Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 1

A basis of a vector space $V$ is a family of vectors that is both linearly independent and a spanning set of $V$.

Write the given finite basis as $\mathcal{B}=\{v_1,\dots ,v_n\}$. First, any other basis ${\mathcal{B}}^{\prime }$ must also be finite. Each $v_i$ is a finite linear combination of elements of ${\mathcal{B}}^{\prime }$, so the union $S$ of the finitely many elements of ${\mathcal{B}}^{\prime }$ occurring in these $n$ expressions is finite. Since $S$ spans every $v_i$, it spans $V$. If some $w\in {\mathcal{B}}^{\prime }\setminus S$ existed, then $w$ would lie in the span of $S$, contradicting the linear independence of ${\mathcal{B}}^{\prime }$. Hence ${\mathcal{B}}^{\prime }=S$; write $|{\mathcal{B}}^{\prime }|=m$.

It remains to prove the elementary Steinitz exchange lemma directly. If independent vectors $w_1,\dots ,w_r$ lie in the span of $v_1,\dots ,v_n$, express $w_1$ in terms of the $v_i$. Some coefficient is nonzero, so the corresponding $v_i$ can be solved for in terms of $w_1$ and the other $v_i$; replacing it by $w_1$ preserves the span. Inductively, after replacing $k$ of the $v_i$ by $w_1,\dots ,w_k$, the expression for $w_{k+1}$ must have a nonzero coefficient on one of the unreplaced $v_i$, since otherwise $w_{k+1}$ would be a linear combination of $w_1,\dots ,w_k$. That $v_i$ can again be replaced. There are only $n$ original vectors to replace, so $r\le n$.

Apply this argument first to the independent family ${\mathcal{B}}^{\prime }$ and the spanning family $\mathcal{B}$ to obtain $m\le n$, and then with the two bases interchanged to obtain $n\le m$. Therefore

A Laurent series about $a$ is an expansionthat converges to $f$ on an annulus $r<|z-a|<R$.

The partial fraction decomposition isFor $0<|z|<1$, expand both denominators by the geometric series:This is actually a Taylor series at zero, since the apparent puncture at zero contains no singularity.

For $1<|z|<2$, the $z+2$ term still expands in nonnegative powers, while the quadratic term must be expanded in negative powers:Hence

The Poincare disc model isThis Riemannian metric has constant Gaussian curvature $-1$.

For $a\in D$ and $|\eta |=1$, the Möbius transformationis an isometry of the disc, and every orientation-preserving disc isometry has this form. In particular, $T_{z_1,1}$ sends $z_1$ to zero and $z_2$ to a point of modulusRotations are also isometries, so it remains only to integrate the metric along a radial geodesic from $0$ to $\rho$:Thus the Hyperbolic distance in the Poincare disc is

Introduce the orthogonal coordinatesThen the constraint and objective becomeThe Lagrange multiplier equations for a stationary point areIf $\lambda =0$, then $u=y=0$ and $v=\pm 1$, givingIf $\lambda \ne 0$, then $v=0$. The first two equations and the constraint implyConverting back to $(x,y,z)$ gives the other four points. Therefore all stationary points arewith matching signs, together withThe middle pair has $\varphi =1/2$, the last pair has $\varphi =-1/2$, and the first pair has $\varphi =0$.

Take $x$ down the plane and $y$ normally away from it, with the solid at $y=0$ and the free surface at $y=h$. A steady unidirectional falling film flow has velocity $\mathbf{u}=u(y){\mathbf{e}}_x$. The streamwise Navier-Stokes equation reduces towhere $\nu$ is the kinematic viscosity. The lower no-slip boundary condition and upper stress-free boundary condition areIntegrating gives the velocity profileThe volume flux per unit width is

If the upper surface is replaced by a stationary solid plane, the second boundary condition becomes $u(h)=0$. The resulting plane-channel profile and flux areThe free surface exerts no tangential shear stress, whereas the stationary upper wall enforces no slip and exerts a retarding shear. This additional drag accounts for the reduced flux.

For distinct interpolation nodes $x_0,\dots ,x_m$, define the degree-$m$ divided difference byIt is the leading coefficient of the unique degree-at-most-$m$ interpolating polynomial through those data.

DefineThe partial polynomial $p_m$ agrees with $p_{m-1}$ at $x_0,\dots ,x_{m-1}$. Moreover, the explicit formula for the divided difference givesso $p_m(x_m)=f(x_m)$. Induction shows that $p_n(x_j)=f(x_j)$ for every $j$. By uniqueness of polynomial interpolation, this is the required Newton interpolation polynomial.

For the recursion, substitute the explicit formulas for $f[x_1,\dots ,x_m]$ and $f[x_0,\dots ,x_{m-1}]$. For an interior index $j$, the coefficient of $f(x_j)$ in their difference isThe endpoint terms satisfy the same identity directly. Dividing by $x_m-x_0$ therefore gives

For observations $x_1,\dots ,x_n$, the log-likelihood of the normal distribution model isDifferentiating first with respect to $\mu$ gives $\hat{\mu }=\overline{x}$. Substituting this value and differentiating with respect to ${\sigma }^2$ gives the normal mean and variance maximum-likelihood estimators

An unbiased estimator $T$ of a parameter $\theta$ satisfies ${\mathbb{E}}_{\theta }[T]=\theta$ for every permissible value of $\theta$. Since the sample mean is linear,so $\hat{\mu }$ is unbiased.

Use the identityThe two expectations on the right are $n{\sigma }^2$ and $n\mathrm{Var}(\overline{X})={\sigma }^2$, respectively. HenceThusThe unbiased sample variance instead divides the same sum of squares by $n-1$.

Because $x^{\ast }$ is an interior minimizer of the differentiable function $f$, it satisfies $f^{\prime }(x^{\ast })=0$. Apply the Taylor theorem with Lagrange remainder to $f^{\prime }$ about $x_n$. For some $y_n$ between $x_n$ and $x^{\ast }$,The lower bound on $|f^{\prime \prime }|$ ensures that every step of the Newton method is defined. Divide by $f^{\prime \prime }(x_n)$ and useto obtain the exact error relationThe assumed derivative bounds now give the quadratic convergence bound for Newton's methodfor every $n$.

The adjugate matrix is the transpose of the cofactor matrix, and it satisfiesPut $M=tI-A$ and substitute the two given expansions into this identity:Comparing coefficients of powers of $t$ givesHere $c_0=1$ because the characteristic polynomial is monic.

Let $M(t_1,\dots ,t_n)=\mathrm{diag}(t_1,\dots ,t_n)-A$. Expanding the determinant along row $i$ shows thatwhere $C_{ii}$ is the corresponding cofactor. Since a diagonal entry of the adjugate matrix is that same cofactor, the multivariable chain rule along the diagonal $t_1=\cdots =t_n=t$ yieldsTherefore

Taking the matrix trace of the adjugate expansion and using part (b) givesHence $\mathrm{Tr}(B_i)=(n-i)c_i$. Taking traces in $B_i=A{B}_{i-1}+c_{i}I$ gives, for $1\le i<n$,For $i=n$, the same formula follows by taking the trace of $-A{B}_{n-1}=c_{n}I$. Iterating the recursion from part (a) also givesConsequently the Faddeev–LeVerrier algorithm isStarting with $c_0=1$, this recursively expresses every $c_i$ using only $\mathrm{Tr}(A),\dots ,\mathrm{Tr}(A^i)$; these are the Newton identities for the characteristic polynomial.

The group $G$ is a nontrivial finite p-group, so the nontrivial center of a finite p-group gives $Z(G)\ne \{e\}$. If $Z(G)\ne G$, then $Z(G)$ is already a nontrivial proper normal subgroup, so $G$ is not a simple group. If $Z(G)=G$, then $G$ is abelian. By Cauchy's theorem for finite groups, it has a subgroup of order $p$; every subgroup of an abelian group is normal, and this subgroup is nontrivial and proper because $|G|=p^4$. Thus in every case

The Sylow theorems state that if $|G|=p^{a}m$ with $p\nmid m$, then:

Interchange the two primes if necessary so that $p<q$. By the Sylow theorems, the number $n_q$ of Sylow $q$-subgroups satisfiesIf $n_q\ne 1$, then $n_q$ is $p$ or $p^2$. The first is impossible because $1<p<q$. The second would implySince $q>p$ is prime, it cannot divide $p-1$, so it must divide $p+1$. But $q>p$ then forces $q=p+1$, impossible because $p$ and $q$ are both odd. Therefore $n_q=1$. The unique Sylow $q$-subgroup is a nontrivial proper normal subgroup, and hence

A map $f:A\to {\mathbb{R}}^m$ is differentiable at $p\in A$ when there is a linear map $Df(p):{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThe linear map $Df(p)$ is the Frechet derivative of $f$ at $p$.

The multivariable chain rule states that if $f$ is differentiable at $p$ and $g$ is differentiable at $f(p)$, thenIndeed, write $f(p+h)=f(p)+Ah+r(h)$ and $g(f(p)+k)=g(f(p))+Bk+s(k)$, where $r(h)=o(|h|)$ and $s(k)=o(|k|)$. Since $k=Ah+r(h)=O(|h|)$,and the last two terms are $o(|h|)$. This proves the formula.

For matrix inversion near the identity matrix,The last term is $O(\Vert h{\Vert }^2)$ because inversion is bounded near $I$. Therefore the inversion map is differentiable at $I$ and

For every invertible matrix $A$,The maps $l_C$ and $r_C$ are linear. Differentiate this identity at $A=I$ and use the chain rule and part (a):Thus $Df(C)(Ch)C=-h$. Given an arbitrary increment $k$, set $h=C^{-1}k$ to obtain the derivative of matrix inversionThis also proves differentiability at every $C\in V$.

Nonnegativity and symmetry are immediate. Moreover, $d(x,y)=0$ exactly when $n_{\min }(x,y)=\infty$, which means $x_n=y_n$ for every $n$.

For three sequences $x,y,z$, if $x$ and $y$ agree through coordinate $r-1$ and $y$ and $z$ agree through coordinate $s-1$, then $x$ and $z$ agree through coordinate $\min (r,s)-1$. Henceand thereforeThus $d$ is a metric; in fact, the stronger first inequality makes it an ultrametric.

A metric space is complete when every Cauchy sequence converges to a point of the space.

Let $x^{(1)},x^{(2)},\dots$ be a Cauchy sequence in $X$, where $x^{(j)}=(x_1^{(j)},x_2^{(j)},\dots )$. For each positive integer $N$, choose $J_N$ such thatThis inequality says that the first $N$ coordinates of $x^{(j)}$ and $x^{(k)}$ agree. Consequently each coordinate $x_n^{(j)}$ is eventually constant. Let $x_n$ be its eventual value and put $x=(x_1,x_2,\dots )\in X$.

For $j\ge J_N$, the first $N$ coordinates of $x^{(j)}$ agree with those of $x$, soAs $N$ is arbitrary, $x^{(j)}\to x$. Therefore

For each $k$, the coordinate projectionis continuous when $\mathbb{Z}$ has the discrete topology: if $d(x,y)<2^{-k}$, then $x_k=y_k$.

If $\gamma :[0,1]\to X$ were a continuous path, then ${\pi }_k\circ \gamma$ would be a continuous map from the connected interval $[0,1]$ to a discrete space, so it would be constant. This holds for every $k$, and therefore $\gamma$ itself must be constant. Since $X$ contains distinct points, no path joins every pair. Henceindeed, every path component is a singleton.

Jordan lemma states that if $a>0$ and $g$ is analytic in the upper half-plane outside a fixed circle, withthen on the upper semicircle $C_R$,Indeed, $|e^{iaR{e}^{i\theta }}|=e^{-aR\sin \theta }$. Using symmetry and $\sin \theta \ge 2\theta /\pi$ on $0\le \theta \le \pi /2$ gives

The residue of $f$ at an isolated singularity $a$ is the coefficient of $(z-a{)}^{-1}$ in its Laurent series, equivalentlyIf $f(z)=g(z)/(z-a{)}^k$, the Taylor expansion of the analytic function $g$ shows that

For the real integral, apply contour integration in the upper half-plane toJordan's lemma removes the semicircular arc. The only enclosed singularity is the double pole at $z=i$, whose residue isThe residue theorem therefore givesIts real part vanishes because $x^3\cos x/(1+x^2{)}^2$ is an odd function, while its imaginary part is the requested integral. Hence

Since $x=r\cos \theta$, the function on the left of the defining expansion is $e^{ix}$. Thereforewhich is the two-dimensional Helmholtz equation.

In polar coordinates,Substitute the defining Fourier cosine series and equate the coefficient of each $\cos (n\theta )$. Its angular second derivative contributes $-n^2$, so every Bessel function $J_n$ satisfiesor equivalently the Bessel differential equation

Expand the exponential through cubic order and useThis givesComparing with $J_0+2{\sum }_{n\ge 1}{i}^n{J}_n\cos (n\theta )$ yields

The Time-dependent Schrodinger equation for the one-dimensional Quantum harmonic oscillator isA stationary state has $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$, so cancellation of the common time factor gives

SetThe equation becomesWith $\psi =f(y)e^{-y^2/2}$,and hence $f$ obeys the Hermite differential equation

If $f$ is monic of degree $N$, the coefficient of $y^N$ in this equation is $ϵ-1-2N$. It must vanish, soWriting $f={\sum }_{k=0}^N{a}_k{y}^k$ gives the recurrenceThe coefficient of $y^{N-1}$ first gives $a_{N-1}=0$, and the recurrence preserves parity. If $N$ is even, every odd coefficient vanishes, so $f$ is an even function. The Gaussian factor is also even; therefore the stationary state $\psi$ has even parity.