Past exam of the mathematics course of the University of Cambridge / 2001 / ib / Paper 3

A norm on a real vector space $V$ is a map $p:V\to [0,\infty )$ satisfying $p(x)=0$ exactly when $x=0$, $p(tx)=|t|p(x)$ for real $t$, and $p(x+y)\le p(x)+p(y)$. Its induced metric is $d_p(x,y)=p(x-y)$. The three axioms respectively give definiteness, compatibility with scaling, and the triangle inequality.

Here equivalent metrics means that the two metrics induce the same topology, and Lipschitz equivalent norms means that there are constants $c,C>0$ such that $cp(x)\le q(x)\le Cp(x)$ for every $x$.

Suppose first that the inequalities hold. Since they also hold for $x-y$, the identity maps between the two metric spaces are Lipschitz continuous. In particular, each is continuous, so the topologies agree.

Conversely, equality of the topologies makes the identity from the $p$-norm to the $q$-norm continuous at zero. There is therefore $\delta >0$ such that $p(x)<\delta$ implies $q(x)<1$. For nonzero $x$, apply this to $\delta x/(2p(x))$ and use the absolute homogeneity of both norms:The reverse identity is also continuous, so there is $\epsilon >0$ with $q(x)<\epsilon \Rightarrow p(x)<1$. Rescaling again gives $p(x)<2q(x)/\epsilon$. ThusThe inequalities also hold at zero. This is topology determines norm equivalence; it works in arbitrary dimension because the rescaling argument needs neither a basis nor compactness of a unit sphere.

Expand using the bilinearity and symmetry of the real inner product:Addition proves the parallelogram law,For the maximum norm on ${\mathbb{R}}^2$, take $x=(1,0)$ and $y=(0,1)$. All four vectors $x,y,x+y,x-y$ have maximum norm one. The left side of the parallelogram law is then $2$, whereas the right side is $4$. The maximum norm cannot come from an inner product.

Use separation of variables in Laplace's equation: writing $\varphi =R(r)\Theta (\theta )$ givesSingle-valuedness requires $2\pi$-periodicity, so the nonconstant angular Fourier modes have integer $n\ge 1$ and are $\cos n\theta ,\sin n\theta$. Their radial solutions are $r^n,r^{-n}$. The zero angular mode has radial solutions $1,\log r$. These give the harmonic Fourier expansions in planar concentric domains used below.

For the annulus with $0<a<b$, constant boundary data suggest a radial harmonic function $A+B\log r$. Enforcing both boundary values givesThis has the required boundary values and satisfies $(r{\varphi }_r{)}_r=0$. It is unique: the difference of two continuous solutions is harmonic in the annulus and zero on both boundary circles, so the maximum principle for harmonic functions, applied to the difference and its negative, makes it zero.

Regularity at $r=0$ excludes every negative radial power and the logarithmic zero mode. Thus the general regular single-valued harmonic function on the disk has the formThe coefficients must give a series converging locally in the disk; if finiteness and continuity on its boundary are required, the boundary Fourier series must represent the prescribed finite continuous boundary values. Conversely every such locally convergent expansion is harmonic, and the Fourier coefficients on circles show that it includes every regular solution. There is no angular term linear in $\theta$, because it would be multivalued.

With the usual exterior-boundary interpretation that the harmonic function remains bounded as $r\to \infty$, positive radial powers and $\log r$ are excluded. HenceThe series must converge locally for $r>1$, with any required finite boundary values at $r=1$. These are the interior harmonic Fourier expansions in planar concentric domains after the inversion $r\mapsto 1/r$.

If “finite” means only finite at each finite point, without a condition at infinity, growing solutions such as $r\cos \theta$ and $\log r$ are allowed. Under that literal reading the general exterior expansion iswith local convergence. Boundedness at infinity is the condition selecting the decaying exterior expansion.

A convenient version of Rouché's theorem is: if $f,g$ are holomorphic functions on a neighbourhood of a closed disk, and $|g|<|f|$ on its boundary, then $f$ and $f+g$ have the same number of zeros, counted with multiplicity, in the disk. In particular neither has a boundary zero under this comparison.

Put $F_a(z)=z^4-a(z-1)(z^2-1)-1/2$ and use the boundary $|z|=\sqrt{2}$. There $|z^4|=4$, $1\le |z^2-1|\le 3$, and $\sqrt{2}-1\le |z-1|\le \sqrt{2}+1$.

For $a=1/3$, compare with $z^4$:Therefore Rouché's theorem gives four zeros.

For $a=12$, compare with $-12(z-1)(z^2-1)$:The comparison polynomial is $-12(z-1{)}^2(z+1)$, which has three zeros in the disk, counting the double zero at $1$.

For $a=5$, take $G(z)=(z^2-1)(z-2)(z-3)$. Expansion gives $F_5=G+1/2$. On the boundary,The last strict inequality follows from $3/2>\sqrt{2}$. Only the simple zeros $1,-1$ of $G$ lie inside the disk. Thus, in the requested order,

For a spherical triangle on a sphere of radius $R$, bounded by shorter great circle arcs and with interior angles $\alpha ,\beta ,\gamma$, the spherical excess formula, the triangular form of the Gauss-Bonnet theorem, isHere is an area proof. On the unit sphere, a spherical lune of angle $\alpha$ has area $2\alpha$: rotating around its vertices shows it occupies a fraction $\alpha /(2\pi )$ of the sphere when one uses the full angular width, or directly integrates ${\int }_0^{\alpha }{\int }_0^{\pi }\sin \theta d\theta d\phi$. The three side great circles of the triangle divide the sphere into eight regions, including the triangle $T$ and its antipodal triangle $-T$. Choose, at each vertex, the lune containing $T$, and also its antipodal lune. Each of $T$ and $-T$ is covered three times by these six lunes; each other region is covered once. Thus, with $A$ the area of $T$,This proves $A=\alpha +\beta +\gamma -\pi$; scaling lengths by $R$ scales area by $R^2$.

Radially project each face of the regular dodecahedron from its centre onto the unit sphere. Each edge and the origin lie in a plane, so its image is a shorter great circle arc. The twelve faces give congruent convex regular spherical polygons partitioning the sphere. At each projected vertex three congruent face angles fill the tangent plane. Hence each spherical pentagon has angle $2\pi /3$ at each of its five vertices. Splitting a pentagon into three spherical triangles and summing the spherical excess formula givesThis also agrees with $4\pi /12$, as required by the spherical partition.

The equality-constrained Lagrangian sufficiency theorem says: if $x^{\ast }$ is feasible and, for some fixed multipliers, it globally minimizes the optimization Lagrangian over all $x$, then it globally minimizes the objective over feasible $x$. On the feasible set the constraint terms vanish, so the two minimizations agree. No convexity assumption is needed for this version once global minimization of the optimization Lagrangian has been established.

Let $S_1={\sum }_i{a}_i$, $S_2={\sum }_i{a}_i^2$, and $\Delta =n{S}_2-S_1^2$. Sincethe two constraints are independent. For the squared Euclidean norm objective in the original PDF, takeCompleting the square shows that its unique global minimizer has $x_i=\lambda +\mu a_i$. The constraints become $n\lambda +S_1\mu =1$, $S_1\lambda +S_2\mu =0$, giving $\lambda =S_2/\Delta$, $\mu =-S_1/\Delta$. The Lagrangian sufficiency theorem therefore givesFor the value, use ${\sum }_i(x_i^{\ast }{)}^2=\lambda {\sum }_i{x}_i^{\ast }+\mu {\sum }_i{a}_i{x}_i^{\ast }=\lambda$. Uniqueness also follows directly: for every feasible $x$, the cross term in ${\sum }_i(x_i-x_i^{\ast }{)}^2$ vanishes, so the difference of objectives equals that strictly positive sum unless $x=x^{\ast }$. This is the minimum squared norm under two affine constraints.

Write $\omega (t)={\prod }_{i=0}^n(t-x_i)$. If $x$ is an interpolation node, the polynomial interpolation error is zero. Otherwise setThe function $F$ has $n+2$ distinct zeros: the $n+1$ interpolation nodes and $x$. Repeated Rolle's theorem therefore gives a point $\xi$ strictly between the smallest and largest of these points with $F^{(n+1)}(\xi )=0$. Since $p^{(n+1)}=0$ and ${\omega }^{(n+1)}=(n+1)!$, this yields the polynomial interpolation errorAll differentiations are justified by $f\in C^{n+1}[a,b]$. At an interpolation node the same displayed equality holds with any $\xi \in [a,b]$, because the product is zero.

Call the four given column vectors $v_1,v_2,v_3,v_4$. Direct componentwise calculation givesThe first two vectors are independent: their first two components give the determinant $1\cdot 2-4\cdot 2=-6\ne 0$. They form a basis of $W$, so $\boxed{\dim W=2}$.

Choose columns $v_1,v_2,0,0,-v_1-v_2$. This gives the matrixIts image is exactly $W$, since all its columns belong to $W$ and its first two span $W$. Its column sum is zero, so $(1,1,1,1,1{)}^T$ belongs to its kernel.

All linear maps with the stated kernel condition and image contained in $W$ factor uniquely through the quotient vector space ${\mathbb{R}}^5/\mathrm{span}\{(1,1,1,1,1{)}^T\}$. This quotient has dimension four. A linear map from it to the two-dimensional $W$ is specified by eight independent real coordinates, so the requested space has dimension $\boxed{8}$. Requiring the image to equal $W$, instead of being contained in $W$, would not define a vector space; the printed containment is essential.

Take positive vortex strength $\kappa$ to mean anticlockwise circulation. A line vortex at $z_0$ has local velocity potential $\kappa \arg (z-z_0)/(2\pi )$. The image vortex at a plane wall must have strength $-\kappa$ to enforce zero normal velocity at $y=0$. At the initial instant, a potential is thereforeThis is a local, multivalued velocity potential: its gradient is single-valued away from the vortex, but the nonzero circulation precludes a globally single-valued potential around it. For a vortex centred at $(X,Y)$, the velocity isThe real vortex and its opposite image give cancelling $u_y$ at the wall. A point vortex moves with the regular velocity induced by the other vortex, not its singular self-field. Evaluating the image field at $(X,b)$ gives $\dot{X}=\kappa /(4\pi b)$ and $\dot{b}=0$. HenceThe time-dependent velocity potential is obtained by replacing $a$ with $X(t)$ above; the image remains at $(X(t),-b)$. Reversing the convention for positive circulation reverses the translation direction.

Use completing the square in the Hermitian form $x^{\ast }Ax$, rather than unitary diagonalization of a normal matrix. The first pivot is one, soA second square completion givesTherefore one answer isThe diagonal entries of $D$ are rational, $\det T=1$, and the displayed identity for every $x$ proves $T^{\ast }DT=A$.

If $T$ were a unitary matrix, this matrix congruence would also be a similarity transformation, so the diagonal entries of $D$ would be the eigenvalues of $A$. Its characteristic polynomial isThus its eigenvalues are $3,3+\sqrt{10},3-\sqrt{10}$, two of which are irrational. A unitary factor cannot give a rational diagonal here. Alternatively, the permitted rational root theorem implies that three rational eigenvalues would all be integers, whereas the quadratic factor has no integer roots.

In the observer's rest frame, $U=(c,0)$ and $P=(E^{\prime }/c,{\mathbf{p}}^{\prime })$. The Lorentz invariance of the Minkowski inner product gives $P\cdot U=E^{\prime }$ andThe relation between four-momentum and velocity is ${\mathbf{p}}^{\prime }=E^{\prime }\mathbf{v}/c^2$, soTaking the nonnegative square root provesFor a future-directed massive particle and observer, $P\cdot U>0$ and the expression lies in $[0,1)$; the massless limit, with nonzero energy, gives $v=c$.

For the differentiability from continuous partial derivatives argument, choose a small box around zero inside the given open set. For $h=(h_1,\dots ,h_p)$ in that box, telescope $f(h)-f(0)$ by changing one coordinate at a time. The one-dimensional mean value theorem on the $j$th coordinate segment giveswhere ${\xi }_j$ lies on that segment, and therefore $\Vert {\xi }_j{\Vert }_2\le \Vert h{\Vert }_2$. A zero-length segment contributes zero and needs no choice of intermediate point. Set $L(h)={\sum }_j{f}_j(0)h_j$. For every $\epsilon >0$, continuity of the finitely many partial derivatives at zero gives $|f_j({\xi }_j)-f_j(0)|<\epsilon /\sqrt{p}$ when $h$ is sufficiently small. Then the Cauchy-Schwarz inequality givesThus $f$ is Fréchet differentiable at zero, with derivative $L$. No continuity of the partial derivatives away from zero was used.

For the supplied one-dimensional function,The derivative at zero follows from $g(s)/s=s\sin (1/s)\to 0$. For nonzero $s$ the derivative is continuous. At zero, the sequences $s_k=1/(2\pi k)$ and $t_k=1/((2k+1)\pi )$ give derivative values $-1$ and $1$. Since $f_x(x,y)=g^{\prime }(x)$, the partial derivative is continuous exactly where $x\ne 0$, irrespective of $y$.

Nevertheless $f$ is differentiable at every point of the plane. The one-dimensional function $g$ is differentiable at every real number. At a fixed $(x_0,y_0)$,Their sum has remainder $o(\sqrt{h^2+k^2})$, which proves Fréchet differentiability with derivative $(h,k)\mapsto g^{\prime }(x_0)h+g^{\prime }(y_0)k$. In particular, at the origin $|f(h,k)|\le h^2+k^2$, an immediate remainder bound. Continuity of all partial derivatives is sufficient, not necessary, for differentiability.

Assume $0<a<b<l$ for the interior pulse; endpoint variants use the same coefficient formula. Orthogonality of the Fourier sine basis givesThe Fourier sine series equals $f$ at its interior continuity points and takes value $1/2$ at the jumps $a,b$; the sine series is zero at both endpoints. Thus it does not retain an arbitrarily assigned value at a jump.

For the string, write $y(x,t)={\sum }_{n\ge 1}{q}_n(t)\sin (n\pi x/l)$. The wave equation on a string givesThe last equality is the Fourier sine series of the Dirac delta at $l/4$. Solving the modal ordinary differential equations givesThis impulsively struck fixed-end string is understood as a distributional weak solution; the initial velocity is a distribution, not a classical continuous function.

To evaluate the requested time, take the odd $2l$-periodic extension $v$ of $\delta (x-l/4)$ and use the D'Alembert formula $y=(2c{)}^{-1}{\int }_{x-ct}^{x+ct}v(s)ds$. At $ct=l/2$, the interval contains the positive impulse at $l/4$. For $x<l/4$, it also contains the negative reflected impulse at $-l/4$, so the contributions cancel. For $l/4<x<3l/4$ only the positive impulse contributes. For $x>3l/4$ neither contributes. Thus the midpoint-valued Fourier series givesThe discontinuity values are a convention for the Fourier representative and do not affect the distribution. The sketch shows the reflected negative front cancelling the left portion of the initially generated plateau.

The product topology on $X\times Y$ consists of arbitrary unions of rectangles $U\times V$ with $U\subseteq X$, $V\subseteq Y$ open. Such rectangles form a basis of a topology: every point lies in one, and intersections of two rectangles are rectangles with open factors.

If $d_X,d_Y$ induce the factor topologies, setDefiniteness and symmetry are immediate. Each factor triangle inequality bounds its distance by the sum of the two relevant product distances; taking the maximum proves the product triangle inequality. A ball of radius $r$ is exactly $B_X(x,r)\times B_Y(y,r)$, so every metric-open set is product-open. Conversely an open rectangle containing $(x,y)$ contains a product ball with radius the minimum of two suitable factor radii. Hence the maximum product metric induces precisely the product topology.

A compact space is one for which every open cover has a finite subcover. Let $\mathcal{U}$ cover $X\times Y$, with both factors compact; if either is empty the result is immediate. For each fixed $x$ and each $y$, choose a rectangle $U_{x,y}\times V_{x,y}$ containing $(x,y)$ and lying in a member of $\mathcal{U}$. The sets $V_{x,y}$ cover $Y$, so finitely many, indexed by $y_1,\dots ,y_{m_x}$, suffice. Their $X$-neighbourhoods have open intersectioncontaining $x$. The finitely many selected members of $\mathcal{U}$ cover all of $U_x\times Y$. Now the sets $U_x$ cover $X$, so select finitely many of these by compactness of $X$. Taking the union of their finite selections from $\mathcal{U}$ gives a finite cover of the whole product. The product of two compact spaces is compact, without requiring the factors to be Hausdorff.

In the upper half-plane model, hyperbolic lines are vertical Euclidean rays and semicircles centred on the real axis, equivalently circle arcs meeting the real boundary orthogonally. A semicircle with endpoints $r<s$ is mapped to the imaginary axis byIts determinant is one, and its real endpoints map to $0,\infty$. A vertical line $x=a$ is mapped to the imaginary axis by $z\mapsto z-a$. Thus every hyperbolic line can be carried to a fixed one; composing one such map with the inverse of another proves transitivity of $\mathrm{PSL}(2,\mathbb{R})$ on the lines.

Reflection in $x=a$ is $R_a(z)=2a-\overline{z}$. Reflection in the unit semicircle is $R_{\circ }(z)=1/\overline{z}$. The latter fixes the semicircle pointwise, is an involution, and preserves the hyperbolic metric because $|dw|=|dz|/|z{|}^2$ and $\mathrm{Im}w=\mathrm{Im}z/|z{|}^2$. The vertical reflection similarly preserves $|dz|/\mathrm{Im}z$. Their composition in the specified order is

Parallel hyperbolic lines have one common ideal endpoint and no interior intersection. Move the common endpoint to infinity by a Möbius transformation. The two lines are then distinct vertical lines $x=a,x=b$. Their hyperbolic reflections have productwhose $m$th power is $z+2m(a-b)$. Since $a\ne b$, no positive power is the identity.

Ultraparallel hyperbolic lines have no common endpoint or intersection. To see their common perpendicular explicitly, first map one line to the imaginary axis. The other has two boundary endpoints on the same side, say $0<r<s$ after possibly reflecting. It is a semicircle of centre $(r+s)/2$ and radius $(s-r)/2$. The circle centred at zero with radius $\sqrt{rs}$ meets it orthogonally, since $((r+s)/2{)}^2=rs+((s-r)/2{)}^2$, and also meets the imaginary axis orthogonally. It supplies the common perpendicular. Map that perpendicular to the imaginary axis. Orthogonality then makes both lines semicircles centred at zero, of distinct radii $r,s$. This normal form can also be obtained by a real Möbius transformation carrying the endpoints of the common perpendicular to $0,\infty$. Their reflections are $z\mapsto r^2/\overline{z}$ and $z\mapsto s^2/\overline{z}$, so their product isNo positive power of this dilation is the identity because $r^2/s^2>0$ and $r\ne s$. Thus the product has infinite order in both nonintersecting cases. Geometrically the ultraparallel product is a translation along the common perpendicular through twice the separation of the lines.

If two hyperbolic lines intersect, move their intersection to the centre of the Poincaré disk model. Lines through the centre are diameters. Reflection in a diameter making Euclidean angle $\alpha$ with the real axis is $z\mapsto e^{2i\alpha }\overline{z}$. The composition of two such reflections is the rotation $z\mapsto e^{2i(\alpha -\beta )}z$. The Poincaré disk model is conformal, so the Euclidean angle difference is the hyperbolic intersection angle $\theta$, up to sign and replacing it by $\pi -\theta$, neither of which changes rationality of $\theta /\pi$.

The rotation has finite order exactly when $m\theta \in \pi \mathbb{Z}$ for some positive integer $m$, equivalently $\theta /\pi \in \mathbb{Q}$. If $\theta /\pi =r/s$ in lowest terms, its order is $s$. Conversely, the nonintersecting cases in part(a) have infinite order. Therefore

Use the original PDF constraints, whose right sides are $8$ and $9$. Introduce nonnegative surplus variables $s_1,s_2$, slack $s_3$, and artificial variables $a_1,a_2$:All seven variables are nonnegative. Phase I of the two-phase simplex method minimizes $w=a_1+a_2$ with initial basis $(a_1,a_2,s_3)$. The initial simplex dictionary isEnter $x_2$ and leave $a_2$: the ratio test gives $\min (8,9/3)=3$. Substituting $x_2=3-x_1/3+s_2/3-a_2/3$ givesNow enter $x_1$ and leave $a_1$. The bounds from $a_1,x_2,s_3$ are $3,9,6$, so the ratio test selects $3$. The new dictionary isThe associated basic solution has $a_1=a_2=0$ and $w=0$, which is a global minimum because artificial variables are nonnegative. Delete their columns for Phase II. In the remaining basis $(x_1,x_2,s_3)$ the objective isFor $2\le p\le 9/4$, both reduced costs are nonnegative, so the simplex method has reached the optimum at $(x_1,x_2)=(3,2)$:For $9/4<p\le 3$, the coefficient of $s_1$ is negative. Enter $s_1$; the ratio test gives $10$ from $x_2$ and $5$ from $s_3$, so $s_3$ leaves. Eliminating $s_1$ givesand henceThe two reduced costs are now positive, so $(6,1)$ is optimal andAt $p=9/4$ the whole feasible edge from $(3,2)$ to $(6,1)$ is optimal; both value formulas agree. These dictionaries explicitly carry out both phases rather than inferring the answer solely from a feasible-region sketch.

The monic orthogonal polynomial $p_n$ is orthogonal to every polynomial of degree less than $n$, since $p_0,\dots ,p_{n-1}$ form a basis of that polynomial space. Let $r_1,\dots ,r_k$ be its distinct interior roots at which its sign changes, and put $q(x)={\prod }_{j=1}^k(x-r_j)$, with $q=1$ if $k=0$. The real polynomial $p_{n}q$ has constant sign throughout $(a,b)$: multiplication by each linear factor removes its corresponding sign change, while even-multiplicity roots do not change sign. It is not identically zero. Since $w>0$, its weighted integral is nonzero.

If $k<n$, orthogonality instead gives ${\int }_a^b{p}_{n}qwdx=0$, a contradiction. Thus $k\ge n$. Degree $n$ also gives $k\le n$, so $k=n$ and all roots have multiplicity one, with no roots left outside the interval. Therefore there are exactly $n$ distinct simple zeros in $(a,b)$. The weighted inner products are understood to exist, as required by the stipulated orthogonality.

Let $d_n(x)=\det (xI-A_n)$, and set $d_0=1$, $d_1=x-a_1$. Expansion of the tridiagonal determinant along its last row givesFor the second term, the last-row entry is $-b_n$, its cofactor has negative sign, and the remaining minor has last-column entry $-b_n$ multiplying $d_{n-2}$; their product is $-b_n^2{d}_{n-2}$. Thus $d_n$ and $p_n$ have identical initial values and recurrence, and induction provesThe eigenvalues of the Jacobi matrix $A_n$ are precisely the roots of this characteristic polynomial. Part(a) therefore gives all $n$ eigenvalues simple and lying strictly in $(a,b)$. This conclusion follows from orthogonality as well as symmetry; symmetry alone would establish real eigenvalues but not their interval location.

Work in the finite-dimensional vector space $V$ implicit in the requested matrix representation and $n=\dim V$. Put $K_j=\ker {\alpha }^j$, with $K_0=0$. These form an increasing sequence ending in $K_m=V$, and $\alpha (K_j)\subseteq K_{j-1}$. Choose a basis of $K_1$, extend it to a basis of $K_2$, and continue until a basis of $V$ is obtained. Each basis vector introduced at stage $j$ maps into the span of vectors chosen at earlier stages. Its matrix column can therefore have nonzero entries only in earlier rows. This proves that the matrix is strictly upper triangular.

For a strictly upper-triangular $n\times n$ matrix $N$, a nonzero contribution to $(N^r{)}_{ij}$ requires an index chain $i<i_1<\cdots <i_{r-1}<j$. No such chain has $n$ links among $n$ indices. Consequently $\boxed{{\alpha }^n=0}$. An example with exact index four is the Nilpotent Jordan blockHere $M^3$ has entry one in position $(1,4)$ and all others zero, while $M^4=0$.

In the adapted basis, $I+A$ is upper triangular with diagonal entries one, so all its eigenvalues are one. This does not persist for the product of two such matrices with unrelated adapted bases. TakeBoth square to zero, buthas characteristic polynomial $t^2-3t+1$ and eigenvalues $(3\pm \sqrt{5})/2$, neither equal to one. The obstruction is that the two nilpotent matrices need not admit a common triangularizing basis.