Please complete in MATLAB:

The Newton's Method code is the following:

function x = newton(f,df,x0,tol,kmax);

% function x = newton(f,df,x0,tol,kmax);

% Given a differentiable function f with df = df/dx, routine,

% when convergent returns an approximate root x obtained via

% Newton-Raphson iteration. Other inputs are an error tolerance

% tol (max over error between successive iterations and abs(f)),

% the max number kmax allowed iterations, and initial iteration

% x0. tol=1e-8 and kmax = 1e5 are defaults, if left unspecified.

switch nargin

case 3,

tol = 1e-8

kmax = 1e5;

case 4,

kmax = 1e5;

case 5,

% Fall through.

otherwise,

error('newton called with incorrect number of arguments')

end

x = x0; err=100; k = 0; % x=x0 here allows return for large tol.

while err >= tol

y = f(x0);

x = x0 - y/df(x0);

err = max(abs(y),abs(x-x0));

x0 = x;

k = k+1;

if k>=kmax

disp(['No convergence after ' num2str(kmax) ' iterations.'])

disp(['Error at this stage is ' num2str(err)])

return

end

end

if isfinite(x)

disp(['Converged in ' num2str(k) ' iterations with tol ' num2str(tol)])

else

disp(['No convergence after ' num2str(k) ' iterations. Iterations' ...

' overflowed to infinity.']);

end

4. Apply Newton's method to find the three roots of the function f(x)=esin3x+x62x4x31 on the interval [2,2]. For each root, print out the sequence of iterates, the errors ei, and the relevant error ratio, ei+1/ei2 or ei+1/ei, that converges to a nonzero limit. Match the limit with the expected value of M=f(r)/(2f(r)) or S=(m1)/m, where m is the order of a multiple root. To measure errors, treat the last iterate as the exact root. Please make sure all tables are well-formatted with numbers reported to enough precision. See page 6 of the lecture root3 for details. Note: for a root of multiplicity Newton's method does not converge quadratically, rather ek+1Sek, with S as above. For example, consider f(x)=(x1)4. Then xf(x)/f(x)=x(x1)4/(4(x1)3)=x41(x1). So Newton's method is xk+1=xk41(xk1). This implies xk+11=43(xk1) and so ek+1=43ek. Linear convergence