Past exam of the short options course of the University of Oxford / 2021 (ASO, A13752W1)

Source: released examination PDF, bundle pages 9–17, paper pages 1–9. Solutions follow every option of the 17 June 2021 Short Options paper. The fixed-length brachistochrone between equal-height endpoints, shortest-path tree and discrete population dynamics answers explain the domain qualifications needed for the literal printed claims.

No. Take $P=2$, $Q=9$. The Jacobi symbol isbut the squares modulo $3$ are $0,1$. A square root of $2$ modulo $9$ would reduce to one modulo $3$, which is impossible. A Jacobi symbol of $1$ records a product of local signs; it does not assert that every local sign is positive, or that a quadratic congruence is soluble.

Factor $Q={\prod }_i{p}_i$ and $Q^{\prime }={\prod }_j{q}_j$, retaining repeated prime factors. If $P$ has a common prime factor with $Q{Q}^{\prime }$, it has one with at least one of $Q,Q^{\prime }$, and both sides of the desired equality are zero. Otherwise the definition of the Jacobi symbol givesAn empty product is $1$, so the argument includes either denominator being $1$.

For a prime number $p$, multiplicativity of the Legendre symbol says $(P{P}^{\prime }/p)=(P/p)(P^{\prime }/p)$, including the case of a zero factor. Therefore, writing $Q={\prod }_i{p}_i$ with repetitions,This product definition still agrees with the stipulated zero when the numerator and denominator are not coprime. For $Q=1$ all three symbols are $1$.

If $P\equiv P^{\prime }(\mathrm{mod}Q)$, then $P\equiv P^{\prime }(\mathrm{mod}p)$ for each prime factor $p$ of $Q$. The Legendre symbol depends only on the numerator's residue class, so the corresponding factors agree. Also $\gcd (P,Q)=\gcd (P^{\prime },Q)$, covering the zero cases. Multiplying givesThe convention for denominator $1$ gives the same conclusion there.

The supplementary law for the Legendre symbol is $(-1/p)=(-1{)}^{(p-1)/2}$ for odd prime numbers $p$. For odd integers $a,b$,and the last term is even. Induction over the prime factors of $Q$ therefore gives ${\sum }_i(p_i-1)/2\equiv (Q-1)/2(\mathrm{mod}2)$. ConsequentlyFor $Q=1$ the exponent is zero.

The second supplementary law for the Legendre symbol is $(2/p)=(-1{)}^{(p^2-1)/8}$. To multiply these signs, observe that for odd $a,b$,Both factors in the numerator are divisible by $8$, so the right side is even. Applying this identity repeatedly to the prime factorization givesAgain the convention at $Q=1$ agrees.

Write $P={\prod }_i{p}_i$, $Q={\prod }_j{q}_j$, counting repetitions. Since $P,Q$ are coprime, every pair $p_i,q_j$ consists of distinct odd prime numbers. Quadratic reciprocity for the Legendre symbol givesMultiplying over all $i,j$ and using multiplicativity of the Jacobi symbol yields the sign exponentby the odd-product parity identity proved in (b)(i). Thus the Jacobi reciprocity law isIf either integer is $1$, its factorization is empty and both sides equal $1$.

Complete the square to obtain the equivalent quadratic congruenceHere $\gcd (1111,8093)=1$: the Euclidean algorithm has successive nonzero remainders $316,163,153,10,3,1$. If this modular congruence had a solution, the Jacobi symbol $(1111/8093)$ would be $1$, since a square coprime to an odd modulus has Legendre symbol $1$ at every prime factor. Calculate instead, using Jacobi reciprocity and reduction of numerators,Therefore there is no solution. This obstruction does not require a separate primality test for $8093$; the Jacobi symbol suffices.

Let $|G|=p^2$. The allowed center of a group theorem gives $|Z(G)|>1$. By Lagrange's theorem its order is $p$ or $p^2$. In the first case $G/Z(G)$ has order $p$ and is a cyclic group. A cyclic quotient by the center forces $G$ to be Abelian: if $G/Z(G)=⟨gZ(G)⟩$, write arbitrary elements as $g^{a}z,g^{b}w$ with $z,w\in Z(G)$, and commute them directly. Hence in every caseFor two examples at prime-cube order, take the dihedral group of a square, $⟨r,s\mid r^4=s^2=1,srs=r^{-1}⟩$, and the quaternion group $Q_8=\{\pm 1,\pm i,\pm j,\pm k\}$. Both have order $8=2^3$. They are nonabelian because $sr\ne rs$ and $ij=k\ne -k=ji$, respectively. They are distinct examples even up to group isomorphism: the dihedral group has five elements of order two and the quaternion group has only one. The request asks for two examples; it does not require the prime to be arbitrary.

Write $|G|=pq$ with $p<q$. The Sylow theorems give $n_q\mid p$ and $n_q\equiv 1(\mathrm{mod}q)$. Since $p<q$, necessarily $n_q=1$. Its unique Sylow subgroup $N$ is a normal subgroup and has prime order $q$, soA Sylow subgroup $H\cong C_p$ has trivial intersection with $N$, and $|NH|=|N||H|=pq$, so $NH=G$. ThusThe semidirect product action is a group homomorphism $C_p\to \mathrm{Aut}(C_q)\cong C_{q-1}$. If the action is trivial, $G\cong C_{pq}$ by the Chinese remainder theorem; if it is nontrivial, $p\mid q-1$. In particular $G$ has both a cyclic normal subgroup and a cyclic quotient, making it a metacyclic group and hence a solvable group.

The Abelian group criterion for a product of two prime squares applies because $11<19$ and $11\nmid 19^2-1=360$. Here its Sylow theorems proof is especially short:The candidates for $n_{19}$ are $1,11,121$, of which only $1$ has the required remainder. The candidates for $n_{11}$ are $1,19,361$, again leaving only $1$. Thus both Sylow subgroups $P,Q$ are normal subgroups. Their group commutators lie in $P\cap Q=\{1\}$, and $PQ=G$, so $G\cong P\times Q$. Both factors are abelian groups of prime-square order by (a).

The Fundamental theorem of finitely generated abelian groups gives preciselyThese four groups are pairwise nonisomorphic: their characteristic Sylow subgroups differ in whether they contain an element of order $11^2$ or $19^2$. The first is also the cyclic group $C_{43681}$.

Apply the Abelian group criterion for a product of two prime squares with the smaller and larger primes in their correct order. For $p<11$, all such groups are Abelian exactly when $p\nmid 11^2-1=120$. Of the primes $2,3,5,7$, only $7$ qualifies. For $p>11$, the criterion is $11\nmid p^2-1$, equivalently $p\not\equiv \pm 1(\mathrm{mod}11)$. Therefore the exact answer isEquivalently, among the stipulated primes $p\ne 11$, exclude $2,3,5$ and the residue classes $\pm 1$ modulo $11$.

For completeness, the necessity in that criterion supplies actual counterexamples at every excluded prime. If $\ell <q$ and $\ell \mid q^2-1$, the multiplicative group of a finite field is cyclic, so choose $\alpha \in {\mathbb{F}}_{q^2}^{\times }$ of order $\ell$. Let a generator of $C_{{\ell }^2}$ act on the additive group of ${\mathbb{F}}_{q^2}$ by multiplication by $\alpha$. This defines a nontrivial group action, hence a nonabelian semidirect product of order ${\ell }^2{q}^2$. Use $(\ell ,q)=(p,11)$ for $p=2,3,5$, and $(\ell ,q)=(11,p)$ for $p\equiv \pm 1(\mathrm{mod}11)$ with $p>11$. Thus no excluded prime can be restored to the answer.

Take $p=5$. The unit $3$ modulo $121$ has order $5$, becauseLet a generator $b$ of $C_{25}$ act on $C_{121}=⟨a⟩$ by $a\mapsto a^3$. The action is well defined since its order $5$ divides $25$. The semidirect producthas order $121\cdot 25=11^2{5}^2$. Its construction as a semidirect product ensures that $a$ retains order $121$. Since $a^3\ne a$, the groups generated by $a$ and $b$ do not commute, so $G$ is nonabelian.

A projective transformation of $\mathbb{P}({\mathbb{F}}^{n+1})$ is induced by an invertible linear map $A:{\mathbb{F}}^{n+1}\to {\mathbb{F}}^{n+1}$:This is well defined because replacing $v$ by a nonzero scalar multiple does not change $[Av]$. Multiplying $A$ by a nonzero scalar also leaves $\tau$ unchanged. Conversely these are precisely the transformations in the projective linear group ${\mathrm{PGL}}_{n+1}(\mathbb{F})$. No field automorphism is included in this definition.

Choose nonzero vectors $p_i\in {\mathbb{F}}^2$ representing $P_i$, and write $[u,v]=\det (u,v)$. Use the cross-ratio conventionDistinctness makes the denominator nonzero. Every rescaling of a representative occurs once in numerator and denominator, so the value depends only on the projective points. Swapping both pairs simply exchanges the numerator's factors and the denominator's factors:Under an invertible linear map $A$, $[Au,Av]=\det (A)[u,v]$. Two factors of $\det (A)$ occur above and below the fraction, proving projective transformation invariance.

Finally choose homogeneous coordinates making $P_0=[1:0]$, $P_1=[0:1]$, $P_2=[1:1]$. This normalization exists by the allowed theorem about a projective frame. A point distinct from $P_0$ is uniquely $[t:1]$, and the displayed cross-ratio is $t$. Thus equality of the two cross-ratios forces equal coordinates andThis also shows that for fixed first three distinct points the fourth point is determined by its cross-ratio.

Conjugate by a projective transformation sending $P$ to $[1:0]$ and $Q$ to $[0:1]$. A matrix representing the conjugated transformation must have zero diagonal, since its first column represents $[0:1]$ and its second represents $[1:0]$. Therefore it has the formIts square is $abI$, a nonzero scalar matrix, which induces the identity on the projective line. Conjugating back provesFor the requested explicit involution, takeMore generally $[x:y]\mapsto [ay:bx]$ works whenever $ab\ne 0$.

The four points form a projective frame: no three are collinear. By the allowed uniqueness theorem and cross-ratio invariance, useIntersecting the indicated projective lines givesA point of $Q_0{Q}_2$ has coordinates $[a:b:a+b]$. Intersecting with $P_0{P}_1$, whose equation is $z=0$, gives $Q_3=[1:-1:0]$. On that line use its first two homogeneous coordinates. The cross-ratio from (a)(ii) isThe characteristic of a field being zero ensures that $Q_3\ne Q_1$. The construction is the diagonal-point property of a complete quadrangle, producing harmonic conjugate points.

Choose a point $A$ off $L$, then a point $B$ on $A{R}_2$ distinct from $A,R_2$. PutThese four points form a projective frame: $A,B$ lie off $L$, and $AB=A{R}_2$ meets $L$ at neither $R_0$ nor $R_1$. Form, using only joins and intersections,Here the point called $Q_1$ in (i) is $R_2$, while $Q_3$ is $R_3$. Thus (i) gives $(R_0{R}_1:R_3{R}_2)=-1$. Interchanging the last two arguments reciprocates the cross-ratio, soAll auxiliary choices are possible over a field of characteristic zero, which is infinite. Uniqueness from (a)(ii) shows that the final point is independent of those choices. The picture uses an affine chart in which every relevant intersection is finite; parallel-line exceptions in another chart are ordinary intersections on the projective line at infinity.

Take $\Omega ={\mathbb{R}}^2$ andIts Jacobian matrix and determinant areThus its total derivative is invertible everywhere, and the inverse function theorem makes it locally invertible. But $F(s,t+2\pi )=F(s,t)$, so it is not injective and has no global inverse on its image. The domain is open and connected as required.

No. On $\Omega ={\mathbb{R}}^2$ letIt is a bijection onto ${\mathbb{R}}^2$, with inverse $(u,v)\mapsto (\sqrt[3]{u},v)$. Nevertheless $D{F}_{(s,t)}=\mathrm{diag}(3s^2,1)$ is singular whenever $s=0$. Set-theoretic invertibility, even together with a continuous inverse, does not force invertibility of the total derivative. If the inverse were assumed differentiable, the chain rule applied to $F^{-1}\circ F$ would force both derivatives to be invertible; that stronger assumption is absent here.

Fix $x\in U$. Since $U$ is open, choose an open ball $B(x,\epsilon )\subset U$. For any $y$ in this ball the segment $x+t(y-x)$ remains in it. For each coordinate $G_j$, the chain rule givesThe ordinary mean value theorem therefore gives $G_j(y)=G_j(x)$ for every coordinate, hence $G(y)=G(x)$. Thus $G$ is a locally constant function.

For each $z\in {\mathbb{R}}^m$, its preimage is a union of such open balls, so $G^{-1}(\{z\})$ is open in ${\mathbb{R}}^n$. Its complement in $U$ is the union of the other fibers and is also open. If $U$ is nonempty and connected, the nonempty fiber through any chosen point is both open and closed in $U$, and must be all of $U$. ThereforeThe empty-domain case is vacuous. Notice that continuity of $DG$ was never needed.

Write $a={\partial }_1{f}_1={\partial }_2{f}_2$ and $b={\partial }_2{f}_1=-{\partial }_1{f}_2$. The Cauchy-Riemann equations giveConsequentlyEquivalently, it is singular exactly when all four partial derivatives vanish. The sum of squares is over $\mathbb{R}$, so no cancellation is possible.

Suppose $D{F}_x$ were invertible at some $x\in \Omega$. The inverse function theorem supplies a neighborhood $V$ of $x$ for which $F(V)$ is a nonempty open subset of ${\mathbb{R}}^2$. But $F(V)\subset M$, and a one-dimensional embedded submanifold of ${\mathbb{R}}^2$ has empty interior. To see this last assertion, a slice chart for an embedded submanifold takes $M$ locally into a coordinate line; a coordinate line cannot contain an open plane neighborhood, and a chart preserves openness. This is a contradiction.

Thus $D{F}_x$ is singular everywhere. Part (c)(i) strengthens singularity to $D{F}_x=0$ everywhere. By (b) and connectedness of $\Omega$,The special form of the Jacobian matrix is essential: an arbitrary differentiable map can have nonconstant image in a curve with derivative of rank one.

Every $a\in \mathbb{R}$ works. Translation preserves a smooth function, and the support of a function becomes the compact interval $[a-1,a+1]$. Hence $\varphi (\cdot -a)\in C_c^{\infty }(\mathbb{R})$, the space of test functions.

Exactly $a\ne 0$ works. For nonzero $a$, composition with $x\mapsto ax$ preserves smoothness, and the support of a function is $[-1/|a|,1/|a|]$, which is compact. For $a=0$ the result is the strictly positive constant $\varphi (0)$ on all of $\mathbb{R}$, so it lacks compact support and is not a test function.

Exactly $|a|>1$ works. Composition with the hyperbolic tangent is smooth for every $a$. If $|a|>1$, set $R=\mathrm{arctanh}(1/|a|)$. For $|x|\ge R$, $|a\tanh x|\ge 1$, so the result vanishes; it is therefore a test function with support of a function $[-R,R]$.

If $|a|\le 1$, every finite $x$ satisfies $|a\tanh x|<1$. The given positivity makes $\varphi (|a|\tanh x)>0$ for every $x$, so its support is all of $\mathbb{R}$. In particular $|a|=1$ is excluded even though the function tends to zero at infinity: decay is not compact support.

Multiplication of a distribution by a smooth function is defined by moving that function onto the test function. The Dirac delta multiplication identity givesIn fact $x^2\delta$ is the zero distribution, independently of the special positivity assumptions on $\varphi$.

Write $F_j(p)=\mathcal{L}{f}_j(p)$, using the square-root branch positive on real $p>0$. Integrating the differential equation and using boundedness at zero givesThe integration constant is zero since $x{f}_1(x)\to 0$. Continuity of $f_0$ then shows $f_1(x)=O(x)$, so $f_1$ extends continuously with $f_1(0)=0$.

The Laplace transform of a derivative and differentiation under the integral sign giveIntegrating and using $F_1(p)\to 0$ as real $p\to \infty$ fixes the constant:The initial value theorem for Laplace transforms gives $f_0(0)={\lim }_{p\to \infty }p{F}_0(p)=1$. It is tempting to apply the derivative rule immediately to $f_0$, but its stated regularity is only continuity. To justify that step, define $h(x)=1-{\int }_0^x{f}_1(t)dt$. ThenUniqueness of the Laplace transform gives $h=f_0$, and hence $f_0$ is continuously differentiable and

Use the causal convolution $(f_0\ast f_1)(x)={\int }_0^x{f}_0(t)f_1(x-t)dt$. The convolution theorem for Laplace transforms and (i) giveSince $\mathcal{L}(\cos x)=p/(p^2+1)$, uniqueness of the Laplace transform yieldsAt $x=0$ the right side is $1-1=0$, agreeing with the empty convolution integral. Identification of $f_0$ as a Bessel function is unnecessary.

Use the Fourier transform convention $\hat{u}(s,t)={\int }_{\mathbb{R}}{e}^{-isx}u(x,t)dx$. Then $\widehat{u_{xx}}=-s^2\hat{u}$ and $\widehat{xu}=i{\partial }_s\hat{u}$, soThus the transformed partial differential equation and initial condition areThe last equality is the Fourier transform of the Dirac delta distribution. The two terms in ${\partial }_x(xu)=u+x{u}_x$ must be transformed together; dropping $u$ would give the wrong equation.

Substitute $\hat{u}=e^{-b(t)s^2}$ into the transformed partial differential equation. Comparing coefficients gives $b^{\prime }=1-2b$, $b(0)=0$, whenceThe stated Gaussian Fourier transform pair, with its inverse factor $1/(2\pi )$, givesThis is a normal distribution density with mean zero and variance $1-e^{-2t}$. Its total integral is one. As $t\downarrow 0$ it converges to $\delta$ in the sense of distributions, while as $t\to \infty$ it tends to the standard normal distribution density. This is the transition density from zero of the Ornstein-Uhlenbeck process with drift $-x$ and diffusion coefficient $\sqrt{2}$.

Let $g>0$ be the gravitational acceleration, and let $y$ denote positive downward depth. For a frictionless particle released from rest, conservation of energy gives $v^2=2gy$. The arc length element of $(x,-y(x))$ is $ds=\sqrt{1+y^{\prime 2}}dx$, so the travel time and length areFor a finite-time single arch, require $y>0$ in $0<x<1$, $y(0)=y(1)=0$, and interpret the endpoint integrals improperly. At a regular constrained minimizer with $S[y]=L$, the Lagrange multiplier rule for an isoperimetric constraint makes $\sqrt{2g}T+\lambda (S-L)$ stationary. Dropping the constant $-\lambda L$ givesFor $L>1$ the constraint is regular under interior variations: if its first variation vanished for every compactly supported variation, the Euler-Lagrange equation for $S$ would force $y^{\prime }$ constant, hence $y\equiv 0$ and $S=1$, a contradiction. Thus the multiplier is justified for a nondegenerate arch.

The printed case $L=1$ is degenerate. Since $\sqrt{1+y^{\prime 2}}\ge 1$, equality of the length with the horizontal separation forces $y^{\prime }=0$ and $y=0$. A particle initially at rest then has no tangential gravitational acceleration and cannot complete the trip. Its travel time is infinite. The finite-time calculation therefore concerns $L>1$, with endpoint regularity qualified below.