[Math] Newton’s Method – Why is there slow convergence with a high multiplicity

Question

I'm using a calculator to observe the sluggishness with which Newton's method converges with $f(x) = (x-1)^8$. I let $x_0 = 1.1$. Clearly it's taking forever to get to the root $x=0$. I'm not completely sure why this is. I know it has to deal with the multiplicity of 8.

I'm looking to reconcile this slow convergence with the theory. Thanks for any help 🙂

Answer 1

Newton's method converges quadratically to the root for any initial approximation provided the root is a simple zero.

For roots that are not simple (higher multiplicity), we do not get quadratic convergence.

We can modify Newton's Method to work with those to:

$$g(x) = x - \dfrac{f(x)f'(x)}{[f'(x)]^2 - [f(x)][f''(x)]}$$

The obvious drawback is the need for the second derivative, but it will converge quadratically if $g$ has the required continuity conditions. This second order term can also cause serious roundoff errors when we have small differences.