CMPSC/Mathematics 455Assignement FiveDue 24 October 2022Fall 2022You are to write a MATLAB script or function based upon problem 16, p.62 of the Ascher and Grief textbook.In that problem,...

Use Matlab to solve this assignment.


CMPSC/Mathematics 455 Assignement Five Due 24 October 2022 Fall 2022 You are to write a MATLAB script or function based upon problem 16, p. 62 of the Ascher and Grief textbook. In that problem, it is pointed out that f(x;α) = α cosh(x/4)− x = 0 (1) has two roots for α = 2 and no roots for α = 10. The question asks whether there is a value of α for which there is only one root. The answer is yes, and that root is a double root. That is, there is a value α∗ where there is a root x∗ such that f(x∗;α) = 0, f ′(x∗;α∗) = 0. The differentiation is with respect to x not α. We can solve for x∗ and α∗ by setting up a nonlinear equation in two variables z1 = x and z2 = α so we let z = ( z1 z2 ) = ( x α ) F(z) = ( F1(z) F2(z) ) = ( f(z1; z2) f ′(z1; z2) ) and then solve F(z) = 0 (2) using Newton’s method. You will try two separate initial guesses z(0) for the root z = ( x∗ α∗ ) of equation (2). They are z(0) = ( 2.5006 6 ) z(0) = ( 4.3944 3 ) You should get approximately the same answer for both initial guesses. 1 With tol = 10−14 = 1e − 14, use Newton’s method to produce the sequence of iterates {z(k)} until either ∥z(k−1) − z(k)∥∞ ≤ tol ∗max{∥z(k)∥∞, tol} or k = 20. You should only have to store the two most recent iterates. For each of the two initial guesses produce your final values of x∗, α∗, and the number of iteations k. Of course, here z(k) = ( x∗ α∗ ) . I strongly recommend producing a function (I will call it prob16 fun) that has the first line function [F, J ] = prob16 fun(z) where F is the 2-vector F(z), the function value at the point z and J is the 2 × 2 Jacobian matrix for F evaluated at z. (Yes, part of the problem is figuring out what the Jacobian is.) You can then build the script around this function. Your program with all scripts, functions, and output should be turned in on Canvas by midnight on the due date. 2