Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 4

Suppose there were only finitely many prime numbers congruent to $2$ modulo $3$, and list them as $p_1,\dots ,p_k$. ConsiderNo $p_i$ divides $N$, and $3$ does not divide $N$. In the prime factorization of $N$, not every prime factor can be congruent to $1$ modulo $3$, since their product would then also be congruent to $1$, whereasThus some prime factor is congruent to $2$ modulo $3$, contradicting the completeness of the list. Hence

Now let $p$ be prime. If $p\ne 3$, then either $p=2$, givingor $p$ is not divisible by $3$, so $p^2\equiv 1(\mathrm{mod}3)$. In the latter caseand is greater than $3$, hence is composite. For $p=3$ the number is $19$, which is prime. Therefore

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LetIf $x$ were rational, thenwould make $\sqrt[3]{6}=(x^3-5)/(3x)$ rational. But if $\sqrt[3]{6}=a/b$ in lowest terms, then $a^3=6b^3$, which is impossible by unique prime factorization: the exponent of $2$ on the two sides is respectively a multiple of three and one more than a multiple of three. Therefore

Use the given convergent seriesSuppose this number were $a/b\in \mathbb{Q}$. Choose $N$ large enough that $b\mid (2N+1)!$. Multiplication by $(2N+1)!$ makes both the rational number and the partial sum through $n=N$ integers. Their differencewould therefore be an integer. It is positive, whilefor sufficiently large $N$, a contradiction. Hence

A transcendental number is a complex number that is not a root of any nonzero polynomial with rational, equivalently integer, coefficients. LetIf $a\ne 0$ and $y$ were an algebraic number, then $e$ would satisfyover the algebraic extension $\mathbb{Q}(y)$. By transitivity of algebraic extensions, $e$ would be algebraic over $\mathbb{Q}$, contradicting its transcendence. If $a=0$, then $b\ne 0$ and $e=b/y$ would again be algebraic. Thus

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Take the scalar product of the Lorentz force equation with the constant magnetic field $\mathbf{B}$. Since $\mathbf{E}\cdot \mathbf{B}=0$ and $(\dot{\mathbf{x}}\times \mathbf{B})\cdot \mathbf{B}=0$,The initial velocity also obeys $\mathbf{v}\cdot \mathbf{B}=0$, so $\mathbf{x}\cdot \mathbf{B}$ remains equal to ${\mathbf{x}}_0\cdot \mathbf{B}$. The trajectory therefore lies in the planethe plane through ${\mathbf{x}}_0$ perpendicular to $\mathbf{B}$.

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Let $\mathbf{E}=E\hat{\mathbf{x}}$ and $\mathbf{B}=B\hat{\mathbf{y}}$, and choose $\hat{\mathbf{z}}=\hat{\mathbf{x}}\times \hat{\mathbf{y}}$. Part (a) gives $y=0$. With the origin at ${\mathbf{x}}_0$, the remaining Cartesian components areThe initial conditions are $x=z=\dot{x}=\dot{z}=0$. Integrating the second equation gives $\dot{z}=\omega x$, soSolving this forced harmonic oscillator and then integrating $\dot{z}=\omega x$ yieldsThus the position vector is ${\mathbf{x}}_0+x(t)\hat{\mathbf{x}}+z(t)\hat{\mathbf{z}}$. The oscillation occurs at the signed cyclotron frequency, superposed on the usual E-cross-B drift.

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The Galilean transformation isTherefore

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The one-dimensional Lorentz transformation givesHence

As $|\beta |\to 0$, $\gamma =1+O({\beta }^2)$. The spatial transformation is therefore $x^{\prime }=x-ut+O({\beta }^2)$, whileUnder the stated condition $|x|<c|t|$, the fractional correction $|ux|/(c^2|t|)$ is smaller than $|\beta |$ and tends to zero. Thus the Lorentz transformation has the nonrelativistic limit given by the Galilean transformation.

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For the special-relativistic matrix,Thus the eigenvalues and corresponding eigenvectors areThe eigenvector equations are $x=ct$ and $x=-ct$. They are the two null directions of the light cone, representing right-moving and left-moving light rays. A Lorentz boost preserves each light-ray direction while rescaling its null coordinate.

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The Bezout identity states that for integers $a,b$, not both zero, there are integers $r,s$ such thatIf the prime number $p$ divides $ab$ but does not divide $a$, then $\gcd (p,a)=1$. Thus $rp+sa=1$ for some integers $r,s$, and multiplication by $b$ givesBoth terms on the right are divisible by $p$, so $p\mid b$. Hence

If $\gcd (m,n)=1$, choose $r,s$ with $rm+sn=1$. Thensatisfies $x\equiv a(\mathrm{mod}m)$ and $x\equiv b(\mathrm{mod}n)$. If $x,x^{\prime }$ are two simultaneous solutions, both $m$ and $n$ divide $x-x^{\prime }$. Coprimality implies $mn\mid x-x^{\prime }$, so the solution is unique modulo $mn$. This proves the two-modulus Chinese remainder theorem.

Let $p$ be odd and supposeThen $p^d\mid (x-1)(x+1)$. Since $\gcd (x-1,x+1)$ divides $2$, the odd prime $p$ cannot divide both factors. The full power $p^d$ must therefore divide one of them, givingThese are distinct, so there are exactly two solutions.

For an odd integerthe Chinese remainder theorem identifies a solution modulo $n$ with independent solutions modulo the $k$ prime powers. Each component has two choices, hence

For $n=2^d$, there is one solution when $d=1$ and two when $d=2$. If $d\ge 3$, a solution $x$ is odd, and of the consecutive even integers $x-1,x+1$, exactly one is divisible by $4$. The 2-adic valuations therefore have minimum one; for their sum to be at least $d$, the other must be at least $d-1$. Thus $x\equiv \pm 1(\mathrm{mod}{2}^{d-1})$, giving the four distinct classesConsequently the answer is

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For $0\le k\le n$, the binomial coefficient isWhen $1\le k\le n-1$,This is Pascal's identity.

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The special binomial theoremfollows by mathematical induction. Multiplication of the formula for $n$ by $1+t$ and collection of the coefficient of $t^k$ uses Pascal's identity from part (a).

Termwise integration from $0$ to $1$ givesReplacing $t$ by $-t$ gives

Finally,Writing $y=1-x$ and using the finite geometric series,turns this into

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PutThe initial values are $S_1=S_2=1$. Using Pascal's identity and the convention that an out-of-range binomial coefficient is zero,Thus $(S_n)$ has the same initial values and linear recurrence relation as the Fibonacci numbers. Mathematical induction gives

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Because $P(\mathbf{x})\subseteq P(\mathbf{x})$, the relation is reflexive. If $\mathbf{x}⪯\mathbf{y}$ and $\mathbf{y}⪯\mathbf{z}$, thenso it is transitive. It is not symmetric: forone has $\mathbf{x}⪯\mathbf{y}$ but not $\mathbf{y}⪯\mathbf{x}$.

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Suppose $\mathbf{x}⪯\mathbf{y}$. Define the nonnegative tuple $\mathbf{z}$ coordinatewise byWhen $y_i=0$, support inclusion forces $x_i=0$, so $x_i=y_i{z}_i$ in every coordinate.

Conversely, if $x_i=y_i{z}_i$ with $y_i,z_i\ge 0$, then $x_i>0$ implies $y_i>0$. Hence $P(\mathbf{x})\subseteq P(\mathbf{y})$ and $\mathbf{x}⪯\mathbf{y}$. This proves the equivalence.

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The definition saysEquality of sets is reflexive, symmetric, and transitive, so $\sim$ is an equivalence relation. Its classes are indexed by all subsets of $\{1,\dots ,n\}$, and every subset occurs as the support of its zero-one indicator vector. The number of classes is therefore

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DefineThen $\mathbf{x}=\mathbf{y}+\mathbf{z}$, the support of $\mathbf{y}$ lies in $P(\mathbf{s})$, and the support of $\mathbf{z}$ is disjoint from $P(\mathbf{s})$. Thus $\mathbf{y}⪯\mathbf{s}$ and $\mathbf{z}\perp \mathbf{s}$.

For uniqueness, if $s_i>0$, orthogonality forces $z_i=0$, hence $y_i=x_i$. If $s_i=0$, the condition $\mathbf{y}⪯\mathbf{s}$ forces $y_i=0$, hence $z_i=x_i$. Every coordinate is therefore forced, proving uniqueness.

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A set is countable when it is finite or its elements can be put in one-to-one correspondence with a subset of $\mathbb{N}$. Equivalently, there is an injection from the set into $\mathbb{N}$.

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List pairs $(m,n)\in {\mathbb{N}}^2$ by increasing value of $m+n$, and within each diagonal by increasing $m$:Every pair occurs after finitely many earlier diagonals. More explicitly, the Cantor pairing functionis an injection ${\mathbb{N}}^2\to \mathbb{N}$. Hence $\mathbb{N}\times \mathbb{N}$ is countable.

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The integers are countable under the enumerationPart (i) then shows that $\mathbb{Z}\times \mathbb{N}$ is countable. The mapis a surjection from $\mathbb{Z}\times \mathbb{N}$ onto the rational numbers. Selecting, for example, the representation in lowest terms with positive denominator gives an injection in the other direction. Thus $\mathbb{Q}$ is countable.

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For a increasing function $F$, define its one-sided limits byThey exist because of monotonicity. The function is discontinuous at $x$ only if the jump intervalis nonempty. By the density of the rational numbers, choose $q_x\in J_x\cap \mathbb{Q}$.

If $x<y$, monotonicity gives $F(x+)\le F(y-)$, so $J_x$ and $J_y$ are disjoint. Consequently $q_x\ne q_y$, and $x\mapsto q_x$ injects the set of discontinuities into $\mathbb{Q}$. Since the rationals are countable by part (ii), the set of discontinuities is countable.

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Suppose first that $|A_n|=1$ for every $n>N$. Then an element of $B$ is determined by its first $N$ coordinates, so $B$ is in bijection withA finite Cartesian product of countable sets is countable by repeated application of the diagonal enumeration in part (b)(i). Hence $B$ is countable.

Conversely, suppose infinitely many factors contain at least two elements. Choose increasing indices $n_1<n_2<\cdots$ and distinct elementsFix one element in every remaining factor. Each binary sequence $\epsilon =({\epsilon }_1,{\epsilon }_2,\dots )$ then defines an element of $B$ by placing $a_j^{{\epsilon }_j}$ in coordinate $n_j$ and the fixed element elsewhere. This map is injective.

The set $\{0,1{\}}^{\mathbb{N}}$ is uncountable by Cantor's diagonal argument: from any proposed list of binary sequences, form a new sequence whose $j$th digit differs from the $j$th digit of the $j$th listed sequence. It is absent from the list. Thus $B$ contains an uncountable subset and cannot be countable.

Therefore

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The center of mass isBy Newton's third law, ${\mathbf{F}}_{21}=-{\mathbf{F}}_{12}$, soThus $\dot{\mathbf{R}}$ is constant. For $\mathbf{r}={\mathbf{x}}_1-{\mathbf{x}}_2$,Thereforewhere $\mu$ is the reduced mass.

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For the proposed circular motion,The separation is ${\mathbf{x}}_1-{\mathbf{x}}_2=2a(\cos \omega t,\sin \omega t,0)$, so the inverse-square force on particle 1 isThe equation $m{\ddot{\mathbf{x}}}_1={\mathbf{F}}_{12}$ holds exactly whenThe equation for particle 2 follows by symmetry.

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At a point $(0,0,z)$, the two particles are at equal distance $\sqrt{a^2+z^2}$. Their force components in the rotating $xy$-plane cancel, while their $z$-components add. Hence a particle initially moving along the $z$-axis remains on it, andCancelling $m_3$ gives

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The equation has the one-dimensional effective potential per unit massThus the conserved specific energy isThe potential has its minimum $-2mk/a$ at $z=0$ and tends to $0$ from below as $|z|\to \infty$. Therefore $-2mk/a<\mathcal{E}<0$ gives bounded oscillation between two turning points; $\mathcal{E}=0$ gives marginal escape with asymptotic speed zero; and $\mathcal{E}>0$ gives escape with nonzero asymptotic speed. The minimum itself is the equilibrium $z=0$.

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For $|z|\ll a$, linearization givesThe leading motion is simple harmonic motion with angular frequencyWith $z(0)=z_0$ and $\dot{z}(0)=0$,and its period is

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The initial specific energy isEscape to infinity is possible exactly when $\mathcal{E}\ge 0$. Thus the escape velocity criterion isEquality corresponds to marginal escape.

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The term $-2\mathbf{\Omega }\times \dot{\mathbf{x}}$ is the Coriolis acceleration, and $-\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{x})$ is the centrifugal acceleration. With $\mathbf{\Omega }=-\Omega \hat{\mathbf{z}}$, the Cartesian components are

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The circular constraint isIts tangent in the $xz$-plane is $(\cos \theta ,0,-\sin \theta )$, while the normal force is radial. Projecting the equations from part (a) onto this tangent eliminates the normal force and givesBecause the ramp is translation-invariant along $y$, $N_y=0$; using $\dot{x}=R\dot{\theta }\cos \theta$ gives

For rest in the rotating frame, $\dot{\theta }=\dot{y}=\ddot{\theta }=\ddot{y}=0$. Assuming $\Omega \ne 0$, the second equation gives $y=0$, and the first givesOn the semicircular ramp the rest points are thereforeand, when ${\Omega }^{2}R\ge g$,

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Neglecting terms of order ${\Omega }^2$ and linearizing in $\theta$ givesThe second equation and the initial data implyIts contribution to the first equation is of order ${\Omega }^2$, so the retained equation isThusIntegrating the associated $y$ velocity and using $y(0)=0$ gives

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The moment of inertia about an axis iswhere $r_{\perp }$ is the perpendicular distance to the axis. The disc mass isTo leading order in $\delta /r$, the distance from the $x$-axis is $|y|$. Using polar coordinates in the disc,

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The axis $y=0,z=h$ is parallel to the $x$-axis and lies a distance $|h|$ from it. The parallel axis theorem therefore givesto leading order in the thickness.

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Measure $z$ from the apex along the cone's symmetry axis. A thin disc at height $z$ has radiusand mass $dM=\rho \pi s^{2}dz$. Its moment about a parallel diameter through its centre is $s^{2}dM/4$, while its centre is distance $z$ from the required axis. Part (a)(ii) givesHenceSince $M=\rho \pi R^{2}H/3$,

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Without friction, rotational kinetic energy is conserved:The constant angular speed is $\omega =\sqrt{2K_0/I}$, so one full rotation takes

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The work law means that friction exerts a constant opposing torque of magnitude $W$. Until the cone stops,Its initial angular speed is ${\omega }_0=\sqrt{2K_0/I}$, and uniform angular deceleration givesTherefore

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A four-vector is an object whose components transform between inertial frames by a Lorentz transformation,Lorentz transformations are defined bywhere $\eta$ is the Minkowski metric. ConsequentlyThus the Minkowski norm of a four-vector is the same in every inertial frame.

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Along the particle's worldline,The boost from $S$ to $S^{\prime }$ givesThereforeand division of the transformed coordinate velocities by this factor gives the relativistic velocity-addition formula

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To first order in $V/c$, $\gamma =1+O(V^2/c^2)$ andThusSince $\mathbf{V}=(V,0,0)$ and $\mathbf{V}\cdot \mathbf{u}=V{u}_x$, these components combine into

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For a photon,Part (ii) gives the first-order changesFor a vector of fixed leading-order magnitude $c$, its small change of angle isSubstitution yields the relativistic aberration formulato leading order.