Timeline for Does there exist another form of the derivative for polynomials?
Current License: CC BY-SA 4.0
3 events
| when toggle format | what | by | license | comment | |
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| Jun 16, 2019 at 14:44 | comment | added | Federico Poloni | @user44191 Good point. And also evaluating the derivative works. | |
| Jun 16, 2019 at 14:00 | comment | added | user44191 | I think any substitution works: for any polynomial $R$, $F_R(P) = P \circ R, H = ux$. | |
| Jun 16, 2019 at 12:11 | history | answered | Federico Poloni | CC BY-SA 4.0 |