Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2

Let $e^2=e$ with $e\ne 0,1$. Then $e(1-e)=0$, and the ideals $eR$ and $(1-e)R$ are nonzero rings with identities $e$ and $1-e$. DefineThis is a ring homomorphism. Its inverse is $(x,y)\mapsto x+y$, because $ex=x$, $(1-e)y=y$, and the cross terms vanish. Hence $R\cong eR\times (1-e)R$, proving (ii).

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Conversely, if $R\cong R_1\times R_2$ with both factors nontrivial, the inverse image of $(1_{R_1},0)$ is an idempotent. It is neither zero nor one, so (i) holds. This completes the ring product decomposition by an idempotent equivalence.

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Forthe parity condition is preserved by componentwise addition, negation, and multiplication, and $(1,1)$ is the identity. Thus $R$ is a ring.

It is not an integral domain, since $(2,0)$ and $(0,2)$ are nonzero elements of $R$ whose product is zero. It is also not a product of two nontrivial rings. Indeed, an idempotent in ${\mathbb{Z}}^2$ has each coordinate in $\{0,1\}$, and the parity condition leaves only $(0,0)$ and $(1,1)$. The proved equivalence then excludes a nontrivial product decomposition.

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Suppose first that $X$ is connected and $f:X\to \mathbb{Z}$ is continuous. Its image is connected by the continuous image of a connected space, but $\mathbb{Z}$ is discrete, so its only connected nonempty subsets are singletons. Hence $f$ is constant.

Conversely, if $X=U\cup V$ is a disconnection into nonempty disjoint open sets, the function equal to zero on $U$ and one on $V$ is continuous and nonconstant. This proves the integer-valued function criterion for connectedness.

Now let $f:X\to \mathbb{Z}$ be continuous under the hypotheses on the family $\mathcal{A}$. Each restriction $f{|}_A$ is constant because $A$ is connected. If $A,B\in \mathcal{A}$, a point of $A\cap B$ shows that their two constants agree. Since the sets cover $X$, $f$ is constant on $X$, and the criterion proves that $X$ is connected. This is the pairwise-intersecting connected cover argument.

Finally, fix $y_0\in Y$. For each $x\in X$, the setis connected: its two connected pieces meet at $(x,y_0)$. The sets $A_x$ cover $X\times Y$ and any two share $X\times \{y_0\}$. The preceding result proves that $X\times Y$ is connected, giving the product of connected spaces result.

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The function is odd, so only sine coefficients occur. For $n\ge 1$,Integration by parts givesandThe terms of order $1/n$ cancel, leavingThus

The Parseval identity givesDirect integration yieldsThereforeas recorded in the Fourier series of x cubed minus pi squared x.

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A capacitor consists of two conductors carrying equal and opposite charges. Its capacitance iswhere $Q$ is the magnitude of the charge on either conductor and $V$ is their potential difference.

For $a<r<b$, a coaxial Gaussian cylinder of length $\ell$ encloses charge $\lambda \ell$. Gauss's law givesTaking $V$ to mean the inner potential minus the outer potential,Since $Q=\lambda L$,

The field energy isThese are the standard coaxial cylindrical capacitor formulas.

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The divergence is the trace of the velocity-gradient tensor:Thus the flow is incompressible exactly whenThe constant vector $U_0$ is unrestricted.

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The vorticity isHence the flow is irrotational exactly whenequivalently when $\Gamma$ is symmetric. Again, $U_0$ is unrestricted.

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A streamline $x(s)=(x(s),y(s),z(s))$ satisfieswith $(x(0),y(0),z(0))=(1,0,0)$. ThereforeFor $\beta \ne 0$ this is a helix, with $x=\cos (\alpha z/\beta )$ and $y=-\sin (\alpha z/\beta )$; for $\beta =0$ it is the unit circle in the plane $z=0$.

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Apart from constants, the log-likelihood isDifferentiating gives ${\ell }^{\prime }(\theta )=n(\overline{X}-\theta )/{\sigma }^2$, so

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Sincean exact 95 percent confidence interval is

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Write the posterior mean and variance asThe central 95 percent posterior credible interval is

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As $n\to \infty$,whileThus the credible interval and confidence interval have asymptotically equal centres and half-widths, and both contract around the true parameter. Their interpretations remain different: the confidence statement concerns repeated samples, whereas the credible statement concerns posterior probability. This is normal-normal conjugacy with known observation variance.

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UseStationarity givesPutting $t=\lambda /2$, this becomesand the constraint requiresThe left side is strictly increasing, and $t=1$ solves the equation. Hence

The objective is convex and the constraint is affine. Its tangent-plane inequality at $(1,1,1)$ gives, for every feasible $(x,y,z)$,so the Lagrange point is globally optimal. Moreover, at the dual value $\lambda =2$, the infimum of the Lagrangian function in constrained optimization is attained at the same point and equals three. The primal and dual values coincide, so strong duality holds.

For the value function, the multiplier convention above gives the derivative of a constrained value functionAt $b=6$, therefore,

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For an $n\times n$ matrix $A$, the characteristic polynomial isThe Cayley-Hamilton theorem states that ${\chi }_A(A)=0$.

Over $\mathbb{C}$, choose a basis in which $A$ is upper triangular, with diagonal entries ${\lambda }_1,\dots ,{\lambda }_n$. For the standard invariant flag $V_j=⟨e_1,\dots ,e_j⟩$,The factors $A-{\lambda }_{j}I$ commute, so applying their product in descending order sends $V_n$ successively into $V_{n-1},\dots ,V_0=0$. Hencewhich proves the theorem.

Direct expansion gives the commutator product rule:

Put $C=[B,A]$. Since $C$ commutes with $A$, repeated use of the product rule givesBy linearity, for every polynomial $\phi$,

Let $D(X)=[B,X]$ and suppose $f(A)=0$. For $k=1$,Assume inductively thatBoth $u$ and $C$ are polynomials in, or commute with, $A$, so $u{C}^m=C^{m}u=0$. Apply the derivation $D$ to $u{C}^m=0$ and multiply on the left by $C^m$:Since $D(u)=f^{(k+1)}(A)C$, this saysThe induction is complete. Taking $f={\chi }_A$ and $k=n$ givesThus $[B,A]$ is nilpotent, which is the Jacobson lemma for a commuting commutator.

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Let $Q$ act by left multiplication on the set $G/P$ of left cosets. Every orbit has size a power of $p$, whileis not divisible by $p$. Therefore at least one orbit has size one. If $gP$ is fixed, then $qgP=gP$ for every $q\in Q$, so $g^{-1}Qg\le P$, equivalentlyThis is Sylow containment from a coset fixed point.

The remaining Sylow theorems state that Sylow $p$-subgroups exist, that any two are conjugate, and that their number $n_p$ satisfies

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By transitivity and the orbit-stabilizer theorem,Inside $G_x\cong S_4$, there are three Sylow $2$-subgroups, each of order eight, and four Sylow $3$-subgroups, each of order three. These are also Sylow subgroups of $G$.

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Count pairs $(x,P)$ where $P$ is a Sylow $2$-subgroup fixing $x$. Each of the seven point stabilizers contains three such subgroups, giving 21 pairs. Conversely, a $2$-group acting on seven points has a fixed point, and a Sylow $2$-subgroup cannot fix two points because a two-point stabilizer has order four. Hence every Sylow $2$-subgroup occurs in exactly one pair, so

Similarly, each point stabilizer contains four Sylow $3$-subgroups, giving 28 pairs. A group of order three acting on seven points has a fixed point, and it cannot fix two because a two-point stabilizer is a $2$-group. Thus

The Sylow congruence and divisibility conditions give $n_7\in \{1,8\}$. Regard the faithful action as an embedding $G\le S_7$. If $n_7=1$, its Sylow $7$-subgroup $P$ is normal, so $G\le N_{S_7}(P)$. The stated fact gives $|N_{S_7}(P)|=7\cdot 6=42$, impossible for a subgroup $G$ of order 168. HenceThese are the Sylow counts in a faithful degree-seven action with S4 point stabilizers.

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Let $N⊴G$ be proper. If $7$ divides $|N|$, then $N$ contains a Sylow $7$-subgroup of $G$, and normality makes it contain all eight of them. Thus $n_7(N)=8$, so the Sylow divisibility theorem givesSince $|N|$ divides 168 and $N$ is proper, this forces $|N|=56$.

Every nontrivial normal subgroup is transitive by normal subgroup orbits in a faithful prime-degree action. HenceBut $N\cap G_x$ is normal in $G_x\cong S_4$, whereas $S_4$ has three, rather than one, Sylow $2$-subgroups. This contradiction proves that no proper normal subgroup has order divisible by seven.

If instead $3$ divides $|N|$, normality makes $N$ contain all 28 Sylow $3$-subgroups of $G$. Hence $n_3(N)=28$, so $28$ divides $|N|/3$ and therefore $84$ divides $|N|$. A proper such subgroup would have order 84, which is divisible by seven and has just been ruled out.

Finally, if $N$ is any nontrivial normal subgroup, the prime-degree orbit argument makes $N$ transitive, so seven divides $|N|$. This is impossible for a proper normal subgroup. Therefore $G$ is simple.

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A sequence $f_n:S\to \mathbb{R}$ converges uniformly to $f$ if, for every $\epsilon >0$, there is $N$ such thatfor every $x\in S$ and every $n\ge N$. A map $h:M\to N$ is uniformly continuous if, for every $\epsilon >0$, there is $\delta >0$ such thatfor all $x,y\in M$.

If $f\in C_0({\mathbb{R}}^d)$, choose a ball outside which $|f|<1$. On the closed ball, continuity gives boundedness, so $f$ is bounded everywhere.

Now let $(f_n)$ be Cauchy in the uniform metric. For each $x$, $(f_n(x))$ is Cauchy in $\mathbb{R}$; let its limit be $f(x)$. Passing to the pointwise limit in the uniform Cauchy estimate shows that $f_n\to f$ uniformly. Hence $f$ is continuous. Given $\epsilon >0$, choose $n$ with $\Vert f-f_n{\Vert }_{\infty }<\epsilon /2$, and then choose $K$ so that $|f_n(x)|<\epsilon /2$ for $\Vert x\Vert >K$. It follows that $f$ also vanishes at infinity. Thus $C_0({\mathbb{R}}^d)$ is complete, as in completeness of continuous functions vanishing at infinity.

Every $f\in C_0({\mathbb{R}}^d)$ is uniformly continuous. Given $\epsilon >0$, choose $R$ so that $|f(x)|<\epsilon /2$ outside the ball of radius $R$. On the compact ball of radius $R+1$, $f$ is uniformly continuous; choose the corresponding $\delta \le 1$. If two points at distance below $\delta$ are not both in that ball, then both lie outside the ball of radius $R$, and their function values differ by less than $\epsilon$.

For the final sequence, continuity of $\epsilon$ at zero gives, for each fixed $x$,Thus pointwise convergence is compulsory. Uniform convergence need not hold: with $\epsilon (t)=t^2$,which is unbounded as a function of $x$ for every fixed $n$.

Under the additional bound $\epsilon (t)\le M|t|$, however,for $x\ne 0$, and the difference is zero at $x=0$. Hence convergence is uniform, by the uniform square-root perturbation under linear growth estimate. The pointwise answer remains yes.

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The tangent vectors ofareHence the first fundamental form has coefficientsPutThe upward unit normal is $N=(-f_u,-f_v,1)/W$, so the second fundamental form has coefficientsThus the two forms are

The Gaussian curvature isSince $EG-F^2=W^2$, the graph formula isas in Gaussian curvature of a graph surface.

For the final claim, fix a point of $\gamma$ and make a rigid motion taking $P$ to the plane $z=0$. The common tangent plane is horizontal, so locally $\Sigma$ is the graph $z=f(u,v)$. Along the projected curve $c(s)$ of tangency,Differentiating the second identity givesBecause $c$ is a smooth curve, $c^{\prime }(s)\ne 0$, so the Hessian is singular. Its determinant is zero, and the graph formula gives $K=0$ at every point of $\gamma$. This is tangency to a plane along a curve forces zero Gaussian curvature.

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Let $|f(z)|\le K$ on $\mathrm{Re}z>0$. If $\mathrm{Re}z>c$, the circle $|w-z|=c$ lies in that half-plane. The Cauchy derivative formula givesThe line segment joining $z_1$ to $z_2$ remains in $\mathrm{Re}z>c$, soThus one may take $M=K/c$, as in the lipschitz bound inside a bounded analytic half-plane.

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The Liouville theorem states that every bounded entire function is constant.

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Suppose $g(\mathbb{C})$ is not dense. Then some open disc $D(w,r)$ is disjoint from the image, sois entire and satisfies $|F(z)|\le 1/r$. Liouville's theorem makes $F$, and hence $g$, constant, a contradiction. Therefore every nonconstant entire $g$ has dense image, as asserted by dense image of a nonconstant entire function.

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Write $h(z)={\sum }_{n=0}^{\infty }{a}_n{z}^n$. Cauchy's coefficient formula on $|z|=R$ givesExcept at the two measure-zero points where $\cos \theta =0$, the hypothesis givesSince $|\cos \theta {|}^{-1/2}$ is integrable,for a constant $C$ independent of $R$. Letting $R\to \infty$ shows that every $a_n=0$. Hence $h\equiv 0$, in particular $h$ is constant. This proves the entire function under a horizontal inverse-square-root bound result.

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Take a variation $y+\epsilon \eta$, where the differentiable function $\eta$ obeysbecause both $y$ and its first derivative have fixed endpoint values. The first variation of the functional isApplying integration by parts once to the second term and twice to the third givesThe endpoint conditions on $\eta$ and ${\eta }^{\prime }$ make every boundary term zero. Since the remaining integral vanishes for every admissible variation, the fundamental lemma of the calculus of variations yields the higher-order Euler-Lagrange equation

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HereSubstitution into the higher-order Euler-Lagrange equation gives the fourth-order ordinary differential equation

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Integrating the ordinary differential equation four times gives a quartic polynomial. It is useful to use the linearity of the equation and split the solution into a gravity part and a force part:wheresatisfies the clamped and torque-free boundary conditions with $F=0$, whilesatisfies the homogeneous equation and contributes the endpoint force. Direct differentiation verifiesThusthe clamped-free beam under uniform load and endpoint force.

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Evaluating the function at the endpoint gives the displacementThereforeThe force-induced displacement $h$ is a linear map of $F$.

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Substitute $y=y_0+y_F$ into the energy functional and integrate the resulting polynomial:The term linear in $F$ cancels. HenceAs required, $E_0$ is independent of the force and $\mathcal{E}$ is a quadratic function of it.

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Taking the derivative of the force-dependent energy givesThus the derivative of the additional minimized internal energy with respect to the applied force equals the resulting endpoint displacement. This is the endpoint force derivative of clamped-free beam energy.

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Apply separation of variables to the Laplace equation in polar coordinates by writing $\varphi (r,\theta )=R(r)\Theta (\theta )$. Division by $R\Theta$ givesThe requirement that $\Theta$ be a real $2\pi$-periodic function restricts the separation constants to $n^2$, with angular factors $\cos (n\theta )$ and $\sin (n\theta )$. For $n\ge 1$, the radial ordinary differential equation is an Euler equation with solutions $r^n$ and $r^{-n}$. For the zero mode, $\Theta$ is constant andso $R=a_0+c_0\log r$. By linearity, superposition givesThis is the separated expansion of a harmonic function in a circular region.

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Regularity at the origin excludes the natural logarithm and every negative radial power. ThusThe coefficients $a_0,a_n,b_n$ remain arbitrary.

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Regularity at infinity excludes the natural logarithm and every positive radial power. ThusThe constant $a_0$ and the decaying coefficients $c_n,d_n$ remain arbitrary.

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An annulus stays away from both the origin and infinity, so every displayed radial mode is regular there. Therefore none of the coefficients is forced to vanish.

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Write $s=\log r$, so the two circular boundaries are $s=0$ and $s=2$. The zero angular mode must interpolate between $-1$ and $1$, giving $s-1=\log r-1$.

For the $n$th cosine mode, the radial factor has equal value $A_n$ at both boundaries. The unique harmonic function with those data isHence the solution of the annular Dirichlet problem isAt $r=1$ and $r=e^2$, the hyperbolic cosine quotient equals one, so the boundary conditions are satisfied term by term.

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The mean of $f$ is zero. Its Fourier cosine series coefficients areThus $A_n=0$ for even $n$ and $A_n=4/(\pi n^2)$ for odd $n$. Substitution into part (i) yields