Timeline for What is a one-parameter Newton's method?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 24, 2015 at 18:49 | vote | accept | Marc Andreson | ||
| Jan 24, 2015 at 9:54 | comment | added | Federico Poloni | I can only answer based on what I read in the paper. If the error is $(p-c_2)e_n^2+O(n^3)$, then you minimize it (asymptotically) with $p=c_2$. Its expression involves second derivatives and may not be readily available, so you could use an approximation. That is the asymptotic convergence speed, though, which is often irrelevant: once you are close to the solution, a couple of iterations get you there anyway in practice. What one would like to minimize is the initial transient phase in which the error does not decrease, and as far as I know it's an open problem for every Newton-like method. | |
| Jan 24, 2015 at 9:35 | comment | added | Marc Andreson | I don't have the access to this [2] paper. But basically, is my assumption correct that $p$ is supposed to make the denominator the largest one? | |
| Jan 24, 2015 at 9:33 | comment | added | Federico Poloni | It's the first time I see that generalization, too, so I can't give you more information than what can be read in the paper. Have you checked reference [2]? | |
| Jan 24, 2015 at 9:14 | comment | added | Marc Andreson | So as to speed up the root finding and get the second-order (quadratic) convergence we need to find a $p$ that will make the denominator the largest one? Could you please include in your response a brief explanation how this is calculated from the error equation (I haven't seen before the asymptotic expressions)? | |
| Jan 24, 2015 at 9:04 | comment | added | Federico Poloni | This is explained in your paper. Check around Equation (2.9), there is an asymptotic expression for the error which involves $p$. | |
| Jan 24, 2015 at 8:51 | comment | added | Marc Andreson | Thank you for the response. How does introducing the $p$ makes the method faster than the classical one? What value should $p$ have to make it work better? | |
| Jan 24, 2015 at 8:48 | history | answered | Federico Poloni | CC BY-SA 3.0 |