# 拉盖尔多项式

${\displaystyle x\,y''+(1-x)\,y'+n\,y=0\,}$

${\displaystyle L_{n}(x)={\frac {e^{x}}{n!}}{\frac {d^{n}}{dx^{n}}}\left(e^{-x}x^{n}\right).}$

${\displaystyle \langle f,g\rangle =\int _{0}^{\infty }f(x)g(x)e^{-x}\,dx.}$

## 前几个拉盖尔多项式

 n ${\displaystyle L_{n}(x)\,}$ 0 ${\displaystyle 1\,}$ 1 ${\displaystyle -x+1\,}$ 2 ${\displaystyle {\scriptstyle {\frac {1}{2}}}(x^{2}-4x+2)\,}$ 3 ${\displaystyle {\scriptstyle {\frac {1}{6}}}(-x^{3}+9x^{2}-18x+6)\,}$ 4 ${\displaystyle {\scriptstyle {\frac {1}{24}}}(x^{4}-16x^{3}+72x^{2}-96x+24)\,}$ 5 ${\displaystyle {\scriptstyle {\frac {1}{120}}}(-x^{5}+25x^{4}-200x^{3}+600x^{2}-600x+120)\,}$ 6 ${\displaystyle {\scriptstyle {\frac {1}{720}}}(x^{6}-36x^{5}+450x^{4}-2400x^{3}+5400x^{2}-4320x+720)\,}$

## 递归定义

${\displaystyle L_{0}(x)=1\,}$
${\displaystyle L_{1}(x)=1-x\,}$

${\displaystyle L_{k+1}(x)={\frac {1}{k+1}}\left((2k+1-x)L_{k}(x)-kL_{k-1}(x)\right).}$

## 广义拉盖尔多项式

${\displaystyle f(x)=\left\{{\begin{matrix}e^{-x}&{\mbox{if}}\ x>0,\\0&{\mbox{if}}\ x<0,\end{matrix}}\right.}$

${\displaystyle E\left[L_{n}(X)L_{m}(X)\right]=0\ {\mbox{whenever}}\ n\neq m.}$

${\displaystyle f(x)=\left\{{\begin{matrix}x^{\alpha }e^{-x}/\Gamma (1+\alpha )&{\mbox{if}}\ x>0,\\0&{\mbox{if}}\ x<0,\end{matrix}}\right.}$

${\displaystyle L_{n}^{(\alpha )}(x)={x^{-\alpha }e^{x} \over n!}{d^{n} \over dx^{n}}\left(e^{-x}x^{n+\alpha }\right).}$

${\displaystyle L_{n}^{(0)}(x)=L_{n}(x).}$

## 广义拉盖尔多项式的性质与应用

• 拉盖尔函数可以由合流超几何函数和Kummer变换得到： ${\displaystyle L_{n}^{(\alpha )}(x):={n+\alpha \choose n}M(-n,\alpha +1,x)={n+\alpha \choose n}\sum _{i=0}(-1)^{i}{\frac {n \choose i}{\alpha +i \choose i}}x^{i}\,}$ ${\displaystyle =e^{x}\cdot {n+\alpha \choose n}M(\alpha +n+1,\alpha +1,-x)}$${\displaystyle ={\frac {e^{x}\sin(n\pi )}{\sin((n+\alpha )\pi )}}L_{-\alpha -n-1}^{(\alpha )}(-x)}$${\displaystyle =e^{x}\cdot \sum _{i=0}(-1)^{i}{\alpha +n+i \choose n}{\frac {x^{i}}{i!}}.}$${\displaystyle n}$为整数时，截断为${\displaystyle n}$阶拉盖尔多项式。
• ${\displaystyle n}$阶拉盖尔多项式可以通过将应用在罗德里格公式上而得到，结果为${\displaystyle L_{n}^{(\alpha )}(x)=\sum _{i=0}^{n}(-1)^{i}{n+\alpha \choose n-i}{\frac {x^{i}}{i!}}}$
• n阶拉盖尔多项式的首项系数为(−1)n/n!；
• 拉盖尔多项式在x=0的取值（常数项）为 ${\displaystyle L_{n}^{(\alpha )}(0)={n+\alpha \choose n}\approx {\frac {n^{\alpha }}{\Gamma (\alpha +1)}};}$
• Ln(α)n的正（应该注意到 ${\displaystyle \left((-1)^{n-i}L_{n-i}^{(\alpha )}\right)_{i=0}^{n}}$构成以施图姆序列），且这些根全部位于区间${\displaystyle (0,n+\alpha +(n-1){\sqrt {n+\alpha }}]}$中。
• ${\displaystyle n}$很大，而${\displaystyle \alpha }$不变，${\displaystyle x>0}$时，拉盖尔多项式的渐近行为如下：
${\displaystyle L_{n}^{(\alpha )}(x)\approx {\frac {n^{{\frac {\alpha }{2}}-{\frac {1}{4}}}}{\sqrt {\pi }}}{\frac {e^{\frac {x}{2}}}{x^{{\frac {\alpha }{2}}+{\frac {1}{4}}}}}\cos \left(2{\sqrt {x\left(n+{\frac {\alpha +1}{2}}\right)}}-{\frac {\pi }{2}}\left(\alpha +{\frac {1}{2}}\right)\right)}$，以及
${\displaystyle L_{n}^{(\alpha )}(-x)\approx {\frac {n^{{\frac {\alpha }{2}}-{\frac {1}{4}}}}{2{\sqrt {\pi }}}}{\frac {e^{-{\frac {x}{2}}}}{x^{{\frac {\alpha }{2}}+{\frac {1}{4}}}}}\exp \left(2{\sqrt {x\left(n+{\frac {\alpha +1}{2}}\right)}}\right)}$[1]
• 前几个广义拉盖尔多项式为：
${\displaystyle L_{0}^{(\alpha )}(x)=1}$
${\displaystyle L_{1}^{(\alpha )}(x)=-x+\alpha +1}$
${\displaystyle L_{2}^{(\alpha )}(x)={\frac {x^{2}}{2}}-(\alpha +2)x+{\frac {(\alpha +2)(\alpha +1)}{2}}}$
${\displaystyle L_{3}^{(\alpha )}(x)={\frac {-x^{3}}{6}}+{\frac {(\alpha +3)x^{2}}{2}}-{\frac {(\alpha +2)(\alpha +3)x}{2}}+{\frac {(\alpha +1)(\alpha +2)(\alpha +3)}{6}}}$
• 根据拉盖尔多项式的定义，可以使用秦九韶算法计算拉盖尔多项式，程序代码如下：
 function LaguerreL(n, alpha, x) {
LaguerreL:= 1; bin:= 1
for i:= n to 1 step -1 {
bin:= bin* (alpha+ i)/ (n+ 1- i)
LaguerreL:= bin- x* LaguerreL/ i
}
return LaguerreL;
}


### 递推关系

${\displaystyle L_{n}^{(\alpha +\beta +1)}(x+y)=\sum _{i=0}^{n}L_{i}^{(\alpha )}(x)L_{n-i}^{(\beta )}(y),}$

${\displaystyle L_{n}^{(\alpha +1)}(x)=\sum _{i=0}^{n}L_{i}^{(\alpha )}(x)}$以及${\displaystyle L_{n}^{(\alpha )}(x)=\sum _{i=0}^{n}{\alpha -\beta +n-i-1 \choose n-i}L_{i}^{(\beta )}(x)}$，或${\displaystyle L_{n}^{(\alpha )}(x)=\sum _{i=0}^{n}{\alpha -\beta +n \choose n-i}L_{i}^{(\beta -i)}(x);}$

{\displaystyle {\begin{aligned}L_{n}^{(\alpha )}(x)-\sum _{j=0}^{\Delta -1}{n+\alpha \choose n-j}(-1)^{j}{\frac {x^{j}}{j!}}&=(-1)^{\Delta }{\frac {x^{\Delta }}{(\Delta -1)!}}\sum _{i=0}^{n-\Delta }{\frac {n+\alpha \choose n-\Delta -i}{(n-i){n \choose i}}}L_{i}^{(\alpha +\Delta )}(x)\\&=(-1)^{\Delta }{\frac {x^{\Delta }}{(\Delta -1)!}}\sum _{i=0}^{n-\Delta }{\frac {n+\alpha -i-1 \choose n-\Delta -i}{(n-i){n \choose i}}}L_{i}^{(n+\alpha +\Delta -i)}(x).\end{aligned}}}

• ${\displaystyle L_{n}^{(\alpha )}(x)=L_{n}^{(\alpha +1)}(x)-L_{n-1}^{(\alpha +1)}(x)}$${\displaystyle =\sum _{j=0}^{k}{k \choose j}L_{n-j}^{(\alpha -k+j)}(x),}$
• ${\displaystyle nL_{n}^{(\alpha )}(x)=(n+\alpha )L_{n-1}^{(\alpha )}(x)-xL_{n-1}^{(\alpha +1)}(x),}$ or ${\displaystyle {\frac {x^{k}}{k!}}L_{n}^{(\alpha )}(x)=\sum _{i=0}^{k}(-1)^{i}{n+i \choose i}{n+\alpha \choose k-i}L_{n+i}^{(\alpha -k)}(x),}$
• ${\displaystyle nL_{n}^{(\alpha +1)}(x)=(n-x)L_{n-1}^{(\alpha +1)}(x)+(n+\alpha )L_{n-1}^{(\alpha )}(x)}$
• ${\displaystyle xL_{n}^{(\alpha +1)}=(n+\alpha )L_{n-1}^{\alpha }(x)-(n-x)L_{n}^{(\alpha )}(x);}$

${\displaystyle L_{n+1}^{(\alpha )}(x)={\frac {1}{n+1}}\left((2n+1+\alpha -x)L_{n}^{(\alpha )}(x)-(n+\alpha )L_{n-1}^{(\alpha )}(x)\right).}$

${\displaystyle i}$${\displaystyle n}$均为整数时，拉盖尔多项式有以下的有趣性质：

${\displaystyle {\frac {(-x)^{i}}{i!}}L_{n}^{(i-n)}(x)={\frac {(-x)^{n}}{n!}}L_{i}^{(n-i)}(x);}$

${\displaystyle {\frac {L_{n}^{(\alpha )}(x)}{n+\alpha \choose n}}=1-\sum _{j=1}^{n}{\frac {x^{j}}{\alpha +j}}{\frac {L_{n-j}^{(j)}(x)}{(j-1)!}}=1-x\sum _{i=1}^{n}{\frac {L_{n-i}^{(-\alpha )}(x)L_{i-1}^{(\alpha +1)}(-x)}{\alpha +i}}.}$

### 拉盖尔多项式的导函数

${\displaystyle {\frac {\mathrm {d} ^{k}}{\mathrm {d} x^{k}}}L_{n}^{(\alpha )}(x)=(-1)^{k}L_{n-k}^{(\alpha +k)}(x)\,;}$

${\displaystyle {\frac {1}{k!}}{\frac {\mathrm {d} ^{k}}{\mathrm {d} x^{k}}}x^{\alpha }L_{n}^{(\alpha )}(x)={n+\alpha \choose k}x^{\alpha -k}L_{n}^{(\alpha -k)}(x),}$

${\displaystyle L_{n}^{(\alpha ')}(x)=(\alpha '-\alpha ){\alpha '+n \choose \alpha '-\alpha }\int _{0}^{x}{\frac {t^{\alpha }(x-t)^{\alpha '-\alpha -1}}{x^{\alpha '}}}L_{n}^{(\alpha )}(t)\,dt.}$

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} \alpha }}L_{n}^{(\alpha )}(x)=\sum _{i=0}^{n-1}{\frac {L_{i}^{(\alpha )}(x)}{n-i}}.}$

${\displaystyle xL_{n}^{(\alpha )\prime \prime }(x)+(\alpha +1-x)L_{n}^{(\alpha )\prime }(x)+nL_{n}^{(\alpha )}(x)=0,\,}$

${\displaystyle xL_{n}^{(k)\prime \prime }(x)+(k+1-x)L_{n}^{(k)\prime }(x)+(n-k)L_{n}^{(k)}(x)=0,\,}$

### 正交性

${\displaystyle \int _{0}^{\infty }x^{\alpha }e^{-x}L_{n}^{(\alpha )}(x)L_{m}^{(\alpha )}(x)dx={\frac {\Gamma (n+\alpha +1)}{n!}}\delta _{n,m},}$

${\displaystyle \int _{0}^{\infty }x^{\alpha '-1}e^{-x}L_{n}^{(\alpha )}(x)dx={\alpha -\alpha '+n \choose n}\Gamma (\alpha ').}$

{\displaystyle {\begin{aligned}K_{n}^{(\alpha )}(x,y)&{:=}{\frac {1}{\Gamma (\alpha +1)}}\sum _{i=0}^{n}{\frac {L_{i}^{(\alpha )}(x)L_{i}^{(\alpha )}(y)}{\alpha +i \choose i}}\\&{=}{\frac {1}{\Gamma (\alpha +1)}}{\frac {L_{n}^{(\alpha )}(x)L_{n+1}^{(\alpha )}(y)-L_{n+1}^{(\alpha )}(x)L_{n}^{(\alpha )}(y)}{{\frac {x-y}{n+1}}{n+\alpha \choose n}}}\\&{=}{\frac {1}{\Gamma (\alpha +1)}}\sum _{i=0}^{n}{\frac {x^{i}}{i!}}{\frac {L_{n-i}^{(\alpha +i)}(x)L_{n-i}^{(\alpha +i+1)}(y)}{{\alpha +n \choose n}{n \choose i}}};\end{aligned}}}

${\displaystyle K_{n}^{(\alpha )}(x,y)={\frac {y}{\alpha +1}}K_{n-1}^{(\alpha +1)}(x,y)+{\frac {1}{\Gamma (\alpha +1)}}{\frac {L_{n}^{(\alpha +1)}(x)L_{n}^{(\alpha )}(y)}{\alpha +n \choose n}}.}$

${\displaystyle y^{\alpha }e^{-y}K_{n}^{(\alpha )}(\cdot ,y)\rightarrow \delta (y-\,\cdot ),}$

${\displaystyle \int _{0}^{\infty }x^{\alpha +1}e^{-x}\left[L_{n}^{(\alpha )}\right]^{2}dx={\frac {(n+\alpha )!}{n!}}(2n+\alpha +1).}$

### 级数展开

${\displaystyle f(x)=\sum _{i=0}f_{i}^{(\alpha )}L_{i}^{(\alpha )}(x).}$

${\displaystyle f_{i}^{(\alpha )}=\int _{0}^{\infty }{\frac {L_{i}^{(\alpha )}(x)}{i+\alpha \choose i}}\cdot {\frac {x^{\alpha }e^{-x}}{\Gamma (\alpha +1)}}\cdot f(x)\,dx.}$

${\displaystyle \|f\|_{L^{2}}^{2}:=\int _{0}^{\infty }{\frac {x^{\alpha }e^{-x}}{\Gamma (\alpha +1)}}|f(x)|^{2}dx=\sum _{i=0}{i+\alpha \choose i}|f_{i}^{(\alpha )}|^{2}<\infty .}$

${\displaystyle f(x)=e^{{\frac {\gamma }{1+\gamma }}x}\cdot \sum _{i=0}{\frac {L_{i}^{(\alpha )}\left({\frac {x}{1+\gamma }}\right)}{(1+\gamma )^{i+\alpha +1}}}\sum _{n=0}^{i}\gamma ^{i-n}{i \choose n}f_{n}^{(\alpha )};}$

${\displaystyle e^{-\gamma x}\cdot L_{n}^{(\alpha )}(x(1+\gamma ))=\sum _{i=n}{\frac {L_{i}^{(\alpha )}(x)}{(1+\gamma )^{i+\alpha +1}}}\gamma ^{i-n}{i \choose n},}$

${\displaystyle L_{n}^{(\alpha )}\left({\frac {x}{1+\gamma }}\right)={\frac {1}{(1+\gamma )^{n}}}\sum _{i=0}^{n}\gamma ^{n-i}{n+\alpha \choose n-i}L_{i}^{(\alpha )}(x).}$

${\displaystyle {\frac {x^{\alpha -\beta }f(x)}{\Gamma (\alpha -\beta +1)}}={\alpha \choose \beta }\sum _{i=0}{\frac {L_{i}^{(\beta )}(x)}{\beta +i \choose i}}\sum _{n=0}^{i}(-1)^{i-n}{\alpha -\beta \choose i-n}{\alpha +n \choose n}f_{n}^{(\alpha )},}$

${\displaystyle {\frac {x^{\alpha -\beta }L_{n}^{(\alpha )}(x)}{\Gamma (\alpha -\beta +1)}}={\alpha \choose \beta }{\alpha +n \choose n}\sum _{i=n}(-1)^{i-n}{\alpha -\beta \choose i-n}{\frac {L_{i}^{(\beta )}(x)}{\beta +i \choose i}}}$

#### 更多的例子

${\displaystyle {\frac {x^{n}}{n!}}=\sum _{i=0}^{n}(-1)^{i}{n+\alpha \choose n-i}L_{i}^{(\alpha )}(x)=(-1)^{n}\sum _{i=0}^{n}L_{i}^{(\alpha -i)}(x){-\alpha \choose n-i},}$

${\displaystyle {n+x \choose n}=\sum _{i=0}^{n}{\frac {\alpha ^{i}}{i!}}L_{n-i}^{(x+i)}(\alpha ).}$

${\displaystyle e^{-\gamma x}=\sum _{i=0}{\frac {\gamma ^{i}}{(1+\gamma )^{i+\alpha +1}}}L_{i}^{(\alpha )}(x)}$（当且仅当 ${\displaystyle \operatorname {Re} {(\gamma )}>-{\frac {1}{2}}}$时收敛）

${\displaystyle {\frac {x^{\beta }e^{-\gamma x}}{\Gamma (\beta +1)}}={\alpha +\beta \choose \alpha }\sum _{i=0}{\frac {L_{i}^{(\alpha )}(x)}{\alpha +i \choose i}}\sum _{j=0}^{i}{\frac {(-1)^{j}}{(1+\gamma )^{\alpha +\beta +j+1}}}{\alpha +\beta +j \choose j}{\alpha +i \choose i-j}.}$

${\displaystyle {\frac {x^{n}e^{-\gamma x}}{n!}}=\sum _{i=0}{\frac {\gamma ^{i}L_{i}^{(\alpha )}(x)}{(1+\gamma )^{i+n+\alpha +1}}}\sum _{j=0}^{n}(-1)^{n-j}\gamma ^{j}{n+\alpha \choose j}{i \choose n-j},}$

${\displaystyle \gamma =0}$时，可以化简为：

${\displaystyle {\frac {x^{\beta }}{\Gamma (\beta +1)}}={\alpha +\beta \choose \alpha }\sum _{i=0}(-1)^{i}{\beta \choose i}{\frac {L_{i}^{(\alpha )}(x)}{\alpha +i \choose i}},}$
${\displaystyle {\frac {x^{\beta }L_{n}^{(\gamma )}(x)}{\Gamma (\beta +1)}}={\alpha +\beta \choose \alpha }\sum _{i=0}{\frac {L_{i}^{(\alpha )}(x)}{\alpha +i \choose i}}\sum _{j=0}^{n}(-1)^{i-j}{n+\gamma \choose n-j}{\beta +j \choose i}{\alpha +\beta +j \choose j}.}$

${\displaystyle \sum _{k\in \mathbb {Z} }e^{-k^{2}\pi x}=\sum _{i=0}L_{i}^{(\alpha )}\left({\frac {x}{t}}\right)\sum _{k\in \mathbb {Z} }{\frac {(k^{2}\pi t)^{i}}{(1+k^{2}\pi t)^{i+\alpha +1}}};}$

${\displaystyle \Gamma (\alpha )=x^{\alpha }\sum _{i=0}{\frac {L_{i}^{(\alpha )}(x)}{\alpha +i}}\qquad \left(\Re (\alpha )<{\frac {1}{2}}\right);}$

${\displaystyle {\frac {\gamma (s;z)}{t^{s}\Gamma (s)}}={\frac {\left({\frac {z}{t}}\right)^{\alpha }}{\Gamma (\alpha +1)}}\sum _{i=0}{\frac {L_{i}^{(\alpha )}\left({\frac {z}{t}}\right)}{\alpha +i \choose i}}\sum _{j=0}^{i}{\frac {(-1)^{j}}{(1+t)^{s+j}}}{s-1+j \choose j}{\alpha -1+i \choose i-j},}$
${\displaystyle {\frac {\gamma (s;z)}{t^{s}\Gamma (s)}}={\alpha +s \choose \alpha +1}\sum _{i=0}{\frac {{\alpha +i+1 \choose i+1}-L_{i+1}^{(\alpha )}\left({\frac {z}{t}}\right)}{\alpha +i+1 \choose i}}\sum _{j=0}^{i}{\frac {(-1)^{j}}{(1+t)^{\alpha +1+s+j}}}{\alpha +s+j \choose j}{\alpha +i+1 \choose i-j}.}$

${\displaystyle \gamma (s,z)={\frac {\gamma ^{s}}{\Gamma (1-s)}}\sum _{i=0}{\frac {L_{i+1}^{(-s)}(0)-L_{i+1}^{(-s)}\left({\frac {z}{\gamma }}\right)}{(1+\gamma )^{i+1}}}\sum _{n=0}^{i}\gamma ^{i-n}{\frac {i \choose n}{n+1-s}};}$

{\displaystyle {\begin{aligned}{\frac {\Gamma (s,z)}{z^{s}e^{-z}}}&=\sum _{k=0}{\frac {L_{k}^{(\alpha )}(z)}{(k+1){k+1+\alpha -s \choose k+1}}}\qquad \left(\Re \left(s-{\frac {\alpha }{2}}\right)<{\frac {1}{4}}\right)\\&=\sum _{k=0}L_{k}^{(\alpha )}(z\,t)\cdot {\frac {_{2}F_{1}\left(1+\alpha +k,1+k;2+\alpha +k-s;{\frac {t-1}{t}}\right)}{t^{k}(k+1){1+\alpha +k-s \choose 1+k}}}\\&=t^{s}\sum _{k=0}L_{k}^{(\alpha )}(z\,t)\cdot {\frac {_{2}F_{1}\left(1-s,1+\alpha -s;2+\alpha +k-s;{\frac {t-1}{t}}\right)}{(k+1){1+\alpha +k-s \choose 1+k}}}\\&=t^{1+\alpha }\sum _{k=0}L_{k}^{(\alpha )}(z\,t)\cdot {\frac {_{2}F_{1}\left(1+\alpha +k,1+\alpha -s;2+\alpha +k-s;1-t\right)}{(k+1){1+\alpha +k-s \choose 1+k}}},\end{aligned}}}

${\displaystyle _{2}F_{1}}$表示超几何函数

## 围道积分表示

${\displaystyle L_{n}^{(\alpha )}(x)={\frac {1}{2\pi i}}\oint {\frac {e^{-{\frac {xt}{1-t}}}}{(1-t)^{\alpha +1}\,t^{n+1}}}\;dt}$

## 与埃爾米特多項式的关系

${\displaystyle H_{2n}(x)=(-1)^{n}\ 2^{2n}\ n!\ L_{n}^{(-1/2)}(x^{2})}$

${\displaystyle H_{2n+1}(x)=(-1)^{n}\ 2^{2n+1}\ n!\ x\ L_{n}^{(1/2)}(x^{2})}$

## 与超几何函数的关系

${\displaystyle L_{n}^{(\alpha )}(x)={n+\alpha \choose n}M(-n,\alpha +1,x)={\frac {(\alpha +1)_{n}}{n!}}\,_{1}F_{1}(-n,\alpha +1,x)}$

${\displaystyle (a)_{n}}$阶乘幂，这里表示升阶乘

## 与贝塞尔函数的关系

{\displaystyle {\begin{aligned}L_{n}^{(\alpha )}(x)=&e^{\frac {x}{2}}\left({\frac {x}{4}}\right)^{n+{\frac {1}{2}}}{\frac {2}{{\sqrt {\pi }}(n+1)!{-{\frac {1}{2}} \choose n+1}}}\cdot \\&\cdot \sum _{k=0}^{n}(-1)^{k+1}{2n+1 \choose n-k}{\frac {{n+\alpha \choose n}{\alpha +2n+1 \choose n-k}}{n-k+\alpha \choose n-k}}\left(k+{\frac {1}{2}}\right)K_{k+{\frac {1}{2}}}\left({\frac {x}{2}}\right)\\=&e^{\frac {x}{2}}\left({\frac {4}{x}}\right)^{n+\alpha +{\frac {1}{2}}}\Gamma \left(\alpha +{\frac {1}{2}}\right){-\alpha -1 \choose n}{-\alpha -{\frac {1}{2}} \choose n}\cdot \\&\cdot n!\sum _{k=n}{\frac {{-2n-1-2\alpha \choose k-n}{-2n-1-\alpha \choose k-n}}{-\alpha -1 \choose k-n}}\left(\alpha +{\frac {1}{2}}+k\right)I_{\alpha +{\frac {1}{2}}+k}\left({\frac {x}{2}}\right)\end{aligned}},}

${\displaystyle L_{n}^{(\alpha )}(x)={\frac {2}{4^{n}(2n+1){-{\frac {1}{2}} \choose n}}}\sum _{k=0}^{n}\left(k+{\frac {1}{2}}\right){\frac {2n+1 \choose n-k}{{n \choose k}^{2}}}{n+\alpha \choose k}{2n+\alpha +1 \choose n-k}{\frac {x^{n-k}}{(n-k)!}}L_{k}^{-2k-1}(x).}$

## 注释

1. ^ Abramowitz, p. 506, 13.3.8

## 参考文献

• Abramowitz, Milton; Stegun, Irene A., eds. (1965), "Chapter 22", Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, New York: Dover, ISBN 0-486-61272-4 .
• B Spain, M G Smith, Functions of mathematical physics, Van Nostrand Reinhold Company, London, 1970. Chapter 10 deals with Laguerre polynomials.
• Eric W. Weisstein, "Laguerre Polynomial", From MathWorld—A Wolfram Web Resource.
• George Arfken and Hans Weber. Mathematical Methods for Physicists. Academic Press. 2000. ISBN 0-12-059825-6.
• S. S. Bayin (2006), Mathematical Methods in Science and Engineering, Wiley, Chapter 3.