Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 3

Since $5y^2\le 9$, we have $y\in \{-1,0,1\}$. For $y=0$, $x^2=9$; for $y=\pm 1$, $x^2=4$. Thus the solutions of the Diophantine equation arewhere the two signs in the second family are independent.

The algebraic norm on $\mathbb{Z}[\sqrt{-5}]$ is $N(a+b\sqrt{-5})=a^2+5b^2$ and satisfies multiplicativity of the algebraic norm. The calculation in part (a) therefore gives all elements of norm $9$:If one of these factored into two nonunits, multiplicativity would force both factors to have norm $3$. But the Diophantine equation $a^2+5b^2=3$ has no solution. Hence all six displayed elements are irreducible. This is a standard example in which an irreducible element need not have prime norm.

Put $I={\int }_0^{1}f(t)dt$. If $I\ne 0$, take the inner product with the unit vector $u=I/\Vert I{\Vert }_2$. Thenby the Cauchy-Schwarz inequality; the case $I=0$ is immediate. This is the triangle inequality for a vector-valued integral.

Equality in the Euclidean argument requires $u\cdot f(t)=\Vert f(t){\Vert }_2$ for every $t$: the nonnegative continuous difference has integral zero. Equality in Cauchy-Schwarz inequality then says $f(t)=\lambda (t)u$ with $\lambda (t)\ge 0$. Conversely, if $f(t)=\lambda (t)v$ for one fixed $v\in {\mathbb{R}}^m$ and a continuous $\lambda \ge 0$, then for every norm,Thus these nonnegative-ray functions, including $f\equiv 0$, are exactly the required functions.

A topological space is connected if it is not the union of two disjoint nonempty open sets, or equivalently has no nontrivial subset that is both an open set and a closed set.

Let $K_x$ be the union of all connected subspaces containing $x$. A union of connected subsets with a common point is connected: if a separation of a topological space existed, every member containing $x$ would lie wholly in the same side. Hence $K_x$ is connected and contains every connected subspace through $x$, so it is the connected component of $x$.

If $K_x\cap K_y\ne \varnothing$, their union is connected and maximality gives $K_x=K_y$. Otherwise they are disjoint. Since every $x$ belongs to $K_x$, the connected components partition $X$.

On $\mathbb{C}\setminus [-1,1]$, the chosen branch of a multivalued function is characterized by $f(z{)}^2=z^2-1$ and $f(x)>0$ for $x>1$; equivalently $f(z)/z\to 1$ as $z\to \infty$. Define $h(z)=-f(-z)$. Then $h(z{)}^2=z^2-1$ and $h(z)/z\to 1$ at infinity, so uniqueness of the normalized holomorphic square root gives $h=f$. Therefore $f(-z)=-f(z)$.

Near infinity, $g(z{)}^2=(z-2)(z^2-1)\sim z^3$, so $g(z)\sim z^{3/2}$. One positive circuit around infinity changes $z^{3/2}$ by the monodromy factor $e^{3\pi i}=-1$. Thus analytic continuation does not return to the original value, and $\infty$ is a branch point.

For $1<x<2$ the specified value is $g(x)=i\sqrt{(2-x)(x^2-1)}$. Continue towards $0$ around the branch point $1$. Along the upper bank, the argument of $(z-2)(z^2-1)$ increases from $\pi$ to $2\pi$, so its square root reaches the negative real value; along the lower bank it decreases from $\pi$ to $0$, giving the positive value. Since the radicand at $0$ is $2$,

The hyperbolic lines $AB$ and $CD$ are both perpendicular to $BC$. They are therefore ultraparallel hyperbolic lines, and $BC$ is their unique common perpendicular of ultraparallel hyperbolic lines.

The common perpendicular realizes the minimum hyperbolic distance between two ultraparallel lines. This follows directly in the Poincare half-plane model after a Möbius transformation sends the two lines to concentric semicircles: the supplied distance formula is minimized along their common radial geodesic. The segment $AD$ also joins the two lines, but it is different from $BC$ in a nondegenerate quadrilateral. Hence

The Euler-Lagrange equations areThe coordinate $\varphi$ is a cyclic coordinate, so its conjugate momentumis conserved. Substitution into the first equation givesMultiplying by $2\dot{\theta }$ shows thatThus $h$ is conserved. Since $l^2/{\sin }^2\theta ={\sin }^2\theta {\dot{\varphi }}^2$, $h$ is the squared speed on the unit sphere, equivalently twice the kinetic energy.

Write $G(x;\xi )=e^{2x}H(x;\xi )$. Since $(D-2{)}^2(e^{2x}H)=e^{2x}{H}^{\prime \prime }$, the Green function must have $H(0;\xi )=H(1;\xi )=0$, be continuous at $x=\xi$, and satisfy the jump condition for an impulse $H_x({\xi }^{+};\xi )-H_x({\xi }^{-};\xi )=e^{-2\xi }$. Solving the two linear pieces gives

The Green-function representation givesThe bracket simplifies to $x(x-1)/2$, soIt vanishes at both endpoints, and applying $(D-2{)}^2$ returns $e^{2x}$.

A Hermitian operator $A$ satisfies $⟨\phi ,A\psi ⟩=⟨A\phi ,\psi ⟩$ on its operator domain. HereTwo integration by parts operations, together with the boundary conditions in the Hamiltonian domain, show that the kinetic term is Hermitian; multiplication by the real function $V$ is Hermitian as well. Thus $H$ is Hermitian.

For a time-independent operator $A$, the Schrodinger equation and its adjoint give the Ehrenfest theoremUsing the canonical commutation relation $[x,p]=i\hslash$,Consequently

In the order $(1,2,3,L,G)$, where $L$ means left without a degree and $G$ means graduated, the transition matrix isHere $p+q+r=1$.

The three year states are transient states: from each there is positive probability of entering an absorbing state, and repeated trials make eventual departure certain because $q<1$. The singleton classes $\{L\}$ and $\{G\}$ are closed absorbing states, hence recurrent. Thus the chain has two closed recurrent classes and three transient singleton classes.

During any year, repeats may be ignored until the first non-repeat outcome. Conditional on that event, the chance of progressing isThe student must progress successfully through all three years. By the Markov property, the required probability is therefore

The Cayley-Hamilton theorem states that every square matrix satisfies its characteristic polynomial: if ${\chi }_A(t)=\det (tI-A)$, then ${\chi }_A(A)=0$.

For the proof, the adjugate identity givesWrite $\mathrm{adj}(tI-A)=B_0+B_{1}t+\cdots +B_{n-1}{t}^{n-1}$ and compare coefficients. Multiplying the resulting recurrence by successive powers of $A$ and summing telescopically yields ${\chi }_A(A)=0$.

By polynomial division, $p=q{\chi }_A+r$ with $\mathrm{deg}r<n$. Therefore $p(A)=r(A)$ by Cayley-Hamilton, while $p(\lambda )=r(\lambda )$ for every eigenvalue $\lambda$, since ${\chi }_A(\lambda )=0$.

For the displayed matrix,The remainder of $t^{1000}$ modulo this polynomial has the form $a{t}^2+c$, because its values at $i$ and $-i$ are both $1$. The equations $-a+c=1$ and $4a+c=2^{1000}$ giveSince $(A^2{)}_{11}=3$, the requested entry is

Gauss lemma for polynomials says that the product of two primitive polynomials in $\mathbb{Z}[X]$ is primitive. Equivalently, a primitive polynomial in $\mathbb{Z}[X]$ is reducible over $\mathbb{Q}$ exactly when it is reducible over $\mathbb{Z}$.

Eisenstein criterion states that ifhas a prime number $p$ such that $p\nmid a_n$, $p\mid a_i$ for every $i<n$, and $p^2\nmid a_0$, then $f$ is irreducible over $\mathbb{Q}$.

After dividing out the polynomial content, suppose for contradiction that the resulting primitive polynomial factors as $f=gh$ with positive-degree $g,h\in \mathbb{Z}[X]$, using Gauss lemma for polynomials. Reduction modulo $p$ gives $\bar{f}=a_n{X}^n$, so in the polynomial ring ${\mathbb{F}}_p[X]$ both $\bar{g}$ and $\bar{h}$ are monomials. Because both factors have positive degree, their constant coefficients are divisible by $p$. Their product $a_0$ is then divisible by $p^2$, a contradiction.

Translate the variable by one:This satisfies Eisenstein criterion with $p=5$: every nonleading coefficient is divisible by $5$, while $25\nmid 30$. Translation $X\mapsto X+1$ is an automorphism of a polynomial ring, so it preserves irreducibility. Hence $f$ is irreducible over $\mathbb{Q}$.

The sequence $(f_n)$ converges uniformly to $f$ on $A$ when, for every $\epsilon >0$, there is $N$ such thatfor every $x\in A$.

The function $f$ is uniformly continuous when, for every $\epsilon >0$, there is $\delta >0$ such that $x,y\in A$ and $\Vert x-y\Vert <\delta$ imply $|f(x)-f(y)|<\epsilon$. The same $\delta$ works throughout the domain.

True. Given $\epsilon >0$, uniform continuity supplies $\delta >0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$. For all sufficiently large $n$, $1/n<\delta$, and thenThis is precisely uniform convergence of $f_n$ to $f$.

False. Take the uniformly continuous function $f(x)=x$. Thenwhich is unbounded as a function of $x$ for every fixed $n$. Thus $g_n$ does not converge uniformly to $f^2$ on $\mathbb{R}$.

The closed bounded set $A\subset {\mathbb{R}}^m$ is compact by the Heine-Borel theorem. Choose a convergent subsequence $x_{n_j}\to x_0$. For fixed $N$ and sufficiently large $j$, monotonicity givesContinuity and passage to the limit give $g_N(x_0)\ge \delta$ for every $N$, hence ${\lim }_N{g}_N(x_0)\ge \delta$.

For the deduction, set $g_n=f_n-f$. These functions are continuous, decrease pointwise, and tend to zero. If convergence were not uniform, some $\delta >0$ would admit $x_n$ with $g_n(x_n)\ge \delta$. The result just proved would produce $x_0$ with $\lim g_n(x_0)\ge \delta$, contradicting pointwise convergence. Therefore $f_n\to f$ uniformly. This is Dini's theorem.

The hypothesis says that $|f|$ attains a global, hence local, maximum at the interior point $a$. By the maximum modulus principle, a nonconstant holomorphic function cannot do this. Therefore $f$ is constant.

This is Schwarz lemma. Define $g(z)=f(z)/z$ for $z\ne 0$ and $g(0)=f^{\prime }(0)$. The removable singularity makes $g$ holomorphic. Applying the maximum modulus principle to $g$ on disks of radius $r<1$, then letting $r\uparrow 1$, gives $|g(z)|\le 1$. Hence

Because $f(D)$ avoids the closed negative real axis, its principal square root $g=\sqrt{f}$ is holomorphic, satisfies $g(0)=1$, and has $\mathrm{Re}g>0$. The Cayley transform between the half-plane and diskmaps the right half-plane into the unit disk and has $h(0)=0$. By Schwarz lemma, $|h(z)|\le |z|$. When $|z|\le 1/2$,Thusso one may take the universal constant $C=9$.

No. The nonvanishing holomorphic functions $f_n(z)=e^{nz}$ satisfy $f_n(0)=1$, butAvoiding zero alone permits an unbounded holomorphic logarithm's real part; the omitted ray in part (c) was what constrained the image after taking a square root.

For a coordinate curve $v=v_0$, the length between $u_1$ and $u_2$ iswhere $E$ is a coefficient of the first fundamental form. Thus $\partial E/\partial v=0$ makes the two $u$-directed sides of every coordinate rectangle equally long. Similarly,so $\partial G/\partial u=0$ makes the two $v$-directed sides equally long.

Conversely, suppose every pair of opposite sides has equal length. Fix $v_0,v_1$. Equality of the first pair for every interval $[u_1,u_2]$ givesThe fundamental theorem of calculus implies that the continuous integrand vanishes, so $E$ is independent of $v$. The same argument shows that $G$ is independent of $u$. Hence the stated equal-length property is equivalent to $E_v=G_u=0$.

For every test function $f$, the change of variables $u=\alpha x$ givesEquality of distributions therefore gives

Using the definition of a distributional derivative,Hence $x{\delta }^{\prime }(x)=-\delta (x)$ as distributions.

Repeated integration by parts, or the definition of the $m$th distributional derivative, gives

Interpreting the displayed function aswe have ${\gamma }_n\ge 0$ and, after $u=nx$, ${\int }_{\mathbb{R}}{\gamma }_n=1$. Moreover its mass outside any fixed neighborhood of zero tends to zero. Therefore $({\gamma }_n)$ is an approximate identity, and for every test function $f$,Thus ${\gamma }_n\to \delta$ in the sense of distributions.

For a test function $f$, substitute $y=nx$ and use the Taylor expansion $f(y/n)=f(0)+(y/n)f^{\prime }(0)+O(n^{-2})$. Oddness removes the first term, while the Gaussian integral gives $\int y^2{e}^{-y^2}dy=\sqrt{\pi }/2$. HenceTherefore

Direct integration givesThe sinc approximate identity converges distributionally to the Dirac delta function, using the supplied integral of $\sin y/y$. Thus the limit is $\delta (x)$.

Since the odd sine part of the complex exponential integrates to zero,Part (e)(ii) therefore yieldsin the sense of distributions. This is the Fourier representation of the Dirac delta function.

The expectation value is the mean outcome over many identically prepared measurements of the observable. For real $\lambda$, positivity of the norm in the Hilbert space givesThis real quadratic has nonpositive quadratic discriminant, soApply this to the centered observables $Q-⟨Q⟩$ and $P-⟨P⟩$ to obtain the Robertson uncertainty principle

For the quantum harmonic oscillator, $[x,p]=i\hslash$ andThe arithmetic-geometric mean inequality and $\Delta x\Delta p\ge \hslash /2$ now give

Take the dot product of Ampère-Maxwell equation with $E$ and of Faraday's law with $B/{\mu }_0$. The divergence of a cross product identitythen gives the local Poynting theoremIntegration and the divergence theorem produce the stated equation. Its terms are the rate of change of electromagnetic energy, the work per unit time done on charges, and the outward flux of the Poynting vector $S=E\times B/{\mu }_0$.

Let the plate separation be $d$ and their radius be $R$, so $A=\pi R^2$. Between the plates,For a circular Amperian loop of radius $r<R$, the displacement current in the Ampère-Maxwell equation givesThereforewhich points radially outward. Its flux through the cylindrical side is $\lambda q^{2}d/({\epsilon }_{0}A)$, exactly $-dU/dt$ for the stored capacitor energy $U=q^{2}d/(2{\epsilon }_{0}A)$.

Let $u(t)$ be the signed air speed along the straw towards the bubble. Conservation of mass in a pipe for the incompressible plug flow givesThe unsteady Bernoulli equation along the straw, whose open end is at atmospheric pressure, givesby the Young–Laplace equation. Substitution yields

Set $w=a^2\dot{a}$. Since $a\dot{w}=-C$, the chain rule gives $wdw/da=-Ca$. The initial condition $\dot{a}(0)=0$ impliesWriting $a=a_0\sin \theta$ and integrating gives the implicit solutionAt disappearance $a=0$ and $\theta =0$, so

Repeated integration by parts inproduces the first $k+1$ terms of the Taylor polynomial and leaves $f(x)$, proving the integral remainder in Taylor theoremApply the linear functional ${\lambda }_k$ to this identity. It annihilates the Taylor polynomial, and commuting it with integration gives the Peano kernel theorem

For the midpoint quadrature rule on $[0,1]$, $e[1]=e[x]=0$, so $k=1$. Its Peano kernel isFor $f(x)=x^2$, the formula givesagreeing with ${\int }_0^1{x}^{2}dx-(1/2{)}^2=1/3-1/4$.

Under the null hypothesis that treatment and recovery are independent, estimate each expected cell count byand form the Pearson chi-squared test of independence statistic ${\chi }^2=\sum (O-E{)}^2/E$. For a $2\times 2$ table its asymptotic null law is a chi-squared distribution with one degree of freedom. A second-order Taylor expansion of the generalized likelihood-ratio test statistic $-2\log \Lambda$ around the fitted counts gives this Pearson statistic, which explains its asymptotic null distribution.

Writing the four residuals over their fitted counts and simplifying givesFor $(a,b,c,d)=(50,10,15,5)$,whose one-degree-of-freedom p-value is about $0.41$. There is no evidence of an effect. One fitted count is only $3.75$, however, so the chi-squared approximation is questionable; Fisher's exact test would be preferable.

The Lagrange sufficiency theorem says that if a feasible point $x^{\ast }$ and nonnegative multipliers satisfy complementary slackness and $x^{\ast }$ globally maximizes the Lagrangian function in constrained optimization, then $x^{\ast }$ globally solves the constrained maximization problem. Indeed, for every feasible $x$, the signs of the constraints give $f(x)\le L(x)\le L(x^{\ast })=f(x^{\ast })$. Concavity and the Karush-Kuhn-Tucker conditions are a common way to verify the global Lagrangian maximum.

For fixed $y$, A maximizessubject to $x_i\ge 0$ and ${\sum }_i{x}_i=X$. This is a concave function of $x$. On every active coordinate, the multiplier equation isThus A's optimal allocation is given by the water-filling algorithm:where $\lambda >0$ is chosen so that ${\sum }_i{x}_i=X$. Coordinates with zero expression receive no resource. This is a water-filling algorithm. The formula gives the optimum when every $y_i>0$. If $y_i=0$, any $x_i>0$ already gives A the whole market $u_i$ in that region, so A assigns such coordinates arbitrarily small positive amounts and applies water filling to the remaining budget; without a convention for $0/0$, the resulting value can be a supremum rather than an attained maximum.