Complete the following steps to show that newtons

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Unformatted text preview: t take to get the estimate for c with an error of at most 10−6 . ANSWER: NUMBER OF ITERATES: h) Using x1 = −1 as your initial estimate, edit the loop provided on page (x) in this lab for Newton’s Method and enter it in Maple, using 10 iterations to estimate the value for c. ANSWER: ESTIMATE FOR C: i) How many iterations of Newton’s Method did it take for your estimate to be accurate to 6 decimal places? ANSWER: j) Which of the two methods, Bisection Method or Newton’s Method, is the most efficient? Explain your answer. ANSWER: xi Copyrighted by B.A. Forrest ([email protected]) Newton’s Method in Maple 3. Complete the following steps to show that Newton’s Method can fail if the choice of the initial estimate for c is poor. In this question, we will use the function f (x) = arctan(x) to show the importance of the initial choice for x1 in Newton’s Method. We know that the only solution to arctan(x) = 0 is x = 0 (look at the graph of arctan(x) to convince yourself!). We will show that if x1 = 0.5, then the Newton’s Method iterates converge quickly to the root c = 0. However, if x1 = 2.0, then Newton’s Method fails to find an accurate approximation for the root c. a) In Maple, create a text region and enter the following sentence: When Newton’s Method Fails. Enter Maple’s restart: command. Insert a new execution group (command prompt). b) In Maple, create the function f (x) = arctan(x). HINT: Use arctan(x) in Maple. c) Plot the graph of f (x) using the ranges x = −10..10. Notice the only root at (0, 0). d) Using x1 = 0.5 as your initial estimate, edit the loop provided in this lab for Newton’s Method and enter it in Maple, using 10 iterations to estimate the value for c. ANSWER: ESTIMATE FOR C: e) Using x1 = 2.0 as your initial estimate, edit the loop provided in this lab for Newton’s Method and enter it in Maple, using 10 iterations to estimate the value for c. ANSWER: ESTIMATE FOR C: f ) Study the output of the iterates for Newton’s Method when you used x1 = 2.0. What is happening to the successive estimates for c when you used x1 = 2.0? ANSWER: ESTIMATE FOR C: g) By studying the graph for f (x) = arctan(x), explain why Newton’s Method failed when you used x1 = 2.0. HINT: What happens to the slope of the tangent lines to arctan(x) as x gets large? ANSWER: 4. Save and print the worksheet for lab 7. Save the Maple worksheet as lab7.mws and print a copy. xii...
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This note was uploaded on 01/09/2013 for the course MATH 127 taught by Professor Prof.smith during the Winter '09 term at Waterloo.

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