Complex Zeroes of Stirling functions of the second kind - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T08:53:04Z http://mathoverflow.net/feeds/question/118007 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118007/complex-zeroes-of-stirling-functions-of-the-second-kind Complex Zeroes of Stirling functions of the second kind Daniel Niv 2013-01-03T23:23:23Z 2013-01-04T18:25:56Z <p>My motivation to the following question stems from the discussion at <a href="http://mathoverflow.net/questions/83999/zeros-of-exponential-function" rel="nofollow">http://mathoverflow.net/questions/83999/zeros-of-exponential-function</a> 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.</p> <p>Define: <code>$S_{(x,n)}=\frac{1}{n!}\sum\limits_{k=1}^{n}{n\choose k}(-1)^{n-k}k^{x}$</code> for integer $n$ and complex $x$.</p> <p>I conjecture that for $n>2$, <code>$S_{(x,n)}$</code> has exactly $n-1$ complex zeroes. I realize this is a much stronger claim than the fact that <code>$S_{(x,n)}$</code> 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 <code>$S_{(x,n)}$</code> seem to alternate for increasing integer $n$ when $x$ is a complex number. </p> <p>When $x$ is a real number, the imaginary part of <code>$S_{(x,n)}$</code> 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 <code>$S_{(x,n)}$</code> for complex $x$ as a linear combination of integer exponential functions that are defined for complex $x$, does it follow that for $n>2$, <code>$S_{(x,n)}$</code> has exactly $n-1$ complex zeroes?</p> <p>$n=2$ is an exception because <code>$S_{(x,2)}$</code> 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 <code>$S_{(x,2)}=0$</code> is equivalent to the equation $2^{x}-2=0$ which only has one exponential term that is non-constant. Since <code>$a_{2}\frac{2\pi}{\log{2}}=a_{3}\frac{2\pi}{\log{3}}=a_{n}\frac{2\pi}{\log{n}}$</code> presumably does not have a solution for <code>$a_{2}, ..., a_{n}$</code> all integers except for when $n=2$ it is unlikely that there are any other complex zeroes of <code>$S_{(x,n)}$</code> in the case that $n\ne 2$ other than $x=1$, ..., $n-1$.</p> <p>Can someone help me think of a way to show that <code>$a_{2}\frac{2\pi}{\log{2}}=a_{3}\frac{2\pi}{\log{3}}=a_{n}\frac{2\pi}{\log{n}}$</code> does not have a solution where <code>$a_{2}, ..., a_{n}$</code> are all integers except for when $n=2$? </p> <p>Furthermore, I want to clarify the fact that I am only considering whether <code>$S_{(x,n)}$</code> is periodic in the imaginary part of $x=a+bi$, because for any complex zero of <code>$S_{(x,n)}$</code>, x=a+bi, $\sum\limits_{k=1}^{n}{n\choose k}(-1)^{n-k}k^{a}\cos{(b\log{k})}=0$ and $\sum\limits_{k=1}^{n}{n\choose k}(-1)^{n-k}k^{a}\sin{(b\log{k})}=0$.</p> <p>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 <code>$a_{2}\frac{2\pi}{\log{2}}=a_{3}\frac{2\pi}{\log{3}}=a_{n}\frac{2\pi}{\log{n}}$</code> does not have a solution where <code>$a_{2}, ..., a_{n}$</code> are all integers except for when $n=2$).</p> <p>Thank you for your help in advance. I hope this makes my line of reasoning more clear.</p> http://mathoverflow.net/questions/118007/complex-zeroes-of-stirling-functions-of-the-second-kind/118013#118013 Answer by Alexandre Eremenko for Complex Zeroes of Stirling functions of the second kind Alexandre Eremenko 2013-01-04T01:43:33Z 2013-01-04T01:43:33Z <p>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 $n=1$.</p> <p>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.</p> <p>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.</p> <p>(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 $0$).</p> http://mathoverflow.net/questions/118007/complex-zeroes-of-stirling-functions-of-the-second-kind/118019#118019 Answer by Daniel Niv for Complex Zeroes of Stirling functions of the second kind Daniel Niv 2013-01-04T03:56:48Z 2013-01-04T03:56:48Z <p>$This \ is \ not \ an \ answer$.</p> <p>This is merely too long to be a comment to Alexandre Eremenko's response.</p> <p>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 <code>$S_{(x,n)}=0$</code>. 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.</p> <p>Is this following logic correct? This is a more plausible fact (which I hope is true).</p> http://mathoverflow.net/questions/118007/complex-zeroes-of-stirling-functions-of-the-second-kind/118076#118076 Answer by Noam D. Elkies for Complex Zeroes of Stirling functions of the second kind Noam D. Elkies 2013-01-04T18:20:45Z 2013-01-04T18:25:56Z <p>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 $\sigma_0 &lt; {\mathop{\rm Re}}(x) &lt; \sigma_1$ (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 $|{\mathop{\rm Im}}(x)| &lt; T$ is asymptotic to $c_n T$ where $c_n := (2\pi)^{-1} \log n$.</p> <p>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 $\sigma_0 &lt; {\mathop{\rm Re}}(x) &lt; \sigma_1$, $|{\mathop{\rm Im}}(x)| &lt; T$ 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 $1$.</p> <p>For $n=3$ one can be still more precise: for each $k=1,2,3,\ldots$, the horizontal strip $k c_3^{\phantom.} &lt; {\mathop{\rm Im}}(x) &lt; (k+1) c_3^{\phantom.}$ 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</p> <pre><code> -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. </code></pre> <p>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 $0 &lt; {\mathop{\rm Im}}(x) &lt; 1000$, 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 $\sigma_0 \doteq -0.3646005647$).</p>