- Please Screenshot Your Code And Your Output Thank You So Much It Would Be Your Biggest Gift For Me If You Can Answer A 1 (159.43 KiB) Viewed 165 times
Please screenshot your code and your output. Thank you so much. It would be your biggest gift for me if you can answer a
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answerhappygod
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Please screenshot your code and your output. Thank you so much. It would be your biggest gift for me if you can answer a
Please screenshot your code and your output. Thank you so much. It would be your biggest gift for me if you can answer all these questions since joining answers is really expensive for me.
Please help me answer all these questions. It would be your biggest gift for me if you can answer all these questions since joining answers is really expensive for me. I would really a thumbs up in return. please screenshot your output. Thank you so much The method will start with computing the midpoint of (a, b), call it po, and use Newton's method with initial guess po to obtain p1. It will then check whether P₁ € (a, b). If p₁ € (a, b), then the code will continue using Newton's method to compute the next iteration P2. If p1 # (a, b), then we will not accept p₁ as the next iteration: instead the code will switch to the bisection method, determine which subinterval among (a, po), (po, b) contains the root, updates the interval (a, b) as the subinterval that contains the root, and sets p₁ to the midpoint of this interval. Once P₁ is obtained, the code will check if the stopping criterion is satisfied. If it is satisfied, the code will return p₁ and the iteration number, and terminate. If it is not satisfied, the code will use Newton's method, with p₁ as the initial guess, to compute p2. Then it will check whether p2 € (a, b), and continue in this way. If the code does not terminate after N iterations, output an error message similar to Newton's method. a.) The function y = log x has a root at x = 1. Run the Python code for Newton's method with po = 2, € = 10-4, N = 20, and then try po = 3. Does Newton's method find the root in each case? If Python gives an error message, explain what the error is. b.) One can combine the bisection method and Newton's method to develop a hybrid method that converges for a wider range of starting values po. and has better convergence rate than the bisection method. Write a Python code for a bisection-Newton hybrid method, as described below. (You can use the Python codes for the bisection and Newton's methods from the lecture notes.) Your code will input f, f', a, b, c. N where f, f' are the function and its derivative, (a, b) is an interval that contains the root (i.e., f(a) f(b) < 0), and , N are the tolerance and the maximum number of iterations. The code will use the same stopping criterion used in Newton's method. Apply the hybrid method to: a polynomial with a known root, and check if the method finds the correct root; y = log x with (a, b) = (0,6), for which Newton's method failed in part (a). c.) Do you think in general the hybrid method converges to the root, provided the initial interval (a, b) contains the root, for any starting value po? Explain.
Please help me answer all these questions. It would be your biggest gift for me if you can answer all these questions since joining answers is really expensive for me. I would really a thumbs up in return. please screenshot your output. Thank you so much The method will start with computing the midpoint of (a, b), call it po, and use Newton's method with initial guess po to obtain p1. It will then check whether P₁ € (a, b). If p₁ € (a, b), then the code will continue using Newton's method to compute the next iteration P2. If p1 # (a, b), then we will not accept p₁ as the next iteration: instead the code will switch to the bisection method, determine which subinterval among (a, po), (po, b) contains the root, updates the interval (a, b) as the subinterval that contains the root, and sets p₁ to the midpoint of this interval. Once P₁ is obtained, the code will check if the stopping criterion is satisfied. If it is satisfied, the code will return p₁ and the iteration number, and terminate. If it is not satisfied, the code will use Newton's method, with p₁ as the initial guess, to compute p2. Then it will check whether p2 € (a, b), and continue in this way. If the code does not terminate after N iterations, output an error message similar to Newton's method. a.) The function y = log x has a root at x = 1. Run the Python code for Newton's method with po = 2, € = 10-4, N = 20, and then try po = 3. Does Newton's method find the root in each case? If Python gives an error message, explain what the error is. b.) One can combine the bisection method and Newton's method to develop a hybrid method that converges for a wider range of starting values po. and has better convergence rate than the bisection method. Write a Python code for a bisection-Newton hybrid method, as described below. (You can use the Python codes for the bisection and Newton's methods from the lecture notes.) Your code will input f, f', a, b, c. N where f, f' are the function and its derivative, (a, b) is an interval that contains the root (i.e., f(a) f(b) < 0), and , N are the tolerance and the maximum number of iterations. The code will use the same stopping criterion used in Newton's method. Apply the hybrid method to: a polynomial with a known root, and check if the method finds the correct root; y = log x with (a, b) = (0,6), for which Newton's method failed in part (a). c.) Do you think in general the hybrid method converges to the root, provided the initial interval (a, b) contains the root, for any starting value po? Explain.