{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

3667930562

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

This preview shows pages 1–2. Sign up to view the full content.

Solution to Class Test held on 4 March 2009 1. (a) B 1 = max(17 , 21 , 15) = 21 , B = max(10 , 6 , 37) = 37 . (b) Using the given value B 1 = 2 / 7, we have cond ( B, ) = B B 1 = 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 find f ( a ) in terms of f ( a ), f ( a + h ) and f ( a + 2 h ), we eliminate F 2 from the above two equations. Using 4(1) (2), we get 4 f ( a + h ) f ( a + 2 h ) = 3 f ( a ) + 2 F 1 4 F 3 + · · · = 3 f ( a ) + 2 f ( a ) h 2 f ( ξ ) h 3 / 3 from which we get f ( a ) = 3 f ( a ) + 4 f ( a + h ) f ( a + 2 h ) 2 h + f ( ξ ) h 2 3 . 4. (a) Let f ( x ) = 5 cos x + x . Newton’s method for finding the root s of f ( x ) = 0 is x n +1 = x n 5 cos x n + x n 5 sin x n + 1 . Starting with x 0 = 2, we get x 1 = 1 . 97724 , x 2 = 1 . 97738 , x 3 = 1 . 97738 . Therefore s = 1 . 97738 correct to 5 decimal places. (b) The root ω lies in the interval [ π, 3 π/ 2]. In this interval, the equation 5 cos x + x = 0 is equivalent to x = 2 π cos 1 ( x/ 5). Starting with x 0 = 4, the fixed-point iteration x n +1 = 2 π cos 1 ( x n / 5) gives x 0 = 4 . 000 , x 1 = 3 . 785 , x 2 = 3 . 854 . 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
3. The given linear system is: 2 3 3 3 4 2 3 2 4 x 1 x 2 x 3 = 14 13 17 [1] [2] [3] Switch row 1 and row 2: R 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}