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This question is exactly the same as the previous one-you can use the same answer! There is a point, however. Please submit your answer and then inspect the test column of the result table. Compare it with the table from the previous question. Can you see the benefit? Write a function called newton_root_finderif, f_prise, xe, tolerance) that takes as parameters: • A function f(x), the root of which is required. • A function f_prime(x), which is the derivative of f(x). An estimate of the value of the root. • The maximum acceptable absolute value of f(x) for x to be considered a valid root. The function should apply the same algorithm as used for the square root exercises and return the first value in the sequence Jo.... that satisfies the required tolerance. For example:

This question is exactly the same as the previous one-you can use the same answer! There is a point, however. Please submit your answer and then inspect the test column of the result table. Compare it with the table from the previous question. Can you see the benefit? Write a function called newton_root_finderif, f_prise, xe, tolerance) that takes as parameters: • A function f(x), the root of which is required. • A function f_prime(x), which is the derivative of f(x). An estimate of the value of the root. • The maximum acceptable absolute value of f(x) for x to be considered a valid root. The function should apply the same algorithm as used for the square root exercises and return the first value in the sequence Jo.... that satisfies the required tolerance. For example:

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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This question is exactly the same as the previous one-you can use the same answer! There is a point, however. Please submit your answer and then inspect the test column of the result table. Compare it with the table from the previous question. Can you see the benefit? Write a function called newton_root_finder(f. f prine, xe, tolerance) that takes as parameters: A function f(x), the root of which is required. A function_prime(x), which is the derivative of f(x). An estimate xe of the value of the root. • The maximum acceptable absolute value of f(x) for x to be considered a valid root. The function should apply the same algorithm as used for the square root exercises and return the first value in the sequence Jo.....that satisfies the required tolerance. For example: Test Result 1.2247452 This first test uses the general root finder to determine the square root of 1.5. This test is exactly equivalent to one used in an earlier question. ans newton_root_finder (lambda x:xx-1.5, lambda x: 2 x, 10, 1.00-06) print(f"(ans:0.713") A trivial solution of a straight line equation 10.0000000 ans newton_root_finder (lambda x: x 10, lambda x: 1, 10000, 0.01) print (f"(ans:0.71)") Answer: (penalty regime: 0, 10, 20,... %)
Transcribed Image Text:This question is exactly the same as the previous one-you can use the same answer! There is a point, however. Please submit your answer and then inspect the test column of the result table. Compare it with the table from the previous question. Can you see the benefit? Write a function called newton_root_finder(f. f prine, xe, tolerance) that takes as parameters: A function f(x), the root of which is required. A function_prime(x), which is the derivative of f(x). An estimate xe of the value of the root. • The maximum acceptable absolute value of f(x) for x to be considered a valid root. The function should apply the same algorithm as used for the square root exercises and return the first value in the sequence Jo.....that satisfies the required tolerance. For example: Test Result 1.2247452 This first test uses the general root finder to determine the square root of 1.5. This test is exactly equivalent to one used in an earlier question. ans newton_root_finder (lambda x:xx-1.5, lambda x: 2 x, 10, 1.00-06) print(f"(ans:0.713") A trivial solution of a straight line equation 10.0000000 ans newton_root_finder (lambda x: x 10, lambda x: 1, 10000, 0.01) print (f"(ans:0.71)") Answer: (penalty regime: 0, 10, 20,... %)
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