Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 2

Suppose first that $M$ is finitely generated. Its quotient $M/N$ is generated by the images of any finite generating set of $M$. A principal ideal domain is Noetherian, so every submodule of the finitely generated $R$-module $M$ is finitely generated; in particular, $N$ is finitely generated.

Conversely, supposeand $M/N$ is generated by the cosets ${\overline{m}}_1,\dots ,{\overline{m}}_s$. For any $m\in M$, its coset is an $R$-linear combination of the ${\overline{m}}_j$, soExpressing this remainder in terms of the $n_i$ shows thatThus $M$ is finitely generated if and only if both the submodule $N$ and the quotient module $M/N$ are finitely generated.

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Let $\Vert K{\Vert }_{\infty }={\max }_{[0,1{]}^2}|K|$. For $f,g\in C([0,1])$,Taking the supremum over $x$ proves that the integral operator $T$ is Lipschitz continuous, hence continuous, in the uniform norm.

It remains to see that $Tf$ is continuous. Since $K$ is uniformly continuous on the compact square and $f$ is bounded,as $x^{\prime }\to x$. Therefore $T:C([0,1])\to C([0,1])$ is well defined and continuous.

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Yes. Fubini's theorem and the triangle inequality giveThus $T$ is Lipschitz continuous with respect to $d_1$ as well.

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Multiplication by the integrating factor $e^{2x}$ givesHence the Sturm-Liouville problem is

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Set $y=e^{-x}u$. Thenso the eigenvalue equation becomesThe Dirichlet eigenvalues and corresponding eigenfunctions are thereforeThey form an infinite discrete increasing set. Their weighted inner products satisfyThus the Sturm-Liouville eigenfunction expansion coefficient is

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Gauss's law in electrostatics statesCylindrical symmetry and a coaxial Gaussian cylinder giveThe field vanishes inside each perfect conductor, and outside the cable because the total enclosed charge per unit length is zero.

Choose the outer conductor's potential to be zero. Since $E=-\nabla V$,The capacitance per unit length is thereforeThe electrostatic energy per unit length isSubstitution of $C$ verifies $U=Q^2/(2C)$.

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If $z(t)$ is the water depth and $A(z)=\pi r(z{)}^2$ is the horizontal cross-sectional area, Torricelli's law gives the volume flux through the hole asConservation of volume givesFor a prescribed constant fall rate $\dot{z}=-\alpha$, $\alpha >0$, this requiresConsequentlyso the container radius must be proportional to the fourth root of height.

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Since $z(t)=h_I-\alpha t$, the free-surface area isuntil the emptying time $t_e=h_I/\alpha$. Equivalently,

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Under the null hypothesis of independence between treatment and outcome, the expected counts in each treatment row are one half of the column totals:The Pearson chi-squared test of independence statistic isThe number of degrees of freedom isAt the 5% level the critical value is $7.81$. Since $8.30>7.81$, we reject the null hypothesis and find statistically significant evidence that the drug and placebo have different effects.

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Introduce slack variables $s_1,s_2,s_3$. In the initial simplex algorithm dictionary, $x_2$ has the largest positive objective coefficient. The ratio test givesso $x_2$ enters and $s_3$ leaves. Solving the third constraint for $x_2$ givesThe objective becomesEvery reduced cost is now nonpositive, so the simplex optimality criterion givesThe other two slacks both equal one.

The dual linear program isThe vectoris dual feasible and has objective value $12$. By weak duality, it and the displayed primal point are optimal; they also satisfy complementary slackness.

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A direct determinant calculation givesMoreover,has rank one, so its kernel has dimension two, while $(A-I{)}^2=0$ and $A\ne I$. Hence the minimal polynomial isThere are two Jordan blocks, because the eigenspace has dimension two, and the largest has size two, because the minimal polynomial has exponent two. Thus

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Choose the least $k\ge 1$ such thatThenis nonzero by minimality, andTherefore $fw=\alpha w$, so $w$ is the required nonzero eigenvector.

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We use induction on $d$. SupposeApplying $f-{\alpha }_{d}I$ givesBy induction, every $c_i({\alpha }_i-{\alpha }_d)=0$ for $i<d$. The eigenvalues are distinct, so $c_i=0$ for $i<d$, and the original relation then gives $c_d=0$. Thus eigenvectors with distinct eigenvalues are linearly independent.

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If $\alpha \ne \beta$, the operator $f-\beta I$ is invertible on $W_{\alpha ,n}$. Indeed, there it equalswhose inverse is the finite geometric series

Now use induction on $d$. If ${\sum }_{i=1}^d{v}_i=0$, apply $(f-{\alpha }_{d}I{)}^n$. The $v_d$ term vanishes, while every transformed vectoris nonzero by the invertibility just proved and still lies in $W_{{\alpha }_i,n}$. The induction hypothesis rules out the resulting relation. Allowing scalar coefficients, and omitting zero terms, gives the same argument. Hence the generalized eigenspaces for distinct eigenvalues form a direct sum, so $v_1,\dots ,v_d$ are linearly independent.

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A subset $S\subseteq M$ freely generates $M$ when every $m\in M$ has a unique expressionwith $r_s\in R$ and only finitely many nonzero coefficients. Equivalently, $S$ is a basis of a module and $M$ is the free module on $S$.

If $S$ freely generates $M$ and $f:S\to N$ is any function, defineUnique coordinates make $\varphi$ well defined; it is an $R$-module homomorphism and is the only extension of $f$. Conversely, apply the proposed universal property to the free module $F=R^{(S)}$ and the inclusion of $S$ into $M$. It gives maps $F\to M$ and $M\to F$ extending the corresponding functions on $S$. Uniqueness makes both composites identity maps, so $M\cong F$ and $S$ freely generates $M$. This is the universal property of a free module.

Now let $T$ generate the free module $M$ and $|T|=m$. Choose a maximal ideal $\mathfrak{m}$ of the nontrivial ring $R$. Then $k=R/\mathfrak{m}$ is a field, andis a $k$-vector space. The images of a basis $S$ of $M$ form a vector-space basis, while the images of $T$ span it. ThereforeIn particular $S$ is finite. Applying this result in both directions to two finite bases shows that they have equal cardinality, the rank of a free module $\mathrm{rk}M$.

A Euclidean domain is a principal ideal domain. By the submodule theorem for free modules over a principal ideal domain, every submodule $N$ of the finite-rank free module $M$ is free. Since a basis of $M$ generates it, the preceding inequality applied in the standard proof gives

The primary decomposition theorem for finitely generated modules over a principal ideal domain states thatwhere $p$ ranges over finitely many nonassociate irreducibles and the positive exponents are uniquely determined up to order. Equivalently, the torsion part decomposes into its primary cyclic summands.

Let $H$ be a finite subgroup of the multiplicative group of a field. It is a finite abelian group, hence the theorem over $\mathbb{Z}$ gives an invariant-factor decompositionIts exponent is $d_r$, so every element of $H$ is a root of $X^{d_r}-1$. A degree-$d_r$ polynomial over a field has at most $d_r$ roots, whenceBut $d_r\le |H|$, and equality forces all earlier factors to be trivial. Thus

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Let $C\subset \mathbb{R}$ be compact and let $M=\Vert f{\Vert }_{\infty }$. Since every $k_n$ is supported in $[-R,R]$, the function $f$ is uniformly continuous on the compact setGiven $\epsilon >0$, choose $\delta >0$ so thatwhenever $x\in C$ and $|t|<\delta$. Since $\int k_n=1$ and $k_n\ge 0$,Property 3 makes the final term smaller than $\epsilon /2$ for all sufficiently large $n$, uniformly in $x\in C$. Hence $f_n\to f$ uniformly on every compact set. The sequence $(k_n)$ is an approximate identity.

For the second part, extend $g$ to a continuous function $\tilde{g}$ on $\mathbb{R}$ by setting it equal to zero outside $[0,1]$; continuity at the endpoints uses $g(0)=g(1)=0$. DefineThese nonnegative kernels have integral one. For every $\delta >0$, their mass outside $[-\delta ,\delta ]$ tends to zero exponentially relative to the mass near zero, so they satisfy property 3.

For $x\in [0,1]$, the convolution isBecause $|x-y|\le 1$ on the square $[0,1{]}^2$, no cutoff remains in this formula. Expanding the $n$th power shows that $p_n$ is a polynomial in $x$. The first part, applied to the compact interval $[0,1]$, givesThis proves the Weierstrass approximation theorem for functions with the stated endpoint values.

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The Poincare half-plane model iswith the orientation inherited from the complex plane. Forone hasThese identities show directly that ${\gamma }^{\ast }d{s}^2=d{s}^2$. The map is holomorphic with nonzero derivative, so it preserves orientation. The matrices $I$ and $-I$ induce the same map, giving the action of PSL2(R).

Conversely, let $F$ be an orientation-preserving isometry. The PSL2(R) action is transitive on $\mathbb{H}$, so compose $F$ with an element taking $F(i)$ back to $i$. The resulting isometry fixes $i$ and acts on $T_i\mathbb{H}$ by an orientation-preserving orthogonal map, hence a rotation. The stabilizer of $i$ in PSL2(R),realizes every such tangent rotation. An isometry is determined by its value and differential at one point because it preserves geodesics and the exponential map. Thus the composed isometry belongs to PSL2(R), and so does $F$.

The map $\tau (z)=-\overline{z}$ is an orientation-reversing isometry. Composing any orientation-reversing isometry with $\tau$ gives an orientation-preserving one, so

A hyperbolic line is a vertical Euclidean line or a semicircle orthogonal to the real axis. Its hyperbolic reflection ${\sigma }_{\ell }$ is the unique orientation-reversing isometry that fixes every point of $\ell$. If $\ell ,{\ell }^{\prime }$ meet at angle $\theta$, thenis the hyperbolic rotation about $A$ through angle $2\theta$. The generators satisfyIf $\theta /\pi =p/q$ in lowest terms, then $\rho$ has order $q$ and the generated group is the finite dihedral group of order $2q$. If $\theta /\pi$ is irrational, $\rho$ has infinite order and the generated group is the infinite dihedral group. In the degenerate case $\ell ={\ell }^{\prime }$, the group has order two.

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Write the Taylor series of the entire function asThe Cauchy estimate on the circle $|z|=R$ givesIf $k>n/2$, letting $R\to \infty$ gives $c_k=0$. Since $n$ is odd,

For the second question, suppose such an $f$ existed. It never vanishes, so $g=1/f$ is analytic on $\mathbb{C}\setminus \{0\}$ andThe Riemann removable singularity theorem extends $g$ analytically across zero with $g(0)=0$. Applying the result just proved with $n=1$, $a=1$, and $b=0$ makes $g$ a polynomial of degree at most zero. It must then be identically zero, contradicting $g=1/f$ away from zero. Therefore no such function exists.

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Liouville theorem states that every bounded entire function is constant.

Because $\mathbb{C}$ is simply connected, the harmonic function $u$ has a global harmonic conjugate $v$, sois entire. Positivity givesLiouville's theorem makes $e^{-F}$ constant. Differentiating this nonzero constant gives $F^{\prime }=0$, so $F$, and hence $u$, is constant.

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SetThen $g$ is entire and never takes a real value. The continuous function $\mathrm{Im}g$ is nowhere zero; because $\mathbb{C}$ is connected, it has one sign everywhere. Thus either $\mathrm{Im}g$ or $-\mathrm{Im}g$ is a positive harmonic function. Part (i) makes it constant. The Cauchy-Riemann equations then force the real part of $g$ to be constant as well. Hence $g$ and therefore $f$ are constant.

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Take a variation $x+\epsilon \xi$ with $\xi (a)=\xi (b)=0$. Expanding the action givesAn integration by parts and the fixed endpoint conditions give the first variationThe fundamental lemma of the calculus of variations therefore yields the Euler-Lagrange equationand the second variation is

Linearizing the equation of motion about $x$ givesso the Jacobi equation isWhen $u$ has no zero on $[a,b]$,The total derivative integrates to zero because $\xi (a)=\xi (b)=0$, hence

For the simple harmonic oscillator, the Jacobi equation is $\ddot{u}+{\omega }^{2}u=0$. Put $t_0=(a+b)/2$ and chooseIf $b-a<\pi /\omega$, then $|\omega (t-t_0)|<\pi /2$ throughout $[a,b]$, so $u$ is positive there. The preceding square identity proves that the classical path is a local minimum of the action whenever the elapsed time is less than half an oscillation period.

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Direct evaluation of the integral givesThe continuous value at $k=0$ is zero, in agreement with the vanishing integral of the odd function $f$. Thus

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Because $e^{-\lambda |k|}$ is an even function, Fourier inversion reduces to a cosine integral:Therefore

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Extend the boundary data oddly to the whole real axis. The extended function is precisely the function $f$ from part (a). Taking the Fourier transform in $x$, the Laplace equation becomesDecay as $y\to \infty$ selectsPart (b), together with the convolution theorem, says that the inverse transform is convolution with the Poisson kernelFor $x,y>0$ this givesThe odd extension enforces $u(0,y)=0$, while the Poisson integral takes the prescribed boundary values at every continuity point and decays at infinity.

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By the superposition principle, interchange $x$ and $y$ in the solution from part (c) and add the two solutions:The first term supplies the required data on the positive $x$-axis and vanishes on the positive $y$-axis; the second does the reverse. Explicitly,It is harmonic in the quarter-plane, has the stated boundary values, and decays in both unbounded directions.

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For a one-dimensional wavefunction,The probability continuity equation isFor a stationary state $\psi (x,t)=\varphi (x)e^{-iEt/\hslash }$, the probability density is independent of time, so ${\partial }_{x}j=0$ and the probability current is constant in space.

For the plane waveone obtainsThis is a momentum eigenstate with momentum $p=\hslash k$ and, when it satisfies the free Schrodinger equation, energy $E={\hslash }^2{k}^2/(2m)$. Its constant density and current describe a spatially uniform beam carrying probability in the sign of $k$. It is not a normalizable wavefunction, so it represents an idealized state rather than a localized particle.

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PutThe Time-independent Schrodinger equation has the formsThe first region contains the incident and reflected waves, while the final region contains only the transmitted wave.

Continuity of $\psi$ and ${\psi }^{\prime }$ at both edges of the finite square well gives the standard transmission coefficientWhen $V_0=3E$, one has $K=2k$, and thereforeHence the transmission probability is

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WriteUsing the Taylor seriesand retaining terms through second order in $|\mathbf{y}|/|\mathbf{x}|$ givesThis is the beginning of the multipole expansion.

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The method of images replaces the earthed plane by an image charge $-q$ at $(-a,0,0)$. The two Coulomb potentials cancel at $x=0$, so uniqueness for the Dirichlet problem makes the resulting field the physical field in $x>0$. With $\mathbf{r}=(x,y,z)$ and ${\mathbf{e}}_x=(1,0,0)$,and