Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 2

Differentiation under the integral sign first givesOn the other hand,Its integral is zero because $\sin \theta$ vanishes at both endpoints. Hence

Differentiating once more,For $x\ne 0$, the preceding identity and ${\sin }^2\theta +{\cos }^2\theta =1$ giveThe equation extends through $x=0$ in its regular limiting form.

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The characteristic polynomial of the linear recurrence relation isThe repeated-root rule therefore givesThe condition $x_0=0$ gives $A=0$. The other two conditions giveso $B+C=2$ and $B+2C=3$. Thus $B=C=1$, and

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A function $f$ on an interval is a convex function ifJensen inequality states that for an integrable random variable $X$, when all terms are defined,

The function $\varphi (x)=x\log x$ is convex on $(0,\infty )$ becauseApplying Jensen's inequality to the uniform distribution on the positive numbers $x_1,\dots ,x_n$ givesWriting $S={\sum }_i{x}_i$ and multiplying by $n/S$ yields

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Expanding around the expected value $\mu$,because $\mathbb{E}[X-\mu ]=0$. Thereforewith equality exactly when $a=\mu$.

For the absolute loss,Leibniz differentiation gives, with $F(a)={\int }_{-\infty }^{a}f(x)dx$,Thus $H$ decreases while $F(a)<1/2$ and increases while $F(a)>1/2$. It is minimized at any median, characterized in the continuous case by

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For a normalized equation$x_0$ is an ordinary point when $P,Q$ are analytic there. It is a regular singular point when $(x-x_0)P(x)$ and $(x-x_0{)}^{2}Q(x)$ are analytic there. These are the ordinary point criterion for a second-order equation and regular singular point criterion for a second-order equation.

Hereso $x=0$ is regular singular. Seek a power-series solution of a differential equationEquating the coefficient of $x^n$ givesand henceThereforeThe series terminates exactly when $\lambda =N$ is a nonnegative integer: the factor with $j=N$ then makes $a_{N+1}=0$. Thus the polynomial solutions occur for

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After division by $x$, the coefficient of $y^{\prime }$ is $P(x)=1/x-1$. The Abel identity for the Wronskian givesso

For $\lambda =1$, direct substitution confirms that $y_1=1-x$. Reduction of order gives a second solution proportional toTo extract the requested coefficients, putFor the differential operatorone findsThe analytic correction $h={\sum }_{n\ge 1}{b}_n{x}^n$ must therefore satisfy $L[h]=3-x$. Its constant and linear coefficients giveHence

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The multivariable chain rule gives, at $(x,y)=(a+t\cos \gamma ,b+t\sin \gamma )$,Sufficient conditions for a strict local minimum at $t=0$ are $g^{\prime }(0)=0$ and $g^{\prime \prime }(0)>0$.

At a stationary point, $g^{\prime }(0)=0$ in every direction. Its second derivative is the quadratic form of the Hessian matrix. Ifthen completing the square givesfor every nonzero direction $(u,v)$. The Hessian is therefore positive definite, and the stationary point is a strict local minimum.

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The stationary equations areThus $y=-x$ and $4x(x^2-2)=0$, givingThe Hessian matrix isAt either nonzero stationary point,so both are strict local minima.

Along a line through the origin,whereWhen $\cos \gamma \ne 0$ and $r=\tan \gamma$,Consequently the origin is a local minimum along the line whenincluding the equality cases because the positive quartic term then leads. It is also a minimum on the vertical line. It is a local maximum along the line whenIn the first case the graph is locally bowl-shaped. In the second it initially bends downward from the origin, but the positive quartic term eventually turns it upward, producing the usual double-well profile.

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The characteristic polynomial isFor ${\lambda }_1=0$ and ${\lambda }_2=1$, corresponding eigenvectors areThey are linearly independent, so the general homogeneous solution is

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Resolve the forcing in the eigenbasis:In the zero-eigenvalue direction, a constant forcing produces the particular integral $(3a-1)t{u}_1$. In the unit-eigenvalue direction, a constant particular integral is $-(1-2a)u_2$. Thus a particular integral depends on time exactly whenAdding the complementary solution from part (a), the general solution isThis is the eigenvector form of the usual variation of parameters calculation.

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For either eigenpair above, ${\lambda }_i\in \{0,1\}$ and hence ${\lambda }_i^n={\lambda }_i$ for every positive integer $n$. Thereforeso both terms in the solution from part (a), and every linear combination of them, solve the higher-order system.

A system of two scalar differential equations of order $n$ has a $2n$-dimensional solution space. The displayed family supplies two independent solutions, so there must befurther linearly independent solutions.

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Write the vector field asThe equilibrium points of a dynamical system in the closed first quadrant areThe Jacobian matrix is

At $(0,0)$ its eigenvalues are $8,-1$, so this is a saddle equilibrium. Its unstable direction is the positive $x$-axis and its stable direction is the positive $y$-axis; to first order, nearby trajectories satisfy $\dot{x}=8x$, $\dot{y}=-y$.

At $(4,0)$ the eigenvalues are $-8,3$, so it is also a saddle. The $x$-axis is the stable direction, while trajectories entering the quadrant in the $y$ direction move away.

At $(1,\sqrt{3})$,whose eigenvalues areIt is therefore a stable spiral. A point immediately to its right moves upward, so nearby trajectories spiral counterclockwise into the equilibrium. These eigendirections and the inward spiral give the requested local phase-portrait sketches.

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The only positive equilibrium solves $1-y=0$ and $x-1=0$, hence isAway from a nullcline,This is separable:Integration gives the first integralThus one convenient conserved quantity is

Its gradient isso $(1,1)$ is its only stationary point in the positive quadrant. Its Hessian matrix iswhich is positive definite everywhere. Hence $(1,1)$ is a strict, indeed global, minimum.

The nearby level curves of $E$ are closed curves surrounding this minimum. Since $E$ is constant along every trajectory, a solution starting on a sufficiently small nearby level set cannot leave the region bounded by a slightly larger level set. This proves Lyapunov stability of the equilibrium: solutions initially close to $(1,1)$ remain close for all time.

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Because $U$ and $V$ are bounded, their exponential series may be integrated term by term for every real $t$. Thus their moment-generating functions satisfyand similarly for $V$. Equality of every moment gives equality term by term, so

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Normalization of the Gaussian distribution givesCompleting the square,Thereforeand hence

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The normalizing constant satisfiesFor an integer $k$,Translation by the integer $k$ merely permutes the summation indices, so the final sum is $B^{-1}$. ConsequentlyThe last equality uses the moment-generating function of the standard normal variable from part (b).

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No. A standard counterexample comes from the log-normal distribution. Letand, for a fixed $0<|\epsilon |\le 1$, letThis is a nonnegative density distinct from $f$. For every nonnegative integer $n$, substituting $z=\log x$ makes the difference of the $n$th moments proportional toIt is the imaginary part ofwhich vanishes because its phase is $2\pi n$. The case $n=0$ also proves that $f_{\epsilon }$ is normalized. Thus the two distributions have every finite moment equal but are different. Unbounded random variables need not be determined by their moments.

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The probability generating function isDifferentiating $n$ times,At $s=0$, only the term $m=n$ remains. Hence

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For $0\le s\le 1$, independence givesTherefore

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For the geometric distribution on $\{1,2,\dots \}$,Summing the geometric series,

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When $r$ red marbles remain, the next draw reduces their number with probability $r/n$. The waiting time $W_r$ for that reduction has a geometric distribution on $\{1,2,\dots \}$ and thereforeThe successive waiting times are independent, andUsing the product rule for probability-generating functions,

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The number of heads has the binomial distribution. Choosing the $k$ head positions gives

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Conditioned on exactly $k$ heads, every $k$-element set of head positions is equally likely. There are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ such sets. A run containing all $k$ heads can begin at any of the positionsand each beginning determines one admissible set. Thus, for $k\ge 1$,

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For the process to stop on toss $n$, the last toss must be the $k$th head, while the first $n-1$ tosses must contain exactly $k-1$ heads. Hence the negative binomial distribution gives

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Let $E_j$ be the expected additional number of tosses when the current terminal run contains $j$ consecutive heads. Then $E_k=0$, and for $0\le j<k$,For $0<p<1$, iterating this recurrence from $j=k-1$ down to $0$ givesSolving,When $p=1$, the waiting time is deterministically $k$, which is also the continuous limit of this formula as $p\to 1$.

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By Tonelli theorem for the nonnegative indicators,If this is finite, then $N$ is finite almost surely: alternatively, Markov inequality givesSince $\{N=\infty \}\subseteq \{N\ge m\}$ for every $m$,This is the first Borel-Cantelli lemma.

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Write $p_n=\mathbb{P}(A_n)$. Independence gives, first for finite partial counts and then by monotone convergence,Using $1-x\le e^{-x}$ factor by factor,

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If ${\sum }_n\mathbb{P}(A_n)=\infty$, part (b) givesThus $2^{-N}=0$ almost surely. By the convention in the question, this occurs exactly when $N=\infty$. ThereforeThis is the second Borel-Cantelli lemma for independent events.

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Divide the keystrokes into disjoint blocks of five, and let $A_n$ be the event that block $n$ is exactly HELLO. The events are independent events, andConsequentlyPart (c) shows that infinitely many of these block events occur almost surely. Each occurrence is an occurrence of HELLO in the full typed sequence, so

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