* Newton's Method 3 solutions submitted (max 10) View my solutions Implement Newton's method as a function: function (p.n] - myNewton(f.fprime, xo, tol) We seek a root of the anonymous function f given an initial guess xo We input f and its derivative fprime We will stop iterating once successive iterates are within tol of each other and will return the final ferate as our approximation p. Also output the number of iterations n it took to reach p As a class convention, we say that the initial guess counts as 1 iteration. So, in a sense, n = 2 would mean we took one step of Newton's method Debugging tips . If test 3 is the only test passed, and the suggested problem in the "code to call function is close but not quite correct, then it is likely that you used the wrong stopping criteria Stop when successive iterations are within fol of each other . Il tests 1 and 7 are the only tests falled, and the suggested problem in the code to call function is close but not quite correct, then it is likely that you stopped correctly, but did not report the most recent iteration If your code failed to complete with this message: Your code could not run to completion, possibly due to an out-of-memory condition. Common causes are other loading large files or a missing ";" from the end of a command.If the problem persists, contact the instructor, it is possible that you are not using the initial condition XO to start iferating in addition to the suggestion made by Matlab of using ** If you number of iterations n is off by one, try changing your initialization of n and for the location of dating n in the loop. We say that the initial guess is n=1. So, in a sense, n = 2 would mean we took one step of Newton's method Function Reset 10 MATLAB Documentation 1 function [p,n] - myNewton(f,fprime , xe tol) 2 Xinput f. anonymous function for root finding problem 3 Xinput fprime, anonymous function, the derivative of f 4 Sinput x, an initial guess s Xinput tol, a tolerance (method will stop when successive iterates are within tol of each other) 6 Youtput p, a root f(p) - 7 Xoutput n, the number of iterations to reach p Code to call your function e Reset 1 xAs a test, the root of f(x) = x+2** is -0,641185744504986 2 % define f, fprime, an initial guess (use 1), and use a tolderance of le-6 2 % send these into the function and store the result as the variable p 4 % the method will converge in 6 iterations 5 format long x to display more digits Secant method 4 solutions submitted (max 10) | View my solutions Implement the secant method as a function: function (p.n] - secant(f,xe x1, tol) We seek a root of the anonymous function f given initial guesses xe and x1. We will stop iterating once successive iterates are within tol of each other and will return the final iterate as our approximation p and the number of iterations n it took to get there As a class convention we count the two initial guesses as iterations. So, n = 3 is a single implementation of the secant method formula. Debugging tips If test 3 is the only test failed, and the suggested problem in the code to call function is close but not quite correct, then it is likely that you stopped correctly, but did not report the most recent iteration . Il test 6 is the only test failed, and the suggested problem in the code to call function is close but not quite correct, then it is likely that you used the wrong stopping criteria Stop when successive iterations are within tol of each other tests 1-3 passed but tests 4-6 failed, it is possible that you are not using the initial conditions xo and xt to start terating . If you number of iterations is off by one, try chaning your intialization of n and/or the location of when you updaten in the loop. We count the two initial guesses as iterations. Son 3 is a single implementation of the secant method formula Function Reset MATLAB Documentation 1 function (P.n) - secant(f,xe,1, tol) 2 Xinput f, root finding function 3 Xinputs x,x1, Initial guesses 4 Xinput tol, a tolerance (stop once successive Iterates are within tol of each other) 5 Xouptput p, a root f(p) - 6 Zoutput, the number of iterations to reach 3 Code to call your function Reset 1 xAs a test, the root of f(x) = x+2"xts -0.641185744455982 2 X check this with initial guesses of 1 and 2 and a tolerance of le-6 3% save the result as the variable p 4 X it takes Biterations to reach this p format long % to display more digits