Quadratic convergence with Newton's method.

Question

Let $g(x)=x^2 \ln(x)$

We have the following equation $g(x)=a$ and let $\alpha(a)$ be the solution for $a>0$

Suppose $a>0$ and $1<\alpha(a)<1+a$.

Show that it converges, except for the first step, to $\alpha(a)$ while using Newton's method, used on $g-a$ started in $x_0=1$

Are you sure this is how the problem is stated? Something seems out of whack. — Amzoti, Jan 08 '14 at 02:25
You are right, I didn't translate correctly. I believe the question is now correct. — DeanTheMachine, Jan 08 '14 at 11:30

score 0 · Accepted Answer · answered Jan 08 '14 at 17:07

Let $\phi(x) = x - \frac{f(x)}{f'(x)}$ and let $\alpha$ be the root we are looking for. The following holds if $f'(\alpha)$ exists and is not equal to $0$ on some interval $I$ and if $f''(\alpha)$ is finite on the same interval $I$. Here $I$ could be something like $[\alpha - y, \alpha + y]$ with $y \geq |\alpha - x_0|$:

$$x_{i+1} - \alpha = \frac{f''(\alpha)}{2f'(\alpha)}(x_i - \alpha)^2$$

From this we can easily conclude that convergence is quadratic.

Quadratic convergence with Newton's method.

1 Answers1