Past exam of the mathematics course of the University of Cambridge / 2020 / ia / Paper 1

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.