Past exam of the mathematics course of the University of Cambridge / 2005 / ib

The minimal polynomial $m_{\beta }(t)$ of the linear map $\beta$ is the unique monic polynomial of smallest degree satisfying $m_{\beta }(\beta )=0$. An annihilating polynomial exists because the powers $I,\beta ,{\beta }^2,\dots$ lie in the finite-dimensional vector space $\mathrm{End}(V)$ and are therefore linearly dependent. If two monic annihilating polynomials had the same least degree, their difference would be an annihilating polynomial of smaller degree, so they coincide.

Write $m_{\beta }(t)=a_0+tq(t)$. If $a_0\ne 0$, evaluation at the linear map gives $\beta q(\beta )=-a_{0}I$. Since $\beta$ commutes with every polynomial in itself, this is a two-sided inverse:Conversely, if $\beta$ is an invertible linear map and $a_0=0$, then $m_{\beta }(t)=tq(t)$ and multiplication of $\beta q(\beta )=0$ by ${\beta }^{-1}$ gives $q(\beta )=0$, contradicting the defining least degree. Thus invertibility is equivalent to a nonzero constant term of the minimal polynomial. In the zero-dimensional case the minimal polynomial is $1$ and the same conclusion holds.

The image is a ring torus: a circle of radius $b$ is rotated about the vertical axis at distance $a$ from that axis. Put $r=\sqrt{x^2+y^2}$. Since $a>b$, every image point has $r>0$, and the image is preciselyConversely, on this level set the pairs $((r-a)/b,z/b)$ and $(x/r,y/r)$ are unit vectors, so choosing their angles gives a preimage.

The specified angular rectangle removes two closed circles from the torus. More explicitly its image isHere $C_u$ is the outer equatorial circle and $C_v$ is a meridian. The given openness property makes $U$ open in $T$. On $U$, the angles of $(r-a)+iz$ and $x+iy$ are uniquely chosen in $(0,2\pi )$; their continuous argument branches give the inverse $(x,y,z)\mapsto (u,v)$. Thus the restriction is a homeomorphism onto $U$, and the inverse is locally smooth.

For the immersion condition, the two derivatives have inner productsThey are linearly independent everywhere, since $a+b\cos u\ge a-b>0$. Hence this is a smooth embedded surface parametrization.

The angular cuts do not represent singularities of the embedded torus of revolution. To give a direct proof covering all points, define $F=(\sqrt{x^2+y^2}-a{)}^2+z^2-b^2$ on the open region $r>0$. On $T$,The implicit function theorem therefore describes $T$ near every point as a smooth graph of two coordinates. The whole torus is a smooth embedded surface.

The Cauchy integral formula states that if $f$ is holomorphic on a neighborhood of a positively oriented simple closed contour $C$ and its interior, then for $w$ inside,For the given counterclockwise circle, divide the rational function:The contour integral of $z$ is zero because it has a primitive, while both points $\pm i$ lie inside the circle. Applying the Cauchy integral formula to the constant function $1$ at each point givesThe sign reverses if the contour orientation is reversed.

Use Minkowski spacetime coordinates $x^{\mu }=(ct,x,y,z)$ and the Minkowski metric of signature $(+,-,-,-)$. For a massive particle with three-velocity $\mathbf{v}=d\mathbf{x}/dt$ and $|\mathbf{v}|<c$, proper time satisfiesThus the four-velocity components areHere upper indices indicate components in the chosen coordinates; lowering with the Minkowski metric changes the signs of the spatial components. Its invariant product defined by the Minkowski metric is

The four-momentum is $p^{\mu }=(E/c,\mathbf{p})=m{U}^{\mu }$, so $E=\gamma m{c}^2$ and $\mathbf{p}=\gamma m\mathbf{v}$. The Taylor expansion of the Lorentz factor at small $|\mathbf{v}|/c$ isConsequently $\mathbf{p}=m\mathbf{v}+O(m|\mathbf{v}{|}^3/c^2)$, recovering ordinary momentum. At rest the four-momentum has energy $E_0=m{c}^2$, so subtracting this rest energy gives the kinetic energyThus the nonrelativistic limit preserves the rest-energy offset as well as the usual kinetic energy.

For a two-dimensional incompressible flow $\mathbf{u}=(u,v)$, the continuity equation is $u_x+v_y=0$. Locally, or globally on a simply connected fluid region, this permits a streamfunction withIndeed, the one-form $-vdx+udy$ is closed by incompressibility and hence is locally the differential of $\psi$. Conversely these expressions automatically give zero divergence by equality of mixed derivatives.

A streamline at a fixed time is tangent to the instantaneous velocity field. Along it,Thus streamlines are connected curves on which the streamfunction is constant; wherever $\nabla \psi \ne 0$, the level curves have exactly the required tangent direction. At a stagnation point, the tangent is degenerate and distinct branches of a level set need not form a single streamline. For unsteady flow this argument fixes time: it does not identify streamlines with particle pathlines.

Integrating ${\psi }_y=x+\sin t$ and imposing $-{\psi }_x=-y$ gives the streamfunctionup to an arbitrary function of time. At the specified time, the streamlines areFor $C\ne 0$, the connected branches are rectangular hyperbolas. The zero level consists of the lines $y=0$ and $x=-1$, with the stagnation point $(-1,0)$ at their intersection; their half-lines and the stationary point are the distinct streamlines.

A particle pathline follows the time-dependent velocity field, so its coordinates satisfy $\dot{x}-x=\sin t$ and $\dot{y}=-y$. The second equation and the initial data give $y=e^{-t}$. Using the integrating factor $e^{-t}$ in the first,Therefore the complete pathline isIf $x_0+\frac{1}{2}\ne 0$, its first coordinate grows without bound in magnitude. If $x_0=-\frac{1}{2}$, the first coordinate remains bounded and periodic while $y\to 0$. The unique release coordinate giving a bounded particle path is $x_0=-1/2$.

For the Cholesky decomposition $A=L{L}^T$, choose $L$ lower triangular with positive diagonal whenever $A$ is positive-definite. Successive entries areThe $(3,2)$ equation gives $l_{31}{l}_{21}+l_{32}{l}_{22}=2+3\lambda$, so $l_{32}=3\sqrt{\lambda +2}$. Finally,Thus for $\lambda >-2$,It is nonsingular, and for every nonzero vector $v$, $v^{T}Av=|L^{T}v{|}^2>0$.

Necessity also follows from the leading principal determinants $2$, $2(\lambda +2)$ and $6(\lambda +2)$, or directly from the second elimination pivot. HenceAt $\lambda =-2$ the displayed limiting factor still satisfies $A=L{L}^T$, but has rank two and zero second diagonal entry: $A$ is positive semidefinite, not positive-definite. For $\lambda <-2$, taking $v=(2,1,0{)}^T$ gives $v^{T}Av=\lambda +2<0$, so no real factor $L{L}^T$ exists.

Under the null hypothesis of equal population probabilities, the three counts have a multinomial distribution with probabilities $1/3$ and expected counts $100$ each. The Pearson chi-squared goodness-of-fit test usesNo probabilities have been estimated from the data, and the three counts sum to a fixed total. Hence the null reference chi-squared distribution has $3-1=2$ statistical degrees of freedom. The supplied $5\%$ upper critical value is $5.99$, and $5.46<5.99$. At the 5% significance level there is insufficient evidence to reject equal popularity.

The asymptotic P-value is $\mathrm{Pr}({\chi }_2^2\ge 5.46)=e^{-5.46/2}\approx 0.0652$. The expected counts are all large, so the approximation is appropriate if the sampled choices are independent and representative of this branch. The conclusion depends on the chosen significance level; for instance, a $10\%$ test would reject. Nonrejection does not establish the null hypothesis, and a survey at one branch does not by itself justify a population claim about every branch.

Associate free real dual variables $u_i$ and $v_j$ with the row and column equality constraints. The Lagrangian for the transportation problem isIts infimum over $x_{ij}\ge 0$ is finite exactly when $u_i+v_j\le c_{ij}$ for every pair. The linear programming duality therefore givesThe dual variables have no sign restrictions because the primal constraints are equalities.

A feasible shipment $x$ is optimal if and only if there exist feasible transportation dual potentials satisfying complementary slackness:Thus every occupied route must have $u_i+v_j=c_{ij}$. To see sufficiency explicitly, the primal minus dual objective isand complementary slackness makes it zero. This is the weak duality certificate. Necessity follows from strong duality for the feasible, bounded linear program. If the total supply $M$ is positive, $x_{ij}=a_i{b}_j/M$ gives feasibility; if $M=0$, $x=0$ does. The feasible shipments form a closed bounded set, so the primal optimum is attained. This also covers zero supplies and demands. The one redundant equality merely leaves the dual gauge freedom $(u,v)\mapsto (u+t,v-t)$.

Let $\dim V=n$ and $\dim W=k$. Extend a basis $w_1,\dots ,w_k$ of the vector subspace $W$ to $w_1,\dots ,w_n$ of $V$, and let $f_1,\dots ,f_n$ be its dual basis. Every linear functional is uniquely ${\sum }_i{c}_i{f}_i$. It vanishes on $W$ exactly when $c_1=\cdots =c_k=0$. Thus the annihilator of a vector subspace has basis $f_{k+1},\dots ,f_n$, and

To apply this to a matrix, regard its rows as elements of $({\mathbb{R}}^n{)}^{\ast }$ and let $R$ be their linear span. Elementary invertible row and column operations reduce a rank-$r$ matrix to a block identity $\mathrm{diag}(I_r,0)$, showing that $\dim R=r$. The equations require that every functional in $R$ vanish on $x$. The canonical identification ${\mathbb{R}}^n\cong (({\mathbb{R}}^n{)}^{\ast }{)}^{\ast }$ sends $x$ to evaluation $f\mapsto f(x)$: it is injective because coordinate functionals separate points, and is bijective by equal dimensions. Consequently the solution vector space is the annihilator of $R$ under this identification. The dimension formula just proved givesA basis of this kernel consists of exactly $n-r$ linearly independent solutions, and every solution is their linear combination.

Let $G$ act on its $n$ left cosets by $g\cdot (xH)=gxH$. This is well-defined and gives a group homomorphism $\rho :G\to S_n$. Its kernel $K$ is a normal subgroup. An element lies in this kernel exactly when it fixes every coset, soThis is the normal core of a subgroup. By the first isomorphism theorem, $G/K\cong \rho (G)$, a subgroup of $S_n$. The Lagrange theorem therefore givesIn particular $K$ has finite index even when $G$ is infinite.

Now let $H$ be generated by an element of order $q$ in a group of order $pq$. Then $|H|=q$ and $[G:H]=p$. The preceding result gives a normal $K\subset H$ with $[G:K]\mid p!$. Since $H$ has prime order, its subgroup $K$ is either $\{e\}$ or all of $H$. The first choice would imply $pq\mid p!$, which is impossible because the prime $q>p$ divides none of the factors of $p!$. Therefore $K=H$, proving that every subgroup generated by an element of order $q$ is normal. No classification of groups of order $pq$ is needed.

Fix $x_0\in E$ and $\epsilon >0$. By uniform convergence, choose $N$ so that $|f_N(x)-f(x)|<\epsilon /3$ for every $x\in E$. By continuity of $f_N$ relative to $E$, there is $\delta >0$ such that $x\in E$ and $|x-x_0|<\delta$ imply $|f_N(x)-f_N(x_0)|<\epsilon /3$. The triangle inequality givesThis proves the uniform limit theorem without requiring $E$ to be open or closed.

For the series, fix an arbitrary $x_0\in (0,1]$ and put $a=x_0/2>0$. On $[a,1]$ each summand $n^{-1-x}$ is continuous andThe Weierstrass M-test gives uniform convergence of the series on $[a,1]$. Its partial sums are continuous, so the uniform limit theorem makes its sum continuous on that interval, in particular at $x_0$ with the appropriate relative topology if $x_0=1$. Since $x_0$ was arbitrary, the sum is continuous on $(0,1]$.

This is local uniform convergence, not uniform convergence on the entire half-open interval. Indeed, for any finite partial sum length $N$, its tail as $x\downarrow 0$ is bounded below by arbitrarily long tails of the divergent harmonic series. There is no continuity claim at the excluded endpoint $0$.

The sum of the two metrics is nonnegative and symmetric. It vanishes exactly when both coordinate distances vanish, hence exactly when the two points agree. Adding the two component triangle inequalities gives the triangle inequality on the Cartesian product, so the displayed function is a metric. Moreover,so the projection map is $1$-Lipschitz and therefore continuous.

To prove sequential compactness directly from compactness, take a sequence $(x_n)$ in $X$. If every point $x$ had a neighborhood containing sequence terms at only finitely many indices, these neighborhoods would form an open cover with a finite subcover, implying that only finitely many sequence terms exist, a contradiction. Thus some $x\in X$ has infinitely many indices in every neighborhood. Choose increasing indices $n_k$ with $d_X(x_{n_k},x)<1/k$; then $x_{n_k}\to x\in X$.

Now let $F\subset X\times Y$ be closed and suppose $y_n\in \pi (F)$ converges to $y$. Choose $x_n$ such that $(x_n,y_n)\in F$. A convergent subsequence $x_{n_k}\to x$ exists by the preceding argument. In the product metric, $(x_{n_k},y_{n_k})\to (x,y)$, and the fact that $F$ is closed gives $(x,y)\in F$. Thus $y\in \pi (F)$, proving that the projection with a compact factor is closed. The sequential criterion for a closed set in a metric space follows by choosing points at distance less than $1/k$ from any point of its closure.

For a counterexample without a compact factor, take $X=Y=\mathbb{R}$ with their usual metrics andThe continuous product function makes $F$ closed, but $\pi (F)=\mathbb{R}\setminus \{0\}$ is not closed. Thus compactness of the discarded factor is essential to the general conclusion.

Use the Möbius transformation $t=(z-1)/(z+1)$, the principal square root $s=\sqrt{t}$ with $\Re s>0$, and then $w=(s-1)/(s+1)$. The resulting conformal map isAll branches here are single-valued: the inverse $z=(1+t)/(1-t)$ shows that the deleted segment corresponds to the nonpositive real axis. However, the exact first image issince $t=1$ would require a source point at infinity. Thus the next image is $\{\Re s>0\}\setminus \{1\}$, andThe strict inequality $|w|<1$ follows from $|s+1{|}^2-|s-1{|}^2=4\Re s>0$. None of the component derivatives vanishes. For completeness the inverse on this punctured disk isGiven $0<|w|<1$, the inverse Cayley transform has positive real part, and reversing the three maps proves both injectivity and surjectivity onto this exact image.

This distinction matters for interpreting the printed request. The displayed function is a conformal map into the unit disk, but no conformal bijection from the specified plane domain onto the full unit disk exists. To prove the obstruction, the loop $z=2e^{i\theta }$ surrounds $0$, a point outside the domain, with winding number $1$. A contraction within the source would remain in $\mathbb{C}\setminus \{0\}$ and preserve winding number, whereas a constant loop has winding number $0$. Hence the source is not simply connected; a disk is simply connected, as straight-line contraction shows, and a conformal bijection is a homeomorphism.

The hinted construction becomes a bijection onto the full disk if the source is instead $\hat{\mathbb{C}}\setminus [-1,1]$, including infinity on the Riemann sphere. Near infinity, $w(z)=-1/(2z)+O(z^{-3})$, so the map extends with $w(\infty )=0$ and nonzero derivative in the coordinate $1/z$. This is the precise filled-in version of the conformal map of a segment complement to a punctured disk.

The Fourier coefficients satisfy $a_0=a_n=0$ andThus the square wave has Fourier seriesIt equals $f$ away from its jump discontinuities, and converges to $0$ at the jumps. The assigned endpoint values do not affect its Fourier coefficients.

For the Poisson equation, use separation of variables $\varphi ={\sum }_{n\text{ odd}}{b}_n{R}_n(r)\sin (n\theta )$. The radial equation is $r^2{R}_n^{\prime \prime }+r{R}_n^{\prime }-n^2{R}_n=r^2$. Its bounded solution with zero Dirichlet boundary condition at $r=1$ isIndeed the negative power is excluded at the centre, and the boundary fixes the coefficient of $r^n$. No resonance at $n=2$ occurs because only odd modes are forced. ThereforeThis is a Poisson equation on a disk with angular forcing. The series converges uniformly on the closed disk: for $n\ge 3$ each term is bounded by a constant times $n^{-3}$. Its first spatial derivatives also converge uniformly. For instance the radial derivative and $r^{-1}$ times the angular derivative of each term are bounded by constants times $n^{-2}$; the $n=1$ harmonic term is smooth in Cartesian coordinates. It defines a function with zero boundary data and the required weak formulation. Applying the Laplacian to finite partial sums gives the Fourier partial sums of $f$, which converge in $L^2$; passing to the limit against smooth compactly supported test functions verifies the Poisson equation. It holds classically off the jump rays and the centre, where the forcing is locally constant. It cannot be interpreted as a global classical equation with continuous second derivatives for discontinuous forcing. Uniqueness follows from the zero-boundary Dirichlet energy identity for the difference of two solutions.

Finally orthogonality gives ${\int }_0^{2\pi }f(\theta )\sin (n\theta )d\theta =\pi b_n$. The radial integral isUniform convergence of $\varphi$ and boundedness of $f$ justify integration term by term, soThe negative sign also agrees with $\int \varphi \Delta \varphi =-\int |\nabla \varphi {|}^2$. The original PDF's radial denominator is $n^2-4$; the converted TeX's $n^2-1$ is a transcription error.

Inside the infinite square well, the time-independent Schrödinger equation is $-{\hslash }^2{u}^{\prime \prime }/(2m)=Eu$ with $u(-a)=u(a)=0$. For $E<0$ the exponential solutions cannot satisfy both conditions except trivially, and for $E=0$ neither can a linear function. For $E>0$ put $k=\sqrt{2mE}/\hslash$. The first boundary condition selects $u=C\sin (k(x+a))$, and the second gives $\sin (2ak)=0$. Thuson $[-a,a]$, with $u_n=0$ outside. Their squared integrals are one, and the sine orthogonality identity gives an orthonormal set. The sine Fourier basis on the interval is complete: odd extension to an interval of length $4a$ reduces this to completeness of the Fourier series basis. Hence these are a complete set of normalized energy eigenstates.

Choosing the harmless sign of the second energy eigenstate so that $v_1=a^{-1/2}\cos (\pi x/(2a))$ and $v_2=a^{-1/2}\sin (\pi x/a)$, the specified wave function is $v_1/\sqrt{5}+2v_2/\sqrt{5}$. The Born rule therefore giveswith no other possible measured energies. The two energy eigenvalues are $E_1={\hslash }^2{\pi }^2/(8m{a}^2)$ and $E_2=4E_1$.

After the first measurement gives the ground state, the old state is $v_1$. Doubling the width puts the walls at $\pm 2a$, and the new normalized ground state is $w_1=(2a{)}^{-1/2}\cos (\pi x/(4a))$. The stipulated unchanged state is zero outside the original interval, so the probability amplitude isSquaring givesThe ground-state overlap after sudden expansion of a square well stays unchanged during subsequent evolution in the fixed new well, because the energy-basis coefficients acquire only phases. The answer is conditional on the already observed old ground state; its earlier probability $1/5$ is not another factor in this conditional question.

Fix the electric potential to vanish at infinity and assume a localized, nonsingular charge density of finite electrostatic energy. Assemble the charge by scaling ${\rho }_s=s\rho$ from $s=0$ to $1$. Linearity gives ${\varphi }_s=s\varphi$, so the work isThe factor $1/2$ prevents double counting of pair interactions. Using Gauss law, $\rho =-{ϵ}_0\Delta \varphi$, and integration by parts,The boundary integral at infinity vanishes for a localized distribution since $\varphi =O(r^{-1})$ and ${\partial }_r\varphi =O(r^{-2})$. Singular point-charge self-energies require a separate regularization and are not finite instances of this identity.

For the uniform solid sphere, spherical symmetry and Gauss law giveIntegrating inward from infinity and matching the continuous electric potential at $R$ givesThe value of the electric field at the centre is the zero vector.

Writing $Q=4\pi \rho R^3/3$, the electrostatic energy of a uniformly charged solid sphere follows either from the charge-potential integral or from the field integral including the exterior:For the nuclear scaling, $Q=Ze$ while constant volume per proton gives $R\propto Z^{1/3}$. Therefore the electric contribution scales as $Q^2/R\propto Z^{5/3}$ in this uniform-density model.

For potential flow $\mathbf{u}=\nabla \phi$ in an inviscid fluid of constant density under the conservative force per unit mass $-\nabla \chi$, the unsteady Bernoulli equation isIt follows by inserting the velocity potential into the Euler equations and integrating the resulting spatial gradient. A time-only change of the velocity potential changes $C(t)$ without altering the velocity field.

For spherical motion in an unbounded liquid, incompressibility gives $r^2{u}_r=A(t)$; the kinematic condition $u_r(R)=\dot{R}$ gives $A=R^2\dot{R}$. Choose $\phi =-R^2\dot{R}/r$, vanishing at infinity. To obtain the displayed bubble relation, take no spatial body-force contribution, neglect viscosity and surface tension, and equate the liquid-side interface pressure to $p_b$. With the far-field pressure $p_{\infty }$, Bernoulli equation evaluated at the surface uses the partial derivative at fixed $r$:Hence the Rayleigh equation for an inviscid spherical bubble is