Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 3

The first isomorphism theorem for rings states that a ring homomorphism $\varphi :R\to S$ induces an isomorphismBecause $J$ is an ideal, sums and products of elements of $R+J$ remain in $R+J$, so it is a subring. The surjective homomorphismhas kernel $R\cap J$. The theorem givesEvaluation at $-2$ givesso this quotient is a field of characteristic zero. The other quotient isIt has characteristic $3$ but is not a field, since the nonzero class of $X-1$ is nilpotent.

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LetBy the regular level set theorem, $F^{-1}(0)$ is a smooth surface wherever $\nabla F\ne 0$. A critical point on the level set would satisfyThese equations imply $4a^3+27b^2=0$. Therefore

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If $4a^3+27b^2=0$ and $(a,b)\ne (0,0)$, there is a nonzero real number $r$ such thatThen $(0,0,r)$ is singular andFor $r>0$, the singular point is locally isolated because all terms inare nonnegative nearby. For $r<0$, the surface is locally a double cone with vertex at that point. Neither neighbourhood is homeomorphic to an open disc, so

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With the stated Fourier transform convention, Fourier inversion givesFor $x>0$, close the contour in the upper half-plane. Jordan lemma removes the semicircle contribution, and the only enclosed pole is $k=ip$. Its residue isThe residue theorem therefore yields

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At a regular constrained stationary point of $F$ on $G=0$, the Lagrange multiplier method solvesFor the first problem, substitute $z=1+xy$ to obtainIts stationary equations imply either $x=y$ or $xy=0$, and in either case the only real stationary point is $x=y=0$, $z=1$. Coercivity ensures that the minimum is attained, soat $(0,0,1)$.

For the second problem, the arithmetic-geometric mean inequality givesConsequentlyEquality occurs at $(0,0,1)$, and hence

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Differentiate the Legendre differential equation and put $Q_n=P_n^{\prime }$. This givesMultiplication by $1-x^2$ puts it in the self-adjoint formThe boundary term vanishes at $x=\pm 1$. Distinct eigenvalues are therefore orthogonal with weight $1-x^2$:

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For a free particle, the Hamiltonian operator isSince ${\chi }^{\prime \prime }=-k^2\chi$,Thuswhere the momentum follows because $-i\hslash d\chi /dx=\hslash k\chi$.

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The probability density and probability current areFor a stationary state, $\psi (x,t)=\chi (x)e^{-iEt/\hslash }$, so $\rho =|\chi {|}^2$ is independent of time. The continuity equation then gives $\partial J/\partial x=0$. Hence

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A plane wave $A{e}^{ikx}$ carries current $(\hslash k/m)|A{|}^2$. The incident, reflected, and transmitted currents are thereforeSpatial constancy of the current givesand henceThis is conservation of probability flux: every incident particle is either reflected or transmitted.

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Choose the polar axis along $\mathbf{d}$ and write $d=|\mathbf{d}|$. Sincethe velocity components areTheir divergence isThus the flow is incompressible for $r\ne 0$.

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For a two-dimensional incompressible flow, the stream function conventionis satisfied by

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Expand according to the hitting time and use detailed balance to reverse each finite path:The event in the final probability is $\{T_A^{+}>n\}$. The tail-sum formula for a nonnegative integer-valued random variable givesTherefore

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Sum the identity from part (a) over $z\in A$. Since $X_{T_A}\in A$ almost surely under positive recurrence,Using the conditional stationary law ${\pi }_A(z)=\pi (z)/\pi (A)$ givesHence the return-time identity

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The minimal polynomial $m_{\alpha }$ is the unique monic polynomial of least positive degree such that $m_{\alpha }(\alpha )=0$. The Cayley-Hamilton theorem says that $\alpha$ satisfies its characteristic polynomial, so such polynomials exist. Dividing any two monic candidates of least degree shows uniqueness.

If $m_{\alpha }(x)=x^m$, then ${\alpha }^m=0$ but ${\alpha }^{m-1}\ne 0$. The least $r$ for which $({\alpha }^3{)}^r=0$ is the least integer satisfying $3r\ge m$. Hence

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If $\alpha v=\lambda v$ for nonzero $v$, thenso ${\lambda }^3$ is an eigenvalue. Let $\omega =e^{2\pi i/3}$. Since $\lambda \ne 0$, the three factors ofare pairwise coprime. The kernel decomposition for coprime polynomials therefore giveswhere an absent eigenspace is interpreted as zero.

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SetFor every $i$,Consequently $q({\alpha }^3)$ contains $m_{\alpha }(\alpha )$ as a polynomial factor and is zero. By the divisibility property of the minimal polynomial,

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If the nonzero ${\lambda }_i$ are real and distinct, their cubes are also distinct. Suppose the multiplicity of $x-{\lambda }_i^3$ in $m_{{\alpha }^3}$ were $d_i<c_i$. Sinceminimality makes $m_{\alpha }(x)$ divide $m_{{\alpha }^3}(x^3)$. At $x={\lambda }_i$, the factor $x^3-{\lambda }_i^3$ has a simple zero because its derivative is $3{\lambda }_i^2\ne 0$. Thus $m_{{\alpha }^3}(x^3)$ has multiplicity only $d_i$ there, contradicting the multiplicity $c_i$ in $m_{\alpha }$. Every exponent is therefore $c_i$, and

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Matrices over a Euclidean domain are equivalent when $B=PAQ$ for invertible matrices $P,Q$, equivalently when one can pass between them by invertible elementary row and column operations.

For a nonzero matrix, move a nonzero entry to the top left and use Euclidean division and row or column operations to replace it by any nonzero remainder. Repeating terminates with an entry $d_1$ dividing every entry; otherwise adding an offending entry into its row would permit one more strict Euclidean reduction. Clear its row and column, then apply induction to the remaining submatrix. This proves equivalence to a diagonal matrix. It is in Smith normal form when, up to units,

Applying these operations over $\mathbb{Z}$ does not change the isomorphism type of the quotient. If the Smith form of $A$ is $\mathrm{diag}(d_1,\dots ,d_n)$, thenIt is finite exactly when every $d_i$ is nonzero, equivalently $\det A\ne 0$, and then

Both displayed matrices have determinant of absolute value $8$. For $A_1$, the gcds of the entries and of the $2$ by $2$ minors are both $1$, giving Smith invariantsFor $A_2$, those gcds are $1$ and $2$, givingThe first group has an element of order eight and the second does not, so

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A map $T:X\to Y$ between metric spaces is a contraction mapping if some $q<1$ satisfies $d(Tx,Ty)\le qd(x,y)$ for all $x,y$.

The Banach fixed-point theorem states that a contraction from a nonempty complete metric space to itself has a unique fixed point. Indeed, for $x_{n+1}=T{x}_n$,so the geometric-series estimate makes $(x_n)$ Cauchy. Completeness gives a limit $x_{\ast }$, continuity gives $T{x}_{\ast }=x_{\ast }$, andproves uniqueness.

Every solution of $x=\cos x$ lies in $[-1,1]$. The cosine maps this complete interval into itself and, by the mean value theorem, has Lipschitz constant at most $\sin 1<1$ there. It therefore has exactly one real fixed point.

The mean value inequality says that on a convex domain, a uniform derivative bound $\Vert Df\Vert \le M$ implies $\Vert f(x)-f(y)\Vert \le M\Vert x-y\Vert$. Equip ${\mathbb{R}}^2$ with the maximum norm and takeFor $(x,y)\in D$, elementary cosine bounds giveso $f(D)\subseteq D$. The maximum absolute row sum ofis at most $2\sin (1/2)<1$. The mean value inequality makes $f$ a contraction on the complete set $D$, so it has a fixed point.

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A topological surface is a Hausdorff, second-countable space in which every point has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^2$.

For $S_1$, interior points and points in the interiors of paired edges plainly have disc neighbourhoods. The corner identifications form two classes; in each class two quarter-discs are glued to make a half-disc, and the adjacent identified edge neighbourhoods complete a disc. Equivalently, this polygon is a cell decomposition of the real projective plane, hence a quotient of $S^2$ by the antipodal group. Thus $S_1$ is a topological surface.

In $S_2$, three distinct sides labelled $a$ are identified. A point in the interior of their common image has a neighbourhood made from three half-discs meeting along their diameters. Removing the common diameter leaves three local sides rather than two, so this neighbourhood is not a disc or half-disc. Therefore

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Cut the octagon for $S_3$ along a diagonal joining the appropriate two vertex classes. The two resulting polygons can be reattached along that diagonal in the opposite order. Reading the new boundary word gives the square wordwhich is exactly the edge identification displayed for $S_4$. The cut-and-paste map is affine on the two pieces and respects every paired edge, so it descends to a homeomorphism $S_3\cong S_4$. Both are the Klein bottle.

After deleting an open disc, the punctured Klein bottle can be realized as a boundary connected sum of two embedded Möbius strips, and hence embeds in ${\mathbb{R}}^3$. The closed Klein bottle cannot embed: every connected closed surface embedded in ${\mathbb{R}}^3$ is two-sided and therefore orientable, by the Jordan-Brouwer separation theorem, whereas the Klein bottle is nonorientable. Thus

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Uniform convergence on compact subsets makes $f$ continuous. For every triangle whose closure lies in $U$, uniform convergence on its boundary permits passage through the contour integral:Morera's theorem implies that $f$ is holomorphic. If a compact set $K⋐U$ is surrounded by a finite union of contours at positive distance from $K$, the Cauchy integral formula for derivatives givesThe uniform bound on the surrounding compact set proves that $f_n^{\prime }\to f^{\prime }$ uniformly on $K$.

If $f$ had distinct zeros $c$ and $d$, choose disjoint small closed discs around them whose boundary contains no zero of $f$. Uniform convergence and Rouché's theorem imply that, for large $n$, $f_n$ has a zero in each disc, contradicting uniqueness of $c_n$. Thus $f$ has at most one zero.

For an example on the unit disc, takeThen $f_n\to f(z)=z-1$, which has no zero in the open disc. In general, Hurwitz's theorem shows that $f$ is zero-free exactly when the unique zeros escape every compact subset of $U$:Indeed, an interior accumulation point of $c_n$ is a zero of $f$, while a zero of $f$ forces the unique $c_n$ into each of its sufficiently small neighbourhoods.

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Apply Green second identity to $u$ and the free-space Green function $G_{fs}$ on $V$ with a small ball about ${\mathbf{r}}_0$ removed. Both functions are harmonic there, so only boundary terms remain. The contribution from the small sphere tends to $u({\mathbf{r}}_0)$ becausehas unit delta source, while the remaining small-sphere term vanishes. Taking the radius to zero gives Green's third identity

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Integrate ${\nabla }^{2}u=0$ over a large half-ball and apply the divergence theorem. The hemispherical contribution vanishes by the assumed decay. On $z=0$, the outward normal is $-{\mathbf{e}}_z$, so $\partial u/\partial n=-p$. Hence compatibility requiresThis is the usual solvability condition for the Neumann problem.

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Reflect the source point ${\mathbf{r}}_0=(x_0,y_0,z_0)$ across the plane to ${\mathbf{r}}_0^{\ast }=(x_0,y_0,-z_0)$. Equal-sign source and image make the normal derivatives cancel on the plane. Thus the method of images gives the Neumann Green function

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Green's identity with $\partial G/\partial n=0$ leaves only the prescribed normal derivative. Since $\partial u/\partial n=-p$ and, on the plane,the decaying solution isThus

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Use the large-$z_0$ expansionThe first term vanishes by the compatibility condition. For the given $p$, symmetry leavesConsequently

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For $X^{\mu }=(ct,x,y,z)$, the stated Lorentz transformation gives

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Transforming the electromagnetic field tensor by $F^{\prime }=\Lambda F{\Lambda }^T$ and reading off its components givesThese are the parallel and transverse field transformation laws for a boost in the $x$ direction.

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Take the wire along the $x$-axis and let $r$ be perpendicular distance from it. Gauss's law and cylindrical symmetry giveFor an observer boosted parallel to the wire,The moving observer sees length contraction and hence line density ${\lambda }^{\prime }=\gamma \lambda$, together with current $I^{\prime }=-\gamma \lambda v$. The transformed magnetic field is exactlywith the direction prescribed by the current. This is Ampère's law for the current seen in the moving frame.

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Axisymmetry and incompressibility giveThus $r{u}_r=C$. Finiteness at the origin forces $C=0$, so

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Using the polar-coordinate curl with $u_{\theta }=r\Omega (r)$ gives

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Take the curl of the Navier-Stokes equation. The pressure term disappears, curl commutes with the Laplacian, and the incompressible vector identitygivesTherefore the material derivative form is

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The vorticity is normal to the plane, so the stretching term vanishes. Axisymmetry also makes angular advection vanish, and part (a) gives no radial advection. Hence the scalar vorticity equation is the radial diffusion equation

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Set the similarity variableThenThe factors of $t$ cancel, leaving the self-similar ordinary differential equation

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Insert the exact data $y(t)=P(t)$ into the linear multistep method. Its local residual is the linear functionalTaylor expansion about $t_n$ shows that a method has order $p$ exactly when this functional annihilates the monomials $1,t,\dots ,t^p$; by linearity this is equivalent to annihilating every polynomial of degree at most $p$. This is precisely

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The Dahlquist equivalence theorem says that a consistent linear multistep method is convergent exactly when it is zero-stable. HereThe consistency conditions hold for every $a$. The two quadratic roots have product one. They are distinct and on the unit circle exactly when $0<a<2$; at $a=2$ the root $-1$ is repeated, while outside this interval one root has modulus greater than one. The root condition for a multistep method therefore givesThe polynomial exactness conditions hold through degree two, while the degree-three residual is $6-a$. Every convergent member thus has(The exceptional value $a=6$ has formal order four but is not zero-stable.)

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The log likelihood, up to constants, isDifferentiation gives the generalized least squares estimatorand maximizing over the scale givesThe divisor $n$ is appropriate for maximum likelihood rather than unbiased estimation.

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The estimator is an affine transformation of the multivariate normal distribution. Its mean is $\beta$, and direct covariance calculation gives

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Write $\hat{\beta }=AY$, whereAny other linear unbiased estimator is $CY$ with $CX=I$, so $C=A+D$ and $DX=0$. The cross covariance vanishes becauseThereforeBy the Gauss-Markov theorem,