Timeline for Can a cubic polynomial in two real variables have three saddle points?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 16, 2023 at 15:13 | history | edited | Peter Mueller | CC BY-SA 4.0 |
even simpler example
|
| May 16, 2023 at 11:46 | comment | added | Peter Mueller | @PavelKocourek Yes, I used Sage for the mere calculation. In your notation, the Hessian determinants in the three critical points are $9\alpha\beta - \gamma^2$, $-9\alpha\beta + 6\alpha\gamma - \gamma^2$ and $-9\alpha\beta + 6\beta\gamma - \gamma^2$. And by just looking at them one sees that one can even have $\alpha=\beta$ and all of them negative. | |
| May 16, 2023 at 8:54 | comment | added | Pavel Kocourek | Thanks for the explanations. Following the procedure you mentioned by hand I found that the general formula for a cubic polynomial $c(x,y)$ with $c(0,0)=0$ and critical points $(0,0),(0,1),$ and $(1,0)$ is $$P(x, y) = \alpha \left(x^3 - \frac{3}{2}x^2\right) + \beta \left(y^3 - \frac{3}{2}y^2\right) + \gamma \left(x^2y + xy^2 - xy\right)$$, so one only need to guess the coefficients $\alpha,\beta,\gamma$ to make the Hessian determinants negative at the three critical points. Did you use Sage for finding the right parameters? | |
| May 15, 2023 at 22:06 | comment | added | Peter Mueller | The edited answer now has a different choice of saddle points yielding a shorter cubic. | |
| May 15, 2023 at 22:02 | history | edited | Peter Mueller | CC BY-SA 4.0 |
simpler example
|
| May 15, 2023 at 21:29 | comment | added | Peter Mueller | @PavelKocourek The three saddle points we are looking for are either on a line (not considered here), or by an affine transformation we may and do assume that they are $(0,0)$, $(1,0)$, $(0,1)$. Furthermore, we may assume that $(0,0)$ is on the curve. All this gives linear equations for the unknown coefficients. Now the Hessian determinants in terms of a parametrization of the linear solution space are simple enough as to see how to choose the parameters to make them negative. | |
| May 15, 2023 at 21:10 | comment | added | Pavel Kocourek | Thanks a lot for this nice example. Could you share some insights on how you found this example? I'd really appreciate some intuition. | |
| May 15, 2023 at 21:08 | vote | accept | Pavel Kocourek | ||
| May 15, 2023 at 20:59 | history | answered | Peter Mueller | CC BY-SA 4.0 |