Why does $\sum a_i \exp(b_i)$ always have root?

Solution 1:

This follows from the theory of entire functions of finite order in complex analysis. Specifically, we have:

Proposition: Suppose $f(z) = \sum_{i=1}^j a_i \exp b_i$ for some polynomials $a_i,b_i$ (which may have complex coefficients, though the question specifies real polynomials). Then if $f\,$ has no complex zeros then there exists a polynomial $P$ such that $f = \exp P$.

Proof: let $d = \max_i\max(\deg a_i,\deg b_i)$. If $d \leq 0$ then $f$ is constant and we may choose for $P$ a constant polynomial. Else there exists a constant $A$ such that $\left|\,f(z)\right| \leq \exp(A\left|z\right|^d)$ for all complex $z$. This makes $f$ an entire function of order at most $d$. If $f$ has no zeros then $f = e^g$ for some analytic function $g$, and it follows that $g$ is a polynomial (by a special case of the Hadamard product for an entire function of finite order). $\Box$

Moreover, once we put the expansion $f(z) = \sum_{i=1}^j a_i \exp b_i$ in normal form by assuming that each $b_i$ vanishes at zero (else subtract $b_i(0)$ from $b_i$ and multiply $a_i$ by $e^{b_i(0)}$), then at least one of the $b_i$ is $P-P(0)$, and we can cancel and combine terms to identify $f$ with $\exp P$. The proof (by considering behavior for large $|z|$) is somewhat tedious, though much easier in the real case [hint: start by writing $f(z) \, / \exp P(z)$ as $\sum_{i=1}^j a_i \exp (b_i-P)$]. In particular, if $j>1$ and no two $b_i$ differ by a constant then $f$ cannot equal $\exp P$ and thus must have complex zeros.