Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 2

The Eisenstein criterion says that a primitive polynomialis irreducible if there is a prime $p$ such thatBy the Gauss lemma for polynomials, irreducibility over $\mathbb{Q}$ and over $\mathbb{Z}$ agree for primitive polynomials.

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LetIf $n=ab$ with $a,b>1$, thenso $P_n$ is reducible.

Conversely, let $n=p$ be prime. Translation by one is an automorphism of $\mathbb{Z}[X]$, andIts leading coefficient is one, every lower coefficient is divisible by $p$, and its constant coefficient is $p$, which is not divisible by $p^2$. The Eisenstein criterion proves that $P_p(X+1)$ is irreducible, hence so is $P_p(X)$. Therefore the geometric-sum irreducibility criterion is

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Regardas a monic quadratic in $X$ over $\mathbb{Q}[Y]$. By the Gauss lemma for polynomials, reducibility in $\mathbb{Q}[X,Y]$ would imply reducibility in $\mathbb{Q}(Y)[X]$, so $1-Y^2$ would be a square in $\mathbb{Q}(Y)$. This is impossible: its zero at $Y=1$ has odd order one, whereas every zero or pole of a square in a rational-function field has even order. Hence

The same odd-valuation obstruction to a rational-function square works over every field of characteristic different from two, since the zeros $Y=1$ and $Y=-1$ are then distinct. It does not work over every field. In characteristic two,so the polynomial is reducible.

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A map $h:X\to X$ is a contraction if there is a constant $q<1$ such thatfor every $x,y\in X$.

The contraction mapping theorem states that a contraction of a nonempty complete metric space has a unique fixed point, and that the iterates from every starting point converge to it. To prove this, choose $x_0\in X$ and put $x_{n+1}=h(x_n)$. Thenso, for $m>n$,Thus $(x_n)$ is Cauchy and converges, by completeness, to some $x_{\ast }$. A contraction is continuous, soIf $y_{\ast }$ is another fixed point, thenforcing $x_{\ast }=y_{\ast }$.

For the Newton mapone has $g(r)=r$ andOn the given neighbourhood,Since $f(r)=0$, choose a closed interval $U^{\prime }$ centred at $r$ and contained in $U$ so small that $M|f(x)|/{\delta }^2\le 1/2$ there. Thenso $g(U^{\prime })\subseteq U^{\prime }$ and $g$ is a contraction on the complete interval $U^{\prime }$. The local contraction proof for Newton iteration therefore shows that $r$ is the unique fixed point of $g$ on $U^{\prime }$.

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Multiplication by $(1-x^2{)}^{-1/2}$ puts the equation in Sturm-Liouville theory form:Multiply the equations for $T_n,T_m$ by $T_m,T_n$, subtract and integrate. The boundary term vanishes because $\sqrt{1-x^2}=0$ at both endpoints and the polynomial derivatives are bounded. Therefore, for $n^2\ne m^2$,This is the usual Chebyshev polynomial orthogonality.

Differentiating the original equation and writing $U_n=T_n^{\prime }$ givesIts self-adjoint form isThe same subtraction argument now yields, for $n^2\ne m^2$,These two relations form the Chebyshev derivative Sturm-Liouville pair.

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Vanishing net charge density does not require vanishing current. Positive and negative charge carriers can cancel in charge density while their oppositely directed motions add to a nonzero current. Charge conservation only requiresfor magnetostatics this becomes $\nabla \cdot J=0$.

Because $\nabla \cdot B=0$, one may introduce a magnetic vector potential withIt is not unique: $A+\nabla \chi$ gives the same field for any scalar $\chi$.

For the stated current,so it is consistent with stationary charge conservation. Direct calculation givesThus $J$ is a Beltrami field. For $\lambda \ne 0$, chooseThen $\nabla \cdot B=0$ and $\nabla \times B={\mu }_{0}J$. A convenient Coulomb-gauge potential issince $\nabla \times A=B$ and $\nabla \cdot A=0$. Gradient gauge terms may of course be added.

If $\lambda =0$, the current is the constant field $J=(0,J_0,0)$. The formulas involving $1/\lambda$ do not apply; one valid choice is

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Writing the velocity as $(u,v)=(y,ax)$ givesso the flow is incompressible. With the conventiona stream function isThe streamlines are its level sets. For $a>0$, they are the hyperbolaswith separatrices $y=\pm \sqrt{a}x$ and a saddle at the origin. For $a<0$, they are concentric ellipsestraversed clockwise. This is the streamline classification of a planar linear saddle or centre.

The scalar vorticity isHence the flow is irrotational exactly whenThen $(u,v)=(y,x)=\nabla \varphi$, and a velocity potential is

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Writing $p_j$ for the probability of face $j$, the hypotheses areagainst

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For observed counts $O_j$ and null expected counts $E_j=n{p}_j$, the Pearson chi-squared goodness-of-fit test usesWhen the null cell probabilities are specified and the expected counts grow, its limiting null distribution isHere $k=6$, so the limit is ${\chi }_5^2$.

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Under fairness every expected count is $90/6=15$. Therefore

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The asymptotic p-value is the upper-tail probabilityNumerically this is approximately $0.392$, so the data provide little evidence against fairness.

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Use multipliers $\lambda ,\mu \ge 0$ forAt an interior point in the nonnegative quadrant, the Karush-Kuhn-Tucker conditions arewith complementary slackness for the two constraints.

For $c=3/e^2$ and $d=2/e$, the square-root constraint is inactive at the answer, so $\mu =0$. The second stationarity equation gives $\lambda =1$, and the first gives $x_1=e^{-2}$. The sum constraint is active, henceIndeed, $\sqrt{x_2}=\sqrt{2}/e<2/e$. The minimum value is

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Now $d=1/e$, and both constraints are active. ThereforeThe first stationarity equation givesand the second givesThus all multiplier and complementary-slackness conditions hold, andwith minimum value

The observation is that lowering $d$ activates the second constraint and moves the optimum to the intersection of the two active boundaries. Rewriting $\sqrt{x_2}\le d$ as $x_2\le d^2$ shows that the feasible set is convex, while $x_1\log x_1-x_2$ is convex. Hence the active-set transition in capped resource allocation and the KKT candidates above give the unique global minima, not merely local stationary points.

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The characteristic polynomial isTo prove triangularizability, use induction on $n$. The result is immediate for $n=1$. Over $\mathbb{C}$, ${\chi }_A$ has a root $\lambda$, so $A$ has an eigenvector $v_1$. Extend it to a basis. In this basis,By induction, a change among the remaining $n-1$ basis vectors makes $B$ upper triangular. Thus the triangularization over an algebraically closed field gives

The minimal polynomial $m_A$ is the monic polynomial of least degree satisfying $m_A(A)=0$. To establish existence without quoting the Cayley-Hamilton theorem, let $T$ be an upper-triangular matrix similar to $A$, with diagonal entries ${\lambda }_1,\dots ,{\lambda }_n$, and putThenThe factors $T-{\lambda }_{k}I$ commute, so applying all of them successively lowers the invariant flag to zero:Similarity gives the same polynomial identity for $A$. Hence a nonzero monic annihilating polynomial of degree $n$ exists, and a least-degree one exists.

If $m$ and $\tilde{m}$ were two monic annihilating polynomials of the same least degree, then $m-\tilde{m}$ would be an annihilating polynomial of smaller degree unless it were zero. Thus the minimal polynomial is unique, and the minimal polynomial bound from a triangular invariant flag gives

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The eigenvalue equation $D_{r}f=\lambda f$ isIf $\lambda =-1$, this forces $f(x+r)=0$ for every $x$, hence $f=0$, so $-1$ is not an eigenvalue. If $c=1+\lambda \ne 0$, choose arbitrary values on representatives of the cosets of $r\mathbb{Z}$ and extend byThis produces nonzero eigenfunctions. Choosing functions supported on distinct cosets gives infinitely many linearly independent eigenfunctions. Thereforeand every eigenspace is infinite-dimensional.

Direct expansion giveswhich is symmetric in $r,s$. Hence

Suppose a degree-$n$ polynomial $p$ were a sum of $n$ periodic functions,Apply the commuting product $D_{r_1}\cdots D_{r_n}$. Every term on the right is killed by its corresponding factor, whereas for leading coefficient $a_n$ the mixed finite difference of a polynomial givesThis contradiction proves

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An ideal of $R/I$ has the form $J/I$ for an ideal $J\supseteq I$ of $R$. If $R$ is Noetherian, writeThenso every ideal of $R/I$ is finitely generated. Hence

The Hilbert basis theorem states that $R[X]$ is Noetherian whenever $R$ is Noetherian. The integers form a principal ideal domain, so every ideal of $\mathbb{Z}$ has one generator and $\mathbb{Z}$ is Noetherian. It follows that $\mathbb{Z}[X]$ is Noetherian.

For a nonsquare integer $d$, evaluation at $\sqrt{d}$ givesIt is surjective and its kernel is $(X^2-d)$. Therefore is Noetherian by the quotient result. This is the noetherianity of a quadratic integer order.

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The coefficient condition says precisely that every nonconstant monomial has positive powers of both variables. ThusIf $c+XYf$ and $d+XYg$ lie in $R$, so do their sum and their productso $R$ is a subring of $K[X,Y]$.

For $n\ge 1$, letThen $I_n\subseteq I_{n+1}$. The containment is strict because $X{Y}^{n+1}\notin I_n$: multiplying a generator $X{Y}^j$ by an element of $R$ produces either a scalar multiple of $X{Y}^j$, or terms divisible by $X^2$. It cannot produce the monomial $X{Y}^{n+1}$ when $j\le n$. Henceis a strictly increasing ideal chain. The constant-plus-ideal non-Noetherian subring therefore satisfies

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A topological space is compact if every open cover has a finite subcover. It is Hausdorff if every pair of distinct points has disjoint open neighbourhoods.

Let $Y$ be closed in compact $X$, and let $\{U_i\}$ be an open cover of $Y$ by sets open in $X$. Adding the open set $X\setminus Y$ gives a cover of $X$, which has a finite subcover. Removing $X\setminus Y$ leaves a finite subcover of $Y$. Thus every closed subspace of a compact space is compact.

Now let $A,B$ be disjoint closed subsets of a compact Hausdorff space. They are compact. For each $a\in A$ and $b\in B$, choose disjoint open sets $U_{a,b}\ni a$ and $V_{a,b}\ni b$. Fixing $a$, finitely many $V_{a,b_j}$ cover $B$. PutThen $U_a$ contains $a$, $V_a$ contains $B$, and they are disjoint. Finitely many $U_{a_i}$ cover $A$. Thereforeare disjoint open neighbourhoods of $A$ and $B$. This proves the normality of a compact Hausdorff space.

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Let $X$ be compact Hausdorff, let $x\in X$, and let $U$ be a neighbourhood of $x$. Choose an open $O$ with $x\in O\subseteq U$. Apply the separation result from part (a) to the disjoint closed sets $\{x\}$ and $X\setminus O$. There is an open $V\ni x$ whose closure lies in $O$. Thenis compact, contains the neighbourhood $V$ of $x$, and lies in $U$. Thus $X$ is locally compact.

Conversely, suppose $X$ is locally compact Hausdorff and $A\cap K$ is closed in every compact $K\subseteq X$. For $x\notin A$, choose a compact neighbourhood $K$ of $x$ and an open set $U$ withSince $A\cap K$ is closed in $K$, there is an open set $W\subseteq X$ such thatThen $U\cap W$ is an open neighbourhood of $x$ disjoint from $A$. Thus $X\setminus A$ is open andThis is the compactly detected closed-set theorem in a locally compact Hausdorff space.

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If $(x,y)\sim (x+a,y+b)$ in the torus, thenThis is Klein-equivalent to $(2x,y)$ by taking $(c,d)=(2a,b)$. Hence $\tilde{\pi }$ descends to a well-defined continuous map

The deck group of the Klein bottle has an index-two subgroup consisting of transformations with even $c$; it is generated by translations $(x,y)\mapsto (x+2,y)$ and $(x,y)\mapsto (x,y+1)$. This is exactly the image under $\tilde{\pi }$ of the torus deck lattice. Therefore the orientation double cover of the Klein bottle has two sheets:

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For $T^2$, take the unit square $[0,1]\times [0,1]$. Identifyso both pairs of opposite edges have matching directions.

For $K$, take $[0,1]\times [-1/2,1/2]$. Identifywith matching directions, andwith reversed directions. These equations specify the arrows on the two fundamental-domain drawings.

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In the Klein fundamental square from part (b), the following straight lines project to the required closed geodesics:this is the one-sided horizontal core, and cutting along it leaves a Möbius strip.

Takethis is a two-sided vertical geodesic, and cutting along it leaves a cylinder.

Finally takeIts endpoints differ by the deck translation $(2,1)$, so it is closed. The parameters $t=1/4$ and $t=3/4$ represent the same Klein-bottle point through an odd-$c$ glide reflection, and no other interior pair does. Thus its image has exactly one transverse self-intersection. This is the flat Klein-bottle geodesic model.

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A loop whose Klein deck transformation has odd $c$ has one closed lift to the orientation double cover; one with even $c$ has two. Thus the preimage of ${\gamma }_1$ contains one closed geodesic, while those of ${\gamma }_2$ and ${\gamma }_3$ contain two each.

In the torus unit square, representatives arewhich maps twice around ${\gamma }_1$;andAll coordinates are taken modulo one. Hence the requested numbers areThe two diagonal lifts of ${\gamma }_3$ meet over its self-intersection point but remain distinct closed geodesics.

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For $0<a<2$, the beta-integral evaluation givesDifferentiation under the integral sign near $a=1$ is justified by domination at zero and infinity. ThereforeWriting $a=1+t$ givesSincethe logarithmic moments of the Cauchy kernel yieldThe same expansion gives $I^{\prime \prime }(1)={\pi }^3/8$, agreeing with the supplied identity.

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For a variation $y+\epsilon \eta$,Integration by parts givesFor fixed endpoints, $\eta (0)=\eta (x_0)=0$. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equationA solution makes the first variation vanish for every admissible variation, so it is a stationary candidate; whether it is a minimum or maximum is decided by higher variations. If endpoint values are free, $\eta$ is arbitrary there, and the boundary term instead vanishes under the natural conditionsThus the same Euler-Lagrange solution is stationary for all free-endpoint variations. These are the natural boundary conditions for a free endpoint.

Forthe two equations areSet $u=y+z$ and $v=y-z$. ThenThe conditions $y(0)=z(0)=0$ giveand hence the most general solution is

Free conditions at $x_0$ are $y^{\prime }(x_0)=z^{\prime }(x_0)=0$. Adding and subtracting them givesThus $A=0$. The zero solution exists for every $x_0$, while nonzero solutions exist precisely whenFor those values they form the one-parameter familywhere $C$ is arbitrary. This is the free-endpoint normal mode of a coupled variational functional.

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The convolution isWe claim that the $n$-fold convolution isThis is true for $n=1$. If it holds for $n-1$ and $x\ge 0$, thenand the convolution vanishes for $x<0$. This is the gamma density from repeated exponential convolution.

The Fourier transform is

The convolution theorem statesIndeed, Fubini and $u=x-y$ giveSince $\hat{F}(k)=(1+ik{)}^{-1}$, induction immediately verifies

For Parseval identity, take $g(x)=\overline{f(-x)}$. ThenUsing Fourier inversion at zero, which follows from the supplied delta identity,and the convolution theorem gives

Apply this to $F_{n+1}$. Sinceone obtains the Even rational Parseval integral

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For a test wavefunction $f$,Thus the canonical commutation relation is

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The time-dependent Schrodinger equation is

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Differentiating $⟨O⟩=⟨\psi |O|\psi ⟩$ and using the Schrodinger equation and its adjoint gives the Ehrenfest theorem

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Using $[A,BC]=[A,B]C+B[A,C]$ and $[x,p_x]=i\hslash I$,

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With $L=x{p}_y-y{p}_x$, the canonical commutators giveIt follows thatTherefore rotational symmetry of the isotropic oscillator gives