Newton's Method for Root Finding - Infinite Loop

[Scootertaj]

1. Construct a function f (x) so that Newton's method gets 'hanging' in an infinite cycle x_n = (-1)ⁿ x0 , no matter how the
start value x₀ is chosen.



2. Homework Equations :
x_n+1 = x_n - f(x_n) / f'(x_n)

The Attempt at a Solution

x_n+1 = x_n - f(x_n) / f'(x_n) = (-1)ⁿ⁺¹x₀ = (-1)ⁿx₀ - f(x_n) / f'(x_n) $\Rightarrow$ f(x_n) / f'(x_n) = 2(-1)ⁿx₀
But, I don't know if that's what I want to do or what to do with it.

Any ideas?

 

[Dick]

If you apply Newtons iteration to any value of x, you want to wind up with -x. Try solving the differential equation -x=x-f(x)/f'(x).

 

[Scootertaj]

Ah, so you would get f(x) / f'(x) = 1/2x $\Rightarrow$ f'(x) = 1/2x * f(x) $\Rightarrow$ f(x) = e^{$\int$ 1/2x}$\Rightarrow$ f(x) = $\sqrt{x}$ ?

 

[Scootertaj]

Dick, you're right that worked perfectly. I had it wrong at first since I had f(x) = e^{-$\int 1/2x$} but we get the double negative and thus get f(x) = e^{$\int 1/2x$}

Thank you!

 

[Dick]

Dick, you're right that worked perfectly. I had it wrong at first since I had f(x) = e^{-$\int 1/2x$} but we get the double negative and thus get f(x) = e^{$\int 1/2x$}

Thank you!

Good! You'll also need to think a little about how to define f(x) for x<0. But drawing a picture should make that pretty clear.

 

[Scootertaj]

Would sqrt(|x|) work?

 

[Dick]

Would sqrt(|x|) work?

Try it. I had been thinking about a different choice, but that works too.

 

[Scootertaj]

Hmm, if you don't mind, what were you thinking of?

 

[Dick]

Hmm, if you don't mind, what were you thinking of?

sqrt(x) for x>=0, -sqrt(-x) for x<0. The graph looks a little 'smoother'. But I don't think there's anything wrong with sqrt(|x|) either.