# Second order Taylor expansion to solve system of equations

Suppose you need to solve $f(\mathbf{x})=\mathbf{0}$ where $f:\mathbb{R}^n \to \mathbb{R}^m$, $m,n>1$. Newton's method relies on first order Taylor expansion of f. Where can I find details of analogous method using second order Taylor expansion? I found at least a dozen numerical analysis books which mention this method, but give no details or applications

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I'd be interested in learning about when using a second order technique might be useful and better than a first order method (Newton or Picard). It seems to me that higher methods are more prone to instabilities. – Deane Yang Oct 2 '10 at 23:17
@Deane: yes, methods with higher-order convergence tend to be finicky in that if you're a teeny bit far away from the solution you want to converge to, there is a great chance of converging to a different solution, or worse, diverge. – J. M. Oct 3 '10 at 0:30
@Yaroslav: Again, as I mentioned in your m.SE question, you need to figure out first what it means to invert a rank-3 tensor before you can figure out how to generalize Halley to multivariate nonlinear systems. Otherwise, there's Householder's method, but since multiplication of vectors, matrices and rank-3 tensors is noncommutative, you need to look at how to multiply the terms to get to your multivariate generalization. – J. M. Oct 3 '10 at 0:39
Halley's method is just one particular form of second order method. Alternative would be to use quadratic formula. Are you sure I just need to take Halley's formula and "plug it in"? – Yaroslav Bulatov Oct 3 '10 at 1:07
Since "order" is so overloaded a word :P , that's the reason I prefer "quadratically/cubically convergent" and "first/second order Taylor expansion" when discussing such iterative methods. :) – J. M. Oct 3 '10 at 14:22