Past exam of the mathematics course of the University of Cambridge / 2001 / ib / Paper 4

The strict distance inequality also gives $d(fx,fy)\le d(x,y)$ when $x=y$, so $f$ is Lipschitz continuous with constant at most one, hence continuous. Put $g(x)=d(x,f(x))$. The triangle inequality givesThus $g$ is also Lipschitz continuous and continuous.

Assume the compact metric space is nonempty, as needed for existence of a point. By the extreme value theorem, $g$ attains its infimum at some $p$. If $h=g(p)>0$, then $p\ne f(p)$, and strict distance decrease givescontradicting minimality. Since distances are nonnegative, $h=0$ and $f(p)=p$. If two different fixed points $p,q$ existed, then $d(p,q)=d(f(p),f(q))<d(p,q)$, another contradiction. Therefore there is exactly one fixed point. This is strict distance decrease on a compact metric space; no uniform contraction factor less than one was assumed.

Differentiate the Legendre differential equation once and put $R_n=P_n^{\prime }$:Multiplying by $1-x^2$ gives its Sturm-Liouville form,Thus the weight is $w(x)=1-x^2$, not one. For $n\ge 1$, $R_n$ is a nonzero polynomial; $R_0=0$ is a trivial solution rather than a nonzero eigenfunction.

Multiply the equations for $n,m$ by $R_m,R_n$ respectively, subtract and integrate. The resulting boundary term is $[(1-x^2{)}^2(R_n^{\prime }{R}_m-R_m^{\prime }{R}_n){]}_{-1}^1=0$, since all these derivatives are finite at the endpoints. For distinct nonnegative indices, ${\lambda }_n-{\lambda }_m\ne 0$, giving the weighted orthogonality of Legendre polynomial derivatives:For $n=0$ the integral is zero simply because $R_0=0$.

Assume $S_{xx}={\sum }_i{x}_i^2>0$, so the slope in this linear regression through the origin is identifiable. Expanding the residual sum of squares givesDifferentiation gives the unique least-squares estimator and its normal distribution:For $n\ge 2$, estimate the unknown variance byThe fitted component and orthogonal residual of the Gaussian vector are independent, and $(n-1)s^2/{\sigma }^2$ has the chi-squared distribution with $n-1$ degrees of freedom. Under the null hypothesis,The Student t test for regression through the origin rejects at significance level $\alpha$ when $|T|>t_{n-1,1-\alpha /2}$, where the subscript denotes the corresponding quantile of Student's t-distribution. If all $x_i=0$, the data contain no information about $\beta$; if $n=1$, no residual degree of freedom is available for this unknown-variance test.

Parametrize the positively oriented circle by $z=r{e}^{i\theta }$. The Cauchy derivative formula at the origin becomesTaking the complex conjugate of the Cauchy integral theorem and using $d\bar{z}=-ir{e}^{-i\theta }d\theta$ givesAdd this zero integral to the derivative integral. Since $f+\bar{f}=2\mathrm{Re}f=2u$, recovering an analytic derivative from the boundary real part yieldsThe complex conjugation on the integrand and the reversed sign in $d\bar{z}$ are both required; the circle's orientation itself has not been reversed.

In a two-player zero-sum game, entry $a_{ij}$ is player I's gain, and player II's loss, when pure strategies $i,j$ are chosen. Player I maximizes the expected payoff, player II minimizes it. A mixed strategy is a probability vector. Sufficient optimality conditions for $p,q$ and value $v$ areIndeed $q$ keeps every opposing payoff at most $v$, while $p$ guarantees at least $v$ against every opposing mixed strategy; their joint payoff must equal $v$.

A pure strategy for the informed player specifies one action after each possible outcome. Order these contingent choices as $(P,P),(P,B),(B,P),(B,B)$, with the first action for a head. The columns are fold and call. Averaging over the fair coin gives the payoff matrix, in currency units,For example, row $(B,P)$ against call averages a gain of two and a loss of one, giving $1/2$.

TakeThen $Aq=(-1,-1,1/3,1/3{)}^T$ and $p^{T}A=(1/3,1/3)$, so the sufficient inequalities hold with game value $v=1/3$ to player I. In behavioral terms, I always bets on a head and bets with probability $1/3$ on a tail; II calls with probability $2/3$. The four rows describe complete contingent plans, not four coin outcomes.

Using ${\chi }_M(t)=\det (tI-M)$, expansion or triangular reordering givesThe eigenspace for $1$ is spanned by $e_1$, so its algebraic multiplicity two requires a size-two Jordan block. A convenient Jordan chain isFor eigenvalue $2$, the eigenspace has basis $w_1=e_1+e_3$, $w_2=e_4$, hence both corresponding blocks have size one. Therefore a Jordan normal form and the minimal polynomial areThe ordered basis $(v_1,v_2,w_1,w_2)$ supplies the change-of-basis matrixIts determinant is one, and the chain equations give $MP=PJ$, proving $P^{-1}MP=J$. The square at $t=1$ in the minimal polynomial is essential: $M$ is not diagonalizable.

For constant density $\rho$, the Euler equations for an inviscid fluid with a conservative force per unit mass areSet $\mathbf{\omega }=\nabla \times \mathbf{u}$. The vector identity $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times \mathbf{\omega }$ and vanishing curl of a gradient givewhere both divergence terms vanish. Thus the vorticity equation isThe partial time derivative measures Eulerian change; $(\mathbf{u}\cdot \nabla )\mathbf{\omega }$ is advection; the right-hand side describes stretching and tilting of vortex lines by velocity gradients. A conservative body force supplies no curl source, and constant density removes any baroclinic pressure source.

For planar velocity independent of $z$, vorticity is perpendicular to the plane, $\mathbf{\omega }=(0,0,\omega )$, and $(\mathbf{\omega }\cdot \nabla )\mathbf{u}=\omega {\partial }_z\mathbf{u}=0$. Hence each material particle conserves its vorticity, $D\omega /Dt=0$. Choose the stream function convention $u={\psi }_y$, $v=-{\psi }_x$. ThenIn a steady flow, ${\psi }_y{\omega }_x-{\psi }_x{\omega }_y=0$, so vorticity is constant along streamlines. Where $\nabla \psi \ne 0$, use $\psi$ as a local transverse coordinate: this equality makes $\omega$ independent of the coordinate along a streamline. ConsequentlyThis is steady planar vorticity as a local function of stream function.

A global single-valued $F$ additionally requires compatible vorticity values on all components of each stream-function level set. That condition is not supplied by steady Euler flow alone. For example, $\psi =x^3-x$ gives $(u,v)=(0,1-3x^2)$, a smooth steady incompressible shear with zero advective acceleration and constant pressure. Yet $\psi (0,y)=\psi (1,y)=0$ while $\Delta \psi (0,y)=0$ and $\Delta \psi (1,y)=6$. Thus the unrestricted global deduction is false; the local result, or a suitable connected-level-set hypothesis, is the valid conclusion.

Multiplying numerator and denominator by the same nonzero constant does not change a Möbius transformation. Since $b\ne 0$, divide by $b$ and relabel the coefficients so that $b=1$. On $|z|=1$, preservation of the circle givesEquating the constant and first Fourier coefficients yieldsIf $c=0$, the second equality forces $a=0$, contradicting the nonzero determinant. Hence $d=\bar{a}/\bar{c}$. With $C=|c{|}^2>0$, the first equality becomesThe root $C=|a{|}^2$ would imply $ad=|a{|}^2/\bar{c}=c$, making the numerator and denominator proportional. It is excluded by nondegeneracy. Thus $|c|=1$ and $d=c\bar{a}$. Writing $1/c=e^{i\psi }$ givesHere $a$ is the normalized coefficient after division by the original $b$. The inequality $|a|\ne 1$ is again the determinant condition. For $|a|<1$ the map preserves the disc; for $|a|>1$ it interchanges its inside and outside. Circle preservation alone does not justify restricting to automorphisms of the unit disk.

The Einstein's principle of relativity says that the laws of physics take the same form in every inertial frame: no inertial frame is physically privileged, and uniform absolute motion cannot be detected by an internal experiment.

Assume $|u|<c$. Direct expansion of the transformed coordinates givesSince $y^{\prime }=y$ and $z^{\prime }=z$, the full Minkowski interval is invariant. Points of the outgoing light shell therefore satisfyAt each fixed $t^{\prime }>0$ this is a sphere of radius $c{t}^{\prime }$. The future branch remains future-directed: for a shell point with $t>0$, $|x|\le ct$ gives $t^{\prime }=\gamma (t-ux/c^2)>0$.

A Lorentz transformation is a linear coordinate transformation preserving the Minkowski metric, equivalently ${\Lambda }^T\eta \Lambda =\eta$. The calculation establishes this property for all events, not only points on the light cone. The displayed map has determinant one and preserves time orientation, so it is the proper orthochronous boost along the $x$ direction.

Set $e_n(x)=|f_n(x)-f(x)|$. The functions $e_n$ are continuous, converge pointwise to zero, and decrease with $n$. This remains true even if the direction of monotonicity of $f_n(x)$ differs between points: at each fixed point a monotone convergent real sequence approaches its limit from one side.

Given $\epsilon >0$, the sets $U_n=\{x:e_n(x)<\epsilon \}$ are relatively open in $[a,b]$, increase with $n$, and cover the interval. Compactness gives a finite subcover; choosing the largest index in it gives $U_N=[a,b]$. For every $n\ge N$ and every $x$, $e_n(x)\le e_N(x)<\epsilon$. Hence $f_n\to f$ uniformly. This proves the required version of Dini's theorem, rather than assuming a common direction of monotonicity everywhere.

On $[0,1)$, take $f_n(x)=x^n$. Each function and its pointwise limit $f=0$ are continuous, and the sequence decreases at every point. NeverthelessThus uniform convergence fails on the noncompact interval although all the other conditions hold.

The radius of the surface of revolution is $|y(x)|$, so its area and the curve's arc length areFor the proposed curve, $y<0$ on the interior. On this branch $A=-2\pi S$ with $S=\int y\sqrt{1+y^{\prime 2}}dx$. Extremizing $S$ subject to $L=l$ is therefore equivalent to making the actual area stationary. Introduce a Lagrange multiplier $\lambda$ and the augmented integrand $F=(y+\lambda )\sqrt{1+y^{\prime 2}}$. The Beltrami identity givesFor the fixed-length stationary surface of revolution, takeIndeed $y^{\prime }=\sinh (kx)$ and $\sqrt{1+y^{\prime 2}}=\cosh (kx)$, so the Beltrami equation holds identically. It also follows directly that $F_y=\cosh (kx)$ and $d(F_{y^{\prime }})/dx=\cosh (kx)$, verifying the Euler-Lagrange equation. The endpoints vanish, and the length condition becomesFor endpoint-fixed variations preserving this length, the augmented first variation is zero; subtracting the zero length variation proves that the area's first variation is zero. Its value, if desired, is $A=\pi [\sinh (2ka)-2ka]/k^2$.

To prove uniqueness, put $s=ka>0$. The equation is $\sinh s/s=l/(2a)>1$. Its derivative is $(s\cosh s-\sinh s)/s^2$. The numerator is zero at $s=0$ and has derivative $s\sinh s>0$, so the ratio increases strictly from one to infinity. The intermediate value theorem gives exactly one positive $k$. The stated upper bound on $l$ is not needed for this uniqueness argument. Stationarity has been proved, without making an unsupported global-minimum claim.

Assume $n\ge 2$ and ${\sigma }^2>0$. Up to a constant, the log-likelihood of the normal sample isFor each $s$, the square is minimized at $\mu =\bar{X}$; optimizing the remaining one-variable likelihood gives the maximum-likelihood estimatorsThey are the global maximum when the residual sum is positive, which occurs with probability one. On the exceptional all-equal sample the likelihood increases without bound as $s\downarrow 0$.

To derive the distributions and independence, write $Z_i=(X_i-\mu )/\sigma$ and perform an orthogonal change of coordinates whose first row is $n^{-1/2}(1,\dots ,1)$. The transformed components $Z_1^{\prime },\dots ,Z_n^{\prime }$ remain independent standard normal random variables, since their joint density depends only on the squared Euclidean norm. ThenThe first expression uses only $Z_1^{\prime }$, the second only the other components. ConsequentlyThis also exhibits why the maximum-likelihood variance uses divisor $n$, even though its chi-squared degrees of freedom are $n-1$.

The independent future observation satisfies $X_0-\bar{X}\sim N(0,{\sigma }^2(1+1/n))$. It is independent of the residual sum of squares, because both $X_0$ and $\bar{X}$ are independent of the residual vector. Combining this normal variable with the independent chi-squared variable givesThe event in the proposed prediction interval is equivalent to $|T|<\gamma \sqrt{(n-1)/n}$. By symmetry and continuity of Student's t-distribution, exact joint coverage $1-\alpha$ therefore requiresThis is the normal prediction interval with maximum-likelihood variance. The factor $\sqrt{n/(n-1)}$ cannot be dropped: the interval uses $\hat{\sigma }$, not the unbiased residual standard deviation. Coverage here is over repeated joint sampling of the old and new observations, not a claim of parameter-free conditional coverage for every realized sample.

Write the Laurent series on the punctured disc as $f(z)={\sum }_{j=-\infty }^{\infty }{a}_j{z}^j$. A removable singularity means that $f$ extends holomorphically to the origin; exactly all negative-index coefficients vanish. A pole of order $k\ge 1$ means that $z^{k}f(z)$ extends with nonzero value at zero; exactly $a_{-k}\ne 0$ and every coefficient below $-k$ vanishes. An essential singularity is neither of these cases; exactly infinitely many negative-index coefficients are nonzero.

Here is a proof of the required Casorati-Weierstrass theorem consequence. If some disc of values around $w$ were missed in a punctured neighbourhood, then $|f(z)-w|\ge \epsilon >0$ there. The holomorphic function $q(z)=1/(f(z)-w)$ would be bounded, and hence have a removable singularity. If its extension had $q(0)\ne 0$, then $f=w+1/q$ would extend holomorphically. If $q(0)=0$, its zero has a finite positive order, since $q$ is not identically zero; thus $f$ would have a pole. Both contradict an essential singularity. Therefore every sufficiently small punctured neighbourhood has values arbitrarily close to every $w$. Choosing $0<|z_n|<\min (r/2,1/n)$ with $|f(z_n)-w|<1/n$ provesThe open mapping theorem states that a nonconstant holomorphic function on a domain maps open sets to open sets. If $f$ were injective with an essential singularity, choose a small open disc $U$ away from zero and a point $z_0\in U$. Its open image contains a disc about $w=f(z_0)$. The sequence just constructed has $z_n\to 0$ outside $U$, but eventually $f(z_n)\in f(U)$. Thus some $y_n\in U$ has $f(y_n)=f(z_n)$ although $y_n\ne z_n$, contradicting injectivity. An injective holomorphic function cannot have an isolated essential singularity.

For an injective entire function $g$, apply this to $f(z)=g(1/z)$. If its singularity at zero were removable, $g$ would be bounded outside a large disc, hence bounded everywhere and constant by Liouville's theorem. This contradicts injectivity. It must therefore be a pole, say of order $k$, so $|g(z)|\le C|z{|}^k$ for large $|z|$. For any Taylor coefficient of index $j>k$, the Cauchy coefficient formula gives $|a_j|\le C{R}^{k-j}$ on arbitrarily large circles; hence $a_j=0$. Thus $g$ is a nonconstant polynomial. Its derivative never vanishes by the allowed injectivity fact. If its degree exceeded one, the fundamental theorem of algebra would give a zero of this nonconstant derivative. Therefore injective entire functions are affine:

Measure quantities in thousands of copies. Use five supply rows: initial inventory, regular production in month 1, overtime in month 1, regular production in month 2, overtime in month 2. Their supplies/capacities are $(10,40,150,40,150)$. The three demand columns are the two due dates and a dummy column for unused capacity, with demands $(40,60,290)$, balancing the total supply $390$.

The transportation problem has cost matrixAn infinite entry prohibits delivering a later month's production to an earlier deadline. The extra $20$ in the second column of the early rows accounts for holding stock through the first month. For production rows, the dummy column represents capacity not used, not copies actually printed. Initial inventory assigned to the dummy column would instead remain in stock through both months, costing $40$; this distinguishes physical inventory from optional production capacity. No gratuitous disposal is needed in the optimum.

Let $x_{ij}\ge 0$ have the stated row and column sums, with the forbidden entries zero, and minimize $\sum c_{ij}{x}_{ij}$. A feasible solution isThus regular production is $40$ in each month; overtime is zero then $10$; first-month closing inventory is $10$, and final inventory is zero. Its total production-plus-holding cost isTo prove optimality, use transportation dual potentials $u=(-430,-30,0,-50,0)$ for rows and $v=(430,450,0)$ for columns. Every allowed edge satisfies $u_i+v_j\le c_{ij}$. Multiplying by any feasible shipment and summing gives the lower boundThe proposed shipment attains it, so the schedule is optimal. The dual calculation certifies the minimum without depending on an unshown transportation-simplex iteration.

Put $C=A+yI$. For every $y$ except the finitely many roots of $\det (A+yI)$, $C$ is invertible and $CB$ is similar to $BC$, since $C^{-1}(CB)C=BC$. HenceBoth sides are polynomials in $y$. Agreement for infinitely many $y$ makes them identical, so evaluating at $y=0$ proves Characteristic polynomials of AB and BA:Their minimal polynomials need not agree. Forwe have $AB=B\ne 0$, $B^2=0$, and $BA=0$. The two minimal polynomials are respectively $t^2$ and $t$, although both characteristic polynomials are $t^2$.

For any $p(t)={\sum }_j{p}_j{t}^j$, the identity $(BA{)}^{j+1}=B(AB{)}^{j}A$ givesTaking $p=m_{AB}$ makes the right-hand side zero. Every annihilating polynomial is divisible by the minimal polynomial, soIf $AB$ is diagonalizable, $p$ has only simple roots. Therefore each nonzero eigenvalue of $BA$ has only size-one Jordan blocks, and the zero-eigenvalue blocks have size at most two. Squaring kills the latter nilpotent blocks, so $(BA{)}^2$ is diagonalizable, as is $(AB{)}^2$.

By their common characteristic polynomial, $AB$ and $BA$ have the same eigenvalues counted with algebraic multiplicity. Squaring produces the same multiset of squared eigenvalues, including combined multiplicities if two eigenvalues have the same square. The two diagonalizable squares therefore have the same diagonal normal form up to permutation. This proves similarity of squared matrix products with one diagonalizable product:It does not imply that $AB$ and $BA$ themselves are conjugate: the possible size-two nilpotent blocks at zero are exactly what squaring removes.

Assume nonzero upstream speed, so $F>0$. For steady constant-density flow, mass conservation gives the constant volume fluxThe Bernoulli equation states that $u^2/2+p/\rho +gz$ is constant along each streamline of a steady inviscid constant-density flow under gravity. It need not have the same value across unrelated streamlines in a general rotational flow. Here uniform upstream velocity supplies the same head, and the slowly varying, nearly horizontal channel permits hydrostatic pressure $p=p_{\mathrm{atm}}+\rho g(h-z)$. Neglecting losses, the Bernoulli equation therefore givesSubstitute $u=UWH/(wh)$ and set $s=h/H$, $F=U^2/(gH)$, the square of the upstream Froude number. Rearrangement givesFor positive depth, this cubic increases to its maximum at $s_{\ast }=(F+2)/3$ and then decreases until it reaches zero at $(F+2)/2$. Its maximum is $G(s_{\ast })=(F+2{)}^3/(27F)$. Consequently the critical width of a rectangular open-channel contraction isAt equality the two positive-depth branches coalesce and the local Froude number equals one.

When $w>w_c$ everywhere, a smooth solution connected to the upstream state cannot change branch without passing through that coalescence. At $w=W$ the upstream depth is $s=1$, with $G^{\prime }(1)=2(F-1)/F$. If $F<1$, this point is on the descending, subcritical branch. At a wider location $w>W$, the left-hand side falls below $1=G(1)$, so the depth on that branch must increase: $h>H$ for subcritical upstream flow $F<1$. If $F>1$, the upstream state is on the ascending, supercritical branch and decreasing the left-hand side decreases $s$: $h<H$ for supercritical upstream flow $F>1$.

For $F=1$, $w_c=W$ and the upstream state is critical. A literally constant-width upstream region then conflicts with the strict assumption $w>w_c$ everywhere. If the upstream state is interpreted only as an asymptotic critical limit, a widened channel has one solution with $h<H$ and one with $h>H$; a further branch condition is required. Merely solving the cubic at one location, without retaining the upstream-connected branch, would not determine the physically continued depth.

Jordan's lemma states that, for $b>0$, if $g(z)\to 0$ uniformly on large upper semicircles, then $\int g(z)e^{ibz}dz\to 0$ on those arcs. Indeed its absolute value is at most $R\max |g|{\int }_0^{\pi }{e}^{-bR\sin \theta }d\theta \le (\pi /b)\max |g|$. For negative $b$, use lower semicircles. The corresponding exponential-decay estimates can also be used on rectangular closures.

First take $a>0$ and $0<\epsilon <a$, and defineUse finite strip rectangles with vertical sides at $\pm R$, $R=N+1/2$, then let $N\to \infty$. Their orientation is positive. At a nonzero integer, the residue isThe origin is removable when $a\ne 0$ and contributes zero. The short vertical-side integrals tend to zero since $|\sin \pi (\pm R+iy)|=\cosh (\pi y)$ and $F=O(R^{-1})$ there. The residue theorem therefore gives the first evaluation,The sum is understood as the limit of symmetric partial sums. It converges for $|x|<\pi$ by the Dirichlet test: the oscillatory partial sums are bounded, and $n/(a^2+n^2)$ decreases eventually to zero.

For the second evaluation, close the top line outward into the upper half-plane and the bottom line outward into the lower half-plane. Because of the strip orientations, both outward closures are clockwise. The upper region contains only the pole $ia$, and the lower one only $-ia$. Their residues areTo justify discarding the added boundaries, use rectangles with far vertical sides at $\pm (N+1/2)$ and outer horizontal sides at imaginary height $\pm R$. The numerator grows at most like $e^{|x||\mathrm{Im}z|}$, while $1/\sin \pi z$ decays like $e^{-\pi |\mathrm{Im}z|}$. Thus the horizontal integral tends exponentially to zero, and the vertical integrals are bounded by a constant times $R^{-1}{\int }_{\epsilon }^R{e}^{-(\pi -|x|)y}dy\to 0$. The strict condition $|x|<\pi$ is essential. The clockwise residue calculation givesEquating the two evaluations proves the alternating sine series with a quadratic denominator:Both sides are even functions of $a$, so this proves the result for every nonzero real $a$. At $a=0$ the displayed zero-index summand and right-hand ratio are literally undefined; under their natural continuous extension, set that summand to zero and take the right-hand limit $-x$. This limit may be passed through the nonzero-index sum, since its difference from the $a=0$ sum is bounded by $2a^2{\sum }_{n\ge 1}{n}^{-3}$. Thus the intended limiting identity at zero is $2{\sum }_{n\ge 1}(-1{)}^n\sin (nx)/n=-x$. The separate contour evaluation above assumes distinct off-axis poles and should not be applied unchanged after they coalesce at the origin.

Conservation of momentum in the parent's rest frame gives $p_A+p_B=0$. Equal daughter masses and the relativistic energy-momentum relation then give equal daughter energies. Energy conservation fixesUsing $E^2=p^2{c}^2+m^2{c}^4$ and the specified direction of A,The velocities follow from $v=p{c}^2/E$:These are spatial momenta in the original inertial frame. For positive daughter rest mass, $m<M/2$ gives $0<V<c$.

Before its second decay A moves uniformly with $x=-Vt$. Its Lorentz factor isThe proper time along its world line satisfies $d\tau =dt/{\gamma }_A$. Since the first decay occurs at the spacetime origin, integration givesEquivalently these are the components of A's constant four-velocity times $\tau$. The assumption $m>0$ is needed for a daughter rest frame and its proper-time clock; a massless daughter has neither.