Advanced Topics (Elite Level)
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
(a) State the Cauchy–Riemann equations and verify that is analytic everywhere except at its poles; locate the poles and classify them. (5 marks)
(b) Compute the residue of 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 by the residue theorem. (7 marks)
(d) Now expand as a Laurent series about the pole , 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 with fixed endpoints, starting from the first variation . State clearly where the fundamental lemma of calculus of variations is invoked. (6 marks)
(b) A cable of fixed length hangs between two points at the same height, minimizing potential energy subject to the length constraint . Set up the isoperimetric problem with a Lagrange multiplier , write the augmented Euler–Lagrange equation, and use the Beltrami identity (since the integrand has no explicit ) to reduce it to a first-order ODE. (8 marks)
(c) Consider the convex optimization problem 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 has transition matrix Find the stationary distribution satisfying , . Prove it is unique here. (6 marks)
(b) Define uniform convergence of a sequence and give the – statement. Show that on 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 and prove by the recurrence that its running time is . Contrast with the naive DFT cost. (6 marks)
Answer keyMark scheme & solutions
Question 1
(a) [5]
- CR equations: for , and . (2)
- is entire; is a polynomial. Quotient analytic wherever denominator . (1)
- Poles where ; both simple poles (denominator has simple zeros, numerator nonzero there). (2)
(b) [5] At (simple pole), (5)
(c) [7]
- Consider over semicircle radius in UHP. (1)
- Jordan's lemma: since and , the arc integral as . (2)
- By residue theorem, (3)
- Taking real parts: (1)
(d) [3] . Let ; then , so Principal part: (single term, simple pole). (3)
Question 2
(a) [6]
- Perturb , . (1)
- . (2)
- Integrate second term by parts: ; boundary term vanishes. (1)
- . (1)
- By fundamental lemma: . (1)
(b) [8]
- Augmented integrand (absorbing constants). (2)
- Euler–Lagrange with no explicit ⇒ Beltrami: . (2)
- , so (3)
- First-order ODE: , solution the catenary . (1)
(c) [8]
- Lagrangian , , constraint . (1)
- Stationarity: . (2)
- Complementary slackness: . If then violates constraint ⇒ constraint active: . (2)
- Then , , feasibility ✓, dual feasibility ✓. (2)
- Objective is strictly convex (Hessian ), feasible set convex ⇒ KKT point is the unique global minimum, value . (1)
Question 3
(a) [6]
- Solve , i.e. . (3)
- Normalize: So . (2)
- Chain irreducible & aperiodic (all entries , finite) ⇒ unique stationary distribution by Perron–Frobenius. (1)
(b) [6]
- Uniform convergence: s.t. . (2)
- Pointwise limit: for , . (2)
- : near , can be made (take ), so sup . Also limit is discontinuous while each continuous — impossible under uniform convergence. Hence not uniform. (2)
(c) [6]
- Recurrence: split DFT into even/odd indexed samples: (combine step = butterfly ops). (2)
- Master theorem / unrolling: . Proof by unrolling: at level there are subproblems of size , each contributing combine work ⇒ per level, levels ⇒ . (3)
- Naive DFT: (each of outputs sums 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)"}
]