Timeline for A question on surjectivity of a bilinear quadratic map
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 21, 2016 at 18:56 | comment | added | Robert Bryant | @YCor: Further, taking $f_a(t) = 1$, one can solve $f'_b(t) = c(t)$ for any given polynomial $c(t)$ to get a polynomial $f_b(t)$, so not bounding the degrees makes the problem trivial. | |
| May 21, 2016 at 17:34 | comment | added | Robert Bryant | @YCor: No, that doesn't happen. For example, if $f_a(t) = 1 + \tfrac13t^4$ and $f_b(t) = -t$, then one has $$f_{ab}(t) = t^4-1.$$ | |
| May 21, 2016 at 12:05 | comment | added | YCor | If $c=c_0+\dots+c_4t^4$ is as above with $12c_0c_4-3c_1c_3+c_2^2<0$ (for instance, $c=t^4-1$) do you know if it still holds that $c$ does not belong to the image for any $n\ge 3$? | |
| May 21, 2016 at 11:53 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Added a sketch of a proof
|
| May 21, 2016 at 11:46 | history | answered | Robert Bryant | CC BY-SA 3.0 |