Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3

For every prime number $p$ with $n<p\le 2n$, the numerator $(2n)!$ contains one factor $p$, whereas neither copy of $n!$ contains one. Consequently the product of these prime numbers divides the binomial coefficient $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$. Since this binomial coefficient is one positive summand of ${\sum }_{j=0}^{2n}\left(\genfrac{}{}{0ex}{}{2n}{j}\right)=2^{2n}$, and other summands are positive,For $x\ge 2$, choose $k$ with $2^{k-1}<x\le 2^k$. The Chebyshev function is increasing, so telescoping givesFor $x=1$ the inequality follows from $\theta (1)=0$. Thus $\boxed{\theta (x)<4(\log 2)x}$ throughout the stated range.

For $n\ge 3$, extend the restriction of the continuous function $f$ to $[1/n,1-1/n]$ to a continuous function $f_n$ on $[0,1]$ by making it constant on each of the two remaining intervals. The Weierstrass approximation theorem supplies a polynomial $P_n$ with $\Vert P_n-f_n{\Vert }_{\infty ,[0,1]}<1/n$. Every compact set $K\subset (0,1)$ is eventually contained in $[1/n,1-1/n]$, where $f_n=f$. Hence ${\sup }_K|P_n-f|<1/n$ eventually. This proves uniform convergence on every compact subset, without assuming any endpoint limits of $f$. The first two polynomials can be chosen arbitrarily.

Write $M={\sup }_{(0,1)}|f|<\infty$ and use the same constant extensions $f_n$ as in part (a). They satisfy $\Vert f_n{\Vert }_{\infty }\le M$. Choose polynomials $Q_n$ with $\Vert Q_n-f_n{\Vert }_{\infty }<1/n$ for $n\ge 3$, and set the first two polynomials equal to zero. Then $\boxed{\underset{n}{\sup }\underset{0<x<1}{\sup }|Q_n(x)|\le M+1}$, while the argument of part (a) proves uniform convergence to $f$ on every compact subset of $(0,1)$. The bound concerns the entire interval, rather than just each separate compact subset.

Consider the Möbius transformationsTheir group lies in ${\mathrm{PSL}}_2(\mathbb{Z})$, which is a discrete subgroup of the Möbius group: integer entries cannot approach the identity matrix except by eventually equalling it, and passing to the quotient by $\{I,-I\}$ preserves this property.

For freeness let $X=\{x\in \mathbb{R}:|x|>1\}$ and $Y=\{x\in \mathbb{R}:|x|<1\}$. For every nonzero integer $n$, $A^n(Y)=(2n-1,2n+1)\subset X$ andA reduced word involving both generators can be conjugated to a cyclically reduced word beginning with $A^n$ and ending with a nonzero power of $B$, after interchanging the generators if necessary. Repeated use of these inclusions shows that this word sends $X$ into the proper subset $A^n(Y)$ of $X$, so it cannot be the identity. Nonzero pure powers of either generator are also nonidentity. This is the ping-pong lemma in this particular action, and proves that $\boxed{⟨A,B⟩\cong F_2}$ is a free group of rank two as well as a discrete subgroup of ${\mathrm{PSL}}_2(\mathbb{C})$.

Use the binary Hamming code $C=\ker H\subset {\mathbb{F}}_2^{15}$, where the $j$th column $h_j$ of the $4\times 15$ parity-check matrix $H$ is the four-digit binary expansion of $j$, for $1\le j\le 15$. Thus the columns are precisely the nonzero vectors of ${\mathbb{F}}_2^4$. They include the four coordinate vectors, so $\mathrm{rank}H=4$ and $\dim C=15-4=11$. Its code rate is therefore $\boxed{11/15}$.

For a received word $y$, compute its syndrome $s=Hy$. If $s=0$, report no error. Otherwise identify the unique column $h_j=s$ and flip bit $j$. Indeed, if $y=c+e_j$ with $c\in C$, then $Hy=Hc+H{e}_j=h_j$, so this procedure recovers $c$ exactly. No column is zero and no two columns coincide; hence no nonzero codeword has Hamming weight one or two. Some three columns sum to zero, for example those indexed by $1,2,3$, so the minimum distance is exactly three. Thus every single-bit error is corrected. The assertion requires at most one error; arbitrary multiple errors need not be detected by this procedure.

Partition the design matrix accordingly as $X=(X_1\cdots X_k)$. The blocks ${\beta }_i,{\beta }_j$ are orthogonal when $X_i^{\mathsf{T}}{X}_j=0$, meaning that their column spaces are orthogonal vectors. Mutual orthogonality means that this holds for every pair of distinct blocks, so $X^{\mathsf{T}}X$ is block diagonal.

For the three scalar coefficients, the corresponding columns are $1,x_1,x_2$. The necessary and sufficient conditions areUnder the assumed full rank, neither explanatory column is zero. The maximum-likelihood estimator is $\hat{\beta }=(X^{\mathsf{T}}X{)}^{-1}{X}^{\mathsf{T}}Y$, and the normal distribution of $Y$ impliesThus the three coefficient estimators are independent normal random variables, not merely uncorrelated. This is a sampling-distribution statement with the true ${\sigma }^2$ fixed.

Let $p_j^n$ be the probability of occupying site $ja$ after $n$ steps, with $p_j^0=1_{\{j=0\}}$ and $\alpha +\beta =1$. The master equation isWrite $p_j^n\approx ap(ja,n\tau )$ so that the limiting $p$ is a probability density. A Taylor expansion of this master equation givesThe weakly biased diffusion limit of a weakly biased random walk keeps $a^2/\tau$ and $(\alpha -\beta )a/\tau$ finite as $a,\tau \to 0$, so $\alpha -\beta =O(a)$. The limiting advection-diffusion equation isIts initial condition is the Dirac delta function at the origin. At finite spacing, the exact variance gained per step is $a^2[1-(\alpha -\beta {)}^2]=4a^2\alpha \beta$. Thus a finite-step central limit theorem gives the variance-matched coefficient $2a^2\alpha \beta /\tau$; its difference from the displayed $D$ vanishes in the stated weak-bias diffusion limit of a weakly biased random walk. Keeping a fixed nonzero bias while taking diffusive space-time scaling would instead make $V$ diverge.

The fixed points satisfy $x[rx(1-x)-1]=0$. There is always $x=0$, and the other branches areThey coincide at $(r,x)=(4,1/2)$. For $0<x<1$, $F(x)>0$ and its maximum occurs at $x=2/3$, with value $4r/27$. Consequently $\boxed{F((0,1))\subset (0,1)⟺0<r<27/4}$; equality at the upper endpoint fails because the maximum is attained inside the interval.

The fixed-point multiplier at zero is zero, so zero is attracting for every $r>0$. On either nonzero branch,The lower branch has multiplier greater than one for $r>4$ and is repelling. The upper branch is attracting for $4<r<16/3$, reaches multiplier $-1$ at $x=3/4$, and is repelling for $r>16/3$. At $r=4$ the multiplier is $+1$, and the two branches are created in a saddle-node bifurcation. The fixed point there is attracting from one side and repelling from the other: $F(x)-x=-4x(x-1/2{)}^2$.

The second change is a supercritical period-doubling bifurcation at $\boxed{(r,x)=(16/3,3/4)}$. To check its type, expand at that fixed point. The local map has quadratic coefficient $a=-20/3$ and cubic coefficient $b=-16/3$ when the multiplier is $-1$. Its second iterate has cubic coefficient $-2(a^2+b)=-704/9<0$. The multiplier decreases through $-1$ as $r$ increases, so a small attracting two-cycle is born on the $r>16/3$ side. These exhaust the fixed-point bifurcations, since their multipliers have no other crossings of $\pm 1$.

First suppose the integral of $g(\xi )/(\xi -z)$ converges, for example if $g\in L^1(\mathbb{R})$. Substitute the complexified coordinates $x=(z+\bar{a})/2$, $y=(z-\bar{a})/(2i)$ into the Poisson integral. The factorization of its denominator givesThe given Schwarz integral formula therefore writes $f(z)$ as $(i\pi {)}^{-1}\int g(\xi )/(\xi -z)d\xi$ plus a constant. For $z=x+iy$, the real part of this integral is exactly the Poisson integral, sinceThe remaining constant has zero real part, giving $\boxed{f(z)=(i\pi {)}^{-1}{\int }_{\mathbb{R}}\frac{g(\xi )}{\xi -z}d\xi +ic}$ with $c\in \mathbb{R}$. The complex substitution is initially understood where the real-analytic harmonic function has its complexification; the holomorphic integral then continues the identity throughout the upper half-plane.

Decay of $u$ alone does not guarantee convergence of the printed unregularized integral. For example, choose a continuous $g$ which vanishes on $(-\infty ,e]$ and equals $1/\log \xi$ for $\xi \ge e^2$, with continuous interpolation. It is bounded and tends to zero at both ends. Its Poisson integral tends to zero as $|z|\to \infty$ in the closed upper half-plane: split the boundary function into a small uniform tail and a compactly supported part. Nevertheless ${\int }^{\infty }g(\xi )/(\xi -z)d\xi$ diverges like $\log \log \xi$, even as a symmetric principal value. With only the stated decay assumptions, the always convergent version isThe subtracted kernel is real on the integration line, so it changes only the imaginary constant and leaves the harmonic function $u$ unchanged. The kernel difference is $O({\xi }^{-2})$, which ensures convergence for bounded $g$.

Put the cylinder's centre at the origin, with $-h/2\le z\le h/2$, $0\le r\le a$, and uniform density $\rho =M/(\pi a^{2}h)$. Symmetry makes the inertia tensor diagonal and gives $I_1=I_2$. Direct integration yieldsFor a transverse principal moment of inertia, the two contributions areTherefore $\boxed{I_1=I_2=M(a^2/4+h^2/12),I_3=M{a}^2/2}$ are the three principal moments of inertia.

Set $I=I_1=I_2$. The third of the Euler equations for a torque-free rigid body gives $I_3{\dot{\omega }}_3=0$, so ${\omega }_3$ is constant. Define $\Omega =(I_3-I){\omega }_3/I$. The first two Euler equations for a torque-free rigid body becomeso ${\omega }_1+i{\omega }_2=C{e}^{i\Omega t}$. Since the body-frame angular momentum has components $(I{\omega }_1,I{\omega }_2,I_3{\omega }_3)$, its transverse components rotate at $\Omega$ while its axial component is constant. Substitution of the principal moments of inertia givesThis is precession relative to the body axes; the angular momentum is constant in an inertial frame. If $h^2=3a^2$, the rate is zero even when ${\omega }_3\ne 0$. If the transverse components vanish, the precession cone degenerates to its axis.

Assemble the star by adding spherical shells. A shell of mass $dm$ at radius $r$ has gravitational potential energy $-Gm(r)dm/r$ relative to infinity, givingMultiplication of the hydrostatic equilibrium equation by $4\pi r^3$ and integration by parts givesFor an isolated star with negligible surface pressure, $P(R)=0$. Using $E_{\mathrm{kin}}=(3/2){\int }_{V}PdV$, the virial theorem becomes $\boxed{2E_{\mathrm{kin}}+E_{\mathrm{grav}}=0}$. If a nonzero external pressure is present, the surface term must be retained: $2E_{\mathrm{kin}}+E_{\mathrm{grav}}=3P(R)V$. Thus the printed zero-surface-term identity implicitly assumes an isolated star.

Use the uniform-density configuration as a trial configuration for the white dwarf. Write $P=C_F{h}^2{n}^{5/3}/m_e$, with $C_F>0$ the numerical coefficient suppressed by the approximate equation of state. Its particle density is $n=3M/(4\pi m_p{R}^3)$, and the nonrelativistic degeneracy pressure has kinetic-energy density $3P/2$. HenceFor uniform density $m(r)=M(r/R{)}^3$, so integration of the shell energy givesBoth constants are positive and independent of $R$. Differentiating $E=\alpha /R^2-\beta /R$ gives its unique stationary radius $R=2\alpha /\beta$. The second derivative there is $2\alpha /R^4>0$, and this is the global minimum since $E\to +\infty$ at zero and $E\to 0^{-}$ at infinity. ThusThe uniform-density approximation is used for this energy estimate; it does not require a spatially constant pressure to satisfy the exact hydrostatic equilibrium equation.

In the multiplicative group ${\mathbb{F}}_p^{\times }$, the squaring map is a group homomorphism with kernel $\{1,-1\}$, since $p$ is odd. Every nonzero square therefore has exactly two square roots. There are $(p-1)/2$ quadratic residues and the same number of nonresidues, and consequently ${\sum }_{a=1}^{p-1}\left(\frac{a}{p}\right)=0$.

For $n{m}_n\equiv 1(\mathrm{mod}p)$, direct multiplication gives $n^2(1+m_n)\equiv n^2+n=n(n+1)$. Multiplicativity of the Legendre symbol, including its value zero on multiples of $p$, impliesInversion permutes ${\mathbb{F}}_p^{\times }$, so $1+m_n$ runs through all residue classes except $1$. Thus

The polynomial form of the Runge approximation theorem says: if $K\subset \mathbb{C}$ is compact, $\mathbb{C}\setminus K$ is connected, and $f$ is holomorphic on a neighborhood of $K$, then for every $\epsilon >0$ there is a polynomial $P$ with ${\sup }_K|f-P|<\epsilon$. Choosing errors tending to zero gives a uniformly convergent sequence of polynomials. Equivalently, on an open domain with connected complement in the Riemann sphere, every holomorphic function is the compact-uniform limit of polynomials. The connected-complement hypothesis excludes enclosed holes containing poles of the function being approximated.

A compact set $K\subset \mathbb{C}$ has exactly one unbounded complementary component, since its complement contains the outside of a sufficiently large disk. The connected, unbounded open set $\Omega$, being disjoint from $K$, lies in that component. In particular $\zeta$ belongs to the unbounded component.

Fill all bounded components of $\mathbb{C}\setminus K$ to obtain the polynomial hull $\hat{K}$. It is a compact set with connected complement, and $\zeta \notin \hat{K}$. Therefore $1/(z-\zeta )$ is holomorphic on a neighborhood of $\hat{K}$. The Runge approximation theorem gives polynomials $P_n$ converging uniformly to this function on $\hat{K}$, hence on $K$. This proves the requested uniform polynomial approximation on $K$, even when $K$ itself has holes. If $K$ is empty, the assertion is vacuous.

Take $\Omega =\{|z|<1\}$, $K=\{|z|=1\}$, and $\zeta =0$. These satisfy the stated connectedness and disjointness conditions. If polynomials $P_n$ converged uniformly to $1/z$ on $K$, convergence of the contour integrals would implya contradiction to the Cauchy integral theorem. Thus boundedness of this $\Omega$ permits the obstruction excluded in part (b).

The species $u$ is the activator, since increasing $u$ increases its own production $u^2/v$; $v$ is the inhibitor, since increasing $v$ reduces that production. The unique positive uniform steady state solves $v=u^2$ and $u^2/v=bu$, giving $\boxed{u_{\ast }=1/b,v_{\ast }=1/b^2}$.

The reaction Jacobian matrix at this state isIts eigenvalues solve ${\lambda }^2+(1-b)\lambda +b=0$. Since $b>0$, both have negative real part precisely for $0<b<1$. This proves stability of the spatially uniform reaction kinetics.

A perturbation proportional to $e^{\lambda t+ikx}$ replaces $J$ by $J-k^2\mathrm{diag}(1,d)$. Put $q=k^2$. Its trace is $b-1-(1+d)q<0$ in the kinetically stable range, and its determinant isA growing spatial mode exists exactly when this determinant is negative for some $q>0$. Its minimum occurs at $q=(bd-1)/(2d)$, which is positive only if $bd>1$, and the minimum is negative precisely if $(bd-1{)}^2>4bd$. Together these conditions reduce to the Turing threshold of the quadratic activator-inhibitor model,The unstable band isAt the Turing instability threshold the band shrinks to its double root, yielding $\boxed{k_c^2=(1+\sqrt{2})/d}$. On a bounded spatial interval, an allowed wavenumber must additionally lie in the unstable band; the displayed parameter condition assumes the continuous spatial spectrum.

Put $s=\sqrt{b}$. The two fixed points on the invariant line $x=0$ are $P_{+}=(0,s)$ and $P_{-}=(0,-s)$. There are also $Q_{\pm }=(\pm \sqrt{b-a^2/4},-a/2)$ when $a^2<4b$. At equality they coalesce with $P_{-}$.

The Jacobian matrix isAt $P_{+}$ its eigenvalues are $-a-2s$ and $2s$, so it is always a saddle equilibrium. At $P_{-}$ they are $2s-a$ and $-2s$, so it is a saddle equilibrium for $a<2s$ and a stable node for $a>2s$, with a zero eigenvalue at $a=2s$.

At $Q_{\pm }$ the characteristic equation is ${\lambda }^2+a\lambda +4b-a^2=0$, givingFor $0<a^2<16b/5$ these points are stable spirals; for $16b/5<a^2<4b$ they are stable nodes. At $a^2=16b/5$ the double negative eigenvalue has only one independent eigenvector, so these are improper stable nodes. When $a=0$ they are center equilibria, as established in part (iii). Thus the collision at $\boxed{a^2=4b>0}$ changes both the number and stability of fixed points; it is the bifurcation classified below.

Set $Y=y+s$ and $\mu =2s-a$, and include $\dot{\mu }=0$ for the extended center manifold. The equations near $P_{-}$ becomeThe stable direction has eigenvalue $-2s$ and the center directions are $(x,\mu )$. Choose a symmetry-preserving center manifold $Y=h(x,\mu )$, even in $x$, containing the equilibrium line $x=0,Y=0$. Its invariance equation isWrite $h=A{x}^2+B\mu x^2+\cdots$. At degree two, $0=(1-2sA)x^2$, so $A=1/(2s)$. At degree three, $2A\mu x^2=-2sB\mu x^2$, so $B=-1/(2s^2)$. ConsequentlyWith $x=\sqrt{s}X$ this is $\dot{X}=\mu X-X^3+O(\mu X^3,X^5)$, the supercritical pitchfork bifurcation. For $\mu <0$, $P_{-}$ is attracting; for $\mu >0$ it becomes a saddle equilibrium and two attracting branches appear, consistent with the exact $Q_{\pm }$.

For $0<a^2<4b$, the vertical invariant line flows from $P_{+}$ towards $P_{-}$ between those points. The two off-axis stable equilibria lie below the horizontal axis, one in each invariant half-plane $x>0$ and $x<0$. For $a^2>4b$, these stable equilibria have merged into the stable node $P_{-}$, while $P_{+}$ remains a saddle equilibrium. Filled green points mark attracting equilibria, green rings mark centers, and red crosses mark saddle equilibria. The plotted arrows and nullclines show the corresponding phase portraits; a focus can become a node within the first parameter range without changing the pitchfork classification.

When $a=0$, define the Hamiltonian functionThen $\dot{x}=-H_y$ and $\dot{y}=H_x$, so $\dot{H}=0$. At $(\sqrt{b},0)$ the Hessian of $H$ is $2\sqrt{b}I$, a strict nondegenerate minimum; at $(-\sqrt{b},0)$ it is $-2\sqrt{b}I$, a strict maximum. Small regular level curves around either point are closed curves, and the vector field is nonzero on them. Motion around each curve therefore returns after a finite period, proving the existence of periodic orbits and classifying both equilibria as center equilibria.

The separatrix level is $H=0$, consisting of $x=0$ and $x^2/3+y^2=b$. The two arcs of this ellipse run from the lower saddle equilibrium to the upper one, while the vertical segment runs from the upper saddle to the lower one. They enclose the two families of periodic orbits, as shown in the last panel of the phase portrait. Outside the loops, regular levels are unbounded.

For $a>0$ the divergence of the original vector field is exactly $-a$. If a periodic orbit existed, uniqueness of trajectories would make it a simple closed curve, tangent everywhere to the vector field. The flux across its boundary would be zero, whereas the divergence theorem would give $-a$ times its positive enclosed area. This contradiction is the Bendixson-Dulac criterion. Hence $\boxed{\text{periodic orbits exist if and only if }a=0}$.

Mass conservation in the Zeldovich approximation gives $\rho d^{3}r=\bar{\rho }{a}^3{d}^{3}q$. Expanding the trajectory Jacobian determinant to first order therefore yields $\boxed{\delta =-{\nabla }_q\cdot \Psi }$. At the same order, ${\nabla }_r=a^{-1}{\nabla }_q$ andThe last equality uses ${\nabla }_q\times \Psi =0$ and the vector calculus $\nabla \mathrm{div}={\nabla }^2+\nabla \times \nabla \times$.

Split the gravitational potential into its homogeneous-background and perturbed parts. The background has ${\nabla }_r{\Phi }_0=(4\pi G\bar{\rho }/3)r$. For the perturbation, Poisson equation and irrotationality givewith a harmonic or uniform-acceleration contribution set to zero by the perturbation boundary conditions. Its divergence is $-4\pi G\bar{\rho }{\nabla }_q\cdot \Psi =4\pi G\bar{\rho }\delta$, as required. Insert this and the pressure gradient into $\ddot{r}=\ddot{a}(q+\Psi )+2\dot{a}\dot{\Psi }+a\ddot{\Psi }$. The background acceleration cancels using $\ddot{a}/a=-4\pi G\bar{\rho }/3$, leavingTake minus the comoving divergence, and expand $\delta (q,t)={\sum }_k{\delta }_k(t)e^{ik\cdot q}$. Since ${\nabla }_q^2$ acts on a Fourier mode by $-k^2$, each mode satisfiesHere the expansion coordinate is comoving; confusing it with the physical coordinate would lose the factor $a^{-2}$.

For $P=\beta {\rho }^{4/3}$, the sound speed obeys $c_s^2=(4\beta /3){\bar{\rho }}^{1/3}$. Using $a=(t/t_0{)}^{2/3}$ and $\bar{\rho }=(6\pi G{t}^2{)}^{-1}$ givesThe mode equation is an Euler differential equation. Substituting ${\delta }_k=t^n$ produces $n(n-1)+(4/3)n-2/3+{\bar{v}}_s^2{k}^2=0$. Thus, for distinct roots,A genuinely growing mode occurs only for ${\bar{v}}_s^2{k}^2<2/3$; at $k=0$ the exponents are $2/3$ and $-1$. At ${\bar{v}}_s^2{k}^2=2/3$ the larger root is zero. Between $2/3$ and $25/36$, both roots are negative. At the repeated-root value ${\bar{v}}_s^2{k}^2=25/36$, the independent solutions are $t^{-1/6}$ and $t^{-1/6}\log t$. Beyond it, write $\nu =\sqrt{{\bar{v}}_s^2{k}^2-25/36}$ to obtain $t^{-1/6}[A\cos (\nu \log t)+B\sin (\nu \log t)]$. Thus the phrase “growing and decaying” applies to the long-wavelength regime; pressure-stabilized modes decay instead.

The induced ordering on $x$ is a well-order. By transfinite recursion define $h(t)=\{h(s):s\in x,s<t\}$ for $t\in x$. Induction shows that each $h(t)$ is an ordinal, that these values strictly increase, and that their range is the ordinal $\mu (x)={\bigcup }_{t\in x}(h(t)+1)$. The map $h$ is an order isomorphism from $x$ to this ordinal. Any order isomorphism between ordinals must fix their elements successively, so this ordinal is unique.

An increasing map $j:\kappa \to \lambda$ between ordinals satisfies $j(\xi )\ge \xi$ by induction: its predecessors include $j(\eta )$ for all $\eta <\xi$. Hence $\kappa \le \lambda$. Apply this to the inclusion $x\subset y$, transported through their order isomorphisms, and then to $y\subset \alpha$. It follows that $\boxed{\mu (x)\le \mu (y)\le \alpha }$.

If each $x_n$ is an initial segment of later $x_m$, uniqueness makes their ordinal enumerations compatible: the enumeration of $x_m$ restricts to that of $x_n$. Their union is consequently an order isomorphism from ${\bigcup }_n{x}_n$ to ${\bigcup }_n\mu (x_n)$. This proves the asserted equality. Without the initial segment condition it fails: in $\alpha =\omega +1$, take $x_n=\{0,\dots ,n-1\}\cup \{\omega \}$. Then $\mu (x_n)=n+1$, whose union is $\omega$, but $\mu ({\bigcup }_n{x}_n)=\omega +1$.