# a comment on the newton’s method posts

Suppose we want to solve for $g(z)=z$, therefore we want to run Newton’s method on $R(z) = g(z) - z$. Therefore, the iteration that I’ve used is

$z_{n+1} = z_{n} - \frac{g(z)-z}{g'(z)-1}$

Which means at points where the derivative is one are going to be messy. In fact, we can see this very easily in

The derivative of cosine is negative sine. Where is the negative sine equal to -1? Since sine is an odd function the negative sine is equal to -1 when the positive sine is equal to plus one. And that happens at $-5\pi/2 = -2.5\pi \approx -7.853981633974483\ldots$ and $-9\pi/2 = -4.5\pi \approx -14.13716694115407\ldots$ which correspond precisely to those weird X shaped things at those locations. But not all of those Xs are on the axes itself, and at the moment I’m not sure what to make of that.