在數學 中，以法國 數學家 埃德蒙·拉蓋爾 拉蓋爾多項式 定義為拉蓋爾方程式 的標準解。
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{\displaystyle x\,y''+(1-x)\,y'+n\,y=0\,}
這是一個二階線性微分方程式 。
這個方程式只有當n 非負時，才有非奇異解。拉蓋爾多項式可用在高斯積分法 中，計算形如
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{\displaystyle \int _{0}^{\infty }f(x)dx}
這些多項式（通常用L 0 , L 1 等表示）構成一個多項式序列 羅德里格公式 遞推得到。
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{\displaystyle L_{n}(x)={\frac {e^{x}}{n!}}{\frac {d^{n}}{dx^{n}}}\left(e^{-x}x^{n}\right).}
在按照下式定義的內積構成的內積空間 中，拉蓋爾多項式是正交多項式 。
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{\displaystyle \langle f,g\rangle =\int _{0}^{\infty }f(x)g(x)e^{-x}\,dx.}
拉蓋爾多項式構成一個Sheffer序列
拉蓋爾多項式在量子力學中有重要應用。氫原子薛丁格方程式 的解的徑向部分，就是拉蓋爾多項式。
物理學家通常採用另外一種拉蓋爾多項式的定義形式，即在上面的形式的基礎上乘上一個n !。
前幾個拉蓋爾多項式的表達式與函數圖像如下：
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{\displaystyle L_{n}(x)\,}
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{\displaystyle 1\,}
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{\displaystyle -x+1\,}
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{\displaystyle {\scriptstyle {\frac {1}{2}}}(x^{2}-4x+2)\,}
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{\displaystyle {\scriptstyle {\frac {1}{6}}}(-x^{3}+9x^{2}-18x+6)\,}
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{\displaystyle {\scriptstyle {\frac {1}{24}}}(x^{4}-16x^{3}+72x^{2}-96x+24)\,}
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{\displaystyle {\scriptstyle {\frac {1}{120}}}(-x^{5}+25x^{4}-200x^{3}+600x^{2}-600x+120)\,}
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{\displaystyle {\scriptstyle {\frac {1}{720}}}(x^{6}-36x^{5}+450x^{4}-2400x^{3}+5400x^{2}-4320x+720)\,}
前六個拉蓋爾多項式 拉蓋爾多項式也可以通過遞迴的方式進行定義。首先，規定前兩個拉蓋爾多項式為：
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{\displaystyle L_{0}(x)=1\,}
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{\displaystyle L_{1}(x)=1-x\,}
然後運用下面的遞迴關係 得到更高階的多項式。
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{\displaystyle L_{k+1}(x)={\frac {1}{k+1}}\left((2k+1-x)L_{k}(x)-kL_{k-1}(x)\right).}
上面提到的拉蓋爾多項式的正交性，也可以用另外一種方式表達。即：如果X 是一個服從指數分布 的隨機變數 （即，機率密度函數 如下式）：
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{\displaystyle f(x)=\left\{{\begin{matrix}e^{-x}&{\mbox{if}}\ x>0,\\0&{\mbox{if}}\ x<0,\end{matrix}}\right.}
那麼：
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{\displaystyle E\left[L_{n}(X)L_{m}(X)\right]=0\ {\mbox{whenever}}\ n\neq m.}
指數分布不是唯一的伽瑪分布 ，對於任意的伽瑪分布（機率密度函數如下，α > −1，參見Γ函數 ）
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{\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.}
相應的正交多項式為形如下式的廣義拉蓋爾多項式 （可以通過羅德里格公式 得到）：
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{\displaystyle L_{n}^{(\alpha )}(x)={x^{-\alpha }e^{x} \over n!}{d^{n} \over dx^{n}}\left(e^{-x}x^{n+\alpha }\right).}
有時也將上面的多項式稱為連帶（聯屬，伴隨）拉蓋爾多項式 。當取α = 0時，就回到拉蓋爾多項式：
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{\displaystyle L_{n}^{(0)}(x)=L_{n}(x).}
拉蓋爾函數可以由合流超幾何函數 和Kummer轉換得到：
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{\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}\,}
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{\displaystyle =e^{x}\cdot {n+\alpha \choose n}M(\alpha +n+1,\alpha +1,-x)}
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{\displaystyle ={\frac {e^{x}\sin(n\pi )}{\sin((n+\alpha )\pi )}}L_{-\alpha -n-1}^{(\alpha )}(-x)}
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{\displaystyle =e^{x}\cdot \sum _{i=0}(-1)^{i}{\alpha +n+i \choose n}{\frac {x^{i}}{i!}}.}
n
{\displaystyle n}
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{\displaystyle n}
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{\displaystyle n}
萊布尼茨乘積求導公式
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{\displaystyle L_{n}^{(\alpha )}(x)=\sum _{i=0}^{n}(-1)^{i}{n+\alpha \choose n-i}{\frac {x^{i}}{i!}}}
n 階拉蓋爾多項式的首項係數為(−1)n n !；拉蓋爾多項式在x=0的取值（常數項 ）為
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{\displaystyle L_{n}^{(\alpha )}(0)={n+\alpha \choose n}\approx {\frac {n^{\alpha }}{\Gamma (\alpha +1)}};}
L n (α ) 有n 個實 的正根 （應該注意到
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{\displaystyle \left((-1)^{n-i}L_{n-i}^{(\alpha )}\right)_{i=0}^{n}}
史特姆序列 ），且這些根全部位於區間
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{\displaystyle (0,n+\alpha +(n-1){\sqrt {n+\alpha }}]}
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{\displaystyle n}
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{\displaystyle \alpha }
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{\displaystyle x>0}
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{\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)}
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{\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]
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{\displaystyle L_{0}^{(\alpha )}(x)=1}
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{\displaystyle L_{1}^{(\alpha )}(x)=-x+\alpha +1}
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{\displaystyle L_{2}^{(\alpha )}(x)={\frac {x^{2}}{2}}-(\alpha +2)x+{\frac {(\alpha +2)(\alpha +1)}{2}}}
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{\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}}}
根據拉蓋爾多項式的定義，可以使用秦九韶算法 計算拉蓋爾多項式，程序代碼如下： 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;
}
拉蓋爾多項式滿足以下的遞迴關係：
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{\displaystyle L_{n}^{(\alpha +\beta +1)}(x+y)=\sum _{i=0}^{n}L_{i}^{(\alpha )}(x)L_{n-i}^{(\beta )}(y),}
特別地，有
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{\displaystyle L_{n}^{(\alpha +1)}(x)=\sum _{i=0}^{n}L_{i}^{(\alpha )}(x)}
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{\displaystyle L_{n}^{(\alpha )}(x)=\sum _{i=0}^{n}{\alpha -\beta +n-i-1 \choose n-i}L_{i}^{(\beta )}(x)}
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{\displaystyle L_{n}^{(\alpha )}(x)=\sum _{i=0}^{n}{\alpha -\beta +n \choose n-i}L_{i}^{(\beta -i)}(x);}
還有
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{\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}}}
運用以上式子可以得到以下四條關係式：
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{\displaystyle L_{n}^{(\alpha )}(x)=L_{n}^{(\alpha +1)}(x)-L_{n-1}^{(\alpha +1)}(x)}
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{\displaystyle =\sum _{j=0}^{k}{k \choose j}L_{n-j}^{(\alpha -k+j)}(x),}
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{\displaystyle nL_{n}^{(\alpha )}(x)=(n+\alpha )L_{n-1}^{(\alpha )}(x)-xL_{n-1}^{(\alpha +1)}(x),}
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{\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),}
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{\displaystyle nL_{n}^{(\alpha +1)}(x)=(n-x)L_{n-1}^{(\alpha +1)}(x)+(n+\alpha )L_{n-1}^{(\alpha )}(x)}
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{\displaystyle xL_{n}^{(\alpha +1)}=(n+\alpha )L_{n-1}^{\alpha }(x)-(n-x)L_{n}^{(\alpha )}(x);}
將它們組合在一起，就得到了最常用的遞迴關係式：
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{\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).}
當
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{\displaystyle i}
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{\displaystyle n}
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{\displaystyle {\frac {(-x)^{i}}{i!}}L_{n}^{(i-n)}(x)={\frac {(-x)^{n}}{n!}}L_{i}^{(n-i)}(x);}
進一步可以得到部分分式分解 ：
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{\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}}.}
將拉蓋爾多項式對自變數x 求導k 次，得到：
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{\displaystyle {\frac {\mathrm {d} ^{k}}{\mathrm {d} x^{k}}}L_{n}^{(\alpha )}(x)=(-1)^{k}L_{n-k}^{(\alpha +k)}(x)\,;}
進一步有：
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)
,
{\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),}
運用柯西多重積分公式
L
n
(
α
′
)
(
x
)
=
(
α
′
−
α
)
(
α
′
+
n
α
′
−
α
)
∫
0
x
t
α
(
x
−
t
)
α
′
−
α
−
1
x
α
′
L
n
(
α
)
(
t
)
d
t
.
{\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.}
將拉蓋爾多項式對參變量
α
{\displaystyle \alpha }
d
d
α
L
n
(
α
)
(
x
)
=
∑
i
=
0
n
−
1
L
i
(
α
)
(
x
)
n
−
i
.
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} \alpha }}L_{n}^{(\alpha )}(x)=\sum _{i=0}^{n-1}{\frac {L_{i}^{(\alpha )}(x)}{n-i}}.}
廣義拉蓋爾多項式滿足下面的微分方程式：
x
L
n
(
α
)
′
′
(
x
)
+
(
α
+
1
−
x
)
L
n
(
α
)
′
(
x
)
+
n
L
n
(
α
)
(
x
)
=
0
,
{\displaystyle xL_{n}^{(\alpha )\prime \prime }(x)+(\alpha +1-x)L_{n}^{(\alpha )\prime }(x)+nL_{n}^{(\alpha )}(x)=0,\,}
可以與拉蓋爾多項式的k階導數所滿足的微分方程式作一比較。
x
L
n
(
k
)
′
′
(
x
)
+
(
k
+
1
−
x
)
L
n
(
k
)
′
(
x
)
+
(
n
−
k
)
L
n
(
k
)
(
x
)
=
0
,
{\displaystyle xL_{n}^{(k)\prime \prime }(x)+(k+1-x)L_{n}^{(k)\prime }(x)+(n-k)L_{n}^{(k)}(x)=0,\,}
僅在此式中，
L
n
(
k
)
(
x
)
≡
d
L
n
(
x
)
d
x
k
{\displaystyle L_{n}^{(k)}(x)\equiv {\frac {dL_{n}(x)}{dx^{k}}}}
於是，當
α
=
0
{\displaystyle \alpha =0}
L
n
(
k
)
(
x
)
=
(
−
1
)
k
d
L
n
+
k
(
x
)
d
x
k
{\displaystyle L_{n}^{(k)}(x)=(-1)^{k}{\frac {dL_{n+k}(x)}{dx^{k}}}}
伴隨拉蓋爾多項式在區間[0, ∞)上以權函數x α e −x 正交：
∫
0
∞
x
α
e
−
x
L
n
(
α
)
(
x
)
L
m
(
α
)
(
x
)
d
x
=
Γ
(
n
+
α
+
1
)
n
!
δ
n
,
m
,
{\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},}
這可由下式得到：
∫
0
∞
x
α
′
−
1
e
−
x
L
n
(
α
)
(
x
)
d
x
=
(
α
−
α
′
+
n
n
)
Γ
(
α
′
)
.
{\displaystyle \int _{0}^{\infty }x^{\alpha '-1}e^{-x}L_{n}^{(\alpha )}(x)dx={\alpha -\alpha '+n \choose n}\Gamma (\alpha ').}
伴隨對稱核多項式可以用拉蓋爾多項式表示為：
K
n
(
α
)
(
x
,
y
)
:=
1
Γ
(
α
+
1
)
∑
i
=
0
n
L
i
(
α
)
(
x
)
L
i
(
α
)
(
y
)
(
α
+
i
i
)
=
1
Γ
(
α
+
1
)
L
n
(
α
)
(
x
)
L
n
+
1
(
α
)
(
y
)
−
L
n
+
1
(
α
)
(
x
)
L
n
(
α
)
(
y
)
x
−
y
n
+
1
(
n
+
α
n
)
=
1
Γ
(
α
+
1
)
∑
i
=
0
n
x
i
i
!
L
n
−
i
(
α
+
i
)
(
x
)
L
n
−
i
(
α
+
i
+
1
)
(
y
)
(
α
+
n
n
)
(
n
i
)
;
{\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}}}
也有下面的遞迴關係：
K
n
(
α
)
(
x
,
y
)
=
y
α
+
1
K
n
−
1
(
α
+
1
)
(
x
,
y
)
+
1
Γ
(
α
+
1
)
L
n
(
α
+
1
)
(
x
)
L
n
(
α
)
(
y
)
(
α
+
n
n
)
.
{\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}}.}
進一步地，在伴L 2 [0, ∞)空間上，有：
y
α
e
−
y
K
n
(
α
)
(
⋅
,
y
)
→
δ
(
y
−
⋅
)
,
{\displaystyle y^{\alpha }e^{-y}K_{n}^{(\alpha )}(\cdot ,y)\rightarrow \delta (y-\,\cdot ),}
在氫原子的量子力學處理中用到了下面的公式：
∫
0
∞
x
α
+
1
e
−
x
[
L
n
(
α
)
]
2
d
x
=
(
n
+
α
)
!
n
!
(
2
n
+
α
+
1
)
.
{\displaystyle \int _{0}^{\infty }x^{\alpha +1}e^{-x}\left[L_{n}^{(\alpha )}\right]^{2}dx={\frac {(n+\alpha )!}{n!}}(2n+\alpha +1).}
設一個函數具有以下的級數展開形式：
f
(
x
)
=
∑
i
=
0
f
i
(
α
)
L
i
(
α
)
(
x
)
.
{\displaystyle f(x)=\sum _{i=0}f_{i}^{(\alpha )}L_{i}^{(\alpha )}(x).}
則展開式的係數由下式給出
f
i
(
α
)
=
∫
0
∞
L
i
(
α
)
(
x
)
(
i
+
α
i
)
⋅
x
α
e
−
x
Γ
(
α
+
1
)
⋅
f
(
x
)
d
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.}
這個級數在Lp空間
L
2
[
0
,
∞
)
{\displaystyle L^{2}[0,\infty )}
若且唯若
‖
f
‖
L
2
2
:=
∫
0
∞
x
α
e
−
x
Γ
(
α
+
1
)
|
f
(
x
)
|
2
d
x
=
∑
i
=
0
(
i
+
α
i
)
|
f
i
(
α
)
|
2
<
∞
.
{\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 .}
一個相關的展開式為：
f
(
x
)
=
e
γ
1
+
γ
x
⋅
∑
i
=
0
L
i
(
α
)
(
x
1
+
γ
)
(
1
+
γ
)
i
+
α
+
1
∑
n
=
0
i
γ
i
−
n
(
i
n
)
f
n
(
α
)
;
{\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 )};}
特別地
e
−
γ
x
⋅
L
n
(
α
)
(
x
(
1
+
γ
)
)
=
∑
i
=
n
L
i
(
α
)
(
x
)
(
1
+
γ
)
i
+
α
+
1
γ
i
−
n
(
i
n
)
,
{\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},}
這可由下式得到：
L
n
(
α
)
(
x
1
+
γ
)
=
1
(
1
+
γ
)
n
∑
i
=
0
n
γ
n
−
i
(
n
+
α
n
−
i
)
L
i
(
α
)
(
x
)
.
{\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).}
還有，當
Re
(
2
α
−
β
)
>
−
1
{\displaystyle \operatorname {Re} {(2\alpha -\beta )}>-1}
x
α
−
β
f
(
x
)
Γ
(
α
−
β
+
1
)
=
(
α
β
)
∑
i
=
0
L
i
(
β
)
(
x
)
(
β
+
i
i
)
∑
n
=
0
i
(
−
1
)
i
−
n
(
α
−
β
i
−
n
)
(
α
+
n
n
)
f
n
(
α
)
,
{\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 )},}
這個結果可以由下式導出，
x
α
−
β
L
n
(
α
)
(
x
)
Γ
(
α
−
β
+
1
)
=
(
α
β
)
(
α
+
n
n
)
∑
i
=
n
(
−
1
)
i
−
n
(
α
−
β
i
−
n
)
L
i
(
β
)
(
x
)
(
β
+
i
i
)
{\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}}}
冪函數 可以展開為：
x
n
n
!
=
∑
i
=
0
n
(
−
1
)
i
(
n
+
α
n
−
i
)
L
i
(
α
)
(
x
)
=
(
−
1
)
n
∑
i
=
0
n
L
i
(
α
−
i
)
(
x
)
(
−
α
n
−
i
)
,
{\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},}
二項式 可以展開為：
(
n
+
x
n
)
=
∑
i
=
0
n
α
i
i
!
L
n
−
i
(
x
+
i
)
(
α
)
.
{\displaystyle {n+x \choose n}=\sum _{i=0}^{n}{\frac {\alpha ^{i}}{i!}}L_{n-i}^{(x+i)}(\alpha ).}
進一步可以得到：
e
−
γ
x
=
∑
i
=
0
γ
i
(
1
+
γ
)
i
+
α
+
1
L
i
(
α
)
(
x
)
{\displaystyle e^{-\gamma x}=\sum _{i=0}{\frac {\gamma ^{i}}{(1+\gamma )^{i+\alpha +1}}}L_{i}^{(\alpha )}(x)}
Re
(
γ
)
>
−
1
2
{\displaystyle \operatorname {Re} {(\gamma )}>-{\frac {1}{2}}}
更一般地
x
β
e
−
γ
x
Γ
(
β
+
1
)
=
(
α
+
β
α
)
∑
i
=
0
L
i
(
α
)
(
x
)
(
α
+
i
i
)
∑
j
=
0
i
(
−
1
)
j
(
1
+
γ
)
α
+
β
+
j
+
1
(
α
+
β
+
j
j
)
(
α
+
i
i
−
j
)
.
{\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}.}
對於非負的整數
β
{\displaystyle \beta }
x
n
e
−
γ
x
n
!
=
∑
i
=
0
γ
i
L
i
(
α
)
(
x
)
(
1
+
γ
)
i
+
n
+
α
+
1
∑
j
=
0
n
(
−
1
)
n
−
j
γ
j
(
n
+
α
j
)
(
i
n
−
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},}
當
γ
=
0
{\displaystyle \gamma =0}
x
β
Γ
(
β
+
1
)
=
(
α
+
β
α
)
∑
i
=
0
(
−
1
)
i
(
β
i
)
L
i
(
α
)
(
x
)
(
α
+
i
i
)
,
{\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}},}
x
β
L
n
(
γ
)
(
x
)
Γ
(
β
+
1
)
=
(
α
+
β
α
)
∑
i
=
0
L
i
(
α
)
(
x
)
(
α
+
i
i
)
∑
j
=
0
n
(
−
1
)
i
−
j
(
n
+
γ
n
−
j
)
(
β
+
j
i
)
(
α
+
β
+
j
j
)
.
{\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}.}
雅可比Theta 函數 有下面的表示：
∑
k
∈
Z
e
−
k
2
π
x
=
∑
i
=
0
L
i
(
α
)
(
x
t
)
∑
k
∈
Z
(
k
2
π
t
)
i
(
1
+
k
2
π
t
)
i
+
α
+
1
;
{\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}}};}
隨意選定參量t，貝索函數 可以表示為：
J
α
(
x
)
(
x
2
)
α
=
e
−
t
Γ
(
α
+
1
)
∑
i
=
0
L
i
(
α
)
(
x
2
4
t
)
(
i
+
α
i
)
t
i
i
!
;
{\displaystyle {\frac {J_{\alpha }(x)}{\left({\frac {x}{2}}\right)^{\alpha }}}={\frac {e^{-t}}{\Gamma (\alpha +1)}}\sum _{i=0}{\frac {L_{i}^{(\alpha )}\left({\frac {x^{2}}{4t}}\right)}{i+\alpha \choose i}}{\frac {t^{i}}{i!}};}
Γ函數 可以展開為：
Γ
(
α
)
=
x
α
∑
i
=
0
L
i
(
α
)
(
x
)
α
+
i
(
ℜ
(
α
)
<
1
2
)
;
{\displaystyle \Gamma (\alpha )=x^{\alpha }\sum _{i=0}{\frac {L_{i}^{(\alpha )}(x)}{\alpha +i}}\qquad \left(\Re (\alpha )<{\frac {1}{2}}\right);}
低階不完全伽瑪函數 可展開為：
γ
(
s
;
z
)
t
s
Γ
(
s
)
=
(
z
t
)
α
Γ
(
α
+
1
)
∑
i
=
0
L
i
(
α
)
(
z
t
)
(
α
+
i
i
)
∑
j
=
0
i
(
−
1
)
j
(
1
+
t
)
s
+
j
(
s
−
1
+
j
j
)
(
α
−
1
+
i
i
−
j
)
,
{\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},}
γ
(
s
;
z
)
t
s
Γ
(
s
)
=
(
α
+
s
α
+
1
)
∑
i
=
0
(
α
+
i
+
1
i
+
1
)
−
L
i
+
1
(
α
)
(
z
t
)
(
α
+
i
+
1
i
)
∑
j
=
0
i
(
−
1
)
j
(
1
+
t
)
α
+
1
+
s
+
j
(
α
+
s
+
j
j
)
(
α
+
i
+
1
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}.}
還有：
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{\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}};}
於是，高階不完全伽瑪函數就是：
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{\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}}}
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{\displaystyle _{2}F_{1}}
超幾何函數 。
拉蓋爾多項式可以用圍道積分 表示，如下式所示：
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{\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}
積分方向逆時針繞原點一周。
廣義拉蓋爾多項式與埃爾米特多項式有下列關係：
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{\displaystyle H_{2n}(x)=(-1)^{n}\ 2^{2n}\ n!\ L_{n}^{(-1/2)}(x^{2})}
以及
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{\displaystyle H_{2n+1}(x)=(-1)^{n}\ 2^{2n+1}\ n!\ x\ L_{n}^{(1/2)}(x^{2})}
這裡的H n x 2 )的埃爾米特多項式 （所謂的「物理學家形式」）。
正因為這樣，廣義拉蓋爾多項式也在量子諧振子 的量子力學處理中出現。
拉蓋爾多項式可以用超幾何函數 來定義，具體地說，是用合流超幾何函數 定義：
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{\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)}
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{\displaystyle (a)_{n}}
階乘冪 ，這裡表示升階乘 。
拉蓋爾多項式與變形貝索函數 之間有以下關係：
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{\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=0}^{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}},}
進一步有：
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{\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).}
^ Abramowitz, p. 506, 13.3.8 （頁面存檔備份 ，存於網際網路檔案館 ）
Abramowitz, Milton; Stegun, Irene A., eds. (1965), "Chapter 22 （頁面存檔備份 ，存於網際網路檔案館 ）", 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 （頁面存檔備份 ，存於網際網路檔案館 ）", 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. 
![{\displaystyle E\left[L_{n}(X)L_{m}(X)\right]=0\ {\mbox{whenever}}\ n\neq m.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5dfce23a3e74e527e067dcbe0fb05ad0cf768a63)
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![{\displaystyle \int _{0}^{\infty }x^{\alpha +1}e^{-x}\left[L_{n}^{(\alpha )}\right]^{2}dx={\frac {(n+\alpha )!}{n!}}(2n+\alpha +1).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb41ec3fea5cdbbd653107cdb8067455a1cbccb9)
