Trigonometric polynomial

A trigonometric polynomial is a finite linear combination of complex exponentials $e^{inx}$, or equivalently of sines and cosines. If it vanishes on an interval, all its coefficients vanish.

Under the displayed diagonal-sum convention, set $w(z)=(1,z^{-1},\dots ,z^{-d}{)}^T$ on the unit circle. A Hermitian matrix $M$ then givesConsequently $M⪰0$ proves nonnegativity. Conversely the Fejér–Riesz theorem produces the rank-one spectral-factor Gram matrix. This gives a semidefinite programming representation of nonnegative trigonometric polynomials. The reversed convention $j-i=k$ instead uses the unconjugated monomial vector.

For coefficient column $q=(q_0,\dots ,q_d{)}^T$, $M_{ij}=q_i\overline{q_j}$ is a Hermitian matrix and a positive semidefinite matrix because $x^{\ast }Mx=|q^{\ast }x{|}^2$. Expanding $|q(z){|}^2$ on the unit circle gives coefficient $p_k={\sum }_{i-j=k}{M}_{ij}$. Thus $M$ has matrix rank one when $q\ne 0$ and zero otherwise. General feasible Gram matrices need not have matrix rank one.

A trigonometric polynomial nonnegative on the unit circle has a polynomial modulus-square factor of polynomial degree at most its trigonometric order. For a nonzero trigonometric polynomial of actual order $d$, conjugate symmetry of trigonometric polynomial coefficients yields reciprocal-conjugate root pairing for $P=z^{d}p$. Even multiplicity of unit-circle roots of a nonnegative trigonometric polynomial permits pairing all $2d$ roots of a polynomial. Select representatives ${\zeta }_1,\dots ,{\zeta }_d$ and useIt follows that $p(z)=c{\prod }_i|z-{\zeta }_i{|}^2$ with $c=(-1{)}^d{p}_d/{\prod }_i\overline{{\zeta }_i}$. Evaluating away from the roots of a polynomial gives $c>0$. Thus $q(z)=\sqrt{c}{\prod }_i(z-{\zeta }_i)$ works. Constant and zero trigonometric polynomials have constant or zero factors. Different selections of roots of a polynomial and constant phases can give different factors.

If $P=z^{d}p$ and $p$ is nonnegative on the unit circle, a root of a polynomial $\zeta =e^{i{\theta }_0}$ there has even multiplicity of a root. Indeed $p(e^{i\theta })$ is a real analytic nonnegative function, whose first nonzero term in its Taylor series at a zero must have even order. The map $\theta \mapsto e^{i\theta }-\zeta$ has a simple zero, and multiplication by $e^{-id\theta }$ is nonvanishing, so the order equals the polynomial multiplicity.

A trigonometric polynomial is real on the unit circle precisely when its finite coefficients have the displayed symmetry. To prove necessity, conjugate its values on the circle, use $\overline{z}=z^{-1}$, and compare coefficients. Multiplying a coefficient difference by $z^d$ gives an ordinary polynomial vanishing at infinitely many points, so it is zero. Symmetry also gives $p(z^{-1})=\overline{p(\overline{z})}$ for every nonzero complex $z$.

For a nonzero real-on-the-circle trigonometric polynomial of actual order $d$, the polynomial $P(z)=z^{d}p(z)$ satisfies $P(z)=z^{2d}\overline{P(1/\overline{z})}$. Its constant term is the nonzero conjugate of its leading coefficient. Hence its nonzero roots of a polynomial occur in pairs $\zeta ,1/\overline{\zeta }$ with the same multiplicity of a root. The coefficient reversal and complex conjugation preserve the exponents of the paired factors. Fixed roots of a polynomial on the unit circle require nonnegativity, rather than this symmetry alone, to have even multiplicity.