Does the Modified Newton Method Converge Quadratically for Specific Functions?

[talolard]

Homework Statement

given $$f \in C^2$$ such that $$f(a)=f'(a)=0 ^f''(a)\neq 0$$ prove that the modified Newton method $$x_{n+1}=x_n-2 \frac{f(x_n){f'(x_n)}$$ coverges with order two.

Homework Equations

if g(x) is an iterative function such that the first m derivatives of g at a are zero and $$g^{(m+1)}\neq 0$$ then the order of convergence is m+2

The Attempt at a Solution

So it seems that i want to show that my iterating function $$g(x)=x-2 \frac{f(x){f'(x)}$$ satisfies $$g(a)=0 ^ g'(a)\neq 0$$
But using le'hospitals rule to find g(a) i have $$g(a)=a-2\frac{f'(a)}{f''(a)}=a \neq 0$$
Whats wrong here?
Thanks
Tal

 

[Mark44]

Fixed your LaTeX. I'm assuming that you were using the symbol ^ to mean "and."

Homework Statement

given $$f \in C^2$$ such that $$f(a)=f'(a)=0 ~\text{and}~ f''(a)\neq 0$$ prove that the modified Newton method $$x_{n+1}=x_n-2 \frac{f(x_n)}{f'(x_n)}$$ coverges with order two.

Homework Equations

if g(x) is an iterative function such that the first m derivatives of g at a are zero and $$g^{(m+1)}\neq 0$$ then the order of convergence is m+2

The Attempt at a Solution

So it seems that i want to show that my iterating function $$g(x)=x-2 \frac{f(x)}{f'(x)}$$ satisfies $$g(a)=0 ~\text{and}~ g'(a)\neq 0$$
But using le'hospitals rule to find g(a) i have $$g(a)=a-2\frac{f'(a)}{f''(a)}=a \neq 0$$
Whats wrong here?
Thanks
Tal

 

[Mark44]

So it seems that i want to show that my iterating function $$g(x)=x-2 \frac{f(x)}{f'(x)}$$ satisfies $$g(a)=0 ~\text{and}~ g'(a)\neq 0$$
But using le'hospitals rule to find g(a) i have $$g(a)=a-2\frac{f'(a)}{f''(a)}=a \neq 0$$
Whats wrong here?

Did you calculate g'(x)? You will need g'(x) so that you can evaluate g'(a). I'm not sure why you think you need L'Hopital's Rule.