Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 2

Differentiation under the integral gives $I\prime \prime ={\int }_0^1{y}^4{e}^{xy}dy$. Integrating the derivative of $e^{xy}(x{y}^3-3y^2)$ and then once more yields $x^2{\int }_0^1{y}^4{e}^{xy}dy-12{\int }_0^1{y}^2{e}^{xy}dy=e^x(x-4)$.

With $y=z/x^2$ the equation becomes $z\prime \prime +z=0$. Thus $y_1=\cos x/x^2$, $y_2=\sin x/x^2$, and $W(y_1,y_2)=x^{-4}\ne 0$.

The transformed inhomogeneous equation is $z\prime \prime +z=x^2$. Taking $z=x^2-2$ gives the particular integral $y_p=1-2/x^2$.

Completing the square gives $X,Y\sim N(0,4/3)$, so each marginal is $\sqrt{3}(2\sqrt{2\pi }{)}^{-1}{e}^{-3x^2/8}$. Their covariance is $2/3$, so they are not independent.

From part (i),The conditional-normal formula therefore givesConsequently the Gaussian conditional expectation is

Writing $S_n=\sum I_i$, each $\mathbb{E}{I}_i=1/n$. Distinct indicators have joint probability $1/[n(n-1)]$ and zero covariance, so $\mathbb{E}{S}_n=1$ and $\mathrm{var}{S}_n=n(1/n)(1-1/n)=1-1/n$.

Cauchy-Schwarz inequality applied to $X{1}_{X>0}$ gives $(\mathbb{E}X{)}^2\le \mathbb{E}(X^2)\mathbb{P}(X>0)$.

Apply part (ii): $\mathbb{P}(S_n>0)\ge 1/\mathbb{E}(S_n^2)=1/(2-1/n)\ge 1/2$.

$y_0\prime \prime =(x^2-1)y_0$, so $\alpha =1$.

Substitution and cancellation give $f\prime \prime -2xf\prime +(\alpha -1)f=0$.

Equating powers gives $(n+2)(n+1)a_{n+2}+(\alpha -1-2n)a_n=0$.

$y\prime (0)=0$ forces the odd series to vanish. The even recurrence terminates at degree $2N$ precisely when $\alpha =4N+1$.

After division by $e^{-x^2/2}$, $f\prime \prime -2xf\prime +8f=2$. The initial data and evenness give $f=1-3x^2+x^4$, hence $y=e^{-x^2/2}(1-3x^2+x^4)$.

$\nabla U=(3x^2-3+y^2,2y(x+1))$, so the critical points are $(\pm 1,0)$. The Hessian matrix at $(1,0)$ is $\mathrm{diag}(6,4)$, making it the local minimum; $(-1,0)$ is degenerate and not a minimum.

$dU/dt=\nabla U\cdot (-\nabla U)=-|\nabla U{|}^2\le 0$. The descending gradient flow trajectory from $(2,1)$ stays in its bounded contour basin and tends to the only critical point there, $(1,0)$.

Writing $\xi =x-1$, $\eta =y$, the linearization of a dynamical system is $\dot{\xi }=-6\xi$, $\dot{\eta }=-4\eta$. Eliminating $t$ gives $\eta =C{\xi }^{2/3}$.

Now $\dot{U}=-|\nabla U{|}^2+U_y\cos \omega t$, whose second term has either sign. Near the minimum, $\xi =C{e}^{-6t}$ and $\dot{\eta }+4\eta =\cos \omega t$, so

Spatial derivatives vanish for $u=y(t)$, giving $\ddot{y}+\gamma \dot{y}+{\omega }^{2}y=f(t)$.

The roots are $r=(-\gamma \pm \sqrt{{\gamma }^2-4{\omega }^2})/2$. They give oscillatory decay for $\gamma <2\omega$, two exponential decays for $\gamma >2\omega$, and $(A+Bt)e^{-\gamma t/2}$ at equality. All tend to zero.

Put $T=\pi /\omega$, $t_0=T/2$, and $\Omega =\omega /2$. A solution satisfying both endpoint conditions iswhere $A=-\sqrt{2}I{\omega }^{-1}{e}^{\sqrt{3}\pi /4}$. The derivative jumps by $I$ at $t_0$, as required.

Substitution gives $({\dot{X}}^2-c^2)\varphi \prime \prime -(\ddot{X}+\gamma \dot{X})\varphi \prime +{\omega }^2\varphi =0$. Multiply by $\varphi \prime$ and integrate over $\mathbb{R}$; decay kills the first and last integrals, leaving $(\ddot{X}+\gamma \dot{X})\int (\varphi \prime {)}^2=0$. Thus a nonconstant profile requires $X=X_{\infty }+C{e}^{-\gamma t}$ and approaches a fixed position.

During warming, $\dot{\theta }=\alpha ({\theta }_1-\theta )$ and $\theta ={\theta }_1-({\theta }_1-{\theta }_0)e^{-\alpha t}$. During cooling, with $\tau =t-T$, $\dot{\theta }=-\alpha (\theta -{\theta }_0)$ and $\theta ={\theta }_0+({\theta }_1-{\theta }_0)(1-e^{-\alpha T})e^{-\alpha \tau }$.

$N(2T)/N(0)=\exp [-{\int }_0^{2T}\beta (\theta (t))dt]$. Evaluation gives the implicit equation

For $x=\alpha T\ll 1$, $e^{-x}(1-e^{-x})=x-\frac{3}{2}{x}^2+O(x^3)$. Hence $\frac{3}{2}{\beta }_{\max }\alpha T^2\sim \log 100$ and

Each of the $\left(\genfrac{}{}{0ex}{}{n}{3}\right)$ triples is a triangle with probability $p^3$, so $\mathbb{E}T=\left(\genfrac{}{}{0ex}{}{n}{3}\right)p^3$.

Markov inequality gives $\mathbb{P}(T>0)\le \mathbb{E}T=O(n^{3-3\alpha })\to 0$ when $\alpha >1$.

Chebyshev inequality gives $\mathbb{P}(T=0)\le \mathbb{P}(|T-\mathbb{E}T|\ge \mathbb{E}T)\le \mathrm{var}(T)/(\mathbb{E}T{)}^2\to 0$.

Only triangle pairs sharing an edge have nonzero covariance, soAfter division by $(\mathbb{E}T{)}^2$ this is $O(n^{-3}{p}^{-3}+n^{-2}{p}^{-1})\to 0$ for $p=n^{-\alpha }$, $0<\alpha <1$; use part (iii).

The waiting time is geometric with mean $1/p$.

First-step equations for states “no trailing H” and “one trailing H” give $E_0=1+q{E}_0+p{E}_1$, $E_1=1+q{E}_0$, hence $E_0=(1+p)/p^2$.

The word $HT$ has no proper self-overlap, so its mean waiting time is the reciprocal of its probability: $1/[p(1-p)]$. This also follows from two first-step equations.

The standard run recursion gives $E_n=1+p^{-1}+\cdots +p^{-n}=(1-p^n)/[(1-p)p^n]$.

For $k\ge n$, $p_k(n)=\left(\genfrac{}{}{0ex}{}{k-1}{n-1}\right)p^n(1-p{)}^{k-n}$. It is the sum of $n$ independent geometric variables, so the central limit theorem applies with mean $n/p$ and standard deviation $\sqrt{n(1-p)}/p$. Take
$k_a(n)=\lceil n/p+a\sqrt{n(1-p)}/p\rceil$ and $k_b(n)=\lfloor n/p+b\sqrt{n(1-p)}/p\rfloor$.

Shift the absorbing interval to $\{0,\dots ,a+1\}$ and start at $1$. The gambler’s-ruin harmonic functions give $\mathbb{P}({\tau }_a<{\tau }_{-1})=1/(a+1)$ and $\mathbb{E}T=1(a+1-1)=a$.

At time $T_k$ the visited vertices form a contiguous arc and the walk is at an endpoint. Exiting an interval of $k$ visited vertices, starting one step from an absorbing endpoint, has mean $k$. Therefore $\mathbb{E}(T_{k+1}-T_k)=k$ and $\mathbb{E}{T}_n={\sum }_{k=1}^{n-1}k=n(n-1)/2$.

For each $j\ne 0$, cut the cycle at $j$. The event that $j$ is last is the event that the lifted walk covers the other $n-1$ residues before crossing that cut. Gambler’s ruin (or cyclic symmetry of the two expanding endpoints) gives $\mathbb{P}(Z=j)=1/(n-1)$.

Conditioning on generation $n$ and using independence gives $F_{n+1}(s)=F_n(F_1(s))$.

Differentiating at $s=1$ gives $\mathbb{E}{X}_{n+1}=\mu \mathbb{E}{X}_n$, hence $\mathbb{E}{X}_n={\mu }^n$.

The extinction probability is the smallest fixed point of the convex generating function $F_1$ in $[0,1]$. Since $F_1\prime (1)=\mu \le 1$ and $p_0,p_2>0$ make strict convexity relevant, the only fixed point is $1$.

Here $F_1(s)=1/(2-s)$ and $\mu =1$, so extinction is certain. Iteration givesand $\mathbb{P}(X_n=0)=F_n(0)=n/(n+1)$.