assignment-2-nonlinear-equations-attempt-review

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Assignment 2 - Nonlinear Equations Attempt review Numerical Methods in Engineering (Concordia University) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Assignment 2 - Nonlinear Equations Attempt review Numerical Methods in Engineering (Concordia University) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Downloaded by NightShade Co (nghtshdeco@gmail.com) lOMoARcPSD|33438865

3/13/2021 Assignment 2 - Nonlinear Equations: Attempt review https://moodle.econcordia.com/mod/quiz/review.php?attempt=19502&cmid=655 1/8 Dashboard / My courses / ENGR 391 (Winter 2021) / Assignments / Assignment 2 - Nonlinear Equations Question 1 Complete Not graded Question 2 Correct Mark 1.00 out of 1.00 Started on Tuesday, 26 January 2021, 6:25 PM State Finished Completed on Tuesday, 2 February 2021, 12:00 AM Time taken 6 days 5 hours Grade 15.67 out of 20.00 ( 78 %) Feedback Well done. However make sure to understand well where you did mistakes and review the corresponding lecture topics. By checking the "yes" answer below I confirm 1. that I have neither given nor received unauthorized aid to answer the questions of this assignment. 2. I agree to follow the rules in regard of online assignments as posted in the course outline 3. I used only octave or Matlab to solve the questions (I am allowed to consult all course material and my own notes) Select one: a. Yes I agree b. No I do not agree Your answer is correct. The correct answer is: Yes I agree Consider the function . We apply Newton's method to find a root of this function. If then the value of is given as Select one: a. 7/3 b. 11/4 c. 10/3  d. 5/2 Your answer is correct. Reference: lecture on Newton's method in topic 2 "Solving nonlinear equations" The correct answer is: 10/3 Downloaded by NightShade Co (nghtshdeco@gmail.com) lOMoARcPSD|33438865

3/13/2021 Assignment 2 - Nonlinear Equations: Attempt review https://moodle.econcordia.com/mod/quiz/review.php?attempt=19502&cmid=655 2/8 Question 3 Correct Mark 1.00 out of 1.00 Question 4 Correct Mark 1.00 out of 1.00 Question 5 Correct Mark 1.00 out of 1.00 Solve the equation using the bisection algorithm starting with and . Conduct three iterations (i.e. find ) and report the approximation of the root computed by the algorithm. Use as many digits as possible. Answer: 1.9375  First iteration: and Second iteration: and Third iteration: and Algorithm estimates the root in the third iteration as Reference: lecture on bisection method in topic 2 "Solving nonlinear equations" The correct answer is: 1.9375 The equation has a solution in . You are using Newton's method to estimate this solution. If the absolute error in iteration 4 is about 0.02, how large do you expect to be the absolute error in iteration 5 ? Select one: a. We can not predict this error with the information given b. 0.001  c. 0.01 d. 0.05 Your answer is correct. Newton's method convergences quadraticaly. The asymptotic error constant is given by . This is . Consequently Reference: Lecture on order of convergence form topic 2 "Nonlinear equations" The correct answer is: 0.001 The root of the equation has a multiplicity of Answer: 2  Indeed is a root as . Define . Then . As the root is at least a double root. As and the root is a double root. Reference: lecture on challenging problems in topic 2 "Solving nonlinear equations" The correct answer is: 2 Downloaded by NightShade Co (nghtshdeco@gmail.com) lOMoARcPSD|33438865

3/13/2021 Assignment 2 - Nonlinear Equations: Attempt review https://moodle.econcordia.com/mod/quiz/review.php?attempt=19502&cmid=655 3/8 Question 6 Incorrect Mark 0.00 out of 1.00 Question 7 Correct Mark 1.00 out of 1.00 Question 8 Correct Mark 1.00 out of 1.00 You are implementing on a computer a numerical algorithm to solve an equation of the form . In order to check your algorithm you decide to test it with the function . Select one: a. This is not a good choice for the function to test the programmed algorithm b. This is a valid choice for the function to test the programmed algorithm  Your answer is incorrect. The function f ( x ) = x 3 is flat around the root r = 0 as it is a triple root. Any numerical algorithm will face problems to find this root accurately. Some of them may even behave differently compared to what they should (see the examples with Newton's algorithm covered in class where a root finding problem with multiple roots becomes problematic). It is not a good idea to check an implementation of an algorithm with a challenging problem. Reference: lecture on challenging problems in topic 2 "Solving nonlinear equations" The correct answer is: This is not a good choice for the function \(f(x)\) to test the programmed algorithm You have to solve an equation \(f(x)=0\) with a precision below a defined value \(TOL\). If \(|f(x_r)|<TOL\) then Select one: a. you have found an approximation \(x_r\) to the specified precision b. you have found an approximation \(x_r\) to the specified precision if \(x_r\) is not a multiple root c. you have found an approximation \(x_r\) to the specified precision as long as you use relative errors d. you can't conclude anything  Your answer is correct. The quantity \(|f(x_r)|\), which is the backward error, doesn't tell you anything about \(|r-x_r|\) which is the error you want to have below the specified tolerance \(TOL\). Reference: lecture on backward and forward errors in topic 2 "Solving nonlinear equations". The correct answer is: you can't conclude anything You are using the bisection method to solve an equation with an initial guess \(a_0=0.5\) and \(b_0=2.5\). It is required to find the approximation of the root with an absolute error less than 0.2. How many iterations do you have to do at least in order to reach the set precision? Answer: 3  You need at least four iterations. Indeed the initial error is about \(\frac{2.5-0.5}{2}=1.0\). After one iteration the error will be \(\frac{1.0}{2}\) which is larger than 0.2 After two iterations the error will be \(\frac{1.0}{2^2}\) which is larger than 0.2 After three iterations you have an error of \(\frac{1.0}{2^3}\) which is less than 0.2. Alternatively you can use the formula of the lecture: \( n \geq \frac{\log(2.5-0.5)-\log 0.2 }{\log 2}-1 \simeq 2.3 \) You find again that at least three iterations will be needed. Reference: lecture on bisection method (error control) in topic 2 "Solving nonlinear equations" The correct answer is: 3 Downloaded by NightShade Co (nghtshdeco@gmail.com) lOMoARcPSD|33438865

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