3667930562 - Solution to Class Test held on 4 March 2009 1....

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Solution to Class Test held on 4 March 2009 1. (a) k B k 1 =max(17 , 21 , 15) = 21 , k B k =max(10 , 6 , 37) = 37 . (b) Using the given value k B 1 k =2 / 7, we have cond ( B, )= k B k k B 1 k =37 × (2 / 7) = 74 / 7 . The system is therefore not ill-conditioned. 2. Let F k = f ( k ) ( a ) h k /k !, ( k =1 , 2 , 3 ,... ). We have Taylor’s series: f ( a + h f ( a )+ 2 F 1 + 4 F 2 + 8 F 3 + 16 F 4 + ··· (1) f ( a +2 h f ( a )+2 F 1 +4 F 2 +8 F 3 +16 F 4 + (2) To Fnd f 0 ( a )intermso f f ( a ), f ( a + h )and f ( a h ), we eliminate F 2 from the above two equations. Using 4(1) (2), we get 4 f ( a + h ) f ( a h )=3 f ( a F 1 4 F 3 + =3 f ( a f 0 ( a ) h 2 f 0 ( ξ ) h 3 / 3 from which we get f 0 ( a 3 f ( a )+4 f ( a + h ) f ( a h ) 2 h + f 0 ( ξ ) h 2 3 . 4. (a) Let f ( x )=5cos x + x . Newton’s method for Fnding the root s of f ( x )=0 is x n +1 = x n 5cos x n + x n 5sin x n +1 . Starting with x 0 ,weget x 1 . 97724 , x 2 . 97738 , x 3 . 97738 . Therefore s . 97738 correct to 5 decimal places. (b) The root ω lies in the interval [ π, 3 π/ 2]. In this interval, the equation x + x = 0 is equivalent to x π cos 1 ( x/ 5). Starting with x 0 = 4, the Fxed-point iteration x n +1 π cos 1 ( x n / 5) gives x 0 =4 . 000 ,x 1 . 785 2 . 854 . 1
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3. The given linear system is: 233 342 324 x 1 x 2 x 3 = 14 13 17 [1] [2] [3] Switch row 1 and row 2: R 2 (2 / 3) R 1 R 3 (1 / 3) R 1 3 42 13 14 17 [2] [1] [3] R 2 (2
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This note was uploaded on 08/27/2010 for the course ME ME4906 taught by Professor Dr.g.p.zheng during the Spring '10 term at NYU Poly.

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3667930562 - Solution to Class Test held on 4 March 2009 1....

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