Timeline for Tell Mathematica to factor something specific
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| S Jun 12, 2015 at 2:58 | history | suggested | Mahdi | CC BY-SA 3.0 |
improved formatting
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| Jun 12, 2015 at 2:32 | review | Suggested edits | |||
| S Jun 12, 2015 at 2:58 | |||||
| Jun 10, 2014 at 8:46 | comment | added | Physics_maths | @Alexei yes it should be built in. I'm surprised it isn't. | |
| Jun 10, 2014 at 8:22 | comment | added | Alexei Boulbitch | @Love Learning. This question has been one of the discussed during the Q&A session of the last Technology Conference. The outcome is that there is no such a function in the main body of Mma, but such a function is badly needed. I can point out to the package "Presentations" of David Park, where there is a function "FactorOut" designed exactly for such a purpose. You may have a look here: home.comcast.net/~djmpark/index.html. Its advantage with respect to the solution of J.W.Perry below, is that the parts are not held. | |
| Jun 10, 2014 at 8:16 | comment | added | Alexei Boulbitch | @Artes The problem is more general than the polynomial one, and the solution of such a problem is of enormous use in itself. I would upvote it. | |
| Jun 10, 2014 at 4:24 | history | tweeted | twitter.com/#!/StackMma/status/476218374658084864 | ||
| Jun 9, 2014 at 20:43 | answer | added | J. W. Perry | timeline score: 14 | |
| Jun 9, 2014 at 19:01 | comment | added | rm -rf♦ |
@LoveLearning People are trying to help you here... please give a concrete example — i.e. actual terms for A1, A2 and A3. We cannot answer for hypothetical and general questions because the strategy will differ depending on what your terms are. I suggest looking at the link in Artes' first comment above.
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| Jun 9, 2014 at 19:00 | comment | added | rm -rf♦ | @Artes (re: deleted comment) Was that really necessary? You can downvote + close if it was really that bad, but you have not done so... | |
| Jun 9, 2014 at 18:15 | comment | added | Physics_maths | The 1+2+3 are not numbers! The represent stuff. | |
| Jun 9, 2014 at 18:14 | history | edited | Physics_maths | CC BY-SA 3.0 |
added 102 characters in body
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| Jun 9, 2014 at 18:04 | comment | added | Daniel Lichtblau |
If the expression is 1+2+3 then, in the absence of Hold or Unevaluated or the like, it will automatically evaluate to 6.
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| Jun 9, 2014 at 18:00 | review | Low quality posts | |||
| Jun 9, 2014 at 18:03 | |||||
| Jun 9, 2014 at 18:00 | comment | added | Yves Klett | Please edit all relevant info, examples etc. into the question, comments are not well-suited for that purpose. | |
| Jun 9, 2014 at 17:48 | comment | added | Physics_maths | For instance if the expression is 1+2+3 and I want to factor 1/(1-x^2) from 1+2 but leave 3 alone. | |
| Jun 9, 2014 at 17:47 | comment | added | Artes |
For such tasks use PolynomialRemainder, however things depend on case by case basis. Your question most likely is a duplicate of this one: Lowering the degree of an polynomial with an assumption that the polynomial has a factor x^2+ax+b.
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| Jun 9, 2014 at 17:42 | history | asked | Physics_maths | CC BY-SA 3.0 |