Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 1

The rank-nullity theorem says that for a linear map $T:V\to W$ with finite-dimensional domain, $\dim V=\dim \ker T+\dim \mathrm{im}T$. Apply it to $\beta {|}_{\mathrm{im}\alpha }$. Its image is $\mathrm{im}(\beta \alpha )$ and its kernel is $\mathrm{im}\alpha \cap \ker \beta$, soSince $\mathrm{im}(\alpha \gamma )\subseteq \mathrm{im}\alpha$,Using the preceding identity for both restrictions and rearranging gives

ChooseIt is holomorphic on the right half-plane with derivative $1/z\ne 0$, and its inverse $z=e^w$ has positive real part precisely for $|\mathrm{Im}w|<\pi /2$. Thus this branch of the complex logarithm is a conformal bijection onto $S$.

For $\mathrm{Re}z>0$,so $w=(z-1)/(z+1)$ lies in the unit disc. Its inverse is $z=(1+w)/(1-w)$, whose real part is $(1-|w{|}^2)/|1-w{|}^2>0$. Both derivatives are nonzero, proving this Möbius transformation is conformal.

Composing the maps givesa conformal bijection from $S$ to the unit disc.

For a geodesic triangle on the unit sphere with angles $\alpha ,\beta ,\gamma$, the Gauss-Bonnet theorem gives

For a geodesic triangulation with $V,E,F$ vertices, edges, and faces, summing the spherical excess gives $4\pi =2\pi V-\pi F$, because all angles around each vertex sum to $2\pi$. Also $3F=2E$. ThereforeThus every geodesic triangulation of the sphere has Euler number $2$.

If six triangles met at every vertex, counting vertex-face incidences would give $3F=6V$, hence $F=2V$. But part b gives $F=2V-4$, a contradiction. Therefore no such triangulation exists.

For $f=x^2+y^2+z^2$ and $g=x^2+2y^2-z^2=1$, the Lagrange multiplier equations areThe four constrained stationary points areOn the constraint, $f=2x^2+3y^2-1$ with $x^2+2y^2\ge 1$. The two $y$-axis points are global minima, while the $x$-axis points are saddles: motion along the waist decreases $f$, whereas motion away from it increases $f$. There is no maximum.

If instead $g=x^2+2y^2-z^2$ is optimized on $x^2+y^2+z^2=1$, the stationary points are all six coordinate-axis points. The $y$-axis points are maxima with value $2$, the $z$-axis points are minima with value $-1$, and the $x$-axis points are saddles with value $1$.

The velocity is $\mathbf{u}=\nabla \varphi =q{\mathbf{e}}_r/(2\pi r)$. Its divergence in polar coordinates is zero for $r>0$, while a circle about the origin has fluxBy contour deformation the same holds for any enclosing contour. This is a two-dimensional point source of strength $q$.

For the second potential,which vanishes at $y=0$. In the full plane this is a pair of equal sources of strength $q$ at $(0,\pm a)$. In $y>0$, it is one source at $(0,a)$ beside an impermeable slip wall, with the lower source serving as its image.

Substituting the exact solution and expanding about $t_n$ gives the local truncation errorThus the method is consistent of order two; as a stable one-step method it has global convergence order $k=2$.

The order cannot be higher. For $y^{\prime }=t^2$, $y(0)=0$, the exact solution is $y=t^3/3$, and direct substitution makes the local truncation error exactly

The distribution function is $F(x)=x^2/{\theta }^2$ on $[0,\theta ]$, so its median is $\theta /\sqrt{2}$. The likelihood function iswhich decreases over its admissible range. Hence $\hat{\theta }=X_{(n)}={\max }_i{X}_i$. By the invariance property of maximum likelihood estimation,

A transportation problem minimizes ${\sum }_{ij}{c}_{ij}{x}_{ij}$ subject to nonnegative shipments, prescribed row supplies, and prescribed column demands.

One optimal shipment table iswhich has the required row sums $(10,8,8)$ and column sums $(3,9,14)$ and costs $4(3)+6(1)+8(3)+3(3)+5(5)=\boxed{76}$.

For an optimality certificate, take row potentials $(1,3,3)$ and column potentials $(0,2,0)$. Their sums equal costs in occupied cells and do not exceed any cost:The dual value is $10+24+24+18=76$, so linear programming duality proves optimality.

A Jordan block $J_m(\lambda )$ has $\lambda$ on its diagonal, ones on its superdiagonal, and zeros elsewhere. A matrix is in Jordan normal form when it is block diagonal with Jordan blocks.

On a block $J_m(\lambda )$, $\ker (\alpha -\lambda I{)}^r$ has dimension $\min (r,m)$. Its increase from $r-1$ to $r$ is one exactly when $m\ge r$. Summing over blocks proves thatcounts the $\lambda$-blocks of size at least $r$.

If $\lambda \ne 0$, the polynomial $z^2$ has nonzero derivative at $\lambda$, soIf $\lambda =0$, squaring the nilpotent shift separates the odd and even basis chains:omitting a zero-size block.

The displayed invertible matrix preserves the subspaces spanned by $(e_1,e_4)$ and $(e_2,e_3)$. Its restrictions have characteristic polynomials $t^2-a_1{a}_4$ and $t^2-a_2{a}_3$. Since invertibility makes both products nonzero, each restriction has two distinct complex eigenvalues. Its Jordan form is thereforefor either choices of the square roots.

Write $|G|=p^{n}m$ with $p\nmid m$. The Sylow theorems state:

Proceed by induction on $|G|$. If $p\mid |Z(G)|$, Cauchy's theorem for finite groups gives a central subgroup $C$ of order $p$. By induction $G/C$ has a subgroup of order $p^{n-1}$; its inverse image in $G$ has order $p^n$.

If $p\nmid |Z(G)|$, the class equation shows that some noncentral conjugacy class has size not divisible by $p$. For a representative $x$, its class size is $[G:C_G(x)]$, so $p^n\mid |C_G(x)|$. This centralizer is proper because $x$ is noncentral, and induction gives it, hence $G$, a subgroup of order $p^n$. This proves Sylow's first theorem.

For $|G|=12$, the third Sylow theorem gives $n_3\equiv 1(\mathrm{mod}3)$ and $n_3\mid 4$, so $n_3=1$ or $4$. In the first case the Sylow $3$-subgroup is unique.

If $n_3=4$, conjugation acts transitively on the four Sylow $3$-subgroups. Each normalizer has order $|G|/4=3$ and is the subgroup it normalizes. The action kernel lies in every such subgroup, whose intersection is trivial, so $G$ embeds in $S_4$. Its image has order $12$ and hence index two. The unique index-two subgroup of $S_4$ is $A_4$, so

The map $f:U\to {\mathbb{R}}^n$ is differentiable at $a$ when there is a linear map $Df(a)$ such thatIt is $C^1$ when it is differentiable everywhere and $a\mapsto Df(a)$ is continuous. The definition immediately gives $\Vert f(a+h)-f(a)\Vert \le (\Vert Df(a)\Vert +o(1))\Vert h\Vert \to 0$, so differentiability implies continuity.

The inverse function theorem says that if $f$ is $C^1$ near $a$ and $Df(a)$ is invertible, then $f$ restricts to a diffeomorphism between neighbourhoods of $a$ and $f(a)$.

For distinct $x,y\in U$, convexity keeps the segment $x+t(y-x)$ in $U$. Continuity of $Df$ makesThe fundamental theorem of calculus along a line segment givesso $\Vert f(y)-f(x)\Vert \ge (1-q)\Vert y-x\Vert >0$. Thus $f$ is injective.

The bound $\Vert Df(a)-I\Vert <1$ makes $Df(a)$ invertible by the Neumann series. The inverse function theorem therefore makes $f$ locally open at every point, so $f(U)$ is open.

Surjectivity need not hold: the identity map on a proper convex open set has proper image. Even for $U={\mathbb{R}}^n$ it can fail. In one dimension, $f(x)=\arctan x$ satisfies $|f^{\prime }(x)-1|=x^2/(1+x^2)<1$, but its image is $(-\pi /2,\pi /2)$.

HereThis matrix acts as multiplication by the complex number $(2x-1)+2iy$, so its operator norm is $\sqrt{(2x-1{)}^2+4y^2}$. The largest set is therefore

A compact space is one for which every open cover has a finite subcover. Let $\mathcal{U}$ cover $[0,1]$, and letThe set is nonempty. If $s=\sup A<1$, choose $U\in \mathcal{U}$ containing $s$. An interval about $s$ lies in $U$; a point of $A$ just to the left of $s$ then extends the finite cover beyond $s$, contradicting the definition of the supremum. Hence $s=1$, and the same neighbourhood argument covers $1$, proving $[0,1]$ compact.

The continuous map $t\mapsto (\cos 2\pi t,\sin 2\pi t)$ maps $[0,1]$ onto $S^1$. A continuous image of a compact space is compact: pull an open cover back, choose a finite subcover, and push the corresponding sets forward. Thus $S^1$ is compact.

In the unusual topology on $X={\mathbb{R}}^2\setminus \{0\}$, a proper closed set is a finite union of punctured lines through the origin. Take any nonempty member $U$ of an open cover. Its complement is such a finite union. Each punctured line has the indiscrete subspace topology, because every other vector subspace meets it only at the removed origin, and is therefore compact. Finitely many finite subcovers, together with $U$, cover $X$. Hence $X$ is compact.

The integrandhas one pole inside the rectangle, at $z=0$, with residue $1$. The residue theorem therefore gives ${\int }_{C}F(z)dz=2\pi i$. The vertical-side integrals tend to zero as $R\to \infty$.

On the horizontal sides, direct substitution $z=x\pm \pi i$, with the upper side oppositely oriented, shows that their sum isTaking the limit and comparing with $2\pi i$ gives

Integrate $z{e}^{i\pi z}/(z^2+a^2)$ over the upper semicircle. Jordan lemma removes the arc, and the pole at $z=ia$ has residue $e^{-\pi a}/2$. HenceThe cosine contribution is odd, while the sine contribution is even. Therefore

The convolution is $(f\ast g)(x)={\int }_{-\infty }^{\infty }f(y)g(x-y)dy$. Under the usual integrability hypotheses, Fubini's theorem and $u=x-y$ give

The supplied transform and the Fourier shift theorem giveSincethe shift and differentiation rules yieldThe coincident boundary values make the endpoint convention immaterial.

For an S-wave, put $u(r)=r\psi (r)$. A bound state has $E=-\epsilon <0$, and the radial Schrödinger equation giveswhereContinuity of $u,u^{\prime }$ at $a$ givesThe first bound state appears at threshold $\kappa \downarrow 0$, where $ka=\pi /2$. The zero-energy threshold itself is not normalizable, so a deuteron bound state exists precisely whenThe wavefunction is $\psi =u/r$, with normalization fixing the remaining constant.

For a localized steady current, defineAssume the external field varies negligibly across the current distribution. Expanding it to first order, the uniform term gives no net force because $\int \mathbf{J}dV=0$. Using the supplied integral identity and $\nabla \cdot \mathbf{B}=0$, the first-order term in $\int \mathbf{J}\times \mathbf{B}dV$ reduces toSince $\mathbf{F}=-\nabla U$, the magnetic dipole potential energy is

Hereincreases away from the $z$-axis. A dipole parallel to $\mathbf{B}$ has energy $-m|\mathbf{B}|$ and is drawn toward stronger field, hence outward from the axis. An antiparallel dipole has energy $+m|\mathbf{B}|$ and is driven toward the weak-field axis. The field therefore spatially separates the two orientations.

Take $x$ vertically downward along the wall and $y$ normally outward, with the wall at $y=0$ and free surface at $y=h$. The expected steady velocity is $\mathbf{u}=u(y){\mathbf{e}}_x$: it vanishes at the wall and is opposed by an upward air shear $ku(h)$ at the surface. Gravity supplies the downward body force $\rho g{\mathbf{e}}_x$, while the surface normal stress is the atmospheric pressure $p_0$.

The steady parallel-flow Navier-Stokes equations reduce toThe boundary conditions areThe last condition balances the liquid's tangential traction with the opposing air stress. Hence $p=p_0$ and

Substitution of $y=h$ gives

The volume flux per unit width isThereforeThe first is the stress-free film; the second has an effectively immobilized free surface.

Let $P=u{u}^T/\Vert u{\Vert }^2$. This is an orthogonal projection matrix, so $P^T=P$ and $P^2=P$. Therefore the Householder transformationsatisfies $H_u^T{H}_u=(I-2P{)}^2=I$. If $\Vert a\Vert =\Vert b\Vert$ and $a\ne b$, thenand hence $H_{a-b}a=a-(a-b)=b$.

Apply a Householder transformation to the first column of $A$ to map it to a multiple of $e_1$, then apply transformations supported on the trailing coordinates to clear each later column below its diagonal. Their product is orthogonal and produces $R$; reversing the product gives $A=QR$.

For the displayed matrix, one resulting factorization isDirect multiplication gives the stated $A$, and $Q^{T}Q=I$.

Minimizing $\Vert Y-X\theta {\Vert }^2$ gives the normal equations $X^{T}X\hat{\theta }=X^{T}Y$. Full column rank makes $X^{T}X$ invertible, soWith the hat matrix $H=X(X^{T}X{)}^{-1}{X}^T$, the residual is $\hat{\epsilon }=(I-H)\epsilon$. The symmetric projection $I-H$ has rank $n-p$, henceand the latter estimator is unbiased.

Withthe model has

Let $D=Z_1+Z_2+Z_3+Z_4-2\pi$ be the observed closure error. Solving the normal equations gives

Every residual equals $D/4$, so the residual sum of squares is $D^2/4$. There is one residual degree of freedom, and therefore

Let the Markov chain state after each toss be $(S,I)$, where $S$ is player $A_1$'s net winnings and $I$ is the next player to toss. From $(S,i)$ it moves to $(S+1,1)$ with probability $p_i$ and to $(S-1,2)$ with probability $q_i=1-p_i$. The initial state is $(0,1)$, and a return means hitting any state with $S=0$ at positive time.

From $(1,1)$, the probability of hitting zero isIndeed, in the positive-drift case the bounded harmonic solution ofuses the decaying root $q_2/p_1$ and boundary value $h_2(0)=1$, yielding $h_1(1)=q_1/p_2$; with nonpositive drift the walk hits zero surely. By the reflected argument, from $(-1,2)$ the probability isConditioning on the first toss, the required return probability isBoth formulas equal one at the boundary $p_1+p_2=1$.