Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 2

Let $f(u)=au(1-u^2)$. A fixed point satisfies $u=f(u)$, soThe nonzero fixed points exist for $a<0$ or $a>1$. Local stability requires $|f^{\prime }(u_{\ast })|<1$, andAway from the boundary values:

At $a=-1,1,2$ a multiplier has modulus one and linear stability is inconclusive; at $a=0$, $0$ is strongly stable.

The chain rule gives $dF(x,y(x))/dx=F_x+F_y{y}^{\prime }$. Equality with $P+Q{y}^{\prime }$ for every $y(x)$ therefore requires $F_x=P$ and $F_y=Q$, which implies $P_y=Q_x$. Conversely, on the plane this compatibility condition makes $Pdx+Qdy$ an exact differential, so a potential $F$ exists.

Here $P=4x^3+3y$, $Q=2y+3x$, and $P_y=Q_x=3$. Integrating gives $F=x^4+3xy+y^2$, hence

By independence and the binomial theorem,Therefore $X+Y\sim \mathrm{Poisson}(\lambda +\mu )$.

For $0\le k\le n$,Thus

The Cauchy-Schwarz inequality is $|\mathbb{E}(XY){|}^2\le \mathbb{E}(X^2)\mathbb{E}(Y^2)$. Markov inequality says that for $X\ge 0$ and $a>0$, $\mathbb{P}(X\ge a)\le \mathbb{E}X/a$.

Jensen inequality states that for convex $\varphi$,Let $\mu =\mathbb{E}X$. A supporting line at $\mu$ gives $\varphi (x)\ge \varphi (\mu )+c(x-\mu )$. Taking expectations makes the linear term vanish.

With $\mu =\mathbb{E}X$,Every summand is nonnegative. If the sum is zero, every value of positive probability equals $\mu$, so $\mathbb{P}(X=\mu )=1$.

After multiplying by the inverse of the derivative matrix, the system isChoose eigenvectors $(1,1{)}^T$ and $(-1/2,1{)}^T$ of eigenvalues $1$ and $4$, and writeThenThe zero initial data giveThe $t{e}^{-4t}$ and $t{e}^{-t}$ terms are resonant responses. When $b=2$ or $c=2$, respectively, the corresponding resonant forcing vanishes and so does that secular factor.

Since ${\partial }_x=\alpha {\partial }_{\xi }+\beta {\partial }_{\eta }$ and ${\partial }_y={\partial }_{\xi }+{\partial }_{\eta }$,If $\alpha ,\beta$ are the distinct real roots of $a{s}^2+bs+c$, then $A=C=0$ and $B=(4ac-b^2)/a\ne 0$. Thus $v_{\xi \eta }=0$. For $s^2+3s+2=(s+1)(s+2)$, take $\xi =y-x$, $\eta =y-2x$, obtainingfor arbitrary twice differentiable functions $f,g$.

If $\alpha =-b/(2a)$ is a repeated root, then $A=0$ and both coefficients of $\beta$ in $B$ vanish, so $B=0$ for every $\beta$. Choose $\beta \ne \alpha$, giving $C\ne 0$ and hence $v_{\eta \eta }=0$. For $(s+1{)}^2$, take $\xi =y-x$, $\eta =y$, and obtain

After division by $x^2$, the equation is $y^{\prime \prime }+x^{-1}{y}^{\prime }-(1+{\alpha }^2{x}^{-2})y=0$. Every $x\ne 0$ is an ordinary point. The origin is singular, but it is a regular singular point because $xP(x)=1$ and $x^{2}Q(x)=-(x^2+{\alpha }^2)$ are analytic there. An ordinary point has analytic normalized coefficients $P,Q$; a singular point failing the displayed regularity test is irregular.

The Frobenius method ansatz $y=x^r{\sum }_{n\ge 0}{a}_n{x}^n$ gives the indicial equation $r^2-{\alpha }^2=0$, $a_1=0$, andFor nonintegral $\alpha$, two independent solutions are

For integral $\alpha$, put $m=|\alpha |$. In the $r=-m$ recurrence the denominator vanishes at $n=2m$, so that series fails or coincides in the exceptional $m=0$ case. Up to scale the single Frobenius series isFor $\alpha =1$, $y_1=x(1+x^2/8+\cdots )$, soIts integral contains both $s^{-2}$ and $\log s$; multiplying by $y_1$ leaves a pole and a logarithmic term. Thus the reduction-of-order solution is not a power series at zero.

Put $\theta =T-T_r$ and $u=U-T_r$. The term $-a\theta$ models oven cooling to the room, $Q$ is heater input, and $-bu=b\theta -bu$ makes the pizza relax toward the oven temperature. Both protocols supply unit total heat because $\int \delta (t)dt=1$ and the rectangular pulse has height $1/\tau$ and width $\tau$.

For the impulse, $\theta$ jumps by one at zero while $u$ remains continuous:for $a\ne b$; when $a=b$, $u_1=bt{e}^{-bt}$.

Define causal functionsThe rectangular-pulse solutions areFinally $T_i=T_r+{\theta }_i$, $U_i=T_r+u_i$. These formulas make $T_2,U_2$ continuous at $0,\tau$; only $T_1$ jumps at the delta impulse. As $\tau \to 0$, the difference quotients tend to $G_a^{\prime }={\theta }_1$ and $K^{\prime }=u_1$, because the rectangular pulse is an approximate identity.

Independence gives $\mathbb{E}E(Z)={\sum }_{z,y}F(y,z)\mathbb{P}(Y=y)\mathbb{P}(Z=z)=\mathbb{E}F(Y,Z)$. Moreover, $\mathbb{E}V(Z)=\mathbb{E}(F(Y,Z{)}^2)-\mathbb{E}(E(Z{)}^2)$, while $\mathrm{Var}E(Z)=\mathbb{E}(E(Z{)}^2)-(\mathbb{E}F(Y,Z){)}^2$. Adding proves the law of total variance

For one Bernoulli variable, direct expansion gives $\mathrm{Var}F(X_1)=p(1-p)(F(1)-F(0){)}^2$. Applying the law of total variance successively to the coordinates gives the variance tensorization

For any $x,z\in \{\pm 1\}$ there is exactly one $y=xz$, so $\mathbb{P}(X=x,Z=z)=1/4=\mathbb{P}(X=x)\mathbb{P}(Z=z)$. Similarly $(Y,Z)$ is independent, and $(X,Y)$ is independent by assumption. They are pairwise independent but not jointly independent, because $XYZ=1$ surely.

Using $\max \{X^2,Y^2\}=(X^2+Y^2+|X^2-Y^2|)/2$ and Cauchy-Schwarz inequality,

At most two of $X_1>X_2$, $X_2>X_3$, and $X_3>X_1$ can hold in any outcome. Taking expectations gives $a_{12}+a_{23}+a_{31}\le 2$, and hence $\min \{a_{12},a_{23},a_{31}\}\le 2/3$.

Let $q_n=\mathbb{P}(X_n=0)$ for a Galton-Watson process starting from one ancestor. Conditional on $X_1=k$, extinction by generation $n+1$ requires $k$ independent descendant processes to be extinct by generation $n$, soBecause zero is absorbing, $q_n\uparrow q$, the eventual extinction probability, and continuity gives $q=F(q)$. If $r\ge 0$ is any fixed point, monotonicity of $F$ and $q_0\le r$ give $q_n\le r$ inductively. Thus $q$ is the smallest nonnegative fixed point.

Here $F(s)=s/4+3s^3/4$. Every individual has at least one child, so $q=0$. The offspring mean is $F^{\prime }(1)=5/2$, hence

Represent $Y_n$ as a sum of $n$ independent $\mathrm{Poisson}(1)$ variables. The central limit theorem givesSince zero is a continuity point of the standard normal distribution,

For $x\ge m$, $X_n=x$ implies $M_n\ge m$. For $x<m$, reflect every step after the first visit to $m$. The reflection principle for simple symmetric random walk bijects such paths ending at $x$ with unrestricted paths ending at $2m-x$. Therefore

Summing part i and reindexing givesSubtracting the formula for $m+1$ yields

Using part ii and symmetry of the simple random walk,for $n\ge 1$. Thus $\mathbb{E}{M}_n^2<\mathbb{E}{X}_n^2=n$.