Timeline for Degree of a rational function [closed]
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 1, 2015 at 8:18 | review | Reopen votes | |||
| Mar 1, 2015 at 11:24 | |||||
| Mar 1, 2015 at 8:02 | history | edited | Jjm | CC BY-SA 3.0 |
more clearly explained, as suggested by the closing votes
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| Feb 27, 2015 at 17:49 | history | closed |
Benoît Kloeckner Stefan Kohl♦ Alex Degtyarev abx Joonas Ilmavirta |
Needs details or clarity | |
| Feb 27, 2015 at 17:40 | vote | accept | Jjm | ||
| Feb 27, 2015 at 17:39 | history | edited | Jjm | CC BY-SA 3.0 |
fixed formula
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| Feb 27, 2015 at 17:21 | review | Close votes | |||
| Feb 27, 2015 at 17:49 | |||||
| Feb 27, 2015 at 16:55 | comment | added | Jjm | @DavidSpeyer But what if $p(z)=q(z)=z$? The common degree is 1, but $f$ is constant and its degree is zero. | |
| Feb 27, 2015 at 16:53 | answer | added | Igor Rivin | timeline score: 2 | |
| Feb 27, 2015 at 16:51 | comment | added | David E Speyer | If you mean degree as a topological map, the correct formula is $\deg(\bar{f}) = \max(\deg(p), \deg(q))$. | |
| Feb 27, 2015 at 16:39 | history | asked | Jjm | CC BY-SA 3.0 |