On the Zeros of Exponential Polynomials

For the first part of the claim, it is clear that \(1,a_1/2\pi ,\ldots ,a_m/2\pi\) are rationally linearly independent and hence, by Kronecker’s Theorem (Theorem 2.5), the set \(\lbrace h(t) \mid t\in \mathbb {N}\rbrace\) is dense in \([0,2\pi)^m\). It follows that \(\lbrace h(t) \mid t\in \mathbb {R}_{\ge 0}\)} is also dense in \([0,2\pi)^m\). For the second part, note that the first coordinate of \(h(t)\) is zero precisely when \(t = -\varphi _1/a_1 + 2n\pi\) for some \(n\in \mathbb {Z}\). At such a time we have \(h(t)=(0,g(n))\), where \(\begin{align*} g(n) & \stackrel{\mathrm{def}}{=}\left\langle \left(n\frac{2\pi a_j}{a_1} + \frac{a_1\varphi _j-\varphi _1a_j}{a_1}\right) \bmod 2\pi : {2\le j \le m} \right\rangle \,. \end{align*}\) As above, we have that \(\lbrace 1, 2\pi a_2 / a_1, \ldots , 2\pi a_m / a_1\rbrace\) are linearly independent over \(\mathbb {Q}\), so applying Kronecker’s Theorem (Theorem 2.5), we have that \(\lbrace g(n) : n \in \mathbb {N}\rbrace\) is dense in \([0,2\pi)^m\).□

Consider a non-zero polynomial \(P \in \mathbb {R}[x_0,\ldots ,x_n]\) whose coefficients are real algebraic numbers. Then we can write \(P(t,e^{r_1t},\ldots ,e^{r_nt})\) in the form \(\begin{equation*} Q_1(t)e^{\beta _1 t} + \cdots + Q_{m}(t) e^{\beta _{m}t} \end{equation*}\) for non-zero univariate polynomials \(Q_1,\ldots ,Q_{m}\) with real algebraic coefficients and real algebraic numbers \(\beta _1\gt \cdots \gt \beta _{m}\). It is clear that for t sufficiently large, \(P(t,e^{r_1t},\ldots ,e^{r_nt})\) has the same sign as the leading term \(Q_1(t)\). The proposition easily follows.□

Since g is semi-algebraic, there is a non-zero polynomial P with real algebraic coefficients such that \(P(\boldsymbol {x},g(\boldsymbol {x}))=0\) for all \(\boldsymbol {x}\in D\) (see [1, Proposition 2.86]). In particular, we have \(P(e^{\boldsymbol {r}t},g(e^{\boldsymbol {r}t}))=0\) for all \(t\gt T_1\). Gathering terms, we can rewrite this equation in the form \(\begin{equation*} Q_1(g(e^{\boldsymbol {r}t}))e^{\beta _1 t} + \cdots + Q_{m}(g(e^{\boldsymbol {r}t}))e^{\beta _{m} t} = 0 \end{equation*}\) for non-zero univariate polynomials \(Q_1,\ldots ,Q_{m}\) with real algebraic coefficients and real algebraic numbers \(\beta _1\gt \cdots \gt \beta _m\). Without loss of generality, assume that \(Q_1\) is monic.

If \(m=1,\) then for all \(t\gt T_1\) we have \(Q_1(g(e^{\boldsymbol {r}t}))=0\). Thus, \(g(e^{\boldsymbol {r}t})\) is equal to some root of \(Q_1\) for all \(t\gt T_1\). Then by Proposition 2.9 there exists \(T_2\) and a root \(g^*\) of \(Q_1\) such that \(g(e^{\boldsymbol {r}t})=g^*\) for all \(t\gt T_2\).

If \(m\gt 1\), since g is a bounded function, for all \(t\gt T_1\) we have (5) \(\begin{eqnarray} \big |Q_1(g(e^{\boldsymbol {r}t}))\big | = \left|Q_2(g(e^{\boldsymbol {r}t})) e^{(\beta _2-\beta _1)t} + \cdots + Q_m(g(e^{\boldsymbol {r}t})) e^{(\beta _m-\beta _1)t} \right| \le M e^{(\beta _2-\beta _1)t} \end{eqnarray}\) for some constant M. Let \(Q_1\) have degree d. Since \(Q_1\) is monic, Equation (5) implies that the distance of \(g(e^{\boldsymbol {r}t})\) to a root of \(Q_1\) is at most \((Me^{(\beta _2-\beta _1)t})^{1/d}\). Hence, there exists a root \(g^*\) of \(Q_1\) and effective constants \(\varepsilon ,T_2\gt 0\) such that \(|g(e^{\boldsymbol {r}t})-g^*| \lt e^{-\varepsilon t}\) for all \(t\gt T_2\).□

We argue that the procedure in Figure 1 determines whether f has a zero in the interval \((a,b)\).

Fig. 1.

View Figure

If the procedure terminates in Line (4), then the output is correct by a direct application of Theorem 3.1. If the procedure terminates in Line (3), then the output is correct by the assumption that the real roots of f are simple and complex roots come in conjugate pairs; in conjunction with Theorem 3.1, this entails that \(N_j\) is odd just in case \(C_j\) contains a real root. It remains to argue that the procedure always terminates.

For a given value of k, Line (2) succeeds if f has no zero on any of the contours \(C_j\). Since f has only finitely many zeros in any bounded region of the complex plane, for k sufficiently large the body of the for loop will terminate. Moreover, for k sufficiently large, each contour \(C_j\) will enclose at most one real root and no complex roots and thus the whole procedure will halt. Thus, termination is assured if we evaluate the loop for all values of k in parallel in a fair way (e.g., divide the computation into phases such that in the mth phase one executes m steps of the loop body for \(k=1,\ldots ,m\)).□

Consider a solution \(t\ne 0\) of Equations (9) and (10). By passing to suitable associates, we may assume without loss of generality that P and Q lie in \(\mathcal {A}\), i.e., that all variables in P and Q appear with non-negative exponent. Moreover, since P and Q are coprime in \(\mathcal {R}\), their greatest common divisor R in \(\mathcal {A}\) is a monomial. In particular, \(\begin{equation*} R(t,e^{a_1 t},\ldots ,e^{a_r t},e^{ib_1 t},\ldots ,e^{i b_s t})\ne 0. \end{equation*}\) Thus, dividing P and Q by R, we may assume that P and Q are coprime in \(\mathcal {A}\) and that Equations (9) and (10) still hold.

Since coprime univariate polynomials cannot have a common root, we may assume without loss of generality that \(r+s\ge 1\). By Schanuel’s Conjecture, the field \(\begin{equation*} \mathbb {Q}(a_1t,\ldots ,a_rt,ib_1t,\ldots ,ib_st,e^{a_1t},\ldots ,e^{a_rt}, e^{ib_1t},\ldots ,e^{ib_st}) \end{equation*}\) has transcendence degree at least \(r+s\) over \(\mathbb {Q}\). Since \(a_1,\ldots ,a_r\) and \(b_1,\ldots ,b_s\) are algebraic over \(\mathbb {Q}\), writing \(\begin{equation*} S:= (t,e^{a_1t},\ldots ,e^{a_rt},e^{ib_1t},\ldots ,e^{ib_st}), \end{equation*}\) it follows that the field \(\mathbb {Q}(S) / \mathbb {Q}\) also has transcendence degree at least \(r+s\) over \(\mathbb {Q}\).

Equations (9) and (10) say that S is a common root of P and Q. Pick some variable \(\sigma \in \lbrace x,y_j,z_j : 1\le i \le r, 1 \le j\le s\rbrace\) that has positive degree in P. Then the entry of S corresponding to \(\sigma\) (where t corresponds to x, \(e^{a_jt}\) to \(y_j\), and \(e^{b_jt}\) to \(z_j\)) is algebraic over the remaining entries of S. We claim that these remaining entries of S are algebraically dependent and thus S comprises at most \(r+s-1\) algebraically independent elements, contradicting Schanuel’s Conjecture. The claim clearly holds if \(\sigma\) does not appear in Q (for then Q gives the desired algebraic relation). On the other hand, if \(\sigma\) has positive degree in Q, then, since P and Q are coprime in \(\mathcal {A}\), the multivariate resultant \(\mathrm{Res}_\sigma (P,Q)\) is a non-zero polynomial in which the variable \(\sigma\) does not appear and which vanishes at S (see, e.g., [9, Page 163]). Thus, the claim also holds in this case. We thus obtain a contradiction and conclude that Equations (9) and (10) have no non-zero solution \(t\in \mathbb {R}\).□

Consider an exponential polynomial (11) \(\begin{gather} f(z)=P(z,e^{a_1z},\ldots ,e^{a_r z},e^{i b_1 z},\ldots , e^{i b_s z}), \end{gather}\) where \(\lbrace a_1,\ldots ,a_r\rbrace\) and \(\lbrace b_1,\ldots ,b_s\rbrace\) are \(\mathbb {Q}\)-linearly independent sets of real algebraic numbers, and \(P\in \mathcal {R}\) is irreducible. We show how to determine whether f has a zero in a bounded interval \(I\subseteq \mathbb {R}_{\ge 0}\) with rational endpoints.

If \(r=s=0,\) then \(f(z)\) is a polynomial with algebraic coefficients and deciding the existence root in I is straightforward.

If r and s are not both zero, then any root of \(f(z)\) must be transcendental by Theorem 2.2 (see [2, Theorem 8] for details). Thus, we may assume without loss of generality that I is an open interval. We continue by considering separately the cases of Type-1 and Type-2 exponential polynomials.

Case (i): f is a Type-1 exponential polynomial. By assumption, P and \(P^\star\) are irreducible and are not associates in \(\mathcal {R}\). Thus, they are coprime in \(\mathcal {R}\). We claim that the equation \(f(z)=0\) has no solution \(z\in \mathbb {R}\). Indeed, if \(z\in \mathbb {R}\) is a zero of f, then \(\begin{equation*} f^\star (z)=\overline{f(\overline{z})} = \overline{f(z)} = 0 \, ; \end{equation*}\) that is, z is a common zero of f and \(f^\star\). It follows that \(\begin{equation*} (z,e^{a_1z},\ldots ,e^{a_r z},e^{i b_1 z},\ldots ,e^{i b_s z}) \end{equation*}\) is a common zero of P and \(P^\star\), contradicting Proposition 3.4.

Case(ii): f is a Type-2 exponential polynomial. We wish to use the zero-finding procedure of Section 3.1 to determine whether f has a zero in the open interval I. To this end we must show that the complex zeros of f come in conjugate pairs and all its real zeros are simple.

Considering complex zeros first, note that by the assumption on f we have \({P}^\star =UP\) for some unit \(U\in \mathcal {R}\). It follows that \(\begin{eqnarray*} f(z) = 0 &\Leftrightarrow & f^\star (z) = 0\qquad \mbox{(since $P^\star =UP$)}\\ & \Leftrightarrow & \overline{f(\overline{z})} = 0 \\ & \Leftrightarrow & f(\overline{z}) = 0. \end{eqnarray*}\) Thus, z is zero of f iff \(\overline{z}\) is a zero of f.

We next show, assuming Schanuel’s Conjecture, that the zeros of f are all simple. We have \(\begin{gather*} f^{\prime }(z) = Q(z,e^{a_1 z},\ldots ,e^{a_r z}, e^{i b_1 z},\ldots ,e^{i b_s z}), \end{gather*}\) for some polynomial \(Q\in \mathcal {R}\). We claim that P and Q are coprime in \(\mathcal {R}\). Since P is irreducible, P and Q can only fail to be coprime if P divides Q.

If P has degree \(k\gt 0\) in X, then Q has degree \(k-1\) in X and thus P cannot divide Q. (Recall that all polynomials in \(\mathcal {R}\) have non-negative degree in the variable X.) On the other hand, if X does not appear in \(P,\) then we can write \(P=\sum _{\boldsymbol {u},\boldsymbol {v}} \alpha _{\boldsymbol {u},\boldsymbol {v}} \boldsymbol {Y}^{\boldsymbol {u}}\boldsymbol {Z}^{\boldsymbol {v}}\), where \(\alpha _{\boldsymbol {u},\boldsymbol {v}} \in \mathbb {C}\) for all \(\boldsymbol {u}\in \mathbb {Z}^r\) and \(\boldsymbol {v}\in \mathbb {Z}^s\). We then have \(Q=\sum _{\boldsymbol {u},\boldsymbol {v}} \alpha _{\boldsymbol {u},\boldsymbol {v}} \beta _{\boldsymbol {u},\boldsymbol {v}} \boldsymbol {Y}^{\boldsymbol {u}}\boldsymbol {Z}^{\boldsymbol {v}}\), where \(\begin{equation*} \beta _{\boldsymbol {u},\boldsymbol {v}}:=\sum _{j=1}^r a_j u_j + i \sum _{j=1}^s b_j v_j. \end{equation*}\) By the rational linear independence of \(\lbrace a_1,\ldots ,a_r\rbrace\) and \(\lbrace b_1,\ldots ,b_s\rbrace\), the numbers \(\beta _{\boldsymbol {u},\boldsymbol {v}}\) are pairwise distinct and non-zero. Since P is not a unit, it has at least two monomials. We conclude that P does not divide Q and hence P and Q are coprime. From Proposition 3.4 we conclude that the equations \(f^{\prime }(z)=f(z)=0\) have no solution \(z\in \mathbb {C}\).□

Write \(\lambda _j=a_j+ib_j\), where \(a_j,b_j\) are real algebraic numbers for \(j=1,\ldots ,k\). By assumption, there is a single real algebraic number b such that each \(b_j\) is an integer multiple of b. For each integer n, both \(\cos (nbt)\) and \(\sin (nbt)\) can be written as polynomials in \(\sin (bt)\) and \(\cos (bt)\) with integer coefficients. If follows that \(\begin{equation*} f(t) = Q(t,e^{a_1t},\ldots ,e^{a_kt},\cos (bt),\sin (bt)), \end{equation*}\) for some polynomial Q with algebraic coefficients.

Write \(R := \lbrace t\ge 0 : \sin (bt) \ge 0\rbrace\) and \(R^{\prime } := \lbrace t\ge 0 : \sin (bt) \le 0\rbrace\). We show how to determine boundedness of \(\lbrace t \in R : f(t) = 0\rbrace\). The procedure to decide boundedness of \(\lbrace t \in R^{\prime } : f(t) = 0\rbrace\) can be obtained with minor modifications.

Consider the semi-algebraic set \(\begin{equation*} E:= \left\lbrace (\boldsymbol {u},x) \in \mathbb {R}^{k+1}\times [-1,1] : Q\left(\boldsymbol u,x,\sqrt {1-x^2}\right)=0) \right\rbrace \end{equation*}\) and note that for \(t \in R\) we have \(f(t)=0\) if and only if \((t,e^{\boldsymbol a t},\cos (bt)) \in E\), where \(e^{\boldsymbol a t}\) denotes \((e^{a_1t},\ldots ,e^{a_kt})\). Let \(E=C_1\cup \cdots \cup C_m\) be a cell decomposition of E, and define \(\begin{equation*} Z_j = \lbrace t \in R : (t,e^{\boldsymbol a t},\cos (bt)) \in C_j\rbrace , \quad j=1,\ldots ,m. \end{equation*}\) Then \(\lbrace t \in R : f(t)=0\rbrace = Z_1 \cup \cdots \cup Z_m\).

Fix \(j\in \lbrace 1,\ldots ,m\rbrace\). We show how to decide whether \(Z_j\) is bounded and, in case \(Z_j\) is bounded, we show how to compute an upper bound on \(Z_j\). This suffices to prove the theorem. To this end, write \(D_j \subseteq \mathbb {R}^{k+1}\) for the projection of \(C_j \subseteq \mathbb {R}^{k+2}\) on the first \(k+1\) coordinates. We consider two cases.

The first case is that \(\lbrace t : (t,e^{\boldsymbol a t}) \in D_j\rbrace\) is bounded. By Proposition 2.9, we can compute an upper bound T of this set. But then T is an effective upper bound on the set \(Z_j\).

The second case is that \(\lbrace t : (t,e^{\boldsymbol a t}) \in D_j\rbrace\) is unbounded. We claim that \(Z_j\) is also unbounded. Indeed, observe that by Proposition 2.9, the set \(\lbrace t : (t,e^{\boldsymbol a t}) \in D_j\rbrace\) contains an unbounded interval \((T,\infty)\). Note also that, by definition of a cell, there exists a continuous semi-algebraic function \(\xi\) with domain \(D_j\) such that \((\boldsymbol u,\xi (\boldsymbol u)) \in C_j\) for all \(\boldsymbol u \in D_j\). Then for all \(t\in R\), \(\begin{eqnarray*} f(t)=0 & \Longleftarrow & (t,e^{\boldsymbol a t},\cos (bt)) \in C_j\\ & \Longleftarrow & (t,e^{\boldsymbol a t}) \in D_j \wedge \xi (t,e^{\boldsymbol a t})=\cos (bt) \\ & \Longleftarrow & {t \in (T,\infty)} \wedge \xi (t,e^{\boldsymbol a t})=\cos (bt). \end{eqnarray*}\) Now \(\xi (t,e^{\boldsymbol at})\) is a continuous function with domain \((T,\infty)\) that takes values in \([-1,1]\). Furthermore, \(R \cap (T,\infty)\) contains infinitely many intervals over which \(\cos (bt)\) runs from \(+1\) to \(-1\). Thus, there are infinitely many \(t\in R\) such that \(\xi (t,e^{\boldsymbol at})=\cos (bt)\) and hence f has infinitely many zeros in \(Z_j\).□

Write \(a_j=\mathrm{Re}(\lambda _j)\) for \(j=1,\ldots ,k\). Let \(b_1,b_2\) be real algebraic numbers, linearly independent over \(\mathbb {Q}\), such that \(\mathrm{Im}(\lambda _j)\) is an integer linear combination of \(b_1\) and \(b_2\) for \(j=1,\ldots ,k\). For each \(n\in \mathbb {Z}\), \(\sin (nb_1t)\) and \(\cos (nb_1t)\) can be written as polynomials in \(\sin (b_1t)\) and \(\cos (b_1t)\) with integer coefficients, and similarly for \(b_2\). It follows that we can write f in the form \(\begin{equation*} f(t) = Q(e^{a_1 t},\ldots , e^{a_kt},\cos (b_1t),\cos (b_2t), \sin (b_1t),\sin (b_2t)) \end{equation*}\) for some polynomial Q with real algebraic coefficients that is computable from f.

Write \(R:= \lbrace t\ge 0:\sin (b_1t)\ge 0 \wedge \sin (b_2t)\ge 0 \rbrace\). We show how to decide boundedness of \(\lbrace t\in R :f(t)=0\rbrace\). The cases for the other three sign conditions on \(\sin (b_1t)\) and \(\sin (b_2t)\) follow mutatis mutandis.

Define a semi-algebraic set \(\begin{equation*} E := \lbrace (\boldsymbol {u},x,y) \in \mathbb {R}^k \times [-1,1]^2: Q (\boldsymbol {u},x,y,\sqrt {1-x^2},\sqrt {1-y^2})=0\rbrace . \end{equation*}\) Then for \(t\in R\) we have \(f(t)=0\) if and only if \((e^{\boldsymbol {a}t},\cos (b_1t),\cos (b_2t)) \in E\), where \(\boldsymbol {a}= (a_1,\ldots ,a_k)\). Now consider a cell decomposition \(E=C_1 \cup \cdots \cup C_m\), and define (12) \(\begin{gather} Z_j := \lbrace t\in R : (e^{\boldsymbol {a}t},\cos (b_1t),\cos (b_2t)) \in C_j \rbrace , \qquad j=1,\ldots ,m. \end{gather}\) Then \(\lbrace t \in R : f(t)=0\rbrace = Z_1 \cup \cdots \cup Z_m\). We now analyze the boundedness of each component \(Z_j\).

Fix \(j\in \lbrace 1,\ldots ,m\rbrace\). We show how to decide boundedness of \(Z_j\). To this end, write \(D_j \subseteq \mathbb {R}^{k}\) for the projection of the corresponding cell \(C_j \subseteq \mathbb {R}^{k+2}\) on the first k coordinates.

First, suppose that \(\lbrace t \in \mathbb {R} : e^{\boldsymbol {a}t} \in D_j \rbrace\) is bounded. Then, by Proposition 2.9, we can compute an upper bound T of this set, entailing that \(Z_j \subseteq [0,T]\). On the other hand, suppose that \(\lbrace t\in \mathbb {R}: e^{\boldsymbol {a}t} \in D_j \rbrace\) is unbounded. Then, by Proposition 2.9, this set contains an unbounded interval \((T,\infty)\) for some \(T\in \mathbb {N}\). Write \(I=[-1,1]\) and define functions \(g_1,g_2,h_1,h_2 : D_j \rightarrow I\) by (13) \(\begin{align} g_1(\boldsymbol {u}) & = \inf \lbrace x\in I : \exists y \, (\boldsymbol {u},x,y) \in C_j\rbrace \qquad & g_2(\boldsymbol {u}) & = \inf \lbrace y\in I : \exists x \, (\boldsymbol {u},x,y) \in C_j\rbrace \end{align}\) (14) \(\begin{align} h_1(\boldsymbol {u}) & = \sup \lbrace x\in I: \exists y \, (\boldsymbol {u},x,y) \in C_j\rbrace \qquad & h_2(\boldsymbol {u}) & = \sup \lbrace y\in I : \exists x \, (\boldsymbol {u},x,y) \in C_j\rbrace . \end{align}\) These functions are all semi-algebraic; hence, by Proposition 2.10, the limits \(g_i^* := \lim _{t\rightarrow \infty } g_i(e^{\boldsymbol {a}t})\) and \(h_i^* := \lim _{t\rightarrow \infty } h_i(e^{\boldsymbol {a}t})\) exist for \(i=1,2\) and are algebraic numbers. Clearly we have \(g_1^*\le h_1^*\) and \(g_2^*\le h_2^*\). We now consider three cases according to the strictness of these inequalities.

Case I: Suppose that \(g_1^*= h_1^*\) and \(g_2^*=h_2^*\). We show that \(Z_j\) is bounded and that we can compute \(T_2\) such that \(Z_j\subseteq [0,T_2]\).

By Proposition 2.10, there exist \(T_1,\varepsilon \gt 0\) such that for all \(t\gt T_1\) and \(i=1,2\), (15) \(\begin{gather} |g_i(e^{\boldsymbol {a}t}) - g_i^*| \lt e^{-\varepsilon t} \mbox{ and } |h_i(e^{\boldsymbol {a}t}) - h_i^*| \lt e^{-\varepsilon t}. \end{gather}\)

Then for \(t \in R\) such that \(t\gt T_1\) we have (16) \(\begin{eqnarray} t\in Z_j & \Longleftrightarrow & \left(e^{\boldsymbol {a}t},\cos (b_1t),\cos (b_2t)\right) \in C_j \;\;{\text{(by Equations }\href{#eq12}{\text{(12))}}}\\ & \Longrightarrow & g_1(e^{\boldsymbol {a}t}) \le \cos (b_1t) \le h_1(e^{\boldsymbol {a}t}) \,\mbox{ and }\, g_2(e^{\boldsymbol {a}t}) \le \cos (b_2t) \le h_2(e^{\boldsymbol {a}t}) \;\;{{\text{(by Equations}} ({{\href{#eq13}{\text{13}}}})\;{{\text {and}\;({\href{#eq14}{\text{14}}}})}}\\ &\Longrightarrow & \left| \cos (b_1t) - g_1^* \right| \lt e^{-\varepsilon t} \mbox{ and } \left| \cos (b_2t) - g_2^* \right| \lt e^{-\varepsilon t} {\text{by Equations }\href{#eq15}{\text{(15)}}}). \end{eqnarray}\)

Write \(g_1^* = \cos (\varphi _1)\) and \(g_2^*=\cos (\varphi _2)\) for some \(\varphi _1,\varphi _2\in [0,\pi ]\). Since \(|\cos (\varphi _1+x)-\cos (\varphi _1)| \ge x^3/3\) for all x sufficiently small (by a Taylor expansion), the inequality in Equation (16) implies that for some \(k_1,k_2 \in \mathbb {Z}\), (17) \(\begin{gather} |b_1t - \varphi _1 - 2k_1\pi | \lt 3e^{-\varepsilon t/3} \,\mbox{ and }\, |b_2t - \varphi _2 - 2k_2\pi | \lt 3e^{-\varepsilon t/3}. \end{gather}\) Combining the upper bounds in Equation (17) with the polynomial lower bounds from Lemma 2.3 (i.e., that either for some N and all t sufficiently large, either \(|b_1t-\varphi _1-2k_1\pi | \gt 1/t^N\) or \(|b_2t-\varphi _2-2k_2\pi | \gt 1/t^N\)), we obtain an effective bound \(T_2\) for which \(t\in Z_j\) implies \(t\lt T_2\).

Case II: Next, suppose that \(g_1^*\lt h_1^*\). In this case we show that \(Z_j\) is unbounded. The geometric intuition is as follows: imagine a particle in the plane whose position at time t is \((\cos (b_1t),\cos (b_2t))\), together with a “moving target” whose extent at time t is \(\Gamma _t:=\lbrace (x,y) : (e^{\boldsymbol {a}t},x,y) \in C_j \rbrace\). Intuitively, such a particle must reach \(\Gamma _t\) for arbitrarily large values of t since its orbit is dense in \([-1,1]^2\) and \(\Gamma _t\) “converges” to a subset of \([-1,1]^2\) that has positive dimension.

Proceeding formally, first notice that \(C_j\) cannot be a \((\ldots ,0,1)\)-cell or a \((\ldots ,0,0)\)-cell, for then we would have \(g_1(\boldsymbol {u}) = h_1(\boldsymbol {u})\) for all \(\boldsymbol {u}\in D_j\) and hence \(g_1^*=h_1^*\). Thus, \(C_j\) must either be a \((\ldots ,1,0)\)-cell or a \((\ldots ,1,1)\)-cell. In either case, \(C_j\) includes a cell of the form \(\lbrace (\boldsymbol {u},x,\xi (\boldsymbol {u},x)) : \boldsymbol {u}\in D_j, g_1(\boldsymbol {u}) \lt x \lt h_1(\boldsymbol {u}) \rbrace\) for some continuous semi-algebraic function \(\xi\).

Let \(c,d\) be real algebraic numbers such that \(g_1^*\lt c\lt d\lt h_1^*\). Write \(c=\cos (\psi ^{\prime })\) and \(d=\cos (\psi)\) for \(0\le \psi \lt \psi ^{\prime } \le \pi\). By Proposition 2.10, the limits \(\lim _{t\rightarrow \infty } \xi (e^{\boldsymbol {a}t},c)\) and \(\lim _{t\rightarrow \infty } \xi (e^{\boldsymbol {a}t},d)\) exist and are algebraic numbers in the interval \([-1,1]\). Let \(\theta ,\theta ^{\prime } \in [0,\pi ]\) be such that \(\cos (\theta)=\lim _{t\rightarrow \infty } \xi (e^{\boldsymbol {a}t},d)\) and \(\cos (\theta ^{\prime }) = \lim _{t\rightarrow \infty } \xi (e^{\boldsymbol {a}t},c)\).

Since \(\cos (\theta)\), \(\cos (\theta ^{\prime })\), \(\cos (\psi)\), and \(\cos (\psi ^{\prime })\) are algebraic, \(e^{i(\theta ^{\prime }-\theta)}\) and \(e^{i(\psi ^{\prime }-\psi)}\) are also algebraic, and hence by the Gelfond-Schneider theorem (Theorem 2.1), the quotient \(\frac{\theta ^{\prime }-\theta }{\psi ^{\prime }-\psi }\) is either rational or transcendental. In particular, we know that it is not equal to \(\frac{b_2}{b_1}\), which is algebraic and irrational. Let us suppose that \(\frac{\theta ^{\prime }-\theta }{\psi ^{\prime }-\psi } \gt \frac{b_2}{b_1}\) (the converse case is almost identical). Then there exists \(\theta ^{\prime \prime }\) with \(\theta \lt \theta ^{\prime \prime }\lt \theta ^{\prime }\), such that (18) \(\begin{gather} \theta \lt \theta ^{\prime \prime } + \frac{b_2}{b_1}(\psi ^{\prime }-\psi) \lt \theta ^{\prime }. \end{gather}\)

Since \(2\pi ,b_1,b_2\) are linearly independent over \(\mathbb {Q}\), it follows from Proposition 2.6 that \(\lbrace (b_1t,b_2t)\bmod 2\pi : t \in \mathbb {R}_{\ge 0} \rbrace\) is dense in \([0,2\pi)^2\). Thus, there is an increasing sequence \(t_1\lt t_2\lt \cdots\), with \(b_1t_n \equiv \psi \bmod 2\pi\) for all n, such that \(b_2t_n \bmod 2\pi\) converges to \(\theta ^{\prime \prime }\). Then, defining \(s_1\lt s_2\lt \cdots\) by \(s_n = t_n + \frac{\psi ^{\prime }-\psi }{b_1}\), we have \(b_1s_n \equiv \psi ^{\prime } \bmod 2\pi\) for all n and, by Equation (18), \(\begin{equation*} \lim _{n \rightarrow \infty } b_2s_n = \lim _{n\rightarrow \infty } b_2t_n +\frac{b_2}{b_1}(\psi ^{\prime }-\psi) = \theta ^{\prime \prime } + \frac{b_2}{b_1}(\psi ^{\prime }-\psi) \lt \theta ^{\prime } \quad (\bmod \; 2\pi). \end{equation*}\)

Let \(\eta (t)=\xi (e^{\boldsymbol {a}t},\cos (b_1t)) - \cos (b_2t)\). Then for \(t\in R\) such that \(g(e^{\boldsymbol {a}t})\lt \cos (b_1t)\lt h(e^{\boldsymbol {a}t})\), \(\begin{eqnarray*} \eta (t)=0&\Longrightarrow & \cos (b_2t)=\xi (e^{\boldsymbol {a}t},\cos (b_1t))\\ &\Longrightarrow & (e^{\boldsymbol {a}t},\cos (b_1t),\cos (b_2t)) \in C_j\\ &\Longrightarrow & t\in Z_j {\text{ (by Equation}}(\href{#eq12}{\text{12}})). \end{eqnarray*}\) Now \(\lim _{n\rightarrow \infty }\eta (t_n)=\cos (\theta)-\cos (\theta ^{\prime \prime })\gt 0\) and \(\lim _{n\rightarrow \infty } \eta (s_n)\lt \cos (\theta ^{\prime })-\cos (\theta ^{\prime })=0\). Moreover, for n sufficiently large we have \([t_n,s_n]\subseteq R\). It follows that \(\eta (t)\) has a zero in every interval \([t_n,s_n]\) for n large enough. We conclude that \(Z_j\) is unbounded.

Case III: Finally, the case \(g_2^* \lt h_2^*\) is symmetric to Case II.□

Suppose that f has dominant term \(At^de^{rt}\) for real algebraic numbers \(A\ne 0\) and r and a nonnegative integer d. Then we can write \(\frac{f(t)}{e^{rt}t^d} = A + g(t)\), where \(|g(t)| = O(1/t)\). Moreover, the constants in the asymptotic notation are effective and hence we can compute the desired threshold T.□

We can assume without loss of generality that \(a/b\not\in \mathbb {Q}\), since otherwise f would have at most one rationally linearly independent frequency and the lemma would follow from Theorem 4.1.

Considering the dominant term \(t(A\cos (at+\varphi _1) + B)\), it is clear that if \(|A|\gt |B|,\) then f changes sign infinitely often, whereas if \(|B|\gt |A|\), then for t large enough, \(f(t)\) has the same sign as B. Thus, we can assume \(|A|=|B|\). Dividing f by B, and replacing \(\varphi _1\) by \(\varphi _1+\pi\) if necessary, we can moreover assume that f has the form (19) \(\begin{gather} f(t) = \underbrace{t(1-\cos (at+\varphi _1))}_{\alpha (t)} + \underbrace{(C\cos (at+\varphi _2)+D)}_{\beta (t)} + \underbrace{e^{r_1t}E\cos (bt+\varphi _3)}_{\gamma (t)}. \end{gather}\)

We now focus on the critical times\(t_j\stackrel{\mathrm{def}}{=}\frac{2j\pi -\varphi _1}{a}\), \(j\in \mathbb {N}\), at which the dominant term \(\alpha (t)\) vanishes. Define \(F \stackrel{\mathrm{def}}{=}C\cos (\varphi _2-\varphi _1)+D\) and notice that \(\beta (t_j)=F\) for all \(j \in \mathbb {N}\).

Case (i). Suppose that \(F\le 0\). By the linear independence of \(a,b\) and Proposition 2.6, we have \(\gamma (t_j)\lt 0\) for infinitely many j. Thus, \(f(t_j) = F+\gamma (t_j)\) is negative for infinitely many j. But, considering the dominant term \(\alpha (t)\), it is also clear that \(\lim \sup _{t\rightarrow \infty }f(t)\gt 0\). We conclude that f has infinitely many zeros.

Case (ii). Suppose that \(F \gt 0\). We claim that f is ultimately positive. Since \(\lim \sup _{t\rightarrow \infty } f(t)\gt 0\), it suffices to show that for t sufficiently large, if \(f^{\prime }(t)=0,\) then \(f(t)\gt 0\).

Since \(\beta\) is uniformly continuous, there exists \(\delta \gt 0\) such that for all \(t\in \mathbb {R}_{\ge 0}\) and \(j\in \mathbb {N}\) such that \(|t-t_j|\lt \delta\) we have \(\beta (t)\gt F/2\). Furthermore, since \(f(t) \ge \beta (t)-e^{r_1t}|E|\), we have that \(f(t)\gt 0\) for all sufficiently large \(t\in \mathbb {R}_{\ge 0}\) and \(j\in \mathbb {N}\) such that \(|t-t_j|\lt \delta\).

Let t be such that \(f^{\prime }(t)=0\) and choose \(j \in \mathbb {N}\) such that \(|at+\varphi _1-j\pi | \le \frac{\pi }{2}\). If j is odd, then \(\cos (at+\varphi _1) \le 0\) and hence \(f(t) \ge t+\beta (t)+\gamma (t)\) is positive for t sufficiently large. Now suppose that j is even—say \(j=2k\) for some k. From the equation \(f^{\prime }(t)=0\) we have (20) \(\begin{gather} |at\sin (at+\varphi _1)| = |\cos (at+\varphi _1) - 1 + \beta ^{\prime }(t)+\gamma ^{\prime }(t) | \le G \end{gather}\) for some positive constant G. It follows that \(\begin{align*} |t-t_j| & \, =\, \frac{1}{a} \left|at+\varphi _1-2k\pi \right| && \lbrace \mbox{defn. of $t_k$, $j=2k$} \rbrace \\ & \,\le \, \frac{2}{a} |\sin (at+\varphi _1-2k\pi)| && \lbrace \mbox{since $|x|\le 2|\sin (x)|$ for all $x\in [-\frac{\pi }{2},\frac{\pi }{2}]$} \rbrace \\ & \, \le \, \frac{2G}{a^2t} && \lbrace {\text{by Equations }{\text(\href{#eq20}{20})}} \rbrace . \end{align*}\) We conclude that for t sufficiently large, \(|t-t_j|\lt \delta\) and hence, as observed above, \(f(t)\gt 0\).□

We can assume without loss of generality that \(a/b\not\in \mathbb {Q}\), since otherwise the claim follows from Theorem 4.1. If \(|A_1| \lt |A_2|\), then f has infinitely many zeros (consider large t such that \(\cos (at+\varphi)=\pm 1\)), whereas if \(|A_1| \gt |A_2|\), then the sign of f will eventually match that of \(A_1\), so f has only finitely many zeros. We are left with the case that \(|A_1| = |A_2|\). Dividing f through by \(A_1\) and replacing \(\varphi\) by \(\varphi +\pi\) if necessary, we can write the exponential polynomial as \(\begin{equation*} f(t) = \underbrace{t(1 - \cos (at + \varphi))}_{\alpha (t)} + \underbrace{A\cos (at + \varphi _1) + B\cos (bt + \varphi _2) + C}_{\beta (t)}. \end{equation*}\) The analysis centers around the behavior of f around the sequence of critical points \(t_j \stackrel{\mathrm{def}}{=}\frac{2j\pi - \varphi }{a}\), \(j\in \mathbb {N}\), at which the dominant term \(\alpha (t)\) is zero. We distinguish three cases, based on comparing \(A\cos (\varphi _1-\varphi) + C\) with \(|B|\).

Case (i). Suppose first that \(A\cos (\varphi _1-\varphi) + C \lt |B|\). We claim that f has infinitely many zeros. Writing \(\varepsilon = |B| - A\cos (\varphi _1-\varphi) - C \gt 0\), we have \(\begin{align*} f(t_j) & = A\cos (\varphi _1-\varphi) + C + B\cos (bt_j + \varphi _2) \\ & = |B| - \varepsilon + B\cos (bt_j + \varphi _2). \end{align*}\) By the linear independence of \(a, b\) and Proposition 2.6, we have that for infinitely many j, \(|B| + B\cos (bt_j+\varphi _2) \in [0, \varepsilon /2]\), say, and hence \(f(t_j) \lt -\varepsilon /2 \lt 0\). Since also \(\lim \sup _{t\rightarrow \infty } f(t)\gt 0\), we conclude that f has infinitely many zeros, as claimed.

Case (ii). Second, suppose that \(A\cos (\varphi _1-\varphi)+C \gt |B|\). We argue that f has finitely many zeros. To start, we note that for all \(j\in \mathbb {N}\), \(\begin{eqnarray*} \beta (t_j)&=& A(\cos (\varphi _1-\varphi)+C+B\cos (bt+\varphi _2) \\ &\ge & A(\cos (\varphi _1-\varphi)+C-|B| \\ & \gt & 0. \end{eqnarray*}\) Since moreover \(\beta\) is uniformly continuous, there exists \(\delta \gt 0\) such that \(\beta (t)\gt 0\) for all \(t\in \mathbb {R}_{\ge 0}\) and \(j\in \mathbb {N}\) such that \(|t-t_j|\lt \delta\). In particular, since \(f(t)\ge \beta (t)\) for all t, we have \(f(t) \gt 0\) for all t and j such that \(|t-t_j|\lt \delta\).

Clearly \(\lim \inf _{t\rightarrow \infty } f(t) \gt 0\). Hence, to show that f is ultimately positive, it suffices to show that \(f(t)\gt 0\) for all sufficiently large t such that \(f^{\prime }(t)=0\). To this end, let t be such that \(f^{\prime }(t)=0\) and choose \(j \in \mathbb {N}\) such that \(|at+\varphi -j\pi | \le \frac{\pi }{2}\). If j is odd, then \(\cos (at+\varphi) \le 0\) and hence \(f(t)\ge t+\beta (t)\) is positive for t sufficiently large. Now suppose that j is even—say \(j=2k\) for some k. From the equation \(f^{\prime }(t)=0\) we get that (21) \(\begin{gather} |at\sin (at+\varphi)| = |\cos (at+\varphi)-1+\beta ^{\prime }(t)| \le D \end{gather}\) for some positive constant D. It follows that \(\begin{align*} |t-t_j| & \, =\, \frac{1}{a} \left|at+\varphi -2k\pi \right| && \lbrace \text{defn. of $t_j$, $j=2k$} \rbrace \\ & \,\le \, \frac{2}{a} | \sin (at+\varphi -2k\pi)| && \lbrace \text{since $|x|\le 2|\sin (x)|$ for all $x\in [-\frac{\pi }{2},\frac{\pi }{2}]$} \rbrace \\ & \, \le \, \frac{2D}{a^2t}. && \lbrace {\text{by Equations }{\text(\href{#eq21}{21})}}\rbrace . \end{align*}\) Thus, for t sufficiently large, \(|t-t_j|\lt \delta\) and hence, as observed above, \(f(t) \gt 0\).

Case (iii). Finally, suppose \(A\cos (\varphi _1-\varphi) + C = |B|\). We claim that f has finitely many zeros if and only if \(\varphi _1-\varphi =k\pi\) for some \(k\in \mathbb {Z}\).

Suppose first \(\varphi _1 - \varphi = k\pi\) for some \(k\in \mathbb {Z}\). Then \(|B|=A\cos (\varphi _1-\varphi) + C=A(-1)^k + C\), so \(\begin{align*} f(t) & = t(1-\cos (at + \varphi)) + A\cos (at+\varphi +k\pi) + B\cos (bt+\varphi _2) +C \\ & = (t - A(-1)^k)(1-\cos (at + \varphi)) + (|B|+B\cos (bt+\varphi _2)). \end{align*}\) For \(t\ge |A|\), \(f(t) = 0\) is equivalent to \(\begin{equation*} \cos (at + \varphi) = 1 \text{ and } \cos (bt + \varphi _2) = -\mbox{sign}(B), \end{equation*}\) which entails that \(e^{iat}, e^{ibt}\) are both algebraic. Then by the Gelfond-Schneider theorem (Theorem 2.1), either \(t=0\) or \(a/b\in \mathbb {Q}\), which is a contradiction. Therefore, \(f(t)\) has no zeros \(t\ge |A|\).

Finally, suppose that \(\varphi _1-\varphi \ne k\pi\) for all \(k\in \mathbb {Z}\) and \(A\cos (\varphi _1-\varphi) + C = |B|\). We claim that f has infinitely many zeros on \([0, \infty)\). Without loss of generality, by replacing \(\varphi _2\) with \(\varphi _2+\pi\) if necessary, assume that \(B \lt 0\). Note also that \(\sin (\varphi _1-\varphi)\ne 0\).

By Theorem 2.7, there exist infinitely many \(k\in \mathbb {N}\) such that there exists a corresponding \(l\in \mathbb {N}\) satisfying (22) \(\begin{align} \left| \frac{b}{a}k - l + \frac{\varphi _2-\varphi \frac{b}{a}}{2\pi }\right| \lt \frac{2}{k}. \mbox{22} \end{align}\) For each such k, we consider \(\begin{equation*} s_k \stackrel{\mathrm{def}}{=}\frac{1}{a} \left(2k\pi -\varphi +\frac{\delta }{k}\right), \end{equation*}\) where \(\delta\) is a real constant chosen such that \(\mathrm{sign}(\delta A \sin (\varphi _1-\varphi)) = 1\) and \(|\delta |\) is sufficiently small (to be specified later). We will show that \(f(s_k)\le 0\) for all sufficiently large k. Since also \(\lim \sup _{t\rightarrow \infty } f(t)\gt 0\), this suffices to show that f has infinitely many zeros. To this end, we will separately bound the terms \(s_k(1-\cos (as_k+\varphi))\), \(A\cos (as_k+\varphi _1)\) and \(B\cos (bs_k + \varphi _2)\) for all k large enough, and hence bound \(f(s_k)\) from above by zero.

First, we consider the dominant term of \(f(s_k)\): \(\begin{align} s_k(1-\cos (as_k+\varphi)) & \, = \, \frac{2k\pi - \varphi + \frac{\delta }{k}}{a}\left(1-\cos \left(\frac{\delta }{k}\right)\right) && \lbrace \text{definition of}\; s_k \rbrace \end{align}\) \(\begin{align} & \,\le \, \frac{2k\pi - \varphi + \frac{\delta }{k}}{a}\frac{\delta ^2}{2k^2} && \lbrace s_k \gt 0 \text{and} 1-\cos (x) \le x^2/2 \rbrace \end{align}\) \(\begin{align} & \, \le \frac{\pi \delta ^2}{a}\frac{1}{k} - \frac{\varphi \delta ^2}{2a}\frac{1}{k^2} + \frac{\delta ^3}{2a}\frac{1}{k^3} && \lbrace \text{rearranging} \rbrace \end{align}\) (23) \(\begin{align} & \, \le \, \frac{2\pi \delta ^2}{a} \frac{1}{k} \, && \lbrace \text{for large enough}\; k \rbrace . \end{align}\)

Next we bound the term \(A\cos (as_k+\varphi _1)\). Here note that for a differentiable function \(f:\mathbb {R}\rightarrow \mathbb {R}\) and \(x_0\in \mathbb {R}\) such that \(f^{\prime }(x_0)\ne 0\), we have \(f(x_0+\varepsilon) \le f(x_0) + \varepsilon \frac{f^{\prime }(x_0)}{2}\) for all \(\varepsilon\) with \(\mathrm{sign}(\varepsilon f^{\prime }(x_0)) = -1\) and \(|\varepsilon |\) sufficiently small. Applying this observation to \(f(x)=A\cos x\), for \(|\delta |\) sufficiently small we have \(\begin{align} A\cos (as_k + \varphi _1) \, = \, & A\cos \left(\varphi _1 - \varphi + \frac{\delta }{k}\right) && \lbrace \mbox{definition of $s_k$} \rbrace \end{align}\) (24) \(\begin{align} \,\le \, & A \cos (\varphi _1 - \varphi) - \frac{\delta }{k} \frac{A\sin (\varphi _1-\varphi)}{2} \, && \lbrace \mbox{$\mathrm{sign}(\delta A \sin (\varphi _1-\varphi)) = 1$} \rbrace . \end{align}\)

Finally, we use Equation () to bound \(B\cos (bs_k+\varphi _2)\): \(\begin{align*} & \cos (bs_k + \varphi _2) & \lbrace \text{definition of $s_k$} \rbrace \\ = & \,\cos \left(\frac{b}{a}\left(2k\pi - \varphi + \frac{\delta }{k}\right) + \varphi _2\right) & \lbrace \text{rearranging} \rbrace \\ = & \,\cos \left(2\pi \left(\frac{b}{a}k + \frac{\varphi _2-\frac{b}{a}\varphi }{2\pi }\right) + \frac{b\delta }{ak}\right) & \lbrace {\text{by Equations }{\text(\href{#eq22}{22})}} \text{for some $l\in \mathbb {N}$ and $\varepsilon \in \mathbb {R}$ with $|\varepsilon |\lt 2/k$} \rbrace \\ = & \,\cos \left(2\pi (l + \varepsilon) + \frac{b\delta }{ak}\right) & \lbrace \text{by the inequality $1-\cos (x) \le x^2 / 2$ for all $x\in \mathbb {R}$} \rbrace \\ \ge & \,1 - \frac{1}{2}\left|2\pi \varepsilon + \frac{b\delta }{ak}\right|^2 & \lbrace \text{by the triangle inequality and $|\varepsilon | \lt 2/k$}\rbrace \\ \ge & \,1 - \frac{1}{2k^2}\left(4\pi + \frac{b|\delta |}{a}\right)^2. \end{align*}\) Hence, noting that \(B\lt 0\), we have (25) \(\begin{align} B\cos (bs_k+\varphi _2) \le B\left(1 - \frac{1}{2k^2}\left(4\pi + \frac{b|\delta |}{a}\right)^2\right). \mbox{32} \end{align}\)

Combining the above three bounds, we have \(\begin{align*} & f(s_k) = s_k(1-\cos (as_k+\varphi)) + A\cos (as_k + \varphi _1) + B\cos (bs_k + \varphi _2) + C \\ & \lbrace {\text{by Equations }{\text(\href{#eq23}{23})}}, {\text(\href{#eq24}{24})}, \text{and} {\text(\href{#eq25}{25})} \rbrace \\ \le & \, \frac{2\pi \delta ^2}{ak} + A \cos (\varphi _1-\varphi) - \frac{\delta }{2k} A \sin (\varphi _1-\varphi) + B\left(1 - \frac{1}{2k^2}\left(4\pi + \frac{b|\delta |}{a}\right)^2\right) + C \\ & \lbrace \text{by $A\cos (\varphi _1-\varphi)+C = |B|$ and $B \lt 0$} \rbrace \\ = &\, \frac{1}{k}\left(\frac{2\pi \delta ^2}{a} - \frac{\delta A \sin (\varphi _1 - \varphi)}{2} \right) - \frac{1}{k^2}\frac{B}{2}\left(4\pi + \frac{b|\delta |}{a}\right)^2. \end{align*}\) Since \(\mathrm{sign}(\delta A \sin (\varphi _1-\varphi))=1\), for \(|\delta |\) small enough we have \(\frac{2\pi \delta ^2}{a} - \frac{\delta A \sin (\varphi _1 - \varphi)}{2} \lt 0\). Then for all large enough k, since \(1/k^2\) shrinks faster than \(1/k\), we will have \(\begin{equation*} \mbox{sign}(f(s_k)) = \mbox{sign}\left(\frac{2\pi \delta ^2}{a} - \delta c \right) = -1. \end{equation*}\) Therefore, f is negative infinitely often and hence has infinitely many zeros, as claimed.□

Since all the coefficient polynomials occurring in f are constant, if \(\lbrace a,b,c\rbrace\) is not a linearly independent set over \(\mathbb {Q},\) then the claim follows from Theorem 4.2. Thus, we may assume that \(a,b,c\) are linearly independent over the rationals.

Next, we argue that if \(D\lt |C|+|B|\), then f has infinitely many zeros. Indeed, by Proposition 2.6, the linear independence of \(a,b,c\) entails that the trajectory \((at+\varphi _1, bt+\varphi _2,ct + \varphi _3)\bmod 2\pi\) is dense in \([0,2\pi)^3\), and moreover the restriction of this trajectory to \(at + \varphi _1 \bmod 2\pi = 0\) is dense in \(\lbrace 0\rbrace \times [0, 2\pi)^2\). Thus, \(\lbrace B\cos (bt + \varphi _2) + C\cos (ct + \varphi _3) + D : t\ge 0\mbox{ and }at + \varphi _1 \bmod 2\pi = 0 \rbrace\) is dense in \([D-|C|-|B|, D+|C|+|B|]\). If \(D\lt |C|+|B|\), then because \(r_2 \lt r_1 \lt 0\), it will happen infinitely often that \(\begin{align*} & 1 - \cos (at + \varphi _1) = 0, \\ & e^{r_1t}(B\cos (bt + \varphi _2) + C\cos (ct + \varphi _3) + D) \lt e^{r_1t}\frac{D-|C|-|B|}{2} \lt -|E|e^{r_2t}, \end{align*}\) and hence \(f(t) \lt 0\). On the other hand, f is also positive infinitely often, so f must have infinitely many zeros.

Suppose now that \(D\ge |B|+|C|\gt 0\). By the premise of the lemma, \(E=0\), so f has the form \(\begin{equation*} f(t) = 1-\cos (at + \varphi _1) + e^{r_1t}(B\cos (bt+\varphi _2) + C\cos (ct+\varphi _3) + D). \end{equation*}\) We argue that f has no zeros, except possibly at \(t=0\). Indeed, we have \(B\cos (bt + \varphi _2) + |B|\ge 0\) and \(C\cos (ct + \varphi _3) + |C|\ge 0\) for all t, so \(f(t)\ge 0\) for all t. Then \(f(t) = 0\) if and only if \(D = |B| + |C|\) and simultaneously \(\begin{align*} & \cos (at + \varphi _1) = 1, \\ & \cos (bt + \varphi _2) = -\mbox{sign}(B), \\ & \cos (ct + \varphi _3) = -\mbox{sign}(C), \end{align*}\) which entails that \(e^{iat},e^{ibt},e^{ict}\) are all algebraic. Therefore, by the Gelfond-Schneider theorem (Theorem 2.1), either \(t=0\) or \(a,b,c\) are rational multiples of one another. The latter possibility contradicts the linear independence of \(a,b,c\), so we conclude that f has at most one zero.□

If \(a/b\in \mathbb {Q}\), then the claim follows from Theorem 4.1, so assume that \(a,b\) are linearly independent over the rationals. Then by Proposition 2.6, it happens infinitely often that \(1-\cos (at+\varphi _1) = 0\) and \(Bt\cos (bt + \varphi _2) \lt -|B|t/2\), say. For large enough such t, we will have \(\begin{equation*} e^{r_1t}(Bt\cos (bt + \varphi _2) + C\cos (bt + \varphi _3) + D) + e^{r_2t}E \lt 0, \end{equation*}\) and hence \(f(t) \lt 0\), so f has infinitely many zeros.□

Suppose first \(a/b\in \mathbb {Q}\). Because of the bound on the order of f, \(f_2\) has order at most 3, so it cannot simultaneously have complex exponents and non-constant coefficients. Thus, either f has at most one linearly independent frequency or f has constant coefficients and at most two linearly independent frequencies. Either way, the result follows from Theorems 4.1 and 4.2.

Therefore, assume now that \(a, b\) are linearly independent over \(\mathbb {Q}\). By Proposition 2.6, the trajectory \((at+\varphi _1 \bmod 2\pi ,bt + \varphi _2\bmod 2\pi)\) is dense in \([0,2\pi)^2\), and moreover the restriction of this trajectory to \(at + \varphi _1 \bmod 2\pi = 0\) is dense in \(\lbrace 0\rbrace \times [0, 2\pi)\).

If \(|C| \lt |B|\), then we argue that f is infinitely often negative, and hence has infinitely many zeros. Indeed, \(|C| \lt |B|\) entails the existence of a non-empty interval \(I\subseteq [0,2\pi)\) such that \(\begin{equation*} bt + \varphi _2\bmod 2\pi \in I \text{ implies } B\cos (bt+\varphi _2) + C \lt 0. \end{equation*}\) What is more, we can in fact find \(\varepsilon \gt 0\) and a subinterval \(I^{\prime } \subseteq I\) such that \(\begin{equation*} bt + \varphi _2\bmod 2\pi \in I^{\prime } \text{ implies } B\cos (bt+\varphi _2)+C \lt -\varepsilon . \end{equation*}\) Thus, by density, \(1 - \cos (at + \varphi _1)=0\) and \(B\cos (bt+\varphi _2) + C \lt -\varepsilon\) will infinitely often hold simultaneously. Then just take t large enough to ensure, say, \(|e^{r_2t}f_2(t)| \lt \varepsilon /2\) at these infinitely many points, and the claim follows.

If \(|C| \gt |B|\), then clearly for all large enough t, we have \(\begin{equation*} \mbox{sign}(e^{r_1t}(B\cos (bt+\varphi _2)+C) + e^{r_2t}f_2(t)) = \mbox{sign}(C). \end{equation*}\) If \(C\lt 0\), then f has infinitely many zeros (consider t such that \(\cos (at+\varphi _1)=1\)), while if \(C\gt 0\), then f is ultimately positive.

Thus, suppose now \(|B|=|C|\). Replacing \(\varphi _2\) by \(\varphi _2+\pi\) if necessary, we can write the function as \(\begin{equation*} f(t) = 1-\cos (at + \varphi _1) + Ce^{r_1t}(1-\cos (bt+\varphi _2)) + e^{r_2t}f_2(t). \end{equation*}\) As \(a,b\) are linearly independent, for all t large enough, \(1-\cos (at+\varphi _1)\) and \(1-\cos (bt+\varphi _2)\) cannot simultaneously be ”too small.” More precisely, by Lemma 2.3, there exist effective constants \(E, T, N\gt 0\) such that for all \(t\ge T\), we have \(\begin{align*} 1-\cos (at+\varphi _1) \gt E / t^N \mbox{ or } 1-\cos (bt+\varphi _2) \gt E / t^N. \end{align*}\) Now, if \(C\lt 0\), it is easy to show that f has infinitely many zeros. Indeed, consider the times t where the dominant term \(1-\cos (at + \varphi _1)\) vanishes. For all large enough such t, since \(|1/t^N|\) shrinks more slowly than \(e^{(r_2-r_1)t}\) and \(|f_2(t)|\) is bounded above by a constant (because its dominant exponents are purely imaginary), we will have \(\begin{align*} f(t) & = e^{r_1t}\left(C(1-\cos (bt + \varphi _2)) + e^{(r_2-r_1)t}f_2(t)\right) \\ & \lt e^{r_1t} \left(ECt^{-N} + e^{(r_2-r_1)t}f_2(t)\right) \\ & \le e^{r_1t}\frac{1}{2}ECt^{-N} \\ & \lt 0, \end{align*}\) so f has infinitely many zeros. Similarly, if \(C\gt 0\), we can show that f is ultimately positive. Indeed, for all t large enough, we have \(\begin{align*} f(t) & \ge e^{r_1t}\left(C(1-\cos (bt+\varphi _2)) + e^{(r_2-r_1)t}f_2(t)\right) \\ & \gt e^{r_1t}\left(CE t^{-N} + e^{(r_2-r_1)t}f_2(t)\right) \\ & \gt \frac{CE}{2}t^{-N} \\ & \gt 0 \end{align*}\)

or

\(\begin{align*} f(t) & \ge 1-\cos (at+\varphi _1) + e^{r_2t}f_2(t) \\ & \gt E t^{-N} + e^{r_2t}f_2(t) \\ & \gt \frac{E}{2}t^{-N} \\ & \gt 0. \end{align*}\) Therefore, f has only finitely many zeros, all occurring up to an effective bound.□

Suppose we are given an exponential polynomial f of order at most 8. By dividing f through by \(e^{rt}\) if necessary, where r is the real part of the dominant exponents of f, we may assume without loss of generality that \(r = 0\) and that all non-dominant exponents have strictly negative real part.

Throughout this proof, we will use the following notational conventions. We denote by \(r_1,r_2,\ldots\) the real parts of the non-dominant exponents of f. These are real algebraic numbers such that \(0 \gt r_1 \gt r_2 \gt \cdots\). The lowercase letters \(a,b,c\) will denote frequencies of f, always real algebraic and strictly positive. Uppercase letters \(A,B,C,\ldots\) will denote real algebraic numbers (of arbitrary sign), while \(\varphi ,\varphi _1,\varphi _2,\varphi _3\) will denote real numbers such that \(e^{i\varphi },e^{i\varphi _1},e^{i\varphi _2},e^{i\varphi _3}\) are algebraic. For an exponential polynomial \(f(t)=\sum _{j=1}^m P_j(t)e^{\lambda _j t}\), we say that exponent \(\lambda _j\) has multiplicity \(\mathrm{mul}(\lambda _j):=\mathrm{deg}(P_j)\).

We will now perform a case analysis on the number of dominant exponents. Throughout, we rely on the equivalent general forms of real-valued exponential polynomials, outlined in Section 2.1. By Lemma 5.2, it is sufficient to confine our attention to exponential polynomials with an odd number of dominant exponents. By Lemma 5.1 above, the claim is already proven for exponential polynomials with a single dominant term, corresponding to a real exponent.

Case I. Suppose first that f has seven dominant exponents: namely 0 and \(\pm ai, \pm bi, \pm ci\) for non-zero real algebraic numbers \(a,b,c\). The dominant complex exponents must all have multiplicity one since otherwise the order of f would exceed 8. If \(\mathrm{mult}(0) \gt 1\), then 0 has strictly greater multiplicity than the other dominant exponents and the claim follows from Lemma 5.1. Thus, we can assume all dominant exponents have multiplicity one.

The sum of the multiplicities of the dominant exponents is 7, so f can have at most one other, necessarily real, exponent without exceeding order 8, and this exponent must necessarily have multiplicity one.

We now consider the dimension of rational vector spaced spanned by \(\lbrace a,b,c\rbrace\). If the dimension is at most 2, then the claim follows from Theorems 4.1 and 4.2, whereas if the dimension is 3, then we are done by Lemma 5.3. This completes Case I.

Case II. Next, suppose f has five dominant exponents: 0 and \(\pm ai, \pm bi\) for non-zero real algebraic numbers \(a,b\). The bound on the order of f guarantees that \((\mathrm{mult}(\pm ai),\mathrm{mult}(\pm bi)) \in \lbrace (1,2), (2,1), (1,1)\rbrace\).

Assume first that \((\mathrm{mult}(\pm ai),\mathrm{mult}(\pm bi)) = (2, 1)\). (The case \((1,2)\) is symmetric.) If \(\mathrm{mult}(0)\ne 2\), then the claim follows from Lemma 5.2, since then f has a single dominant term that corresponds to a non-real exponent. Thus, assume that \(\mathrm{mult}(0)=2\), so f is given by \(\begin{equation*} f(t) = t(A_1 + A_2\cos (at + \varphi)) + A\cos (at + \varphi _1) + B\cos (bt + \varphi _2) + C. \end{equation*}\) The claim is proven for exponential polynomials of this form in Lemma 5.5.

We can now assume that the complex dominant exponents all have multiplicity one: \(\mathrm{mult}(\pm ai) = \mathrm{mult}(\pm bi) = 1\). If \(\mathrm{mult}(0) \gt 1\), then we are done by Lemma 5.1, so assume \(\mathrm{mult}(0)=1\). If \(a/b\not\in \mathbb {Q}\), the claim follows from Lemma 5.3, so assume \(a/b\in \mathbb {Q}\). If f has no complex exponents other than \(\pm ai\) and \(\pm bi\), then the \(\mathbb {Q}\)-span of the frequencies of f has dimension one and the claim follows by Theorem 4.1. On the other hand, if f has another conjugate pair of complex exponents, say \(r_1\pm ci\) with \(r_1 \lt 0\), then by the order bound of 8, f has constant coefficients and the \(\mathbb {Q}\)-span of the frequencies has dimension at most two, so the claim follows by Theorems 4.1 and 4.2.

Case III. Finally, we can assume that f has three dominant exponents: one real 0 and a complex pair \(\pm ai\). By Lemmas 5.1 and 5.2, the claim is proven when \(\mathrm{mult}(0)\ne \mathrm{mult}(\pm ai)\). Taking into account the order bound on f, we have \(\mathrm{mult}(0)=\mathrm{mult}(\pm ai)\le 2\). If \(\mathrm{mult}(0)=\mathrm{mult}(\pm ai)=2\), then f has the form \(\begin{equation*} f(t) = t(A\cos (a t+\varphi _1) + B) + (C\cos (a t + \varphi _2) + D) + f_1(t), \end{equation*}\) where \(f_1(t)\) is an exponential polynomial or order at most 2 whose exponents have strictly negative real parts. If the exponents of \(f_1\) are both real, then the claim follows by Theorem 4.1; otherwise, the claim follows from Lemma 5.4.

Suppose now \(\mathrm{mult}(0)=\mathrm{mult}(\pm ai)=1\), so f has the form \(\begin{equation*} f(t) = A_1 + A_2\cos (at + \varphi _1) + f_1(t), \end{equation*}\) where \(f_1\) is an exponential polynomial of order at most 5 whose exponents all have strictly negative real part, so that \(f_1(t)\) tends to 0 exponentially quickly as t grows to infinity. Notice that if \(|A_1| \gt |A_2|\), then f cannot have infinitely many zeros. Indeed, for t sufficiently large we will have \(|f_1(t)| \lt |A_1| - |A_2|\), and the sign of f will match that of \(A_1\). On the other hand, if \(|A_2| \gt |A_1|\), then f will change sign infinitely often. It remains to consider the case that \(|A_1| = |A_2|\). Moreover, dividing through by \(A_1\) and, if necessary, replacing \(\varphi _1\) by \(\varphi _1 + \pi\), we can assume without loss of generality that \(A_1 = 1\) and \(A_2 = -1\), so that \(\begin{equation*} f(t) = 1 - \cos (at + \varphi _1) + f_1(t). \end{equation*}\) Since \(f_1(t)\) converges to zero as t grows to infinity, it is clear that \(\lim \sup _{t\rightarrow \infty } f(t)\gt 0\). Thus, f has infinitely many zeros if also \(\lim \inf _{t\rightarrow \infty } f(t)\lt 0\).

From here onward, we proceed by performing a case split on the dominant exponents of \(f_1\), that is, the exponents of f of the second-greatest real part. (Note in passing that if \(f_1\) is identically zero, then f has infinitely many zeros, occurring at t such that \(\cos (at + \varphi _1) = 1\).) Let the exponents of \(f_1\) have real part \(r_1 \lt 0\).

Case III.1. Suppose first that the exponents of f of the second-largest real part are five in number, that is, a real frequency \(r_1\) and two pairs of complex exponents \(r_1\pm bi\), \(r_1\pm ci\). By the order bound on f, \(\mathrm{mult}(r_1 \pm bi) = \mathrm{mult}(r_1 \pm ci) = \mathrm{mult}(r_1) = 1\), so that f has the form \(\begin{equation*} f(t) = 1-\cos (at+\varphi _1) + e^{r_1t}(B\cos (bt + \varphi _2) + C\cos (ct + \varphi _3) + D). \end{equation*}\) Then the claim follows from Lemma 5.6.

Case III.2. Next, suppose that the exponents of f of the second-largest real part are four in number, that is, two pairs of complex exponents \(r_1\pm bi\), \(r_1\pm ci\). Then f is of the form \(\begin{equation*} f(t) = 1 - \cos (at + \varphi _1) + e^{r_1t}(B\cos (bt + \varphi _2) + C\cos (ct + \varphi _3)) + e^{r_2t}D. \end{equation*}\) The claim follows from Lemma 5.6.

Case III.3. Next, suppose that the exponents of f of the second-largest real part are three in number, a real \(r_1\) and a complex pair \(r_1\pm bi\) with \(\mathrm{mult}(r_1\pm bi) \ge 2\). By the order bound on f, we must have \(\mathrm{mult}(r_1\pm bi) = 2\) and \(\mathrm{mult}(r_1) = 1\), so that f has the form \(\begin{equation*} f(t) = 1 - \cos (at + \varphi _1) + e^{r_1t}(Bt\cos (bt + \varphi _2) + C\cos (bt + \varphi _3) + D). \end{equation*}\) In this case, the claim follows from Lemma 5.7.

Case III.4. Next, suppose that the exponents of f of the second-largest real part are two in number, a complex pair \(r_1\pm bi\) with \(\mathrm{mult}(r_1\pm bi)\ge 2\). By the order bound on f, \(\mathrm{mult}(r_1\pm bi) = 2\) and f can have at most one other exponent, so that \(\begin{equation*} f(t) = 1 - \cos (at + \varphi _1) + e^{r_1t}(Bt\cos (bt + \varphi _2) + C\cos (bt + \varphi _3)) + De^{r_2t}. \end{equation*}\) The claim follows from Lemma 5.7.

Case III.5. Next, suppose that the exponents of f of the second-largest real part are three in number, a simple complex pair \(r_1 \pm bi\) and a (possibly repeated) real exponent \(r_1\).

If \(\mathrm{mult}(r_1) \gt 1\), then by Lemma 5.1, \(f_1\) is ultimately positive or ultimately negative, depending on the sign of the leading term of the polynomial associated with \(e^{r_1t}\). In the former case f is also ultimately positive, while in the latter case f has infinitely many zeros since f is negative for arbitrarily large values of t (consider large t for which \(1-\cos (at+\varphi _1) = 0\)).

Therefore, assume \(\mathrm{mult}(r_1) = \mathrm{mult}(r_1\pm bi)=1\), so that \(\begin{equation*} f(t) = 1 - \cos (at + \varphi _1) + e^{r_1t}(B\cos (bt + \varphi _2) + C) + f_2(t), \end{equation*}\) where \(f_2\) is an exponential polynomial of order at most 2 whose dominant exponents have real part \(r_2\) such that \(r_2 \lt r_1\). The claim follows from Lemma 5.8.

Case III.6. Next, suppose that the exponents of f of the second-largest real part are a simple complex pair \(r_1\pm bi\), so that \(\begin{equation*} f(t) = 1 - \cos (at + \varphi _1) + e^{r_1t}B\cos (bt + \varphi _2) + f_2(t), \end{equation*}\) where \(f_2\) is an exponential polynomial of order at most 3 whose dominant exponents have real part strictly smaller than \(r_1\). Then the result follows from Lemma 5.8.

Case III.7. Finally, if the exponents of f of the second-largest real part comprise a single real exponent \(r_1\), then by Lemma 5.1, \(f_1\) is ultimately positive or ultimately negative, depending on the sign of the leading coefficient of the polynomial associated with \(e^{r_1t}\). In the former case, f is also ultimately positive, while in the latter, f has infinitely many zeros.

This completes the proof.□