# Exam 2 - Exam 2 Solution Accuracy of Linear Systems Curve...

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Unformatted text preview: Exam 2: Solution Accuracy of Linear Systems; Curve Fitting MATH 3315 / CSE 3365 : 801C – Spring Semester 2007 Total points: 100 + 10 (extra credits) Thursday 22 March The SMU honor code applies. Don’t forget to write your name and sign. An answer should include necessary steps, unless it requires only one step. Question 1 (21 points) Consider the following linear system (let’s call it the linear system I): 1000 x 1 + 999 x 2 = 1999 999 x 1 + 998 x 2 = 1997 (1) Write down the expressions of the residuals r 1 and r 2 for a solution x 1 , x 2 . (4 points) Answer: r 1 = 1999- (1000 x 1 + 999 x 2 ) = 1999- 1000 x 1- 999 x 2 r 2 = 1997- (999 x 1 + 998 x 2 ) = 1997- 999 x 1- 998 x 2 (2) Calculate the residuals r 1 and r 2 for a solution x 1 = 1, x 2 = 1 of this linear system. (2 points); Is this solution the exact solution? (2 points) Answer: r 1 = 1999- 1000 × 1- 999 × 1 = 0 r 2 = 1997- 999 × 1- 998 × 1 = 0 This solution is exact. (3) It can be shown that a computed solution x 1 = 20 . 97, x 2 =- 18 . 99 gives residuals r 1 = 0 . 01, r 2 =- . 01, which are of small magnitudes. Can we conclude that this solution must be accurate? (2 points) Answer: No. 1 (4) In (2), we know the solution x 1 = 20 . 97, x 2 =- 18 . 99 gives residuals r 1 = 0 . 01, r 2 =- . 01. If we view the residuals as perturbations of the right hand side of the linear system I, then we know x 1 = 20 . 97, x 2 =- 18 . 99 is the exact solution of the following linear system (let’s call it the linear system II). 1000 x 1 + 999 x 2 = b 1 999 x 1 + 998 x 2 = b 2 Treat the residuals as perturbations of the right hand side of the linear system I to obtain the values of b 1 and b 2 . You should be able to do so without a calculator. (4 points) Answer: r 1 = 1999- (1000 x 1 + 999 x 2 ) = 1999- b 1 = 0 . 01 ⇒ b 1 = 1999- . 01 = 1998 . 99 r 2 = 1997- (999 x 1 + 998 x 2 ) = 1997- b 2 =- . 01 ⇒ b 2 = 1997 + 0 . 01 = 1997 . 01 (5) By comparing the linear systems I and II and by comparing their exact solutions, can we conclude that the linear system I is ill-conditioned? (2 points); Answer: Yes. (6) Can we also conclude that the linear system II is ill-conditioned? (2 points) Why? (3 points) Answer: Yes, the change from the linear system II to the linear system I is very small (only very small change to the right hand side), but the change in the solution is very large (from x 1 = 20 . 97, x 2 =- 18 . 99 to x 1 = 1, x 2 = 1). Question 2 (15 points) (1) Find the power series form of the interpolating polynomial p 2 ( x ) to the data (1 , 2), (2 , 3), (3 , 6). (12 points) noindent Answer: We are given the data (1 , 2) , (2 , 3) , (3 , 6). The quadratic interpolating polynomial written in power series form is p 2 ( x ) = a + a 1 x + a 2 x 2 So, the interpolating conditions are p 2 (1) = a + a 1 (1) + a 2 (1 2 ) = 2 p 2 (2) = a + a 1 (2) + a 2 (2 2 ) = 3 p 2 (3) = a + a 1 (3) + a 2 (3 2 ) = 6 The solution to these three linear equations in three unknowns is a = 3, a 1 =- 2, a 2 = 1. So the interpolating polynomial is= 1....
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## This test prep was uploaded on 04/07/2008 for the course MATH 3337 taught by Professor Xu during the Fall '07 term at Southern Methodist.

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Exam 2 - Exam 2 Solution Accuracy of Linear Systems Curve...

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