Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 1

The mapis a nonzero linear functional: otherwise $v$ would lie in the radical of a bilinear form, contradicting nondegeneracy. Its kernel is $v^{\perp }$, so the rank-nullity theorem gives $\dim v^{\perp }=n-1$. Antisymmetry gives $⟨v,v⟩=-⟨v,v⟩=0$, hence $v\in v^{\perp }$.

Suppose $w\in W$ is orthogonal to every vector of $W$. Since $W\subseteq v^{\perp }$, it is also orthogonal to $v$, and therefore toFor a nondegenerate bilinear form, $(v^{\perp }{)}^{\perp }=\mathbb{R}v$: both sides have dimension one and the latter is contained in the former. Thus $w\in W\cap \mathbb{R}v=\{0\}$, proving that the restriction to $W$ is nondegenerate.

The space $W$ has dimension $n-2$ and again carries a nondegenerate antisymmetric form. Induction, starting from the zero-dimensional space, shows that $\dim W$ is even. Therefore $n=\dim W+2$ is even. Equivalently, every finite-dimensional symplectic vector space has even dimension.

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The condition $D{f}_p\ne 0$ says that $0$ is a regular value of $f$. The regular level set theorem therefore makes $f^{-1}(0)$ a smooth submanifold of codimension one in ${\mathbb{R}}^3$, hence a smooth surface.

Forthe gradient isOn $F^{-1}(0)$, $x^2+y^2=\cosh (z^2)\ge 1$, so $x$ and $y$ cannot both vanish. Thus $\nabla F\ne 0$ there, and the given set is a smooth surface.

Not every smooth surface in ${\mathbb{R}}^3$ is a global zero set. Every set $f^{-1}(0)$ is closed because $f$ is continuous, whereas the open unit diskis a smooth surface but is not closed in ${\mathbb{R}}^3$.

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The integrand has an order-two pole at $0$ and a simple pole at $2$. Write it near zero as $h(z)/z^2$, whereIts residue at zero isAt $z=2$ the residue isThe residue theorem now gives

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Introduce a Lagrange multiplier $\lambda$ for the normalization and varyFor a smooth variation $u+\epsilon \eta$ with $\eta =0$ on $\partial D$, integration by parts givesThe fundamental lemma of the calculus of variations therefore yields the Euler-Lagrange equation

Multiplying by $u$ and integrating, while usingthe divergence theorem and $u=0$ on the boundary giveThe normalization is one, so the multiplier equals the stationary value:

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Apply symmetric Gaussian elimination without row exchanges. At step $k$, let $d_k$ be the leading diagonal entry of the remaining symmetric Schur complement. If $d_k\le 0$, stop and report that $A$ is not positive definite. If $d_k>0$, use it to eliminate the rest of its row and column. If all steps succeed, this constructs an LDL decompositionwhere $L$ is unit lower triangular and $D=\mathrm{diag}(d_1,\dots ,d_n)$ has positive diagonal.

The test is correct from first principles. If all $d_k>0$, then for every nonzero $x$,because $L$ is invertible. Conversely, if $A$ is positive definite, its first pivot is $a_{11}>0$, and completing the square givesChoosing $s=-b^{T}y/a$ shows that the Schur complement is positive definite. Induction forces every pivot to be positive. This is also Sylvester's criterion.

At step $k$, updating the remaining matrix costs $O((n-k{)}^2)$ arithmetic operations. Hence the total iswhich proves the existence of the required algorithm.

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A sufficient statistic $T$ is one for which the conditional distribution of the full sample $(X_1,\dots ,X_n)$ given $T$ does not depend on $p$. TakeFor a binary sample $x$ with ${\sum }_i{x}_i=t$, its likelihood function iswhich depends on the data only through $t$. By the Fisher-Neyman factorization theorem, $T$ is sufficient. Equivalently, conditionally on $T=t$, the sample is uniform over the $\left(\genfrac{}{}{0ex}{}{n}{t}\right)$ binary vectors containing $t$ ones, independently of $p$.

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The Rao-Blackwell theorem states that if $T$ is sufficient and $U$ is an estimator with finite variance, thenhas the same expectation as $U$ and no larger variance; it preserves unbiasedness. The tower property of conditional expectation givesThe law of total variance givesSufficiency ensures that $U^{\ast }$ is a statistic whose definition does not depend on the unknown parameter. The inequality is strict exactly when the conditional variance is positive with positive probability.

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The estimator $U=X_1{X}_2$ is unbiased for $p^2$ because the independent random variables $X_1,X_2$ satisfy $\mathbb{E}[X_1{X}_2]=p^2$. Given $T=t$, all placements of the $t$ successes are equally likely, soThus the Rao-Blackwellized estimator isIt is unbiased by the tower property. Since $n\ge 3$ and $p\in (0,1)$, the event $T=2$ has positive probability, and conditionally on it $X_1{X}_2$ takes both zero and one with positive probability. Hence $\mathbb{E}[\mathrm{Var}(X_1{X}_2\mid T)]>0$, so the new estimator has strictly smaller variance.

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For $0\le t\le 1$ and $x,y\in {\mathbb{R}}^d$, convexity givesTaking the maximum over $i$ and then bounding each term by the corresponding endpoint maxima yieldsThus the finite pointwise maximum of convex functions is convex. Summing the original inequalities over $i$ proves that ${\sum }_i{f}_i$ is convex as well.

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Because $x\mapsto c^{T}x$ is a linear map,Hence composition of a convex function with an affine map is convex, and $g$ is convex.

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The softplus functionis convex becausePart (b) therefore shows that every $\beta \mapsto h(a_i^T\beta )$ is convex. The absolute value function is convex, so $\beta \mapsto |{\beta }_j|$ is convex for every coordinate $j$. Finally, part (a) says that a finite sum of convex functions is convex. Thereforeis convex. The second sum is the L1 norm regularizer.

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The Nilpotent Jordan block $J_d$ shifts each standard basis vector one place toward the first coordinate. ThereforeThus $J_d^0=I$, for $1\le n<d$ the matrix has ones precisely on its $n$th superdiagonal, and

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Since $\lambda I$ commutes with $J_d$, the binomial theorem for commuting matrices and part (i) giveEquivalently, its $k$th superdiagonal is constant with value $\left(\genfrac{}{}{0ex}{}{n}{k}\right){\lambda }^{n-k}$ for $0\le k<d$, with the convention that this value is zero when $k>n$.

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If $\varphi v=\mu v$ for a nonzero eigenvector $v$, thenHence every eigenvalue satisfies $\mu =0$. Since a complex endomorphism has an eigenvalue, zero is the only possible eigenvalue.

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The Jordan normal form can contain only blocks $J_r(0)$ with eigenvalue zero. Moreover,Thus the possible blocks are precisely the nilpotent Jordan blocks of sizes $1\le r\le n$.

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Assume ${\varphi }^2=0$. ThenSet $W_1=\mathrm{im}\varphi$, choose a direct-sum complement $U$ such that $\ker \varphi =U\oplus W_1$, and choose a complement $W_2$ such that $V=\ker \varphi \oplus W_2$.

The restriction $\varphi {|}_{W_2}:W_2\to W_1$ is injective because $W_2\cap \ker \varphi =0$, and it is surjective because every image $\varphi (v)$ equals $\varphi (w_2)$ after decomposing $v=k+w_2$. It is therefore a isomorphism, so $\dim W_2=\dim W_1$ and $\varphi (W_2)=W_1$. Also $\varphi (U)=\varphi (W_1)=0$. Hencehas all the required properties.

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Suppose first that every ideal of $R$ is finitely generated. For an ascending chainthe union $I={\bigcup }_n{I}_n$ is an ideal. Its finite generating set lies in one $I_N$, so $I_n=I_N$ for all $n\ge N$. Thus $R$ satisfies the ascending chain condition and is a Noetherian ring.

Conversely, if an ideal $I$ is not finitely generated, choose $a_1\in I$, and after choosing $a_1,\dots ,a_n$, chooseThis creates a strictly ascending chain of ideals, contradicting Noetherianity. Hence every ideal is finitely generated.

If $\phi :R\to S$ is surjective and $J⊲S$, then ${\phi }^{-1}(J)$ is an ideal of $R$. If it is generated by $a_1,\dots ,a_m$, then $J$ is generated by $\phi (a_1),\dots ,\phi (a_m)$. Thus every ideal of $S$ is finitely generated, so $S$ is Noetherian.

The Hilbert basis theorem states that if $R$ is a commutative Noetherian ring, then $R[X]$ is Noetherian. To prove it, let $I⊲R[X]$. The leading coefficients of polynomials in $I$ generate an ideal of $R$; choose generators that occur as leading coefficients of $f_1,\dots ,f_s\in I$, and let $N={\max }_i\mathrm{deg}{f}_i$. For each degree below $N$, the leading coefficients of members of $I$ of at most that degree likewise form a finitely generated ideal; choose finitely many corresponding polynomials. Any $f\in I$ of degree at least $N$ can have its leading term cancelled by a linear combination of monomial multiples of the $f_i$. Repeating lowers its degree below $N$, where the second finite list completes the reduction. These finitely many selected polynomials generate $I$, proving the theorem.

For every $\alpha \in \mathbb{C}$, the ring $\mathbb{Z}[\alpha ]$ is Noetherian. If $\alpha$ is transcendental, evaluation identifies it with $\mathbb{Z}[X]$. If $\alpha$ is algebraic, it is a quotient of $\mathbb{Z}[X]$. The integers are Noetherian, the Hilbert basis theorem handles $\mathbb{Z}[X]$, and quotients preserve Noetherianity.

For a unique factorization domain that is not Noetherian, takeEvery polynomial and every factorization uses only finitely many variables, so existence and uniqueness of factorization reduce to a finite-variable polynomial ring, which is a UFD. Butis a strictly ascending chain, so the ring is not Noetherian.

Finally, the ring $C(\mathbb{R},\mathbb{R})$ is not Noetherian. For $n\ge 1$, letThese are ideals and $I_n⊊I_{n+1}$: the distance function to $[-1/(n+1),1/(n+1)]$ belongs to $I_{n+1}$ but not to $I_n$. This strict ascending chain proves the claim.

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Assume inverse images of open sets are open. If $x_n\to a$ and $V$ is any open neighbourhood of $f(a)$, then $f^{-1}(V)$ is an open neighbourhood of $a$. By the definition of convergence in a metric space, $x_n\in f^{-1}(V)$ eventually, and hence $f(x_n)\in V$ eventually. Thus $f(x_n)\to f(a)$.

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Assume the sequential condition, and let $A\subseteq Y$ be open. If $f^{-1}(A)$ were not open, some $a\in f^{-1}(A)$ would have no ball contained in $f^{-1}(A)$. For each $n$, chooseThen $x_n\to a$, so $f(x_n)\to f(a)\in A$. Because $A$ is open, this forces $f(x_n)\in A$ eventually, a contradiction. Therefore $f^{-1}(A)$ is open. This proves the sequential characterization of continuity in metric spaces.

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This is always true: it is the Heine-Cantor theorem. If uniform continuity failed, there would be an $\epsilon >0$ and sequences $x_n,y_n\in X$ such thatBy sequential compactness of a compact metric space, some subsequence $x_{n_k}$ converges to $x\in X$. The triangle inequality gives $y_{n_k}\to x$. Continuity then makes both image subsequences converge to $f(x)$, contradicting their separation by $\epsilon$.

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This may be false because continuity only forces the continuous image of a compact space $f(X)$ to be compact, not the whole codomain. For example, let $X=\{0\}$, let $Y=\mathbb{R}$, and set $f(0)=0$. The domain is compact and $f$ is continuous, but $\mathbb{R}$ is not compact.

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This is always true. The continuous image $f(X)$ of a connected space is connected. Its closure is also connected: if the closure were separated into disjoint nonempty relatively open sets, connectedness would put $f(X)$ inside one of them, preventing its closure from meeting the other. Since $f(X)$ is dense in $Y$, its closure is $Y$, so $Y$ is connected.

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This is always true for metric spaces. Let $x_n\to x$. If $f(x_n)$ did not converge to $f(x)$, some subsequence would remain at least $\epsilon >0$ from $f(x)$. Compactness of $Y$ gives a further subsequenceThen $(x_{n_k},f(x_{n_k}))\to (x,y)$. The graph is closed, so $(x,y)$ belongs to it and $y=f(x)$, contradicting the $\epsilon$ separation. Thus $f(x_n)\to f(x)$ for every convergent sequence, and the equivalence proved above makes $f$ continuous. This is the closed-graph criterion with compact codomain.

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For an oriented smooth surface $S$, the Gauss map sends $p$ to the chosen unit normal vector $N(p)\in S^2$. Since $|N{|}^2=1$, differentiation shows that $D{N}_p(X)$ is perpendicular to $N(p)$ and hence lies in $T_{p}S$.

In a local parametrization $\varphi (u,v)$ with $n=N\circ \varphi$, differentiatinggivesThus the bilinear form $(X,Y)\mapsto D{N}_p(X)\cdot Y$ is symmetric, so $D{N}_p$ is self-adjoint. The Gaussian curvature isWriting the coefficients of the first fundamental form asand those of the second fundamental form asone obtains

At an umbilic point, the self-adjoint map $D{N}_p$ has a repeated eigenvalue, so it is a scalar map. If every point is umbilic, there is a function $\lambda$ withEquality of mixed partial derivatives givesThe two tangent vectors are linearly independent, hence ${\lambda }_u={\lambda }_v=0$. Since ${\mathbb{R}}^2$ is connected, $\lambda$ is constant.

If $\lambda =0$, then $n$ is constant andso $S$ lies in a plane. If $\lambda \ne 0$, thenThus $n-\lambda \varphi =c$ is constant, andTherefore $S$ lies in a sphere of radius $1/|\lambda |$. This proves that the surface is part of a plane or part of a sphere.

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The Laurent theorem says that if $f$ is analytic on an annulusthen it has a unique Laurent seriesconverging locally uniformly on that annulus, wherefor any positively oriented circle $C$ in the annulus around $a$.

An isolated singularity at $a$ is a point at which $f$ is not analytic although it is analytic on some punctured neighbourhood. It is removable when every $c_n$ with $n<0$ vanishes; it is a pole of order $m$ when $c_{-m}\ne 0$ and $c_n=0$ for $n<-m$; and it is essential when infinitely many negative-index coefficients are nonzero.

For $0<|z|<1$,For $|z|>1$,The coefficients are unique after the annulus is fixed; these expansions differ because they represent the function on different annuli. At zero the first expansion has principal part $-z^{-1}$, so zero is a simple pole with residue $-1$.

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Put $g=1/f$. The hypothesis $|f(z)|\to \infty$ gives $g(z)\to 0$ as $z\to a$, so defining $g(a)=0$ makes $g$ continuous on $U$ and analytic there by the stated assumption. Its zero at $a$ has some finite order $m\ge 1$, and henceThereforehas a pole of order $m$: its Laurent series has $c_{-m}\ne 0$ and $c_n=0$ for $n<-m$.

Now let $f$ be entire and tend to infinity at infinity. The functiontends to infinity as $z\to 0$, so the preceding argument says that $h$ has a pole at zero. If the Taylor series of $f$ is $f(w)={\sum }_{n\ge 0}{a}_n{w}^n$, thenA pole has only finitely many negative powers, so $a_n=0$ for all sufficiently large $n$. Thus $f$ is a polynomial.

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Forboth $e^z-1$ and $\log (1+z)$ have a simple zero at zero. Hence $g$ has a simple pole, and its residue is

For $h(z)=\sin z\sin (1/z)$, multiplication of the two Laurent series shows that all powers are even:There are infinitely many negative powers, so zero is an essential singularity. There is no $z^{-1}$ term, and therefore

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With $\xi =x+ct$ and $\eta =x-ct$,so the wave equation becomesThus $u=F(\xi )+G(\eta )$. At $t=0$ the initial data giveSolving for $F^{\prime }$ and $G^{\prime }$ and integrating gives d'Alembert's formula>

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Extend the forcing oddly across the boundary:The initial displacement $\sin x$ is already odd, so its homogeneous evolution on the line is $\sin x\cos (ct)$. Applying Duhamel's principle to the odd extension givesThe odd-reflection method makes $u(0,t)=0$. At $t=0$ the double integral vanishes together with its first time derivative, so the prescribed initial displacement and velocity are also satisfied.

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For a normalized wavefunction obeying the infinite-wall Dirichlet boundary condition, integration by parts gives the energy expectationThus every energy eigenvalue is nonnegative.

For $0<E<U_0$, the Time-independent Schrodinger equation and the wall conditions givewhere $k=\sqrt{2mE}/\hslash$ and $l=\sqrt{2m(U_0-E)}/\hslash$. Continuity of $\psi$ and ${\psi }^{\prime }$ at the finite potential step givesDividing and rearranging yields the bound-state quantization condition

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Before the change, the normalized ground state of the infinite square well isFor an allowed post-quench energy $E\in (0,U_0)$ satisfying part (i), define the unnormalized eigenfunctionIts normalization factor isThe Born rule therefore givesFor a value of $E$ that is not an eigenvalue, this probability is zero. The sudden change leaves the wavefunction fixed, while the energy eigenbasis changes.

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Since the magnetic field is $B=\nabla \times A$, Stokes theorem gives, for any oriented surface $S$ bounded by $C$,Under a gauge transformation $A\mapsto A+\nabla \chi$, the integral changes bybecause $C$ is closed. The flux expression is therefore gauge independent.

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The magnetostatic Ampère-Maxwell equation is $\nabla \times B={\mu }_{0}J$. Substituting $B=\nabla \times A$ and using the Coulomb gauge $\nabla \cdot A=0$ givesThe free-space Green function of the Laplacian therefore givesFor the thin wire current stated in the question this becomes

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Substituting the wire potential from part (b) into the flux formula from part (a) gives Neumann's mutual-inductance formulaInterchanging the two curves proves $L_{12}=L_{21}$.

Parametrize the coaxial circles byThen, with $\theta =\varphi -\psi$ and $R=\sqrt{a^2+b^2+c^2}$,where $q=2ab/R^2$. One angular integration contributes $2\pi$, soSince $ab=q{R}^2/2$, this is

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