Question

Problem 1 (Matlab): One of the most fundamental root finding algorithms is Newton's Method. Given a real-valued, differentiable function f, Newton's method is given by 1. Initialization: Pick a point xo which is near the root of f Iteratively define points rn+1 for n = 0,1,2,..., by 2. Iteration: f(xn) nt1 In 3. Termination: Stop when some stopping criterion occurs said in the literature). For the purposes of this problem, the stopping criterion will be 100 iterations (This sounds vague, but is often what is Under mild conditions, choice of xo is made well. Write a MATLAB script which performs Newton's method for the polynomial one can show that Newton's method will converge to a root of f, provided that the function 5л + 6 %3 (а — 3) (a — 1)(г + 2) f(x) 322 where the initial point xo can when o-0.3396, To = easily be specified. Then find the roots which Newton's Method converges to -0.3386, To-0.3376, and TO = -0.3366

Answer 1

les MATLAB R2016a bin Command Window Editor Newtor Newton.m 1 x0-input ('Enter the initial point: '); A n 0; % Number of iterations 2 while n<-100 3 - f-x0 3-2*x0 2-5*x0+6 4 fd 3*x0^ 2-4*x0-5; derivative of f xx0-(f/fd) x0 x; 7 n-n+1; end , x) fprintf('The required root of f is: %f\n 10

Files MATLAB R2016a bin Command Window Editor Newton.m >>Newton Enter the initial1 point: -0.3396 The required root of f is:3.000000 >>Newton Enter the initial1 point:-0.3386 The required root of f is:-2.000000 >>Newton Enter the initial1 point:-0.3376 The required root of f is: 3.000000 >>Newton Enter the initial point:-0.3366 The required root of f is: 1.000000