一个渐近级数问题

数学星空

试给出下面式子的渐近级数展开(精确到\(O(n^{-6})\)

\[\frac{(n!)^2}{(n-k)!(n+k)!}\]

设\(w=\frac{k}{\sqrt{n}}\)

进一步，我们知道伯努利分布可以简化为泊松分布来计算：

\[\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k} \approx \frac{e^{-np} (np)^k}{k!}\]

现在的问题是给出上式更精确的渐近表示？

数学星空

在《Analytic Combinatorics》中看到如下结果：

\(e^{-w^2}(1-\frac{w^4-3w^2}{6n}+\frac{5w^8-54w^6+135w^4-60w^2}{360n^2}+\ldots)\)


为了输入方便，记

\(w=\frac{k}{\sqrt{n}}=\frac{k}{t},T=\frac{w}{\sqrt{n}}=\frac{w}{t}\)

\(\frac{(2n)!}{(n-k)!(n+k)!}\frac{(n!)^2}{(2n)!}\)

\(=\frac{(n!)^2}{(n-wt)!(n+wt)!}\)

\(=\sqrt{\frac{n^2}{n^2-w^2t^2}}(\frac{n^2}{n^2-w^2t^2})^n(\frac{n-wt}{n+wt})^{wt}(\frac{M^2}{N_1N_2})\)

\(=(1-T^2)^{-\frac{1}{2}-n}(-1+\frac{2}{1+T})^{Tw^2}(\frac{M^2}{N_1N_2})\)

\(=e^{-w^2}(1-\frac{w^4-3w^2}{6n}+\frac{5w^8-54w^6+135w^4-60w^2}{360n^2}-\frac{w^4(35w^8-819w^6+5967w^4-15435w^2+11340)}{45360n^3}+\frac{w^2(175w^{14}-7140w^{12}+103086w^{10}-650340w^8+1775655w^6-1701000w^4+75600w^2+181440)}{5443200n^4}\)

\(+\frac{w^4(385w^{16}-24255w^{14}+581526w^{12}-6756618w^{10}+40015485w^8-115246395w^6+131808600w^4-10478160w^2-35925120)}{359251200n^5}+\ldots)\)

或者

\(\frac{(2n)!}{(n-k)!(n+k)!}\frac{(n!)^2}{(2n)!}\)

\(=\frac{(n!)^2}{(n-k)!(n+k)!}\)

\(=\sqrt{\frac{n^2}{n^2-k^2}}(\frac{n^2}{n^2-k^2})^n(\frac{n-k}{n+k})^k(\frac{M^2}{N_1N_2})\)

\(=e^{-\frac{k^2}{n}}(1+\frac{k^2}{2n^2}-\frac{k^2(k^2+1)}{6n^3}+\frac{3k^4}{8n^4}-\frac{k^2(9k^4+15k^2-2)}{60n^5}+\frac{k^4(2k^4+49k^2+2)}{144n^6}\)

\(-\frac{k^2(221k^6+525k^4-168k^2+40)}{1680n^7}-\frac{k^4(104k^6+1879k^4+168k^2-32)}{5760n^8}-\frac{k^2(70k^{10}+10839k^8+33285k^6-18956k^4+11160k^2-30240)}{90720n^9}+\ldots)\)


注:\(M,N_1,N_2\)分别为\(n!,(n-wt)!,(n+wt)!\)渐近展开余项

\(M=1+\frac{1}{12n}+\frac{1}{288n^2}-\frac{139}{51840n^3}-\frac{571}{2488320n^4}-\frac{163879}{209018880n^5}+\ldots\)

数学星空

对于伯努利分布渐近于泊松分布的分析：

\[ \begin{align}
&\mathrel{\phantom{=}}
\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k}\\
&=\frac{(np)^k(1-p)^n}{k!}\frac{\prod_{i=1}^{k-1} (1-\frac{i}{n})}{(1-p)^k}\\
&=\frac{(np)^k e^{-np}M_1}{k!}\frac{M_2}{N}\\
&=\frac{(np)^k e^{-np}}{k!}(1+S_0+S_1n^2+S_2n+\frac{S_3}{n}+\frac{S_4}{n^2}+\ldots)\\
\label{eq:cubesum}
\end{align} \]

并且：

\(S_0=(p+\frac{3p^2}{4}+\frac{p^3}{6}+\frac{35p^4}{96}+\frac{41p^5}{360}+\ldots)k+( \frac{p^2}{4}+\frac{11p^3}{12}+\frac{65p^4}{192}+\frac{91p^5}{160}+\ldots)k^2+(-\frac{p^3}{12}+\frac{35p^4}{96}+\frac{187p^5}{576}+\frac{461p^6}{5760}+\ldots)k^3+\ldots\)

\(S_1=\frac{p^4}{8}-\frac{p^5}{6}+(\frac{p^5}{8}-\frac{5p^6}{48}-\frac{p^7}{24}-\frac{7p^8}{288}+\ldots)k+(\frac{p^6}{16}-\frac{p^7}{48}-\frac{5p^8}{192}+\ldots)k^2+(\frac{p^7}{48}+\frac{p^8}{288}-\frac{p^9}{192}+\ldots)k^3+\ldots\)        

\(S_2=\frac{p^2}{2}-\frac{p^3}{3}+\frac{p^4}{4}-\frac{p^5}{5}+\ldots+( \frac{p^3}{2}-\frac{p^4}{48}+\frac{p^5}{6}-\frac{11p^6}{180}-\frac{17p^8}{240}+\ldots)k+(\frac{3p^4}{16}+\frac{11p^5}{48}+\frac{13p^6}{96}+\frac{43p^7}{720}+\ldots)k^2+(\frac{p^5}{48}+\frac{11p^6}{72}+\frac{11p^7}{96}-\frac{397p^7}{5760}+\ldots)k^3+\ldots\)

\(S_3=( \frac{1}{2}-\frac{p^2}{24}+\frac{p^3}{36}-\frac{p^4}{48}+\frac{p^5}{60}+\ldots)k+(-\frac{1}{2}+\frac{p}{2}+\frac{7p^2}{16}+\frac{121p^4}{576}+\frac{p^5}{40}+\ldots )k^2+(-\frac{p}{2}-\frac{5p^2}{24}-\frac{p^4}{128}+\frac{587p^5}{2880}+\ldots)k^3+\ldots\)

\(S_4=(-\frac{1}{12}+\frac{p^4}{960}-\frac{p^5}{720}+\ldots)k+(\frac{3}{8}-\frac{p}{12}-\frac{5p^2}{48}+\frac{p^3}{72}-\frac{119p^4}{2304}+\frac{19p^5}{2160}-\frac{p^6}{1152}+\ldots)k^2+(-\frac{5}{12}+\frac{3p}{8}+\frac{31p^2}{96}-\frac{7p^3}{72}+\frac{311p^4}{2304}-\frac{407p^5}{11520}+\ldots)k^3+\ldots\)

注：\(M_1,M_2,N\)分别为\( (1-p)^n，\prod_{i=1}^{k-1} (1-\frac{i}{n})，(1-p)^{k} \)渐近展开的余项

\(M_1=1+\frac{np^2}{2}-\frac{np^3}{3}+(\frac{n}{4}+\frac{n^2}{8})p^4+(-\frac{n}{5}-\frac{n^2}{6})p^5+(\frac{n}{6}+\frac{13n^2}{72}+\frac{n^3}{48})p^6+(-\frac{n}{7}-\frac{11n^2}{60}-\frac{n^3}{24})p^7+(\frac{n}{8}+\frac{29n^2}{160}+\frac{17n^2}{288}+\frac{n^4}{384})p^8+\ldots\)

\(M_2=1-\frac{k(k-1)}{2n}+\frac{k(k-1)(k-2)(3k-1)}{24n^2}-\frac{k^2(k-2)(k-3)(k-1)^2}{48n^3}+\frac{k(k-1)(k-2)(k-3)(k-4)(15k^3-30k^2+5k+2}{5760n^4}-\frac{k^2(k-2)(k-3)(k-4)(k-5)(3k^2-7k-2)(k-1)^2}{11520n^5}+\frac{k(k-1)(k-2)(k-3)(k-4)(k-5)(k-6)(k-7)(63k^5-315k^4+315k^3+91k^2-42k-16}{290340n^6}+\ldots\)

\(\frac{1}{N}=1+kp-\frac{k(-k-1)p^2}{2}+\frac{k(-k-1)(-k-2)p^3}{6}-\frac{k(-k-1)(-k-2)(-k-3)p^4}{24}+\frac{k(-k-1)(-k-2)(-k-3)(-k-4)p^5}{120}-\frac{k(-k-1)(-k-2)(-k-3)(-k-4)(-k-5)p^6}{720}+\frac{k(-k-1)(-k-2)(-k-3)(-k-4)(-k-5)(-k-6)p^7}{5040}-\frac{k(-k-1)(-k-2)(-k-3)(-k-4)(-k-5)(-k-6)(-k-7)p^8}{40320}+\ldots\)

显然上面的表达式过于复杂（由于含有3个参数\(n,p,k\)），研究更简洁精确的形式更有应用价值。

282842712474

第一个问题相当于研究
$$\frac{n(n-1)...(n-k+1)}{(n+1)(n+2)...(n+k)}$$
我倾向于研究更加对称的
$$\frac{(n-1)(n-2)...(n-k)}{(n+1)(n+2)...(n+k)}$$

数学星空

对于楼上更一般的形式的渐近分析：

\(\frac{(n-m)(n-2m)(n-3m)\dots(n-km)}{(n+m)(n+2m)(n+3m)\dots(n+km)}\)

\(=1-\frac{km}{n}-\frac{k(2k-1)(k-1)m^2}{6n^2}+\frac{k^2(2k-1)(k-1)m^3}{6n^3}+\frac{k(k-1)(k-2)(2k-1)(2k-3)(5k+1)m^4}{360n^4}-\frac{k^2(k-1)(k-2)(2k-1)(2k-3)(5k+1)m^5}{360n^5}-\frac{k(k-1)(k-2)(k-3)(2k-1)(2k-3)(2k-5)(35k^2+21k+4)m^6}{45360n^6}+\frac{k^2(k-1)(k-2)(k-3)(2k-1)(2k-3)(2k-5)(35k^2+21k+4)m^7}{45360n^7}+\frac{k(k-1)(k-2)(k-3)(k-4)(5k+2)(2k-1)(2k-3)(2k-5)(2k-7)(35k^2+28k+9)m^8}{5443200n^8}-\frac{k^2(k-1)(k-2)(k-3)(k-4)(5k+2)(2k-1)(2k-3)(2k-5)(2k-7)(35k^2+28k+9)m^9}{5443200n^9}+\ldots\)

当\(m=1\)时，化为楼上的情形

\(\frac{(n-1)(n-2)(n-3)\dots(n-k)}{(n+1)(n+2)(n+3)\dots(n+k)}\)

\(=1-\frac{k}{n}-\frac{k(2k-1)(k-1)}{6n^2}+\frac{k^2(2k-1)(k-1)}{6n^3}+\frac{k(k-1)(k-2)(2k-1)(2k-3)(5k+1)}{360n^4}-\frac{k^2(k-1)(k-2)(2k-1)(2k-3)(5k+1)}{360n^5}-\frac{k(k-1)(k-2)(k-3)(2k-1)(2k-3)(2k-5)(35k^2+21k+4)}{45360n^6}+\frac{k^2(k-1)(k-2)(k-3)(2k-1)(2k-3)(2k-5)(35k^2+21k+4)}{45360n^7}+\frac{k(k-1)(k-2)(k-3)(k-4)(5k+2)(2k-1)(2k-3)(2k-5)(2k-7)(35k^2+28k+9)}{5443200n^8}-\frac{k^2(k-1)(k-2)(k-3)(k-4)(5k+2)(2k-1)(2k-3)(2k-5)(2k-7)(35k^2+28k+9)}{5443200n^9}+\ldots\)