Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 2

An integral domain is a nonzero commutative ring with identity and no zero divisors. A ring is Noetherian when every ideal is finitely generated, equivalently every ascending ideal chain stabilizes. A principal ideal domain is an integral domain in which every ideal is generated by one element.

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Equality of constant terms is preserved by componentwise addition, subtraction, and multiplication, and $(1,1)$ belongs to the set, so $R$ is a subring. The projection is surjective andThe nonzero elements $(x,0)$ and $(0,y)$ have product zero, so $R$ is not an integral domain and hence not a PID. In fact $R\cong \mathbb{C}[x,y]/(xy)$, which also shows that it is Noetherian.

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Pointwise, $\tanh (nx)$ tends to $-1,0,1$ according as $x<0,x=0,x>0$. This limit is discontinuous, whereas a uniform limit of continuous functions is continuous. Convergence is therefore not uniform.

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The limit is $\sin x$, andso convergence is uniform.

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The pointwise limit is $f(x)=e^x/\cosh (ax)$ andThe final factor is bounded on $\mathbb{R}$ exactly when $|a|\ge 1$; when $|a|<1$ it diverges as $x\to +\infty$. Thus convergence is uniform exactly for $|a|\ge 1$.

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The fundamental theorem of calculus givesThus $f_n(x)\to 4h^{\prime }(x)$. Uniform continuity of $h^{\prime }$ makes the difference from $4h^{\prime }(x)$ uniformly small over every interval of radius $1/n$, so convergence is uniform on all of $\mathbb{R}$.

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Let $r_1,r_2$ be the roots of $\alpha r^2+\beta r+\gamma =0$. The causal Green function is $G(t,\tau )=H(t-\tau )g(t-\tau )$, wherefor distinct roots, and $g(s)=s{e}^{rs}/\alpha$ for a repeated root. These formulas enforce $g(0)=0$ and $\alpha g^{\prime }(0)=1$, hence the required delta jump.

For the oscillator,If $\lambda \ne \omega$, this isAt resonance $\lambda =\omega$, the limiting expression is

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Maxwell equations areTaking the divergence of the Ampère-Maxwell equation and using Gauss's law givesThereforebecause $J$ vanishes on the boundary. Componentwise integration by parts also giveswith the boundary term again zero.

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For incompressible inviscid flow without body force,Taking curl givesequivalently ${\partial }_t\omega =\nabla \times (u\times \omega )$.

Direct differentiation of the stated field gives $\omega =u$. Hence $u\times \omega =0$, so the vorticity equation gives ${\partial }_t\omega =0$ and the field is steady. FinallyEuler's equation therefore impliesso $H$ is spatially uniform.

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The likelihood is $\theta e^{-\theta X}$. Multiplication by the prior gives a density proportional to ${\theta }^m{e}^{-(\lambda +X)\theta }$, soin shape-rate parametrisation.

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Squared-error posterior risk is minimized by the posterior mean. Hence

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Put $k=m+1$ and $L=\lambda +X$. The posterior risk iswhere $M(s)=(L/(L-s){)}^k$. Differentiating in $a$ givesThus

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A set is convex when it contains every segment joining two of its points. A function on a convex set is convex whenfor all $x,y$ and $0\le t\le 1$.

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Take $(t,x),(s,y)$ and $0\le \alpha \le 1$, and put $T=\alpha t+(1-\alpha )s$. ThenConvexity of $f$, followed by multiplication by $T$, proves convexity of its perspective $g(t,x)=tf(x/t)$.

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Let $z=tx+(1-t)y$. Apply the hypothesis with first argument $z$ and second argument successively $x,y$:Multiply these inequalities by $t$ and $1-t$ respectively and add. The right sides cancel, leaving $f(z)\le tf(x)+(1-t)f(y)$.

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A Jordan block is $J_m(\alpha )=\alpha I+N$, where $N$ has ones on the superdiagonal and zeros elsewhere. Since $N^2=0$ for $m=2$, Taylor expansion givesThe Jordan normal form is the block-diagonal matrix, unique up to block order, to which $A$ is similar and whose blocks are Jordan blocks.

A block of size at least two for eigenvalue $\lambda$ contributes a generalized eigenvector in $\ker (A-\lambda I{)}^2\setminus \ker (A-\lambda I)$. Thus the displayed equality for every $\lambda$ is equivalent to every block having size one, which is diagonalizability.

The transpose of each Jordan block is similar to it by reversing its basis. Hence $A^T$ has the same Jordan form as $A$.

Write $A=PJ{P}^{-1}$. Let $R$ be block diagonal with the reversal matrix on each Jordan block. Both $R$ and $RJ$ are symmetric and $R$ is invertible. Thereforewhere $C,D$ are symmetric and $C$ is nonsingular.

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A euclidean domain is an integral domain with a function $d$ on nonzero elements such that for $a,b\ne 0$ there are $q,r$ with $a=bq+r$ and either $r=0$ or $d(r)<d(b)$.

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For a nonzero ideal $I$, choose $a\in I$ of minimal Euclidean value. Divide any $x\in I$ by $a$: $x=qa+r$. Then $r\in I$, and minimality forces $r=0$. Hence $I=(a)$; the zero ideal is also principal.

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The class $a$ is a unit modulo $p^e$ exactly when $\varphi (a)\ne 0$. This says a representative is not divisible by $p$, so it is coprime to $p^e$ and Bézout gives an inverse.

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The class $a$ is nilpotent exactly when $\varphi (a)=0$. Divisibility by $p$ makes $a^e$ zero modulo $p^e$. Conversely, the image of a nilpotent in the field $R/(p)$ is nilpotent and therefore zero.

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The Chinese remainder theorem identifiesThus a class is a unit exactly when its image modulo every $p_i$ is nonzero, equivalently when no $p_i$ divides a representative.

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A unit modulo $d$ has nonzero image modulo every $p_i$. Any representative therefore satisfies the criterion in part (c), so it is already a unit modulo $I$. Its image is the given unit modulo $d$, proving surjectivity on unit groups.

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The product topology has basis $U\times V$ with $U$ open in $X$ and $V$ open in $Y$. Since ${\Pi }_X^{-1}(U)=U\times Y$ and similarly for ${\Pi }_Y$, both projections are continuous.

If $Y$ is Hausdorff, a point $(y_1,y_2)$ off the diagonal has disjoint product neighborhoods, so the diagonal is closed. Conversely, if the diagonal is closed, its complement supplies disjoint neighborhoods of any two distinct points.

The graph is the inverse image of the diagonal under $(x,y)\mapsto (f(x),y)$, so a continuous map into a Hausdorff space has closed graph. Conversely, if $X,Y$ are compact Hausdorff and $C\subseteq Y$ is closed, then ${\Gamma }_f\cap (X\times C)$ is compact. Its projection onto $X$ is $f^{-1}(C)$, compact and hence closed. Thus $f$ is continuous.

The function $f(0)=0$, $f(x)=1/x$ for $x\ne 0$ is discontinuous but has closed graph: a convergent graph sequence approaching $x=0$ cannot have $x\ne 0$ and bounded second coordinate unless it eventually reaches the isolated graph point $(0,0)$.

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A topological surface is a Hausdorff second-countable space locally homeomorphic to ${\mathbb{R}}^2$. Pairing opposite sides of the hexagon gives neighborhoods modelled on discs, including at edge and vertex classes; compactness follows from the compact polygon quotient. Cutting and rearranging the polygon gives the usual opposite-side square, hence a torus.

The full double cone is not a surface at the origin: deleting the vertex from a small neighborhood leaves two components, unlike a punctured disc. The upper cone with $z>0$ excludes that vertex and has ordinary annular charts, so it is a surface.

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Writing the upper cone as $(r\cos \theta ,r\sin \theta ,r)$ gives metricOn any polar branch, the mappulls this back to $d{s}^2+s^{2}d{\phi }^2$, so it is a local isometry. Strictly, because $\sqrt{2}\phi$ is not $2\pi$-periodic, this formula is globally single-valued on the universal polar cover and locally on ${\mathbb{R}}^2\setminus \{0\}$; that is the developing map used here.

An isometry fixing $p$ lifts locally to a Euclidean isometry fixing a lift of $p$. Compatibility with the cone's angular identification leaves only the identity and reflection across a radial line. Descending to the cone, these are the identity and reflection in the plane through the cone axis and $p$.

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Apply the residue theorem to $f(z)/(z-w{)}^{n+1}$; its residue at $w$ is $f^{(n)}(w)/n!$, which proves Cauchy's derivative formula.

For a compact subdisc choose a slightly larger contour still inside $D(a,R)$. Cauchy's derivative formula bounds the supremum of $g_k^{(n)}-g^{(n)}$ on the smaller disc by a fixed constant times the supremum of $g_k-g$ on the contour. Local uniform convergence therefore passes to every derivative.

On $|z-a|=\epsilon$, $g$ has a positive minimum modulus. For large $k$, $|g_k-g|<|g|$ there. Rouché's theorem then says $g_k$ and $g$ have the same number of zeros inside, namely $m$, counting multiplicity.

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If $f$ had a pole at one of the finitely many candidate points, choose a polynomial vanishing to sufficiently high order at all the other candidates. Hermite interpolation lets its remaining Taylor coefficients be chosen so that the residue of $p^{2}f$ at the selected pole is nonzero. A small contour around that pole would then have nonzero integral by the residue theorem, contradicting the hypothesis. Hence none of the candidate points is a pole and $f$ is entire.

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Schwarz reflection shows that the Taylor coefficients along the real axis are real. Since $v(x,0)=0$ and $v(x,y)>0$ for $y>0$, one has $v_y(x,0)\ge 0$. The Cauchy–Riemann equations give $h^{\prime }(x)=u_x(x,0)=v_y(x,0)\ge 0$.

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If $h^{\prime }(0)=0$, let $a_k{z}^k$ be the first nonzero Taylor term. Then $k\ge 2$ and $a_k$ is real. But $\mathrm{Im}(a_k{r}^k{e}^{ik\theta })$ changes sign as $0<\theta <\pi$, and for small $r$ the leading term controls the sign. This contradicts positivity in the upper half-plane. Hence $h^{\prime }(0)\ne 0$.

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Put $W=1+{\rho }^{\prime 2}+{\rho }^2{\varphi }^{\prime 2}$. The Euler–Lagrange equations areWhen $n=n(\rho )$, the helical ansatz $\rho =R$, $\varphi ={\varphi }_0+\omega z$ automatically satisfies the second equation. The first becomesso $a=1$.

The endpoints have the same cylindrical angle exactly when $\omega L=2\pi k$ for some integer $k$. Thus a positive $L$ with $L=2\pi k/\omega (R)$ admits the two special solutions of opposite handedness, using $\omega$ and $-\omega$.

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Transverse force balance on a short element givesWith fixed ends,Here $A_n=0$. If $b_n$ are the sine coefficients of initial velocity, thenThe total deposited energy is $E=\rho v^2/2$, whileThe seventh mode is eliminated by striking at a node $l=jL/7$, $j=1,\dots ,6$. For a narrow hammer, each fixed-mode fraction is $O(\epsilon )$; beyond the inverse-width cutoff the envelope falls as $n^{-2}$, modulated by the two sine factors.

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The commutator follows directly on a test function:Expansion givesTo make $V_{-}=2m$, takeThenwhich tends to $2m$ at both ends.

Applying $p+if$ to $e^{ikx}$ gives a partner scattering state with asymptotic amplitudes $\hslash k-2mi$ at $-\infty$ and $\hslash k+2mi$ at $+\infty$, and no reflected wave. Their moduli agree. Subtracting the common constant $2m$ therefore shows that the displayed $-4m{\mathrm{sech}}^2$ potential is reflectionless:

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Taking curls givesBoth have zero divergence and curl, so they satisfy the static vacuum equations. Their upward fluxes through the loop areFor clockwise current $I$, the net force is zero in $B_1$, while in $B_2$

Put $K=2\pi b_2{r}^2$. Motion in $B_1$ changes no flux, so no current is induced and $z=z_0-g{t}^2/2$. In $B_2$, using counterclockwise current as positive,ThusWith $\Gamma =K^2/R$, the velocity tends to $-mg/\Gamma$ andAt terminal speed, $I^{2}R=mg|\dot{z}|$, so Joule loss equals gravitational-energy loss. For $R=0$, Ohm's quasistatic formula is singular: flux is conserved, persistent current stores magnetic energy, and ideal motion has no resistive terminal state (inductance must be retained).

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Write the internal stage asThe two methods have $\theta =1/2$ and $3/4$. Expansion gives stability functionBoth therefore have order two; order three would require $\theta =1/3$.

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The linear stability domain is $\{z\in \mathbb{C}:|R(z)|\le 1\}$ for the test equation $y^{\prime }=\lambda y$, $z=h\lambda$. A method is A-stable when this domain contains the entire closed left half-plane.

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For $\theta =1/2$,so $|R(z)|\le 1$ exactly when $\mathrm{Re}z\le 0$. Thus the first method is A-stable with stability domain the closed left half-plane. The $\theta =3/4$ method is not A-stable: along the negative real axis its quadratic numerator makes $|R(z)|$ grow without bound as $z\to -\infty$.

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If $S_k<n\le S_{k+1}$, the construction gives $X_n=S_{k+1}-n$. This is exactly the stated maximum over the next renewal endpoint.

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From state $i>0$ the chain moves deterministically to $i-1$. From zero it jumps to $j$ when the next lifetime is $j+1$. Thuswith all other probabilities zero.

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The balance recurrence isWorking down from $N-1$ gives ${\pi }_i={\pi }_{0}P(T_1>i)$. Since the sum of these tails is $E[T_1]$, the unique invariant law is

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Take $N\ge 2$ and $q_2=1$. The chain alternates deterministically between zero and one, so $P(X_n=0)$ alternates between one and zero and does not converge.

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Adding $\epsilon I$ leaves every invariant distribution unchanged. On the unique closed communicating class it also supplies a self-loop, making the chain aperiodic; transient states do not affect the limit. Thereforeindependently of $\epsilon \in (0,1)$.

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