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Pavel Kocourek
Pavel Kocourek
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By Bézout's theorem a quartic polynomial $p(x,y)$ can have at most 9 isolated critical points. Can all of them be saddle points?

In case of a cubic polynomial there is a mechanical way to answer this type of questions: One can find a general form of such polynomials with critical points at three given points, e.g. at $(0,0),(0,1),$ and $(1,0)$, and then play with the remaining free parameters until a polynomial with the desired properties if found, as there is only one remaining critical point whose position was not normalized by affine transformation [1], [2].

However, the above approach won't work in case of a quartic polynomial, as then affine transformation normalizes position of only 3 out of the 9 potential critical points.

Note that a quartic polynomial in two real variables can have at most 5 minima out of its 9 potential critical points [3], [4]. Can it reversely have no extreme points, so that all 9 of the potential critical points would be saddle points?

By Bézout's theorem a quartic polynomial $p(x,y)$ can have at most 9 isolated critical points. Can all of them be saddle points?

In case of a cubic polynomial there is a mechanical way to answer this type of questions: One can find a general form of such polynomials with critical points at three given points, e.g. at $(0,0),(0,1),$ and $(1,0)$, and then play with the remaining free parameters until a polynomial with the desired properties if found, as there is only one remaining critical point whose position was not normalized by affine transformation [1], [2].

However, the above approach won't work in case of a quartic polynomial, as then affine transformation normalizes position of only 3 out of the 9 potential critical points.

Note that a quartic polynomial in two real variables can have at most 5 minima out of its 9 potential critical points [3]. Can it reversely have no extreme points, so that all 9 of the potential critical points would be saddle points?

By Bézout's theorem a quartic polynomial $p(x,y)$ can have at most 9 isolated critical points. Can all of them be saddle points?

In case of a cubic polynomial there is a mechanical way to answer this type of questions: One can find a general form of such polynomials with critical points at three given points, e.g. at $(0,0),(0,1),$ and $(1,0)$, and then play with the remaining free parameters until a polynomial with the desired properties if found, as there is only one remaining critical point whose position was not normalized by affine transformation [1], [2].

However, the above approach won't work in case of a quartic polynomial, as then affine transformation normalizes position of only 3 out of the 9 potential critical points.

Note that a quartic polynomial in two real variables can have at most 5 minima out of its 9 potential critical points [3], [4]. Can it reversely have no extreme points, so that all 9 of the potential critical points would be saddle points?

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Pavel Kocourek
Pavel Kocourek
  • 291
  • 1
  • 6

How many saddle points can a quartic polynomial in two real variables have? All 9?

By Bézout's theorem a quartic polynomial $p(x,y)$ can have at most 9 isolated critical points. Can all of them be saddle points?

In case of a cubic polynomial there is a mechanical way to answer this type of questions: One can find a general form of such polynomials with critical points at three given points, e.g. at $(0,0),(0,1),$ and $(1,0)$, and then play with the remaining free parameters until a polynomial with the desired properties if found, as there is only one remaining critical point whose position was not normalized by affine transformation [1], [2].

However, the above approach won't work in case of a quartic polynomial, as then affine transformation normalizes position of only 3 out of the 9 potential critical points.

Note that a quartic polynomial in two real variables can have at most 5 minima out of its 9 potential critical points [3]. Can it reversely have no extreme points, so that all 9 of the potential critical points would be saddle points?