# How many local extrema can a multidimensional polynomial have?

So say I have a $n$ dimensional polynomial of degree $m$. Assume that $n\geq m$. Now for a degree $m$ polynomial in one dimension, we know that there can be at most $m-1$ local extrema. Is there a similar rule for multidimensional polynomials?

One important thing to note is that the polynomial I am working with is only linear terms of each dimension, so there will never be a $x_k^y$ for any $y>1$, for example

Good: $f(\vec x) = x_1x_2x_3 + x_1x_2 - x_3$

Bad: $f(\vec x) = x_1^2x_2 + x_3^3 - x_2$

• Just as a start, if your exponents are all one, then that will be true of every partial derivative as well. So finding critical points is just finding zero sets of first degree polynomials, aka algebraic varieties. That's a place you can start looking. (eg: en.wikipedia.org/wiki/Algebraic_variety) – dbx Jun 21 '18 at 20:19

If $f(x_1,\ldots,x_n)$ is a polynomial of degree $m$, then the critical points are the intersection of the $n$ sets $$V_i = \left\{\vec x \in \mathbb{R}^n \middle| \frac{\partial f}{\partial x_i}(\vec x) = 0\right\}$$ for $i = 1,\ldots,n$. Now $\partial f/\partial x_i$ has degree at most $m-1$, so Bézout's theorem tells you that the intersection has either infinitely many points or at most $$\prod_{i=1}^n \deg \frac{\partial f}{\partial x_i} \le (m-1)^n$$ points. If there are infinitely many points, then either $\partial f/\partial x_i \equiv 0$ for some $i$ or $\partial f_i/\partial x_i$ and $\partial f_j/\partial x_j$ have a common factor (i.e. $V_i$ and $V_j$ have a common component) for some $i\ne j$.
So $f$ has either infinitely many critical points or at most $(m-1)^n$ critical points.
• No, not even for $n=1$. – arkeet Jun 23 '18 at 18:56
• Follow up question, so I was looking at this faculty.csuci.edu/brian.sittinger/2nd_DerivTest.pdf, which was talking about various types of second derivative tests for multivariables. Since in my situations any double partial derivative is always 0 ($f_{xx}, f_{yy}, etc.$) , does that imply that all critical points in my functions are saddle points? – wjmccann Jun 24 '18 at 5:07
• @arkeet Say I was to then take $f(\vec x)$ and square it. Since there are no local maxima nor minima, does that mean that the only minima of the squared variant would be the zeros of the original function? – wjmccann Jun 24 '18 at 18:57