= Paper 2 {scope} https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2021/paperia_2_2021.pdf = 1A {parent=Paper 2} {scope} {title2=Differential Equations} = Solution {parent=1A} The homogeneous $$ y_{n+2}-4y_{n+1}+4y_n=0 $$ has characteristic polynomial $(r-2)^2$, so $$ y_n^{(h)}=(C+Dn)2^n. $$ For a particular solution, substitute $y_n=an+b$. The left-hand side becomes $$ an+(b-2a), $$ so matching $n$ gives $a=1$ and $b=2$. Hence $$ y_n=(C+Dn)2^n+n+2. $$ The conditions $y_0=1$ and $y_1=0$ give $C=-1$ and $D=-1/2$. Therefore $$ \boxed{y_n=n+2-\left(1+\frac n2\right)2^n}. $$ Solved by gpt-5.6-sol high. = 2A {parent=Paper 2} {scope} {title2=Differential Equations} = Solution {parent=2A} Differentiating the gives $$ \begin{aligned} W' &=(y_1y_2'-y_2y_1')'\\ &=y_1y_2''-y_2y_1''\\ &=y_1(-py_2'-qy_2)-y_2(-py_1'-qy_1)\\ &=-pW. \end{aligned} $$ Thus $W'+pW=0$, the . Also, $$ \left(\frac{y_2}{y_1}\right)' =\frac{y_1y_2'-y_2y_1'}{y_1^2} =\frac W{y_1^2}. $$ If $y_2(x_0)=0$, integration gives the formula $$ \boxed{ y_2(x)=y_1(x)\int_{x_0}^x\frac{W(t)}{y_1(t)^2}\,dt}. $$ For the specified equation, division by $x^2$ gives $p(x)=-1/x$, so $W=Cx$. With $y_1=x^3$, an independent solution is $$ x^3\int x^{-5}\,dx, $$ which is a nonzero multiple of $x^{-1}$. Hence $$ y=Ax^3+\frac Bx. $$ The conditions at $x=1$ give $A+B=0$ and $3A-B=1$, so $$ \boxed{y(x)=\frac14\left(x^3-\frac1x\right)}. $$ Solved by gpt-5.6-sol high. = 3D {parent=Paper 2} {scope} {title2=Probability} = Solution {parent=3D} 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. Therefore $$ q_{n+1}=(1-p)q_n+p(1-q_n) =p+(1-2p)q_n. $$ Since $q_0=1$, subtracting the fixed point $1/2$ gives $$ q_{n+1}-\frac12 =(1-2p)\left(q_n-\frac12\right). $$ Thus $$ \boxed{ q_n=\frac{1+(1-2p)^n}{2}}. $$ The formula also covers the endpoint cases $p=0$ and $p=1$. Solved by gpt-5.6-sol high. = 4F {parent=Paper 2} {scope} {title2=Probability} = Solution {parent=4F} Normalization of the gives $$ 1=c\int_0^{\sqrt3}\frac{dx}{1+x^2} =c\arctan(\sqrt3) =c\frac\pi3, $$ so $$ \boxed{c=\frac3\pi}. $$ The is $$ \mathbb EX =\frac3\pi\int_0^{\sqrt3}\frac{x}{1+x^2}\,dx =\boxed{\frac{3\log2}{\pi}}. $$ The second moment is $$ \mathbb EX^2 =\frac3\pi\int_0^{\sqrt3}\frac{x^2}{1+x^2}\,dx =\frac{3\sqrt3}{\pi}-1. $$ Therefore the is $$ \boxed{ \operatorname{var}X =\frac{3\sqrt3}{\pi}-1 -\frac{9(\log2)^2}{\pi^2}}. $$ For the $m$, $$ \frac3\pi\arctan m=\frac12, $$ so $$ \boxed{m=\tan\frac\pi6=\frac1{\sqrt3}}. $$ Solved by gpt-5.6-sol high. = 5A {parent=Paper 2} {scope} {title2=Differential Equations} = Solution {parent=5A} For $$ y''+P(x)y'+Q(x)y=0, $$ $x_0$ 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 $(x-x_0)^2Q(x)$ are analytic there. These are the ordinary-point and [regular-singular] criteria. For [Kummer's equation], substitute $$ y=\sum_{m=0}^{\infty}c_mx^m. $$ Equating the coefficient of $x^m$ gives $$ (m+1)(m+b)c_{m+1}=(m+a)c_m. $$ Taking $c_0=1$, $$ \boxed{ c_m(a,b)=\frac{(a)_m}{(b)_m\,m!}}, $$ where $(a)_m=a(a+1)\cdots(a+m-1)$ is the rising factorial. Thus $$ y_1=M(x,a,b) =\sum_{m=0}^{\infty}\frac{(a)_m}{(b)_m\,m!}x^m. $$ Putting $y=x^{1-b}u$ and simplifying gives $$ xu''+(2-b-x)u'-(a-b+1)u=0. $$ Therefore $$ \boxed{ y_2=x^{1-b}M(x,a-b+1,2-b)}. $$ For nonintegral $b$, the powers $x^0$ and $x^{1-b}$ at zero are distinct, so these solutions are linearly independent. When $b\to1$, 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, $$ \begin{aligned} \lim_{b\to1}\frac{y_2-y_1}{b-1} &=-M(x,a,1)\log x\\ &\quad-M_a(x,a,1)-2M_b(x,a,1). \end{aligned} $$ Hence at $b=1$ one may take $$ \boxed{ M(x,a,1),\qquad -M(x,a,1)\log x-M_a(x,a,1)-2M_b(x,a,1)} $$ as two linearly independent solutions. Their independence follows from the logarithmic term in the second solution. Solved by gpt-5.6-sol high. = 6A {parent=Paper 2} {scope} {title2=Differential Equations} = Solution {parent=6A} With $\eta=x-t$ and $\xi=x+t$, the gives $$ \partial_x=\partial_\eta+\partial_\xi, \qquad \partial_t=-\partial_\eta+\partial_\xi. $$ Hence $$ u_{xx}-u_{tt}=4U_{\eta\xi}, $$ so the is equivalent to $U_{\eta\xi}=0$. Integrating in each variable gives $$ u(x,t)=F(x-t)+G(x+t). $$ The initial conditions imply $$ F(x)+G(x)=f(x), \qquad -F'(x)+G'(x)=g(x). $$ Solving and integrating yields the $$ \boxed{ u(x,t)=\frac12[f(x-t)+f(x+t)] +\frac12\int_{x-t}^{x+t}g(y)\,dy}. $$ 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. Thus $$ \boxed{u(x,t)=0\quad\text{when }|x-x_0|>r+t}, $$ which is finite propagation speed. For the finite string, differentiate its energy and use $y_{tt}=y_{xx}$: $$ \begin{aligned} \frac{dE}{dt} &=\int_0^L(y_xy_{xt}+y_ty_{tt})\,dx\\ &=\int_0^L(y_xy_{xt}+y_ty_{xx})\,dx\\ &=[y_ty_x]_0^L. \end{aligned} $$ 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 $$ \boxed{E(t)=E(0)}. $$ Solved by gpt-5.6-sol high. = 7A {parent=Paper 2} {scope} {title2=Differential Equations} = Solution {parent=7A} Put $\omega=\dot\theta$. The first-order system is $$ \boxed{ \dot\theta=\omega,\qquad \dot\omega=-\sin\theta \left(\lambda-2\mu+\frac{2\mu}{\sqrt{5+4\cos\theta}}\right)}. $$ Write $$ h(\theta)=\lambda-2\mu +\frac{2\mu}{\sqrt{5+4\cos\theta}}. $$ On $[0,\pi]$, $h$ is strictly increasing. Moreover, $$ h(0)=\lambda-\frac{4\mu}{3}<0, \qquad h(\pi)=\lambda>0 $$ when $3\lambda<4\mu$. Thus there is one $\theta_*\in(0,\pi)$ with $h(\theta_*)=0$, in addition to the fixed points at $\theta=0,\pi$. Explicitly, $$ \boxed{ \cos\theta_* =\frac{\mu^2}{(2\mu-\lambda)^2}-\frac54}. $$ Let $g(\theta)=\sin\theta\,h(\theta)$. Linearization at $(\theta_0,0)$ has eigenvalues satisfying $$ \sigma^2=-g'(\theta_0). $$ At zero, $g'(0)=h(0)<0$, and at $\pi$, $g'(\pi)=-h(\pi)<0$, so both endpoints are saddles. At the interior point, $$ g'(\theta_*)=\sin\theta_*h'(\theta_*)>0, $$ so $(\theta_*,0)$ is a center. On the full angular interval there is a second center at $(-\theta_*,0)$. The system has the conserved energy of a , $$ \boxed{ E=\frac{\omega^2}{2} -(\lambda-2\mu)\cos\theta -\mu\sqrt{5+4\cos\theta}}. $$ For $\lambda=1$, $\mu=3/2$, $$ \cos\theta_*=-\frac{11}{16}. $$ 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. Solved by gpt-5.6-sol high. = 8A {parent=Paper 2} {scope} {title2=Differential Equations} = Solution {parent=8A} Termwise differentiation of the gives $$ \begin{aligned} \frac d{dt}e^{tA} &=\sum_{m=1}^{\infty}\frac{m t^{m-1}A^m}{m!}\\ &=A\sum_{j=0}^{\infty}\frac{t^jA^j}{j!} =Ae^{tA}. \end{aligned} $$ Thus $y(t)=e^{tA}y_0$ satisfies $y'=Ay$ 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 $e^{-tA}$: $$ \frac d{dt}(e^{-tA}x)=e^{-tA}f(t). $$ Integration and multiplication by $e^{tA}$ give the variation-of-constants formula $$ \boxed{ x(t)=e^{tA}x_0 +\int_0^te^{(t-s)A}f(s)\,ds}. $$ For the stated matrix, $$ A^2= \begin{pmatrix} 12&-4&-4\\ 12&-4&-4\\ 24&-8&-8 \end{pmatrix}, \qquad A^3=0, $$ so $$ e^{tA}=I+tA+\frac{t^2}{2}A^2. $$ Writing $f(t)=v\sin t$ with $v=(1,3,0)^T$, one finds $$ Ax_0=0,\qquad Av=(8,8,16)^T,\qquad A^2v=0. $$ Therefore $$ x(t)=x_0+v(1-\cos t)+Av(t-\sin t), $$ or explicitly $$ \boxed{ x(t)= \begin{pmatrix} 8t-8\sin t-\cos t+2\\ 8t-8\sin t-3\cos t+4\\ 16t-16\sin t+2 \end{pmatrix}}. $$ Solved by gpt-5.6-sol high. = 9E {parent=Paper 2} {scope} {title2=Probability} = a {parent=9e} {scope} = i {parent=a} {scope} = Solution {parent=i} For $\mathbb P(B)>0$, the is $$ \mathbb P(A\mid B) =\frac{\mathbb P(A\cap B)}{\mathbb P(B)}. $$ Hence $$ \mathbb P(B_j\mid A) =\frac{\mathbb P(A\mid B_j)\mathbb P(B_j)} {\mathbb P(A)}. $$ Because the $B_k$ form a partition, the law of total probability gives $$ \mathbb P(A) =\sum_{k=1}^n\mathbb P(A\mid B_k)\mathbb P(B_k). $$ Substitution proves in the required form: $$ \boxed{ \mathbb P(B_j\mid A) =\frac{\mathbb P(A\mid B_j)\mathbb P(B_j)} {\sum_{k=1}^n\mathbb P(A\mid B_k)\mathbb P(B_k)}}. $$ Solved by gpt-5.6-sol high. = ii {parent=a} {scope} = Solution {parent=ii} Urn $r$ has $n-r$ blue balls among $n-1$ balls. Averaging over the uniformly chosen urn, $$ \mathbb P(\text{first blue}) =\frac1n\sum_{r=1}^n\frac{n-r}{n-1} =\boxed{\frac12}. $$ The probability that both removed balls are blue is $$ \begin{aligned} \mathbb P(B_1\cap B_2) &=\frac1n\sum_{r=1}^n \frac{(n-r)(n-r-1)}{(n-1)(n-2)}\\ &=\frac13. \end{aligned} $$ Therefore $$ \boxed{ \mathbb P(B_2\mid B_1) =\frac{\mathbb P(B_1\cap B_2)}{\mathbb P(B_1)} =\frac23}. $$ The first blue draw makes urns with more blue balls more likely, which explains why the conditional probability exceeds one half. Solved by gpt-5.6-sol high. = b {parent=9e} {scope} = i {parent=b} {scope} = Solution {parent=i} Events $A$ and $B$ are [independent] when $$ \mathbb P(A\cap B)=\mathbb P(A)\mathbb P(B). $$ For a possible sum $2\leq t\leq12$, let $N_t$ be the number of ordered die pairs with sum $t$. Then $$ \mathbb P(A_t)=\frac{N_t}{36}, \qquad \mathbb P(B_i)=\frac16. $$ If $t-i\in\{1,\ldots,6\}$, then $\mathbb P(A_t\cap B_i)=1/36$; otherwise it is zero. Independence for a possible sum therefore requires $$ \frac1{36}=\frac{N_t}{216}, $$ so $N_t=6$. This occurs only for $t=7$, and then $7-i$ is valid for every $i=1,\ldots,6$. Thus $$ \boxed{A_t\text{ and }B_i\text{ are independent exactly when } t=7,\ 1\leq i\leq6}. $$ If impossible sums are admitted as events of probability zero, they are trivially independent as well. Solved by gpt-5.6-sol high. = ii {parent=b} {scope} = Solution {parent=ii} 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 therefore $$ \frac25+\frac35\frac25 =\frac{16}{25}>\frac12. $$ Thus the preferences form a nontransitive cycle: $$ 1\text{ beats }2,\qquad 2\text{ beats }3,\qquad 3\text{ beats }1. $$ The second chooser can always select a coin that has winning probability greater than one half against the first choice. Therefore $$ \boxed{\text{it is preferable to choose second}.} $$ Solved by gpt-5.6-sol high. = 10E {parent=Paper 2} {scope} {title2=Probability} = a {parent=10e} {scope} = Solution {parent=a} 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 $$ \boxed{ \mathbb P(\text{first roll is }1\mid A) =\frac{1/6}{1/2}=\frac13}. $$ Solved by gpt-5.6-sol high. = b {parent=10e} {scope} = i {parent=b} {scope} = Solution {parent=i} Let $h(x)$ be the probability that the [simple symmetric random walk] starting at $x\in\{0,\ldots,n\}$ hits zero before $n$. The first-step recurrence is $$ h(x)=\frac12h(x-1)+\frac12h(x+1), \qquad h(0)=1,\quad h(n)=0. $$ The recurrence says that $h$ is affine, and the boundary values determine it: $$ \boxed{h(x)=\frac{n-x}{n}}. $$ Solved by gpt-5.6-sol high. = ii {parent=b} {scope} = Solution {parent=ii} Part (i) gives $\mathbb P(A)=2/3$. Since $\{X_1=0\}\subseteq A$ and has probability $1/2$, $$ \boxed{\mathbb P(X_1=0\mid A)=\frac{1/2}{2/3}=\frac34}. $$ 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 $A$. If $$ e_i=\mathbb E_i(T\mid A), $$ first-step analysis gives $$ e_1=1+\frac14e_2, \qquad e_2=1+e_1. $$ Solving, $$ \boxed{\mathbb E(T\mid A)=e_1=\frac53}. $$ Solved by gpt-5.6-sol high. = iii {parent=b} {scope} = Solution {parent=iii} For hitting zero before four, the success probabilities are $$ h_i=\frac{4-i}{4}. $$ Conditioning each transition on eventual success gives $$ \begin{array}{c|cc} \text{state}&\text{down}&\text{up}\\ \hline 1&2/3&1/3\\ 2&3/4&1/4\\ 3&1&0. \end{array} $$ Thus the conditional expected hitting times satisfy $$ e_1=1+\frac13e_2, \qquad e_2=1+\frac34e_1+\frac14e_3, \qquad e_3=1+e_2. $$ Solving this linear system gives $$ \boxed{\mathbb E(T\mid B)=e_1=\frac73}. $$ Solved by gpt-5.6-sol high. = 11D {parent=Paper 2} {scope} {title2=Probability} = Solution {parent=11D} Uniform area density in the unit disk is $1/\pi$. The Jacobian gives the joint density $$ f_{R,\Theta}(r,\theta) =\frac r\pi =\left(2r\right)\left(\frac1{2\pi}\right), \qquad 0\leq r\leq1,\quad0\leq\theta<2\pi. $$ Thus the coordinates are independent and the has $$ \boxed{ f_R(r)=2r\mathbf1_{[0,1]}(r), \qquad f_\Theta(\theta)=\frac1{2\pi}\mathbf1_{[0,2\pi)}(\theta)}. $$ For independent points $A,B$, the area of triangle $OAB$ is $$ \frac12R_AR_B|\sin(\Theta_A-\Theta_B)|. $$ Now $$ \mathbb ER=\int_0^1 2r^2\,dr=\frac23, \qquad \mathbb E|\sin(\Theta_A-\Theta_B)|=\frac2\pi. $$ Independence therefore gives $$ \boxed{\mathbb E\,\operatorname{Area}(OAB)=\frac4{9\pi}}. $$ Conditioned on $A,B$, the probability that $C$ lies inside $OAB$ is its area divided by the disk area $\pi$. Hence $$ \boxed{\mathbb P(C\in\triangle OAB)=\frac4{9\pi^2}}. $$ 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 , this has probability $1/4$. Therefore $$ \boxed{ \mathbb P(O,A,B,C\text{ form a convex quadrilateral}) =1-\frac14-\frac{4}{3\pi^2} =\frac34-\frac4{3\pi^2}}. $$ Solved by gpt-5.6-sol high. = 12F {parent=Paper 2} {scope} {title2=Probability} = i {parent=12f} {scope} = Solution {parent=i} The states that for a random variable $Y$ with finite variance and every $\delta>0$, $$ \mathbb P(|Y-\mathbb EY|\geq\delta) \leq\frac{\operatorname{var}Y}{\delta^2}. $$ Indeed, applied to the nonnegative variable $(Y-\mathbb EY)^2$ gives $$ \mathbb P((Y-\mathbb EY)^2\geq\delta^2) \leq\frac{\mathbb E[(Y-\mathbb EY)^2]}{\delta^2}. $$ The sum $S_n=X_1+\cdots+X_n$ has the , so $$ \boxed{ B_n(p)= \sum_{k=0}^n f(k/n)\binom nkp^k(1-p)^{n-k}}. $$ For fixed $n$ this is a finite sum of polynomial functions of $p$, hence is a polynomial. Solved by gpt-5.6-sol high. = ii {parent=12f} {scope} = Solution {parent=ii} The sample mean $S_n/n$ has expectation $p$ and variance $$ \frac{p(1-p)}n\leq\frac1{4n}. $$ The sum in the question is exactly $$ \mathbb P\left(\left|\frac{S_n}{n}-p\right|>\delta\right). $$ Chebyshev's inequality therefore gives, uniformly in $p$, $$ \boxed{ \sum_{k\in K_\delta}\binom nkp^k(1-p)^{n-k} \leq\frac1{4n\delta^2}}. $$ Solved by gpt-5.6-sol high. = iii {parent=12f} {scope} = Solution {parent=iii} Because $f$ is continuous on the compact interval $[0,1]$, it is uniformly continuous and bounded; write $|f|\leq M$. Given $\epsilon>0$, choose $\delta>0$ such that $$ |x-y|\leq\delta \quad\Longrightarrow\quad |f(x)-f(y)|\leq\epsilon. $$ Then $$ \begin{aligned} |B_n(p)-f(p)| &\leq \mathbb E\left|f(S_n/n)-f(p)\right|\\ &\leq \epsilon +2M\mathbb P\left(\left|S_n/n-p\right|>\delta\right)\\ &\leq \epsilon+\frac{M}{2n\delta^2}. \end{aligned} $$ The bound is independent of $p$. Taking the supremum, then $n\to\infty$, gives a limit superior at most $\epsilon$. Since $\epsilon$ is arbitrary, $$ \boxed{ \sup_{p\in[0,1]}|f(p)-B_n(p)|\longrightarrow0}. $$ This is the proof of the . Solved by gpt-5.6-sol high.