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Unsw Maths2089 June 2011 Exam Paper

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FAMILY NAME: . . . . . . . . . . . . . . . . . . . . . . . . . . . . OTHER NAME(S):. . . . . . . . . . . . . . . . . . . . . . . . . . STUDENT NUMBER: . . . . . . . . . . . . . . . . . . . . . . SIGNATURE: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . THE UNIVERSITY OF NEW SOUTH WALES SCHOOL OF MATHEMATICS AND STATISTICS June 2011 MATH2089 Numerical Methods and Statistics (1) TIME ALLOWED – 3 Hours (2) TOTAL NUMBER OF QUESTIONS – 6 (3) ANSWER ALL QUESTIONS (4) THE QUESTIONS ARE OF EQUAL VALUE (5) THIS PAPER MAY NOT BE RETAINED BY THE CANDIDATE (6) ONLY CALCULATORS WITH AN AFFIXED “UNSW APPROVED” STICKER MAY BE USED (7) STATISTICAL FORMULAE ARE ATTACHED AT END OF PAPER STATISTICAL TABLES ARE ATTACHED AT END OF …show more content…

ii) Find an initial value problem (IVP) satisfied by y. Please see over . . . June 2011 MATH2089 Page 4 c) When using N equal intervals of width h, the error in estimating the integral τ f (t)dt 0 by the Trapezoidal rule is Etrap = O(h2 ) and by Simpson’s rule is ESimp = O(h4 ). i) For each of these numerical integration rules, what conditions are required on the integrand f so these error estimates are valid? ii) Suppose that the error using h = 5 × 10−3 is E0 = 1.19 × 10−4 when using either the Trapezoidal rule or Simpson’s rule. For both rules, ˆ estimate the error if an interval width of h = 1 × 10−3 is used. iii) The Matlab command [z, w] = gauleg(N); calculates the N Gauss-Legendre nodes z and weights w for the interval [−1, 1]. Show how z and w can be used to numerically calculate τ f (t)dt. 0 Please see over . . . June 2011 3. MATH2089 Page 5 Answer in a separate book marked Question 3 Fick’s second law states how the concentration of a chemical changes with time because of diffusion. Let c(x, y) denote the concentration at position (x, y) ∈ Ω. The steady state version of Fick’s second law (without interior sources of the chemical) is Laplace’s equation ∂2c ∂2c + 2 = 0. ∂x2 ∂y Consider a problem with domain Ω = {(x, y) : 0 ≤ x ≤ 1, xi = ih, i = 0, . . . , n 0 ≤ y ≤ 1.5} . j = 0, . . . , (3.1) Ω is discretised using h = 1/n (where n is even) and 3n . 2 The domain is illustrated in Figure 3.1 for n = 6. The

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