Can someone look over my newton's method script and see if it looks ok?

for F(X)=x^3-2x and F'(x)=3x^2-2 with initial guess of x=1

if true
  % code
endx = 1;
Tol = 0.0000001;
count = 0;
dx=1;   
f=-1;    
fprintf('step      x           dx           f(x)\n')
fprintf('----  -----------  ---------    ----------\n')
fprintf('%3i %12.8f %12.8f %12.8f\n',count,x,dx,f)
xVec=x;fVec=f;
while (dx > Tol || abs(f)>Tol) 
  count = count + 1;
  fprime = 3*x^2 - 2;
  xnew = x - (f/fprime);   
  dx=abs(x-xnew);          
  x = xnew;
  f = x^3 - 2*x;      
  fprintf('%3i %12.8f %12.8f %12.8f\n',count,x,dx,f)
end

Answer 1

[John D'Errico]

Several obvious things.

First of all, you never evaluate f to start the method out. Yes, since you know the function f AND the starting value, f(1) = -1, so your code looks like it will work since you hard coded f initially as -1, but it is poor coding style anyway.

Second, I strongly suggest you learn not to hard code in your functions, like f(x) and f'(x). Learn to set up such a function up front, so you could use your code to solve a more general problem. For example, you could set up function handles up front...

f_x = @(x) x.^3 - 2*x;
fp_x = @(x) 3*x.^2 - 2;

I'd also suggest putting in a test for a MaxFunEvals or MaxIterations to prevent some problems. You also use the same Tol parameter for both a test on dx and on abs(f). While that might work on some problems, it is a bad idea in general.

[Pam]

Oh ok thank you for the advice I'm still new to all this. Just as it is know though do u think it will work?

[John D'Errico]

It looks like you missed one other thing to make it work. You never defined an initial value for x. Ah, but as I write this, you have defined a variable endx, which I realize is a paste issue.

With that, I ran your code. The function you have has several roots. We can solve for them using roots, a numerical solver...

roots([1 0 -2 0])
ans =
            0
       1.4142
      -1.4142

Or the symbolic way, as...

syms x
solve(x^3 - 2*x)
ans =
        0
  2^(1/2)
 -2^(1/2)

In either case, the roots are 0 and +/-sqrt(2).

Running your Newton's method code, it yields a nice approximation to sqrt(2), one of the roots. As expected, it is quadratically convergent near the solution.

step      x           dx           f(x)
----  -----------  ---------    ----------
  0   1.00000000   1.00000000  -1.00000000
  1   2.00000000   1.00000000   4.00000000
  2   1.60000000   0.40000000   0.89600000
  3   1.44225352   0.15774648   0.11551761
  4   1.41501064   0.02724288   0.00319099
  5   1.41421424   0.00079640   0.00000269
  6   1.41421356   0.00000067   0.00000000
  7   1.41421356   0.00000000   0.00000000

So your code worked sucessfully as is.

[Pam]

ok thank you so much for the feedback