f(x) = ((x 1)^2) e^x

Apply the Newton iteration formula to f(x) using the following code. Use the default convergence tolerance of 10^6 and record the iterations starting from an initial guess of x0 = 2.

Code is as follows:-

function [root, iter, xlist] = newton( func, pfunc, xguess, tol, mult ) %NEWTON Newton's method for solving a nonlinear equation. % % Sample usage: % [root, niter, xlist] = newton( func, pfunc, xguess, [tol], [mult] ) % % Input: % func - function to be solved % pfunc - derivative of 'func' % xguess - initial guess at root % tol - convergence tolerance (OPTIONAL, defaults to 1e-6) % mult - factor for multiple roots (OPTIONAL, defaults to 1.0) % % Output: % root - final estimate of the root % niter - number of iterations until converged % xlist - list of iterates, an array of length 'niter' % First, do some error checking on parameters. if nargin < 3 fprintf(1, 'NEWTON: must be called with at least three arguments' ); error( 'Usage: [root, niter, xlist] = newton( func, pfunc, xguess, [tol], [mult] )' ); end if nargin < 4, tol = 1e-6; end if nargin < 5, mult = 1.0; end % fcnchk(...) converts function parameters to the correct type % to allow evaluation by feval(). func = fcnchk( func ); pfunc= fcnchk( pfunc ); x = xguess; fx = feval( func, x ); fpx = feval( pfunc, x ); if( fx == 0 | fpx == 0 ) error( 'NEWTON: both f and f'' must be non-zero at the initial guess' ); end xlist= [ x ]; done = 0; iter = 0; while( ~done ) x0 = x; x = x0 - fx / fpx * mult; fx = feval( func, x ); fpx = feval( pfunc, x ); if( abs(x-x0) < tol ) % absolute tolerance on x done = 1; else xlist = [ xlist; x ]; % add to the list of x-values iter = iter + 1; end end root = x; %END newton.