Past exam of the mathematics course of the University of Cambridge / 2008 / ii

For a real number $\alpha$, set ${\alpha }_0=\alpha$ and $a_j=\lfloor {\alpha }_j\rfloor$. If ${\alpha }_j-a_j\ne 0$, define ${\alpha }_{j+1}=({\alpha }_j-a_j{)}^{-1}$. The resulting simple continued fraction is $[a_0;a_1,a_2,\dots ]$, where $a_0$ is an integer and subsequent $a_j$ are positive integers. It terminates when a complete quotient is integral; otherwise its convergents tend to $\alpha$.

For the continued fraction of the square root of nineteen, the complete quotients after $a_0=4$ areTheir integer parts are $2,1,3,1,2,8$. Taking the reciprocal after subtracting $8$ from the last quotient returns the first one, so

The nonzero leading coefficients and distinct degrees make $P_0,\dots ,P_{n-1}$ a basis of the polynomials of degree at most $n-1$. Consequently the given orthogonality implies ${\int }_a^{b}f(x)Q(x)dx=0$ for every such polynomial $Q$.

If $f$ has infinitely many distinct zeros in $(a,b)$, there is nothing to prove. Otherwise suppose it has fewer than $n$. Among those zeros, list the sign-changing ones as $t_1<\cdots <t_m$, with $m<n$, and put $Q(x)={\prod }_{j=1}^m(x-t_j)$, using $Q=1$ when $m=0$. On each interval between zeros, $f$ has a fixed sign by continuity. Crossing a sign-changing zero flips both $f$ and $Q$; crossing any other zero flips neither sign. Thus $fQ$ has one constant sign wherever nonzero. Since $f$ is not identically zero, continuity gives an open interval on which $fQ$ is strictly of that sign. Hence its integral is nonzero, contrary to orthogonality to $Q$. Therefore $f$ has at least $n$ distinct zeros in $(a,b)$. This is the reusable orthogonality forces many interior zeros argument.

Let $T$ be an isometry, and set $S(x)=T(x)-T(0)$. Then $S(0)=0$, and preservation of distances gives $|S(x)|=|x|$. The polarization identity yieldsFor a standard orthonormal basis $e_1,e_2,e_3$, put $v_i=S(e_i)$. These three vectors are an orthonormal basis. Since $⟨S(x),v_i⟩=x_i$, we have $S(x)={\sum }_i{x}_i{v}_i=Qx$ for an orthogonal matrix $Q$. Thusan affine map.

For a finite isometry group $G$, take $c=|G{|}^{-1}{\sum }_{g\in G}g(0)$. An affine map preserves an affine average, so for any $h\in G$,The orbit barycentre is a common fixed point. This proves the fixed point of a finite Euclidean isometry group result.

If the possible values have probabilities $p_1,\dots ,p_N$, allowing zeros, the Shannon entropy in bits is $H(X)=-{\sum }_i{p}_i{\log }_2{p}_i$, with $0{\log }_{2}0=0$. Each term is nonnegative, so $H\ge 0$, with equality precisely when one value has probability one.

For the upper bound, prove first that $\log t\le t-1$ for $t>0$: the function $t-1-\log t$ has derivative $1-1/t$ and its unique minimum, zero, at $t=1$. Let $m$ be the number of positive probabilities. Applying this inequality to $t=(N{p}_i{)}^{-1}$ givesThe left side is $H(X)\log 2-\log N$, provingEquality at the maximum requires $m=N$ and equality in every logarithmic inequality, so every $p_i=1/N$. Conversely the uniform distribution attains it. For $N=1$ the maximum and minimum coincide at zero.

The inverse logit link is ${\mu }_i(\beta )=e^{{\beta }^T{x}_i}/(1+e^{{\beta }^T{x}_i})$. independence therefore gives the Likelihood function and log-likelihoodDifferentiation yields the score ${\nabla }_{\beta }\ell ={\sum }_i{x}_i(y_i-{\mu }_i)$. The matrix whose rows are $x_i^T$ has dimension $n\times p$, not the PDF's printed $p\times n$. With this necessary dimension correction, any finite maximum Likelihood function estimate satisfiesThe Hessian matrix is $-X^{T}WX$, with $W_{ii}={\mu }_i(1-{\mu }_i)$, so the log-likelihood is concave. These score equations apply to a finite interior estimate; complete or quasi-complete separation can instead make its supremum occur only as coefficients diverge.

The deviance is twice the difference between maximized saturated and fitted log-likelihoods. A saturated Bernoulli distribution has fitted means $y_i\in \{0,1\}$ and log-likelihood zero, interpreted by limits. Hence the binomial deviance here isZero-multiplier logarithms are interpreted as zero.

Continue in the bistable regime $k>2$. At $s=0$ the reaction function factors as $f(g,0)=g(kg-1-g^2)/(1+g^2)$. Besides $g=0$, its positive equilibria areAt zero, $f_g=-1$, so it is stable. At a positive equilibrium, use $k=(1+g^2)/g$ to obtain $f_g=(1-g^2)/(1+g^2)$. Therefore $g_1$ is unstable and $g_2$ stable. The sign of $f$ is negative on $(0,g_1)$, positive on $(g_1,g_2)$ and negative beyond $g_2$; these signs determine the requested reaction-rate sketch.

The stationary points of $f(g,0)$ obey $(1+g^2{)}^2=2kg$. For $k>2$ its smaller solution $g_c<1$ is the local minimum, and its larger solution lies above one and is the local maximum. Since increasing $s$ translates the whole graph upward, the low and middle equilibria collide whenFor $s>s_c$, both stationary values of $f$ are positive, while $f(g,s)\to -\infty$ as $g\to \infty$; there is exactly one nonnegative equilibrium, on the descending high branch. At equality a double low equilibrium remains, so the uniqueness assertion requires strict exceedance.

For completeness, if the later positive-input request is considered for arbitrary $k>0$, the derivative is $f_g=2kg/(1+g^2{)}^2-1$. Its maximum is $9k/(8\sqrt{3})-1$, attained at $g=1/\sqrt{3}$. For $k\le 8\sqrt{3}/9$, the reaction graph is decreasing and every $s>0$ already has a unique steady state, so take $s_c=0$. For $k>8\sqrt{3}/9$, the smaller positive stationary point gives the same $s_c=g_c(1-g_c^2)/2$; above it the graph crosses zero only on its final descending branch. Persistent robust switching after input is removed uses the previously assumed bistable regime $k>2$.

Starting near zero, raise $s$ past $s_c$ and keep it there long enough for $g$ to pass the unstable threshold and approach the high branch. Decrease $s$ back to zero slowly, or after ensuring $g>g_1$. The trajectory then stays in the basin of the high stable equilibrium and ends at the displayed $g_2$. The persistent switched state is $g_2=(k+\sqrt{k^2-4})/2$. This is the saturating autocatalytic switch; the conclusion presupposes $k>2$ and enough forcing duration, not merely an arbitrarily short pulse of large amplitude.

For $k=0$, the first integralhas $\dot{E}=0$. The equilibrium points are $(0,0)$ and $(1,0)$. The linearization has eigenvalues $\pm i$ at the origin and $\pm 1$ at $(1,0)$. The origin is a center equilibrium: the sufficiently small positive level sets of $E$ are closed curves, since its Hessian matrix there is positive-definite. The other equilibrium is a saddle equilibrium.

At the saddle, $E=1/6$, and the separatrix satisfiesIts two branches with $-1/2\le x\le 1$ form a homoclinic orbit through the saddle; its remaining branches extend to $x>1$. Inside the loop, the closed trajectories run clockwise: above the horizontal axis $\dot{x}=y>0$. The saddle directions have slopes $+1$ and $-1$. Outside the loop there are unbounded trajectories. The left panel shows the closed orbits and all saddle branches.

For general $k$, the same energy gives $\dot{E}=-k{y}^2$. For $k=0.1$ the eigenvalues at the origin are $-0.05\pm i\sqrt{3.99}/2$, so it is a stable focus. At $(1,0)$ they are $(-0.1\pm \sqrt{4.01})/2$, so it remains a saddle equilibrium. The vector field has constant divergence $-0.1$: there are no periodic orbits, by the planar divergence criterion. Nor can the former homoclinic orbit survive, since $E$ strictly decreases along every nonconstant orbit except at isolated crossings of $y=0$.

The stable manifold of the saddle separates escape to positive $x$ from convergence to the origin. Its two branches approach the saddle in forward time with tangent slope $(-0.1-\sqrt{4.01})/2$. The left unstable branch spirals into the origin, whereas the right unstable branch escapes. The shaded component on the origin's side of the stable manifold in the right panel is the basin of attraction. The boundary itself is excluded: its trajectories converge to the saddle. This boundary is the damped stable manifold, rather than the conservative energy contour. energy decrease prevents bounded recurrent motion, and every bounded orbit off the saddle's stable manifold consequently converges to the origin. The plotted manifold was obtained by integrating backwards from the saddle's stable eigendirection.

Near zero the magnitude of the integrand is of order $t^{\mathrm{Re}z-1}$, and at infinity it is of order $t^{\mathrm{Re}z-3}$. Thus the improper integral converges exactly in the stripAt a boundary of this strip the logarithmic-variable oscillation has no limiting primitive, so conditional convergence does not enlarge it.

Use a keyhole contour about the positive real axis for $w^{z-1}/(1+w{)}^2$, taking $0<\arg w<2\pi$. The circular arcs vanish in the indicated strip. The upper and lower banks contribute $(1-e^{2\pi iz})F(z)$. The only enclosed pole is the double pole at $w=-1$, whose residue is $(z-1)e^{i\pi z}$. The residue theorem therefore givesAt $z=1$ the quotient has a removable singularity, with limit one, agreeing with ${\int }_0^{\infty }(1+t{)}^{-2}dt=1$.

For an arbitrary smooth variation $q_i\mapsto q_i+\epsilon {\eta }_i$ vanishing at both endpoints,The endpoint term in integration by parts is zero. stationarity for every variation and the fundamental lemma of the calculus of variations imply the Euler-Lagrange equations $d(L_{{\dot{q}}_i})/dt=L_{q_i}$. Least action here means stationary action, not necessarily a global minimum. The conjugate momentum is $p_i=L_{{\dot{q}}_i}$; a cyclic coordinate, $L_{q_i}=0$, makes $p_i$ constant.

For the symmetric top, $\psi$ and $\varphi$ are cyclic. Thusare constant, and consequently so is ${\omega }_3$. Conservation of energy from time-translation invariance also supplies the conserved energyThese give the requested additional conserved quantities $p_{\varphi }$ and $E$. The $\theta$ equation isFor constant $\theta$ and constant precession rate $\dot{\varphi }$, it reduces toConversely, choose constants satisfying this balance and set $\dot{\psi }={\omega }_3-\dot{\varphi }\cos \theta$. The cyclic equations and $\theta$ equation are then all satisfied, so this constructs uniform precession wherever the Euler-angle chart is regular.