Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 2

The homogeneous linear recurrence relationhas characteristic polynomial $(r-2{)}^2$, soFor a particular solution, substitute $y_n=an+b$. The left-hand side becomesso matching $n$ gives $a=1$ and $b=2$. HenceThe conditions $y_0=1$ and $y_1=0$ give $C=-1$ and $D=-1/2$. Therefore

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Differentiating the Wronskian givesThus $W^{\prime }+pW=0$, the Abel identity.

Also,If $y_2(x_0)=0$, integration gives the reduction of order formula

For the specified equation, division by $x^2$ gives $p(x)=-1/x$, so $W=Cx$. With $y_1=x^3$, an independent solution iswhich is a nonzero multiple of $x^{-1}$. HenceThe conditions at $x=1$ give $A+B=0$ and $3A-B=1$, so

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After $n$ tosses, the next total is even if either the current total is even and a tail occurs, or the current total is odd and a head occurs. ThereforeSince $q_0=1$, subtracting the fixed point $1/2$ givesThusThe formula also covers the endpoint cases $p=0$ and $p=1$.

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Normalization of the probability density function givessoThe expected value isThe second moment isTherefore the variance is

For the median $m$,so

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For$x_0$ is an ordinary point when $P$ and $Q$ are analytic there, and a singular point otherwise. A singular point is regular singular when $(x-x_0)P(x)$ and $(x-x_0{)}^{2}Q(x)$ are analytic there. These are the ordinary-point and regular-singular criteria.

For Kummer's equation, substituteEquating the coefficient of $x^m$ givesTaking $c_0=1$,where $(a{)}_m=a(a+1)\cdots (a+m-1)$ is the rising factorial. Thus

Putting $y=x^{1-b}u$ and simplifying givesThereforeFor nonintegral $b$, the powers $x^0$ and $x^{1-b}$ at zero are distinct, so these solutions are linearly independent.

When $b\to 1$, both solutions tend to $M(x,a,1)$. Differentiate their difference with respect to $b$. Writing $M_a,M_b$ for derivatives with respect to the second and third arguments,Hence at $b=1$ one may takeas two linearly independent solutions. Their independence follows from the logarithmic term in the second solution.

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With $\eta =x-t$ and $\xi =x+t$, the chain rule givesHenceso the wave equation is equivalent to $U_{\eta \xi }=0$. Integrating in each variable givesThe initial conditions implySolving and integrating yields the D'Alembert formula with initial velocity

If $|x-x_0|>r+t$, then both $x-t$ and $x+t$ lie outside the support interval $[x_0-r,x_0+r]$, and every point between them does as well. All three terms therefore vanish. Thuswhich is finite propagation speed.

For the finite string, differentiate its energy and use $y_{tt}=y_{xx}$:The fixed-end conditions hold for every $t$, so differentiation gives $y_t(0,t)=y_t(L,t)=0$. The boundary term vanishes and

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Put $\omega =\dot{\theta }$. The first-order system isWriteOn $[0,\pi ]$, $h$ is strictly increasing. Moreover,when $3\lambda <4\mu$. Thus there is one ${\theta }_{\ast }\in (0,\pi )$ with $h({\theta }_{\ast })=0$, in addition to the fixed points at $\theta =0,\pi$. Explicitly,

Let $g(\theta )=\sin \theta h(\theta )$. Linearization at $({\theta }_0,0)$ has eigenvalues satisfyingAt zero, $g^{\prime }(0)=h(0)<0$, and at $\pi$, $g^{\prime }(\pi )=-h(\pi )<0$, so both endpoints are saddles. At the interior point,so $({\theta }_{\ast },0)$ is a center. On the full angular interval there is a second center at $(-{\theta }_{\ast },0)$.

The system has the conserved energy of a conservative planar phase portrait,For $\lambda =1$, $\mu =3/2$,The phase portrait on the cylinder consists of centers at $\theta =\pm \arccos (-11/16)$ surrounded by closed periodic-energy curves, with saddles at $\theta =0$ and the identified point $\theta =\pi =-\pi$. The saddle-energy contours form the separatrices between librations in the potential wells and trajectories crossing the lower barrier.

If $3\lambda >4\mu$, then $h(0)>0$ and monotonicity gives no interior zero on $(0,\pi )$. The point $(0,0)$ becomes a center, while $(\pi ,0)$ remains a saddle. At $3\lambda =4\mu$, the two off-axis centers coalesce with the origin and the linearization there is degenerate.

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Termwise differentiation of the matrix exponential givesThus $y(t)=e^{tA}{y}_0$ satisfies $y^{\prime }=Ay$ and $y(0)=y_0$. If two solutions existed, multiplying their difference by $e^{-tA}$ would give a vector with zero derivative and zero initial value, proving uniqueness.

For the inhomogeneous equation, multiply by the integrating factor $e^{-tA}$:Integration and multiplication by $e^{tA}$ give the variation-of-constants formula

For the stated matrix,soWriting $f(t)=v\sin t$ with $v=(1,3,0{)}^T$, one findsThereforeor explicitly

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For $\mathbb{P}(B)>0$, the conditional probability isHenceBecause the $B_k$ form a partition, the law of total probability givesSubstitution proves Bayes theorem in the required form:

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Urn $r$ has $n-r$ blue balls among $n-1$ balls. Averaging over the uniformly chosen urn,The probability that both removed balls are blue isThereforeThe first blue draw makes urns with more blue balls more likely, which explains why the conditional probability exceeds one half.

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Events $A$ and $B$ are independent whenFor a possible sum $2\le t\le 12$, let $N_t$ be the number of ordered die pairs with sum $t$. ThenIf $t-i\in \{1,\dots ,6\}$, then $\mathbb{P}(A_t\cap B_i)=1/36$; otherwise it is zero. Independence for a possible sum therefore requiresso $N_t=6$. This occurs only for $t=7$, and then $7-i$ is valid for every $i=1,\dots ,6$. ThusIf impossible sums are admitted as events of probability zero, they are trivially independent as well.

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Call the coins $1,2,3$ in the order given. Coin 1 beats the constant score of coin 2 exactly when it shows a head, so coin 1 wins with probability $3/5$. Coin 2 beats coin 3 exactly when coin 3 shows a head and scores three, again with probability $3/5$.

Coin 3 beats coin 1 whenever coin 3 shows a tail, or when coin 3 shows a head and coin 1 shows a tail. Its winning probability is thereforeThus the preferences form a nontransitive cycle:The second chooser can always select a coin that has winning probability greater than one half against the first choice. Therefore

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Ignore rolls showing $2,3,4,5$ and consider the first roll among $\{1,6\}$. By symmetry, the event $A$ that a 1 occurs before a 6 has probability $1/2$. The event that the very first roll is 1 has probability $1/6$ and is contained in $A$. Hence

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Let $h(x)$ be the probability that the simple symmetric random walk starting at $x\in \{0,\dots ,n\}$ hits zero before $n$. The first-step recurrence isThe recurrence says that $h$ is affine, and the boundary values determine it:

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Part (i) gives $\mathbb{P}(A)=2/3$. Since $\{X_1=0\}\subseteq A$ and has probability $1/2$,Conditioned on $A$, a walk at one moves to zero with probability $3/4$ and to two with probability $1/4$. A walk at two must next move to one, since a move to three would violate $A$. Iffirst-step analysis givesSolving,

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For hitting zero before four, the success probabilities areConditioning each transition on eventual success givesThus the conditional expected hitting times satisfySolving this linear system gives

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Uniform area density in the unit disk is $1/\pi$. The polar coordinates Jacobian gives the joint densityThus the coordinates are independent and the uniform random point in a disk has

For independent points $A,B$, the area of triangle $OAB$ isNowIndependence therefore givesConditioned on $A,B$, the probability that $C$ lies inside $OAB$ is its area divided by the disk area $\pi$. Hence

Four points in general position fail to form a convex quadrilateral exactly when one lies inside the triangle formed by the other three. Each of $A,B,C$ lies inside the triangle formed by $O$ and the other two with probability $4/(9{\pi }^2)$, and these three events are disjoint. The remaining possibility is that $O$ lies inside $ABC$. By origin in a triangle of three radial random points, this has probability $1/4$. Therefore

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The Chebyshev inequality states that for a random variable $Y$ with finite variance and every $\delta >0$,Indeed, Markov inequality applied to the nonnegative variable $(Y-\mathbb{E}Y{)}^2$ gives

The sum $S_n=X_1+\cdots +X_n$ has the binomial distribution, soFor fixed $n$ this is a finite sum of polynomial functions of $p$, hence is a polynomial.

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The sample mean $S_n/n$ has expectation $p$ and varianceThe sum in the question is exactlyChebyshev's inequality therefore gives, uniformly in $p$,

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Because $f$ is continuous on the compact interval $[0,1]$, it is uniformly continuous and bounded; write $|f|\le M$. Given $ϵ>0$, choose $\delta >0$ such thatThenThe bound is independent of $p$. Taking the supremum, then $n\to \infty$, gives a limit superior at most $ϵ$. Since $ϵ$ is arbitrary,This is the Bernstein polynomial proof of the Weierstrass approximation theorem.

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