My motivation to the following question stems from the discussion at Zeros of "exponential" function about the real zeroes of Stirling numbers of the second kind, I am curious in exploring the complex zeroes of Stirling functions of the second kind.

Define: $S_{(x,n)}=\frac{1}{n!}\sum\limits_{k=1}^{n}{n\choose k}(-1)^{n-k}k^{x}$ for integer $n$ and complex $x$.

I conjecture that for $n>2$, $S_{(x,n)}$ has exactly $n-1$ complex zeroes. I realize this is a much stronger claim than the fact that $S_{(x,n)}$ has $n-1$ real zeroes. However, I have noticed after examining a large amount of mathematical data that for $x>n$, both the real parts and the imaginary parts of $S_{(x,n)}$ seem to alternate for increasing integer $n$ when $x$ is a complex number.

When $x$ is a real number, the imaginary part of $S_{(x,n)}$ is zero and we can say that the imaginary part alternates as well for real $x$ from $+0=0$ to $-0=0$. My question becomes the following. If we consider $S_{(x,n)}$ for complex $x$ as a linear combination of integer exponential functions that are defined for complex $x$, does it follow that for $n>2$, $S_{(x,n)}$ has exactly $n-1$ complex zeroes?

$n=2$ is an exception because $S_{(x,2)}$ is periodic in the imaginary part $b$ of $x=a+bi$ with period $\frac{2\pi}{\log{3}}$. This is an exception because the equation $S_{(x,2)}=0$ is equivalent to the equation $2^{x}-2=0$ which only has one exponential term that is non-constant. Since $a_{2}\frac{2\pi}{\log{2}}=a_{3}\frac{2\pi}{\log{3}}=a_{n}\frac{2\pi}{\log{n}}$ presumably does not have a solution for $a_{2}, ..., a_{n}$ all integers except for when $n=2$ it is unlikely that there are any other complex zeroes of $S_{(x,n)}$ in the case that $n\ne 2$ other than $x=1$, ..., $n-1$.

Can someone help me think of a way to show that $a_{2}\frac{2\pi}{\log{2}}=a_{3}\frac{2\pi}{\log{3}}=a_{n}\frac{2\pi}{\log{n}}$ does not have a solution where $a_{2}, ..., a_{n}$ are all integers except for when $n=2$?

Furthermore, I want to clarify the fact that I am only considering whether $S_{(x,n)}$ is periodic in the imaginary part of $x=a+bi$, because for any complex zero of $S_{(x,n)}$, x=a+bi, $\sum\limits_{k=1}^{n}{n\choose k}(-1)^{n-k}k^{a}\cos{(b\log{k})}=0$ and $\sum\limits_{k=1}^{n}{n\choose k}(-1)^{n-k}k^{a}\sin{(b\log{k})}=0$.

We note that in each equation $a$ and $b$ are real. Therefore, as Elkies already argued, the maximum number of real zeroes of an equation $\sum\limits_{k=1}^{n}{n\choose k}(-1)^{n-k}k^{a}\cos{(b\log{k})}=0$ or $\sum\limits_{k=1}^{n}{n\choose k}(-1)^{n-k}k^{a}\sin{(b\log{k})}=0$ with $n-1$ sign changes is $n-1$. In each case, there are at maximum $n-1$ values of $a$ that satisfy the equation. For each value of $a$, it can be observed that there is at most $1$ value of $b$ that satisfies both equations (this is equivalent to the fact that $a_{2}\frac{2\pi}{\log{2}}=a_{3}\frac{2\pi}{\log{3}}=a_{n}\frac{2\pi}{\log{n}}$ does not have a solution where $a_{2}, ..., a_{n}$ are all integers except for when $n=2$).

Thank you for your help in advance. I hope this makes my line of reasoning more clear.


Your function has infinitely many complex zeros. Indeed, it is of the form $$\sum_{k=1}^n a_k e^{\lambda_k z},$$ where $a_k$ are constants, and $\lambda_k=\log k$ are all distinct. A function of this form always has infinitely many zeros, unless $n=1$.

Proof. This is an entire function of order $1$, normal type. So, if it has finitely many zeros, then by Hadamard's factorization theorem, it must be of the form $P(z)e^{cz}$, where $P$ is a polynomial, and $c$ is a constant.

Thus we have $$\sum_{k=1}^n a_k e^{(\lambda_k-c)z}\equiv P(z).$$ By differentiating this sufficiently many times to kill $P$, we will obtain that exponentials with distinct exponents are linearly dependent, while everyone knows that this is not so.

(For the proof, consider the Wronski determinant, after division by an exponential it will become a Wandermonde determinant) and we know that Wandermonde determinant of distinct numbers is never $0$).

  • $\begingroup$ Yes, this second question seems to be equivalent to the arithmetic question on logs of integers. I thought it is known that logs of integers are linearly independent over rationals, but I really do not know the literature in number theory. $\endgroup$ – Alexandre Eremenko Jan 4 '13 at 4:03
  • 1
    $\begingroup$ Well, independent modulo obvious relations like $\log 4 = 2 \log 2$. This much is elementary (equivalent to unique factorization). Getting good irrationality measures is much harder, but is not needed here. $\endgroup$ – Noam D. Elkies Jan 4 '13 at 17:24
  • $\begingroup$ Okay. I guess I was getting confused with irrationality measure. Thank you. $\endgroup$ – Daniel Niv Jan 4 '13 at 22:51

Alexandre Eremenko already showed that for each $n>1$ the function $$ S_n(x) := \frac1{n!} \sum_{k=1}^n {n \choose k} (-1)^{n-k} k^x $$ has infinitely many zeros $x \in {\bf C}$. One can still say more: for each $n$ (including $n=2$, and for that matter $n=1$) the zeros are limited to a vertical strip $\sigma_0 < {\mathop{\rm Re}}(x) < \sigma_1$ (this is elementary: if the real part of $x$ is too positive or too negative then the $k=n$ or $k=1$ term dominates), and the number of complex conjugate pairs with $|{\mathop{\rm Im}}(x)| < T$ is asymptotic to $c_n T$ where $c_n := (2\pi)^{-1} \log n$.

The proof is similar to the standard proof of the asymptotic vertical distribution of zeros of the Riemann zeta function, but easier because $S_n$ is an elementary function. Use the argument principle to express the number of zeros in the rectangle $\sigma_0 < {\mathop{\rm Re}}(x) < \sigma_1$, $|{\mathop{\rm Im}}(x)| < T$ as a contour integral over its boundary. The integrals over the vertical $\sigma_0$ and $\sigma_1$ edges contribute $O_n(1)$ and $2 c_n T + O_n(1)$ respectively. For the horizontal edges: use the Hadamard factorization of $S_n$, take its logarithmic derivative, show that the number of zeros with $|{\mathop{\rm Im}}(x) - T| \leq 2$ is bounded, and show that the integral of $|S_n^{\phantom.\prime}/S_n|$ over the rectangle $[\sigma_0,\sigma_1] + i [T-1,T+1]$ is $O_n(1)$ and thus that we make the horizontal $S_n^{\phantom.\prime}/S_n$ integral also $O_n(1)$ by changing $T$ by at most $1$.

For $n=3$ one can be still more precise: for each $k=1,2,3,\ldots$, the horizontal strip $k c_3^{\phantom.} < {\mathop{\rm Im}}(x) < (k+1) c_3^{\phantom.}$ contains a zero $x_k$ of $S_3$, and the full set of zeros consists of these $x_k$, their complex conjugates, and the real zeros $x=1$ and $x=2$. This is obtained by applying Rouché's theorem to $6 S_3(x) = 3^x - 3\cdot 2^x + 3$ with comparison function $3^x + 3$. This $x_k$ can be approximated numerically by integrating $(2\pi i)^{-1} z (S_3^{\phantom.\prime}(z) / S_3(z)) dz$ around the boundary of this rectangle and then applying Newton's formula to get even closer. The first five complex zeros $x_1,x_2,x_3,x_4,x_5$ are approximately

 -0.3397375 +  8.9137244 i,
  2.8692517 + 15.2110263 i,
  0.0637801 + 18.6324632 i,
 -0.1248035 + 26.7730278 i, and
  2.9811739 + 31.1087024 i.

Note that ${\mathop{\rm Re}}(x_5)$ is almost $3$. It so happens that $\sigma_1$ is exactly $3$ (since $3^3 = 2 \cdot 2^3 + 3$). Of the $174$ zeros with $0 < {\mathop{\rm Im}}(x) < 1000$, the one with largest real part is $x_{116} \doteq 2.99976958 + 666.32539172i$. The least real part in that range is attained by $x_{158} \doteq -0.36455251 + 906.47874219i$ (while $\sigma_0 \doteq -0.3646005647$).

  • $\begingroup$ I am not sure how I am supposed to acknowledge both answers. However, I voted you up. So thank you. Anyway, I had another question that is actually very central to why I asked this question. This might not be true even in the real case, but my intuition is hoping that this is true even in the complex case. I conjecture that $\frac{S_{(x,n+1)}}{S_{(x,n)}}=f_{n}(x)$ is a monotonically increasing function for all increasing real parts of $x$. I believe this is true in the case where $x$ is real. However, since there is a maximum of 1 zero of $\frac{d}{dx}S_{(x,n+1)}$ it makes me hopeful. $\endgroup$ – Daniel Niv Jan 4 '13 at 22:50
  • $\begingroup$ I expanded on that first comment in my question on this page if you are interested. mathoverflow.net/questions/118096/… $\endgroup$ – Daniel Niv Jan 4 '13 at 23:56
  • $\begingroup$ When I expanded on that comment. I clarified that what I mean by a complex valued function increasing monotonically is that $f_{n}(x)$ is increasing monotonically in both its real part and its imaginary part. $\endgroup$ – Daniel Niv Jan 5 '13 at 19:56

$This \ is \ not \ an \ answer$.

This is merely too long to be a comment to Alexandre Eremenko's response.

Let me rephrase my comment. I managed to show before that by using Vandermonde matrices that the vectors $(1^{x}, ..., n^{x})^{T}, (1^{1}, ..., n^{1})^{T}, (1^{2}, ..., n^{2})^{T}, ..., (1^{n-1}, .., n^{n-1})$ are linearly dependent precisely when $S_{(x,n)}=0$. Therefore, $\sum\limits_{k=1}^{n}{n\choose k}(-1)^{n-k}k^{a}\cos{(b\log{k})}=0$ and $\sum\limits_{k=1}^{n}{n\choose k}(-1)^{n-k}k^{a}\sin{(b\log{k})}=0$ for at most $n-1$ values of $a$ for a fixed $b$. This can be seen because by acknowledging the fact that $(1^{x}, ..., n^{x})^{T}, (1^{1}, ..., n^{1})^{T}, (1^{2}, ..., n^{2})^{T}, ..., (1^{n-1}, .., n^{n-1})^{T}$ are linearly dependent for all real $x$ $x\ne 1, ..., n-1$, then we can conclude that $\sum\limits_{k=1}^{n}{n\choose k}(-1)^{n-k}k^{a}\cos{(b\log{k})}=0$ and $\sum\limits_{k=1}^{n}{n\choose k}(-1)^{n-k}k^{a}\sin{(b\log{k})}=0$ has at most $n-1$ zeroes because it can be written $({n\choose 1}(-1)^{n-1}\cos{(b\log{1})}, ...,{n\choose n}(-1)^{n-n}\cos{(b\log{n})})^{T} \cdot (1^{a}, ..., n^{a})^{T}$ and $a$ is a real number.

Is this following logic correct? This is a more plausible fact (which I hope is true).


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