Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 3

Lagrange theorem for polynomial congruences says that a nonzero polynomial of degree $d$ over a field has at most $d$ distinct roots of a polynomial. It follows by repeatedly factoring out $x-a$ at a root. Over the finite field ${\mathbb{F}}_p$, the only solutions of $x^2=1$ are $1$ and $-1$: these are distinct because $p$ is odd, and the root bound excludes any others.

Every nonzero element has a unique multiplicative inverse. Pair each element of ${\mathbb{F}}_p^{\times }$ with its inverse. All pairs of distinct elements contribute $1$ to the product; the only unpaired elements are $1$ and $-1$. Thus Wilson theorem follows:If Lagrange's theorem is formulated for finite groups, it states that the order of a subgroup divides the order of the group; in particular every element of ${\mathbb{F}}_p^{\times }$ has order dividing $p-1$. The inverse-pairing proof above establishes the required congruence directly.

Put $m=(p-1)/2$ and let $O=1\cdot 3\cdot 5\cdots (p-2)$. The even residues $2,4,\dots ,p-1$ are precisely the negatives, in reverse order, of the odd residues $1,3,\dots ,p-2$. Hence their product is $(-1{)}^{m}O$ in the finite field ${\mathbb{F}}_p$. Applying Wilson theorem givesThe initial factor $1^2$ can be omitted, soFor $p=3$ the product is empty and equals $1$, consistently with the formula.

Let $K=\{z:|\mathrm{Re}z|+|\mathrm{Im}z|\le 1\}$. Choose a unit complex number $u$ normal to a supporting side of this convex set, withSuch a choice exists explicitly: take $u=(s+it)/\sqrt{2}$, where $s,t\in \{-1,1\}$ are the signs of the real and imaginary parts of $\zeta$; then $h=1/\sqrt{2}$ and the strict inequality is exactly the hypothesis. Set $c=-Ru$. Since $|z|\le 1$ on $K$,For sufficiently large $R$, the former is strictly smaller than the latter uniformly on $K$. Consequently $q={\max }_K|z-c|/|\zeta -c|<1$.

The geometric series now supplies actual approximating polynomials:Choose $N$ so the right-hand side is below $ϵ$. This proves the required uniform approximation without assuming a general approximation theorem.

Yes. Take exterior poles ${\zeta }_n=1+1/n$. By part (i), choose a polynomial $p_n$ withFor any fixed $z\in \Gamma$, $z\ne 1$, and therefore $(z-{\zeta }_n{)}^{-1}\to (z-1{)}^{-1}$. The triangle inequality provesThis is pointwise convergence. It is not uniform convergence on $\Gamma$: every polynomial is bounded on the closed diamond, whereas $(z-1{)}^{-1}$ is unbounded as $z\to 1$ along $\Gamma$.

A Kleinian group is a discrete subgroup of ${\mathrm{PSL}}_2(\mathbb{C})$, acting by Möbius transformations on the Riemann sphere, equivalently by orientation-preserving isometries of hyperbolic space ${\mathbb{H}}^3$.

Consider the classes ofThey generate a discrete subgroup because their matrix representatives have integer entries: a sequence of such matrices converging to $I$ or $-I$ is eventually constant. On the real projective line set $X=\{x:|x|>1\}\cup \{\infty \}$ and $Y=\{x:|x|<1\}$. For every nonzero integer $m$,The last inclusion follows from $|2mx+1|\ge 2|x|-1>|x|$ when $|x|>1$; also $B^m(\infty )=1/(2m)\in Y$. The ping-pong lemma therefore says that the subgroup is the free product $⟨A⟩\ast ⟨B⟩$. Both generators have infinite order, soa free group on two generators. The ping-pong inclusions ensure that no nonempty reduced word can act as the identity, which is the essential freeness argument.

For a binary linear code $C\le {\mathbb{F}}_2^n$, its rank is its dimension of a vector space $r$. Use the weight enumerator conventionwhere $w(c)$ is the Hamming weight. This convention is essential here: $W_C(1,0)$ counts words of weight $n$, rather than zero words. There are $2^r$ codewords, so $\boxed{W_C(1,1)=2^r}$.

There is only one binary word of weight $n$, namely $1=(1,\dots ,1)$. Thus $W_C(1,0)=1$ precisely when $1\in C$. If this holds, the linear map $c\mapsto c+1$ is a bijection of $C$ and replaces weight $w$ by $n-w$; hence $W_C(s,t)=W_C(t,s)$. Conversely this symmetry gives $W_C(1,0)=W_C(0,1)=1$, because the zero word is the unique word of weight zero.

The repetition code consists of $0,1$, givingThe single parity-check code consists of all words of even weight. Exactly $\left(\genfrac{}{}{0ex}{}{n}{j}\right)$ words have weight $j$, so the binomial theorem gives

A generalized linear model has independent responses whose distributions belong to an exponential family:Here the expected value and variance are ${\mu }_i=b^{\prime }({\theta }_i)$ and $\mathrm{Var}(Y_i)=a_i(\varphi )b^{\prime \prime }({\theta }_i)$. A specified one-to-one link function $g$ connects the mean to the linear predictor: $g({\mu }_i)=x_i^T\beta$. The canonical link function is $g({\mu }_i)={\theta }_i$.

In binomial regression, $Y_i\sim \mathrm{Bin}(m_i,p_i)$ independently, with known trial counts $m_i$ and probabilities related to the predictor. The logistic regression and probit regression specifications are respectivelyHere $\Phi$ is the standard normal distribution function. The binomial log-density contains $y_i\log (p_i/(1-p_i))+m_i\log (1-p_i)$, so its natural parameter is the log-odds: the logistic link is canonical. If the response mean is written as ${\mu }_i=m_i{p}_i$, the canonical link is $g_i({\mu }_i)=\log ({\mu }_i/(m_i-{\mu }_i))$.

The physical state space is the positively invariant set $X,Y,Z\ge 0$, $X+Y+Z=N$: adding the equations gives zero, and the vector field points inward at each coordinate boundary. At an equilibrium point, either $Y=0$, giving $(X,Y,Z)=(N,0,0)$, or $\beta X=b+r$. ThereforeA positive endemic equilibrium exists exactly when $N>N_c$.

For $N<N_c$, put $d=b+r-\beta N>0$. Since $X\le N$,The last limit follows either by evaluating the exponential integral (with a $t{e}^{-bt}$ term when $d=b$), or by splitting it into an early decaying part and a late part with uniformly small $Y$. Hence $X\to N$ for every physical initial condition. Although immune numbers need not decrease monotonically initially, both infected and immune numbers tend to zero.

At the endemic equilibrium, the Jacobian matrix of the $X,Y$ subsystem isA disturbance proportional to $e^{pt}$ therefore obeysBoth coefficients are positive. Real roots have negative sum and positive product; complex conjugate roots have real part $-(b+\beta Y_s)/2<0$. Thus the endemic equilibrium is asymptotically stable by linear stability analysis.

For the iteration of a map $f(x)=\lambda x(1-x^2)$, a fixed point satisfies $x[1-\lambda +\lambda x^2]=0$. If $0<\lambda <1$, the bracket is strictly positive, so the sole fixed point is $0$. Its multiplier of a periodic orbit of an iteration is $f^{\prime }(0)=\lambda$, whose modulus is below one. Thus $\boxed{x=0\text{ is locally asymptotically stable}}$ by fixed point stability for an iteration.

For $\lambda >1$, the multiplier at the origin exceeds one, so it is unstable. The other fixed points areUsing fixed point stability for an iteration, $|3-2\lambda |<1$ exactly when $1<\lambda <2$, and $|3-2\lambda |>1$ when $\lambda >2$. HenceThe endpoint $\lambda =2$ has multiplier $-1$ and requires nonlinear analysis, which is not part of the asserted open intervals.

On $(0,1)$ the map is positive. Its maximum occurs at $x=1/\sqrt{3}$ and equals $2\lambda /(3\sqrt{3})$. Thus if $\lambda <3\sqrt{3}/2$, then $f((0,1))\subset (0,1)$. Since $f$ is odd, also $f((-1,0))\subset (-1,0)$. Induction givesThe case $x_0=0$ is constant. These intervals are positively invariant sets for the discrete dynamical system.

Put $a=2/\sqrt{3}$. On $[-a,a]$, the extrema of $|x(1-x^2)|$ occur at $\pm 1/\sqrt{3}$ or the endpoints. The corresponding values are $2/(3\sqrt{3})$, including the endpoint value $a(a^2-1)$. ConsequentlyThus $[-a,a]$ is a trapping region andIt contains a neighbourhood of the origin. Sign changes are possible because $f(x)$ has opposite sign to $x$ for $1<|x|<a$.

The numbers $z_j$ are the nonzero zeros of a function $f$, repeated according to their multiplicity. The displayed Weierstrass factorization theorem formula uses genus-one Weierstrass elementary factors and a linear exponential factor; it applies under the corresponding growth assumptions, rather than to every arbitrary entire function.

Apply it to the entire function $g(z)=\sin (\pi z)/(\pi z)$, with its removable value $g(0)=1$. Its zeros are $\pm 1,\pm 2,\dots$, each simple, and $g^{\prime }(0)=0$ because $g$ is even. Pairing the factors at $n$ and $-n$ givessoThe product converges uniformly on compact subsets because $\sum n^{-2}<\infty$. Away from the integers its logarithmic derivative can be taken term by term: the resulting series also converges locally uniformly, its terms being $O(n^{-2})$. HenceThis is an identity of meromorphic functions, understood away from their poles.

Noether theorem associates a conserved quantity ${\sum }_i{p}_i\delta q_i$ to a continuous symmetry of the Lagrangian (with the usual boundary-term correction if the Lagrangian changes by a total derivative). The angles $\psi$ and $\varphi$ are cyclic coordinates, so for the heavy symmetric topTime-translation symmetry gives conserved energy, whose Hamiltonian expression isIt is quadratic in the momenta; $p_{\theta }=I_1\dot{\theta }$. Eliminating the cyclic velocities and putting $\mu =\cos \theta$, so ${\dot{\mu }}^2={\sin }^2\theta {\dot{\theta }}^2$, yieldsThe first term has degree three and the second degree two, displaying the cubic governing nutation.

As $T\downarrow 0$, the Fermi-Dirac distribution tends to a step function: $\bar{n}(p)=1$ for $ϵ(p)<{\mu }_0$, and $0$ for $ϵ(p)>{\mu }_0$. The value at equality is immaterial to the integrals. The Fermi momentum is defined by $ϵ(p_F)={\mu }_0$: the Pauli exclusion principle fills the lowest available states and forbids multiple occupation of a state with a fixed spin. Thus the sketch is a horizontal line at one up to $p_F$, followed by a jump to zero.

In the ultrarelativistic, $ϵ(p)\simeq cp$, givingEliminate $p_F$ to obtainThe supplied expression for $ϵ$ includes rest energy; in this limit its leading term is also the leading kinetic energy.

Electrical neutrality gives $n_e\simeq n_p$ at the accuracy of the proposed estimate. With $V\sim R^3$ and $n_p\sim M/(m_p{R}^3)$, the degenerate electron gas result givesThe gravitational potential energy is $E_{\mathrm{grav}}\sim -G{M}^2/R$. Both scale as $R^{-1}$, so in the strictly ultrarelativistic approximationFor sufficiently large mass the bracket is negative, and contraction lowers the energy without bound in this model: electron degeneracy pressure cannot halt collapse. Equating the scales gives the Chandrasekhar massBelow this scale, the strictly ultrarelativistic expression alone has no finite-radius energy minimum; including the transition to nonrelativistic electrons produces the stable white dwarf branch. Composition and the stellar density profile supply the numerical factors omitted by this scaling estimate.

For $\sigma =\mathrm{Re}s>1$, the Dirichlet series $\zeta (s)={\sum }_{n\ge 1}{n}^{-s}$ is absolutely convergent. A finite product over primes $p\le P$ can be expanded into absolutely convergent geometric series. By unique prime factorization,The omitted integers are all greater than $P$, so the difference from $\zeta (s)$ has modulus at most ${\sum }_{n>P}{n}^{-\sigma }\to 0$. ThusThis defines the infinite Euler product as the limit of finite products. Its absolute convergence also follows from ${\sum }_p{p}^{-\sigma }<\infty$, with $\log (1-p^{-s}{)}^{-1}=p^{-s}+O(p^{-2\sigma })$ for sufficiently large $p$.

Since $n^{-s}-(n+1{)}^{-s}=s{\int }_n^{n+1}{x}^{-s-1}dx$, the assumed identity becomesfor $\mathrm{Re}s>1$, where $\{x\}$ is the fractional part. ConsequentlyTo justify the analytic continuation, on any compact subset of this half-plane choose $ϵ>0$ with $\mathrm{Re}s\ge ϵ$. The integrand is bounded by $x^{-1-ϵ}$, an integrable function. Each derivative with respect to $s$ is bounded by a constant times $(1+\log x{)}^j{x}^{-1-ϵ}$, also integrable. Differentiation under the integral is valid, so the right-hand side is holomorphic throughout the half-plane, including $s=1$.

Let $E=\Vert f-p{\Vert }_{\infty }$. If $E=0$, minimality is immediate. Otherwise suppose a polynomial $q$ of degree at most $n$ has strictly smaller error. At each alternation point,Thus $q-p$ changes sign between every pair of consecutive $x_j$. By the intermediate value theorem it has at least $n+1$ distinct roots of a polynomial, contradicting its degree bound. ThereforeThis is the sufficiency direction of the Chebyshev alternation theorem.

Let $E_{\ast }={\inf }_{\mathrm{deg}q\le n}\Vert f-q{\Vert }_{\infty }$, and choose a minimizing sequence $p_m$ with $\Vert f-p_m{\Vert }_{\infty }\to E_{\ast }$. Since $E_{\ast }\le \Vert f{\Vert }_{\infty }$, discarding finitely many terms ensures $\Vert p_m{\Vert }_{\infty }\le 2\Vert f{\Vert }_{\infty }+1$.

Fix $n+1$ distinct nodes $a_0,\dots ,a_n$. The Lagrange interpolation formula expressesThe vector of node values is bounded in ${\mathbb{R}}^{n+1}$, so the Bolzano-Weierstrass theorem supplies a subsequence converging to $(v_0,\dots ,v_n)$. Put $p=\sum v_j{\ell }_j$, of degree at most $n$. Because the ${\ell }_j$ are fixed bounded functions,along that subsequence. The reverse triangle inequality implies $\Vert f-p{\Vert }_{\infty }=E_{\ast }$. Thus a best uniform approximation exists, with no appeal to an unproved abstract existence result.

Here ${\mathcal{P}}_n$ is the vector space of real polynomials of degree at most $n$.

Enumerate the finite set increasingly, $y_1<\cdots <y_{n+2}$, so alternating signs refer to spatial order. This is the enumeration needed for the interpolation construction below. The Lagrange cardinal polynomials satisfy $r_j(y_k)={\delta }_{jk}$ and have degree $n+1$.

If the printed indexing were interpreted as an arbitrary fixed permutation, the assertion in (a) would need this qualification. For example with $n=2$, $(y_1,y_2,y_3,y_4)=(0,1/3,1,2/3)$ gives $r(x)=-1+9x(1-x)$ of degree two; for $f=0$ every $\lambda$ works. Ordering the nodes resolves the issue and yields the requested minimizer for every set $Y$.

The coefficient of $x^{n+1}$ in $r_j$ isWith the increasing enumeration, its sign is $(-1{)}^{n+2-j}$. Therefore all terms $(-1{)}^j{c}_j$ have the same sign $(-1{)}^{n+2}$, and $R:={\sum }_j(-1{)}^j{c}_j\ne 0$. The leading coefficient of $t$ is $T:={\sum }_{j}f(y_j)c_j$. Canceling the coefficient of $x^{n+1}$ gives the unique valueBoth interpolants have degree at most $n+1$, so this single cancellation is equivalent to $\mathrm{deg}(t-\lambda r)\le n$.

Set $p=t-\lambda r$ using part (a). At every node,If $\lambda =0$, the error vanishes. Otherwise any degree-at-most-$n$ polynomial $q$ with strictly smaller error would make $q-p$ have alternating nonzero signs at the increasingly ordered nodes (reversed globally if $\lambda <0$). It would then have at least $n+1$ distinct roots by the intermediate value theorem, impossible. Hence

The reaction–diffusion system for amoebae contains local reproduction $\alpha n_0^{2}n$ and density-dependent loss $-\alpha n^3$, giving carrying density $n_0$. The term ${\nabla }^{2}n$ describes diffusion with coefficient one. The chemotaxis flux is $J_{\mathrm{chem}}=\chi (n)n\nabla a$, so its contribution to the conservation equation is $-\nabla \cdot J_{\mathrm{chem}}$. Attraction towards greater chemical concentration requires $\chi (n)>0$ at the relevant positive densities; negative values would mean repulsion, and zero removes the chemotactic response.

In the chemical equation, $\beta n$ is secretion proportional to amoeba density, $-\gamma a$ is first-order chemical degradation, and $D{\nabla }^{2}a$ is chemical diffusion. The cell reaction term permits births and deaths, so the total number of amoebae is not conserved.

For a spatially homogeneous positive steady state, $\alpha n(n_0^2-n^2)=0$ and $\beta n-\gamma a=0$, givingWrite $n=n_0+\eta$, $a=a_0+\xi$, and define $A=2\alpha n_0^2>0$. Homogeneous linearization givesThe eigenvalues are $-A,-\gamma$, both negative. Hence this steady state is stable to homogeneous disturbances, including the repeated-eigenvalue case, where a possible $t{e}^{-At}$ factor still decays.

The background has $\nabla a_0=0$, so the linear chemotaxis term is $-{\chi }_0{n}_0{\nabla }^2\xi$; derivatives of $\chi$ multiply products of disturbances and do not enter at first order. For a normal mode with ${\nabla }^2=-k^2$ and time dependence $e^{pt}$, put $q=k^2$. Thenand the characteristic polynomial isIts linear coefficient is positive for every $q\ge 0$. Thus instability occurs exactly when its constant term $Q(q)$ is negative: the roots are then real with opposite signs. The positive quadratic $Q$ is negative somewhere on $q>0$ precisely whenThereforeAt equality the neutral wavenumber is $q_c=\sqrt{A\gamma /D}$ and $p=0$. Above threshold the unstable band is $q_{-}<k^2<q_{+}$, where