Timeline for Numerical differentiation. What is the best method?
Current License: CC BY-SA 4.0
12 events
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Mar 19 at 13:34 | comment | added | Martin Sleziak | The link to the NAG numerical libraries no longer works - would it perhaps be reasonable to link to Wikipedia instead? I'd guess that it is reasonable assumption that dead links in the Wikipedia articles are updated quite often. | |
Mar 19 at 13:32 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
added 193 characters in body
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Jan 23, 2020 at 12:55 | comment | added | F. Jatpil | @user541686. There is a hidden issue: if you perform the integration using large number of divisions you can (partially) cancel numerical (round-off) errors and get significantly more precise result! (As I point to it in my poor-rated answer here lower). Please try! Take a test function, program Lanczos formula with some integration method (I used trapezoidal rule), increase progressively the number of divisions (10,100,1000,10000,..) .. and observe! I ensure you that the increase of precision is due to statistical cancellation of numerical errors and not due to smaller discretization error. | |
Jan 4, 2020 at 21:40 | comment | added | F. Jatpil | Warning: self promotion. Concerning Lanczos-type derivatives, higher order approach is to be used for significantly improved total error vixra.org/pdf/1912.0340v1.pdf | |
Mar 1, 2015 at 13:02 | comment | added | user541686 | I don't understand how the Lanczos derivative is any better than the finite-difference approximation. In order to evaluate the integral you need to perform a subtraction, which seems to be exactly equivalent to the finite-difference formula in terms of accuracy/stability... am I missing something? How are you supposed to do that integration? | |
May 12, 2011 at 4:33 | vote | accept | Yrogirg | ||
May 10, 2011 at 16:24 | comment | added | J. M. isn't a mathematician | @Yrogirg: As I said, they're definitely not foolproof, and the process of numerical differentiation itself is unstable to begin with; what I've listed are some of the best ones you can do even with this severe handicap. But indeed having symbolic derivatives is much better than any of my proposals. | |
May 10, 2011 at 15:18 | comment | added | Yrogirg | At the end I have abandoned all the numeric techniques and decided to stick to symbolic differentiation. I had to rewrite the program but now at least I have exact differentiation. May be later I will have to return to the issue. | |
May 8, 2011 at 17:57 | comment | added | J. M. isn't a mathematician | "As for using complex numbers I don't actually see the point in it." - the differentiation problem tends to be a bit more stable if you do complex evaluation versus constraining yourself to evaluating only at real values, but that's why I gave the Richardson and Lanczos methods since I do know that sometimes you can only evaluate at reals... | |
May 8, 2011 at 17:41 | comment | added | Yrogirg | Thank you very much for the respond, now I'm studying the techniques connected to Lanczos derivative, at least at the moment they look promising for me. I will report later on the results. As for using complex numbers I don't actually see the point in it. I can't use them to find derivatives of already implemented functions of type $\text{Double} \to \text{Double}$. If I was to reimplement these functions I would rather use dual numbers to have exact (i.e. exact as symbolical ones) derivatives. | |
May 8, 2011 at 16:04 | comment | added | J. M. isn't a mathematician | This algorithm from the ACM might also be of interest: dx.doi.org/10.1145/362759.362820 (@Netlib: netlib.org/toms/413). See also dx.doi.org/10.1137/0704019 , ams.org/journals/mcom/1968-22-102/S0025-5718-1968-0230468-5/… , dx.doi.org/10.1137/S003614459631241X , and dx.doi.org/10.1145/838250.838251 . You've quite a lot to choose from, if you are able to evaluate your function to differentiate at complex arguments (as with using the Cauchy differentiation formula). | |
May 8, 2011 at 15:33 | history | answered | J. M. isn't a mathematician | CC BY-SA 3.0 |