Past exam of the mathematics course of the University of Cambridge / 2020 / ia

Using Euler's formula, the complex exponential function satisfiesHence

The complex sine function obeysThus

Since $z\overline{z}=|z{|}^2=x^2+y^2$, the complex conjugate givesTherefore

Factor the expression using the complex conjugate:It is real, so

All values of the complex logarithm areBecause $x>0$, one may take the complex argument $\arg z=\arctan (y/x)$. Consequently

Invert the equation and regard $x$ as a function of $y$:This is a first-order linear differential equation. Multiplication by the integrating factor $e^{-y}$ givesso $x=e^{2y}+C{e}^y$. The initial condition $x=1$ at $y=0$ gives $C=0$. Hence $x=e^{2y}$ and

Define $F(x)={\int }_a^{x}f(t)dt$. By the fundamental theorem of calculus, $F^{\prime }=f$. The chain rule givesIntegrating from $\alpha$ to $\beta$ and using $g(\alpha )=a$, $g(\beta )=b$ yields the change of variables formulaThe argument also covers a decreasing $g$ because both integrals are oriented.

For $0<\epsilon <e^{-1}$, set $u=\log (1/x)$. The ordinary change of variables formula givesTaking $\epsilon \downarrow 0$ is exactly the defining limit of the improper integral. By the improper power integral, it converges for $\theta >1$, and

Let $Z_n$ be the size of generation $n$. This is a Galton-Watson process with offspring meanThe expected generation size in a Galton-Watson process is thereforeIts probability generating function isBy the Galton-Watson extinction fixed point, the extinction probability $q$ is the smallest solution in $[0,1]$ of $G(q)=q$. Factoring givesso $q=(\sqrt{29}-5)/2$. The probability of production continuing forever is

Expanding the squared Euclidean distances givesThus one may takeThe normal is parallel to $\overrightarrow{AB}$, so $L_{AB}$ is its perpendicular bisector. If $\mathbf{x}$ lies on both $L_{AB}$ and $L_{BC}$, then $|\mathbf{x}-\mathbf{a}|=|\mathbf{x}-\mathbf{b}|=|\mathbf{x}-\mathbf{c}|$, hence it also lies on $L_{CA}$. Geometrically, the three perpendicular bisectors of the noncollinear triangle concur at its circumcenter.

Assume the direction vectors are nonzero. The cross product equation is equivalent toso $\mathbf{x}-\mathbf{a}$ is parallel to $\mathbf{u}$ and the locus is the line $\mathbf{x}=\mathbf{a}+\lambda \mathbf{u}$.

For nonparallel directions, ${\mathbf{u}}_1\times {\mathbf{u}}_2$ is normal to both lines. Projecting their displacement onto the corresponding unit vector gives the distance between skew linesIf the directions are parallel, the corresponding formula is

A matrix is Hermitian when $A^{†}=A$, and unitary when $U^{†}U=U{U}^{†}=I$.

If $A\mathbf{u}=\lambda \mathbf{u}$ with $\mathbf{u}\ne 0$, thenso every eigenvalue is real. If $A\mathbf{u}=\lambda \mathbf{u}$ and $A\mathbf{v}=\mu \mathbf{v}$, thenDistinct eigenvalues therefore have orthogonal eigenvectors.

The normalized eigenvectors form an orthonormal basis, so the matrix $U=({\mathbf{u}}_1\cdots {\mathbf{u}}_n)$ satisfies $U^{†}U=I$ and is unitary. Withthe equations $A{\mathbf{u}}_j={\lambda }_j{\mathbf{u}}_j$ say $AU=UD$, hence $A=UD{U}^{†}$. This is the finite-dimensional unitary diagonalization of a normal matrix.

A unitary conjugate of an arbitrary diagonal matrix need not be Hermitian: $U=(1)$ and $D=(i)$ give $UD{U}^{†}=(i)$. It is Hermitian exactly when all diagonal entries of $D$ are real.

For the displayed matrix, one valid decomposition isThe columns are orthonormal eigenvectors, so direct multiplication gives $UD{U}^{†}=A$.

For the heat kerneldirect partial differentiation givesThus $K$ solves the heat equation. For bounded continuous $f$, differentiation under the integral sign givesWith $Y=(x-y)/\sqrt{4t}$,The Gaussian integral makes the weight have total mass one, and the dominated convergence theorem gives $u(x,t)\to f(x)$. This is the Gaussian approximate identity.

For the viscous Burgers equation with unit viscosity, the Cole-Hopf transformation $w=-2u_x/u$ reduces the equation to $u_t=u_{xx}$. To obtain initial value $g$, choose(up to an irrelevant positive constant), and set $u=K_t\ast f$. ThereforeEquivalently,and the approximate-identity limit gives $w(x,t)\to g(x)$.

The first and third equations form a constant-coefficient linear system of differential equationsThe eigenvectors $(1,1)$ and $(1,-1)$ have eigenvalues $2$ and $-4$. Applying the initial data givesThe remaining equation becomesAn integrating factor, or the usual exponential and trigonometric particular solutions, givesThese values satisfy all three initial conditions.

Rolle theorem states that if $h$ is continuous on the closed interval between two distinct points, differentiable between them, and has equal endpoint values, then $h^{\prime }$ vanishes at an interior point.

Let $P_N(t)={\sum }_{j=0}^N{f}^{(j)}(0)t^j/j!$ and, for $x\ne 0$, putThen $h(x)=0$ and $h^{(j)}(0)=0$ for $0\le j\le N$. Applying Rolle's theorem successively $N+1$ times gives a point $\xi =\theta x$, $0<\theta <1$, where $h^{(N+1)}(\xi )=0$. Hencewhich is the stated Taylor theorem with Lagrange remainder.

If $f^{\prime }=0$, the $N=0$ case gives $f(x)=f(0)+x{f}^{\prime }(\theta x)=f(0)$. Thus $f$ is constant; equivalently this is the immediate corollary of the mean value theorem.

DifferentiateUsing $s^{\prime }=c$ and $c^{\prime }=-s$ givesBy the zero derivative on a connected open set, $F$ is independent of $x$.

The constant from part (i) is found by setting $x=0$:Taking $a=x+y$ therefore gives the sine addition formula

Differentiation givesThe initial conditions set the constant to one, sothe Pythagorean trigonometric identity.

Since $|c|\le 1$, Taylor theorem with Lagrange remainder givesfor some $0<\xi <1$, whilefor some $0<\eta <2$. The intermediate value theorem supplies $k\in (1,2)$ with $c(k)=0$. The sine addition formula gives $s(2k)=2s(k)c(k)=0$. The analogous cosine addition formula gives $c(2k)=c(k{)}^2-s(k{)}^2=-1$, soThus $s$ is a periodic function.

Choose a closed interval containing the bounded sequence. Bisect it and retain a half containing infinitely many terms; repeat. The resulting intervals are nested and their lengths tend to zero. By the nested interval theorem they have one common point $x$, and choosing successively indexed terms from these intervals produces a subsequence converging to $x$. This proves the Bolzano-Weierstrass theorem.

Boundedness of $(z_n)$ makes both real sequences $(x_n)$ and $(y_n)$ bounded. The Bolzano-Weierstrass theorem first gives a subsequence $(z_{n_j})$ for which $(x_{n_j})$ converges. Applying it again to $(y_{n_j})$ gives a subsubsequence for which the imaginary parts also converge. Since convergence of complex numbers is equivalent to convergence of their real parts and imaginary parts, this subsubsequence converges in $\mathbb{C}$.

Construct the nested index sequences inductively. Start with $N^{(0)}=(1,2,\dots )$. Having chosen $N^{(i-1)}$, apply the Bolzano-Weierstrass theorem to the bounded sequence of $i$th coordinates along those indices, and let $N^{(i)}$ be an increasing subsequence along which that coordinate converges. Because each new sequence is nested inside all earlier ones, coordinates $1,\dots ,i$ all converge along $N^{(i)}$.

Now take the diagonal subsequence argumentNestedness impliesso $(m_j)$ is strictly increasing. For every fixed $i$, the tail $(m_j{)}_{j\ge i}$ is a subsequence of $N^{(i)}$; hence $(x_{m_j}^{(i)})$ converges. This single diagonal subsequence works for every coordinate.

For events $A_1,\dots ,A_n$, the probabilistic inclusion-exclusion principle isTo prove it, fix an outcome lying in exactly $r$ events. Its total coefficient on the right isby the binomial theorem; outcomes in no event contribute zero. Taking expectations of this pointwise indicator identity proves the formula. Truncating after the pair terms gives the Bonferroni inequalities

For the final bound, let $X={\sum }_i{1}_{A_i}$ and $S=\mathbb{E}X={\sum }_i\mathbb{P}(A_i)$. The intersection assumption givesThe Cauchy-Schwarz inequality applied to $X{1}_{X>0}$ yields $S^2\le \mathbb{E}[X^2]\mathbb{P}(X>0)\le S+n-1$. Thereforeso the required universal constant may be taken as $\boxed{c=2}$. This is a second moment method estimate.

The standard normal density isIt is nonnegative, and the Gaussian integral gives ${\int }_{-\infty }^{\infty }\varphi (z)dz=1$, so it is a probability density function. Completing the square gives the moment-generating function of a standard normal variableThus $\mathbb{E}Z=M_Z^{\prime }(0)=0$ and

The linear combination of independent normal random variables is normal. HenceThis also follows by multiplying the moment-generating functions of the independent and identically distributed random variables.

Since $Y_1=X_1^2$ and a standard normal variable has second and fourth moments $1$ and $3$,Thus $T_n$ has the chi-squared distribution with $n$ degrees of freedom and

Convergence in distribution $W_n\Rightarrow W$ means that the cumulative distribution functions satisfy $F_{W_n}(x)\to F_W(x)$ at every continuity point of $F_W$. The Lévy continuity theorem states that this is equivalent to pointwise convergence of characteristic functions to the characteristic function of $W$, provided the limiting function is continuous at zero.

For fixed $t$, independence and a Gaussian integral giveUsing $\log (1-z)=-z-z^2/2+O(z^3)$,Therefore ${\phi }_{V_n}(t)\to e^{-t^2/2}$, the characteristic function of the standard normal distribution. Lévy's theorem now givesThis derives the limit directly, without assuming the central limit theorem.

A transposition exchanges exactly two letters and fixes the rest. A permutation cycle $(a_1{a}_2\dots ,a_k)$ sends each displayed letter to the next, sends $a_k$ to $a_1$, and fixes all other letters.

After imposing $\sigma (1)=2$, the remaining $n-1$ domain elements may be sent bijectively to the remaining $n-1$ values in arbitrary order. Hence the number is

There are $\lfloor n/2\rfloor$ even letters. The stated condition and injectivity force the even letters to permute among themselves, which can be done in $\lfloor n/2\rfloor !$ ways. The remaining odd letters must then permute among themselves. Thus the number is

Choose the support of the four-cycle and then cyclically order it. A chosen set of four letters has $(4-1)!=6$ distinct cycles, so the answer is

The product of two transpositions has three possible forms. Equal transpositions produce the identity, transpositions meeting in one letter produce a three-cycle, and disjoint transpositions produce two disjoint two-cycles. There are two three-cycles on each chosen triple and three pairings of each chosen four-set. Therefore the number of distinct permutations representable this way is

An equivalence relation is reflexive, symmetric, and transitive. HereThe relation is reflexive and symmetric, but it is not transitive. For example, takeThe real parts of $x\overline{y}$ and $y\overline{z}$ are positive, whereas that of $x\overline{z}$ is negative. Hence this is not an equivalence relation.

By the equality condition in the complex triangle inequality,for nonzero $x,y$ exactly when $x/y$ is a positive real number. Having the same complex argument is reflexive, symmetric, and transitive, so this is an equivalence relation. Its classes are the open rays from the origin.

This relation is reflexive because choosing $n=1$ gives $|x/x|=1$. It need not be symmetric: take positive real numbers $y=e$ and $x=e^2$. Then $|x/y^2|=1$ is rational, so $x\sim y$. Butis transcendental, hence irrational, for every integer $m\ge 1$. Thus $y\nsim x$, and the relation is not an equivalence relation.

Set $F(x)=|x^3-x|$. The relation is exactlyIt is therefore the equivalence relation induced by a function: equality of the two function values is reflexive, symmetric, and transitive. Hence this is an equivalence relation.

Parametrize the straight segment by $x=t$, $y=1+t$, $0\le t\le 1$. The line integral becomes

For the parabola, $x=t$, $y=1+t^2$, $dx=dt$, and $dy=2tdt$. HenceThe difference from the straight-path value reflects that the one-form is not exact.

Green theorem states that for a positively oriented planar region $D$,Here $P=x{y}^2-\lambda y$ and $Q=x^{2}y$, soThe stated orientations make $C_1\cup C_2$ the positive boundary of the region inside the radius-$a$ circle and outside the radius-$b$ circle. Its area is $\pi (a^2-b^2)$ regardless of the inner circle's center. Therefore

The central force conserves angular momentum. The initial impact parameter gives $L=ub$, while at closest approach the velocity is transverse and $L=wp$. Thus $u=wp/b$. Conservation of energy givesUnder $u^2\gg 2k/R$, neglect the initial potential term and substitute $u=wp/b$:Consequently

Now $E=u^2/2-k/R<0$, so the orbit is bound. Since $L=ub$, the initial energy isAt the farthest central-force radial turning point $q$, the radial speed vanishes and the transverse speed is $L/q$, givingEquating these expressions and multiplying by $2q^2$ yieldsThus $A=L^2/b^2-2k/R$, $B=2k$, and $C=-L^2$.