Timeline for Precise algorithm for finding higher order derivatives
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Sep 19, 2023 at 14:18 | comment | added | m4r35n357 | BTW, print(~(a**9)) also works if you can bear the domain restriction a>0. | |
| Sep 19, 2023 at 12:25 | comment | added | m4r35n357 | I have reproduced correct results in the Python version of my software. github.com/m4r35n357/ODE-Playground/tree/python. This is from an ipython session (I have used vertical bars in place of carriage returns): Context.places = 6 | a = Series.get(10, 1.0).var | print(~(a * a * a * a * a * a * a * a * a)) | +1.000000e+00 +9.000000e+00 +7.200000e+01 +5.040000e+02 +3.024000e+03 +1.512000e+04 +6.048000e+04 +1.814400e+05 +3.628800e+05 +3.628800e+05 | |
| Sep 19, 2023 at 11:11 | comment | added | m4r35n357 | Non-direct PDF link aimsciences.org/article/doi/10.3934/proc.2013.2013.587 from author's home page: neidinger.net/publicat.html | |
| Sep 19, 2023 at 10:42 | comment | added | m4r35n357 | Are you asking about "numerically defined" functions, or those described by differentiable "component" functions. I ask because if the latter, you can determine higher derivative fairly trivially, without numerical errors caused by small time steps. This technique is inexplicably (to me) relatively unknown in the numerical analysis world. See here for an example: pdfs.semanticscholar.org/fb1d/… | |
| Jul 12, 2021 at 20:22 | vote | accept | Natrium | ||
| Jul 12, 2021 at 1:39 | answer | added | Pseudonym♦ | timeline score: 3 | |
| Jul 9, 2021 at 18:05 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jun 9, 2021 at 17:51 | history | edited | D.W.♦ |
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| Jun 9, 2021 at 17:51 | answer | added | D.W.♦ | timeline score: 1 | |
| Jun 9, 2021 at 17:21 | review | First posts | |||
| Jun 23, 2021 at 17:20 | |||||
| Jun 9, 2021 at 17:18 | history | asked | Natrium | CC BY-SA 4.0 |