NATIONAL UNIVERSITY OF SINGAPORE PC5215 NUMERICAL RECIPES WITH APPLICATIONS. (Semester I: AY ) Time Allowed: 2 Hours

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1 NATIONAL UNIVERSITY OF SINGAPORE PC55 NUMERICAL RECIPES WITH APPLICATIONS (Semester I: AY 0-) Time Allowed: Hours INSTRUCTIONS TO CANDIDATES. This examination paper contains FIVE questions and comprises THREE printed pages.. Answer ALL the questions; questions carry equal marks.. Answers to the questions are to be written in the answer books. 4. This is a CLOSED BOOK examination. 5. Non-programmable calculators are allowed.

2 . The IEEE 754 standard for floating point representation of the double precision number has a sign bit, followed by bits for the biased exponent (with a bias of 0) and the rest 5 bits for the fractional part. a. Give the bit patterns for the 64-bit double precision representation of and /. b. Discuss what impact reducing the biased exponent field to 8 bits has, in terms of the precision and magnitude of the floating point numbers. a. - is (in hexadecimal notation) BFF FFFFFFFFFFFFF, / is FD b. If the exponent field is reduced and fractional part increased, then the precision increases (machine epsilon is smaller) but the magnitude decreases.. Consider numerical integration formula by Gaussian quadrature method. a. State the main idea of Gaussian quadrature. b. Consider a three-point Gaussian quadrature formula in the interval [0,], f ( x ) dx w f ( x ) + w f ( x ) + w f ( x ). For polynomials,, x, x,, x n, 0 the formula is exact. What is the largest integer n that this is true? c. Determine the three-point formula, i.e., determine the abscissas x, x, x, and the weights w, w, w. a. by changing both the weights and abscissas, a maximum of accuracy is obtained, i.e., the formula is exact for polynomials of order N- for N point Gaussian quadrature. b. n=5. c. P 0 =, P = x-/, P =x -x+/6, P =x -/x +/5x -/0. This is obtained by the orthogonal condition <P i P j > = 0 (i < j). The root of P (x)=0 gives x =/, x = ( / 5) /, x = (+ / 5) /. The weights are determined by the condition that the polynomials, x, x can be integrated exactly. Given, w+ w + w =, xw + xw + xw =, x w+ x w + x w =. The solution is w =4/9, w =w =5/8.

3 . Consider a Monte Carlo simulation to generate Boltzmann distribution using the Metropolis algorithm for the following statistical-mechanical model: H({ σ, σ, σ }) = Jσσ Jσσ J( σ + σ ), where the spins σ i =±, i =,,, and J>0 is some constant. a. Assuming that the spins are chosen at random with equal probability, determine the transition matrix W. b. What is the equilibrium/invariant distribution P associated with W? c. Outline pseudo-code that computes σ, i.e., the equilibrium average value of the first spin, in a Metropolis Monte Carlo simulation. a. The energy of the 8 states are J J The transition matrix is (in the same order as the state/energy list above) r r r r / / r 0 0 r / 0 0 / 0 / r 0 r 0 / 0 / / / 0 4 J /( kt ), r = e. 0 / / / 0 / 0 / / 0 0 / / / r / / / r b. By design, P is the equilibrium (canonical) distribution, P =exp(-h(σ)/kt)/z, ie., P = [exp(4j/kt),,,, exp(-4j/kt),,, ]/Z, Z = 6 + exp(-4j/kt) + exp(4j/kt), satisfies P=PW (But we don t need to solve the eigenvalue problem to get P, as the purpose of the Metropolis algorithm is to design W such that P is the equilibrium one). c. set σ =σ =σ =. M=0. for(i= to 0 6 ) { j = floor(ξ) +, Compute H(σ),

4 } σ j = σ j, compute again H(σ ) (where the spin j is flipped), ΔE = H(σ )-H(σ), if (ξ > exp(-δe/(kt)) { σ j = σ j, (flip back) }, M = M + σ, M = M/ The algorithm of the conjugate-gradient method for locating minimum is as follows: i. Initialize n 0 = g 0 = - f(x 0 ), i = 0, ii. Find λ that minimizes f(x i +λn i ), let x i+ =x i +λn i iii. Compute new negative gradient g i+ = - f(x i+ ) gi+ gi+ iv. Compute γ i = gi gi v. Update new search direction as n i+ = g i+ + γ i n i ; ++i, go to ii a. If f is a function of two scalar variables x and y, f = x + xy+ y x+ y, how many times do we need to iterate the steps? b. Starting from the point (x,y)=(0,0), determine n 0 and n according to the algorithm above and show that n T 0 A n =0, where A is the coefficient matrix of the quadratic term in f. c. Work out the steps to find the minimum (x min, y min ) of f. a. in two steps since it is two dimensional. b. n 0 = g 0 = (-x-y+,-(x+y+)) = (, -), f = (/)λ - λ, min is at λ=, (x, y ) = (,-), g =(,), γ =, n = (,) + γ(,-) = (, 0). A =, n T 0 A n = 0. c. (x,y ) = (,-) is the minimum location. 5. Consider the problem of damped nonlinear oscillator governed by the second order differential equation, 4

5 d x dx γ x Ft () = +, dt dt where γ is a damping constant and F(t) some external driving force which depends on time explicitly. a. Rewrite the equation as a set of first order differential equations equivalent to the original one. b. Give the numerical algorithm of the midpoint method, and determine the local truncation error, i.e., the order n in error O(h n ) where h is step size. c. Give a version of symplectic algorithm, if possible. If not possible, explain why not? dx = y, dt a. dy γ y x Ft ( ). dt = + b. the midpoint algorithm is, in general: K = hf(, t Yn ) h K K = hf( tn +, Yn + ) Yn+ = Yn + K + Oh ( ) where errors are of rd order, and vectors are x y Y=, (, txy, ) y F = γ y x + Ft () In component form for this problem, we have h xn+ = xn + h yn + ( γ yn + xn + Ft ( n) h h h yn+ = yn + h γ yn + ( γ yn + xn + Ft ( n) + ( xn + yn) + Ft ( n + ) c. No, since the system cannot be derived from a Hamiltonian due to the damping or time-dependent forces. [It has nothing to do with linear or nonlinear]. - the end - [JSW] 5