Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 3

The dual space is $V^{\ast }={\mathrm{Hom}}_{\mathbb{R}}(V,\mathbb{R})$, the vector space of real-valued linear maps on $V$, with pointwise addition and scalar multiplication. Choose a basis $e_1,\dots ,e_n$ of $V$. Every $v$ has unique coordinates $v={\sum }_i{x}_i{e}_i$, so $e^i(v)=x_i$ defines a linear functional, with $e^i(e_j)={\delta }_{ij}$.

If ${\sum }_i{a}_i{e}^i=0$, evaluation on $e_j$ gives $a_j=0$, proving linear independence. For any $\ell \in V^{\ast }$, linearity gives $\ell (v)={\sum }_i{x}_i\ell (e_i)$, hence $\ell ={\sum }_i\ell (e_i)e^i$. Thus the dual basis spans the dual space as well, proving

Write $\tau =(1+\sqrt{-3})/2$. The parity condition gives $R=\mathbb{Z}[\tau ]$, since $(a+b\sqrt{-3})/2=(a-b)/2+b\tau$. For any such element,If $a,b$ are even the numerator is divisible by four; if they are odd it is congruent to $1+3=0$ modulo four. Therefore $N(z)$ is a nonnegative integer. It is zero only at zero, and the Eisenstein-integer norm is multiplicative because complex conjugation respects multiplication.

If $z$ is a unit, $N(z)N(z^{-1})=1$, so $N(z)=1$. Conversely $N(z)=1$ makes $\overline{z}$ its inverse in $R$. The equation $a^2+3b^2=4$ permits exactly $(a,b)=(\pm 2,0)$ or $(\pm 1,\pm 1)$. HenceIndeed ${\tau }^2=\tau -1$ and ${\tau }^3=-1$, so this unit group is cyclic of order six.

For the prime-ideal assertion, ${\tau }^2-\tau +1=0$ and the polynomial $t^2-t+1$ has distinct roots $3,5$ modulo seven. Consequently evaluation at either root defines a surjective ring map $R\to {\mathbb{F}}_7$. Its kernelsare distinct maximal ideals, hence prime ideals. If $A+B\tau$ belongs to both, then $A+3B\equiv A+5B\equiv 0(\mathrm{mod}7)$. Subtraction gives $2B\equiv 0$, so both $A$ and $B$ are divisible by seven. The converse inclusion is immediate, provingEquivalently, the two distinct factors of the reduced polynomial give the Chinese remainder theorem decomposition $R/7R\cong {\mathbb{F}}_7\times {\mathbb{F}}_7$. This proof does not require the permitted unique-factorization assumption.

The Euler formula for a sphere, applied to the surface of a convex polyhedron, is $V-E+F=2$. In a regular polyhedron let $q\ge 3$ be the common number of edges incident at a vertex. Counting vertex-edge and face-edge incidences, pentagonal faces give $qV=2E$ and $5F=2E$. Hencewhose right-hand side can be positive only if $q<10/3$. Thus $q=3$. Substitution then yields $2=E/15$, provingThese are the counts of the dodecahedron. Here regular polyhedron is understood in the convex, ordinary-face sense used by the stated Euler formula; self-intersecting star constructions do not share these hypotheses.

Nonnegativity and symmetry of the maximum product metric follow from those of the two factor metrics. Its value is zero precisely when both coordinates agree. For three points $u=(x,x^{\prime })$, $v=(y,y^{\prime })$, $w=(z,z^{\prime })$, the two triangle inequalities giveAll metric axioms therefore hold.

When the two factors are the same metric space, consider an off-diagonal point $(x,y)$ and set $\delta =d(x,y)>0$. A product-metric ball of radius $\delta /3$ cannot contain any $(z,z)$: such membership would imply $d(x,z),d(y,z)<\delta /3$ and then $d(x,y)<2\delta /3$, a contradiction. The complement of the diagonal is thus open, soThis proves the closed diagonal of a metric space. Equality of the factors here means equality as metric spaces, as intended; merely sharing an underlying set while allowing unrelated metrics would be a different assertion.

The contour has winding number one around the single pole at zero. By the Cauchy integral formula, or the residue of $1/z$,Rotation by $z\mapsto iz$ carries each oriented edge to the next oriented edge and leaves $dz/z$ unchanged. The four edge integrals are therefore equal, each $\pi i/2$.

On the right edge use $z=1+it$, $-1\le t\le 1$, soThe real integrand $t/(1+t^2)$ is odd and integrates to zero. Dividing by $i$ gives ${\int }_{-1}^1(1+t^2{)}^{-1}dt=\pi /2$ without needing its real antiderivative.

For an endpoint-fixed variation $\eta =\delta x$,The boundary term vanishes. Thus every solution of the harmonic oscillator equation is stationary for this action functional. For $\sin \omega T\ne 0$, the endpoint-matching solution is the linear combinationwhich directly satisfies $x_c(0)=a$, $x_c(T)=b$ and ${\ddot{x}}_c=-{\omega }^2{x}_c$. This proves stationarity, rather than assuming the explicit path is a minimum.

On that path, $(x_c{\dot{x}}_c{)}^{\prime }={\dot{x}}_c^2-{\omega }^2{x}_c^2$, so integration by parts reduces the entire action to an endpoint term. The endpoint derivatives areTherefore the fixed-endpoint harmonic-oscillator action isThe displayed formula assumes nonresonant endpoints. If $\omega \ne 0$ and $\omega T=k\pi$, a solution exists only for $b=(-1{)}^{k}a$; then $x=a\cos \omega t+B\sin \omega t$ is a family and its action is zero. For $\omega =0$ the limiting stationary path is linear, with action $(b-a{)}^2/(2T)$.

Choose positive circulation anticlockwise as viewed from positive $z$. The linked magnetic flux is $\Phi =2\ell XB$, so Faraday law, including the motional contribution, givesThe conducting circuit consists of two arms of length $X$ and two transverse pieces of length $2\ell$. Its resistance is $\mathcal{R}=2R(X+2\ell )$, where $R$ is resistance per unit length. Neglecting self-inductance as in the quasistatic circuit model, the current isOn the moving rod this positive current points along $+y$, so the Lorentz force is $I(2\ell {\mathbf{e}}_y)\times (B{\mathbf{e}}_z)=2\ell IB{\mathbf{e}}_x$. Its mass is $2\ell M$, not the mass of the entire circuit. Newton's equation consequently givesFor constant $B$, the force opposes $\dot{X}$, as expected from Lenz's law. The prescribed time dependence of the external field can also supply energy to the rod.

Take $S^{\prime }$ to move along $+x$ at velocity $V$, with the origins coincident at zero time. The Lorentz transformation isDifferentiating along the particle trajectory gives the relativistic velocity-addition formulaFor a timelike particle, $|u|<c$ and $|V|<c$. Direct algebra givesso $|u^{\prime }|<c$, in particular $u^{\prime }<c$. Strictness requires a massive, subluminal particle; for a photon with $u=\pm c$, the transformation preserves $u^{\prime }=\pm c$ instead.

With the proton treated as fixed and electron mass $m$, the classical energy and angular momentum areFor a circular orbit, $L=mrv$ and Coulomb's law determines its centripetal acceleration:Eliminating $v$ gives $r=4\pi {ϵ}_0{n}^2{\hslash }^2/(m{e}^2)$. Thus the Bohr radius and orbit radii areThe kinetic energy is half the magnitude of the attractive potential energy. HenceThis is the circular-orbit Bohr model; including proton recoil replaces $m$ by the reduced mass.

For velocity $\mathbf{u}=\nabla \varphi$ and a conservative body force $-\nabla \chi$, the unsteady Bernoulli equation isIt follows by integrating the inviscid Euler equations in space; a purely time-dependent potential gauge can absorb $C(t)$.

Use a coordinate along the thin U-tube from the left surface to the right, and let the right surface rise by $\zeta$ while the left falls by $\zeta$. Incompressibility and uniform area give velocity $\dot{\zeta }$ along the whole fluid column. If its total length is $L$, the velocity-potential difference between the surfaces is $L\dot{\zeta }$; the column length stays constant as their displacements cancel. Both surfaces have atmospheric pressure and the same speed. Subtracting their Bernoulli balances, with $\chi =gy$, therefore yieldsIn the long-leg approximation underlying the requested formula, the base bend contributes negligible length and $L=2h$. ConsequentlyIf the horizontal connector or bend has nonnegligible length, it belongs in $L$ and the general frequency is $\sqrt{2g/L}$; the stated height alone would then not determine the inertia.

Let $B$ denote the zero-diagonal matrix with off-diagonal entries appearing in the iteration, and $G=-B/\alpha$ its iteration matrix. Its row sums are three, so $B1=31$. The other eigenvalues are $2\omega +{\omega }^2$ and $2{\omega }^2+\omega$, where $\omega =e^{2\pi i/3}$, namely $-3/2\pm i\sqrt{3}/2$, both of modulus $\sqrt{3}$. Thus the spectral radius is $\rho (G)=3/|\alpha |$.

For $|\alpha |>3$, the infinity-norm estimate $\Vert G{\Vert }_{\infty }=3/|\alpha |<1$ already gives a contraction, so the iterates converge from every initial vector to the unique solution of $(\alpha I+B)x=b$. Conversely, two iterates whose initial difference is a nonzero multiple of $1$ retain the difference $(-3/\alpha {)}^{k}1$. This fails to decay when $|\alpha |\le 3$, so convergence cannot be guaranteed for arbitrary starts. ThereforeAt $\alpha =3$ the offending mode oscillates; at $\alpha =-3$ it is stationary and the coefficient matrix is singular. A specially chosen starting vector can suppress a bad mode, but that is not guaranteed convergence.

Let $(q_R,q_S,q_P)$ be player two's mixed strategy in Rock paper scissors. The respective expected payoffs from choosing rock, scissors and paper are $1-p$, $-p$ and $2p-1$. Thus player two solves the linear programming problemA linear objective on this probability simplex is maximized by a largest-payoff pure response, or by any mixture of tied maximizers. Scissors is never a maximizing response: rock exceeds it by one. Comparing the other two gives $1-p⋛2p-1$ according as $p⋚2/3$. ThereforeThe optimal expected payoff is $1-p$ below the tie and $2p-1$ above it, with value $1/3$ at the tie.

Let $m_{\alpha }(x)$ be the monic minimal polynomial. The Cayley-Hamilton theorem says ${\chi }_{\alpha }(\alpha )=0$. Divide ${\chi }_{\alpha }$ by $m_{\alpha }$; its remainder also annihilates $\alpha$, and a nonzero remainder of smaller degree would contradict minimality. Thus $m_{\alpha }$ divides ${\chi }_{\alpha }$.

Every characteristic root ${\lambda }_i$ is an eigenvalue, with a nonzero eigenvector $v_i$. Since $m_{\alpha }(\alpha )v_i=m_{\alpha }({\lambda }_i)v_i=0$, it must be a root of the minimal polynomial as well. This proves there are no possibilities beyondEvery such choice really occurs. For each $i$, take one Jordan block of size $r_i$ at ${\lambda }_i$, together with $n_i-r_i$ scalar blocks at that eigenvalue, and form their direct sum. The characteristic polynomial has the prescribed multiplicities. The nilpotent part of a block of size $r_i$ has its $r_i$th power zero and its $(r_i-1)$st power nonzero, so this component's minimal exponent is exactly $r_i$. A polynomial annihilates the direct sum precisely when it annihilates every component, establishing the displayed minimal polynomial without any extra restrictions.

Take a size-three Jordan block at one and a scalar block at three:Triangularity gives ${\chi }_A(x)=(x-1{)}^3(x-3)$. On the first block, $(A-I{)}^3$ is zero but $(A-I{)}^2$ is not; on the last component the eigenvalue is three. Thus both factors with exactly those powers are needed, and their product annihilates $A$. Its minimal polynomial is therefore also $(x-1{)}^3(x-3)$.

An example in dimension four isBoth are invertible, so each has rank four. Both have characteristic polynomial $(x-1{)}^4$, and each has minimal polynomial $(x-1{)}^2$, since its nonzero nilpotent part squares to zero. However $\mathrm{rank}(A-I)=2$ and $\mathrm{rank}(B-I)=1$. If $B=PA{P}^{-1}$, then $B-I=P(A-I)P^{-1}$, which would preserve these ranks. The contradiction proves that the matrices are not similar, despite all three specified invariants agreeing.

Put the generators into the rows of the integer matrixIts determinant is $5-2-21=-18$. A full-rank integer lattice generated by its rows has index equal to the absolute determinant, henceFor a direct verification, $u,v,e_3$ form a unimodular basis because their row determinant is $-1$, and $w=-7u+5v+18e_3$. In these integer coordinates $M=\mathbb{Z}u\oplus \mathbb{Z}v\oplus 18\mathbb{Z}{e}_3$, visibly leaving eighteen cosets.

A proper finite-index subgroup of a torsion-free group cannot be a direct summand when the complement is a subgroup of that same free abelian group. Indeed, if $L=M\oplus K$, every $k\in K$ has $18k\in M$ because the quotient has order eighteen. But also $18k\in K$, so $18k=0$ by the trivial intersection. The ambient group ${\mathbb{Z}}^3$ is torsion-free, forcing $k=0$ and hence $K=0$. This would give $M=L$, contrary to index eighteen. Thus $M$ is not a direct summand.

The matrix with rows $u,v,e_3$ has determinantIts inverse has integer entries by the adjugate formula, so these rows form an integer basis of $L$. The unimodular basis test for an integer direct summand now givesThus $N$ is a direct summand, with the displayed complement. This proves integral splitting, not just real or rational linear independence.

In the integer basis $u,v,e_3$, direct substitution gives $w=-7u+5v+18e_3$. ThereforeThe first two basis coordinates disappear in the quotient and the third is taken modulo eighteen. Equivalently the Smith normal form has diagonal entries $1,1,18$.

Work on the unit sphere, so side lengths are central angles. Place the right-angle vertex at the north pole and choose perpendicular meridians through it. The other vertices can then be writtenTheir dot product is the cosine of the central angle between them, namely the hypotenuse length $c$. Hence the spherical Pythagorean relation isFor the small triangle let its hypotenuse be $c_{\lambda }$. The same formula gives $\cos c_{\lambda }=\cos (\lambda a)\cos (\lambda b)$, so $c_{\lambda }\to 0$. Expanding each cosine about zero yieldsIn particular $c_{\lambda }=O(\lambda )$, and thus $c_{\lambda }^2={\lambda }^2(a^2+b^2)+O({\lambda }^4)$. Consequently $(c_{\lambda }/\lambda {)}^2\to a^2+b^2$, the ordinary Pythagorean theorem in the locally flat limit.

The contraction mapping theorem states: if $(X,d)$ is a nonempty complete metric space and $f:X\to X$ satisfies $d(fx,fy)\le qd(x,y)$ for a fixed $0\le q<1$, then it has exactly one fixed point and the iteration from every starting point converges to it.

To prove existence, choose $x_0$ and let $x_{n+1}=f(x_n)$. Induction gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$. The triangle inequality and geometric sum then imply, for $m>n$,The iterates are a Cauchy sequence; completeness supplies their limit $x_{\ast }$. The contraction is continuous, so $f(x_{\ast })=\lim f(x_n)=\lim x_{n+1}=x_{\ast }$. If $y_{\ast }$ is another fixed point, $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$ forces equality of the points. The same estimate gives geometric convergence from every initial point.

For the particular map, the original PDF gives $X=[\sqrt{a/2},\infty )$, which is a closed, complete subset of the real line. For positive $x$, the arithmetic-geometric mean inequality gives $f(x)=(x+a/x)/2\ge \sqrt{a}$, proving $f(X)\subset X$. FurthermoreThe mean value theorem therefore gives $|f(x)-f(y)|\le |x-y|/2$, a genuine contraction. The fixed-point equation is $x^2=a$, and only the positive root belongs to $X$, soThe printed square-root endpoint is essential: the TeX transcription loses it.

Let $K\subset \mathbb{C}\setminus \mathbb{Z}$ be compact and contained in $|z|\le R$. For $|n|>2R$, $|z-n|\ge |n|/2$, hence $|(z-n{)}^{-2}|\le 4/n^2$ uniformly on $K$. The tail is therefore absolutely and uniformly convergent by the Weierstrass M-test. Its finitely many remaining summands are holomorphic on a neighborhood of $K$.

The theorem that a locally uniform limit of holomorphic functions is holomorphic now proves that the sum is analytic on $\mathbb{C}\setminus \mathbb{Z}$. Absolute convergence also permits a shift of the integer index:Thus the function has period one.

Fix $z_0\ne 0,1$ and take a small disk about it that avoids zero and one. There is a holomorphic logarithm $L$ on this disk. The function $L(z)/(2\pi i)$ cannot be an integer there: if it were an integer, exponentiation would give $z=1$. Thus $f(L(z)/(2\pi i))$ is holomorphic on the disk.

Where the prescribed principal logarithm and $L$ are both evaluated, they differ by $2\pi ik$ for an integer $k$. By the period-one identity, their compositions with $f$ coincide. This remains true on either side of the principal branch cut, and at a point on that cut one may instead choose the logarithm with argument between zero and $2\pi$. The local holomorphic expressions therefore agree with the stated pointwise definition and with each other. This is periodic holomorphic descent through the exponential map, provingThe principal logarithm itself is not analytic across its cut; periodicity of the outer function removes that obstruction.

Near zero the $n=0$ summand supplies the pole and the remaining series is holomorphic, so $f(w)=w^{-2}+F(w)$ with $F$ holomorphic near zero. Choose the local logarithm with $L(1)=0$. Writing $h=z-1$ gives $L(z)=h-h^2/2+O(h^3)$ and henceTherefore $z=1$ is a double pole, with leading coefficient ${\lim }_{z\to 1}(z-1{)}^{2}g(z)=-4{\pi }^2$. Its residue, if desired, is $-4{\pi }^2$.

For an integrand $I(y,y^{\prime })$ with no explicit $x$ dependence, the Euler-Lagrange equation is $I_y-d(I_{y^{\prime }})/dx=0$. Differentiate the proposed first integral:Thus $I-y^{\prime }{I}_{y^{\prime }}$ is constant along an extremal, the Beltrami identity.

For the optical path the Fermat principle uses the travel-time integrand $I=e^{-\lambda y}\sqrt{1+y^{\prime 2}}$, in the stated speed units. Its first integral isSet $u=k{e}^{\lambda y}$. Then $u^{\prime 2}={\lambda }^2(1-u^2)$, and integration gives $u=\cos [\lambda (x-x_0)]$ on a positive arch. The symmetric endpoint conditions require $x_0=0$ and $k=\cos \lambda a$. For $\lambda >0$ and $0<\lambda a<\pi /2$, the resulting ray in an exponential-speed medium isIt is above zero inside the endpoints. Substitution also verifies the Euler-Lagrange equation $y^{\prime \prime }=-\lambda (1+y^{\prime 2})$, so the first-integral construction does not accidentally select a spurious constant-height path.

Here $y^{\prime }=-\tan \lambda x$. For $\lambda a$ close to $\pi /2$, the endpoint slopes are steep and the central height $-{\lambda }^{-1}\log \cos \lambda a$ is large. The travel time evaluates directly toIt tends to $2/\lambda$ as the arch height diverges. Positive $\lambda$ is the increasing-speed interpretation needed for a ray arching into $y>0$; for $\lambda <0$ the same algebraic expression bends below that region and is not an admissible interior path. At zero $\lambda$ the limiting straight boundary path has travel time $2a$.

Use the Maxwell curl equationsDot the first with $\mathbf{E}/{\mu }_0$ and the second with $\mathbf{B}/{\mu }_0$, then combine them. The identity $\nabla \cdot (\mathbf{E}\times \mathbf{B})=\mathbf{B}\cdot \nabla \times \mathbf{E}-\mathbf{E}\cdot \nabla \times \mathbf{B}$ givesThis proves the Poynting theorem $W_t+\nabla \cdot \mathbf{S}+\mathbf{J}\cdot \mathbf{E}=0$, with the stated energy density and Poynting vector. The last term is work transferred from the electromagnetic field to matter.

For the specified vacuum wave, substitution into the same curl equations gives $k{E}_0=\omega B_0$ and $k{B}_0={ϵ}_0{\mu }_0\omega E_0$. Thus ${\omega }^2=k^2/{ϵ}_0{\mu }_0$, and for propagation along positive $z$,Putting $\vartheta =kz-\omega t$,The equal electric and magnetic contributions account for the factor in $W$. Energy propagates at speed $c$, whose square is $1/({ϵ}_0{\mu }_0)$; the reciprocal product is not itself the speed.

The scattering interpretation assumes $0<ϵ<U$, so $k=\sqrt{ϵ}$ is real and $K=\sqrt{U-ϵ}>0$. The incoming amplitudes are $a$ from the left and $b$ from the right; the outgoing amplitudes are $c,d$. The conserved probability current iswhich gives $j=\hslash k(|a{|}^2-|c{|}^2)/m$ on the left and $j=\hslash k(|d{|}^2-|b{|}^2)/m$ on the right. Equality proves $|c{|}^2+|d{|}^2=|a{|}^2+|b{|}^2$. Equal exterior wavenumbers make the flux weights equal.

Continuity of the wavefunction and its derivative at the finite potential steps yields the four relations, with $C=\cosh KL$ and $S=\sinh KL$,Set $\lambda =K/(ik)$. Substituting $e=a+c$, $f=(a-c)/\lambda$ and subtracting the last two relations after dividing the fourth by $ik$ giveswhere $D=C-(\lambda +{\lambda }^{-1})S/2$. HenceReflection about the barrier midpoint interchanges the left and right incidence problems and, with the chosen phase origins, exchanges $a\leftrightarrow b$ and $c\leftrightarrow d$. ThereforeFor incidence from the left alone, the transmission coefficient is the transmitted-to-incident probability flux ratio, $T=|d/a{|}^2=|D{|}^{-2}$. Since $\lambda =-iK/k$, direct calculation givesFor fixed energy inside the barrier range and $KL\gg 1$, ${\sinh }^{2}KL\sim e^{2KL}/4$, soThis is the thick-barrier quantum tunnelling limit, not a uniform approximation as energy approaches the barrier top while $KL$ remains small.

Let the fluid be at rest at infinity, and write its velocity potential as $\varphi =f(r)\cos \theta$, the angular dependence selected by translation along the polar axis. The separated Laplace equation iswith radial solutions $r$ and $r^{-2}$. Decay at infinity eliminates the former. The no-penetration condition on the moving sphere is $u_r(a)=U\cos \theta$, giving $f(r)=-U{a}^3/(2r^2)$. Thus the potential flow around a translating sphere isFor density $\rho$, integrate its kinetic energy over the exterior. The angular factor satisfies $\int ({\cos }^2\theta +{\sin }^2\theta /4)d\Omega =2\pi$, and ${\int }_a^{\infty }{r}^{-4}dr=1/(3a^3)$. HenceThe added mass of a sphere is consequently $M_f/2$. For a heavy sphere, $M>M_f$, falling through distance $h$, the net gravitational work after buoyancy is $(M-M_f)gh$. Its own kinetic energy plus that of the fluid is $(M+M_f/2)U^2/2$. Conservation of energy from rest therefore givesThis is the unbounded inviscid-fluid model; walls or viscous dissipation would alter the flow and its energy.