Timeline for Second order Taylor expansion to solve system of equations
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Dec 22, 2010 at 15:57 | history | edited | J. M. isn't a mathematician |
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| Dec 22, 2010 at 15:56 | answer | added | J. M. isn't a mathematician | timeline score: 1 | |
| Oct 3, 2010 at 14:22 | comment | added | J. M. isn't a mathematician | 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. :) | |
| Oct 3, 2010 at 11:52 | comment | added | Suvrit | A minor caution about a difference in terminology: in optimization, Newton-type methods are generally called second-order, while here it seems they are called first-order, because essentially instead of solving $\min g(x)$, one is considering the associated problem of solving $f(x) := \nabla g(x) = 0$. | |
| Oct 3, 2010 at 2:48 | comment | added | J. M. isn't a mathematician | Both Halley and Householder are cubically convergent methods that are derived from the truncation up to the quadratic term of the multivariate Taylor series. There's another formula due to Halley that requires square roots in the univariate case, so I won't talk more about it because generalizing that is a whole new can of worms. | |
| Oct 3, 2010 at 1:07 | comment | added | Yaroslav Bulatov | 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"? | |
| Oct 3, 2010 at 0:39 | comment | added | J. M. isn't a mathematician | @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. | |
| Oct 3, 2010 at 0:30 | comment | added | J. M. isn't a mathematician | @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. | |
| Oct 2, 2010 at 23:17 | comment | added | Deane Yang | 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. | |
| Oct 2, 2010 at 22:44 | answer | added | Mike Spivey | timeline score: 3 | |
| Oct 2, 2010 at 21:37 | history | asked | Yaroslav Bulatov | CC BY-SA 2.5 |