When studying local extreme points of functions of several (real) variables, the textbook asks in one of the exercises to consider the polynomial

Here we have and , so the only critical point of is Since and the *Hessian* of is , it follows that is a local minimum of and, since it is the only critical point, it is in fact an absolute minimum with

being a polynomial, it is reasonable to expect that there is an algebraic explanation as for why is its minimum, and why it lies at . After all, this is what happens in one variable: If and , then

and obviously has a minimum at , and this minimum is

The polynomial of the example above can be analyzed this way as well. A bit of algebra shows that we can write

and it follows immediately that has a minimum value of , achieved precisely when both and , i.e, at

(One can go further, and explain how to go in a systematic way about the `bit of algebra’ that led to the representation of as above, but I will leave that for a future occasion.)

What we did with is not a mere coincidence. **Hilbert**’s 17th of the 23 problems of his famous address to the *Second International Congress of Mathematicians* in Paris, 1900, asks whether every polynomial with real coefficients which is non-negative for all (real) choices of is actually a sum of squares of *rational* functions. (A rational function is a quotient of polynomials.) A nonnegative polynomial is usually called positive definite, but I won’t use this notation here.

If Hilbert’s problem had an affirmative solution, this would provide a clear explanation as for why is non-negative.

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