1/(2\pi )}$ on the angle interval. The joint density factors as the product of these marginals, so $\boxed{R\text{ and }\Theta \text{ are independent}}$. The arbitrary angle assigned at $R=0$ is irrelevant because that event has probability zero.

Condition on the state after $m$ steps and use the Markov property and time-homogeneous Markov property:This is the Chapman-Kolmogorov equation, or $P^{m+n}=P^m{P}^n$. A stationary distribution is a probability row vector with nonnegative entries summing to one and $\boxed{\pi P=\pi }$. Inductively $\pi P^n=\pi$, so a chain started with that distribution retains it at every time.

For the given matrix, its first stationary equation is ${\pi }_1={\pi }_1/2+({\pi }_2+{\pi }_3)/3$. Normalization gives ${\pi }_1=2/5$. The third equation gives ${\pi }_3=2{\pi }_2/3+{\pi }_3/3$, hence ${\pi }_3={\pi }_2$. ThereforeDirect multiplication verifies the second equation as well. The finite chain is irreducible, so this stationary probability distribution is unique.

For equal one-second observation windows, the sample mean estimates the Poisson distribution mean:This matches the expected mean count; it is also the maximum-likelihood estimator for independent Poisson distribution observations. With $N=12169$ windows, fitted expected frequencies are $E_j=N{e}^{-\hat{\lambda }}{\hat{\lambda }}^j/j!$ for $j=0,\dots ,4$, and $E_{\ge 5}=N[1-{\sum }_{j=0}^4{e}^{-\hat{\lambda }}{\hat{\lambda }}^j/j!]$.

Each last-column entry is $(O_j-E_j{)}^2/E_j$, and the Pearson statistic is $\boxed{X^2=\sum _{j\in \{0,1,2,3,4,\ge 5\}}(O_j-E_j{)}^2/E_j}$. The displayed expected frequencies and contributions are rounded, so their printed values need not reproduce every final digit of a calculation using unrounded probabilities.

Separate high-count classes have very small expected frequencies, making the chi-square approximation unreliable. Pooling the upper tail gives a class with expected count about $20.9$, comfortably above the customary small-cell threshold. The pooling protects the asymptotic test approximation, while retaining the tail count as part of the goodness-of-fit comparison.

There are six pooled classes, one normalization constraint, and one fitted Poisson distribution parameter. Thus the reference degrees of freedom are $\boxed{6-1-1=4}$. Under the null model the Pearson statistic has approximately the Chi-squared distribution with four degrees of freedom. Its supplied 5% upper critical value is $9.49$.

Because $\boxed{2.07<9.49}$, do not reject the Poisson count distribution at the 5% level. This is consistency with that distribution, not proof that it is exactly true. A test of the marginal count frequencies also does not by itself verify independence of successive time windows.

For nondegenerate design let $S_{xx}={\sum }_i{x}_i^2>0$. Maximizing the Gaussian likelihood over the mean parameters is equivalent to minimizing ${\sum }_i(Y_i-\alpha -\beta x_i{)}^2$. Since ${\sum }_i{x}_i=0$, the normal equations separate:Its error is the linear combination $\hat{\beta }-\beta ={\sum }_i{x}_i{ϵ}_i/S_{xx}$. Independence and normality therefore giveIf every design point is zero, the slope is not identifiable and no such slope estimator can be inferred; the standard regression answer assumes the stated nondegeneracy.

For $n>2$, the usual unbiased estimator of variance isThe fitted mean lies in the two-dimensional span of the constant design vector and $x$. Orthogonal projection of the Gaussian error onto its complement gives $n-2$ independent standardized normal coordinates. HenceThis residual projection is independent of the fitted-coefficient projection. If maximum likelihood for variance is intended instead, its estimator is ${\hat{\sigma }}_{\mathrm{ML}}^2=\mathrm{RSS}/n$, with $n{\hat{\sigma }}_{\mathrm{ML}}^2/{\sigma }^2\sim {\chi }_{n-2}^2$; the denominator differs, not the residual degrees of freedom.

Independence of the slope estimate and residual variance, together with their normal and chi-square laws, gives the Student pivotal quantityThus an exact Gaussian-model two-sided 95% confidence interval isUsing a normal quantile without estimating the variance uncertainty would not give this exact finite-sample coverage.

The first-column largest pivot element is four, so eliminate with multipliers $1/2,0,1/4$ below it. The augmented matrix becomesPartial pivoting in the second column swaps rows two and three because two is largest in modulus. The next elimination multipliers are $1/4$ and $1/2$. This givesThe third pivot element is already largest and the final multiplier below it is zero. Back substitution gives $z=1$, $y=3$, $x=-6$, $w=-1/2$. ThereforeThe row swap must also reorder the previously stored first-column multipliers in $L$. The factors areMultiplying these factors verifies $PA=LU$. This is the required elimination with partial pivoting, not an alternative solution method.

For any $q$ in the polynomial trial space, $f-q=(f-p)+(p-q)$ and $p-q$ is in that same space. The assumed orthogonality therefore gives the Pythagorean identityThus $p$ is the best approximation; equality is possible only when $q=p$ because the inner product is positive-definite.

Write the linear polynomial as $p=A+Bx$. Orthogonality of $x^2-p$ to $1$ and $x$ in the weighted inner product givesSolving yieldsThe weighted squared error is ${\int }_0^{1}x(x^2-6x/5+3/10{)}^{2}dx=1/600$, providing an additional exact check. The weight's zero value at one endpoint does not destroy positive definiteness on continuous functions: a nonzero continuous function is nonzero on an interval of positive weight.