= Paper 1 {scope} https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2024/paperib_1_2024.pdf = 1G {parent=Paper 1} {scope} {title2=Linear Algebra} = Solution {parent=1G} For a $T:V\to W$, its rank is $\dim\operatorname{im}T$ and its nullity is $\dim\ker T$. The states that, when $V$ is finite-dimensional, $$ \dim V=\dim\ker T+\dim\operatorname{im}T. $$ To prove it, take a $v_1,\ldots,v_k$ of $\ker T$ and extend it to a $$ v_1,\ldots,v_k,v_{k+1},\ldots,v_n $$ of $V$. The $Tv_{k+1},\ldots,Tv_n$ span $\operatorname{im}T$. They are also linearly independent: if $\sum_{j=k+1}^na_jTv_j=0$, then $\sum_{j=k+1}^na_jv_j\in\ker T$, and independence of the chosen 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 gives $$ \begin{pmatrix} 1&0&3&0&43/3\\ 0&1&3&0&-5/3\\ 0&0&0&1&-11/3 \end{pmatrix}. $$ Taking $x_3=s$ and $x_5=t$ therefore gives $$ (x_1,x_2,x_3,x_4,x_5) =s(-3,-3,1,0,0)+\frac t3(-43,5,0,11,3). $$ Hence $$ \boxed{\dim W=2}, \qquad \boxed{\{(-3,-3,1,0,0),(-43,5,0,11,3)\}} $$ is a of $W$. Solved by gpt-5.6-sol high. = 2E {parent=Paper 1} {scope} {title2=Geometry} = Solution {parent=2E} After the $2n$ sides are identified in pairs, the gives one face, $n$ edges and some number $V\geq1$ of vertices. The of a closed orientable genus-$g$ surface is $2-2g$, so $$ 2-2g=V-n+1, \qquad n=V-1+2g\geq2g. $$ For genus two, take a regular octagon in the and identify each side with its opposite side, with the orientation reversed along the boundary. One cyclic labelling is $$ a,b,c,d,a^{-1},b^{-1},c^{-1},d^{-1}. $$ The 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 consequently $$ \boxed{\frac{2\pi}{8}=\frac\pi4}. $$ Equivalently, the formula shows that the has area $$ (8-2)\pi-8\left(\frac\pi4\right)=4\pi, $$ as required by the for a genus-two surface of curvature $-1$. Solved by gpt-5.6-sol high. = 3B {parent=Paper 1} {scope} {title2=Complex Analysis OR Complex Methods} = Solution {parent=3B} Since $$ f'(z)=\pi\sinh(\pi z), $$ the critical points are exactly $$ \boxed{z=ki\quad(k\in\mathbb Z)}. $$ The shows that $f$ is conformal locally everywhere else. For $z=x+iy$, $$ f(z)=\cosh(\pi x)\cos(\pi y) +i\sinh(\pi x)\sin(\pi y). $$ Its imaginary part is positive throughout $S$. On the three boundary pieces, $$ \begin{aligned} L_1&:\quad f(x)=\cosh(\pi x)\in(1,\infty),\\ L_2&:\quad f(iy)=\cos(\pi y)\in(-1,1),\\ L_3&:\quad f(x+i)=-\cosh(\pi x)\in(-\infty,-1). \end{aligned} $$ These intervals traverse the boundary of the upper half-plane. The inverse branch $$ z=\frac1\pi\operatorname{arcosh}\zeta, \qquad \Re z>0,\quad0<\Im z<1, $$ exists for $\Im\zeta>0$, proving that the image is precisely $$ \boxed{f(S)=\{\zeta:\Im\zeta>0\}}. $$ This is the . Solved by gpt-5.6-sol high. = 4C {parent=Paper 1} {scope} {title2=Variational Principles} = Solution {parent=4C} At a regular constrained extremum of $f$ on $g=0$, the tangent of $f$ vanish. Since $\nabla g$ is normal to the constraint surface, the condition is $$ \nabla f=\lambda\nabla g, \qquad g=0. $$ One 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 is $$ A=3xy+2xz+4yz, $$ because 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 are $$ 3y+2z=\lambda yz, \qquad 3x+4z=\lambda xz, \qquad 2x+4y=\lambda xy. $$ They imply $$ 3xy=2xz=4yz. $$ Thus $x=2y$ and $z=3y/2$; imposing $xyz=3$ gives $y=1$. Therefore $$ \boxed{x=2,\qquad y=1,\qquad z=\frac32}. $$ This is the global minimum: by the , $$ A\geq3\sqrt[3]{(3xy)(2xz)(4yz)} =3\sqrt[3]{24(xyz)^2}=18, $$ and equality holds at these dimensions. This is an instance of . Solved by gpt-5.6-sol high. = 5A {parent=Paper 1} {scope} {title2=Numerical Analysis} = Solution {parent=5A} For distinct nodes, define the divided difference by $$ f[x_0,\ldots,x_k] =\sum_{j=0}^k \frac{f(x_j)}{\prod_{\substack{0\leq i\leq k\\i\ne j}}(x_j-x_i)}. $$ This is the coefficient of $x^k$ in the through the first $k+1$ data points. Let $p_k$ denote that interpolating . The difference $p_k-p_{k-1}$ vanishes at $x_0,\ldots,x_{k-1}$, so $$ p_k(x)-p_{k-1}(x)=c_k\prod_{i=0}^{k-1}(x-x_i). $$ The Lagrange formula shows that the leading coefficient of $p_k$ is $f[x_0,\ldots,x_k]$, whereas $p_{k-1}$ has degree at most $k-1$. Hence $c_k=f[x_0,\ldots,x_k]$. Starting with $p_0=f(x_0)$ and iterating gives the $$ \boxed{ p_n(x)=f(x_0)+\sum_{k=1}^n f[x_0,\ldots,x_k] \prod_{i=0}^{k-1}(x-x_i)}. $$ The divided-difference recurrence is $$ \boxed{ f[x_0,\ldots,x_k] =\frac{f[x_1,\ldots,x_k]-f[x_0,\ldots,x_{k-1}]} {x_k-x_0}}. $$ For three nodes, the triangular table is $$ \begin{array}{ccc} f[x_0]&\longrightarrow&f[x_0,x_1]\longrightarrow f[x_0,x_1,x_2]\\ f[x_1]&\longrightarrow&f[x_1,x_2]\\ f[x_2] \end{array} $$ where 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 $$ \boxed{n+(n-1)+\cdots+1=\frac{n(n+1)}2}. $$ Solved by gpt-5.6-sol high. = 6H {parent=Paper 1} {scope} {title2=Statistics} = a {parent=6h} {scope} = Solution {parent=a} 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 says this is equivalent to a factorization $$ p_\theta(x)=g_\theta(T(x))h(x), $$ where $h$ is independent of $\theta$. Solved by gpt-5.6-sol high. = b {parent=6h} {scope} = Solution {parent=b} A sufficient statistic $T$ is minimal sufficient if it is a of every other sufficient statistic: whenever $U$ is sufficient, there is a $h$ 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 says that $T$ is minimal sufficient when $$ T(x)=T(y) \quad\Longleftrightarrow\quad \frac{p_\theta(x)}{p_\theta(y)} \text{ is independent of }\theta. $$ Solved by gpt-5.6-sol high. = c {parent=6h} {scope} = Solution {parent=c} For $\sigma>0$, the density is $$ p_\sigma(x)=\frac1{\sqrt{2\pi}\sigma} \exp\left(-\frac{x^2}{2\sigma^2}\right) =g_\sigma(|x|), $$ so the proves that $|X|$ is sufficient. Moreover, $$ \frac{p_\sigma(x)}{p_\sigma(y)} =\exp\left(-\frac{x^2-y^2}{2\sigma^2}\right) $$ is independent of $\sigma$ exactly when $x^2=y^2$, equivalently $|x|=|y|$. The therefore proves that $$ \boxed{|X|\text{ is minimal sufficient}.} $$ Solved by gpt-5.6-sol high. = d {parent=6h} {scope} = Solution {parent=d} The joint density depends on the sample through $$ \sum_{i=1}^nX_i^2, $$ not merely through $\sum_i|X_i|$. For example, after padding with zeros when $n>2$, take $$ x=(1,1,0,\ldots,0), \qquad y=(2,0,0,\ldots,0). $$ Both samples have sum of absolute values $2$, but $$ \frac{p_\sigma(x)}{p_\sigma(y)} =\exp\left(\frac1{\sigma^2}\right), $$ which depends on $\sigma$. The likelihood-ratio necessary condition for sufficiency fails, so $$ \boxed{\sum_{i=1}^n|X_i|\text{ is not sufficient}.} $$ Solved by gpt-5.6-sol high. = 7H {parent=Paper 1} {scope} {title2=Optimisation} = a {parent=7h} {scope} = Solution {parent=a} Introduce nonnegative dual variables $y$ and $z$ for the two constraints. The is $$ \boxed{ \begin{aligned} \text{maximise}\quad&ty+rz\\ \text{subject to}\quad&b_i y+c_i z\leq a_i \quad(1\leq i\leq n),\\ &y,z\geq0. \end{aligned}} $$ Indeed, if $x$ and $(y,z)$ are feasible, then $$ ty+rz \leq y\sum_i b_ix_i+z\sum_i c_ix_i =\sum_i(b_i y+c_i z)x_i \leq\sum_i a_ix_i, $$ which is . Solved by gpt-5.6-sol high. = b {parent=7h} {scope} = Solution {parent=b} The dual from part (a) is $$ \begin{aligned} \text{maximise}\quad&7y+11z\\ \text{subject to}\quad &y+z\leq3,\qquad y\leq2,\qquad z\leq2,\\ &y+2z\geq2,qquad y,z\geq0. \end{aligned} $$ The point $(y,z)=(1,2)$ is dual feasible and has value $29$. The primal point $$ (x_1,x_2,x_3,x_4)=(7,0,4,0) $$ is feasible and also has objective value $$ 3(7)+2(4)=29. $$ By , neither point can be improved, so $$ \boxed{\min=29,\qquad(x_1,x_2,x_3,x_4)=(7,0,4,0)}. $$ The strict dual inequalities for $x_2$ and $x_4$, together with , 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. Solved by gpt-5.6-sol high. = 8G {parent=Paper 1} {scope} {title2=Linear Algebra} = a {parent=8g} {scope} = Solution {parent=a} The dual space is $$ V^*=\operatorname{Hom}_F(V,F), $$ the of linear functionals on $V$. If $e_1,\ldots,e_n$ is a of a finite-dimensional $V$, define $e_i^*(e_j)=\delta_{ij}$. Every $f\in V^*$ has the unique expansion $$ f=\sum_{i=1}^nf(e_i)e_i^*, $$ so the $e_1^*,\ldots,e_n^*$ proves $$ \boxed{\dim V^*=\dim V=n}. $$ For $U\leq V$, its is $$ U^\circ=\{f\in V^*:f(u)=0\text{ for every }u\in U\}. $$ If $u_1,\ldots,u_k$ is a of $U$ and is extended to a $u_1,\ldots,u_n$ of $V$, then $$ U^\circ=\operatorname{span}\{u_{k+1}^*,\ldots,u_n^*\}. $$ Consequently $$ \boxed{\dim U^\circ=\dim V-\dim U}. $$ If $U\ne V$, this dimension is positive, giving a nonzero functional that vanishes on $U$. For a $\alpha:V\to W$, the is $$ \alpha^*:W^*\to V^*, \qquad \alpha^*(g)=g\circ\alpha. $$ Now $$ g\in\ker\alpha^* \Longleftrightarrow g(\alpha v)=0\text{ for every }v \Longleftrightarrow g\in(\operatorname{im}\alpha)^\circ, $$ so $$ \boxed{\ker\alpha^*=(\operatorname{im}\alpha)^\circ}. $$ Every $g\circ\alpha$ vanishes on $\ker\alpha$, hence $$ \operatorname{im}\alpha^*\subseteq(\ker\alpha)^\circ. $$ The two spaces have the same dimension, since $$ \begin{aligned} \dim\operatorname{im}\alpha^* &=\dim W-\dim(\operatorname{im}\alpha)^\circ =\dim\operatorname{im}\alpha,\\ \dim(\ker\alpha)^\circ &=\dim V-\dim\ker\alpha =\dim\operatorname{im}\alpha. \end{aligned} $$ Thus $$ \boxed{\operatorname{im}\alpha^*=(\ker\alpha)^\circ}. $$ Let $Q:V\to V/U$ be the quotient map. Since $Q$ is onto, $Q^*$ is injective, and the preceding identity gives $$ \operatorname{im}Q^*=(\ker Q)^\circ=U^\circ. $$ Therefore $$ \boxed{(V/U)^*\cong U^\circ}. $$ For the inclusion $J:U\hookrightarrow V$, the map $J^*:V^*\to U^*$ is restriction to $U$. It is onto and has kernel $$ \ker J^*=(\operatorname{im}J)^\circ=U^\circ. $$ The first isomorphism theorem now gives the other : $$ \boxed{U^*\cong V^*/U^\circ}. $$ Solved by gpt-5.6-sol high. = b {parent=8g} {scope} = Solution {parent=b} Suppose that $Q$ were not surjective. Its image would be a proper subspace of $F^n$, so there would be a nonzero functional $$ \ell(y_1,\ldots,y_n)=\sum_{j=1}^nc_jy_j $$ vanishing on $\operatorname{im}Q$. Then $$ 0=\ell(Qx)=\sum_{j=1}^nc_jq_j(x) $$ for every $x\in V$, contradicting the linear independence of the $q_j$. Hence the gives $$ \boxed{Q:V\to F^n\text{ is surjective}}. $$ Let $K=\bigcap_j\ker q_j=\ker Q$ and suppose $K\subseteq\ker f$. Define $$ g:F^n\to F, \qquad g(Qx)=f(x). $$ This 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 $a_1,\ldots,a_n$ such that $g(y)=\sum_ja_jy_j$. Therefore $$ f(x)=g(Qx)=\sum_{j=1}^na_jq_j(x), $$ and hence $$ \boxed{f\in\operatorname{span}\{q_1,\ldots,q_n\}}. $$ Solved by gpt-5.6-sol high. = 9E {parent=Paper 1} {scope} {title2=Groups, Rings and Modules} = a {parent=9e} {scope} = Solution {parent=a} Let $|G|=p^am$ with $p\nmid m$. The state: * $G$ has a of order $p^a$; * every $p$-subgroup of $G$ is contained in a Sylow $p$-subgroup; * all Sylow $p$-subgroups are conjugate; * their number $n_p$ satisfies $n_p\equiv1\pmod p$ and $n_p\mid m$. Solved by gpt-5.6-sol high. = i {parent=a} {scope} = Solution {parent=i} Since $|S_3|=6$, its Sylow $2$-subgroups have order two and its Sylow $3$-subgroups have order three. They are $$ \boxed{\langle(12)\rangle,\quad\langle(13)\rangle,\quad\langle(23)\rangle} $$ and $$ \boxed{A_3=\langle(123)\rangle}, $$ respectively. The latter is the unique Sylow $3$-subgroup. Solved by gpt-5.6-sol high. = ii {parent=a} {scope} = Solution {parent=ii} A Sylow $2$-subgroup of $S_4$ has order $8$. For example, $$ P=\langle(1234),(13)\rangle\cong D_8. $$ It is the normalizer of $\langle(1234)\rangle$. The six $4$-cycles form three inverse pairs, and a dihedral of order eight has a unique cyclic of order four. Hence the three Sylow $2$-subgroups are $$ \boxed{ N_{S_4}(\langle(1234)\rangle),\quad N_{S_4}(\langle(1243)\rangle),\quad N_{S_4}(\langle(1324)\rangle)}. $$ Solved by gpt-5.6-sol high. = iii {parent=a} {scope} = Solution {parent=iii} 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 $i\in\{1,\ldots,5\}$, take the identity and the three double transpositions fixing $i$. For instance, the fixing $5$ is $$ \{1,(12)(34),(13)(24),(14)(23)\}. $$ These five point stabilizers are precisely the Sylow $2$-subgroups. Indeed, the $15$ double transpositions occur three to each such , so there are $$ \boxed{n_2=5} $$ of them. Together with the previous parts, this gives the . Solved by gpt-5.6-sol high. = b {parent=9e} {scope} = i {parent=b} {scope} = Solution {parent=i} Let $G$ act by left multiplication on the coset set $G/H$. Composing the resulting homomorphism $$ G\longrightarrow S_{[G:H]} $$ with the sign homomorphism gives a homomorphism $G\to\{\pm1\}$. 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. Therefore $$ [G:H]=[G:P][P:H]=2[G:P]\equiv2\pmod4. $$ An 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. Hence $$ \boxed{x\text{ fixes at least one coset of }H}. $$ This is the . Solved by gpt-5.6-sol high. = ii {parent=b} {scope} = Solution {parent=ii} If $gH$ is a fixed coset, then $$ xgH=gH. $$ Multiplying on the left by $g^{-1}$ gives $$ g^{-1}xgH=H, $$ so $g^{-1}xg\in H$. Thus $$ \boxed{x\text{ is conjugate in }G\text{ to an element of }H}. $$ Solved by gpt-5.6-sol high. = 10F {parent=Paper 1} {scope} {title2=Analysis and Topology} = a {parent=10f} {scope} = Solution {parent=a} A metric space is complete when every Cauchy 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 proves that $Y$ is complete. Solved by gpt-5.6-sol high. = b {parent=10f} {scope} = Solution {parent=b} The proposed implication is not symmetric. In $X=\mathbb R$, take $$ A=\{0,2\},\qquad B=\{0\},\qquad r=1. $$ Then $B\subseteq E_1(A)$, but $2\notin E_1(B)$, so $A\nsubseteq E_1(B)$. For a point and a nonempty set, write $d(x,B)=\inf_{b\in B}d(x,b)$. The given is equivalently $$ d_H(A,B)=\max\left\{ \sup_{a\in A}d(a,B), \sup_{b\in B}d(b,A) \right\}. $$ It 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 gives $$ d(a,C)\leq d(a,B)+\sup_{b\in B}d(b,C). $$ Taking suprema over $a\in A$, and then repeating with $A$ and $C$ interchanged, yields $$ d_H(A,C)\leq d_H(A,B)+d_H(B,C). $$ Thus $d_H$ 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, $$ d_H(\{x\},\{y\})=d(x,y), $$ 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 $\varepsilon>0$ and all sufficiently large $n$, $$ A\subseteq E_\varepsilon(\{x_n\}). $$ Hence any $a,b\in A$ satisfy $d(a,b)\leq d(a,x_n)+d(x_n,b)\leq2\varepsilon$. Letting $\varepsilon\downarrow0$ gives $a=b$, so the nonempty set $A$ is a singleton. This is the . If $H(X)$ is complete, its closed subspace $\theta(X)$ is complete. Since $\theta$ is an isometry, $X$ is complete. Hence . Solved by gpt-5.6-sol high. = 11E {parent=Paper 1} {scope} {title2=Geometry} = i {parent=11e} {scope} = Solution {parent=i} A regularly parametrized curve $\gamma$ on an embedded surface $\Sigma$ is a geodesic when its covariant vanishes. Equivalently, in the ambient Euclidean space, $$ \ddot\gamma(t)\perp T_{\gamma(t)}\Sigma $$ for an affine geodesic parameter. A is one whose maximal affine parameter interval is all of $\mathbb R$. Solved by gpt-5.6-sol high. = ii {parent=11e} {scope} = Solution {parent=ii} Tangency along $\gamma$ means $$ T_{\gamma(t)}\Sigma_1=T_{\gamma(t)}\Sigma_2 $$ for 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 : $$ \boxed{\gamma\text{ is geodesic in }\Sigma_1 \Longleftrightarrow \gamma\text{ is geodesic in }\Sigma_2}. $$ Solved by gpt-5.6-sol high. = iii {parent=11e} {scope} = Solution {parent=iii} For fixed $u$, varying $v$ traces a unit circle in the radial-vertical plane centred at $$ (2\cos u,2\sin u,u). $$ Thus $\Sigma_1$ is a unit tube whose centre winds helically around the $z$-axis. Since $$ \sigma(t,\pi)=(\cos t,\sin t,t)=\gamma(t), $$ the given curve is its inner helix. Take $\Sigma_2$ to be the circular cylinder $x^2+y^2=1$. Along $v=\pi$, $$ \sigma_u=(-\sin u,\cos u,1), \qquad \sigma_v=(0,0,-1), $$ so their span is the tangent plane of that cylinder. The two surfaces are therefore tangent along $\gamma$. Moreover, $$ \ddot\gamma(t)=(-\cos t,-\sin t,0), $$ 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 . Solved by gpt-5.6-sol high. = iv {parent=11e} {scope} = Solution {parent=iv} Set $$ \Gamma(t)=\sigma(t,0)=(3\cos t,3\sin t,t), \qquad t\in\mathbb R. $$ Along $v=0$, $$ \sigma_u=(-3\sin u,3\cos u,1), \qquad \sigma_v=(0,0,1), $$ whose span is the tangent plane of the cylinder $x^2+y^2=9$. Also $$ \ddot\Gamma(t)=(-3\cos t,-3\sin t,0) $$ is 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 $1$. Solved by gpt-5.6-sol high. = 12F {parent=Paper 1} {scope} {title2=Complex Analysis OR Complex Methods} = a {parent=12f} {scope} = Solution {parent=a} The can be stated through the Laurent expansion of $f$ about $a$: * The singularity is removable when $f$ extends holomorphically to $a$, equivalently when every negative-power Laurent coefficient is zero. * It is a pole of order $k$ when $(z-a)^kf(z)$ extends holomorphically to a nonzero at $a$. Equivalently, the starts with a nonzero $(z-a)^{-k}$ term and has no more negative power. * It is essential when it is neither removable nor a pole, equivalently when infinitely many negative-power coefficients are nonzero. Solved by gpt-5.6-sol high. = b {parent=12f} {scope} = Solution {parent=b} Write the Laurent expansion on the punctured disc as $$ f(z)=\sum_{n=-\infty}^{\infty}a_nz^n. $$ For every integer $k\geq1$, the Laurent coefficient formula and the give $$ \begin{aligned} |a_{-k}| &=\left|\frac{r^k}{2\pi} \int_0^{2\pi}f(re^{i\theta})e^{ik\theta}\,d\theta\right|\\ &\leq\frac{r^k}{2\pi} \left(\int_0^{2\pi}|f(re^{i\theta})|^2d\theta\right)^{1/2} \left(\int_0^{2\pi}1\,d\theta\right)^{1/2}\\ &\leq\frac{r^k}{\sqrt{2\pi}}. \end{aligned} $$ Letting $r\downarrow0$ shows that $a_{-k}=0$. Every coefficient in the principal part vanishes, so the proves that $$ \boxed{f\text{ has a removable singularity at }0}. $$ Solved by gpt-5.6-sol high. = c {parent=12f} {scope} = i {parent=c} {scope} = Solution {parent=i} Since $\tan z=\sin z/\cos z$, its singularities occur at the simple zeros of the denominator, $$ a_k=\frac{(2k+1)\pi}{2}, \qquad k\in\mathbb Z. $$ At each one, $\sin a_k\ne0$ and $(\cos z)'|_{a_k}=-\sin a_k\ne0$, so all are simple poles. There are no other finite singularities: $$ \boxed{\tan z\text{ is meromorphic with simple poles at }a_k}. $$ Solved by gpt-5.6-sol high. = ii {parent=c} {scope} = Solution {parent=ii} Put $$ z_k=\frac1{a_k}=\frac{2}{(2k+1)\pi}, \qquad w=z-z_k. $$ Near $a_k$, writing $\delta=1/z-a_k$ gives $$ \tan(a_k+\delta)=-\cot\delta=-\frac1\delta+\frac\delta3+O(\delta^3). $$ But $$ \delta=\frac1z-\frac1{z_k}=-\frac{w}{z_k(z_k+w)}, $$ and hence, exactly, $$ -\frac1\delta=\frac{z_k(z_k+w)}w=\frac{z_k^2}{w}+z_k. $$ Since $\delta=O(w)$, the first two Laurent terms are $$ \boxed{ \tan\frac1z =\frac{z_k^2}{z-z_k}+z_k+O(z-z_k)}. $$ Equivalently, substitute $z_k=2/((2k+1)\pi)$ in this expression. Solved by gpt-5.6-sol high. = iii {parent=c} {scope} = Solution {parent=iii} At every $z_k$, the $\tan(1/z)$ has a simple pole. The , so $$ \boxed{z_k=\frac{2}{(2k+1)\pi}\quad(k\in\mathbb Z)} $$ are 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 : zero is a , specifically an accumulation point of essential singularities, rather than an isolated essential singularity. There are no other singularities in $\mathbb C$. Solved by gpt-5.6-sol high. = 13B {parent=Paper 1} {scope} {title2=Methods} = Solution {parent=13B} Continuity at $x=\xi$ is automatic in the proposed expression. Integrating the differential equation through $x=\xi$ gives the jump $$ G_x(\xi^+,\xi)-G_x(\xi^-,\xi)=1. $$ Thus $$ c(\xi)\left[y_1(\xi)y_2'(\xi)-y_1'(\xi)y_2(\xi)\right]=1. $$ In terms of the $$ W(x)=y_1(x)y_2'(x)-y_1'(x)y_2(x), $$ one has $$ \boxed{c(\xi)=\frac1{W(\xi)}}. $$ The hold because $y_1'(0)=0$ and $y_2'(1)=0$. The gives $W'=-\alpha W$. If $\alpha=0$, the Wronskian and hence $c$ are constant. The two branches of the displayed formula are then interchanged by $x\leftrightarrow\xi$, proving $$ \boxed{G(x,\xi)=G(\xi,x)}. $$ This is the symmetry of the in the self-adjoint case. For $G''-G=\delta(x-\xi)$, choose $$ y_1(x)=\cosh x, \qquad y_2(x)=\cosh(1-x). $$ Their Wronskian is $$ W=-\cosh x\sinh(1-x)-\sinh x\cosh(1-x)=-\sinh1. $$ Writing $x_<=\min(x,\xi)$ and $x_>=\max(x,\xi)$ gives $$ \boxed{G(x,\xi)=-\frac{\cosh x_<\cosh(1-x_>)}{\sinh1}}. $$ The solution of the inhomogeneous problem is $y(x)=\int_0^1G(x,\xi)\xi\,d\xi$. Equivalently, solving directly gives $$ y(x)=A\cosh x+B\sinh x-x. $$ The two Neumann conditions yield $B=1$ and $$ A=\frac{1-\cosh1}{\sinh1}=-\tanh\frac12. $$ Therefore $$ \boxed{y(x)=\sinh x-\tanh\left(\frac12\right)\cosh x-x}. $$ Solved by gpt-5.6-sol high. = 14A {parent=Paper 1} {scope} {title2=Quantum Mechanics} = i {parent=14a} {scope} = Solution {parent=i} The anti-muon has the same charge as a proton, so the Coulomb potential seen by the electron is the same as in the . The main difference is the source mass. Separating the centre-of-mass motion replaces $m_e$ by the $$ \mu=\frac{m_em_\mu}{m_e+m_\mu}. $$ Since $m_\mu\gg m_e$, 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. Solved by gpt-5.6-sol high. = ii {parent=14a} {scope} = Solution {parent=ii} For fixed $l$, the radial equation is $$ -\frac{\hbar^2}{2\mu r^2}\frac{d}{dr} \left(r^2\frac{dR}{dr}\right) +\frac{\hbar^2l(l+1)}{2\mu r^2}R -\frac{e^2}{4\pi\epsilon_0r}R =ER. $$ Substitute $$ R(r)=Cr^l e^{-r/b}. $$ Cancellation of the coefficient of $1/r$ requires $$ b=(l+1)a, \qquad \boxed{a=\frac{4\pi\epsilon_0\hbar^2}{\mu e^2}}, $$ the reduced-mass . The remaining constant term gives $$ \boxed{E_l=-\frac{\hbar^2}{2\mu a^2(l+1)^2} =-\frac{e^2}{8\pi\epsilon_0a(l+1)^2}}. $$ Each such no-radial-node state is a with principal quantum number $n=l+1$. Solved by gpt-5.6-sol high. = iii {parent=14a} {scope} = Solution {parent=iii} Comparing the prepared energy with the formula in part (ii), $$ \frac1{8(l+1)^2}=\frac1{72}, $$ so $l+1=3$ and $l=2$. The radial probability density, apart from normalization, is $$ r^2|R(r)|^2=r^{2l+2}e^{-2r/[a(l+1)]}. $$ Using the given factorial , its mean is the ratio $$ \langle r\rangle =\frac{\int_0^\infty r^{2l+3}e^{-2r/[a(l+1)]}\,dr} {\int_0^\infty r^{2l+2}e^{-2r/[a(l+1)]}\,dr} =\frac{(2l+3)(l+1)}2a. $$ For $l=2$ this gives the $$ \boxed{\langle r\rangle=\frac{21}{2}a}, \qquad \boxed{t=\frac{21}{2}}. $$ Solved by gpt-5.6-sol high. = iv {parent=14a} {scope} = Solution {parent=iv} The energy fixes $l=2$, irrespective of the magnetic quantum number $m$. Therefore every prepared state is an eigenstate of $\hat L^2$ with $$ L=\sqrt{l(l+1)}\hbar=\sqrt6\hbar. $$ An immediate measurement of the total orbital-angular-momentum magnitude cannot yield $\hbar$, so the requested probability is $$ \boxed{0}. $$ Solved by gpt-5.6-sol high. = 15C {parent=Paper 1} {scope} {title2=Electromagnetism} = Solution {parent=15C} Differentiate under the and use $\nabla_x|x-x'|^{-1}=-\nabla_{x'}|x-x'|^{-1}$: $$ \begin{aligned} \nabla\cdot A &=-\frac{\mu_0}{4\pi}\int J_i(x') \partial_i'\frac1{|x-x'|}\,d^3x'\\ &=\frac{\mu_0}{4\pi}\int \frac{\nabla'\cdot J(x')}{|x-x'|}\,d^3x'. \end{aligned} $$ The omitted boundary term vanishes if the current is localized sufficiently rapidly. A steady current obeys charge conservation $\nabla'\cdot J=0$, hence $$ \boxed{\nabla\cdot A=0}. $$ For $r=|x|$ much larger than the source size, $$ \frac1{|x-x'|}=\frac1r+\frac{x\cdot x'}{r^3}+O(r^{-3}|x'|^2). $$ Localization and $\nabla'\cdot J=0$ imply $\int J\,d^3x'=0$. They also imply $$ \int(x_i'J_j+x_j'J_i)\,d^3x'=0, $$ by integrating $\partial_k'(x_i'x_j'J_k)$. Thus the first nonzero moment is antisymmetric and can be written using the $$ \boxed{m=\frac12\int x'\times J(x')\,d^3x'}. $$ Consequently the has far field $$ \boxed{A(x)=\frac{\mu_0}{4\pi}\frac{m\times x}{r^3}+\cdots}. $$ The dimensions are $$ \boxed{[J]=\mathrm{A\,m^{-2}}}, \qquad \boxed{[m]=\mathrm{A\,m^2}}. $$ Solved by gpt-5.6-sol high. = i {parent=15c} {scope} = Solution {parent=i} The total hoop charge is $Q=2\pi R\eta$. One revolution takes $2\pi/\omega$, so the current is $$ I=\frac{Q\omega}{2\pi}=\eta R\omega. $$ Multiplying by the enclosed area gives $$ \boxed{m=I\pi R^2\hat n =\pi\eta\omega R^3\hat n}. $$ Solved by gpt-5.6-sol high. = ii {parent=15c} {scope} = Solution {parent=ii} An annulus of radius $s$ and width $ds$ has charge $dQ=2\pi s\sigma\,ds$, current $$ dI=\frac{\omega}{2\pi}dQ=\sigma\omega s\,ds, $$ and dipole moment $dm=\pi s^2dI\,\hat n$. Therefore the is $$ \boxed{ m=\pi\sigma\omega\int_0^R s^3\,ds\,\hat n =\frac{\pi\sigma\omega R^4}{4}\hat n}. $$ Solved by gpt-5.6-sol high. = 16D {parent=Paper 1} {scope} {title2=Fluid Dynamics} = Solution {parent=16D} Let $x$ point downslope and $z$ point normally away from the plane, with $0\leq z\leq h$. Write the parallel as $u(z)e_x$ and take $\tau>0$ to be the magnitude of the air's upslope stress. The steady equations are $$ 0=-p_x+\rho g\sin\alpha+\mu u'', \qquad 0=-p_z-\rho g\cos\alpha, $$ with $$ u(0)=0, \qquad p(h)=p_0, \qquad \mu u'(h)=-\tau. $$ The free-surface condition makes $p_x=0$, and integration gives $$ \boxed{p(x,z)=p_0+\rho g\cos\alpha\,(h-z)}, $$ $$ \boxed{u(z)=\frac{\rho g\sin\alpha}{\mu} \left(hz-\frac{z^2}{2}\right)-\frac{\tau z}{\mu}}. $$ The surface , downslope shear exerted by the fluid on the plane, and volume flux per unit width are $$ \boxed{u_h=\frac h\mu\left(\frac{\rho gh\sin\alpha}{2}-\tau\right)}, $$ $$ \boxed{\tau_0=\mu u'(0)=\rho gh\sin\alpha-\tau}, $$ $$ \boxed{q=\int_0^hu(z)\,dz =\frac{\rho g h^3\sin\alpha}{3\mu} -\frac{\tau h^2}{2\mu}}. $$ Thus the reverses in the three senses when $$ \begin{array}{c|c} \text{quantity directed upslope}&\text{condition}\\ \hline u_h&\tau>\frac12\rho gh\sin\alpha\\ q&\tau>\frac23\rho gh\sin\alpha\\ \tau_0&\tau>\rho gh\sin\alpha. \end{array} $$ The order of increasing required air stress is therefore $$ \boxed{\text{surface velocity},\quad\text{volume flux},\quad\text{stress on the plane}}. $$ Solved by gpt-5.6-sol high. = 17A {parent=Paper 1} {scope} {title2=Numerical Analysis} = Solution {parent=17A} A $\Omega^{[p,q]}$ is the identity except in rows and columns $p,q$, where it has the block $$ \begin{pmatrix}c&s\\-s&c\end{pmatrix}, \qquad c=\cos\theta, \quad s=\sin\theta. $$ For $a=A_{pj}$ and $b=A_{qj}$, choose $$ c=\frac{a}{\sqrt{a^2+b^2}}, \qquad s=\frac{b}{\sqrt{a^2+b^2}} $$ when $(a,b)\ne(0,0)$. Then $$ (\Omega^{[p,q]}A)_{qj}=-sa+cb=0. $$ If both entries vanish, any angle works. For the given , first use $$ \Omega^{[1,3]}= \begin{pmatrix} 1/\sqrt2&0&1/\sqrt2\\ 0&1&0\\ -1/\sqrt2&0&1/\sqrt2 \end{pmatrix}. $$ It gives $$ \Omega^{[1,3]}A= \begin{pmatrix} 2&1&1\\ 0&\sqrt3&0\\ 0&1&\sqrt3 \end{pmatrix}. $$ Then use $$ \Omega^{[2,3]}= \begin{pmatrix} 1&0&0\\ 0&\sqrt3/2&1/2\\ 0&-1/2&\sqrt3/2 \end{pmatrix}. $$ Thus $R=\Omega^{[2,3]}\Omega^{[1,3]}A$ is $$ R= \begin{pmatrix} 2&1&1\\ 0&2&\sqrt3/2\\ 0&0&3/2 \end{pmatrix}. $$ All leading row entries are positive. The resulting is $$ \boxed{A=QR}, $$ where $$ \boxed{ Q=(\Omega^{[2,3]}\Omega^{[1,3]})^T =\begin{pmatrix} \sqrt2/2&-\sqrt2/4&-\sqrt6/4\\ 0&\sqrt3/2&-1/2\\ \sqrt2/2&\sqrt2/4&\sqrt6/4 \end{pmatrix}} $$ and $R$ is the above. Since it is a product of transposed rotations, $Q$ is orthogonal. Solved by gpt-5.6-sol high. = 18H {parent=Paper 1} {scope} {title2=Statistics} = a {parent=18h} {scope} = Solution {parent=a} Write $X_i=\theta+U_i$, where the $U_i$ are independent $\operatorname{Uniform}(0,1)$ variables. Convolution gives the : $$ f_{X_1+X_2}(s)= \begin{cases} s-2\theta,&2\theta\leq s\leq2\theta+1,\\ 2\theta+2-s,&2\theta+1\leq s\leq2\theta+2,\\ 0,&\text{otherwise}. \end{cases} $$ Its plot is a triangle with vertices $$ (2\theta,0), \qquad(2\theta+1,1), \qquad(2\theta+2,0). $$ Solved by gpt-5.6-sol high. = b {parent=18h} {scope} = Solution {parent=b} For Test 1, the rejection probability is increasing in $\theta$, so its size over $\theta\leq0$ is attained at $\theta=0$ and equals $$ \mathbb P_0(X_1>0.95)=0.05. $$ Test 2 is also monotone under shifts in $\theta$, so its size is $\mathbb P_0(U_1+U_2>C)$. For $1\leq C\leq2$, the triangular upper tail is $$ \mathbb P(U_1+U_2>C)=\frac{(2-C)^2}{2}. $$ Equating this to $0.05$ gives $$ \boxed{C=2-\frac1{\sqrt{10}}}. $$ Solved by gpt-5.6-sol high. = c {parent=18h} {scope} = Solution {parent=c} The for Test 1 is $$ \boxed{ \pi_1(\theta)= \begin{cases} 0,&\theta\leq-0.05,\\ \theta+0.05,&-0.05<\theta<0.95,\\ 1,&\theta\geq0.95. \end{cases}} $$ Put $a=1/\sqrt{10}$. Test 2 rejects when $$ U_1+U_2>2-a-2\theta. $$ Using the two branches of the triangular distribution gives $$ \boxed{ \pi_2(\theta)= \begin{cases} 0,&\theta\leq-a/2,\\ \frac12(a+2\theta)^2, &-a/2<\theta<(1-a)/2,\\ 1-\frac12(2-a-2\theta)^2, &(1-a)/2\leq\theta<(2-a)/2,\\ 1,&\theta\geq(2-a)/2. \end{cases}} $$ Both equal $0.05$ at the least favourable null value $\theta=0$. Solved by gpt-5.6-sol high. = d {parent=18h} {scope} = Solution {parent=d} The two tests have the same size, but their powers cross. At $\theta=0.1$, $$ \pi_1(0.1)=0.15, \qquad \pi_2(0.1)=\frac12\left(\frac1{\sqrt{10}}+0.2\right)^2<0.15. $$ At $\theta=0.5$, $$ \pi_1(0.5)=0.55, \qquad \pi_2(0.5)=1-\frac12\left(1-\frac1{\sqrt{10}}\right)^2>0.55. $$ Thus Test 2 is a same-size test that beats Test 1 at one alternative, while Test 1 beats Test 2 at another. By the , $$ \boxed{\text{neither Test 1 nor Test 2 is uniformly most powerful}.} $$ Solved by gpt-5.6-sol high. = 19H {parent=Paper 1} {scope} {title2=Markov Chains} = a {parent=19h} {scope} = Solution {parent=a} Starting from $i$, the probability of reaching $j$ before returning to $i$ is $\alpha$. Conditional on reaching $j$, each visit to $j$ is followed by a hit on $i$ before the next return to $j$ with probability $\alpha$, by symmetry and the . For $\alpha>0$, the number of visits after entry is therefore geometric on $\{1,2,\ldots\}$ with mean $1/\alpha$. Hence the gives $$ \boxed{\mathbb E_iN=\alpha\frac1\alpha=1} \qquad(\alpha>0). $$ In the degenerate case $\alpha=0$, the chain never reaches $j$ before its return to $i$, so $N=0$ almost surely and $\mathbb E_iN=0$. Solved by gpt-5.6-sol high. = b {parent=19h} {scope} = Solution {parent=b} There are no visits when the initial excursion returns to $i$ before hitting $j$, so $$ \boxed{\mathbb P_i(N=0)=1-\alpha}. $$ For $k\geq1$, the chain must first hit $j$, return to $j$ before $i$ exactly $k-1$ times, and then hit $i$ before another return to $j$. The gives $$ \boxed{ \mathbb P_i(N=k)=\alpha^2(1-\alpha)^{k-1}, \qquad k\geq1}. $$ The formula also covers the boundary cases $\alpha=0$ and $\alpha=1$. Solved by gpt-5.6-sol high. = c {parent=19h} {scope} = Solution {parent=c} For the converse as well as the stated symmetric case, set $$ a=\mathbb P_i(T_j0$. During one return cycle from $i$ to $i$, the chain enters $j$ with probability $a$ and, after entry, makes a geometric number of visits to $j$ with success parameter $b$. Therefore $$ \mathbb E_iN=\frac ab. $$ On the other hand, the gives $$ \mathbb E_iN=\frac{\pi(j)}{\pi(i)}. $$ Consequently $$ \frac ab=\frac{\pi(j)}{\pi(i)}. $$ It follows that $$ \boxed{a=b\quad\Longleftrightarrow\quad\pi(i)=\pi(j)}, $$ which is precisely the claimed equivalence between symmetry of $i,j$ and equality of their invariant masses. Solved by gpt-5.6-sol high.