Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 1

For a linear map $T:V\to W$, its rank is $\dim \mathrm{im}T$ and its nullity is $\dim \ker T$. The rank-nullity theorem states that, when $V$ is finite-dimensional,To prove it, take a basis $v_1,\dots ,v_k$ of $\ker T$ and extend it to a basisof $V$. The vectors $T{v}_{k+1},\dots ,T{v}_n$ span $\mathrm{im}T$. They are also linearly independent: if ${\sum }_{j=k+1}^n{a}_{j}T{v}_j=0$, then ${\sum }_{j=k+1}^n{a}_j{v}_j\in \ker T$, and independence of the chosen basis forces every $a_j$ to vanish. Thus the rank is $n-k$ and the nullity is $k$.

For the given subspace, row reduction of the coefficient matrix givesTaking $x_3=s$ and $x_5=t$ therefore givesHenceis a basis of $W$.

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After the $2n$ sides are identified in pairs, the polygonal-schema Euler count gives one face, $n$ edges and some number $V\ge 1$ of vertices. The Euler characteristic of a closed orientable genus-$g$ surface is $2-2g$, so

For genus two, take a regular octagon in the Poincare disc model and identify each side with its opposite side, with the orientation reversed along the boundary. One cyclic labelling isThe quotient has $V=1$, $E=4$, and $F=1$, hence Euler characteristic $-2$ and genus two. All eight vertices become one point, so smoothness requires their angles to sum to $2\pi$. Each interior angle is consequentlyEquivalently, the hyperbolic polygon area formula shows that the regular hyperbolic octagon fundamental polygon has areaas required by the Gauss-Bonnet theorem for a genus-two surface of curvature $-1$.

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Sincethe critical points are exactlyThe holomorphic inverse function theorem shows that $f$ is conformal locally everywhere else.

For $z=x+iy$,Its imaginary part is positive throughout $S$. On the three boundary pieces,These intervals traverse the boundary of the upper half-plane. The inverse branchexists for $\Im \zeta >0$, proving that the image is preciselyThis is the hyperbolic-cosine half-strip map.

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At a regular constrained extremum of $f$ on $g=0$, the tangent derivatives of $f$ vanish. Since $\nabla g$ is normal to the constraint surface, the Lagrange multiplier condition isOne solves these equations and then compares the resulting candidates, including any boundary or singular cases.

Let the base have dimensions $x$ and $y$, and let the height be $z$. Measure cardboard relative to the thickness of the front and back. The weighted amount used isbecause the bottom has triple thickness, the two front and back faces have ordinary thickness, and the two side faces have double thickness. The constraint is $xyz=3$.

The multiplier equations areThey implyThus $x=2y$ and $z=3y/2$; imposing $xyz=3$ gives $y=1$. ThereforeThis is the global minimum: by the arithmetic-geometric mean inequality,and equality holds at these dimensions. This is an instance of weighted open-box minimization.

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For distinct nodes, define the divided difference byThis is the coefficient of $x^k$ in the Lagrange interpolation polynomial through the first $k+1$ data points.

Let $p_k$ denote that interpolating polynomial. The difference $p_k-p_{k-1}$ vanishes at $x_0,\dots ,x_{k-1}$, soThe Lagrange formula shows that the leading coefficient of $p_k$ is $f[x_0,\dots ,x_k]$, whereas $p_{k-1}$ has degree at most $k-1$. Hence $c_k=f[x_0,\dots ,x_k]$. Starting with $p_0=f(x_0)$ and iterating gives the Newton interpolation polynomial

The divided-difference recurrence isFor three nodes, the triangular table iswhere each entry in a new column uses the two adjacent entries to its left. There are $n$ first differences, $n-1$ second differences, and so on, each requiring one division. The exact total is

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A statistic $T(X)$ is sufficient for a parameter $\theta$ when the conditional distribution of the full sample $X$ given $T$ does not depend on $\theta$. For a dominated model, the Fisher-Neyman factorization theorem says this is equivalent to a factorizationwhere $h$ is independent of $\theta$.

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A sufficient statistic $T$ is minimal sufficient if it is a function of every other sufficient statistic: whenever $U$ is sufficient, there is a function $h$ such that $T=h(U)$ almost surely, up to the usual null-set qualification. Thus its level sets give the coarsest sufficient partition of the sample space. In a dominated family with positive densities, the likelihood-ratio criterion for minimal sufficiency says that $T$ is minimal sufficient when

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For $\sigma >0$, the density isso the Fisher-Neyman factorization theorem proves that $|X|$ is sufficient.

Moreover,is independent of $\sigma$ exactly when $x^2=y^2$, equivalently $|x|=|y|$. The likelihood-ratio criterion for minimal sufficiency therefore proves that

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The joint density depends on the sample throughnot merely through ${\sum }_i|X_i|$. For example, after padding with zeros when $n>2$, takeBoth samples have sum of absolute values $2$, butwhich depends on $\sigma$. The likelihood-ratio necessary condition for sufficiency fails, so

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Introduce nonnegative dual variables $y$ and $z$ for the two constraints. The dual of a minimization linear program in inequality form isIndeed, if $x$ and $(y,z)$ are feasible, thenwhich is weak duality.

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The dual from part (a) isThe point $(y,z)=(1,2)$ is dual feasible and has value $29$. The primal pointis feasible and also has objective valueBy weak duality, neither point can be improved, soThe strict dual inequalities for $x_2$ and $x_4$, together with complementary slackness, force $x_2=x_4=0$ at any optimum; the two tight primal constraints then force $x_1=7$ and $x_3=4$. Thus the displayed minimizer is unique.

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The dual space isthe vector space of linear functionals on $V$. If $e_1,\dots ,e_n$ is a basis of a finite-dimensional $V$, define $e_i^{\ast }(e_j)={\delta }_{ij}$. Every $f\in V^{\ast }$ has the unique expansionso the dual basis $e_1^{\ast },\dots ,e_n^{\ast }$ proves

For $U\le V$, its annihilator of a vector subspace isIf $u_1,\dots ,u_k$ is a basis of $U$ and is extended to a basis $u_1,\dots ,u_n$ of $V$, thenConsequentlyIf $U\ne V$, this dimension is positive, giving a nonzero functional that vanishes on $U$.

For a linear map $\alpha :V\to W$, the dual map isNowsoEvery $g\circ \alpha$ vanishes on $\ker \alpha$, henceThe two spaces have the same dimension, sinceThus

Let $Q:V\to V/U$ be the quotient map. Since $Q$ is onto, $Q^{\ast }$ is injective, and the preceding identity givesThereforeFor the inclusion $J:U\hookrightarrow V$, the map $J^{\ast }:V^{\ast }\to U^{\ast }$ is restriction to $U$. It is onto and has kernelThe first isomorphism theorem now gives the other duals of a subspace and its quotient:

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Suppose that $Q$ were not surjective. Its image would be a proper subspace of $F^n$, so there would be a nonzero functionalvanishing on $\mathrm{im}Q$. Thenfor every $x\in V$, contradicting the linear independence of the $q_j$. Hence the surjectivity of independent linear functionals gives

Let $K={\bigcap }_j\ker q_j=\ker Q$ and suppose $K\subseteq \ker f$. DefineThis is well-defined: if $Qx=Qy$, then $x-y\in K$, so $f(x)=f(y)$. It is linear, and surjectivity of $Q$ means it is defined on all of $F^n$. Thus there are scalars $a_1,\dots ,a_n$ such that $g(y)={\sum }_j{a}_j{y}_j$. Thereforeand hence

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Let $|G|=p^{a}m$ with $p\nmid m$. The Sylow theorems state:

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Since $|S_3|=6$, its Sylow $2$-subgroups have order two and its Sylow $3$-subgroups have order three. They areandrespectively. The latter is the unique Sylow $3$-subgroup.

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A Sylow $2$-subgroup of $S_4$ has order $8$. For example,It is the normalizer of $⟨(1234)⟩$. The six $4$-cycles form three inverse pairs, and a dihedral group of order eight has a unique cyclic subgroup of order four. Hence the three Sylow $2$-subgroups are

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A Sylow $2$-subgroup of $A_5$ has order four. Since $A_5$ has no element of order four, each is a Klein four-group. For every $i\in \{1,\dots ,5\}$, take the identity and the three double transpositions fixing $i$. For instance, the subgroup fixing $5$ isThese five point stabilizers are precisely the Sylow $2$-subgroups. Indeed, the $15$ double transpositions occur three to each such subgroup, so there areof them. Together with the previous parts, this gives the Sylow subgroups of S3, S4 and A5.

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Let $G$ act by left multiplication on the coset set $G/H$. Composing the resulting homomorphismwith the sign homomorphism gives a homomorphism $G\to \{\pm 1\}$. If it were nontrivial, its kernel would have index two, contrary to the hypothesis. Thus every element of $G$, and in particular $x$, induces an even permutation.

Since $P$ is a Sylow $2$-subgroup, $[G:P]$ is odd. ThereforeAn involution acts as disjoint transpositions and fixed points. If $x$ fixed no coset, it would be a product of $[G:H]/2$ transpositions, an odd number, contradicting evenness. HenceThis is the even-involution coset fixed-point lemma.

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If $gH$ is a fixed coset, thenMultiplying on the left by $g^{-1}$ givesso $g^{-1}xg\in H$. Thus

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A metric space is complete when every Cauchy sequence in it converges to a point of the space. Let $(y_n)$ be Cauchy in a closed subspace $Y$ of a complete metric space $X$. It is also Cauchy in $X$, so $y_n\to x$ for some $x\in X$. Since $Y$ is closed and every $y_n$ lies in $Y$, one has $x\in Y$. Thus the closed-subspace completeness theorem proves that $Y$ is complete.

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The proposed implication is not symmetric. In $X=\mathbb{R}$, takeThen $B\subseteq E_1(A)$, but $2\notin E_1(B)$, so $A⊈E_1(B)$.

For a point and a nonempty set, write $d(x,B)={\inf }_{b\in B}d(x,b)$. The given Hausdorff distance is equivalentlyIt is nonnegative and symmetric. If $d_H(A,B)=0$, then every $a\in A$ has $d(a,B)=0$, so $a\in \overline{B}=B$ because $B$ is closed; hence $A\subseteq B$, and symmetry gives $A=B$.

For the triangle inequality, the ordinary triangle inequality givesTaking suprema over $a\in A$, and then repeating with $A$ and $C$ interchanged, yieldsThus $d_H$ is a metric on $H(X)$. Closedness is essential: if it is omitted, then the distinct bounded sets $(0,1)$ and $[0,1]$ have Hausdorff distance zero.

For singleton sets,so $\theta (x)=\{x\}$ is an isometry. Its image is closed. Indeed, if $\{x_n\}$ converges in Hausdorff distance to $A\in H(X)$, then for every $\epsilon >0$ and all sufficiently large $n$,Hence any $a,b\in A$ satisfy $d(a,b)\le d(a,x_n)+d(x_n,b)\le 2\epsilon$. Letting $\epsilon \downarrow 0$ gives $a=b$, so the nonempty set $A$ is a singleton. This is the closed singleton embedding in a Hausdorff hyperspace.

If $H(X)$ is complete, its closed subspace $\theta (X)$ is complete. Since $\theta$ is an isometry, $X$ is complete. Hence completeness is reflected by the Hausdorff hyperspace.

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A regularly parametrized curve $\gamma$ on an embedded surface $\Sigma$ is a geodesic when its covariant acceleration vanishes. Equivalently, in the ambient Euclidean space,for an affine geodesic parameter. A complete geodesic is one whose maximal affine parameter interval is all of $\mathbb{R}$.

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Tangency along $\gamma$ meansfor every $t$. Consequently $\ddot{\gamma }(t)$ is perpendicular to the first tangent plane exactly when it is perpendicular to the second. The ambient characterization of a geodesic therefore proves the tangency invariance of an ambient-surface geodesic:

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For fixed $u$, varying $v$ traces a unit circle in the radial-vertical plane centred atThus ${\Sigma }_1$ is a unit tube whose centre winds helically around the $z$-axis. Sincethe given curve is its inner helix.

Take ${\Sigma }_2$ to be the circular cylinder $x^2+y^2=1$. Along $v=\pi$,so their span is the tangent plane of that cylinder. The two surfaces are therefore tangent along $\gamma$. Moreover,which is normal to the cylinder, so $\gamma$ is a geodesic of ${\Sigma }_2$. Part (ii) now proves that it is a geodesic of ${\Sigma }_1$. This is the cylindrical-helix tangency construction.

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SetAlong $v=0$,whose span is the tangent plane of the cylinder $x^2+y^2=9$. Alsois normal to that cylinder. The same tangency argument proves that $\Gamma$ is a geodesic of ${\Sigma }_1$. It is defined for every $t\in \mathbb{R}$, hence complete, and it is disjoint from $\gamma$ because their distances from the $z$-axis are respectively $3$ and $1$.

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The classification of isolated singularities can be stated through the Laurent expansion of $f$ about $a$:

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Write the Laurent expansion on the punctured disc asFor every integer $k\ge 1$, the Laurent coefficient formula and the Cauchy-Schwarz inequality giveLetting $r\downarrow 0$ shows that $a_{-k}=0$. Every coefficient in the principal part vanishes, so the uniform L2 circle bound for a removable singularity proves that

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Since $\tan z=\sin z/\cos z$, its singularities occur at the simple zeros of the denominator,At each one, $\sin a_k\ne 0$ and $(\cos z{)}^{\prime }{|}_{a_k}=-\sin a_k\ne 0$, so all are simple poles. There are no other finite singularities:

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PutNear $a_k$, writing $\delta =1/z-a_k$ givesButand hence, exactly,Since $\delta =O(w)$, the first two Laurent terms areEquivalently, substitute $z_k=2/((2k+1)\pi )$ in this expression.

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At every $z_k$, the function $\tan (1/z)$ has a simple pole. The exponential of a pole is an essential singularity, soare essential singularities of $\exp (\tan (1/z))$.

These points accumulate at $0$. Therefore no punctured neighbourhood of zero is a domain of holomorphy for the function: zero is a non-isolated singularity, specifically an accumulation point of essential singularities, rather than an isolated essential singularity. There are no other singularities in $\mathbb{C}$.

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Continuity at $x=\xi$ is automatic in the proposed expression. Integrating the differential equation through $x=\xi$ gives the derivative jumpThusIn terms of the Wronskianone hasThe boundary conditions hold because $y_1^{\prime }(0)=0$ and $y_2^{\prime }(1)=0$.

The Abel identity gives $W^{\prime }=-\alpha W$. If $\alpha =0$, the Wronskian and hence $c$ are constant. The two branches of the displayed formula are then interchanged by $x\leftrightarrow \xi$, provingThis is the symmetry of the Neumann Green function for a second-order ordinary differential equation in the self-adjoint case.

For $G^{\prime \prime }-G=\delta (x-\xi )$, chooseTheir Wronskian isWriting $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$ givesThe solution of the inhomogeneous problem is $y(x)={\int }_0^{1}G(x,\xi )\xi d\xi$. Equivalently, solving directly givesThe two Neumann conditions yield $B=1$ andTherefore

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The anti-muon has the same charge as a proton, so the Coulomb potential seen by the electron is the same as in the hydrogen atom. The main difference is the source mass. Separating the centre-of-mass motion replaces $m_e$ by the Coulomb two-body reduced massSince $m_{\mu }\gg m_e$, this is close to $m_e$, though less close than the electron-proton reduced mass. The smaller reduced mass gives a slightly larger Bohr radius and energies of slightly smaller magnitude. Under the assumptions in the question, the Coulomb form, angular eigenfunctions and quantum numbers are otherwise unchanged; anti-muon decay and spin effects are being neglected.

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For fixed $l$, the radial equation isSubstituteCancellation of the coefficient of $1/r$ requiresthe reduced-mass Bohr radius. The remaining constant term givesEach such no-radial-node state is a circular Coulomb bound state with principal quantum number $n=l+1$.