Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 2

The complementary solution comes from the characteristic equation $r^4-a^4=0$, whose roots are $a,-a,ia,-ia$. It is thereforeFor the resonant exponential particular solution, the forcing is already a homogeneous exponential, so the method of undetermined coefficients needs the resonant trial $y_p=Kx{e}^{-ax}$. If $P(r)=r^4-a^4$, the identity $P(D)[x{e}^{rx}]=P^{\prime }(r)e^{rx}$ when $P(r)=0$ gives $P^{\prime }(-a)=-4a^3$. Hence $K=-1/(4a^3)$ andThe decay condition first eliminates the growing exponential, $C_1=0$. A nonzero combination of the sine and cosine has a fixed nonzero amplitude and cannot tend to zero, so $C_3=C_4=0$ as well. Both remaining exponential terms decay because $a>0$. Finally $y(0)=C_2=1$, giving the unique answerThis is an instance of resonance in a differential equation: multiplication by $x$ supplies the particular solution when the forcing exponent is a simple characteristic root.

For the cubic population iteration, put $g(u)=\lambda u(1-u^2)$. Its fixed points solveThus $u_{\ast }=0$ always exists. For $\lambda \ne 0$ there are alsowhenever the expression under the square root is positive, namely $\lambda <0$ or $\lambda >1$. At $\lambda =1$ these coincide with zero; at $\lambda =0$ only zero exists.

The criterion for fixed point stability for an iteration is $|g^{\prime }(u_{\ast })|<1$. To justify it, continuity of $g^{\prime }$ gives a small interval around the fixed point where $|g^{\prime }|\le q<1$. The mean value theorem then gives $|g(u)-u_{\ast }|\le q|u-u_{\ast }|$, so this interval is invariant and the error decreases geometrically. This proves local asymptotic stability rather than merely linear boundedness.

Here $g^{\prime }(u)=\lambda (1-3u^2)$. At zero its value is $\lambda$, so zero is locally asymptotically stable for $-1<\lambda <1$. At a nonzero fixed point it isThe strict inequality $|3-2\lambda |<1$ is equivalent to $1<\lambda <2$. Both nonzero real fixed points are therefore locally asymptotically stable on that interval. This proves the two requested existence ranges. The strict derivative criterion alone makes no conclusion at the endpoint multipliers of modulus one.

The sampling without replacement makes all $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ unordered pairs equally likely. The count of red-red pairs is $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$, while the count of mixed pairs is $3(n-3)$. Thus their probabilities satisfyand cancellation gives $n-3=3$. Hence $\boxed{n=6}$, with three black socks. The count of black-black pairs is also $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$, soThe mixed, red-red and black-black probabilities are respectively $3/5,1/5,1/5$, which add to one and verify the given ratio.

Let $X$ be one score. It has the discrete uniform distribution on $1,\dots ,6$, so its expected value and second moment areThereforeFor independent throws, linearity of expectation and additivity of variance give $\mathbb{E}{Y}_n=7n/2$ and $\mathrm{Var}(Y_n)=35n/12$. The sample average consequently has expectation $7/2$ and variance $35/(12n)$.

The Chebyshev inequality follows by applying the Markov inequality to the nonnegative square of a centered random variable. It givesThis is at most $1/10$ if $n\ge 350/27$. Thus $n=13$ suffices, as does every larger integer. This is the smallest integer guaranteed by this particular Chebyshev inequality bound; it need not be the smallest integer satisfying the actual probability requirement.

Measure temperature relative to ambient by ${\theta }_i=T_i-T_{\infty }$. Newton's law of cooling models ordinary cooling by ${\theta }_i^{\prime }=-a{\theta }_i$, since heat loss is proportional to the excess temperature. The additional negative forcing represents the cooling from the milk, with the temperature drop or rate normalized to one in the chosen units.

The Dirac delta function has unit integral and is supported at the instant of addition. It therefore imposes a sudden unit drop for cup 1. The difference $H(t-1)-H(t-2)$ of Heaviside step functions is one during the addition interval and zero outside it, giving a unit cooling rate for cup 2. The second temperature is continuous; only its derivative changes at the switching times. The value assigned to a Heaviside step function exactly at its jump does not change the integrated solution. This is the impulse versus finite-duration cooling comparison.

Before time one, both forcing terms vanish. The separable differential equation ${\theta }_i^{\prime }=-a{\theta }_i$ gives ${\theta }_i(t)=C_i{e}^{-at}$. The initial values give $C_1=C_2=T_0-T_{\infty }$, and henceIn particular the common left-hand limit at time one is $T_{\infty }+(T_0-T_{\infty })e^{-a}$.

Integrate the first equation over $(1-\epsilon ,1+\epsilon )$. The integral of the ordinary cooling term tends to zero as $\epsilon \downarrow 0$, while the Dirac delta function integrates to one. Thus the jump condition for an impulse isAfter the jump there is no further forcing, so Newton's law of cooling givesEquivalently,The last term has jump minus one at time one and then decays at the same exponential rate as the unforced temperature.

On the active interval the excess temperature satisfies ${\theta }_2^{\prime }+a{\theta }_2=-1$. Multiplication by the integrating factor $e^{at}$ and integration from time one givesThere is no impulse in this equation, so continuity supplies ${\theta }_2(1)=(T_0-T_{\infty })e^{-a}$. ThereforeThe equivalent correction to the common unforced temperature is $-(1-e^{-a(t-1)})/a$, which vanishes continuously at the beginning of the interval.

At time two, continuity of the second temperature givesThe forcing then turns off, so Newton's law of cooling again gives exponential decay:Subtracting this from the first temperature and expressing both corrections with the same exponential factor givesBecause $e^a>1+a$ for $a>0$, this difference is positive. Thus the cup cooled instantaneously is warmer after the addition interval: its cooling occurred earlier and has undergone more subsequent exponential attenuation.

During the addition interval, subtracting the formulas in (ii) and (iii) cancels the common unforced temperature:This is zero precisely when $e^{-a(t-1)}=1/(1+a)$, soThe inequalities $0<\log (1+a)<a$ give $1<t^{\ast }<2$, so using the active-interval formula was justified. The difference has derivative $(1+a)e^{-a(t-1)}>0$, making this crossing unique within that interval. Immediately after time one it is minus one; after time two it is strictly positive by (iv). Hence there is no other equality at any finite time after time one. Both temperatures approach the same ambient limit, but that limiting equality is not an additional crossing.

With the indicated order of the two solutions, define the Wronskian byWhere $y_1\ne 0$, this gives the first-order linear differential equationIntegrating produces the reduction of order formulaThis formula is local on an interval where the known solution does not vanish; an arbitrary nonzero scaling of the linearly independent solution changes the normalization of $W$.

Differentiate the Wronskian and use the differential equation for each solution:Thus $\boxed{W^{\prime }+pW=0}$, proving the Abel identity, and $W=C\exp (-\int pdx)$. For continuous coefficients on a nonsingular interval, linearly independent solutions have $C\ne 0$.

For the particular equation, write $s=x-1$. The proposed solution is $y_1=s^2$, with $y_1^{\prime }=2s$ and $y_1^{\prime \prime }=2$, so substitution gives $2s^2+2s^2-4s^2=0$. Away from the singular point $x=1$, division by $s^2$ gives $p=1/s$. The Abel identity therefore giveswith a separate constant on each interval lying on one side of the singular point. The reduction of order integral is $\int W/y_1^{2}dx=C\int s^{-5}ds=-C/(4s^4)$, soChoosing $C_2=0$, $C=-4$ givesThe nonzero Wronskian proves linear independence. Direct substitution also verifies the second solution, since $s^2(6s^{-4})+s(-2s^{-3})-4s^{-2}=0$. These solutions form a basis on either nonsingular interval; the second solution is not defined at $x=1$.

Write a power series $y={\sum }_{n\ge 0}{c}_n{x}^n$. Substitution givesThe lowest coefficients give $c_1=0$, $c_3=0$, with $c_0,c_2$ free. For $n\ge 4$, matching the coefficient of $x^{n-1}$ gives the recurrenceAll odd coefficients consequently vanish. For the first solution, $c_0=a$, $c_2=y_1^{\prime \prime }(0)/2=0$, so the first three nonzero terms, when $a\ne 0$, areFor the second, $c_0=0$, $c_2=b/2$, givingIf its prescribed constant is zero, the corresponding solution is identically zero and has no nonzero terms.

For the change of variable, put $y(x)=Y(u)$, $u=x^{\alpha }$, with $\alpha \ne 0$. The chain rule givesAfter division by $x^{2\alpha -1}$, the coefficient of $Y_u$ is $\alpha (\alpha -2)x^{-\alpha }$ and that of $Y$ is $4x^{4-2\alpha }$. A genuine constant-coefficient equation requires $\alpha =2$, which both removes $Y_u$ and makes the final coefficient constant. The transformed equation is $4Y_{uu}+4Y=0$.

This quadratic-variable reduction of a singular oscillator yieldsIt holds on each interval away from zero and both solutions extend smoothly through zero. Here $y(0)=A$ and $y^{\prime \prime }(0)=2B$, so the specified solutions are exactly $a\cos (x^2)$ and $(b/2)\sin (x^2)$. Their Taylor series agree with the recurrences above. Although the original equation has a regular singular point at zero, these two linearly independent smooth solutions supply the required data there.

For this quartic potential with fourfold symmetry, the first derivatives factor:The critical points are the origin, the four axis points $(\pm 1,0),(0,\pm 1)$, and four points with both coordinates nonzero. At the latter, $x^2+b{y}^2=y^2+b{x}^2=1$. Subtracting and using $b\ne 1$ gives $x^2=y^2=1/(1+b)$. Thus the complete list iswhere the two signs in the final group are chosen separately.

The Hessian isAt the origin it is $2I$, so there is a strict local minimum with value zero. At an axis point its eigenvalues are $-4$ and $2(1-b)$, giving a saddle point of a scalar function if $b<1$ and a strict local maximum if $b>1$. All four axis values are $1/2$.

At a diagonal point the Hessian has eigenvaluesHence these four points are strict local maxima for $b<1$ and saddle points of a scalar function for $b>1$. Their value is $1/(1+b)$. None of these critical points is degenerate when $b\ne 1$.

To describe the level curves precisely, use polar coordinates and writeFor a level $f=c$, the possible radii satisfyretaining only real, nonnegative values. This formula, together with reflection in each axis and interchange of the coordinates, determines the level curves. The radial maximum in a direction is $1/(2q)$, attained at $r^2=1/q$. It also proves that the local maxima identified above are global: the largest directional value is $1/(1+b)$ on the diagonals if $b<1$, and $1/2$ on the axes if $b>1$.

For negative $c$ only the plus sign gives a positive radius, producing one outer closed level curve. At $c=0$ there is the isolated origin and the outer curve $r^2=2/q$. For $0<c<c_s$, where $c_s$ is the saddle value, both radial branches exist in every direction, giving an inner loop around the origin and an outer loop. At $c=c_s$ they meet at the four saddle points of a scalar function. Above the saddle value, four separate loops surround the four maxima; they shrink to those points at the maximum value. There are no level curves above it.

For $0<b<1$, the saddle value is $c_s=1/2$ and the four high-level loops lie along the diagonals. For $b>1$, $c_s=1/(1+b)$ and the four high-level loops lie along the coordinate axes. The sketches show representative values in both regimes; the highlighted level curve is the saddle level.

Finally, at $b=1$ the expression depends only on radius:Consequently $\boxed{\max f=1/2\text{, attained exactly on }{x}^2+y^2=1.}$ The maximum set is the whole unit circle, not merely four isolated points.

Let $G(s)=\mathbb{E}[s^{\xi }]$ be the offspring probability generating function, and let $G_n(s)=\mathbb{E}[s^{Z_n}]$, for $0\le s\le 1$. Given $Z_n=k$, the next generation is a sum of $k$ independent offspring counts. Multiplication of their probability generating functions givesTaking conditional expectation and then averaging provesThus $G_n$ is the $n$-fold composition of $G$ with itself. Induction makes this precise: it is true for $n=0$, and composing once more gives the next generation. In particular $G_n\circ G=G\circ G_n$, since both are the same iterated function. This is the generating-function recursion for a Galton-Watson process.

Write $m=\mathbb{E}{Z}_1=G^{\prime }(1^{-})$, initially assuming it is finite. The law of total expectation givesStarting at one therefore yields $\boxed{\mathbb{E}{Z}_n=m^n}$, with the value at $n=0$ understood to be one even if $m=0$. Equivalently, differentiating the composition at one gives the same recurrence. If the offspring expectation is infinite, every positive generation has infinite expectation: a positive chance of at least one parent remains at each finite generation, and conditional on any positive parent count the next expected count is infinite.

For a Poisson branching process, $G(s)=e^{\lambda (s-1)}$. Extinction is absorbing because an empty generation has no parents, so extinction by generation $n$ is precisely $Z_n=0$. Hence $x_n=G_n(0)$. Using the composition in the order $G\circ G_n$ givesFor example $x_1=e^{-\lambda }$ is the chance of no offspring in the first generation. The recurrence includes the degenerate case $\lambda =0$, where extinction occurs in the first generation with certainty.

Put $q=1-p$ and track the difference $S_n$ between the numbers of wins. It is a biased random walk with increments $+1$ and $-1$, stopped on first reaching $3$ or $-3$.

To finish at $+3$ on game five, there must be four wins for $A$ and one for $B$, with the final game won by $A$. The unique loss for $A$ can be in position one, two or three. If it were in position four or five, the first three games would already have stopped the contest. Each of the three allowed paths remains strictly inside the barriers until the last step and has probability $p^{4}q$. By interchanging players, there are also three paths ending at $-3$, each with probability $p{q}^4$. ThereforeFor a fair game this is $3/16$. Counting all five-game sequences with final win difference three would overcount paths on which the contest ended earlier.

This is a gambler's ruin hitting problem. For $0<p<1$, let $h_j$ be the probability of reaching $+3$ before $-3$ when the current difference is $j$. Conditioning on the next independent game givesStopping occurs almost surely: conditional on being inside the barriers, three consecutive wins by either player force absorption within the next three games. This event has probability $p^3+q^3>0$, so the chance of surviving $m$ such blocks is at most $(1-p^3-q^3{)}^m$, which tends to zero.

If $p\ne q$, the characteristic roots of the interior recurrence are $1$ and $r=q/p$, so $h_j=A+B{r}^j$. Imposing the two boundary values yieldsIn particular,For $p=q=1/2$ the recurrence instead has affine solutions, and the boundary values give $h_j=(j+3)/6$, hence $h_0=1/2$. The deterministic cases $p=0,1$ give zero and one respectively. The formula covers all of them continuously:The stopping argument and the boundary recurrence show why the answer concerns the first barrier reached rather than the win difference after a predetermined number of games.

Write $s=\sqrt{1-{\rho }^2}>0$. The inverse change of variables is $y=\rho x+sz$, whose Jacobian determinant has absolute value $s$. Completing the square givesThe change of variables formula therefore transforms the joint probability density function toIt factors into two normalized standard normal distribution densities. Thus $X$ and $Z$ are independent $N(0,1)$ variables; factorization proves the requested independence, rather than simply zero covariance.

The standard Gaussian random vector $(X,Z)$ has a rotation-invariant density. In polar coordinates, $X=r\cos \theta$, $Z=r\sin \theta$, its density with the area factor is $(2\pi {)}^{-1}r{e}^{-r^2/2}$, so the angular coordinate is uniform on an interval of length $2\pi$.

Put $\alpha =\arcsin \rho \in (-\pi /2,\pi /2)$, so $s=\cos \alpha$ andIn the right half-plane, choose $-\pi /2<\theta <\pi /2$. The additional condition $Y>0$ then requires $-\alpha <\theta <\pi /2$. This wedge has angle $\pi /2+\alpha$; its boundary has probability zero. ConsequentlyAt $\rho =0$ this is $1/4$, as required by independence. The limits as $\rho \to 1$ and $\rho \to -1$ are $1/2$ and zero, respectively. The angular calculation is also the geometric basis of the Gaussian sign-correlation identity.

Use the positive-support convention for the geometric distribution and put $q=1-p$. Differentiating the convergent geometric series ${\sum }_{k\ge 0}{q}^k=(1-q{)}^{-1}$ givesConsequentlySubtracting the square of the expectation from the second moment yields

For the die, this is the coupon collector problem. When $r$ scores are still missing, let $W_r$ count rolls up to and including the next newly observed score. Every fresh roll has success probability $r/6$, so $W_r$ has a geometric distribution of parameter $r/6$. These six successive waiting times are independent: conditional on the entire past at the start of a stage, its waiting-time distribution depends only on the number $r$, not on the identities of the missing scores. Iterated conditioning therefore factors their joint distribution. The total number of rolls is $N={\sum }_{r=1}^6{W}_r$.

Using linearity of expectation givesUsing variance additivity for independent random variables, or the coupon collector waiting-time variance formula just derived by this decomposition, givesHence the requested standard deviation isThe hint rounds the sum of squares from $53.69$ to $53.7$; using that rounded value gives $\sqrt{39}\approx 6.245$, consistent at the precision of the hint. The exact computation preserves the slightly more precise value above.