Is there a theorem which provides conditions under which a power series satisfies the reciprocal root sum law?

Question

Kalman - Six ways to sum a series discusses Euler's original proof for the Basel problem $\sum\limits_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6} $:

$$\frac{\sin(\sqrt x)}{\sqrt x} = 1- \frac{x}{3!}+ \frac{x^2}{5!}- \frac{x^3}{7!}+\dotsb.$$ The roots of $\frac{\sin(\sqrt x)}{\sqrt x}$ are the numbers $\pi^2, 4\pi^2, 9\pi^2, 16\pi^2, \dots$. Now Euler knew that adding up the reciprocals of all the roots of a polynomial results in the negative of the ratio of the linear coefficient to the constant coefficient. In symbols, if $$(x − r_1)(x − r_2)\dotsm(x − r_n) = x^{n} + a_{n−1}x^{n−1} + \dots + a_1x + a_0$$ then $$\sum_{k=1}^n \frac{1}{r_k}= \frac{-a_1}{a_0}.$$ Assuming that the same law must hold for a power series expansion, he applied it to $\frac{\sin(\sqrt x)}{\sqrt x}$ , concluding that $$\frac{1}{6}=\sum\limits_{n=1}^\infty \frac{1}{(\pi n)^2}.$$ Why is this not considered a valid proof today? The problem is that power series are not polynomials, and do not share all the properties of polynomials. $$\frac{1}{1 − x}=1+ x + x^2 + x^3 + \dotsb$$ holds for all $x$ of absolute value less than $1$. Now consider the function $g(x)=2 − \frac 1 {1 − x}$. Clearly, $g$ has a single root, $\frac 1 2$. The power series expansion for $g(x)$ is $ 1−x−x^2−x^3−\dotsb$, so $a_0 = 1$ and $a_1 = −1$. The sum of the reciprocal roots does not equal the ratio $−\frac{a_1}{a_0}$. While this example shows that the reciprocal root sum law cannot be applied blindly to all power series, it does not imply that the law never holds. Indeed, the law must hold for the function $\frac{\sin(\sqrt x)}{\sqrt x}$ because we have independent proofs of Euler’s result. Notice the differences between $\frac{\sin(\sqrt x)}{\sqrt x}$ and the $g$ of the counterexample. The function $\frac{\sin(\sqrt x)}{\sqrt x}$ has an infinite number of roots, where $g$ has but one. And $\frac{\sin(\sqrt x)}{\sqrt x}$ has a power series that converges for all $x$, where the series for $g$ only converges for $−1 <x< 1$.

Is there a theorem which provides conditions under which a power series satisfies the reciprocal root sum law? I don’t know.

Is there a known theorem that specifies conditions under which a power series representation satisfies the reciprocal root sum law? My search hasn't yielded any such theorem. Does such a theorem exist, and if so, what are the conditions it outlines?

Answer 1

A sufficient condition is that the function $$f(z)=\sum_0^\infty a_nz^n$$ is entire, of order less than $1$. The order $\rho$ can be determined from the coefficients: $$\rho=\limsup_{n\to\infty}\frac{n\log n}{\log(1/|a_n|)}.$$ Once you know that $\rho<1$, Hadamard's factorization theorem gives that $$f(z)=z^m\prod_n\left(1-\frac{z}{c_n}\right),$$ where $$\sum\frac{1}{|c_n|}<\infty,\quad\quad\quad\quad (1)$$ and your "reciprocal roots formula" follows since the product is convergent, so you can formally perform the multiplication and obtain the usual connection between zeros and coefficients.

In your example, the order is $1/2$. The condition that order is less than $1$ can be slightly relaxed, that is the formula still holds for some functions of order $1$, see, for example Levin, Distribution of zeros of entire functions, AMS 1980.

ADDED: Using arguments from this book one can derive a necessary and sufficient condition in terms of $a_n$ for the series (1) to converge (absolutely) and for validity of your formula. Let $$m(r)=\sum_n|a_n|r^n,\quad \mu(r)=\max_n|a_n|r^n$$ Then, if $\rho<1$, then $$\log m(r)\sim\log\mu(r),\quad r\to\infty$$ (Polya-Szego, Problems and theorems... IV, I 54) and the necessary and sufficient condition is $$\int_1^\infty\frac{\log m(r)}{r^2}dr<\infty,$$ or the equivalent condition $$\int_1^\infty\frac{\log\mu(r)}{r^2}dr<\infty.$$ One can also discuss the validity of your formula in the generalized sense, when the series $$\sum_n\frac{1}{c_n}:=\lim_{r\to\infty}\sum_{|c_n|\leq r}\frac{1}{c_n}$$ is not absolutely convergent.

A generalization to any finite order can be obtained by using the full strength of the Hadamard's theorem.