Level 5 — MasteryAdvanced Topics (Elite Level)

Advanced Topics (Elite Level)

90 minutes60 marksprintable — key stays hidden on paper

Level 5 — Cross-Domain Mastery (Complex Analysis · Variational Calculus · Optimization · Stochastic Processes)

Time limit: 90 minutes Total marks: 60

Instructions: Answer all three questions. Full rigour and justification required. Use ...... / ...... for mathematics. Partial credit awarded for correct method.


Question 1 — Complex Analysis meets Physics (20 marks)

A damped oscillator's response requires the real integral

I=cos(ωx)x2+a2dx,a>0, ω>0.I = \int_{-\infty}^{\infty} \frac{\cos(\omega x)}{x^2 + a^2}\, dx, \qquad a>0,\ \omega>0.

(a) State the Cauchy–Riemann equations and verify that f(z)=eiωzz2+a2f(z)=\dfrac{e^{i\omega z}}{z^2+a^2} is analytic everywhere except at its poles; locate the poles and classify them. (5 marks)

(b) Compute the residue of f(z)f(z) at the pole lying in the upper half-plane. (5 marks)

(c) Using a semicircular contour in the upper half-plane, apply Jordan's lemma to justify vanishing of the arc integral, then evaluate II by the residue theorem. (7 marks)

(d) Now expand g(z)=1z2+a2g(z)=\dfrac{1}{z^2+a^2} as a Laurent series about the pole z=iaz=ia, valid in a punctured neighbourhood, and state its principal part. (3 marks)


Question 2 — Calculus of Variations & Constrained Optimization (22 marks)

(a) Derive the Euler–Lagrange equation for a functional J[y]=x1x2F(x,y,y)dxJ[y]=\int_{x_1}^{x_2} F(x,y,y')\,dx with fixed endpoints, starting from the first variation δJ=0\delta J = 0. State clearly where the fundamental lemma of calculus of variations is invoked. (6 marks)

(b) A cable of fixed length LL hangs between two points at the same height, minimizing potential energy y1+y2dx\propto \int y\sqrt{1+y'^2}\,dx subject to the length constraint 1+y2dx=L\int \sqrt{1+y'^2}\,dx = L. Set up the isoperimetric problem with a Lagrange multiplier λ\lambda, write the augmented Euler–Lagrange equation, and use the Beltrami identity (since the integrand has no explicit xx) to reduce it to a first-order ODE. (8 marks)

(c) Consider the convex optimization problem minxR2 12(x12+x22)s.t.x1+x22.\min_{x\in\mathbb{R}^2}\ \tfrac12(x_1^2+x_2^2)\quad\text{s.t.}\quad x_1+x_2\ge 2. Write the KKT conditions, solve them, and verify the solution is the global minimum by convexity. (8 marks)


Question 3 — Stochastic Processes & Numerical Analysis (18 marks)

(a) A Markov chain on states {1,2}\{1,2\} has transition matrix P=(0.70.30.40.6).P=\begin{pmatrix} 0.7 & 0.3 \\ 0.4 & 0.6 \end{pmatrix}. Find the stationary distribution π\pi satisfying πP=π\pi P = \pi, iπi=1\sum_i\pi_i=1. Prove it is unique here. (6 marks)

(b) Define uniform convergence of a sequence (fn)(f_n) and give the ε\varepsilonNN statement. Show that fn(x)=xnf_n(x)=x^n on [0,1][0,1] converges pointwise but not uniformly, identifying the limit function and the obstruction. (6 marks)

(c) State the divide-and-conquer recurrence for the Cooley–Tukey FFT of length N=2kN=2^k and prove by the recurrence that its running time is Θ(NlogN)\Theta(N\log N). Contrast with the naive DFT cost. (6 marks)


Answer keyMark scheme & solutions

Question 1

(a) [5]

  • CR equations: for f=u+ivf=u+iv, ux=vyu_x=v_y and uy=vxu_y=-v_x. (2)
  • eiωze^{i\omega z} is entire; z2+a2z^2+a^2 is a polynomial. Quotient analytic wherever denominator 0\ne0. (1)
  • Poles where z2+a2=0z=±iaz^2+a^2=0 \Rightarrow z=\pm ia; both simple poles (denominator has simple zeros, numerator nonzero there). (2)

(b) [5] At z=iaz=ia (simple pole), Res=eiωzddz(z2+a2)z=ia=eiω(ia)2ia=eωa2ia.\text{Res} = \dfrac{e^{i\omega z}}{\frac{d}{dz}(z^2+a^2)}\Big|_{z=ia} = \dfrac{e^{i\omega(ia)}}{2ia} = \dfrac{e^{-\omega a}}{2ia}. (5)

(c) [7]

  • Consider Cfdz\oint_C f\,dz over semicircle radius RR in UHP. (1)
  • Jordan's lemma: since ω>0\omega>0 and 1/(z2+a2)0|1/(z^2+a^2)|\to0, the arc integral 0\to0 as RR\to\infty. (2)
  • By residue theorem, eiωxx2+a2dx=2πieωa2ia=πaeωa.\int_{-\infty}^{\infty}\frac{e^{i\omega x}}{x^2+a^2}dx = 2\pi i\cdot\frac{e^{-\omega a}}{2ia} = \frac{\pi}{a}e^{-\omega a}. (3)
  • Taking real parts: I=cosωxx2+a2dx=πaeωa.I=\displaystyle\int_{-\infty}^{\infty}\frac{\cos\omega x}{x^2+a^2}dx = \dfrac{\pi}{a}e^{-\omega a}. (1)

(d) [3] g(z)=1(zia)(z+ia)g(z)=\dfrac{1}{(z-ia)(z+ia)}. Let w=ziaw=z-ia; then z+ia=w+2iaz+ia = w+2ia, so g=1w1w+2ia=12iaw11+w2ia=12iawn0(w2ia)n.g=\frac{1}{w}\cdot\frac{1}{w+2ia}=\frac{1}{2ia\,w}\cdot\frac{1}{1+\frac{w}{2ia}}=\frac{1}{2ia\,w}\sum_{n\ge0}\left(-\frac{w}{2ia}\right)^n. Principal part: 12ia1zia\dfrac{1}{2ia}\cdot\dfrac{1}{z-ia} (single term, simple pole). (3)


Question 2

(a) [6]

  • Perturb yy+εηy\to y+\varepsilon\eta, η(x1)=η(x2)=0\eta(x_1)=\eta(x_2)=0. (1)
  • δJ=x1x2(Fyη+Fyη)dx=0\delta J=\int_{x_1}^{x_2}\big(F_y\eta+F_{y'}\eta'\big)dx=0. (2)
  • Integrate second term by parts: Fyηdx=[Fyη]x1x2ddxFyηdx\int F_{y'}\eta'dx=[F_{y'}\eta]_{x_1}^{x_2}-\int\frac{d}{dx}F_{y'}\,\eta\,dx; boundary term vanishes. (1)
  • δJ=(FyddxFy)ηdx=0 η\delta J=\int(F_y-\frac{d}{dx}F_{y'})\eta\,dx=0\ \forall\eta. (1)
  • By fundamental lemma: FyddxFy=0F_y-\dfrac{d}{dx}F_{y'}=0. (1)

(b) [8]

  • Augmented integrand G=(yλ)1+y2G=(y-\lambda)\sqrt{1+y'^2} (absorbing constants). (2)
  • Euler–Lagrange with no explicit xx ⇒ Beltrami: GyGy=CG-y'G_{y'}=C. (2)
  • Gy=(yλ)y1+y2G_{y'}=(y-\lambda)\dfrac{y'}{\sqrt{1+y'^2}}, so GyGy=(yλ)(1+y2y21+y2)=yλ1+y2=C.G-y'G_{y'}=(y-\lambda)\left(\sqrt{1+y'^2}-\frac{y'^2}{\sqrt{1+y'^2}}\right)=\frac{y-\lambda}{\sqrt{1+y'^2}}=C. (3)
  • First-order ODE: yλ1+y2=C\dfrac{y-\lambda}{\sqrt{1+y'^2}}=C, solution the catenary y=λ+Ccosh ⁣(xx0C)y=\lambda+C\cosh\!\big(\frac{x-x_0}{C}\big). (1)

(c) [8]

  • Lagrangian L=12(x12+x22)μ(x1+x22)L=\tfrac12(x_1^2+x_2^2)-\mu(x_1+x_2-2), μ0\mu\ge0, constraint x1+x22x_1+x_2\ge2. (1)
  • Stationarity: xiμ=0x1=x2=μx_i-\mu=0\Rightarrow x_1=x_2=\mu. (2)
  • Complementary slackness: μ(x1+x22)=0\mu(x_1+x_2-2)=0. If μ=0\mu=0 then x=0x=0 violates constraint ⇒ constraint active: x1+x2=2x_1+x_2=2. (2)
  • Then 2μ=2μ=12\mu=2\Rightarrow\mu=1, x1=x2=1x_1=x_2=1, feasibility 0\ge0 ✓, dual feasibility μ=10\mu=1\ge0 ✓. (2)
  • Objective is strictly convex (Hessian I0I\succ0), feasible set convex ⇒ KKT point is the unique global minimum, value =1=1. (1)

Question 3

(a) [6]

  • Solve π1=0.7π1+0.4π2\pi_1=0.7\pi_1+0.4\pi_2, i.e. 0.3π1=0.4π2π1=43π20.3\pi_1=0.4\pi_2\Rightarrow \pi_1=\frac{4}{3}\pi_2. (3)
  • Normalize: π1+π2=143π2+π2=1π2=37, π1=47.\pi_1+\pi_2=1\Rightarrow \frac{4}{3}\pi_2+\pi_2=1\Rightarrow\pi_2=\frac37,\ \pi_1=\frac47. So π=(4/7,3/7)\pi=(4/7,\,3/7). (2)
  • Chain irreducible & aperiodic (all entries >0>0, finite) ⇒ unique stationary distribution by Perron–Frobenius. (1)

(b) [6]

  • Uniform convergence: ε>0 N\forall\varepsilon>0\ \exists N s.t. nN, x: fn(x)f(x)<ε\forall n\ge N,\ \forall x:\ |f_n(x)-f(x)|<\varepsilon. (2)
  • Pointwise limit: f(x)=0f(x)=0 for x[0,1)x\in[0,1), f(1)=1f(1)=1. (2)
  • supx[0,1]xnf(x)\sup_{x\in[0,1]}|x^n-f(x)|: near x=1x=1^-, xnx^n can be made 12\ge\frac12 (take x=21/nx=2^{-1/n}), so sup ↛0\not\to0. Also limit is discontinuous while each fnf_n continuous — impossible under uniform convergence. Hence not uniform. (2)

(c) [6]

  • Recurrence: split DFT into even/odd indexed samples: T(N)=2T(N/2)+Θ(N)T(N)=2T(N/2)+\Theta(N) (combine step = NN butterfly ops). (2)
  • Master theorem / unrolling: T(N)=Θ(NlogN)T(N)=\Theta(N\log N). Proof by unrolling: at level jj there are 2j2^j subproblems of size N/2jN/2^j, each contributing Θ(N/2j)\Theta(N/2^j) combine work ⇒ Θ(N)\Theta(N) per level, log2N\log_2 N levels ⇒ Θ(NlogN)\Theta(N\log N). (3)
  • Naive DFT: Θ(N2)\Theta(N^2) (each of NN outputs sums NN terms). FFT is asymptotically superior. (1)

[
  {"claim":"Residue of e^{i w z}/(z^2+a^2) at z=ia is e^{-wa}/(2ia)","code":"z,w,a=symbols('z w a',positive=True); f=exp(I*w*z)/(z**2+a**2); r=residue(f,z,I*a); result=simplify(r-exp(-w*a)/(2*I*a))==0"},
  {"claim":"Integral of cos(wx)/(x^2+a^2) = pi/a * e^{-wa}","code":"x,w,a=symbols('x w a',positive=True); val=integrate(cos(w*x)/(x**2+a**2),(x,-oo,oo)); result=simplify(val-pi/a*exp(-a*w))==0"},
  {"claim":"KKT solution x=(1,1) has objective value 1","code":"x1,x2=1,1; result=(Rational(1,2)*(x1**2+x2**2)==1) and (x1+x2>=2)"},
  {"claim":"Stationary distribution of P is (4/7,3/7)","code":"P=Matrix([[Rational(7,10),Rational(3,10)],[Rational(4,10),Rational(6,10)]]); pi=Matrix([[Rational(4,7),Rational(3,7)]]); result=simplify(pi*P-pi)==zeros(1,2)"}
]