Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 1

The characteristic polynomial of the given matrix is $(t-1)(t-2{)}^2$. Solving the two eigenvalue equations givesFor example, multiplying the matrix by these three vectors gives respectively $1$, $2$ and $2$ times the original vector. The eigenvectors form a basis: the matrix with these columns has nonzero determinant $-1$. Thus an eigenbasis and its diagonal matrix are

Write $J_r(\lambda )$ for a size-$r$ Jordan block. The characteristic polynomial fixes the total dimensions at $2$ and $-7$, while the minimal polynomial fixes the largest Jordan block at each eigenvalue. Consequently, the blocks for $2$ have total size four, largest size two, and at least one size-two block. The blocks for $-7$ both have size one. The complete list, up to similarity, isThese Jordan normal forms are inequivalent because their $2$-eigenspaces have different dimensions, two and three.

At the origin, the Taylor series $\sin z=z+O(z^3)$ and $1-\cos z=z^2/2+O(z^4)$ giveThus $f_1$ has a pole of order three and $f_2$ a pole of order four at zero. For the third function, the Laurent serieshas infinitely many nonzero negative-power terms, so zero is an essential singularity of $f_3$.

To classify the point at infinity on the Riemann sphere, use the coordinate $w=1/z$. Thenso $f_3$ has a simple pole at infinity.

For the other two functions, an important qualification is necessary. The poles of $f_1$ at nonzero integer multiples of $\pi$, and those of $f_2$ at nonzero integer multiples of $2\pi$, approach infinity. Equivalently, their images in the $w$ coordinate accumulate at zero. Hence infinity is a non-isolated singularity of $f_1$ and $f_2$. Neither function is holomorphic on any punctured neighbourhood of infinity, so the classification of isolated singularities into removable singularities, poles and essential singularities does not apply there.

The Poincare disc model is the open unit disc $D=\{z=x+iy:|z|<1\}$ equipped with the Riemannian metricThis normalization gives Gaussian curvature $-1$. In particular, a Euclidean curve has hyperbolic length $\int 2|dz|/(1-|z{|}^2)$; the boundary circle is infinitely far away in hyperbolic distance.

A suitable Poincare disc automorphism is the Möbius transformationIt sends $a$ to zero and has inverse $(w+a)/(1+\overline{a}w)$. To verify the isometry directly, note thatThese identities show both that $f$ maps the unit disc onto itself and that $2|df|/(1-|f{|}^2)=2|dz|/(1-|z{|}^2)$, preserving the Riemannian metric.

The radial segment is a geodesic of the Poincare disc model. Its hyperbolic length isOne can also verify that it minimizes length of a curve: along any piecewise smooth curve from zero to radius $r$, the Euclidean speed is at least the absolute radial speed. Integrating $2|dr|/(1-r^2)$ is at least the absolute change in $\log ((1+r)/(1-r))$, with equality on the radial segment. Thus the answer is independent of $\theta$.

Use an isometry of the Poincare disc model to move the centre to zero. A hyperbolic disc of radius $\rho$ then has Euclidean radius $R=\tanh (\rho /2)$ by the hyperbolic distance formula. The area element of a surface of the Riemannian metric is $4rdrd\theta /(1-r^2{)}^2$, soTherefore the area is independent of the centre, and

For a twice differentiable function, convexity on ${\mathbb{R}}^2$ is equivalent to its Hessian matrix being positive semidefinite everywhere. Here the Hessian matrix is constant:Its diagonal entries are positive, so positive semidefiniteness is equivalent to $\det H=8-{\alpha }^2\ge 0$. The required parameter range isThe endpoints give convex functions with one flat direction; strict inequalities give a Hessian matrix defining a positive-definite quadratic form and a strictly convex function.

Completing the square in the optimization constraint givesThe left side is nonnegative, so every feasible point has $x_1^2\ge 4$. Equality is feasible precisely when $x_2=3x_1/2$ and $x_1=\pm 2$. Thus the global minimum isThere is no maximum: for every $|x_1|\ge 2$, the values $x_2=3x_1/2\pm \sqrt{x_1^2/4-1}$ satisfy the optimization constraint, allowing $x_1^2$ to be arbitrarily large. This direct argument also establishes that the two stationary points are the only local minima on the two branches of the constraint curve.

Put $r^2=x^2+y^2$. On the specified domain, $r>0$, and direct differentiation gives the divergence and vorticityFor example, the third component of the curl is $A(r^2-2x^2)/r^4+A(r^2-2y^2)/r^4=0$; the other two components vanish because $U$ depends only on $z$ and the transverse velocity is independent of $z$. Hence the flow is both incompressible and irrotational exactly whenThe exclusion of the axis matters: these are ordinary derivatives on the domain, with no assertion of regularity on the removed axis.

In cylindrical coordinates, the transverse velocity is $A{\mathbf{e}}_{\theta }/r$, so its contribution to the circulation is $Ad\theta$. The vertical contribution is $U(z)dz$, the differential of an antiderivative of $U$, and therefore has zero integral around any closed curve. If $k$ is the oriented winding number of the curve's projection about the origin, thenIn particular, a single positive winding gives $2\pi A$; reversing orientation reverses the sign. The nonzero circulation is compatible with irrotational flow because the domain is not simply connected. For $A\ne 0$, an enclosing curve cannot bound a surface contained entirely in the domain to which Stokes theorem would apply.

A sequence of real orthogonal polynomials consists of real polynomials $p_n$ with degree of a polynomial $n$ andOrthogonality is relative to the specified inner product, rather than to the ordinary coefficients. A monic polynomial has leading coefficient one, so a monic orthogonal polynomial has the form $p_n(x)=x^n+\text{terms of lower degree}$. In particular, the monic degree-zero polynomial is $p_0=1$.

Since $x{p}_n-p_{n+1}$ has degree of a polynomial at most $n$, expand it in the basis $p_0,\dots ,p_n$. Multiplication by $x$ is self-adjoint for this weighted inner product: $⟨xp,q⟩=⟨p,xq⟩$. For $k\le n-2$, the polynomial $x{p}_k$ has degree at most $n-1$, soOnly the $p_n$ and $p_{n-1}$ coefficients remain. Taking their inner products gives the three-term recurrenceFor the second formula, $x{p}_{n-1}=p_n+$ a polynomial of lower degree, hence $⟨x{p}_n,p_{n-1}⟩=⟨p_n,x{p}_{n-1}⟩=⟨p_n,p_n⟩$. At $n=0$, take $p_{-1}=0$, ${\beta }_0=0$ and the same formula for ${\alpha }_0$. Positivity of the weight ensures that all the squared norms in the denominators are nonzero.

Let $S={\sum }_{i=1}^n{X}_i$. The likelihood ratio of rate two to rate one is $2^n{e}^{-S}$, which decreases strictly with $S$. By the Neyman-Pearson lemma, the most powerful test rejects for small values of the sum:Under the null, $S$ has the gamma distribution with shape $n$ and rate one. Thus $c_{\alpha }$ is its $\alpha$-quantile, equivalently half the $\alpha$-quantile of the chi-squared distribution with $2n$ degrees of freedom. The distribution is continuous, so boundary randomization is unnecessary for $0<\alpha <1$.

For every fixed alternative $\lambda >1$, the likelihood ratio against rate one is ${\lambda }^n{e}^{-(\lambda -1)S}$, again strictly decreasing. Therefore the same region is most powerful against every such alternative. Moreover, if $G$ has the shape-$n$, rate-one gamma distribution, then $S$ at rate $\lambda$ has the same distribution as $G/\lambda$. It follows that ${\mathbb{P}}_{\lambda }(S<c_{\alpha })=\mathbb{P}(G<\lambda c_{\alpha })$ increases with $\lambda$, so the maximum rejection probability over $\lambda \le 1$ is $\alpha$ at one. Any competing size-$\alpha$ test for the composite null is also of level at most $\alpha$ at rate one, and is bounded in power by the Neyman-Pearson lemma for each alternative. The test is uniformly most powerful for the composite hypotheses as well. This is the uniformly most powerful test for an exponential rate based on the sample sum.

Let $p$ be the row player's mixed strategy and $q$ the column player's mixed strategy. The fourth row is strictly dominated by the first. A useful candidate mixes the first two rows so that their first and third column payoffs agree. If the first row has probability $t$, these payoffs are $7t-2$ and $2-7t$, giving $t=2/7$ and common value zero.

To prove that these are optimal mixed strategies, use the two explicit mixed strategiesTheir payoff vectors satisfyThus the row player guarantees a payoff at least zero against every column mixed strategy, and the column player guarantees a payoff at most zero against every row mixed strategy. This matching pair proves that the value of the zero-sum game is zero.

These optimal mixed strategies are also unique. Since $Aq$ is strictly negative in rows three and four, any optimal row strategy gives them probability zero. Against columns one and three it must then satisfy both $7t-2\ge 0$ and $2-7t\ge 0$, so $t=2/7$. Similarly, $p^{T}A$ is strictly positive in column two, forcing an optimal column strategy to avoid that column; its constraints against rows one and two then force equal weights on columns one and three.

For matrices $A,B$ and scalars $r,s\in \mathbb{F}$, the definition of the trace givesHence the trace is a linear functional on the vector space of $n\times n$ matrices.

The definition of matrix multiplication givesThe middle equality uses commutativity of scalar multiplication in $\mathbb{R}$ or $\mathbb{C}$, together with interchange of finite sums. Thus the trace is unchanged by cyclically moving the two factors.

To prove that trace is the unique normalized cyclic linear functional, let $E_{ij}$ be the matrix units. For $i\ne j$, $E_{ii}{E}_{ij}=E_{ij}$ whereas $E_{ij}{E}_{ii}=0$, so the assumed identity for the linear functional gives $f(E_{ij})=0$. Also, $E_{ij}{E}_{ji}=E_{ii}$ and $E_{ji}{E}_{ij}=E_{jj}$, giving $f(E_{ii})=f(E_{jj})$. Write this common value as $c$. Since the matrix units form a basis,For $n=1$ the same conclusion follows immediately from the one-dimensional basis $E_{11}$. Finally, $f(I)=nc$; if this equals $n$, then $c=1$ over either specified field. The normalization makes $f$ exactly the trace.

Every commutator $[A,B]=AB-BA$ has zero trace, by the cyclic identity. Hence its linear span $W$ is contained in $\ker \mathrm{Tr}$. For the reverse inclusion, the matrix units satisfyThe off-diagonal matrix units and the differences $E_{ii}-E_{nn}$ for $i<n$ span all traceless matrices: a traceless diagonal part is ${\sum }_{i<n}{a}_{ii}(E_{ii}-E_{nn})$. Thus every element of the kernel is a linear combination of commutators, provingThis establishes that traceless matrices are spanned by commutators. For $n=1$, both vector subspaces are zero. Notice that the argument needs a span of commutators; it does not assume that every such sum has already been written as a single commutator.

We prove prime avoidance by contradiction. Remove any prime ideal contained in another one on the list; this leaves the same union. If the ideal $I$ is contained in none of the remaining prime ideals, choose $a_i\in I\setminus P_i$. For every $j\ne i$, incomparability lets us choose $b_{ij}\in P_j\setminus P_i$. SetFor a one-element list, use the empty product $b_1=1$. By the defining property of a prime ideal, $b_i\notin P_i$, and also $a_i{b}_i\notin P_i$. But $b_j\in P_i$ for every $j\ne i$. Thus all summands except $a_i{b}_i$ lie in $P_i$, and consequently $c\notin P_i$ for every $i$. On the other hand, $c\in I$ because each $a_i\in I$. This contradicts the assumed containment in the union. Therefore

The evaluation map $\phi :\mathbb{R}[x]\to \mathbb{C}$, $p\mapsto p(i)$, is a ring homomorphism and is onto because $a+bx$ maps to any $a+bi$. By polynomial division, writeThen $p(i)=0$ precisely when $a=b=0$. Thus $\ker \phi =(x^2+1)$, and the first isomorphism theorem givesSince the quotient ring is a field, $(x^2+1)$ is a maximal ideal.

Evaluation at a real point defines a surjective ring homomorphism $p\mapsto p(a,b)$ from $\mathbb{R}[x,y]$ to $\mathbb{R}$. Its kernel is $(x-a,y-b)$: subtracting the constant $p(a,b)$ expresses the remainder as a multiple of $x-a$ plus a multiple of $y-b$, by successive polynomial division. The first isomorphism theorem therefore identifies the quotient ring with the field $\mathbb{R}$, proving each $I_{a,b}$ is maximal.

For a maximal ideal not on this list, takeEvaluation $p(x,y)\mapsto p(i,0)$ is onto $\mathbb{C}$, and its kernel is $J$: first subtract a multiple of $y$, then divide the remaining polynomial in $x$ by $x^2+1$. Hence $\mathbb{R}[x,y]/J\cong \mathbb{C}$, a field. Finally, if $J=I_{a,b}$ for real $a,b$, the member $x^2+1\in J$ would vanish at $(a,b)$, requiring $a^2+1=0$, which is impossible over $\mathbb{R}$.

A Cauchy sequence in a metric space satisfies: for every $\epsilon >0$ there is $N$ such that $d(x_m,x_n)<\epsilon$ whenever $m,n\ge N$. Suppose a subsequence $x_{n_k}$ converges to $x\in X$. Choose $N$ so that the Cauchy bound is $\epsilon /2$, and then choose $k$ with $n_k\ge N$ and $d(x_{n_k},x)<\epsilon /2$. For every $n\ge N$, the triangle inequality givesThus the whole Cauchy sequence converges to the subsequential limit.

For fixed $m$, the reverse triangle inequality givesThe right side tends uniformly to zero as $n,k\to \infty$ because $(x_n)$ is a Cauchy sequence. Hence $(d(x_m,x_n){)}_n$ is a real Cauchy sequence. By completeness of $\mathbb{R}$, it has a finite, nonnegative limit:This conclusion uses completeness only of the real line, without assuming that $X$ is complete.

Given $\epsilon >0$, the Cauchy sequence property gives $N$ with $d(x_m,x_n)<\epsilon /2$ for all $m,n\ge N$. Fixing $m\ge N$ and letting $n\to \infty$ yields $0\le d_m\le \epsilon /2<\epsilon$. Therefore

Write the subsequence as $y_n=x_{k_n}$ with strictly increasing indices $k_n$. If $y_{\ell }=x_m$, then $d(y_{\ell },y_n)=d(x_m,x_{k_n})$. This is a subsequence of the real sequence converging to $d_m$, soThe conclusion depends on the point $x_m$ itself, even if that point appears at several indices.

Apply the triangle inequality with a third point $x_k$:Letting $k\to \infty$ gives the desired bounds. Interchanging $m,n$ also gives the sharper symmetric lower bound:

First handle possible zero values. If $d_m=d_n=0$, the previous upper bound forces $x_m=x_n$. Thus, if infinitely many $d_m$ vanish, choosing those indices gives a constant subsequence, which satisfies both inequalities.

Otherwise, discard the finitely many zero values and use $d_m\to 0$ to choose increasing indices $k_j$ withSet $y_j=x_{k_j}$. The preceding bounds give $|{\delta }_j-{\delta }_{\ell }|\le d(y_j,y_{\ell })\le {\delta }_j+{\delta }_{\ell }$. For $m\ge 2$,For $m<n$, use ${\delta }_n\le {\delta }_m/8$ and ${\delta }_{n+1}\le {\delta }_m/64$ to obtainThe same holds when $n<m$ by symmetry; when $m=n$, both distances are zero. This proves both required inequalities. The adjacent-distance inequality is understood for $m\ge 2$, since this subsequence starts at index one and has no $y_0$.

We argue by contradiction using the previous subsequence construction. If $X$ is not complete, choose a Cauchy sequence with no limit in $X$. None of its $d_m$ can be zero: if $d_m=0$, then $d(x_m,x_n)\to 0$, giving a limit in $X$. The construction therefore produces pairwise distinct $y_j$ withLet $Y=\{y_1,y_2,\dots \}$. This subset is closed in $X$. Indeed, if a point $x\notin Y$ lay in its closure, there would be points of $Y$ tending to $x$. Their indices must be unbounded, because the distance from $x$ to any finite subset of $Y$ is positive. Passing to increasing indices gives a convergent subsequence of the original Cauchy sequence. Part (a) would then make the whole sequence converge to $x$, a contradiction.

Now define $F:Y\to Y$ by $F(y_j)=y_{j+1}$. Distinctness makes this well-defined, and the displayed inequality makes it a contraction mapping with constant at most $1/2$. It has no fixed point, contradicting the assumed property of closed subsets. ConsequentlyThis is the converse direction of contraction property of closed subsets characterizes completeness.

For $q\in \mathbb{Q}\setminus \{0\}$ write $q=p^{k}u/v$, where $k\in \mathbb{Z}$ and neither integer $u$ nor integer $v$ is divisible by $p$. Define the P-adic valuation $v_p(q)=k$ and the p-adic absolute value $|q{|}_p=p^{-k}$, with $|0{|}_p=0$. The p-adic metric isFactoring out the smaller power of $p$ from two rational numbers shows $v_p(q+r)\ge \min \{v_p(q),v_p(r)\}$, with the usual convention $v_p(0)=+\infty$. Consequently the ultrametric inequality givesTogether with symmetry, positivity and separation of points, this verifies that $d_p$ is a metric.

For the given geometric partial sums, use the finite geometric series identity. Since $p$ does not divide $p-1$,Thus the limit is the rational number $\boxed{1/(1-p)}$, even though the partial sums diverge in the ordinary real metric.

If $|a{|}_p>|b{|}_p$, the ultrametric inequality gives $|a+b{|}_p\le |a{|}_p$. Applying it to $a=(a+b)-b$ gives $|a{|}_p\le \max \{|a+b{|}_p,|b{|}_p\}$. Since $|b{|}_p<|a{|}_p$, this forces $|a+b{|}_p=|a{|}_p$. Interchanging $a,b$ handles the other unequal case, provingFinally, let $b$ be outside the open ball $B(a,\delta )$, so $d_p(a,b)\ge \delta$. If $d_p(b,c)<\delta$, the preceding unequal-norm identity gives $d_p(a,c)=d_p(a,b)\ge \delta$. Therefore $B(b,\delta )$ also lies outside $B(a,\delta )$. Its complement is open, so every p-adic open ball is also closed, that is, a clopen set.

Take the principal square root $w=u+iv$, which has $u>0$ on the slit domain. The inverse relation $z=w^2$ gives $x=u^2-v^2$ and $y=2uv$. Hence the defining inequality for the parabolic domain becomesThus the image is the strip $\boxed{0<\mathrm{Re}w<1}$. Conversely, every point of this strip squares to a point of the parabolic domain outside the removed nonpositive real axis, and its principal square root is the original $w$. This proves that the map is onto the strip and one-to-one. Its derivative $1/(2\sqrt{z})$ is nonzero there, so it is a conformal map.

It helps to separate the tangent and the squaring map. Put $t=\tan (z/2)$, with $z=x+iy$. The identityshows that $0<x<\pi /2$ gives $\mathrm{Re}t>0$ and $|t|<1$. To see that every point of this right half-disc occurs, setThen $z=-i\mathrm{Log}R$ has $0<\mathrm{Re}z<\pi /2$ and satisfies $\tan (z/2)=t$. The principal complex logarithm is unambiguous because $R$ is in the first quadrant.

Squaring maps this right half-disc bijectively onto the unit disc with the interval $(-1,0]$ removed: the argument of $t$ lies between $-\pi /2$ and $\pi /2$, and doubles into $(-\pi ,\pi )$. Thus the desired image isNeither derivative vanishes in the domain, so the composition is also a conformal map onto that slit disc.

On the parabolic domain with the nonpositive real axis removed, first apply the principal square root, then scale by $\pi /2$, and then apply the map of part (b). This gives the stated composite $C(z)$ and maps bijectively onto the slit unit disc.

There is no genuine branch cut in the composite. The function ${\tan }^2(\pi s/4)$ is even in $s$, so changing the local square root has no effect. Near zero its Taylor series starts asgiving a holomorphic extension with $C(0)=0$. There are no poles within the parabolic domain: the possible squared locations are $z=(2+4k{)}^2$, all on the excluded real portion $x\ge 4$.

For $z=-r^2$ with $r\ge 0$, the continuation givesAs $r$ ranges from zero to infinity, these values run once through $(-1,0]$. They fill precisely the slit omitted by the previous mapping, and no point away from the negative real axis has an image in that slit. Therefore the extended map is a bijection onto the whole unit disc:It is holomorphic throughout, so this bijection is the desired conformal map from a parabolic domain to the unit disc; in particular $C^{\prime }(0)={\pi }^2/16\ne 0$.