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Information about math 417: numerical analysis course at the university of california, berkeley, focusing on problem 1-4 related to newton's method. Students will explore the convergence properties of newton's method for finding roots of functions, including the conditions for a single-step convergence and slow convergence. They will also write a program to implement newton's method and analyze its convergence order and error constants.
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Instructors: Prof. Wolfgang Bangerth, Prof. Guido Kanschat [email protected], [email protected] Teaching Assistants: Seungil Kim, Yan Li [email protected], [email protected]
Problem 1 (Newton’s method). For certain functions, Newton’s method will always converge in a single step, no matter where we start. What functions are these, and why is a single step enough? (Hint: think about the graphical interpretation of Newton’s method, and when it will produce a new iteration that falls exactly onto the true root of the function.) (2 points)
Problem 2 (Newton’s method). For functions f (x) of one variable x, New- ton’s method almost always converges very quickly (in a matter of a few itera- tions). However, almost always is not always, and we can find examples where Newton’s method converges rather slowly. Write a program to find the zero x = 1 of the function f (x) = x^25 − 1
that uses Newton’s method and starts at x 0 = 20.
(a) How many iterations do you need to achieve an accuracy of 10−^8? (b) You will observe very slow convergence. Can you explain from the formulas that express the error en as a function of en− 1 why convergence is so slow? (c) Does the method still converge of second order? (6 points)
Problem 3 (Newton’s method for convex functions). Prove that New- ton’s method converges from any point p 0 with f ′(p 0 ) 6 = 0 to a root of the function f , provided that f has at least one zero, and is globally convex, i.e. f ′′(x) > 0 for all x. (4 points)
Problem 4 (Convergence order). Determine the order of convergence and the asymptotic error constant for the following sequences:
(a) an = 5. 0625 , 2. 25 , 1 , 49 , (^1681) (b) bn = 2. 718 , 2. 175 , 1. 740 , 1. 392 , 1. 113 , 0. 8907 (c) cn = 0. 318 , 0. 180 , 0. 0761 , 0. 021 , 3. 04 · 10 −^3 , 1. 68 · 10 −^4 , 2. 17 · 10 −^6. (3 points)