Timeline for Defining a function which is numerical on all vectors of arguments
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 28, 2020 at 6:57 | history | edited | pisco | CC BY-SA 4.0 | added 520 characters in body |
| May 28, 2020 at 6:52 | vote | accept | pisco | ||
| May 28, 2020 at 6:04 | history | edited | pisco | CC BY-SA 4.0 | edited title |
| May 28, 2020 at 5:50 | history | edited | pisco | CC BY-SA 4.0 | added 182 characters in body |
| May 20, 2020 at 6:00 | vote | accept | pisco | ||
| May 28, 2020 at 5:50 | |||||
| May 16, 2020 at 18:43 | answer | added | Michael E2 | timeline score: 2 | |
| May 14, 2020 at 17:18 | answer | added | swish | timeline score: 2 | |
| May 14, 2020 at 15:27 | comment | added | swish | @youyou Isn't it because f evaluates to 1 first? NumericQ doesn't have any Hold* attributes. | |
| May 14, 2020 at 15:00 | history | tweeted | twitter.com/StackMma/status/1260948046529126401 | ||
| May 14, 2020 at 13:21 | comment | added | youyou | I don't have a rigorous answer, but it appears that it is linked to the fact f is not defined. If you define f (e.g. f[l___]:=1 the result of NumericQ@f[{1},2,3] become True | |
| May 14, 2020 at 12:08 | history | edited | pisco | CC BY-SA 4.0 | added 19 characters in body |
| May 14, 2020 at 12:03 | history | asked | pisco | CC BY-SA 4.0 |