Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 3

Fordefine its action on the projective line $F\cup \{\infty \}$ by the Möbius transformationwith the usual conventions when the denominator vanishes or $x=\infty$. Matrix multiplication agrees with composition, so this is a group action. If $A$ fixes every projective point, then it fixes $0$ and $\infty$, forcing $b=c=0$, and fixing $1$ then gives $a=d$. Thus the kernel consists exactly of the nonzero scalar matrices $Z$. The induced group homomorphismis therefore injective. This is the projective general linear group action on the projective line.

For $F={\mathbb{F}}_4$,The scalar subgroup has order $|Z|=3$, soThe four classes represented byform a subgroup $P$ of order four, hence a Sylow subgroup for the prime two. Its action is translation $x\mapsto x+b$ on ${\mathbb{F}}_4$ and fixes $\infty$. For $b\ne 0$, characteristic two makes this permutation a product of two disjoint transpositions on the four finite points. It is therefore even, andThis realizes the Sylow 2-subgroup of PGL2 over F4.

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In the Poincare half-plane model, the Riemannian metricassigns to a smooth curve $\gamma (t)=(x(t),y(t))$ the lengthThe metric is conformal to the Euclidean metric, so hyperbolic and Euclidean angles agree. Its area element is

The two geodesics from $e^{i\alpha }$ and $e^{i\beta }$ to infinity are the vertical lines $x=\cos \alpha$ and $x=\cos \beta$, while the third side is the unit semicircle. The triangle lies above that semicircle, so its area isIts interior angles are $\alpha$, $\pi -\beta$, and zero at the ideal vertex, and henceTriangulating a geodesic polygon with $n$ sides into $n-2$ triangles givesthe area of a hyperbolic geodesic polygon and the polygonal Gauss-Bonnet theorem.

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Taking the complex derivative along the real direction givesTaking it along the imaginary direction and equating the two expressions gives the Cauchy-Riemann equations

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Differentiate the affine relation with respect to $x$ and $y$:Using the Cauchy-Riemann equations, the second equation becomesThusThe determinant is $-(a^2+b^2)\ne 0$, so $u_x=v_x=0$ and therefore $f^{\prime }=0$. A holomorphic function with zero derivative on a connected open set is constant. Hence $f$ is constant, as in analytic function with image in an affine real line.

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A function $f:D\to \mathbb{R}$ on a convex set $D$ is convex whenfor all $x,y\in D$ and $0\le t\le 1$.

If $f$ is once differentiable, this is equivalent to monotonicity of its gradient:Equivalently, every tangent hyperplane supports the graph:If $f$ is twice differentiable, convexity is equivalent to the hessian matrix being positive semidefinite throughout $D$.

Forthe Hessian isA real symmetric two-by-two matrix is positive semidefinite exactly when its two diagonal entries and its determinant are nonnegative. ThusThe largest convexity domain is thereforeFor $A\ne 0$, its boundary is the hyperbola $y=A^2/(36x)$ in the first quadrant and the domain lies above it. For $A=0$, it is the closed first quadrant. This is the convexity domain of x cubed plus y cubed plus Axy.

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Away from $x=\xi$, the Green function solves $G^{\prime \prime }-G=0$. The solution satisfying the left boundary condition is proportional to $\sinh x$, while the solution decaying at infinity is proportional to $e^{-x}$. Continuity at $\xi$ and the unit derivative jumpgiveor $G=-\sinh (\min \{x,\xi \})e^{-\max \{x,\xi \}}$, the dirichlet half-line Green function for d2 minus 1.

The required solution isEvaluation of the two elementary integrals, split at $\xi =x$, givesIndeed, direct differentiation gives $y^{\prime \prime }-y=e^{-2x}$, and both boundary conditions are immediate.

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The spatial wavefunction $\chi (x)$ is the position amplitude of a stationary quantum state, so $|\chi (x){|}^2$ is its probability density after normalization. The real function $U(x)$ is the potential energy, $E$ is the energy eigenvalue, $m$ is the particle mass, and $\hslash$ is the reduced Planck constant.

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As $|x|\to \infty$, the equation approachesIf $E>U_0$, the asymptotic solutions are oscillatory and do not decay, so there are no bound states. If $0<E<U_0$, the asymptotic solutions are exponential; one decaying branch can be selected at each end. Bound states are therefore possible, although only energies satisfying the global matching conditions actually occur.

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Suppose ${\chi }_1$ and ${\chi }_2$ are two bound-state solutions with the same energy. Their Wronskiansatisfies $W^{\prime }=0$ because the Schrodinger equation contains no first-derivative term. Hence $W$ is constant. Both wavefunctions and their derivatives decay at infinity, so $W\to 0$ there and therefore $W\equiv 0$. The two solutions are linearly dependent. Thus every one-dimensional bound-state energy is nondegenerate.

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A velocity potential satisfying uniform flow at infinity and no radial flow through $r=a$ isThereforeOn the cylinder, $u_r=0$ and $u_{\theta }=-2U\sin \theta$. The steady Bernoulli equation, compared with the uniform flow at infinity, givesHencethe pressure in potential flow around a circular cylinder.

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Let $\pi =({\pi }_A,{\pi }_B,{\pi }_C)$. Solving the stationary distribution equations $\pi P=\pi$ together with ${\pi }_A+{\pi }_B+{\pi }_C=1$ gives

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Yes. The three nontrivial detailed balance identities areandThus the chain is a reversible Markov chain.

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By stationarity,The two-step return probability isTherefore the requested probability is

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A linear operator $\alpha$ on a real inner-product space is self-adjoint whenfor all $v,w$.

To prove the finite-dimensional spectral theorem, first note that the continuous quadratic function $⟨\alpha v,v⟩$ has a maximum on the unit sphere. The Lagrange multiplier equation at a maximizing vector $v$ gives $\alpha v=\lambda v$, so $\alpha$ has a real unit eigenvector. Its orthogonal complement is invariant becauseInduction on the dimension supplies an orthonormal eigenbasis of that complement and hence of $V$.

For $f\ne 0$,while symmetry and bilinearity are immediate, so the displayed formula defines an inner product on $P_n$. Sinceintegration by parts givesThe boundary term vanishes for polynomials, proving that $\alpha$ is self-adjoint.

On the monomial $x^k$,Thus the matrix of $\alpha$ in the monomial basis is triangular with diagonalThese are therefore its eigenvalues. For $n=2$, corresponding eigenvectors arefor eigenvalues $0,-1,-2$, respectively. They are the first Laguerre polynomials up to normalization, as described by the Laguerre differential operator on polynomials.

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A polynomial over a unique factorization domain is primitive when the greatest common divisor of its coefficients is a unit.

Assume that the primitive polynomial $f$ factors in $F[X]$ as $f=gh$, with both factors of positive degree. Clearing denominators and removing contents writeswhere $a,b\in F^{\times }$ and $g_0,h_0\in R[X]$ are primitive. ThenBy Gauss lemma for polynomials, $g_0{h}_0$ is primitive. Since $f$ is also primitive, comparison of contents forces $ab$ to be a unit of $R$. Absorbing that unit into one factor gives a nontrivial factorization of $f$ in $R[X]$, contrary to its assumed irreducibility. Hence $f$ is irreducible in $F[X]$.

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Suppose $f$ is prime in $F[X]$ and divides $gh$ in $R[X]$. It divides one factor in $F[X]$, say $g=fq$ for $q\in F[X]$. Write $q=a{q}_0$ with $q_0\in R[X]$ primitive and $a\in F$. The product $f{q}_0$ is primitive. Sincehas all coefficients in $R$, a denominator of $a$ would divide every coefficient of the primitive polynomial $f{q}_0$; it must therefore be a unit. Thus $a\in R$, so $q\in R[X]$ and $f$ divides $g$ in $R[X]$. Hence $f$ is prime in $R[X]$.

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To deduce the final assertion, use the characterization that an integral domain is a unique factorization domain when it is atomic and every irreducible element is prime. Factoring the contents in $R$ and then the primitive parts by degree shows that $R[X]$ is atomic.

An irreducible constant of $R$ is prime in $R[X]$. Every irreducible polynomial of positive degree is, up to a constant unit, primitive. Part (i) makes it irreducible in $F[X]$; because $F[X]$ is a principal ideal domain, it is prime there, and part (ii) makes it prime in $R[X]$. Thus every irreducible of $R[X]$ is prime, andThis is the polynomial ring over a unique factorization domain theorem.

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Choose a coprime presentationFor a polynomial $r$, write $r^{\ast }(X,Y)=r(Y,X)$. Symmetry of $f$ givesSince $\mathbb{C}[X,Y]$ is a unique factorization domain and $\gcd (p,q)=1$, one has $p\mid p^{\ast }$ and $q\mid q^{\ast }$. Exchanging the variables preserves total degree, sofor nonzero constants $\lambda ,\mu$. The displayed identity gives $\lambda =\mu$, and applying the exchange twice gives ${\lambda }^2=1$.

If $\lambda =-1$, then $p(X,X)=q(X,X)=0$, so $X-Y$ divides both polynomials, contradicting coprimality. Therefore $\lambda =1$, and both $p$ and $q$ are symmetric. Taking $g=p$ and $h=q$ proves the claim, which is the symmetric rational function in two variables result.

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For $a\in U$, the function $f:U\to {\mathbb{R}}^n$ is differentiable at $a$ if there is a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$ such thatThe linear map is unique and is the Frechet derivative $D{f}_a=A$. The function is continuously differentiable at $a$ if it is differentiable on a neighbourhood of $a$ and the map $x\mapsto D{f}_x$, with values in the space of linear maps equipped with the operator norm, is continuous at $a$.

If $T:{\mathbb{R}}^m\to {\mathbb{R}}^n$ is linear, thenConsequently $D{T}_a=T$ at every $a$. The derivative is a constant function of $a$, hence is continuous, so every linear map is continuously differentiable everywhere.

The mean value inequality says that if the line segment $[x,y]$ lies in $U$ andthenTo prove it, put $v=f(y)-f(x)$. The claim is immediate if $v=0$. Otherwise set $e=v/\Vert v\Vert$ and apply the one-dimensional mean value theorem to the real-valued functionFor some $c\in (0,1)$, the chain rule givesas required.

Now suppose that $U$ is open and connected and that $D{f}_a=0$ for every $a\in U$. Every point has an open ball contained in $U$. The mean value inequality with $M=0$ shows that $f$ is constant on each such ball, so $f$ is locally constant. Fix $a_0\in U$. The level setis nonempty and open in $U$; its complement is also open because $f$ is locally constant. Since $U$ is connected, $E=U$. This proves the zero derivative on a connected open set result: $f$ is constant.

The inverse function theorem states that if $f:U\to {\mathbb{R}}^m$ is continuously differentiable, $a\in U$, and $D{f}_a$ is an invertible linear map, then there are open neighbourhoods $V$ of $a$ and $W$ of $f(a)$ such that $f{|}_V:V\to W$ is a bijection whose inverse is continuously differentiable.

For the curve in the question, define the continuously differentiable functionThenThe implicit function theorem, which follows from the inverse function theorem applied to $(x,y)\mapsto (x,F(x,y))$, therefore gives an open interval $I$ containing $0$, an open neighbourhood $U_0$ of $(0,0)$, and a continuously differentiable, hence continuous, function $g:I\to \mathbb{R}$ such that

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Write $q^1=u$, $q^2=v$, ${\sigma }_i={\partial }_i\sigma$, and let $g_{ij}={\sigma }_i\cdot {\sigma }_j$ be the first fundamental form. Sincedifferentiating ${\sigma }_u\cdot \dot{\gamma }$ and ${\sigma }_v\cdot \dot{\gamma }$, or equivalently taking the two inner products of the second formula with ${\sigma }_k$, givesThe matrix $(g_{k\ell })$ is invertible because $\sigma$ is a regular embedded surface parametrization. Consequently the two geodesic equations hold exactly when $\ddot{\gamma }$ is orthogonal to both tangent vectors ${\sigma }_u$ and ${\sigma }_v$, which says precisely that $\ddot{\gamma }$ is a normal vector. This is the ambient acceleration criterion for a surface geodesic.

Because $\dot{\gamma }$ is tangent and $\ddot{\gamma }$ is normal,Thus an affinely parametrized geodesic has constant speed, as recorded by constant speed of an affinely parametrized geodesic.

For a surface of revolution, use profile arc length $s$ and azimuth $\varphi$. Its Riemannian metric isThe azimuth is an ignorable coordinate, so the corresponding geodesic equation has the first integralIf $v=|\dot{\gamma }|$ and $\theta$ is the oriented angle with the parallel, then the component of velocity along the parallel isSince $v$ is constant,This is the Clairaut first integral for a surface of revolution.

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Take the parallel at height one,It has constant speed $\sqrt{2}$. Its distance from the axis is the constant $\rho =\sqrt{2}$, and its tangent is parallel to the parallel, so $\theta =0$ and it satisfies the Clairaut first integral for a surface of revolution.

However,whereas a normal vector to $x^2+y^2-z^2=1$ at $\gamma (t)$ isThese vectors are not parallel. By the ambient acceleration criterion for a surface geodesic, $\gamma$ is not a geodesic. Thus constant speed together with Clairaut's relation is not sufficient for the geodesic equations.

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The planemeets the one-sheet hyperboloid in the two disjoint curvesThey are the two opposite meridians of the surface of revolution. Parametrize the generating profile by arc length. Its acceleration lies in the meridian plane and is orthogonal to the profile tangent; it is also orthogonal to the azimuthal tangent. It is therefore normal to the surface, so both meridians are geodesics. This is the disjoint case in plane-section geodesics of the unit one-sheet hyperboloid.

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The tangent planecuts the hyperboloid inso its intersection is the pair of straight linesTheir accelerations vanish, hence are normal to the surface, so both lines are geodesics. Their tangent vectors at their intersection $(1,0,0)$ are $(0,1,1)$ and $(0,-1,1)$, whose inner product is zero. They therefore intersect at a right angle, giving the second case of plane-section geodesics of the unit one-sheet hyperboloid.

There are also geodesics entirely contained in $z>0$. In the coordinatesthe metric isFor a unit-speed geodesic, the Clairaut first integral for a surface of revolution is $c=(1+z^2)\dot{\varphi }$, and the constant-speed equation becomesChoose $c>1$ and initial height $z_0=\sqrt{c^2-1}$. The geodesic initially tangent to the parallel has $\dot{z}=0$, and the displayed identity prevents it from entering $0<z<z_0$. Since the parallel at $z_0>0$ is not itself a geodesic, the curve turns there and otherwise has $z>z_0$. Hence it remains entirely in $z>0$; this is a geodesic trapped in one half of the unit one-sheet hyperboloid.

Finally, the waistis a geodesic and is preserved setwise by every isometry of the hyperboloid. Indeed, its Gaussian curvature isThe value $K=-1$ occurs exactly at $z=0$. Since Gaussian curvature is intrinsic and therefore preserved by isometries, every isometry preserves the waist. Thus the answer to the final question is also yes, with the isometry-invariant waist geodesic of the unit one-sheet hyperboloid as an example.

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Rouché's theorem states that if $g$ and $h$ are holomorphic functions on a neighbourhood of the closure of a bounded domain andon its positively oriented boundary, then $g$ and $g+h$ have the same number of zeros in the domain, counted with multiplicity.

The Open mapping theorem states that a nonconstant holomorphic function on a domain maps every open subset to an open subset. To prove it, fix $z_0$ and put $w_0=f(z_0)$. By the identity theorem, the zeros of $f-w_0$ are isolated. Choose $r>0$ so that the closed disc $\overline{D(z_0,r)}$ lies in the domain and $z_0$ is the only zero of $f-w_0$ in that disc. ThenWhenever $|w-w_0|<\delta$, the constant $w_0-w$ has modulus smaller than $f-w_0$ on the circle. The Rouche theorem therefore says thathas the same positive number of zeros in the disc as $f-w_0$. Thus every $w\in D(w_0,\delta )$ lies in the image of $f$, proving that the image is open.

If $|f|$ had a local maximum at $z_0$, take a small open disc $D$ on which $|f(z)|\le |f(z_0)|$. The restriction of $f$ to $D$ cannot be constant, since the identity theorem would then make $f$ constant on the whole domain. The complex open mapping theorem says that $f(D)$ is an open neighbourhood of $f(z_0)$, and such a neighbourhood contains points of modulus greater than $|f(z_0)|$, a contradiction. This proves the maximum modulus principle from the complex open mapping theorem.

Now let $\Omega$ be bounded. Its closure is compact, so the continuous function $|f|$ attains a maximum there. If $f$ is nonconstant, the maximum modulus principle excludes an interior maximum; if $f$ is constant, the claimed bound is immediate. Hencewhich is the maximum modulus principle on a bounded domain.

Finally let $\Omega =\{z:\mathrm{Re}z>1\}$ and suppose $|f|\le K$ throughout $\overline{\Omega }$. Fix $z_0\in \Omega$. The principal complex logarithm is holomorphic on the right half-plane, so for each positive integer $n$ the functionis holomorphic on $\Omega$ and continuous on its closure. Given $\epsilon >0$, choose $R>|z_0|$ so large thatApply the bounded-domain result to $g_n$ on $\Omega \cap D(0,R)$. On the vertical part of the boundary, $|z|\ge 1$ and hence $|g_n(z)|\le M$. On the circular part,It follows thatLetting $n\to \infty$ and then $\epsilon \to 0$ gives $|f(z_0)|\le M$. Since $z_0$ was arbitrary, this proves the bounded half-plane maximum principle.

The boundedness assumption is necessary. The functionis holomorphic on $\Omega$ and continuous on its closure, and $|f(1+iy)|=e$ on the boundary, but $|f(x)|=e^x$ is unbounded for real $x>1$. Thus the boundary estimate does not control an unbounded holomorphic function on this unbounded domain.

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By the Fourier inversion theorem and Fubini's theorem,Thus multiplication of Fourier transforms corresponds to the convolution $h=f\ast g$, which is the convolution theorem in the convention used here.

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The Fourier transform of a derivative givesHencefor $k\ne 0$. The assumed existence of the transforms excludes an arbitrary nonzero constant in $p$.

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Using Euler's formula and the Inverse Fourier transforms of phase factors,andThese are distributional identities, with $\delta$ the Dirac delta function.

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Take the Fourier transform in $x$. The wave equation becomes the ordinary differential equationwith initial data $\tilde{u}(k,0)=\tilde{f}(k)$ and ${\partial }_t\tilde{u}(k,0)=\tilde{g}(k)$. Therefore

Part (c) and the Translation property of the Fourier transform invert the first term toFor the second term choose an antiderivative $p$ of $g$. By part (b), $\tilde{g}=ik\tilde{p}$, and henceIts inverse transform isCombining the terms gives the D'Alembert formula

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Here the D'Alembert formula with initial velocity becomesWhen $t>1$ andthe integration interval contains the whole support $[-1,1]$. Since $g$ is odd,Thus the requested region containing the line $x=0$ isThis is the central zero region for odd compactly supported initial velocity. The initial disturbance splits into left-moving and right-moving waves; after they pass, their opposite signed impulses cancel and leave an expanding undisturbed region around the origin.

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The magnetic field is $B=\nabla \times A$. The magnetostatic Ampère-Maxwell equation is $\nabla \times B={\mu }_{0}J$, so the curl of the curl identity givesA gauge transformation $A\mapsto A+\nabla \chi$ does not change $B$. We may therefore choose the Coulomb gauge $\nabla \cdot A=0$ by solving the Poisson equation ${\nabla }^2\chi =-\nabla \cdot A$. It follows component by component in a Cartesian coordinate system thatThe free-space Green function of the Laplacian then givesFor a localized steady current density, conservation of electric charge gives ${\nabla }^{\prime }\cdot J=0$; integration by parts confirms directly that this integral has zero divergence and hence obeys the Coulomb gauge.