Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 2

The zero function is a solution. On an interval where $y\ne 0$, putThenThe differential equation becomesHence $z=Cx$, and integration givesThus all nonzero solutions aretogether with $\boxed{y\equiv 0}$.

Multiplyingby $y^{-n}$ and setting $z=y^{1-n}$ givesTherefore the Bernoulli differential equation becomes

For the stated equation,Set $z=\sqrt{x}$. ThenThe integrating factor is $e^{t^3}$, soHenceThe condition $x(1)=4$ gives $C=2e-\frac{1}{2}$. Consequently

Set $x=y=0$ in the functional equation:The denominator is defined. If $f(0)\ne 0$, cancellation would give $f(0{)}^2=-1$, impossible for a real-valued function. Hence

Using the addition law with $y=h$,Differentiability makes $f$ continuous, so $f(h)\to f(0)=0$, andWriting $C=f^{\prime }(0)$ and taking the limit gives

Sinceintegration givesPart (i) gives $D=0$, so the tangent addition functional equation has the solutionsfor constants $C$ such that no pole lies in $(-a,a)$. Conversely, the tangent addition formula verifies the functional equation. If $f^{\prime }(0)=2$, then $C=2$, andon any such interval, necessarily with $a\le \pi /4$.

The Markov inequality states that for a nonnegative random variable $Y$ and $a>0$,For $t>0$, the event $X\ge x$ is the same as $e^{tX}\ge e^{tx}$. Applying Markov's inequality to the nonnegative variable $e^{tX}$ gives the exponential Markov bound

Independence givesThis is the moment generating function of a Poisson distribution with mean $n$. Hence

The exponential Markov bound gives, for every $t>0$,The exponent is minimized when $e^t=2$, so take $t=\log 2$. Then

The bivariate normal distribution has density

Completing the square in $x_1$ in the joint density shows that the conditional distribution of a bivariate normal variable isThus its conditional density is

LetThey are jointly normal because they are linear combinations of a Gaussian vector. When ${\sigma }_1={\sigma }_2=\sigma$,Hence uncorrelated jointly normal variables are independent. Their means and variances areandThereforeindependently.

After multiplying either solution by $-1$ if necessary, suppose both are positive on $(x_1,x_2)$. SetThe equations giveSince ${\phi }_1$ is positive between consecutive zeros,ThereforeBut $W$ is nondecreasing, so both endpoint values coincide at zero and $W^{\prime }\equiv 0$. Since both solutions are strictly positive inside,This is the equality case in the wronskian proof of Sturm comparison.

Suppose ${\phi }_2$ has no zero in $(x_1,x_2)$. It has a constant sign there, so replace it by its negative if needed and apply part (i). The conclusion is $q_1\equiv q_2$ on that interval. Taking the contrapositive proves the Sturm comparison theorem: unless the coefficients agree identically, ${\phi }_2$ has a zero between the consecutive zeros of ${\phi }_1$.

TakeIts consecutive zeros are $-1$ and $\pi -1$. For the stated equation,and $q_2$ is not identically equal to $1$ on the interval. The Sturm comparison theorem therefore shows that every nontrivial solution has at least one zero in

Rewrite the equation asOn $[2,6]$, one has $q(x)\le 1/3$. If a nontrivial solution had two zeros, choose two consecutive ones $\alpha <\beta$ in that interval. Then $\beta -\alpha \le 4<\pi \sqrt{3}$.

The solutionof ${\phi }^{\prime \prime }+\frac{1}{3}\phi =0$ is strictly positive on $[\alpha ,\beta ]$. Since $q\le 1/3$ and the coefficients are not identical on this interval, the Sturm comparison theorem requires $\phi$ to have a zero between $\alpha$ and $\beta$, a contradiction. Hence every nontrivial solution has

The homogeneous characteristic equation isA linear particular solution $u_n=an+b$ givesso $a=-1/4$ and $b=-11/16$. ThusThe initial data yield $A=1$ and $B=-5/16$, hence

For the ordinary generating function of a recurrence, multiply by $x^n$ and sum for $n\ge 2$. Using $u_0=0,u_1=2$ givesEquivalently,Expanding each geometric series gives exactly the displayed formula for $u_n$, verifying consistency.

The floor-half generating function isSumming $u_n-2u_{n-1}=\lfloor n/2\rfloor$ for $n\ge 1$ givesso

Sincea particular solution isAdding the homogeneous term $C{2}^n$ and imposing $u_0=1$ gives $C=5/3$. Therefore

TakeThenWith $F(0)=1$, separation givesand henceon its maximal interval of existence.

The only equilibrium is $(0,0)$. Its linearization iswhose eigenvalues are $\pm i$, so the linearized system has a centre.

The exact radial identity from part (i) givesThus the origin is asymptotically stable for $k<0$, neutrally stable for $k=0$, and unstable for $k>0$. Moreover,so trajectories rotate clockwise.

The radially symmetric planar dynamical system formulas giveWith $r(0)=1$ and $\theta (0)=0$,again on the maximal interval for which the denominator is positive.

For the modified system, $F=x^2+y^2$ satisfiesWith $F(0)=1$, the logistic solution isConsequently,

Write $\psi =a{e}^{iS}$ with $a=\sqrt{\rho }$ and $v=S_x$. ThenwhileEquating imaginary parts givesand equating real parts and differentiating in $x$ givesThusThis is the hydrodynamic form of the Defocusing nonlinear Schrödinger equation.

For a stationary gas with constant phase, $\psi =a(x)$ is real and time independent. The field equation reduces toA positive constant solution obeys $d(1-d^2)=0$, so

The travelling-wave substitution $\xi =x-Ut$ givesWriteIts real part satisfiesFor $U>0$, the imaginary part also givesWith $\alpha =\sqrt{1-2U^2}$ and a translation chosen so that the notch is centred at zero,Its limiting modulus is $\sqrt{{\alpha }^2+2U^2}=1=d$, as required. The same expression at $U=0$ is obtained directly from the second-order equation.

The dark soliton density isFor $U=0$,a stationary notch reaching zero at its centre. For $U=1/2$,a shallower notch with minimum $1/2$. Since $\xi =x-Ut$, the first profile remains fixed and the second translates rigidly to the right at speed $1/2$.

Put $Q={\sum }_{j=1}^n{q}_j$. Since the variables have continuous distributions, the minimum is unique almost surely. Integrating over its possible value givesThis is the joint tail calculation for competing exponential clocks.

Setting $t=0$ in part (i) givesAlsosoPart (i) now factors aswhich proves that $K$ and $T$ are independent.

This is the unit-rate Erlang distribution. For completeness, use convolution. The result is clear for $n=1$. Ifthen independence and the unit exponential density give, for $x>0$,Induction therefore proves

Conditioning on $N$ and using the moment-generating function of a unit exponential variable,where the geometric series converges precisely when $\theta <p$. This is the moment-generating function of an exponential variable of rate $p$, so the geometric sum of exponential variables satisfies

The $n$ possible matching events are independent, each with probability $1/n$. Henceand therefore

When exactly $n-j$ faces have appeared, there are $j$ unseen faces, so each new roll discovers one with probability $j/n$. Let $G_j$ be the number of further rolls needed at that stage. ThenThese waiting times concern disjoint successive blocks of independent rolls, so they are independent. Thus the coupon collector problem has the decomposition

Using the mean $1/p$ of a geometric distribution and the decomposition in part (ii),

Independence and the geometric variance formula yieldSince $n(n-j)\le n^2$,

Part (iii) and the given harmonic-sum asymptotic show thatFor any fixed $\epsilon >0$, this deterministic ratio lies within $\epsilon /2$ of $1$ for all sufficiently large $n$. The Chebyshev inequality and part (iv) then giveThus

Let $B_n$ count the $+1$ steps. ThenConsequentlyand $S_n$ has probability zero at integers of the other parity. This is the finite-time law of a simple random walk on the integer line.

The preceding representation and the central limit theorem giveSinceone may takeFor $0<p<1$ this gives the required convergence to $\Phi (x)$ at every real $x$.

Let $m_j$ be the expected remaining time when the walk is at $j$. The first-step recurrence for gambler's ruin iswith $m_{-a}=m_b=0$. The quadraticsatisfies both the recurrence and the boundary conditions, and their finite linear system has a unique solution. Starting from zero therefore gives

Only even times contribute, and at time $2k$ the walk is at zero exactly when it has made $k$ steps in each direction. By linearity of expectation,The Stirling formula implies thatHence there is a constant $c_0>0$ such that this term is at least $c_0/\sqrt{k}$ for every $k\ge 1$, after decreasing $c_0$ to cover the finitely many small values. ThereforeThe final expression is $2c_0(\sqrt{n+1}-1)$, so, after another adjustment for small $n$, there is a constant $c>0$ withfor every $n$. This is the lower bound recorded by expected visits to the origin by a simple random walk.