关于和式 $\sum_{k=0}^{n} k! {n\choose k}^2 (-1)^{k}$ 的显式估计
数论
2024-08-20 v3
摘要
我们致力于对二项式和 Q n ( x ) = ∑ k = 0 n k ! ( n k ) 2 ( − x ) k Q_n(x)=\sum_{k=0}^{n} k! {n\choose k}^2 (-x)^{k} Q n ( x ) = ∑ k = 0 n k ! ( k n ) 2 ( − x ) k x = 1 x=1 x = 1 n = 0 , 1 , 2 , … n=0,1,2,\ldots n = 0 , 1 , 2 , … Q n ( x ) Q_n(x) Q n ( x ) E ( x ) = ∑ k = 0 ∞ k ! x k E(x) = \sum_{k=0}^{\infty} k! x^k E ( x ) = ∑ k = 0 ∞ k ! x k L n ( x ) = 1 n ! e x ( d d x ) n ( e − x x n ) L_n(x) = \frac{1}{n!} e^x \left(\frac{d}{dx}\right)^n (e^{-x}x^n) L n ( x ) = n ! 1 e x ( d x d ) n ( e − x x n ) n = 0 , 1 , 2 , … n=0,1,2,\ldots n = 0 , 1 , 2 , … ∣ L n ( 1 ) − e π ⋅ cos ( 2 n − π 4 ) n 1 / 4 + 17 48 e π sin ( 2 n − π 4 ) n 3 / 4 ∣ < 0.51 n \left|L_n(1)-\sqrt{\frac{e}{\pi}}\cdot \frac{\cos (2\sqrt{n}-\frac{\pi}{4})}{n^{1/4}} +\frac{17}{48}\sqrt{\frac{e}{\pi}}\frac{\sin(2\sqrt{n}-\frac{\pi}{4})}{n^{3/4}}\right|<\frac{0.51}{n} L n ( 1 ) − π e ⋅ n 1/4 cos ( 2 n − 4 π ) + 48 17 π e n 3/4 sin ( 2 n − 4 π ) < n 0.51 ∣ Q n ( 1 ) ∣ ≤ n ! |Q_{n}(1)| \le n! ∣ Q n ( 1 ) ∣ ≤ n !
引用
@article{arxiv.2310.11468,
title = {Explicit estimates for the sum $\sum_{k=0}^{n} k! {n\choose k}^2 (-1)^{k}$},
author = {Anne-Maria Ernvall-Hytönen and Tapani Matala-aho},
journal= {arXiv preprint arXiv:2310.11468},
year = {2024}
}