Timeline for Replace function call with expression resulting from the symbolic evaluation of the function when defining a second function?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 16, 2020 at 9:23 | history | edited | CommunityBot | Commonmark migration | |
| Sep 9, 2013 at 13:10 | vote | accept | user565739 | ||
| Aug 10, 2013 at 18:30 | answer | added | Jens | timeline score: 4 | |
| Aug 10, 2013 at 13:48 | history | edited | m_goldberg | CC BY-SA 3.0 | General cleanup |
| Aug 10, 2013 at 9:39 | answer | added | Mr.Wizard | timeline score: 7 | |
| Aug 10, 2013 at 9:11 | comment | added | andre314 | about = and := there is reference.wolfram.com/mathematica/tutorial/… | |
| Aug 10, 2013 at 8:57 | history | tweeted | twitter.com/#!/StackMma/status/366121158866771968 | ||
| Aug 10, 2013 at 8:50 | comment | added | user565739 | @andre: Sorry for my typo in the post, but what I mean is that g[x_, y_] := ...f[x,y], not g[x_, y_] =: .... But thank you for your reply, because I don't know the difference between := and =. Would you mind explain a bit or a source? | |
| Aug 10, 2013 at 8:35 | comment | added | user565739 | @Kuba, when use h in g, then when I evaluate g, would f will be evaluated? | |
| Aug 10, 2013 at 8:32 | comment | added | Kuba | Without any insight into the code, for example what f is, we can not help you. For example, maybe f can be defined with Set instead of SetDelayed. You can always h[x_,y_]=f[x,y] and then work on h with g. | |
| Aug 10, 2013 at 7:58 | history | asked | user565739 | CC BY-SA 3.0 |