Problems and Theorems in Analysis · 6

Let $a_1, a_2, \ldots , a_n, \lambda_1, \lambda_2, \ldots , \lambda_n$ be real constants, $\lambda_1 < \lambda_2 < \lambda_3 < \cdots < \lambda_n$. Denote by $Z$ the number of real zeros of the entire function

and by $C$ the number of changes of sign in the sequence of numbers $a_1, a_2, \ldots , a_n$. Then $C-Z$ is a non-negative even integer. (A generalization of 36, 37 different from 41: Replace $e^x$ by $x$.)

Proof. We may assume without loss of generality that $a_1, a_2, \ldots a_n$ are all different from $0$. That $C-Z$ is even is evident from the fact that as $x \to -\infty$ the term $a_1 e^{\lambda_1 x}$ is the dominant term, and as $x \to +\infty$ the term $a_n e^{\lambda_n x}$ is the dominant term. [8, 9, 37.] That $C-Z \geq 0$ may be proved by mathematical induction; we deduce the relation for $C$ from that for $C-1$, with the aid of Rolle's theorem [solution 75]. Indeed, if there are no changes of sign, evidently $Z$ is also $=0$ and the statement holds. Assume that it holds in the case that there are $C-1$ changes of sign. $F(x)$ has $C$ changes of sign, $C \geq 1$; e.g. let $\alpha +1$ be an index of change, $1 \leq \alpha < n$, $a_{\alpha }a_{\alpha +1} < 0$. Choose a number $\lambda$ in the interval $\lambda_{\alpha } < \lambda < \lambda_{\alpha +1}$ and consider the function

Let the number of zeros of $F^*(x)$ be $Z^*$. We have [6, 12]

Let the number of changes of sign of the sequence of coefficients

be $C^*$. Clearly

By the assumption of the mathematical induction argument we have

From the three relations above it follows that

q.e.d.---Was it necessary to assume from the outset that $a_1, a_2, \ldots , a_n$ are all $\neq 0$? Could we not also have chosen $\lambda =\lambda_{\alpha }$ or $\lambda =\lambda_{\alpha +1}$?

The entire function

has simple zeros at the points

and no other real zeros. (Interpret $k^x$ as $e^{x \log k}$, with the principal value of $\log k$.)