Writting Code for newtons method

The professor asked us the following problems:

PROBLEM1. The function f(x)=sin(x) has a zero on the interval (3,4), namely, x*=pi. Perform three iterations of newton mathod to aproxximate this zero, using x1=4. Determine the absolute error in each of the computed approximations. What is the apparent order of convergence?

PROBLEM2. Apply the Newton Method to find the solution to (x^3)-x-3=0 starting with x1=0. COmpute x2, x3, x4,x5,x6,x7 and x8 compare numbers (x1,x5), (x2,x6), (x3,x7), (x4,x8). What can you conclude from this computations (use your computer code)?

he have us the following code:

% Sept/2016  
%  
% Solve f(x) = 0 using bisection method. 
% 
% The function is defined in func(x);  
% Input 
tol = 1.e-10; 
a = 1.0;  
b = 2.0; 
nmax = 100; 
% Initialization  
itcount = 0; 
error = 1.0; 
% Graph of the function 
xval = linspace(a,b,100); 
for i=1:100 
fval(i) = func(xval(i)); 
end plot(xval,fval); 
grid on;  
hold on; 
% iteration begins here 
while (itcount <= nmax && error >= tol) 
itcount = itcount + 1; 
% Generate and save iteratres 
x = a + (b-a)/2; 
z(itcount) = x; 
fa = func(a);  
fb = func(b); 
fx = func(x);  
error = abs(fx); 
%  error = abs(x - xold); 
if (error < tol) 
x_final = x; 
else 
if (fa*fx < 0) 
% root is between a and x 
b = x; 
else 
% root is between x and b 
a = x; 
end 
end 
plot(z(1:itcount),zeros(itcount,1),'r+'); 
pause(5)  
end 
if (itcount < nmax); 
val = func(x); 
fprintf(1,'Converged solution after %5d iterations',itcount); 
fprintf(1,' is %15.7e, %e \n',x_final, val); 
else 
fprintf(1,'Not converged after %5d iterations',nmax); 
end 
function val = func(x) 
%val = x^3 + 4 * x^2 - 10; 
val = x^3 - x - 3; 
%val = sin(x);  
end 

[John D'Errico]

Odd that you are asked how to write Newton's method, yet you show us code to do bisection.

[Jim Riggs]

I'm guessing that he wants you to compare the bisection method with Newton's method.

E.g. which one converges faster?

[Rik]

@Emmanuel: The point is: it is unclear what your question is and what you have tried so far on your own to solve this homework question.

Answer 1

[David Hill]

f=@sin;
fp=@cos;
function x = newtonMethod(f,fp,x,i)
for j=1:i
    er=-f(x)/fp(x);
    x=x+er;
end
end
x=newtonMethod(f,fp,4,3);

Then just change functions and call the newtonMethod again.

f=@(x)x^3-x-3;
fp=@(x)3*x^2-1;
x=newtonMethod(f,fp,0,8)