Past exam of the mathematics course of the University of Cambridge / 2020 / ib / Paper 2

Fix $\alpha \in \Omega$. If $\beta =g\alpha$, then normality of $H$ givesThus the transitive group action of $G$ permutes the orbits of a group action of $H$, so all $H$-orbits have the same cardinality $d$. They partition the prime-sized set $\Omega$, hence $d$ divides $|\Omega |$. The only possibilities are $d=1$ and $d=|\Omega |$. The first would make every element of $H$ fix every point, contrary to the assumption that $H$ acts nontrivially. Therefore $d=|\Omega |$ and

Now suppose $H\cap G_{\alpha }=\{1\}$. Define the orbit mapIt is surjective by transitivity of $H$. If $\theta (h_1)=\theta (h_2)$, then $h_2^{-1}{h}_1\in H\cap G_{\alpha }$, so $h_1=h_2$; hence $\theta$ is bijective. For $k\in G_{\alpha }$,Thus $\theta$ is an equivariant map: conjugation by $G_{\alpha }$ on $H$ corresponds exactly to the given action of $G_{\alpha }$ on $\Omega$.

Write $D(f)=\mathbb{C}\setminus f^{-1}(0)$. The zero and constant-one polynomials give $D(0)=\varnothing$ and $D(1)=\mathbb{C}$. Also,so $\tau$ is closed under finite intersections. Every nonzero one-variable complex polynomial has only finitely many roots, and every finite subset $\{a_1,\dots ,a_m\}$ is the zero set of ${\prod }_j(z-a_j)$. Consequently $\tau$ is exactly the cofinite topology: its open sets are the empty set and the complements of finite sets. An arbitrary union of such sets is again empty or has finite complement, so $\tau$ is a topology.

The product topology on $X\times Y$ is the topology with basisThe proposed complement need not be open. TakeThen $g^{-1}(0)$ is the diagonal. If its complement were open, a point such as $(0,1)$ would have a basic neighbourhood $U\times V$ contained in that complement. But $U$ and $V$ are nonempty cofinite subsets of $\mathbb{C}$, so $U\cap V\ne \varnothing$. For $t\in U\cap V$, the point $(t,t)$ belongs both to $U\times V$ and to the diagonal, a contradiction. Hence

Because the constraint determinesthe constrained function isIts stationary points satisfyWhen $yz=0$, these equations giveWhen $yz\ne 0$, subtracting the two bracketed equations gives $z^2=a^2{y}^2$, and substitution then giveswith the two signs chosen independently. These are exactly the constrained stationary points; equivalently they follow from the Lagrange multiplier equations.

Under the additional restriction $x\ge 0$, the constraint gives the compact elliptical disc $a^2{y}^2+z^2\le b^2$. On its boundary $x=0$, the objective is zero. At the four nonzero interior stationary points,with the sign determined by $yz$. Therefore

Use the angular-frequency Fourier transformThenwith the continuous value $\hat{f}(0)=2A$.

The convolution $(f\ast f)(x)$ is $A^2$ times the length of the overlap of the intervals $[-1,1]$ and $[x-1,x+1]$. Thus

The convolution theorem states, for this convention, thatSince $g=(B/A^2)(f\ast f)$, it follows thatagain interpreted continuously at $k=0$, where $\hat{g}(0)=4B$.

Let $r$ be distance from the common centre. By Gauss's law, the electric field is radial and equalsTaking the electric potential to vanish at infinity and requiring continuity across each shell gives

The electrostatic energy is the integral of the electric-field energy density:Only the two nonzero-field regions contribute, soTherefore

Let the upper plate in flow A move at speed $U$, with the lower plate fixed. The Couette flow profile isso its mid-plane speed is $U/2$ and its total viscous dissipation per unit plate area is

Write $G=-dp/dx>0$ for the constant pressure-gradient magnitude in flow B. The plane Poiseuille flow profile isEquality of the mid-plane speeds givesSince $u_B^{\prime }(y)=G(h-2y)/(2\mu )$,Hence

SetConditioning on $X_n$ gives the affine recurrenceIts fixed point is the first component of the stationary distribution,Thus $p_n-{\pi }_1=(1-\alpha -\beta {)}^n(p_0-{\pi }_1)$, and $p_0=1$ gives

Starting instead from state two gives the same recurrence with initial value zero. Therefore

An endomorphism $\alpha$ is a projection onto its image precisely when it is the identity on $\mathrm{im}\alpha$ and kills a complementary subspace. If ${\alpha }^2=\alpha$, thenso $\alpha$ is the identity on its image. Moreover,and $\mathrm{im}\alpha \cap \ker \alpha =\{0\}$. Henceand $\alpha$ is the projection onto its image along its kernel. Conversely, every projection is the identity after one application, so applying it twice has the same effect: ${\alpha }^2=\alpha$.

Statement (i) is false. On a two-dimensional vector space, take the nonzero nilpotent endomorphismThen ${\alpha }^2={\alpha }^3=0$, but ${\alpha }^2\ne \alpha$, so $\alpha$ is not idempotent.

Statement (ii) is false. The condition is necessary for an idempotent, but it is not sufficient. For example,satisfies $(I-\alpha {)}^2=0$, hence $\alpha (I-\alpha {)}^2=0$, whileover, for example, $\mathbb{Q}$.

Statement (iii) is false over an arbitrary field. Expanding the assumed idempotence givesMultiplying this relation on the left and right by $\alpha$ showsThus $2\alpha \beta =0$, so the conclusion would be true in characteristic different from two. In characteristic two, however, take $\alpha =\beta =I$. Then $\alpha$, $\beta$, and $\alpha +\beta =0$ are idempotent, but $\alpha \beta =I\ne 0$.

Statement (iv) is false because $\alpha \beta =0$ need not imply $\beta \alpha =0$. For example, over any field takeBoth maps are idempotent and $\alpha \beta =0$, butConsequently

Gauss lemma for polynomials states that the product of two primitive polynomials over a unique factorization domain is primitive. In particular, a primitive polynomial in $\mathbb{Z}[X]$ is reducible over $\mathbb{Q}$ exactly when it is reducible over $\mathbb{Z}$.

Eisenstein criterion says that a primitive polynomialis irreducible over $\mathbb{Q}$ if there is a prime $p$ such that $p\nmid a_n$, $p\mid a_j$ for every $j<n$, and $p^2\nmid a_0$. To prove it, suppose $F=GH$ with both factors of positive degree. After removing contents, Gauss's lemma lets us take $G,H\in \mathbb{Z}[X]$ primitive. Reduction modulo $p$ givesso both reduced factors are monomials of positive degree. Their constant terms are therefore divisible by $p$, making $p^2$ divide $a_0=G(0)H(0)$, a contradiction.

An algebraic integer is a complex number satisfying a monic polynomial in $\mathbb{Z}[X]$. Let $\alpha$ be one and let $m_{\alpha }\in \mathbb{Q}[X]$ be its monic minimal polynomial. The usual proof using polynomial content shows that $m_{\alpha }\in \mathbb{Z}[X]$. The setis the kernel of the polynomial-evaluation ring homomorphism, hence an ideal. Every $f\in I_{\alpha }$ is divisible by $m_{\alpha }$ in $\mathbb{Q}[X]$. Because $m_{\alpha }$ is monic, polynomial division of $f$ by $m_{\alpha }$ takes place in $\mathbb{Z}[X]$, and the remainder must vanish by minimality. Thereforewhere $m_{\alpha }$ is monic and irreducible.

Fortranslation givesThis is Eisenstein at $p=3$, so $f(X+1)$ and therefore $f(X)$ are irreducible over $\mathbb{Q}$. HenceIrreducibility and Gauss's lemma also show that $(f)$ is a prime ideal of $\mathbb{Z}[X]$, soIt is not a field. The class of $2$ is nonzero but not a unit: reducing further modulo $2$ gives a nonzero quotientin which the image of $2$ is zero, whereas a unit must remain a unit under every unital quotient map. Thus

HereThe L1 norm and uniform norm satisfyso the identity from the $d_{\infty }$ metric to the $d_1$ metric is one-Lipschitz and therefore continuous. The two metrics do not induce the same topology on all of $C[0,1]$. Forone has $d_1(f_n,0)=1/(2n)\to 0$ but $d_{\infty }(f_n,0)=1$. Thus $d_1$ convergence need not imply uniform convergence.

A map $F:(X,d_X)\to (Y,d_Y)$ is Lipschitz continuous if there is $L<\infty$ such thatfor all $x,x^{\prime }\in X$. Evaluation at $a\in [0,1]$ is one-Lipschitz because

Choose distinct $a_0,\dots ,a_n\in [0,1]$ and defineThe Vandermonde determinant is nonzero, so $T$ is a bijection. It is one-Lipschitz for the two uniform metrics. If ${\ell }_i$ are the associated Lagrange cardinal polynomials, thenand henceThus $T^{-1}$ is also Lipschitz.

Let ${\hat{\mathcal{P}}}_n$ be the polynomials whose values lie in $[-1,1]$. Its image under $T$ lies in the bounded cube $[-1,1{]}^{n+1}$. It is closed: if $T(p_j)\to v$, then the Lipschitz inverse gives uniform convergence $p_j\to p=T^{-1}(v)$, and passing to the limit pointwise preserves $|p(x)|\le 1$. By the Heine-Borel theorem, $T({\hat{\mathcal{P}}}_n)$ is compact, and the continuous inverse carries compactness back to ${\hat{\mathcal{P}}}_n$. Therefore

The identity mapis a continuous bijection by $d_1\le d_{\infty }$. Its domain is compact by part (i), and its codomain is a Hausdorff space because every metric space is Hausdorff. The compact-to-Hausdorff continuous bijection theorem therefore makes the identity a homeomorphism. Consequently

For a continuously differentiable curve $\gamma (t)=x(t)+iy(t)$ in the Poincare half-plane model, its hyperbolic length isThe hyperbolic lines are the geodesics: vertical Euclidean lines and Euclidean semicircles whose centres lie on the real axis.

The unique hyperbolic line through distinct $z,w\in H$ can be carried by an upper-half-plane Möbius transformation to the imaginary axis. Such transformations are isometries by hypothesis, so it is enough to consider endpoints $i{y}_0$ and $i{y}_1$. Every joining curve satisfiesEquality holds for the monotone vertical parametrizationThus the appropriately parametrized hyperbolic-line segment $[z,w]$ attains the infimum defining $\rho (z,w)$.

Now let $l,m$ have positive infimum distance. They neither intersect nor share an ideal endpoint: an intersection would give distance zero directly, while lines sharing an ideal endpoint approach one another arbitrarily closely near that endpoint. Hence they are ultraparallel. An upper-half-plane isometry puts them into the formThis is the normal form underlying the common perpendicular of ultraparallel hyperbolic lines. Put $z=e^{u+i\theta }$. Sincethe metric becomesEvery path from $l$ to $m$ therefore hasEquality requires $\theta =\pi /2$ and monotone $u$, so it is attained on the imaginary axis, which meets both semicircles orthogonally. Its intersection points $ir$ and $iR$ realize the distance

For a nonattained example, take the vertical lines $l=\{iy:y>0\}$ and $m=\{1+iy:y>0\}$. They share the ideal endpoint at infinity. The horizontal segment at height $y$ has hyperbolic length $1/y$, so their infimum distance is zero, but distinct points of $H$ always have positive distance. Hence the infimum is not attained.

The partial fraction decomposition isFor $0<|z|<2$, expand the second term as a geometric series:Hence

For $|z|>2$, expand directly in inverse powers:Therefore the outer Laurent series is

Put $w=z-1$. ThenThusThis is actually a Taylor series, because $z=1$ is a regular point. The point $z=0$ is a simple pole. At infinity,extends analytically to $w=0$, so infinity is a removable singularity of $f$, with extended value zero.

For the real integral, first writeLetWith $z=e^{i\theta }$,The denominator has roots $1/2$ and $2$, so only $1/2$ lies inside the contour. Its residue is $1/(3i)$, and the residue theorem givesConsequently

LetTheir Bessel equations in self-adjoint form areMultiply the first by $v$, the second by $u$, subtract, and integrate. The left side becomes a boundary term:Regularity at zero makes the lower boundary term vanish, whileTherefore

If $\alpha ={\gamma }_k$ and $\beta ={\gamma }_{\ell }$ with $k\ne \ell$, both endpoint values of $J_0$ vanish, so the integral is zero. This is the weighted orthogonality of distinct eigenfunctions in a Sturm-Liouville problem. For the norm, hold $\alpha ={\gamma }_k$ and let $\beta \to {\gamma }_k$. Sincethe quotient tends to

By separation of variables, the axisymmetric normal modes satisfying regularity at $r=0$ and the fixed-edge condition at $r=1$ arewhere $J_0({\gamma }_k)=0$. The zero initial displacement removes all cosine terms, soDifferentiating at $t=0$ gives the weighted Fourier-Bessel expansionMultiply by $r{J}_0({\gamma }_{\ell }r)$ and integrate from zero to one. The orthogonality and norm from part (i) giveUsingthe remaining integral isHenceand therefore

For a one-dimensional wavefunction, the probability current isA plane wave $D{e}^{ikx}$ carries current $(\hslash k/m)|D{|}^2$ to the right, while $D{e}^{-ikx}$ carries the negative of this current. Thus the asymptotic terms are interpreted as an incident wave of amplitude $A$, a reflected wave of amplitude $B$, and a transmitted wave of amplitude $C$. Their currents areCurrent conservation for a real potential gives $j_{\mathrm{in}}+j_{\mathrm{ref}}=j_{\mathrm{tr}}$, and thereforewith $P_{\mathrm{ref}}+P_{\mathrm{tr}}=1$.

A normalizable bound state has $E<0$. Writingits decaying asymptotic behaviour is

Write $T=\tanh (ax)$ and $S={\mathrm{sech}}^2(ax)$. Usingtwo differentiations ofgiveThus it solves the time-independent Schrodinger equation in the stated Pöschl-Teller potential, with

As $x\to -\infty$ and $x\to +\infty$, respectively,There is no left-moving term, so this is a reflectionless potential:The incident and transmitted amplitudes have equal modulus becauseHence

Now set $k=i\lambda$ with $\lambda >0$. ThenAt $+\infty$ this decays for every positive $\lambda$. At $-\infty$, however, the factor tends to $\lambda -a$ while $e^{-\lambda x}$ grows. Normalizability therefore requires $\lambda =a$. In that casewhich decays at both ends. The bound-state energy is