This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Exercise 3 – Fix step iteration (4 marks) Consider following equation: (1) 1 3 2 = +x x and following two functions: (2) ( 29 ( 29 1 3 1 2 1 + = x x g and ( 29 x x g 1 3 2= a) Prove that (1) is equivalent to ( 29 x x g = 1 and as well equivalent to ( 29 x x g = 2 b) Plot the functions ( 29 x g 1 and ( 29 x g 2 for x between 0 and 5. x x y y c) Considering your figures from b), which one of the two functions ( 29 x g 1 and ( 29 x g 2 could you use to solve equation (1) using the fix point iteration method starting with 5 . 1 = o x ? Justify your answer by showing your iterations on your plots from figure b). d) For the function that you have chosen in c), solve equation (1) using the fix point iteration until you reach an answer with 4 significant digits ( 5 . 1 = o x ). Present your calculations in a table like : i i x ( 29 1 + i x g Error estimation 1.5 Answer:...
View
Full Document
 Spring '09
 HOIDICK
 Industrial Engineering, Numerical Analysis, XO, 1.5 g, Department of Mechanical and Industrial Engineering, fix point iteration

Click to edit the document details