Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 1

The Leibniz definition isThe adjugate is the transpose of the cofactor matrix. Cofactor expansion givesFor the displayed tridiagonal matrix, expansion along the last row gives $D_n=2D_{n-1}-D_{n-2}$, with $D_1=2,D_2=3$. Induction yields

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In local coordinates $u=(u^1,u^2)$ with metric $g_{ij}={\partial }_i\sigma \cdot {\partial }_j\sigma$, the energy isThe Euler–Lagrange equations are equivalentlywhere ${\Gamma }_{ij}^k=\frac{1}{2}{g}^{k\ell }({\partial }_i{g}_{j\ell }+{\partial }_j{g}_{i\ell }-{\partial }_{\ell }{g}_{ij})$.

For the plane-section curve $\eta$, constant speed gives ${\eta }^{\prime \prime }\perp {\eta }^{\prime }$. Both ${\eta }^{\prime \prime }$ and the surface normal lie in $P$, while $P$ is spanned by ${\eta }^{\prime }$ and that normal along the intersection. Hence ${\eta }^{\prime \prime }$ is normal to the surface. Its tangential acceleration vanishes, which is the geodesic equation.

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Jordan's lemma states that if $f$ is holomorphic in the upper half-plane apart from finitely many poles and $|f(z)|\le M/|z|$ on sufficiently large upper semicircles, then for $a>0$ the integral of $e^{iaz}f(z)$ over those arcs tends to zero. Indeed, split the arc away from its endpoints, where exponential decay is uniform, and bound the two short endpoint arcs using $|e^{iaR{e}^{i\theta }}|=e^{-aR\sin \theta }$ and $\sin \theta \ge 2\theta /\pi$ on $[0,\pi /2]$.

Apply the upper semicircle to $z{e}^{iz}/(1+z^2)$. The sole enclosed pole is $i$, with residue $e^{-1}/2$. Therefore the contour integral is $\pi i/e$; taking imaginary parts gives

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If $f$ is meromorphic inside and on a positively oriented simple closed contour $C$, with no pole on $C$, then

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Integrate $e^{inz}/(z^4+1)$ over the upper semicircle. Jordan's lemma removes the arc and the upper poles are $e^{i\pi /4}$ and $e^{3i\pi /4}$. Summing their residues and taking real parts gives

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For every tangent vector $v\perp x_0$, stationarity of $Q$ on the sphere gives $2v^{T}A{x}_0=0$. Hence $A{x}_0$ is parallel to $x_0$, say $A{x}_0=E{x}_0$. Taking the inner product with $x_0$ gives $E=Q(x_0)$. The spectral theorem shows that this is the largest eigenvalue.

For $A=\left(\begin{array}{cc}1& t\\ t& 1\end{array}\right)$ the eigenvalues are $1\pm t$, soIts graph is a V translated upward and is convex.

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The Dahlquist equivalence theorem says that a consistent linear multistep method is convergent exactly when it is zero-stable.

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The first characteristic polynomial factors asThe root condition holds precisely for $0<\alpha <2$: in this range the two reciprocal roots are distinct and on the unit circle; at either endpoint a unit root is repeated, and outside it one root has modulus greater than one. Consistency is given, so these and only these values are convergent. The exceptional order-three value $\alpha =6$ is outside this interval; consequently every convergent case has order two.

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The Neyman–Pearson lemma says that among tests of size at most $\alpha$ for two simple hypotheses, a likelihood-ratio test rejecting where $p(x;{\theta }_1)/p(x;{\theta }_0)>k$ is most powerful, with boundary randomisation if required. If $\phi$ is that test and $\psi$ any competing test, choose $k$ so the sizes agree. Pointwise,Integration and the size inequality yield $E_1\phi \ge E_1\psi$.

Herewhich decreases with $|x|$. Thus reject for $|X|\le c$, where

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Multiplying $Ax\le b$ by $y\ge 0$ gives the upper bound $c^{T}x\le y^{T}Ax\le y^{T}b$ whenever $A^{T}y\ge c$. Minimising the bound yields the dual

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The primal point $x=(1/2,1/2,0)$ is feasible and has value $5/2$. The dual point $y=(1/2,3/2)$ is feasible becauseand has value $2(1/2)+3/2=5/2$. Weak duality proves both are optimal. Thus the requested optimal primal solution is

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For a linear map $T:V\to W$ with $V$ finite-dimensional,Nowwhich is the required inequality after rank-nullity for $\alpha$.

If $X$ and $Y$ represent the same map, then $Y=C^{-1}XB$, where $B$ changes new domain coordinates to old ones and $C$ does the same in the codomain.

Invertible block row and column operations reduceto $\mathrm{diag}(P,S-R{P}^{-1}Q)$, proving the rank formula. Apply it to $\left(\begin{array}{cc}{I}_n& Q\\ R& I_m\end{array}\right)$ first with the upper-left block and then with the lower-right block. Equating the results gives

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Let $k=|G/H|$. The coset action gives a nontrivial homomorphism $G\to S_k$ whose kernel is normal. Simplicity makes it injective. The sign map is trivial on the nonabelian simple image, so $G$ embeds in $A_k$. For $k\le 4$, $A_k$ is solvable, as are its subgroups, whereas a nonabelian simple group is not. Hence $k\ge 5$.

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Let $S$ act by conjugation on the Sylow $p$-subgroups. The only fixed point is $S$: if $S$ normalizes another $T$, then $ST$ is a $p$-subgroup, forcing $S=T$. For $T=gS{g}^{-1}\ne S$, a stabilizer element in $S$ normalizes $T$, hence lies in $T$ by the same argument; the hypothesis then makes it trivial. Every nontrivial orbit therefore has size $|S|$, so

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Sylow gives $n_7\mid 24$ and $n_7\equiv 1(\mathrm{mod}7)$. Simplicity excludes $n_7=1$, so $n_7=8$. Distinct order-seven subgroups intersect trivially, hence there areelements of order seven.

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If all distinct Sylow 2-subgroups met trivially, part (b) would give $n_2\equiv 1(\mathrm{mod}8)$. But Sylow gives $n_2\mid 21$ and $n_2$ odd, so $n_2\in \{3,7,21\}$ after excluding normality, none congruent to one modulo eight. Thus some distinct pair has nontrivial intersection. That intersection is a nontrivial finite 2-group, so Cauchy's theorem supplies an element of order two.

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Uniform continuity means that for every $\epsilon >0$ there is a single $\delta >0$ such that $d(x,y)<\delta$ implies $|f(x)-f(y)|<\epsilon$ for all $x,y$. A $d^{\prime }$-Cauchy sequence converges uniformly pointwise to a bounded function; passing a uniform-continuity estimate through a sufficiently close member proves that the limit is uniformly continuous. Thus $C_{b,u}(X)$ is complete.

If $f\in C_0({\mathbb{R}}^n)$, decay at infinity and compactness of a ball make it bounded. Uniform continuity on a sufficiently large compact ball, together with small values outside it, proves global uniform continuity. Hence $C_0\subset C_{b,u}$.

A uniform limit of functions vanishing at infinity also vanishes at infinity, so $C_0$ is closed. It is not compact: translate a fixed compactly supported bump of height one so that the supports are disjoint. The resulting sequence has pairwise sup distance one and no convergent subsequence.

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An allowable parametrisation is a smooth homeomorphism from an open subset of ${\mathbb{R}}^2$ onto an open subset of $S$, with derivative of rank two. Away from the axis, the rotation orbit has nonzero tangent. A transverse curve supplied by the submanifold theorem, followed by the rotation action, giveswith $|v|<\pi$ and $(f^{\prime },g^{\prime })\ne (0,0)$.

For a ruled parametrisation, regularity is exactly

Rotate and translate along the axis so the specified ruling isHere $s\ne 0$ because the line is not parallel to the axis, and $c\ne 0$: if $c=0$, rotating the horizontal tangent line makes the ruled parametrisation singular at its closest point. Rotating $L$ givesa one-sheet hyperboloid. Rotation invariance puts this whole surface in $\Sigma$. Connectedness and the fact that a complete embedded hyperboloid cannot be a proper subset of another connected embedded surface force equality. Rescaling radial and axial coordinates gives a diffeomorphism with $x^2+y^2=1+z^2$.

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For any circle $|\zeta |=\rho$ with $r<\rho <R$, Cauchy's formula applied inside and outside the circle gives the locally uniformly convergent Laurent expansionAt zero, the singularity is removable exactly when all $a_n$ with $n<0$ vanish; it is a pole of order $k$ when $a_{-k}\ne 0$ and $a_n=0$ for $n<-k$; it is essential when infinitely many negative coefficients are nonzero.

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Since ${\csc }^{2}z=z^{-2}+1/3+O(z^2)$, the principal parts cancel and $f_1=-1/3+O(z^2)$. The singularity is removable.

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Termwise integration of the exponential series givesThere are infinitely many negative Laurent coefficients, so zero is an essential singularity.

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For distinct $z,w$ in the convex disc, integrate along the segment:The second term has modulus strictly below $|f^{\prime }(z_0)||z-w|$, so the sum cannot vanish. Thus $f$ is one-to-one.

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Harmonic means twice continuously differentiable with $u_{xx}+u_{yy}=0$. On the simply connected plane, choose an entire harmonic conjugate so $F=u+iv$ is entire. If $u\ge 0$, then $e^{-F}$ is bounded; Liouville's theorem makes it, and hence $u$, constant.

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Choose an entire $F$ with real part $u$. The bound implies that $e^F$ has at most polynomial growth, so Cauchy's estimates make it a polynomial. It has no zeros, hence that polynomial is constant. Therefore $F$, and in particular $u$, is constant.

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Legendre's equation isWith $⟨f,g⟩={\int }_{-1}^{1}f\bar{g}dx$, integration by parts has no endpoint term because $1-x^2=0$ there, so the operator is self-adjoint. Sturm–Liouville eigenvalues are real, can be ordered increasingly, and their eigenfunctions are orthogonal and complete under standard regularity assumptions.

Substitution of $y=\sum a_n{x}^n$ givesThe series terminates at degree $\ell$ when $\lambda =\ell (\ell +1)$. Normalizing at one gives

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The separated axisymmetric solution isSince $x(1-x^2)=\frac{2}{5}(P_1-P_3)$, regularity at the origin and the boundary condition give

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Write $E=-{\hslash }^2{\kappa }^2/(2m)$. An even bound state is proportional to $\cos (kx)$ inside the well and to $e^{-\kappa |x|}$ outside, whereContinuity of the logarithmic derivative at $a$ givesFor sufficiently small $a$, $ka$ lies in the first monotone branch, so this equation has exactly one positive root and the even state is unique up to scale.

As $a\downarrow 0$, $\kappa \sim k^{2}a\to m/{\hslash }^2$. HenceThe wells converge distributionally to $V_0(x)=-\delta (x)$, whose normalized even bound state isIts derivative jump is precisely the one imposed by the delta potential.

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Using $\rho ={\epsilon }_0\nabla \cdot E$, $E=-\nabla \varphi$, integration by parts, and $\varphi =0$ on the boundary givesPut $C=q/(4\pi {\epsilon }_0)$. Gauss's law gives the radial fieldTaking zero potential outside,The charge formula givesThe field formula gives the same result after integrating $4\pi r^2{E}_r^2$ over the two nonzero annuli.

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For steady inviscid flow, Bernoulli's equation along a streamline isWriting steady Euler flow in divergence form and integrating over $V$ givesby incompressibility and the divergence theorem.

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The speeds are $U=q/A$ upstream and $v=q/(2a)$ in either daughter vessel. Bernoulli therefore gives

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Take the downstream tissue pressure as gauge zero. The two transverse momentum fluxes cancel. The force of the fluid on the junction is axial and equalswhere $U=q/A$, $v=q/(2a)$ and the pressure difference is that in part (i).

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For a unit vector $v$, a Householder reflection is $H=I-2v{v}^T$. Clearly $H^T=H$, and $H^{T}H=(I-2v{v}^T{)}^2=I$.

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The vector $v$ has eigenvalue $-1$, while every vector in $v^{\perp }$ has eigenvalue $1$. Their multiplicities are one and $n-1$, respectively.

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At step $k$, choose a Householder reflection acting only on coordinates $k,\dots ,n$ that maps the trailing part of column $k$ to a multiple of the first coordinate in that block. It zeros every entry below the diagonal without changing earlier columns. Induction produces $H_n\cdots H_{1}A=R$ upper triangular.

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For symmetric $A$, apply at stage $k$ the same reflection on both sides, choosing it to zero entries below the first subdiagonal in column $k$. Orthogonal similarity preserves symmetry, so the corresponding row entries vanish too and previous zeros remain. After finitely many stages $QA{Q}^T$ is symmetric tridiagonal. Constructing each reflector uses only arithmetic and one square root.

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Let $S_{xx}=\sum X_i^2$. The likelihood equations giveThese are orthogonal projections of a Gaussian vector onto the span of $X$ and its orthogonal complement, hence are independent.

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With $s^2=\sum (Y_i-X_i\hat{\beta }{)}^2/(n-1)$,Thus the interval is

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Unbiasedness requires $\sum c_i{X}_i=1$. Cauchy–Schwarz gives $1\le (\sum c_i^2)S_{xx}$, soEquality holds exactly for $c_i=X_i/S_{xx}$.

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The reverse-regression estimate is $\hat{b}=\sum X_i{Y}_i/\sum Y_i^2$. Thereforeby Cauchy–Schwarz. Equality means the two data vectors are proportional, which is exactly when both fitted residual sums, and hence both variance MLEs, vanish.

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A chain is reversible with respect to $\pi$ when ${\pi }_i{P}_{ij}={\pi }_j{P}_{ji}$ for all states. For random walk on a finite connected undirected graph, ${\pi }_i=\mathrm{deg}(i)/(2|E|)$ satisfies this because both sides equal $1/(2|E|)$ on an edge and zero otherwise.

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The graph has six edges, and $A$ has degree one. Hence ${\pi }_A=1/12$. Kac's return-time formula gives

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After the forced first step $A\to B$, let $h_i=P_i(T_A<T_F)$. Symmetry gives $h_C=h_D=x$, whileSolving gives $h_B=2/3$. Thus the requested probability is $2/3$.

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Let $g_i=E_i[T_A{1}_{\{T_A<T_F\}}]$ for the chain absorbed at $A$ or $F$. With $g_C=g_D=y$, First-step analysis givesThus $y=3$ and $g_B=8/3$. Including the initial step from $A$, the conditional mean is

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