Past exam of the mathematics course of the University of Cambridge / 2001 / ib / Paper 1

A function $f:I\to \mathbb{R}$ is uniformly continuous ifThe crucial point is that $\delta$ works throughout the interval, not just near one chosen point.

Yes, it is bounded on the closed unit interval. Choose $\delta$ for $\epsilon =1$, and choose finitely many grid points $t_j$ such that every $x\in [0,1]$ is within $\delta$ of some $t_j$. Then $|f(x)|\le 1+{\max }_j|f(t_j)|$, a finite bound. Equivalently, uniform continuity implies continuity, and a continuous real function on a compact set is bounded.

Yes, a bounded derivative gives uniform continuity on the half-line. If $|f^{\prime }|\le K$, the mean value theorem gives $|f(x)-f(y)|\le K|x-y|$. This is a Lipschitz condition, so for $K>0$ use $\delta =\epsilon /K$. If $K=0$, the function is constant. As usual for a derivative on this closed half-line, continuity at its endpoint is included.

Not uniformly continuous. Take $x_n=n$, $y_n=n+1/n$. Then $|x_n-y_n|\to 0$, butThis contradicts the defining implication for uniform continuity, for example with $\epsilon =1$. The sequential criterion for uniform continuity makes explicit why arbitrarily close arguments at large distances from the origin matter.

Not uniformly continuous. ChooseTheir separation is $\pi /(x_n+y_n)\to 0$, while $\sin (x_n^2)=1$ and $\sin (y_n^2)=-1$. Thus the function values stay distance two apart, contradicting uniform continuity. Boundedness of the function does not prevent increasingly rapid oscillations.

Uniformly continuous. For $x\ge 1$, the derivative satisfiesThe mean value theorem therefore gives a Lipschitz condition with constant two, hence uniform continuity.

For the real even function, let $s_N=a_0/2+{\sum }_{n=1}^N{a}_n\cos nx$. Fourier orthogonality gives ${\int }_{-\pi }^{\pi }\cos nxdx=0$ for $n\ge 1$ and ${\int }_{-\pi }^{\pi }\cos nx\cos mxdx=\pi {\delta }_{nm}$. ConsequentlyFor an even square-integrable function, the complete Fourier cosine series converges in the L2 norm; hence its norms converge. Taking $N\to \infty$ proves Parseval's identity in the required normalization:This limit argument avoids multiplying an infinite series without a justification of L2 norm convergence.

For the quadratic function, the Fourier coefficients areThe last equality follows from two integrations by parts, using $\sin n\pi =0$. ThusThis Fourier series is absolutely convergent; the periodically extended quadratic is continuous, so it represents its values also at the endpoints. Substituting its coefficients into Parseval's identity gives

Put $S_1={\sum }_{i=1}^n{X}_i$, $S_2={\sum }_{i=1}^n{X}_i^2$. The joint normal distribution density can be written asThe Fisher-Neyman factorization theorem says that, for a family dominated by a common measure, a statistic $T$ is a sufficient statistic when its densities have the form $g_{\mu }(T(\mathbf{x}))h(\mathbf{x})$ with $h$ independent of the parameter. The displayed factorization therefore shows that $(S_1,S_2)$ is a two-dimensional sufficient statistic.

For maximum likelihood estimation, discard parameter-independent terms from the log-likelihood:If $S_2>0$, the quadratic equation has exactly one positive root. The numerator is positive before that root and negative after it, while the likelihood tends to zero at both ends of $\mu >0$. Hence the normal likelihood with variance equal to squared mean gives the global maximizerThis holds almost surely. The exceptional all-zero sample has $S_2=0$ and likelihood proportional to ${\mu }^{-n}$, so its supremum is approached as $\mu \downarrow 0$ and there is no positive maximizer.

The curvature-minus-one Poincaré disk model has Riemannian metricThe geodesics through zero are Euclidean diameters. Indeed, for a path written in polar coordinates, its length is at least $\int 2|\dot{r}|/(1-r^2)dt$, and hence at least $2\mathrm{artanh}r$ between zero and a point of modulus $r$. The radial segment attains this lower bound. By uniqueness of a geodesic with given initial tangent, these radial paths describe all geodesics through the origin. The radial hyperbolic distance isThus a hyperbolic circle in the Poincare disc is the indicated Euclidean circle.

An isometry from the Poincaré half-plane model to the Poincaré disk model is the Cayley transform between the half-plane and disk,Writing $z=x+iy$, one has $1-|w{|}^2=4y/|z+i{|}^2$ and $|dw{|}^2=4|dz{|}^2/|z+i{|}^4$. Substitution into the disk Riemannian metric gives $d{s}^2=|dz{|}^2/y^2$, the half-plane metric, and $i$ maps to zero.

Put $r=\tanh (\rho /2)$. The inverse image of $|w|=r$ obeysSince $(1+r^2)/(1-r^2)=\cosh \rho$, completing the square yieldsThis proves that the hyperbolic circle in the upper half-plane has Euclidean centre $i\cosh \rho$ and radius $\sinh \rho$.

Expansion of the determinant along the first row givesThus the matrix is invertible exactly when $x\ne 1$. The minor formed from the first two rows and the last two columns is $-1$, independently of $x$, so the matrix rank is always at least two. ThereforeTaking cofactors and transposing them gives the adjugate matrixThe adjugate identity $M\mathrm{adj}M=(\det M)I$ then givesIn particular, the singular case has rank two, rather than a further exceptional rank-one value.

Take $z$ positive downwards, and let the surface pressure be $p_0$. Static force balance gives the hydrostatic pressure equation $dp/dz=\rho g$. HenceOn a submerged body's surface, pressure traction is $-p\mathbf{n}$, where $\mathbf{n}$ points out of the body. The divergence theorem gives the resultantThus the buoyancy is upward with magnitude $\rho gV$, proving Archimedes' principle. The constant surface pressure makes no contribution because ${\int }_{\partial V}\mathbf{n}dS=0$.

For a floating iceberg of total volume $V$, the displaced seawater volume is $V-V_I$. Neglecting air buoyancy, equality of weight and buoyancy gives ${\rho }_{I}gV={\rho }_{w}g(V-V_I)$. Consequentlywith $0<{\rho }_I<{\rho }_w$, as required for partial flotation.

The Cauchy integral formula states that if $f$ is holomorphic on a neighbourhood of a simple closed positively oriented contour and its interior, then, for an interior point $z$,Fix $0<r<1$ and apply it on $|\xi |=r$. For $|z|<r$, the geometric seriesconverges uniformly in $\xi$ and locally uniformly in $z$. Indeed, for $|z|\le s<r$ its absolute terms are bounded by $s^n/r^{n+1}$. Since $f$ is bounded on the contour, termwise contour integration is valid, givingTermwise differentiation of this locally convergent power series gives $f^{(n)}(0)=n!c_n$. ThereforeThis also derives the needed Taylor series representation rather than assuming an unjustified interchange at the boundary of the analytic disk.

The discriminant of a binary quadratic form is $d=b^2-4ac$. If the form is positive-definite, evaluating it at $(1,0)$ gives $a>0$. For $a\ne 0$, completing the square givesEvaluating along $x=-by/(2a)$ shows that positivity also requires $d<0$. Conversely these two inequalities make both displayed coefficients positive, provingA reduced positive definite binary quadratic form satisfies $|b|\le a\le c$, with $b\ge 0$ if $|b|=a$ or $a=c$. The last convention avoids counting both boundary representatives. For fixed negative discriminant, these inequalities implyThere are finitely many such integers $a$, finitely many integers $b$ with $|b|\le a$, and then $c=(b^2-d)/(4a)$ is determined. Hence the number of reduced forms is finite. If no admissible integer coefficients have the given discriminant, this number is zero.

For $d=-39$, the bound gives $a=1,2,3$. The integrality of $c=(b^2+39)/(4a)$ and the boundary sign convention give respectivelyThese are all the reduced positive definite binary quadratic forms of discriminant minus thirty-nine. Thusrepresented by $x^2+xy+10y^2$, $2x^2+xy+5y^2$, $2x^2-xy+5y^2$, and $3x^2+3xy+4y^2$.

For an incident particle from the left, take $E>0$ and $E>V_0$, and defineThe time-independent Schrödinger equation gives an incident plus reflected plane wave on the left and an outgoing transmitted plane wave on the right:Continuity of the wavefunction and its derivative at a finite potential jump yields $1+r=t$ and $k(1-r)=qt$. Hence $r=(k-q)/(k+q)$ and $t=2k/(k+q)$. The incident and reflected probability currents have equal wave-number magnitude, soAs a check, the transmitted current fraction is $(q/k)|t{|}^2=4kq/(k+q{)}^2$, and the two probabilities sum to one.

For a positive step, $E\downarrow V_0$ makes $q\downarrow 0$, so $P\to 1$. For a negative step with fixed positive incident energy, $V_0\to -\infty$ makes $q\to \infty$, so again $P\to 1$. The latter is quantum reflection from an increasingly large wave-number mismatch, even though the step is downward.

Assume $a<b$. On continuous functions on this compact interval, both proposed distances are finite and symmetric, and vanish on identical functions. For the supremum norm, zero distance plainly implies pointwise equality. For the L2 norm, if a continuous difference is nonzero at any point, it stays bounded away from zero on a subinterval of positive length. Its squared integral is then positive, so zero distance also implies equality.

For the first metric, the pointwise triangle inequality followed by the supremum gives $d_1(x,z)\le d_1(x,y)+d_1(y,z)$. For the second, put $u=x-y$, $v=y-z$. The Cauchy-Schwarz inequality givesTaking square roots proves the required triangle inequality. Thus both are metrics.

The continuous-function space is complete in the uniform metric. If $(x_n)$ is a Cauchy sequence in $d_1$, at each $t$ its scalar values have a limit $x(t)$, by completeness of the scalar field. Given $\epsilon >0$, choose $N$ with $\Vert x_n-x_m{\Vert }_{\infty }<\epsilon$ for $m,n\ge N$. Letting $m\to \infty$ gives $|x_n(t)-x(t)|\le \epsilon$ uniformly in $t$. Hence there is uniform convergence. To see that the limit is continuous at $t_0$, approximate it uniformly by one continuous $x_n$ and useThe outer terms can be made small uniformly, and the middle term is small near $t_0$. This proves the needed uniform limit theorem and completeness.

For the second metric, let $H$ be zero on $(-1,0)$ and one on $(0,1)$; its value at zero is immaterial to the integral. The supplied ramp functions are continuous and satisfyAlthough $H$ is not in the continuous-function space, the integral triangle inequality yields $d_2(x_m,x_n)\le (3m{)}^{-1/2}+(3n{)}^{-1/2}\to 0$. Thus the sequence is Cauchy. If it converged in $d_2$ to a continuous $x$, the same inequality would give $\Vert x-H{\Vert }_2=0$. Continuity would then force $x=0$ on the negative half-interval and $x=1$ on the positive half-interval, contradicting continuity at zero. Therefore the continuous-function space is not complete in $d_2$; its limit in the larger L2 space is the discontinuous step.

Substituting $y=x^p$ in the homogeneous ordinary differential equation gives $p(p-1)-2=(p-2)(p+1)=0$. ThusFor a Green's function decaying at the two ends, use $u(x)=x^2$ on the side of zero and $v(x)=x^{-1}$ on the side of infinity. Their Wronskian is $u{v}^{\prime }-u^{\prime }v=-3$. The function must be continuous at $x=\xi$, or its second derivative would contain a derivative of a Dirac delta. Its derivative jump must be one, to give the specified positive delta source. These two conditions yield the boundary-decaying Green kernel for an inverse-square differential operator:Its derivatives at the join are $-2/3$ on the left and $1/3$ on the right, confirming the jump one. Away from the join each branch solves the homogeneous equation, and both endpoint conditions hold.

Integrate this Green's function against the forcing, whose support is between zero and one. For $0<x\le 1$,For $x\ge 1$, only the right branch contributes. ThereforeOn the first branch, $y^{\prime }=(2x/3)\log x+x/9$ and $y^{\prime \prime }=(2/3)\log x+7/9$, so $y^{\prime \prime }-2y/x^2=1$. The second branch is a multiple of the homogeneous solution $x^{-1}$, so its forcing is zero. At $x=1$, both $y$ and $y^{\prime }$ agree, with values $-1/9$ and $1/9$, so there is no additional delta source. Finally $x^2\log x\to 0$ as $x\downarrow 0$ and $x^{-1}\to 0$ as $x\to \infty$. A homogeneous solution obeying both boundary conditions has $C=D=0$, establishing uniqueness.

A simple hypothesis specifies a single probability law, with no unknown parameter remaining. For a possibly randomized hypothesis test, let $\phi (x)\in [0,1]$ be its conditional rejection probability. Its test size is $E_0\phi (X)$ and its statistical power against the specified alternative is $E_1\phi (X)$.

The Neyman-Pearson lemma states that for two densities $f_0,f_1$ with respect to a common measure, a likelihood-ratio test rejecting when $f_1/f_0>k$, accepting when it is below $k$, and randomizing on equality to obtain size $\alpha$, maximizes power among all tests of size at most $\alpha$.

Here the likelihood ratio isWriting $r=|x|$, its logarithm apart from the constant is $(r^2-r)/2$. This decreases until $r=1/2$ and then increases. In particular it is nonpositive for $0\le r\le 1$ and positive for $r>1$. The supplied normal-tail value and $\alpha <1/4$ imply that the quantile $t$ defined by $2[1-\Phi (t)]=\alpha$ satisfies $t>1$. Hence the threshold $\Lambda (t)$ is above every central value on $r\le 1$; outside that interval the ratio is strictly increasing. The Neyman-Pearson lemma therefore selects precisely the two tails:The equality boundary has probability zero under both continuous laws, so no randomization is needed. Under the alternative density, the statistical power isThe central dip in the likelihood ratio is why checking the threshold against the entire central region, rather than assuming monotonicity in $|x|$ everywhere, is necessary.

Project from the north pole $N=(0,0,1)$ onto the equatorial plane, identifying that plane with the complex plane. The line from $N$ to $P=(X,Y,Z)$ meets the plane inThis is stereographic projection. Its inverse isRotation through $\theta$ about the $z$-axis changes $X+iY$ to $e^{i\theta }(X+iY)$ and preserves $Z$, so it becomes the Möbius transformation $u\mapsto e^{i\theta }u$.

Let $B$ be the rotation matrix supplied in the question. It sends the $z$-axis to the $x$-axis, and therefore $R_x(\theta )=B{R}_z(\theta )B^{-1}$. If $b=\varphi B{\varphi }^{-1}$ is its given Möbius transformation, thenwhich is again a Möbius transformation. Such $z$- and $x$-axis rotations can send any chosen point of the sphere to the south pole: first rotate its horizontal projection onto the $y$-axis, then rotate in the $yz$-plane. This supplies the rotations needed below.

The antipodal stereographic coordinate relation is $\varphi (-P)=-1/\overline{\varphi (P)}$, as follows immediately from the inverse formula. Fix the cross-ratio conventionWe claim the required ordering isFor the antipodal cross-ratio and spherical distance identity, rotate $P$ to the south pole. Its coordinate becomes zero and its antipode becomes infinity. A point at angular spherical distance $d$ from the south pole has vertical coordinate $-\cos d$ and horizontal magnitude $\sin d$, so its new projected coordinate $v^{\prime }$ has modulus $\sin d/(1+\cos d)=\tan (d/2)$. By Möbius invariance of the cross-ratio, the displayed expression equalsFor finite nonsingular coordinates the same ratio is $-|u-v{|}^2/|1+u\bar{v}{|}^2$. Zero or infinite antipodal coordinates are interpreted by limits. When $Q=-P$, $d=\pi$ and the ratio has the extended value infinity; the finite real formula applies for $0<d<\pi$.

Write $J_k(\lambda )$ for the Jordan block with diagonal $\lambda$ and ones on the superdiagonal. Its characteristic polynomial and minimal polynomial are both $(x-\lambda {)}^k$. Therefore one answer to the first request isFor the second request, takeThe algebraic multiplicities are five and two, and the largest block at each eigenvalue has size two, so both prescribed polynomials hold. They are not similar matrices: the eigenspace at three has dimensions three and four respectively, and similarity preserves eigenspace dimension.

There is no third similarity class. At eigenvalue one, the total size two and maximal block size two force the single block $J_2(1)$. At eigenvalue three, the blocks partition five, each has size at most two, and at least one has size two. The only partitions are $2+2+1$ and $2+1+1+1$. The uniqueness of Jordan normal form up to block order therefore leaves exactly the two classes already exhibited.

Let $I$ denote the identity endomorphism. The two-root minimal polynomial gives $(\alpha -{\lambda }_{1}I)(\alpha -{\lambda }_{2}I)=0$. Every vector has the decompositionThe first summand lies in the eigenspace $E_1=\ker (\alpha -{\lambda }_{1}I)$, and the second in $E_2$. If a vector belongs to both, subtracting its two eigenvalue equations gives $({\lambda }_1-{\lambda }_2)v=0$, hence $v=0$. Thus $V=E_1\oplus E_2$.

For distinct real roots ${\lambda }_1,\dots ,{\lambda }_m$, define polynomial projectorsThe Lagrange interpolation polynomials in this expression sum to one: their sum has degree at most $m-1$ and equals one at the $m$ distinct roots. Hence ${\sum }_i{P}_i=I$. Also $(\alpha -{\lambda }_{i}I)P_i=0$, because its numerator is the minimal polynomial evaluated at $\alpha$. Thus every vector is a sum of eigenvectors, $v={\sum }_i{P}_{i}v$. On $E_j$, $P_i$ acts as ${\delta }_{ij}I$, so applying $P_i$ to any zero sum of vectors from the eigenspaces shows that every summand is zero. ThereforeChoosing a basis in each eigenspace proves diagonalizability over the reals. Conversely, if a real matrix is similar over the reals to a diagonal matrix, a polynomial annihilates it exactly when it vanishes at each distinct real diagonal entry. Its minimal polynomial is therefore the product of those distinct linear factors. We have proved real diagonalizability is equivalent to a minimal polynomial with distinct real roots.

Put $\theta =x-t$. Taking the gradient of the velocity potential givesThe motion is not generally incompressible; a velocity potential ensures irrotationality, not zero divergence. Both components have zero time average at each fixed point because their sine and cosine averages vanish.

The fluid acceleration is the material derivative $D\mathbf{u}/Dt={\partial }_t\mathbf{u}+(\mathbf{u}\cdot \nabla )\mathbf{u}$. Direct differentiation givesAveraging over one period at fixed $(x,y)$ therefore yieldsThis need not be the derivative of the mean velocity: the nonlinear advective term remains after averaging.

The dyed particle obeys the Lagrangian trajectory equationsExpand $x=ϵx_1+{ϵ}^2{x}_2+O({ϵ}^3)$ and $y=ϵy_1+{ϵ}^2{y}_2+O({ϵ}^3)$, for bounded times. The first-order equations give $x_1=0$, ${\dot{y}}_1=\cos t$, so $y_1=\sin t$. At second order,Integrating with zero initial values verifiesThe velocity averaged along this particle over $0\le t\le 2\pi$ is its displacement divided by $2\pi$. HenceThis Stokes drift is a Lagrangian mean; it does not contradict the zero Eulerian mean velocity at a fixed point. The secular term is a small-parameter expansion, not an assertion of uniform accuracy as $t\to \infty$.

Use the Fourier transform convention $\hat{F}(k)={\int }_{\mathbb{R}}F(x)e^{-ikx}dx$. First compute the Fourier coefficients of the periodized sum. Absolute integrability givesThus Fubini's theorem justifies termwise integration. For integer $m$, the change of variable $x=2\pi j+\tau$ and $e^{2\pi imj}=1$ giveReindexing a convergent periodization gives periodicity; the integral argument always gives its periodic representative almost everywhere. This proves the Fourier-coefficient content of the Poisson summation formula.

There is a genuine qualification to the printed pointwise claim: the periodization of an integrable function need not agree pointwise with its Fourier series under the stated hypotheses alone. Set $F(0)=1$ and $F(x)=0$ for $x\ne 0$. This is absolutely integrable, and on one closed period the periodization has only two possibly nonzero summands, so the series is uniformly convergent. Yet $f(0)=1$, while $\hat{F}(m)=0$ for every $m$. The claimed pointwise equality at zero is therefore false. This counterexample also works for Riemann integrability, not only for an almost-everywhere convention.

The coefficient identity gives a rigorous general replacement: the weighted Fejér sumsconverge to the periodization in the integral norm, and at every continuity point. Indeed they are its convolution with the normalized Fejér kernel; that kernel is nonnegative, has integral one, and has integral tending to zero outside every fixed neighbourhood of zero. For convergence in the L1 norm, split the convolution error into small translations, controlled by continuity of translation in the integral norm, and the remaining tail. The same split gives pointwise convergence at a continuity point. If the periodization is piecewise continuously differentiable, the ordinary Fourier series has the usual midpoint limit; if it is continuous and its coefficients are absolutely summable, the displayed unweighted formula holds pointwise. These are sufficient interpretations of the intended identity.

For the specified exponential example all the needed Fourier series convergence properties do hold. Split its Fourier transform at zero to obtainIts periodization is continuous, with uniformly geometrically bounded tails, and the sampled coefficients are absolutely summable. The absolutely summable coefficients define a uniformly convergent Fourier series with the computed coefficients. Its continuous sum equals the continuous periodization: their difference has every Fourier coefficient zero, hence is zero in the L2 norm by Fourier completeness, and continuity upgrades that to equality everywhere. Therefore its ordinary Poisson summation formula is valid. Evaluate it at zero:Thus the fully justified requested sum is

First work over a field of characteristic different from two. If a nonzero symmetric bilinear form had $\varphi (v,v)=0$ for every vector, polarization would givea contradiction. Hence choose $e$ with $\varphi (e,e)\ne 0$. Every $v$ decomposes uniquely aswith the second summand orthogonal to $e$. Induct on the orthogonal complement to obtain a diagonal basis. If the restricted form becomes zero, any basis of that remaining space completes the diagonalization. In this basis the rank of a quadratic form is exactly the number of nonzero diagonal coefficients; the remaining basis vectors span its radical of a bilinear form.

Over the reals, rescale the nonzero basis vectors so the diagonal entries are $p$ copies of $+1$, $q$ copies of $-1$ and $z=n-p-q$ zeros. Define the signature of a quadratic form by $\sigma =p-q$, so $r=p+q$. To prove it is basis independent, note that $p$ is the largest dimension of a linear subspace on which the form is positive-definite. The positive coordinate linear subspace attains dimension $p$; on any other positive linear subspace, projection to those positive coordinates is injective, since its kernel has nonpositive quadratic value. Thus its dimension is at most $p$. The same argument for the negative form characterizes $q$. This proves Sylvester's law of inertia and that the signature is well-defined.

Let $e_i$ and $f_i$ be positive and negative normalized basis vectors, and let $Z$ be the radical of a bilinear form. The span of $Z$ and $e_i+f_i$ for $1\le i\le \min (p,q)$ is a totally isotropic subspace: all mutual bilinear pairings vanish. Its dimension is $z+\min (p,q)$. Conversely, project any totally isotropic subspace $U$ to the nondegenerate positive-plus-negative quotient. The projected linear subspace is still null, and its projections to both the positive and negative coordinate spaces are injective: a vector with one component zero would have strictly signed quadratic value unless it vanished. Its dimension is therefore at most $\min (p,q)$, while the kernel of the quotient map on $U$ has dimension at most $z$. HenceThis proves both construction and sharpness of the maximum dimension of a totally isotropic subspace.

For the five-variable example, the symmetric coefficient matrix isIts cyclic structure makes the vectors $(1,{\zeta }^k,{\zeta }^{2k},{\zeta }^{3k},{\zeta }^{4k})$, $\zeta =e^{2\pi i/5}$, eigenvectors with eigenvalues ${\zeta }^k+{\zeta }^{-k}=2\cos (2\pi k/5)$. Thus the eigenvalues areTheir product is $2$, and three are positive and two negative. ThereforeThe real and imaginary parts of each conjugate pair of eigenvectors give the corresponding real two-dimensional eigenspaces, consistent with the real inertia calculation.

Infinite walls impose the Dirichlet boundary conditions $\psi (-a)=\psi (a)=0$. Inside the infinite square well, the time-independent Schrödinger equation is $-{\hslash }^2{\psi }^{\prime \prime }/(2m)=E\psi$. Its nonzero solutions satisfying both endpoint conditions have wave numbers $k_n=n\pi /(2a)$, $n\ge 1$. Sincethe normalized energy eigenstates are the sine modes with prefactor $1/\sqrt{a}$. They are orthonormal by sine orthogonality and complete by the Fourier sine series on an interval of length $2a$. Differentiating twice gives