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

The characteristic polynomial of the homogeneous equation isso $y_h=C_0+(C_1+C_{2}x)e^{2x}$. For a particular integral put $y=e^{2x}v$. ThenTaking $v^{\prime \prime }=x/2-1/4$ gives $v=x^3/12-x^2/8$ after discarding homogeneous terms. Therefore

Since $1+\cos 2y=2{\cos }^{2}y$, separation givesThe limiting condition gives $C=1$. Taking ${\tan }^{-1}$ with range $(-\pi /2,\pi /2)$ and selecting the branch approaching $9\pi /4$ yieldsfor $x>0$.

Write $u=xv$ on an interval not containing zero. The equation becomes $x^2{v}^{\prime }=x$, so $v^{\prime }=1/x$. Hence

For a nonnegative random variable $Y$ and $a>0$, the Markov inequality isIndeed, $Y\ge a{1}_{\{Y\ge a\}}$ pointwise. Taking expectations and dividing by $a$ proves the claim.

Completing the square gives

For any $t>0$, the exponential Markov bound and part (b) giveThe minimum occurs at $t=v$, so the Gaussian tail bound is

The first $k$ digits must be distinct and may be chosen inways; the remaining digits are then forced in reverse order. Thus, for $k\le 10$,and the probability is zero for $k>10$.

Let $E_{\ell }$ be the event that the block beginning at position $\ell$, $1\le \ell \le k+1$, is a spalindrome. Each has probability $(10{)}_k/10^{2k}$. If two starting positions differ by $d<k$, the first half of the later block contains the middle digit of the earlier palindrome twice, contradicting distinctness. Thus the only possible overlap is $E_1\cap E_{k+1}$.

That intersection consists of words $A{A}^{\mathrm{rev}}A$, where $A$ has $k$ distinct digits, and hence has probability $(10{)}_k/10^{3k}$. This overlap structure of digit spalindromes makes inclusion--exclusion give

LetThe integrating factor identity $(Iy{)}^{\prime }=If$ givesIf $y_1,y_2$ are distinct solutions, their difference is a nonzero homogeneous solution, so every solution is

For $x>0$, division by $x$ givesAn integrating factor is $e^{-x^2}/x$, soEvery choice exceptgrows like $x{e}^{x^2}$. The bounded solution selected by a terminal condition is thereforeIntegration by parts, or the Gaussian-tail asymptotic, givesIt starts from $0$ at $x=0$, remains negative, and decreases monotonically toward the horizontal asymptote $-1/2$.

The parameter sensitivity equation follows by differentiating with respect to $\mu$ and writing $u=y_{\mu }$:At $\mu =0$, $y_x=y$, $y(0)=1$, so $y(x,0)=e^x$. Thenand the integrating factor $e^{-x}$ yields

Equating powers in $y=y_0+\mu y_1+{\mu }^2{y}_2+\cdots$ givesConsequently

For every test function $\phi$ and nonzero $p$, substitution $u=pt$ with the orientation handled by $|p|$ givesHence the Dirac delta scaling identity is $\delta (pt)=\delta (t)/|p|$.

The derivative of a distribution is defined by integration by parts. Therefore

For a compactly supported test function $f$,Thus this is the distributional derivative of the Heaviside step function: $H^{\prime }=\delta$.

For $t<0$ both inputs vanish. Eliminating $y$ gives $x^{\prime \prime }+4x=0$, and the data at $t=-\pi$ giveFor all $t$, differentiating the first equation and using the second givesThus $x$ is continuous at zero while $x^{\prime }$ jumps by one. For $t>0$ the resulting solution isSince $\cos 2t=1-2{\sin }^{2}t$, this iswhereThus $a=1/2$, $b=1$, and the sketch of $q$ is a downward unit step at $t=\pi$.

The system matrix has eigenvalues $2$ and $3$, with eigenvectors in the directions $y=-x$ and $y=-x/2$, respectively. Both eigenvalues are positive, so the origin is an unstable node. The two eigendirections are straight trajectories directed away from the origin. Every other trajectory also moves outward, tangent near the origin in backward time to the faster eigendirection and asymptotic in forward time to the eigenvalue-$3$ direction.

With $y=\dot{x}$, the system isAt equilibrium $y=0$ and $x^2=1$, so the equilibrium points are $(1,0)$ and $(-1,0)$.

The Jacobian isAt $(1,0)$ its eigenvalues are $1,-4$, so this point is a saddle. At $(-1,0)$ they are $(-3\pm i\sqrt{7})/2$, so this point is a stable spiral.

The sign of $\dot{x}$ is the sign of $y$. The $\dot{y}=0$ nullcline is$\dot{y}$ is positive below this curve and negative above it. Combining the upper and lower half-planes with the regions above and below this even curve labels the four requested sign combinations.

Draw the even $\dot{y}=0$ nullcline through $(\pm 1,0)$ and use the arrows from part (iii). Trajectories near $(-1,0)$ spiral clockwise inward because, immediately to its right on the $x$-axis, $\dot{y}<0$. At $(1,0)$ the stable and unstable eigendirections have slopes $-4$ and $1$. Their four separatrix branches divide the portrait; the only trajectory actually reaching $(1,0)$ at finite time is the constant one, while these branches approach it as $t\to \pm \infty$. All remaining curves follow the sign field, with those in its basin spiralling into $(-1,0)$.

Conditional on $N=n$, there are $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ pairs, and each is both friendly and birthday-matched with probability $1/(2\cdot 365)$. Since $\mathbb{E}[N(N-1)]=365^2$ for a Poisson variable,

By Poisson thinning, allocating a Poisson$(365)$ population independently and uniformly among $365$ days gives independent Poisson$(1)$ counts. Hence

Given $Z_i=n$, the probability of no friendship among people sharing day $i$ is $2^{-n(n-1)/2}$. The day counts are independent, so

This is the Erlang distribution with shape $n$ and rate one. Direct integration gives, for $t<1$,

Differentiating the moment-generating function at zero gives

An Erlang$(n,1)$ variable is a sum of $n$ independent rate-one exponential variables, each with mean and variance one. The central limit theorem therefore givesConsequently, for every fixed $q\ge 0$,

The competing exponential clocks calculation givesso $V$ is exponential with rate ${\lambda }_1+{\lambda }_2$. Also

The exponential memoryless property gives

After the first of two equal-rate clocks rings, the residual lifetime of the other clock is, by memorylessness, a fresh exponential variable of rate $\lambda$, independent of the first ringing time and of which clock rang. Henceand these variables are independent.

Take the spatial Fourier transform of the wave equation. The transformed initial-value problem isand hence

Part (c) and the convolution theorem turn the first term intoFor the second term, choose $p$ with $p^{\prime }=g$. Part (b) says $\tilde{g}=ik\tilde{p}$, soPart (c) now makes its inverse transformCombining the two terms gives the D'Alembert formula

Let $A=2V$ and $B=W-V$. Part (c) makes $A$ and $B$ independent identically distributed exponential variables of rate $\lambda$. The change of variables $(A,B)=(rs,(1-r)s)$ has Jacobian $s$, and integrating ${\lambda }^{2}s{e}^{-\lambda s}$ over $s>0$ gives density one for $0<r<1$. Therefore the uniform ratio of independent exponential variables gives

The probability generating function isTermwise differentiation givesThe coefficients are nonnegative, and $\mathbb{P}(X\ge 2)>0$, so both derivatives are positive and nondecreasing on $(0,1]$.

Let $d_n$ be the probability of extinction by generation $n$. Conditioning on the initial individual's offspring givesThe events increase to eventual extinction, so $d_n\uparrow d$ and continuity gives $d=G_X(d)$. If $r\ge 0$ is any other fixed point, induction from $d_0=0\le r$ and monotonicity of $G_X$ gives $d_n\le r$ for every $n$. Hence $d\le r$, proving the Galton-Watson extinction fixed point characterization.

Since $G_X(1)=1$ and $G_X^{\prime }(1)=\mu >1$, for $t<1$ sufficiently close to one,At zero, $G_X(0)-0\ge 0$. Continuity therefore gives a fixed point in $[0,1)$, and part (b) implies $d<1$.

Taylor's theorem about $1$ gives, for some $\xi \in [t,1]$,Since $G_X^{\prime \prime }$ is nondecreasing andthe stated inequality follows. At the fixed point $d<1$,Therefore the quadratic Galton-Watson extinction bound isThe denominator is at least $2(\mu -1)$ because $X$ is integer-valued and $\mathbb{E}[(X-1)(X-2)]\ge 0$, so $d^{\ast }\ge 0$.