exam1s - 1 2 ( x 2-3 x + 2), L 1 ( x ) =-( x 2-2 x ), L 2 (...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Numerical Analysis/Exam #1/October 15, 2009 SOLUTIONS 1. (a) f (0) = - 3 < 0, f (2) = 5 > 0, so there is a root in (0 , 2). (b) f 0 ( x ) = 3 x 2 0 on (0 , 2), so f is increasing and root is unique. (c) a b m 0 2 1 1 2 1.5 1 1.5 1.25 1.25 1.5 1.375 1.375 1.5 1.4375 (d) ln 2 × 10 5 ln 2 17 . 6: at least 18 iterations required. (e) The root cannot be bracketed since g ( x ) 0 for all x . 2. (a) 1.000000 1.666667 1.471111 1.442812 1.442250 1.442250 (b) 0: multiple root, linear convergence; 1: simple root, quadratic con- vergence. 3. (a) L 0 ( x ) =
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 2 ( x 2-3 x + 2), L 1 ( x ) =-( x 2-2 x ), L 2 ( x ) = 1 2 ( x 2-x ), L ( x ) = 2 x 2 + 3. (b) f ( x ) L ( x ) = 4 x , f (1 . 8) L (1 . 8) = 7 . 2. 4. (a) 11 10 1 . 1 (b) 10 7 180 1 / 4 15 . 35: n 16. 5. x f ( x ) f ( x ) 2 . 3 . 00 . 05 2 . 2 3 . 02 . 15 2 . 4 3 . 06 . 25 2 . 6 3 . 12 . 35 6. 41 8 29 4 41 8 6 17 2 6 7 10 7 8 12 8 8 16 8...
View Full Document

This note was uploaded on 10/10/2010 for the course MATH 01:640:111 taught by Professor Carey during the Spring '10 term at Saint Louis.

Ask a homework question - tutors are online