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2 Fix typo: sigma_1 (not sigma_3) is 3

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 RouchÃ©'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 $\sigma_3$ \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$\sigma_0 \doteq -0.3646005647$). 1 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 RouchÃ©'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$\sigma_3$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\$).