Problems and Theorems in Analysis · 1

If we have only $\alpha >1$, then the integral still represents an entire function. If $\alpha$ is not an even integer, this function has only finitely many real zeros.

Let $\alpha_1, \alpha_2, \ldots , \alpha_s$ denote the zeros of $G(z)$ in $[0,m]$,

and set

Then argue as follows:

(1)

has precisely $m$ changes of sign (by 47 and the definition of $G^*(z)$) and precisely $m$ positive zeros (by a slight modification of 167---make use of 67 more completely).

(2)

has precisely $m-1$ changes of sign (attention to the various possible cases) and so by 36 at most $m-1$ positive zeros. Yet it has (by a slight modification of solution 66) at least $m-1$, and so finally exactly $m-1$, positive zeros;

(3) Introduce successively $\alpha_2, \alpha_3, \ldots , \alpha_s$ and repeat appropriately the arguments of (2) (attention at each step to the changes wrought by the preceding steps) to arrive at the desired proposition.

We make the following assumptions: The entire function

is a limit of polynomials with only positive zeros, but not itself a polynomial [161].

The entire function $G(z)$ is either defined as in 165 or it is equal to the function defined in 165 multiplied by $z$; it has precisely $s$ non-negative zeros which are simple and the distance between any two of them is not less than $1$. (We obtain the present $G(z)$ from the $G(z)$ of 171.1 by making $m$ arbitrarily large.)

Then the entire function

has precisely $s$ non-positive zeros.

If $\nu \geq 0$ the entire function

has precisely $[\nu ]$ non-positive zeros. (This contains the result of 159.)

Let $k, l, m$ and $q$ be non-negative integers,

If all the zeros of the polynomial

are negative, then the zeros of

are also all negative. (This implies, for $n \to \infty$, the result of 160.)

If all the zeros of the polynomial

are negative, then the zeros of

are also all negative; $k, l, m$ and $q$ are subject to the same conditions as in 171.4.