Suppose that f:\mathbb R \to \mathbb R is continuously differentiable and that x_0 is a simple root of f. (This just means that f(x_0)=0 but f'(x_0)\neq0. Show that x_0 is a super-attractive fixed point of the Newton’s method iteration function for f.
Given that Newton’s Method is x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}, by the definitions it’s clear that all x_n = x_0. Heuristically, since we’ll never leave x_0, this fixed point is super-attractive. This is demonstrated by the fact that x_0 is a constant, so the derivative is 0, thus Newton’s method is super-attractive for x_0.
I’m not sure I follow. Can you state precisely what properties of f and x_0 you are using here?
By that argument, every fixed point is super-attractive. I think we’ll need to apply the definition of super-attractive here, which will involve the derivative N' of the Newton’s method iteration function N.
Here I’m using the fact that f(x_0) = 0 and f'(x_0) \not = 0. As Newton’s is x_0 - \frac{f(x_0)}{f'(x_0)}, the top function is zero, so x_1 = x_0 - 0. Continuing this iteration, if my understanding is correct, it seems that all x_n=x_0.
If N(x) is our Newton’s iteration function, but all our iterates are x_0, then N'(x) should be the derivative of a constant, and thus zero, super-attractive. Is my logic faulty here?
Yes, I agree with this. I just think it should be made explicit in the solution.
Yes, there is some faulty logic here. I still maintain that this logic implies that every fixed point is super-attractive.
Let’s look at a very simple example, say f(x)=x^2. This function has two fixed points:
If we start the iteration from either of those points, then the resulting orbit will be constant in both cases. The sequence will be constant with x_n=0 in the first case and constant with x_n=1 in the second case.
In reality, though, only x_0=0 is a super-attractive fixed point. This computation is based on the value of the derivative of f at the fixed point. We have f'(x)=2x, thus: