Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 3

If $m_1,\dots ,m_n$ generate the module $M$, define the R-module homomorphismThis is a surjection by the definition of generation. Conversely, an R-module homomorphism that is a surjection sends the standard basis vectors of the free module $R^n$ to a generating family, since every vector in $R^n$ is their linear combination. ThusThis is the finite generation as a quotient of a finite free module criterion. The zero module causes no exception: it is generated by the empty family, or by one zero element.

For the requested example, take $M=(\mathbb{Z}/2\mathbb{Z}{)}^2$, with the usual integer scalar action. Reducing both coordinates modulo two gives an R-module homomorphism that is a surjection ${\mathbb{Z}}^2\to M$ with nonzero kernel $2{\mathbb{Z}}^2$, so it is not an isomorphism. Every image of $\mathbb{Z}$ is a cyclic module, whereas every element of $M$ has additive order at most two and $M$ has four elements. Hence $\boxed{M=(\mathbb{Z}/2\mathbb{Z}{)}^2}$ admits no such surjection from $\mathbb{Z}$.

A metric space is a complete metric space when every Cauchy sequence converges to a point of that space: for each $\epsilon >0$, the condition $d(x_m,x_n)<\epsilon$ for all sufficiently large $m,n$ must imply existence of $x$ with $d(x_n,x)\to 0$.

A Cauchy sequence in $[0,1]$ is also Cauchy in $\mathbb{R}$, so it converges there by the assumed completeness. Its limit remains between zero and one, since inequalities are preserved under limits. ThereforeThis is an instance of the fact that a closed subset of a complete metric space is complete with its induced metric.

The rational sequence $q_n=10^{-n}\lfloor 10^n\sqrt{2}\rfloor$ satisfies $0\le \sqrt{2}-q_n<10^{-n}$. It is therefore a Cauchy sequence in the induced metric on $\mathbb{Q}$. Any limit in $\mathbb{Q}$ would also be its real limit, which is $\sqrt{2}\notin \mathbb{Q}$. ThusCompleteness concerns limits lying in the space, so being a dense subset of $\mathbb{R}$ does not suffice.

Let $x_n=(\pi /2+2\pi n{)}^{-1}$. The points $p_n=(x_n,1)$ lie on the specified graph and converge in ${\mathbb{R}}^2$ to $(0,1)$. They form a Cauchy sequence in the induced metric, but their only possible limit is absent from the space. Adding $(0,0)$ does not supply that missing limit. Hence the specified space is not complete.

For the unheaded request on $\mathbb{R}$, use the arctangent pullback metricPositivity, symmetry and the triangle inequality follow from the ordinary distance, and injectivity of $\arctan$ gives $d(x,y)=0$ only if $x=y$. The sequence $n$ is Cauchy for this metric, because $\arctan n\to \pi /2$, but cannot converge to any finite real $x$: that would require $\arctan x=\pi /2$. Thus this metric makes $\mathbb{R}$ incomplete, even though it induces its usual topology.

The product topology has as a basis of a topology all sets $U\times V$, where $U$ is open in $X$ and $V$ is open in $Y$. Open sets are arbitrary unions of these rectangles. The projection maps are $p(x,y)=x$ and $q(x,y)=y$. For any open $U\subset X$ and $V\subset Y$,These are open rectangles, including the empty case. Taking inverse images of open sets therefore proves that both projection maps are continuous functions.

If $X$ is Hausdorff, every pair $x\ne y$ has disjoint open neighborhoods $U$ and $V$. The rectangle $U\times V$ is an open neighborhood of $(x,y)$ disjoint from the diagonal. Hence every point of its complement is interior, so the diagonal is a closed set.

Conversely, if the diagonal is closed, its complement is open in the product topology. For $x\ne y$, the rectangle-basis property gives open $U\ni x$ and $V\ni y$ with $U\times V$ disjoint from the diagonal. If $z\in U\cap V$, then $(z,z)$ lies in that rectangle, a contradiction. Thus $U\cap V=\varnothing$, giving the Hausdorff separation condition. ConsequentlyThe argument includes spaces with zero or one point, for which both conditions hold.

Let $Y(s)$ be the Laplace transform of $y$. The initial values give $\mathcal{L}(y^{\prime })=sY$ and $\mathcal{L}(y^{\prime \prime })=s^{2}Y-1$, while $\mathcal{L}(t{e}^{-3t})=(s+3{)}^{-2}$. Therefore, for $\mathrm{Re}s$ sufficiently large,The partial fraction decomposition isInverting the elementary Laplace transforms givesThese coefficients give $y(0)=0$ and $y^{\prime }(0)=1$; the transformed equation, or direct differentiation, verifies the forcing. Uniqueness follows from uniqueness for the corresponding linear initial-value ordinary differential equation.

Write $w$ for the complex coordinate. The inverse stereographic projection iswith $\infty$ mapped to the north pole. A rotation $R_z(\theta )$ about the third axis sends $w$ to $e^{i\theta }w$, represented byFor the rotation $R_y(\theta ):(x,y,z)\mapsto (x\cos \theta +z\sin \theta ,y,-x\sin \theta +z\cos \theta )$, substitution in the stereographic projection givesBoth displayed matrices are unitary with determinant one, hence belong to the special unitary group.

Use the rotation-generation result that every $R\in SO(3)$ can be written $R_z(\alpha )R_y(\beta )R_z(\gamma )$, with angles chosen suitably. This is the Euler-angle decomposition, which is allowed here. Composition of Möbius transformations corresponds to multiplication of their matrix representatives, soThe formulas extend at poles in the Riemann sphere. This realizes sphere rotations as special-unitary Möbius transformations. The two representatives $U$ and $-U$ define the same Möbius transformation, as expected for the double covering of the special orthogonal group.

For the boundary constraint $g(x,y)=x^2+y^2-1=0$, the Lagrange multiplier equations for $z=xy$ areNeither coordinate can vanish at a stationary point. Substitution gives $4{\lambda }^2=1$. For $\lambda =1/2$, $x=y=\pm 1/\sqrt{2}$ and $z=1/2$; for $\lambda =-1/2$, $x=-y=\pm 1/\sqrt{2}$ and $z=-1/2$. The constraint circle is compact, its gradient is nowhere zero, and every extremum must occur among these stationary points. ThusThe paired signs in each expression are linked. The inequality $2|xy|\le x^2+y^2=1$ also confirms the global values.

Let $c>0$ denote the stated unit cost of cylindrical wall, and assume $V>0$. The circular cylinder and hemisphere give the volume constraint and total costOnly the walls and cap have specified costs, so no separate base cost is included. Eliminating $h$ giveswith $0<r\le (3V/(2\pi ){)}^{1/3}$ from $h\ge 0$. Now $C^{\prime }(r)=-2cV/r^2+(16/3)\pi cr$ vanishes at $r^3=3V/(8\pi )$. Moreover $C^{\prime \prime }(r)=4cV/r^3+(16/3)\pi c>0$ and the stationary point is feasible, so it is the unique global minimum. The optimal dimensions are

The substitution removes the zero-order term: $u_x=e^{-x^2}(v_x-2xv)$ and $u_y=e^{-x^2}{v}_y$, so the partial differential equation becomes $v_x+x{v}_y=1$. Its method of characteristics givesThe invariant is $\eta =y-x^2/2$. Therefore $v=x+F(\eta )$. The initial line $x=0$ is noncharacteristic, and the boundary data give $F(\eta )=\eta e^{-{\eta }^2}$. HenceEach characteristic meets the data line once, ensuring uniqueness among differentiable solutions, and substituting the result verifies both the equation and the data.

Inside the infinite square well, the time-independent Schrodinger equation is $-({\hslash }^2/(2m)){\psi }^{\prime \prime }=E\psi$. The infinite exterior potential imposes $\psi (0)=\psi (a)=0$. For $E>0$, the first condition leaves a sine and the second requires $ka=n\pi$. A zero or negative energy gives only the zero solution under these conditions. The normalized energy eigenstates and eigenvalues areThe wavefunctions vanish outside the well, and a constant phase is immaterial. Normalization uses ${\int }_0^a{\sin }^2(n\pi x/a)dx=a/2$.

The Born rule gives the probability of $E_n$ as the squared coefficient of the initial state in the energy eigenstate basis. Direct integration givesThe strict energy cutoff corresponds to $n^2<5$, so only $n=1,2$ contribute. Both coefficients equal $2/\pi$, and thereforeThe discontinuous initial wavefunction is a legitimate normalized $L^2$ state for this spectral probability calculation, although it is not in the Hamiltonian's differentiability domain; finite mean energy is not required to compute this probability.

A reversible Markov chain has a stationary law $\pi$ for which every finite path has the same distribution as its time-reversed path:For a transition matrix $P$, the equivalent detailed balance condition is ${\pi }_i{P}_{ij}={\pi }_j{P}_{ji}$ for all $i,j$. It implies $\pi P=\pi$ by summing over $i$. Multiplying the balance identities along a path shows equality of its forward and reverse probabilities when $X_0\sim \pi$; conversely, taking paths of length one gives those identities.

For the usual undirected graph with at least two vertices, a simple random walk chooses a neighbor uniformly at each step, so $P_{ij}=1/d_i$ when $i$ and $j$ are adjacent and zero otherwise, where $d_i$ is the degree of a vertex. The handshaking lemma gives ${\sum }_i{d}_i=2|E|$. Thus the stationary distribution of a graph random walk isFor adjacent $i,j$, ${\pi }_i{P}_{ij}=1/(2|E|)={\pi }_j{P}_{ji}$, and for nonadjacent pairs both sides vanish. By detailed balance, this is an invariant distribution and the walk started from it is a reversible Markov chain. Connectedness makes the finite walk an irreducible Markov chain, giving uniqueness of its invariant distribution. No aperiodicity assumption is needed for uniqueness, so a bipartite graph is allowed.

If the connected graph has a single isolated vertex, the usual neighbor-choice formula is undefined. With the natural absorbing convention $P_{11}=1$, its unique invariant distribution is ${\pi }_1=1$ and it is reversible. The degree-normalization formula presupposes at least one edge.

Associate to $f$ the symmetric bilinear form $B(u,v)=\frac{1}{2}(f(u+v)-f(u)-f(v))$. If $f$ is identically zero, every basis is diagonal. Otherwise choose $v$ with $f(v)\ne 0$. Every $w$ has the decompositionwhose second term lies in $v^{\perp }$. Thus $V=\mathbb{R}v\oplus v^{\perp }$, with no cross term in $f$. Induction on dimension gives a diagonal basis. Rescaling its nonzero entries reduces the form towith $k$ additional zero coordinates. Define the rank of a quadratic form $r=p+q$ and its signature $s=p-q$. Rank is intrinsic, since it is the rank of $v\mapsto B(v,\cdot )$, equivalently $n-\dim \mathrm{rad}B$. Also $p$ is the greatest dimension of a subspace on which the restriction of $f$ is a positive-definite quadratic form: any such subspace intersects the negative-plus-zero coordinate space only at zero, so its dimension is at most $p$, while the positive coordinate space attains $p$. The corresponding negative statement identifies $q$. This proves the basis independence of $r$ and $s$, the Sylvester's law of inertia conclusion.

For a totally isotropic subspace $W$ on which $f$ vanishes, polarization identity also makes $B$ vanish on $W\times W$. Quotient by the radical of a bilinear form. In the resulting nondegenerate form, projection of an isotropic subspace to either the positive or negative coordinate space is injective, so its dimension is at most $\min (p,q)$. Consequently $\dim W\le k+\min (p,q)$. The radical of a bilinear form together with the vectors $e_i^{+}+e_i^{-}$, $1\le i\le \min (p,q)$, attains this bound. The maximum dimension of a totally isotropic subspace is thereforeThe criterion for membership in a diagonal basis follows similarly: a nonzero vector can belong to a diagonal basis if it has nonzero quadratic value, by the same orthogonal splitting, or if it lies in the radical of a bilinear form, by choosing a radical basis containing it. An isotropic vector in a diagonal basis must be orthogonal to every basis vector, and hence must lie in that radical. For the specified form, its radical is the $z$-axis, giving exactlyThe zero vector never belongs to a basis.

A Euclidean domain is an integral domain $R$ with a function $\nu :R\setminus \{0\}\to \mathbb{N}$ such that for each $a\in R$ and $b\ne 0$ there are $q,r\in R$ with $a=qb+r$ and either $r=0$ or $\nu (r)<\nu (b)$. Some conventions add a multiplicative monotonicity condition; it is unnecessary for this implication.

Let $I$ be a nonzero ideal. Choose $d\in I\setminus \{0\}$ minimizing $\nu (d)$, possible because the values lie in the nonnegative integers. For $a\in I$, division by $d$ gives $a=qd+r$, with $r=a-qd\in I$. A nonzero remainder would contradict minimality, so $r=0$ and $a\in (d)$. Since $d\in I$, the reverse inclusion also holds. The zero ideal is $(0)$. Thus

Reducing coefficients modulo $p$ gives the ring isomorphismIf $d=\mathrm{deg}f>0$, its monic leading coefficient remains one modulo $p$, so $\mathrm{deg}\bar{f}=d$. Polynomial division gives a unique representative of degree less than $d$: existence follows by division, and the difference of two such representatives cannot be a nonzero multiple of a degree-$d$ monic polynomial. There are $p$ choices for each of the $d$ coefficients, provingIrreducibility of $\bar{f}$ is not needed for finiteness; without it the quotient ring need not be a field.

Write the Gaussian integer $\alpha =a+bi$, with $a,b\in \mathbb{Z}$, and set $R=\mathbb{Z}[\alpha ]$. Complex conjugation preserves $R$, because $\bar{\alpha }=2a-\alpha$. Choose a nonzero $\beta \in P$. Then $N=\beta \bar{\beta }$ is a positive integer in $P$. It cannot be one because $P$ is a proper prime ideal. Factoring $N$ into rational primes and applying primality repeatedly gives a prime $p\in P$.

The element $\alpha$ satisfies the monic polynomial $F(t)=t^2-2at+a^2+b^2$. Evaluation induces a ring homomorphism that is a surjectionThe source has $p^2$ elements by part (b), soIf $b=0$, use $F(t)=t-a$ instead and get at most $p$ elements. Moreover $R/P$ is an integral domain, so its finiteness makes it a field. This proves the finite residue-field theorem for Gaussian integer orders without incorrectly assuming that every Gaussian integer order is a principal ideal domain.

A contraction mapping satisfies $d(Tx,Ty)\le qd(x,y)$ for all $x,y$, with one fixed $0\le q<1$. The contraction mapping theorem states that such a map on a nonempty complete metric space has a unique fixed point and that every sequence of iterates converges to it.

Choose $x_0$ and put $x_{n+1}=T{x}_n$. Iteration gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$, so for $m>n$,Thus the iterates form a Cauchy sequence and converge by completeness to $x_{\ast }$. The Lipschitz bound gives continuity of $T$, hence $T{x}_{\ast }=x_{\ast }$. Two fixed points satisfy $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing equality of the points. Passing $m\to \infty$ also gives the usual geometric error estimate. For $q=0$, convergence occurs after one application.

If $f^k$ is a contraction mapping, let $x_{\ast }$ be its unique fixed point. The commutation identity $f^{k}f=f{f}^k$ gives $f^k(f{x}_{\ast })=f{x}_{\ast }$, so uniqueness implies $f{x}_{\ast }=x_{\ast }$. Any fixed point of $f$ is a fixed point of $f^k$, giving uniqueness, without even assuming $f$ continuous. On the complete discrete metric space $\{0,1,2\}$, the map $f(0)=f(1)=0$, $f(2)=1$ has constant second iterate but is not a contraction: $d(f2,f0)=d(2,0)=1$.

For the specified function, $x+\sin x\ge 0$ on $[0,\infty )$ because it starts at zero and has derivative $1+\cos x\ge 0$. Therefore $f$ maps the complete half-line into itself. Its derivative satisfiesThe mean value theorem therefore makes it a contraction of constant at most $2/3$. Hence

Fix $R>|z|$. The Cauchy integral formula and the uniformly convergent geometric series on $|\zeta |=R$ giveDefiningthe remainder after degree $N$ has modulus at most $M_R(|z|/R{)}^{N+1}/(1-|z|/R)$, which tends to zero uniformly on each smaller disk. Differentiating the integral formula at zero identifies $c_j=f^{(j)}(0)/j!$, independently of $R$. Since $f$ is an entire function, $R$ can be chosen larger than any given $|z|$, proving the global Taylor series.

For the requested radius $r$, the same coefficient integral gives the Cauchy estimateTo characterize equality in the Cauchy coefficient estimate, parametrize its circle:If $M_r=0$, the identity theorem gives $f=0$. Otherwise equality in the modulus bound requires the continuous integrand to have both constant modulus $M_r$ and constant complex argument. To see this directly, rotate its average to the positive real axis: its real part is bounded above by $M_r$, and equality of its average with $M_r$ forces that real part to equal $M_r$ everywhere. Thus $f(r{e}^{i\theta })=c_n{r}^n{e}^{in\theta }$ on the circle. The identity theorem extends that equality to the plane. Conversely every such monomial attains the bound. Thereforeincluding the zero function and, when $n=0$, arbitrary constant functions.

At a regular point, choose the unit normal $\mathbf{n}=({\sigma }_x\times {\sigma }_y)/|{\sigma }_x\times {\sigma }_y|$. The first fundamental form is the induced squared lengthThe second fundamental form isThe Gaussian curvature is the determinant of the shape operator, $K=(eg-f^2)/(EG-F^2)$, independent of the normal orientation.

For the given graph, ${\sigma }_x=(1,0,y)$ and ${\sigma }_y=(0,1,x)$, so with $S=1+x^2+y^2$,ThereforeThe surface is regular everywhere. The printed description “hyperboloid” is a naming error: this graph is a hyperbolic paraboloid, and the computation uses the actual parametrization.

Assume the parametrization is $C^2$ and regular, so its curvature is defined. Write $\mathbf{v}={\mathbf{a}}^{\prime }+y{\mathbf{b}}^{\prime }$ and $W=|\mathbf{v}\times \mathbf{b}|>0$. Since ${\sigma }_x=\mathbf{v}$ and ${\sigma }_y=\mathbf{b}$, its first fundamental form isFor the unit normal $\mathbf{n}=(\mathbf{v}\times \mathbf{b})/W$, the mixed coefficient of the second fundamental form iswhere the term containing ${\mathbf{b}}^{\prime }$ twice vanishes. Also $g=\mathbf{n}\cdot {\sigma }_{yy}=0$, while $EG-F^2=W^2$. ThusThe Gaussian curvature vanishes exactly when the displayed scalar triple product is zero. This is the Gaussian curvature of a ruled surface criterion; at a singular point with $W=0$, curvature and the claimed equivalence are not defined.

Interpret the three conditions at zero as an initial-value problem on $x\ge 0$, with interior source $\xi >0$. A causal Green function satisfies ${\mathcal{L}}_{x}G=\delta (x-\xi )$ and vanishes for $x<\xi$. Since the leading coefficient of the third derivative is one, the jump conditions for a third-order Green function require $G$ and $G^{\prime }$ continuous and $G^{\prime \prime }$ to jump by one. For $t=x-\xi >0$, the homogeneous roots are $0,2,-1$. Solving $g(0)=g^{\prime }(0)=0$, $g^{\prime \prime }(0)=1$ yieldsFor $\xi >0$ it vanishes near zero, so all three initial boundary conditions hold. The matching conditions ensure that ${\partial }_x^{3}G$ supplies precisely a Dirac delta, without unwanted delta derivatives.

The homogeneous solution carrying the nonzero initial data is $y_h=\frac{1}{2}-\frac{1}{6}{e}^{2x}+\frac{2}{3}{e}^{-x}$. The Green-function representation therefore givesFor $x<1$ the integral vanishes. For $x\ge 1$, set $t=x-1$ and integrate to obtain $\frac{e}{6}(e^{2t}-3e^t-e^{-t}+3)$. ThusThe bracket and its first two derivatives vanish at $x=1$, so $y,y^{\prime },y^{\prime \prime }$ match continuously there; its third derivative jumps by the stated forcing. The value assigned to $\Theta (0)$ is immaterial. The source at the initial endpoint $\xi =0$ would need a separate endpoint-delta convention, which is not used in this integral.

The orbital angular momentum operators areIn particular ${\hat{L}}_3=-i\hslash (x_1{\partial }_2-x_2{\partial }_1)$. For smooth compactly supported functions $\varphi ,\psi$, integration by parts produces no boundary term, and ${\partial }_2{x}_1={\partial }_1{x}_2=0$. HenceThis proves Hermiticity on the common test-function domain, and the same calculation applies on suitable decaying domains. As an unbounded operator, it is not defined on every $L^2$ function: the usual self-adjoint operator realization is the generator of unitary rotations about the third axis. Thus the integration statement includes its necessary domain/boundary assumptions.

Take a normalized state for which the indicated variances and operator pairings exist. Set $A={\hat{L}}_3-⟨{\hat{L}}_3⟩$ and $B={\hat{x}}_1-⟨{\hat{x}}_1⟩$. For real $\lambda$, positivity of the squared norm givesThe canonical commutation relations yield $[{\hat{L}}_3,{\hat{x}}_1]=i\hslash {\hat{x}}_2$. Thus the quadratic polynomial in $\lambda$ is nonnegative everywhere only if its discriminant is nonpositive. This gives the Robertson uncertainty principleIf either variance is zero, the same polynomial argument forces the commutator expectation to vanish, so that edge case is included. Centering the operators is essential to obtain variances rather than uncentered second moments.

Use the radial exponential $e^{-\alpha r}$ shown in the original PDF; the TeX's $e^{-\alpha x}$ is a transcription defect. Every orbital angular momentum operator annihilates a radial function, so it acts only on $P=x_1+x_2+2x_3$. The angular momentum of a radial function times a linear polynomial is determined by its degree-one angular factor: direct application of the rotation derivatives givesEquivalently its restriction to the sphere is a degree-one spherical harmonic. ThusFurthermore ${\hat{L}}_3\psi =-i\hslash K(x_1-x_2)e^{-\alpha r}$. Its expected value is proportional to the integral of $(x_1+x_2+2x_3)(x_1-x_2)e^{-2\alpha r}$. The cross terms are odd in a coordinate, and the integrals of $x_1^2$ and $x_2^2$ agree by rotational symmetry. HenceThe state need not be an eigenstate of ${\hat{L}}_3$. If normalization is desired, $\int P^2{e}^{-2\alpha r}{d}^{3}x=6\pi /{\alpha }^5$, so $K={\alpha }^{5/2}/\sqrt{6\pi }$; the zero expectation holds after normalization for any nonzero initial $K$.

Let the smaller charge be $q$ at $\mathbf{a}$ and the other charge be $-Q$ at $\mathbf{b}$, where $0<q<Q$; an overall sign reversal changes no zero-potential locus. Put $\lambda =q/Q\in (0,1)$. The electrostatic potential vanishes exactly when $|\mathbf{x}-\mathbf{a}|=\lambda |\mathbf{x}-\mathbf{b}|$. Completing the square gives the Apollonius sphereThe distances from its center to the charge positions are $|\mathbf{c}-\mathbf{a}|=\lambda R<R$ and $|\mathbf{c}-\mathbf{b}|=R/\lambda >R$. Thus it encloses exactly the charge of smaller magnitude. Unequal magnitudes are essential: the equal-magnitude case instead gives the perpendicular bisector plane.

Write $R=|\mathbf{p}|>a$. In SI units, the exterior electrostatic potential solveswith the point-charge singularity $q/(4\pi {\epsilon }_0|\mathbf{x}-\mathbf{p}|)$. The image charge for a grounded conducting sphere choiceslie inside the sphere. On $|\mathbf{x}|=a$, direct squaring gives $|\mathbf{x}-{\mathbf{p}}^{\prime }|=(a/R)|\mathbf{x}-\mathbf{p}|$. Thereforesatisfies the source equation, grounded boundary value and decay at infinity. The Uniqueness of the Dirichlet problem, applied to the harmonic difference with its singularity removed, identifies this with the physical exterior potential. Thus its gradient gives the required exterior electric field.