生成函数 二项式

hbghlyj

$$\left [\sum_{k=0}^n \binom nk x^k \right ] \left [\sum_{k=0}^n \binom nk (-x)^k \right ] = (1+x)^n (1-x)^n= \left (1-x^2 \right )^n = \sum_{k=0}^n \binom nk (-1)^k x^{2k}.$$
比较$x^n$系数：$$\sum_{k=0}^n (-1)^k \binom nk ^2 = \left \{ \begin{array}{cl} 0 & \text{for odd }n \\ (-1)^\frac n2 \binom{n}{n/2} & \text{for even } n. \end{array} \right.$$
但WolframAlpha算出：
\[{\binom{n}{0}}^2 - {\binom{n}{1}}^2 + {\binom{n}{2}}^2 - \cdots + (-1)^n {\binom{n}{n}}^2=\frac{\sqrtπ 2^n}{\Gamma(\frac{1 - n}2)\Gamma(\frac{n + 2}2)}\]
为什么是这个呢😯

hbghlyj

en.wikipedia.org/wiki/Reflection_formula
Euler's reflection formula $\displaystyle \Gamma (z)\Gamma (1-z)={\frac {\pi }{\sin {(\pi z)}}},\;z\not \in \mathbb {Z}$

hbghlyj

加起来是$(\frac{1 - n}2)+(\frac{n + 2}2)=\frac32$，不是1啊，无法直接用reflection formula🤔

hbghlyj

mathworld.wolfram.com/GammaFunction.html
In general, for $n$ a positive integer $n=1, 2, \dots$
$\Gamma(\frac12+n)=\frac{(2n-1)!!}{2^n}\sqrt\pi$
$\Gamma(\frac12-n)=\frac{(-1)^n2^n}{(2n-1)!!}\sqrt\pi.$

hbghlyj

当$n$为奇数，
$\frac{1-n}2\inZ_{-}\Rightarrow\Gamma(\frac{1 - n}2)=\infty,$
$\frac{n + 2}2\inZ+\frac12\Rightarrow\Gamma(\frac{n + 2}2)<\infty,$
$\Rightarrow\frac{\sqrtπ 2^n}{\Gamma(\frac{1 - n}2)\Gamma(\frac{n + 2}2)}=0$
当$n$为偶数，
$\frac{1-n}2\inZ+\frac12\Rightarrow\Gamma(\frac{1-n}2)=\sqrt\pi\frac{(-1)^{n/2}2^{n/2}}{(n-1)!!},$
$\frac{n+2}2\inZ_+\Rightarrow\Gamma(\frac{n + 2}2)=\frac n2!,$
$\Rightarrow\frac{\sqrtπ 2^n}{\Gamma(\frac{1 - n}2)\Gamma(\frac{n + 2}2)}=\frac{\sqrtπ 2^n}{\sqrt\pi\frac{(-1)^{n/2}2^{n/2}}{(n-1)!!}\frac n2!}=(-1)^{\frac n2}\frac{2^{n/2}(n-1)!!}{\frac n2!}=(-1)^{\frac n2}\frac{2^{n/2}\frac n2!(n-1)!!}{\frac n2!\frac n2!}=(-1)^{\frac n2}\frac{n!!(n-1)!!}{\frac n2!\frac n2!}=(-1)^{\frac n2}\frac{n!}{\frac n2!\frac n2!}=(-1)^\frac n2 \binom{n}{n/2} $

hbghlyj

hbghlyj 发表于 2024-3-9 09:22
In general, for $n$ a positive integer $n=1, 2, \dots$
$\Gamma(\frac12+n)=\frac{(2n-1)!!}{2^n}\sqrt\pi$
$\Gamma(\frac12-n)=\frac{(-1)^n2^n}{(2n-1)!!}\sqrt\pi.$

补一个证明：
首先$\frac12+\frac12=1$，由reflection formula得$\Gamma(\frac12)^2=\frac{\pi}{\sin\frac\pi2}=\pi\Rightarrow\Gamma(\frac12)=\sqrt\pi$
由递推式$\Gamma(\frac12+n)=(\frac12+n-1)\Gamma(\frac12+n-1)=\dots=(\frac12+n-1)(\frac12+n-2)\dots\frac12\Gamma(\frac12)$
即$\Gamma(\frac12+n)=\frac{(2n-1)!!}{2^n}\sqrt\pi$

hbghlyj

hbghlyj 发表于 2024-3-9 09:08
比较$x^n$系数：$$\sum_{k=0}^n (-1)^k \binom nk ^2 = \left \{ \begin{array}{cl} 0 & \text{for odd }n \\ (-1)^\frac n2 \binom{n}{n/2} & \text{for even } n. \end{array} \right.$$

当$n$为奇数，很容易证明$\sum_{k=0}^n (-1)^k \binom nk ^2 =\sum_{k=0}^{\frac{n-1}2} (-1)^k \binom nk ^2+(-1)^{n-k} \binom n{n-k}^2=\sum_{k=0}^{\frac{n-1}2} (-1)^k\left(\binom nk ^2-\binom n{n-k}^2\right)=0$
当$n$为偶数，见kuing.cjhb.site/forum.php?mod=viewthread&tid=12081
很神奇，这个的部分和是有组合意义的。