> Is your conjecture that Newton's method would always work at least as well as gradient descent when the vector field has nonzero rotation? > You could easil

> Is your conjecture that Newton's method would always work at least as well as gradient descent when the vector field has nonzero rotation? > You could easily come up with examples where gradient descent would work better despite the rotation, and I expect you could also find lots of rotation cases where Newton's method would get completely thrown off (whereas GD wouldn't). I don't think I understand the meaning of the questions. It isn't gradient descent unless you have a potential function. You cannot do gradient descent on a vector field that is not a gradient. You can solve an ODE. You can, of course, turn a root-finding problem (F(x) = 0) into a minimization problem by defining a potential like g(x) = (F(x))^2. You can do gradient descent on g. So in that sense, both of these problems reduce to each other, though there are better methods for solving F(x) = 0 than minimizing g. I still don't understand this sense of optimization being easier than root finding. What I do understand is some sense in which there are more vector fields than there are gradients, since only some vector fields are gradients of potential functions. My comment was intended to point that out, since I thought that is maybe what you meant.