拉盖尔多项式

在数学中，以法国数学家埃德蒙·拉盖尔（英语：Edmond Laguerre）命名的拉盖尔多项式定义为拉盖尔方程的标准解。 x y ″ + ( 1 − x ) y ′ + n y = 0 {\displaystyle x\,y''+(1-x)\,y'+n\,y=0\,} 这是一个二阶线性微分方程。 这个方程只有当n非负时，才有非奇异解。拉盖尔多项式可用在高斯积分法中，计算形如 ∫ 0 ∞ f ( x ) d x {\displaystyle \int _{0}^{\infty }f(x)dx} 的积分。 这些多项式（通常用L0, L1等表示）构成一个多项式序列（英语：polynomial sequence）。这个多项式序列可以用罗德里格公式递推得到。 L n ( x ) = e x n ! d n d x n ( e − x x n ) . {\displaystyle L_{n}(x)={\frac {e^{x}}{n!}}{\frac {d^{n}}{dx^{n}}}\left(e^{-x}x^{n}\right).} 在按照下式定义的内积构成的内积空间中，拉盖尔多项式是正交多项式。 ⟨ f , g ⟩ = ∫ 0 ∞ f ( x ) g ( x ) e − x d x . {\displaystyle \langle f,g\rangle =\int _{0}^{\infty }f(x)g(x)e^{-x}\,dx.} 拉盖尔多项式构成一个Sheffer序列（英语：Sheffer sequence）。 拉盖尔多项式在量子力学中有重要应用。氢原子薛定谔方程的解的径向部分，就是拉盖尔多项式。 物理学家通常采用另外一种拉盖尔多项式的定义形式，即在上面的形式的基础上乘上一个n!。 前几个拉盖尔多项式的表达式与函数图像如下： 拉盖尔多项式也可以通过递归的方式进行定义。首先，规定前两个拉盖尔多项式为： L 0 ( x ) = 1 {\displaystyle L_{0}(x)=1\,} L 1 ( x ) = 1 − x {\displaystyle L_{1}(x)=1-x\,} 然后运用下面的递推关系得到更高阶的多项式。 L k + 1 ( x ) = 1 k + 1 ( ( 2 k + 1 − x ) L k ( x ) − k L k − 1 ( x ) ) . {\displaystyle L_{k+1}(x)={\frac {1}{k+1}}\left((2k+1-x)L_{k}(x)-kL_{k-1}(x)\right).} 上面提到的拉盖尔多项式的正交性，也可以用另外一种方式表达。即：如果X是一个服从指数分布的随机变量（即，概率密度函数如下式）： f ( x ) = { e − x if   x > 0 , 0 if   x < 0 , {\displaystyle f(x)=\left\{{\begin{matrix}e^{-x}&{\mbox{if}}\ x>0,\\0&{\mbox{if}}\ x<0,\end{matrix}}\right.} 那么： E [ L n ( X ) L m ( X ) ] = 0   whenever   n ≠ m . {\displaystyle E\left[L_{n}(X)L_{m}(X)\right]=0\ {\mbox{whenever}}\ n\neq m.} 指数分布不是唯一的伽玛分布，对于任意的伽玛分布（概率密度函数如下，α > −1，参见Γ函数） f ( x ) = { x α e − x / Γ ( 1 + α ) if   x > 0 , 0 if   x < 0 , {\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.} 相应的正交多项式为形如下式的广义拉盖尔多项式（可以通过罗德里格公式得到）： L n ( α ) ( x ) = x − α e x n ! d n d x n ( e − x x n + α ) . {\displaystyle L_{n}^{(\alpha )}(x)={x^{-\alpha }e^{x} \over n!}{d^{n} \over dx^{n}}\left(e^{-x}x^{n+\alpha }\right).} 有时也将上面的多项式称为连带（联属，伴随）拉盖尔多项式。当取α = 0时，就回到拉盖尔多项式： L n ( 0 ) ( x ) = L n ( x ) . {\displaystyle L_{n}^{(0)}(x)=L_{n}(x).} 拉盖尔函数可以由合流超几何函数和Kummer变换得到： L n ( α ) ( x ) := ( n + α n ) M ( − n , α + 1 , x ) = ( n + α n ) ∑ i = 0 ( − 1 ) i ( n i ) ( α + i i ) x i {\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}\,} = e x ⋅ ( n + α n ) M ( α + n + 1 , α + 1 , − x ) {\displaystyle =e^{x}\cdot {n+\alpha \choose n}M(\alpha +n+1,\alpha +1,-x)} = e x sin ⁡ ( n π ) sin ⁡ ( ( n + α ) π ) L − α − n − 1 ( α ) ( − x ) {\displaystyle ={\frac {e^{x}\sin(n\pi )}{\sin((n+\alpha )\pi )}}L_{-\alpha -n-1}^{(\alpha )}(-x)} = e x ⋅ ∑ i = 0 ( − 1 ) i ( α + n + i n ) x i i ! . {\displaystyle =e^{x}\cdot \sum _{i=0}(-1)^{i}{\alpha +n+i \choose n}{\frac {x^{i}}{i!}}.} 当 n {\displaystyle n} 为整数时，截断为 n {\displaystyle n} 阶拉盖尔多项式。 n {\displaystyle n} 阶拉盖尔多项式可以通过将莱布尼茨乘积求导公式（英语：Leibniz's theorem for differentiation of a product）应用在罗德里格公式上而得到，结果为 L n ( α ) ( x ) = ∑ i = 0 n ( − 1 ) i ( n + α n − i ) x i i ! {\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 L n ( α ) ( 0 ) = ( n + α n ) ≈ n α Γ ( α + 1 ) ; {\displaystyle L_{n}^{(\alpha )}(0)={n+\alpha \choose n}\approx {\frac {n^{\alpha }}{\Gamma (\alpha +1)}};} Ln(α)有n个实的正根（应该注意到 ( ( − 1 ) n − i L n − i ( α ) ) i = 0 n {\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 ( 0 , n + α + ( n − 1 ) n + α ] {\displaystyle (0,n+\alpha +(n-1){\sqrt {n+\alpha }}]} 中。 当 n {\displaystyle n} 很大，而 α {\displaystyle \alpha } 不变， x > 0 {\displaystyle x>0} 时，拉盖尔多项式的渐近行为如下： L n ( α ) ( x ) ≈ n α 2 − 1 4 π e x 2 x α 2 + 1 4 cos ⁡ ( 2 x ( n + α + 1 2 ) − π 2 ( α + 1 2 ) ) {\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 L n ( α ) ( − x ) ≈ n α 2 − 1 4 2 π e − x 2 x α 2 + 1 4 exp ⁡ ( 2 x ( n + α + 1 2 ) ) {\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)} 。 前几个广义拉盖尔多项式为： L 0 ( α ) ( x ) = 1 {\displaystyle L_{0}^{(\alpha )}(x)=1} L 1 ( α ) ( x ) = − x + α + 1 {\displaystyle L_{1}^{(\alpha )}(x)=-x+\alpha +1} L 2 ( α ) ( x ) = x 2 2 − ( α + 2 ) x + ( α + 2 ) ( α + 1 ) 2 {\displaystyle L_{2}^{(\alpha )}(x)={\frac {x^{2}}{2}}-(\alpha +2)x+{\frac {(\alpha +2)(\alpha +1)}{2}}} L 3 ( α ) ( x ) = − x 3 6 + ( α + 3 ) x 2 2 − ( α + 2 ) ( α + 3 ) x 2 + ( α + 1 ) ( α + 2 ) ( α + 3 ) 6 {\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; } 拉盖尔多项式满足以下的递推关系： L n ( α + β + 1 ) ( x + y ) = ∑ i = 0 n L i ( α ) ( x ) L n − i ( β ) ( y ) , {\displaystyle L_{n}^{(\alpha +\beta +1)}(x+y)=\sum _{i=0}^{n}L_{i}^{(\alpha )}(x)L_{n-i}^{(\beta )}(y),} 特别地，有 L n ( α + 1 ) ( x ) = ∑ i = 0 n L i ( α ) ( x ) {\displaystyle L_{n}^{(\alpha +1)}(x)=\sum _{i=0}^{n}L_{i}^{(\alpha )}(x)} 以及 L n ( α ) ( x ) = ∑ i = 0 n ( α − β + n − i − 1 n − i ) L i ( β ) ( x ) {\displaystyle L_{n}^{(\alpha )}(x)=\sum _{i=0}^{n}{\alpha -\beta +n-i-1 \choose n-i}L_{i}^{(\beta )}(x)} ，或 L n ( α ) ( x ) = ∑ i = 0 n ( α − β + n n − i ) L i ( β − i ) ( x ) ; {\displaystyle L_{n}^{(\alpha )}(x)=\sum _{i=0}^{n}{\alpha -\beta +n \choose n-i}L_{i}^{(\beta -i)}(x);} 还有 L n ( α ) ( x ) − ∑ j = 0 Δ − 1 ( n + α n − j ) ( − 1 ) j x j j ! = ( − 1 ) Δ x Δ ( Δ − 1 ) ! ∑ i = 0 n − Δ ( n + α n − Δ − i ) ( n − i ) ( n i ) L i ( α + Δ ) ( x ) = ( − 1 ) Δ x Δ ( Δ − 1 ) ! ∑ i = 0 n − Δ ( n + α − i − 1 n − Δ − i ) ( n − i ) ( n i ) L i ( n + α + Δ − 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}}} 运用以上式子可以得到以下四条关系式： L n ( α ) ( x ) = L n ( α + 1 ) ( x ) − L n − 1 ( α + 1 ) ( x ) {\displaystyle L_{n}^{(\alpha )}(x)=L_{n}^{(\alpha +1)}(x)-L_{n-1}^{(\alpha +1)}(x)} = ∑ j = 0 k ( k j ) L n − j ( α − k + j ) ( x ) , {\displaystyle =\sum _{j=0}^{k}{k \choose j}L_{n-j}^{(\alpha -k+j)}(x),} n L n ( α ) ( x ) = ( n + α ) L n − 1 ( α ) ( x ) − x L n − 1 ( α + 1 ) ( x ) , {\displaystyle nL_{n}^{(\alpha )}(x)=(n+\alpha )L_{n-1}^{(\alpha )}(x)-xL_{n-1}^{(\alpha +1)}(x),} or x k k ! L n ( α ) ( x ) = ∑ i = 0 k ( − 1 ) i ( n + i i ) ( n + α k − i ) L n + i ( α − k ) ( x ) , {\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),} n L n ( α + 1 ) ( x ) = ( n − x ) L n − 1 ( α + 1 ) ( x ) + ( n + α ) L n − 1 ( α ) ( x ) {\displaystyle nL_{n}^{(\alpha +1)}(x)=(n-x)L_{n-1}^{(\alpha +1)}(x)+(n+\alpha )L_{n-1}^{(\alpha )}(x)} x L n ( α + 1 ) = ( n + α ) L n − 1 α ( x ) − ( n − x ) L n ( α ) ( x ) ; {\displaystyle xL_{n}^{(\alpha +1)}=(n+\alpha )L_{n-1}^{\alpha }(x)-(n-x)L_{n}^{(\alpha )}(x);} 将它们组合在一起，就得到了最常用的递推关系式： L n + 1 ( α ) ( x ) = 1 n + 1 ( ( 2 n + 1 + α − x ) L n ( α ) ( x ) − ( n + α ) L n − 1 ( α ) ( 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).} 当 i {\displaystyle i} 与 n {\displaystyle n} 均为整数时，拉盖尔多项式有以下的有趣性质： ( − x ) i i ! L n ( i − n ) ( x ) = ( − x ) n n ! L i ( n − i ) ( x ) ; {\displaystyle {\frac {(-x)^{i}}{i!}}L_{n}^{(i-n)}(x)={\frac {(-x)^{n}}{n!}}L_{i}^{(n-i)}(x);} 进一步可以得到部分分式分解： L n ( α ) ( x ) ( n + α n ) = 1 − ∑ j = 1 n x j α + j L n − j ( j ) ( x ) ( j − 1 ) ! = 1 − x ∑ i = 1 n L n − i ( − α ) ( x ) L i − 1 ( α + 1 ) ( − x ) α + i . {\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}}.} 将拉盖尔多项式对自变量x求导k次，得到： d k d x k L n ( α ) ( x ) = ( − 1 ) k L n − k ( α + k ) ( x ) ; {\displaystyle {\frac {\mathrm {d} ^{k}}{\mathrm {d} x^{k}}}L_{n}^{(\alpha )}(x)=(-1)^{k}L_{n-k}^{(\alpha +k)}(x)\,;} 进一步有： 1 k ! d k d x k x α L n ( α ) ( x ) = ( n + α k ) x α − k L n ( α − 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),} 运用柯西多重积分公式（英语：Cauchy formula for repeated integration）可以得到： 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.} 将拉盖尔多项式对参变量 α {\displaystyle \alpha } 求导，得到下面的有意思的结果： 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}}.} 广义拉盖尔多项式满足下面的微分方程： 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,\,} 可以与拉盖尔多项式的k阶导数所满足的微分方程作一比较。 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,\,} 仅在此式中， L n ( k ) ( x ) ≡ d L n ( x ) d x k {\displaystyle L_{n}^{(k)}(x)\equiv {\frac {dL_{n}(x)}{dx^{k}}}} （后面这个符号又有了新的含义）。 于是，当 α = 0 {\displaystyle \alpha =0} 时，广义拉盖尔多项式可以用拉盖尔多项式的导数表示： 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}}}} 式中的上标(k)容易与求导k次混淆。 伴随拉盖尔多项式在区间[0, ∞)上以权函数xα 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},} 这可由下式得到： ∫ 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 ').} 伴随对称核多项式可以用拉盖尔多项式表示为： 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}}} 也有下面的递推关系： 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}}.} 进一步地，在伴L2[0, ∞)空间上，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 ),} 在氢原子的量子力学处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).} 设一个函数具有以下的级数展开形式： f ( x ) = ∑ i = 0 f i ( α ) L i ( α ) ( x ) . {\displaystyle f(x)=\sum _{i=0}f_{i}^{(\alpha )}L_{i}^{(\alpha )}(x).} 则展开式的系数由下式给出 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.} 这个级数在Lp空间 L 2 [ 0 , ∞ ) {\displaystyle L^{2}[0,\infty )} 上收敛，当且仅当 ‖ 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 .} 一个相关的展开式为： 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 )};} 特别地 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},} 这可由下式得到： 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).} 还有，当 Re ⁡ ( 2 α − β ) > − 1 {\displaystyle \operatorname {Re} {(2\alpha -\beta )}>-1} 时， 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 )},} 这个结果可以由下式导出， 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}}} 幂函数可以展开为： 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},} 二项式可以展开为： ( 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 ).} 进一步可以得到： 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)} （当且仅当 Re ⁡ ( γ ) > − 1 2 {\displaystyle \operatorname {Re} {(\gamma )}>-{\frac {1}{2}}} 时收敛） 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}.} 对于非负的整数 β {\displaystyle \beta } ，可以化简为： 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},} 当 γ = 0 {\displaystyle \gamma =0} 时，可以化简为： 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}},} 或 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 函数有下面的表示： ∑ 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}}};} 随意选定参量t，贝塞尔函数可以表示为： 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函数可以展开为： 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);} 低阶不完全伽玛函数可展开为： 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}.} 还有： γ ( s , z ) = γ s Γ ( 1 − s ) ∑ i = 0 L i + 1 ( − s ) ( 0 ) − L i + 1 ( − s ) ( z γ ) ( 1 + γ ) i + 1 ∑ n = 0 i γ i − n ( i n ) n + 1 − s ; {\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}};} 于是，高阶不完全伽玛函数就是： Γ ( s , z ) z s e − z = ∑ k = 0 L k ( α ) ( z ) ( k + 1 ) ( k + 1 + α − s k + 1 ) ( ℜ ( s − α 2 ) < 1 4 ) = ∑ k = 0 L k ( α ) ( z t ) ⋅ 2 F 1 ( 1 + α + k , 1 + k ; 2 + α + k − s ; t − 1 t ) t k ( k + 1 ) ( 1 + α + k − s 1 + k ) = t s ∑ k = 0 L k ( α ) ( z t ) ⋅ 2 F 1 ( 1 − s , 1 + α − s ; 2 + α + k − s ; t − 1 t ) ( k + 1 ) ( 1 + α + k − s 1 + k ) = t 1 + α ∑ k = 0 L k ( α ) ( z t ) ⋅ 2 F 1 ( 1 + α + k , 1 + α − s ; 2 + α + k − s ; 1 − t ) ( k + 1 ) ( 1 + α + k − s 1 + k ) , {\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}}} 2 F 1 {\displaystyle _{2}F_{1}} 表示超几何函数。 拉盖尔多项式可以用围道积分表示，如下式所示： L n ( α ) ( x ) = 1 2 π i ∮ e − x t 1 − t ( 1 − t ) α + 1 t n + 1 d t {\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} 积分方向逆时针绕原点一周。 广义拉盖尔多项式与埃爾米特多項式有下列关系： H 2 n ( x ) = ( − 1 ) n   2 2 n   n !   L n ( − 1 / 2 ) ( x 2 ) {\displaystyle H_{2n}(x)=(-1)^{n}\ 2^{2n}\ n!\ L_{n}^{(-1/2)}(x^{2})} 以及 H 2 n + 1 ( x ) = ( − 1 ) n   2 2 n + 1   n !   x   L n ( 1 / 2 ) ( x 2 ) {\displaystyle H_{2n+1}(x)=(-1)^{n}\ 2^{2n+1}\ n!\ x\ L_{n}^{(1/2)}(x^{2})} 这里的Hn表示乘上了exp(−x2)的埃爾米特多項式（所谓的“物理学家形式”）。 正因为这样，广义拉盖尔多项式也在量子谐振子的量子力学处理中出现。 拉盖尔多项式可以用超几何函数来定义，具体地说，是用合流超几何函数定义： L n ( α ) ( x ) = ( n + α n ) M ( − n , α + 1 , x ) = ( α + 1 ) n n ! 1 F 1 ( − n , α + 1 , x ) {\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)} ( a ) n {\displaystyle (a)_{n}} 是阶乘幂，这里表示升阶乘。 拉盖尔多项式与变形贝塞尔函数之间有以下关系： L n ( α ) ( x ) = e x 2 ( x 4 ) n + 1 2 2 π ( n + 1 ) ! ( − 1 2 n + 1 ) ⋅ ⋅ ∑ k = 0 n ( − 1 ) k + 1 ( 2 n + 1 n − k ) ( n + α n ) ( α + 2 n + 1 n − k ) ( n − k + α n − k ) ( k + 1 2 ) K k + 1 2 ( x 2 ) = e x 2 ( 4 x ) n + α + 1 2 Γ ( α + 1 2 ) ( − α − 1 n ) ( − α − 1 2 n ) ⋅ ⋅ n ! ∑ k = 0 n ( − 2 n − 1 − 2 α k − n ) ( − 2 n − 1 − α k − n ) ( − α − 1 k − n ) ( α + 1 2 + k ) I α + 1 2 + k ( x 2 ) , {\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}},} 进一步有： L n ( α ) ( x ) = 2 4 n ( 2 n + 1 ) ( − 1 2 n ) ∑ k = 0 n ( k + 1 2 ) ( 2 n + 1 n − k ) ( n k ) 2 ( n + α k ) ( 2 n + α + 1 n − k ) x n − k ( n − k ) ! L k − 2 k − 1 ( x ) . {\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).} 氢原子量子力学处理中的拉盖尔多项式的快速求导法（页面存档备份，存于互联网档案馆） Abramowitz, p. 506, 13.3.8 （页面存档备份，存于互联网档案馆） Abramowitz, Milton; Stegun, Irene A., eds. (1965), "Chapter 22 （页面存档备份，存于互联网档案馆）", 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 （页面存档备份，存于互联网档案馆）", 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.