# Nemirovski

### From Wikimization

the answer is no, there is no necessity to sample .

The construction is like this. The nonnegativity of a trigonometric polynomial on a segment
is equivalent to the nonnegativity of an algebraic polynomial on another segment since and are rational functions of ; you take this as your new variable and get rid of denominators (they are powers of which does not affect positivity of the polynomial).

Now, nonnegativity of an algebraic polynomial on a segment reduces to nonnegatiivity
of another algebraic polynomial on the entire axis. Indeed, w.l.o.g. we can assume that the segment in question is [-1,1], and nonnegativity of on this segment is equivalent to the nonnegativity of
on the entire axis, and you again can get rid of denominators.

The bottom line is that nonnegativity of trigonometric polynomial of degree d on a given segment is equivalent to nonnegativity of another polynomial, of degree not exceeding a known bound 2D, on the entire axis. The crucial fact is that the coefficients of the resulting polynomial, as follows from the construction, are affine functions of the coefficients of the initial polynomial.

Now, the vector of coefficients of a polynomial of degree 2D come from nonnegative polynomial if and only if there exists a positive semidefinite matrix of size D+1 such that
identically in one has
(this is an immediate corollary of the fact that a polynomial which is nonnegative on the entire axis is sum of squares of other polynomials). Thus, comes from a nonnegative polynomial if and only if
for all , where is positive semidefinite.

The bottom line is that is a vector of coefficients of nonnegative on a given segment trigonometric polynomial if and only if there esists a positive semidefinitite matrix such that for certain known matrix and linear map . In other words, the system of constraints
expresses equivalently the fact that the trigonometric polynomial with coefficients is nonnegative on a given segment. Note that this construction is due to Yurii Nesterov.