Timeline for root solving without analytic derivative
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Dec 20, 2018 at 13:31 | history | edited | Federico Poloni |
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| Dec 21, 2016 at 19:31 | vote | accept | Sujay | ||
| Dec 20, 2016 at 15:33 | comment | added | Deane Yang | Does Newton's method necessarily work if the function is only piecewise differentiable? Also, in what form do you know f in between knots? You might want to look into automatic differentiation, which can compute $f'$, if $f$ can be computed by your software using only elementary operations and functions, even if you don't know the explicit formula yourself. | |
| Dec 20, 2016 at 6:54 | answer | added | David Ketcheson | timeline score: 2 | |
| Dec 19, 2016 at 22:13 | comment | added | Amir Sagiv | Ok - (a) If you know the position of the knot points - why do you need to algorithmically find them? (b) Answering only about the numerical differentiation - there is a lot to be done to avoid this precision error. First see Wikipedia - "Finite Difference", then you can read a lot about it, e.g. in Conte and De Boor book. | |
| Dec 19, 2016 at 21:42 | comment | added | Sujay | @Amir I know the exact position of knot points, so that wont be a problem. But if d above is chosen very small, precision error happens ( dividing two near zero quantities). | |
| Dec 19, 2016 at 21:33 | comment | added | Amir Sagiv | Ok. Do you have any estimation on the number of knots? If you don't, then any finite difference scheme won't work well, because you can never know if your differentiation is small enough. The other method, unfortunately, I don't know. | |
| Dec 19, 2016 at 21:30 | comment | added | Sujay | @AmirSagiv I meant that I dont have analytic expression for df/dx , only numerical approximation like (f(x+d) -f(x-d))/2d . I understand that Newtons method works, but is it better or worse or equal to (say) secant method when derivative is calculated numerically like this ? | |
| Dec 19, 2016 at 21:21 | comment | added | Amir Sagiv | When you say you can not derive - do you mean numerically or analytically? Conversrely, do you sample $f$ on as many grid points as you want? If the answer is positive, then the Newton method will work, with the right choice of differentiation scheme. | |
| Dec 19, 2016 at 19:01 | review | First posts | |||
| Dec 19, 2016 at 20:05 | |||||
| Dec 19, 2016 at 18:59 | comment | added | Igor Rivin | I am not sure what the theoretical answer is, but in my experience, the second method performs terribly. | |
| Dec 19, 2016 at 18:56 | history | asked | Sujay | CC BY-SA 3.0 |