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Sep 7, 2020 at 12:17 comment added Robert Bryant @DorianoBrogioli: Yes, that's true. In fact, any $f(x_1,y_1)$ satisfies the equation, but the generic one is not of the desired form.
Sep 7, 2020 at 12:16 comment added Doriano Brogioli It seems that this very simple example works for every $n$.
Sep 7, 2020 at 12:15 vote accept Doriano Brogioli
Sep 7, 2020 at 12:11 comment added Robert Bryant @DorianoBrogioli: Yes, for example, let $f = x_1^2y_1+x_1y_1^2$.
Sep 7, 2020 at 11:56 comment added Doriano Brogioli Could you please write an explicit example of $f$ which meets the first equation but not the second, for the 2D case?
Sep 7, 2020 at 11:45 history answered Robert Bryant CC BY-SA 4.0