Past exam of the mathematics course of the University of Cambridge / 2007 / ib

In the given basis, the matrix is $A=\left(\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right)$. Let $\omega =e^{2\pi i/3}$, a primitive cube root of unity. For each $\xi \in \{1,\omega ,{\omega }^2\}$,Thus these are eigenvectors with distinct eigenvalues $1,{\omega }^2,\omega$, respectively, and form an eigenbasis. SetThe columns are mutually orthogonal with squared norm three, so $P^{-1}=P^{\ast }/3$. The diagonalization and characteristic polynomial areScaling $P$ by $1/\sqrt{3}$ gives a unitary diagonalizing matrix.

For a geodesic spherical triangle on a sphere of radius $R$, the Gauss-Bonnet theorem giveswhere $\alpha ,\beta ,\gamma$ are its interior angles. Indeed, its geodesic sides have zero geodesic curvature, its Gaussian curvature is $R^{-2}$, and its three exterior turning angles sum to $3\pi -(\alpha +\beta +\gamma )$.

Choose a point strictly inside a full-dimensional convex polyhedron and radially project its boundary onto the unit sphere. Triangulate each projected face by diagonals, using only its existing vertices. Convexity ensures that the resulting spherical triangles cover the sphere once, with disjoint interiors. A face with $k$ edges gives $k-2$ triangles, so their number is $T={\sum }_{\mathrm{faces}}(k-2)=2E-2F$, since every original edge belongs to two faces.

Sum the spherical excess formula over all triangles. The total area is $4\pi$, and the angles incident to each projected vertex sum to $2\pi$. No additional vertices were introduced by the face triangulations. HenceThis is the spherical-triangle proof of the polyhedron Euler formula.

The partial fraction decomposition is $f(z)=1/(z-i)+1/(z+i)$. Put $w=z-1$ and expand each reciprocal by a geometric series:Adding gives the Taylor seriesEquivalently, since $1\pm i=\sqrt{2}{e}^{\pm i\pi /4}$, its coefficient is $(-1{)}^n{2}^{(1-n)/2}\cos ((n+1)\pi /4)$. The first terms are $1-\frac{1}{2}{w}^2+\frac{1}{2}{w}^3-\frac{1}{4}{w}^4+\cdots$.

The nearest nonremovable poles are $i$ and $-i$, both at distance $\sqrt{2}$ from the center. The function is holomorphic throughout the open disk of that radius and cannot be holomorphic across either pole. Therefore the largest radius of convergence is $r=\sqrt{2}$.

Use the position four-vector $X^a=(ct,\mathbf{x})$ and Minkowski signature $(+,-,-,-)$. Along a massive particle's timelike path,Thus its four-velocity is $U^a=d{X}^a/d\tau =\gamma (c,\mathbf{v})$, and multiplying by the invariant rest mass givesIts temporal component is $E/c$, and the invariant relation is $(E/c{)}^2-|\mathbf{p}{|}^2=M^2{c}^2$.

For a massless particle, use the null energy–momentum relation rather than substituting $M=0$ into $M\gamma$: a massless particle has no proper-time parametrization of a timelike rest frame. With positive energy, $p^a{p}_a=0$ implies $E/c=|\mathbf{p}|$. HenceThe original PDF retains the spatial vector in this expression; the TeX conversion incorrectly replaced it by its magnitude.

The vorticity is the curl of the velocity, perpendicular to the flow plane:The planar rotation rate is half this scalar vorticity, not the vorticity itself.

The velocity divergence is ${\partial }_x{u}_x+{\partial }_y{u}_y=\alpha -\alpha =0$, satisfying the constant-density incompressible flow condition. Choose the stream function convention $(u_x,u_y)=({\psi }_y,-{\psi }_x)$. Integration then givesIndeed, ${\psi }_y=\alpha x-\beta y$ and $-{\psi }_x=\beta x-\alpha y$. Streamlines are contours of this function. The vector field is also compatible with steady inviscid momentum balance without body forces: $(\mathbf{u}\cdot \nabla )\mathbf{u}=({\alpha }^2-{\beta }^2)(x,y)$ is balanced by pressure $p=p_0-\rho ({\alpha }^2-{\beta }^2)(x^2+y^2)/2$.

Rotate coordinates to $s=(x+y)/\sqrt{2}$ and $t=(x-y)/\sqrt{2}$. The stream function becomesFor $\alpha <\beta$, both coefficients are negative, so every nonzero contour is a bounded ellipse surrounding the stationary origin. For $\alpha >\beta$, the form is indefinite: nonzero contours are unbounded hyperbolas, and the zero contour consists of straight separatrices. Consequently all nonstationary streamlines are bounded exactly when $\alpha <\beta$. The origin is itself a stationary bounded trajectory in every case; the assertion about boundedness concerns the nonstationary streamlines. The equality case is handled separately below. These are the linear planar strain and rotation streamlines.

If $\alpha =\beta$, $\psi =-\alpha (x-y{)}^2/2$. Its contours are the parallel lines $x-y=d$. On each such line,For $d\ne 0$ this is an unbounded straight streamline. Every point on $x=y$ has zero velocity, so that diagonal is a line of equilibria, rather than one moving streamline.

The first matrix column is already the unit vector $e_1$, so the first QR elimination needs no reflection. In the remaining three coordinates, the second column is $v=(2,2,1{)}^T$, with norm three. To map it to $(-3,0,0{)}^T$, use the Householder transformationIt is symmetric and orthogonal and satisfies $Hv=(-3,0,0{)}^T$. Extend it to $Q^T=\mathrm{diag}(1,H)$. The QR decomposition and transformed right side areFor the lower right-side block, $w^T(1,4,-1{)}^T=12$, so its transformed value is $(1,4,-1{)}^T-(12/15)w=(-3,12/5,-9/5{)}^T$.

Orthogonality preserves the least-squares residual norm. Thus the first two squared residual terms can and must be set to zero, while the last two are independent of the unknowns:The minimum squared residual is $(12/5{)}^2+(-9/5{)}^2=9$. The solution is unique since the triangular factor has nonzero diagonal entries. No normal equations are used.

Write $\bar{X}=n^{-1}{\sum }_i{X}_i$, $Q={\sum }_i(X_i-\bar{X}{)}^2$, and $Q_0={\sum }_i(X_i-{\mu }_0{)}^2=Q+n(\bar{X}-{\mu }_0{)}^2$. The normal likelihood is proportional to $({\sigma }^2{)}^{-n/2}\exp [-{\sum }_i(X_i-\mu {)}^2/(2{\sigma }^2)]$. Its unrestricted maximizers are $\hat{\mu }=\bar{X}$, ${\hat{\sigma }}^2=Q/n$. Under the null the variance maximizer is $Q_0/n$. Substitution gives the generalized likelihood-ratio test statisticThe ratio decreases as $|T|$ increases, so the likelihood-ratio rejection region is two-sided in $T$. Under $H_0$, $\sqrt{n}(\bar{X}-{\mu }_0)/\sigma$ is standard normal and $Q/{\sigma }^2$ is independent chi-squared with $n-1$ degrees of freedom. Independence follows by resolving the isotropic Gaussian sample vector into its projection on the constant vector and its orthogonal complement. Hence $T$ has the exact Student t-distribution with $n-1$ degrees of freedom.

For a chosen significance level $\alpha$, compute the sample mean and the sample standard deviation with denominator $n-1$, and reject precisely whenEquivalently, report the two-sided $p$-value $2[1-F_{t_{n-1}}(|T|)]$. This requires $n\ge 2$ and positive population variance; $S>0$ holds almost surely under that model. The nuisance variance is removed by the exact null distribution, so one should not substitute a normal critical value or a large-sample chi-squared approximation. This is the normal-mean likelihood ratio with unknown variance.

For a finite directed flow network with nonnegative finite edge capacities, a source $s$ and a distinct sink $t$, the max-flow min-cut theorem states that the largest feasible flow value equals the smallest capacity of a cut separating $s$ from $t$. A feasible flow has $0\le f_e\le c_e$ on each directed edge and flow conservation at all other vertices. Its value is the net outflow at $s$.

For any cut $S$ containing $s$ but not $t$, sum conservation over vertices of $S$. Internal edges cancel, givingThus every flow value is bounded by every cut capacity.

A maximum flow exists: the feasible set is nonempty, closed and bounded in a finite-dimensional edge-coordinate space, hence compact, and the flow value is continuous. For such a flow, form its residual network: each original edge has forward residual capacity $c_e-f_e$ and backward residual capacity $f_e$. Keep these as edge-labelled arcs if the original network already has edges in both directions. If a directed positive-residual path joined $s$ to $t$, increasing the flow along its forward arcs and decreasing it along its backward arcs by the path's positive bottleneck capacity would preserve feasibility and increase the value. That contradicts maximality.

Let $S$ be the vertices reachable from $s$ by positive residual arcs. Then $t\notin S$. Every original edge from $S$ to $S^c$ is saturated, otherwise its forward residual arc would extend reachability. Every original edge from $S^c$ to $S$ has zero flow, otherwise its backward residual arc would extend reachability. ThereforeTogether with the bound for every flow and cut, this provesThe compactness step makes the proof valid for arbitrary real capacities; it does not assume that a particular augmenting-path algorithm terminates after finitely many augmentations.

Sylvester's law of inertia states that a real quadratic form can be brought by an invertible real change of coordinates towith $r$ remaining zero coordinates, and that the integers $(p,q,r)$ are uniquely determined. To prove existence, represent it by a symmetric matrix. The allowed orthogonal diagonalization yields real eigenvalues, and scaling each nonzero eigen-coordinate by the square root of its absolute eigenvalue changes its coefficient to $+1$ or $-1$.

For uniqueness, $p$ is the largest dimension of a subspace on which the form is positive definite. The positive coordinate subspace attains $p$. Any subspace of dimension greater than $p$ intersects the $(q+r)$-dimensional span of the negative and zero coordinates nontrivially, by the dimension formula. A nonzero vector in that intersection has nonpositive quadratic value, contradicting positive definiteness. Similarly, $q$ is the maximal dimension of a negative-definite subspace. These descriptions are independent of coordinates, and $r=\dim V-p-q$ is then fixed. This proves existence and uniqueness, including degenerate forms.

For the final application, nonsingularity gives $p+q=2m$. In diagonal coordinates write $v=(v_{+},v_{-})$, with $Q(v)=|v_{+}{|}^2-|v_{-}{|}^2$. If $v\in U$ projects to zero in the positive coordinates, $Q(v)=0$ forces $v_{-}=0$, so projection $U\to {\mathbb{R}}^p$ is injective. The same argument applies to the negative coordinates. Hence $m\le p$ and $m\le q$, and their sum is $2m$. ThereforeThe inertia signature is $(m,m)$, or signature difference $p-q=0$ if that convention is used. This is the isotropic dimension bound from real inertia.

The Fundamental theorem of finitely generated abelian groups says that every finitely generated abelian group is isomorphic toThe rank of an abelian group $r$ and the invariant factors $d_j$ are uniquely determined, with the finite summand allowed to be absent. Equivalently, one can decompose the torsion subgroup into cyclic groups of prime-power orders. The infinite cyclic part and torsion part must both be included in the statement.

A nonzero polynomial of degree $n$ over a field has at most $n$ distinct roots in that field. For a proof, the factor theorem gives $(t-a)\mid f(t)$ whenever $f(a)=0$. Distinct linear factors are coprime in $K[t]$, a unique factorization domain, so for distinct roots $a_1,\dots ,a_k$ their product divides $f$. Its degree is $k$, giving $k\le n$. One may equivalently remove one linear factor at a time and argue by induction.

If the field has at least $n$ elements, this bound is attained by a product of $n$ distinct linear factors. For a fixed finite field with fewer than $n$ elements, the largest possible number of distinct roots is $\min (n,|K|)$ instead; multiplicities do not create more distinct roots. The zero polynomial is excluded since it has no finite degree.

Let $G$ be a finite subgroup of $K^{\times }$. It is abelian since multiplication in the field commutes. The structure theorem therefore writes $G$ as a product of cyclic groups with invariant factors $d_1\mid \cdots \mid d_s$, omitting trivial factors. Every element satisfies $g^{d_s}=1$, so $G$ consists of roots of $t^{d_s}-1$. The root bound from part (ii) giveswhile the decomposition gives $|G|={\prod }_j{d}_j\ge d_s$. Equality forces every factor except the last to be trivial. Hence $G$ is cyclic. The trivial subgroup is cyclic as well. This is the finite multiplicative subgroup of a field is cyclic theorem, proved here using the preceding two parts.

The infinite case is false. In ${\mathbb{C}}^{\times }$, the group of all roots of unity is infinite, but every element has finite order. If it were cyclic, its generator would have finite order and the whole group would be finite, a contradiction.

Differentiability at $p$ means there is a linear map $A=Df{|}_p$ such thatThis is the Fréchet derivative definition; the same linear approximation must work in every direction, not merely along coordinate axes.

The chain rule says that if $f$ is differentiable at $p$ and $g$ is differentiable at $f(p)$, then $g\circ f$ is differentiable at $p$ andFor its proof, put $A=Df{|}_p$, $B=Dg{|}_{f(p)}$, and $k=f(p+h)-f(p)=Ah+r(h)$. Since a linear map between these finite-dimensional normed spaces is bounded, $\Vert k\Vert =O(\Vert h\Vert )$. Differentiability of $g$ gives $g(f(p)+k)=g(f(p))+Bk+s(k)$ with $s(k)=o(\Vert k\Vert )$. ThereforeIf $k=0$, take $s(0)=0$; otherwise $\Vert s(k)\Vert /\Vert h\Vert =(\Vert s(k)\Vert /\Vert k\Vert )(\Vert k\Vert /\Vert h\Vert )\to 0$. This proves the required full linear approximation.

For the Euclidean norm, at every $x\ne 0$ the ordinary one-coordinate differentiation givesAt the origin its difference quotient in direction $e_i$ is $|t|/t$, whose right and left limits are $1$ and $-1$. Thus none of its coordinate partial derivatives exists at the origin.

The product topology on $X\times Y$ has as a basis the rectangles $U\times V$, where $U$ is open in $X$ and $V$ is open in $Y$. Equivalently, it is the coarsest topology making both coordinate projections continuous.

To show that the product of two Hausdorff spaces is Hausdorff, take distinct points $(x_1,y_1)$ and $(x_2,y_2)$. If $x_1\ne x_2$, choose disjoint open neighborhoods $U_1,U_2$ in $X$; the rectangles $U_1\times Y$ and $U_2\times Y$ separate the product points. If the first coordinates agree, the second coordinates differ, and the same argument uses disjoint open neighborhoods in $Y$. Thus every pair of distinct product points has disjoint open neighborhoods.

The three implications below prove the requested equivalence without assuming metrizability.

Assume $X$ is Hausdorff. For any $(x_1,x_2)$ outside the diagonal, the points differ, so choose disjoint open neighborhoods $U_1,U_2$. Their rectangle contains $(x_1,x_2)$ and misses the diagonal. Thus the complement of the diagonal is open in the product topology, proving (i) implies (ii). This is one direction of the closed diagonal theorem.

Assume the diagonal $\Delta \subset X\times X$ is closed. For continuous $f,g:Y\to X$, the map $h:Y\to X\times X$, $h(y)=(f(y),g(y))$, is continuous: the inverse image of a basic rectangle $U\times V$ is $f^{-1}(U)\cap g^{-1}(V)$. Thereforeis closed as the inverse image of a closed set under a continuous map. This proves (ii) implies (iii).

Assume (iii). Take $Y=X\times X$ and let $f,g$ be the two coordinate projections, continuous by the definition of the product topology. Their equality set is the diagonal, so it is closed. For distinct $x_1,x_2$, the point $(x_1,x_2)$ lies in its open complement. Choose a basic rectangle $U_1\times U_2$ containing it and contained in that complement. If $U_1\cap U_2$ were nonempty, a point $u$ in their intersection would give $(u,u)$ in the rectangle, impossible. Thus $U_1,U_2$ are disjoint open neighborhoods of $x_1,x_2$, proving that $X$ is Hausdorff. Hence (iii) implies (i), completing all three equivalences.

Integrate the entire function $e^{i{z}^2}$ around the positively oriented sector boundary: the real segment $0\to R$, the arc $R{e}^{i\theta }$ for $0\le \theta \le \pi /4$, and the radial segment back to zero. Cauchy integral theorem givesThe last term follows from $z=r{e}^{i\pi /4}$ and $i{z}^2=-r^2$; its minus sign is the return orientation.

On the arc, $|e^{i{z}^2}|=e^{-R^2\sin (2\theta )}$. The supplied sine bound gives $\sin (2\theta )\ge 4\theta /\pi$ throughout this sector, soThe Gaussian integral then yields ${\int }_0^{\infty }{e}^{i{x}^2}dx=e^{i\pi /4}\sqrt{\pi }/2$. Taking real and imaginary parts proves both Fresnel integrals:The contour estimate proves existence of the improper oscillatory integrals as well as their values; absolute convergence is neither asserted nor required.

Use the Fourier transform conventionDecay alone does not guarantee existence of the ordinary integrals, so impose the usual integrability and regularity assumptions. One sufficient version of the Fourier inversion theorem is that $f$ and $\tilde{f}$ are integrable and $f$ is continuous, in which case the inversion integral recovers $f$. Under the standard piecewise-smooth Dirichlet assumptions, inversion instead gives the midpoint of the two one-sided limits at a jump, with a suitable improper integral interpretation.

For integrable $f,g$, define $(f\ast g)(x)={\int }_{\mathbb{R}}f(x-y)g(y)dy$. The convolution theorem statesTo prove it, absolute integrability permits Fubini's theorem. Set $t=x-y$ in the inner integral:The absolute double integral is $\Vert f{\Vert }_1\Vert g{\Vert }_1$, justifying the interchange. The computations below assume $a,b>0$ and use the source PDF's factor $e^{ikx}$ in the final integral, omitted in the converted TeX.

Split the exponential kernel at zero. For $a>0$,Thus its Fourier transform isBoth half-line integrals converge absolutely because $a>0$.

The interval indicator has transformwith continuous value $2b$ at $k=0$. By the convolution theorem, $h=f\ast g$ has transform $4a\sin (bk)/[k(a^2+k^2)]$. Fourier inversion therefore expresses the requested integral asFor $x>b$, $|x-y|=x-y$, so $h(x)=2e^{-ax}\sinh (ab)/a$. Symmetry gives the same expression with $|x|$ for $x<-b$. For $|x|\le b$, splitting the integral at $y=x$ givesConsequently the value for every real $x$ isThe two expressions agree at $x=\pm b$. This is the exponential kernel convolved with an interval indicator. The spectral integral is absolutely convergent: its apparent singularity at zero is removable, and the tail decays as $O(|k{|}^{-3})$.

For a stationary spherically symmetric state, set $u(r)=r\psi (r)$. The Schrödinger equation becomesA bound state must have $E<0$, while the well's lower bound gives $E>-U$ for a nonzero square-integrable eigenstate. Define $k=\sqrt{2m(U+E)}/\hslash$ and $\kappa =\sqrt{-2mE}/\hslash$. Regularity at the origin requires $u(0)=0$, so the interior cosine solution is excluded; square integrability at infinity excludes the growing exterior exponential. Matching $u$ and $u^{\prime }$ at $r=a$ therefore givesThe limit at the origin is the finite value $Ck$. Continuity of $u,u^{\prime }$ is equivalent here to continuity of $\psi ,{\psi }^{\prime }$ at the finite potential jump. If desired, the normalized state hasThe equation for $k$ implicitly determines the energy, with $k^2+{\kappa }^2=2mU/{\hslash }^2$.