拉盖尔多项式

拉盖尔多项式满足以下的递推关系：

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

特别地，有

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

还有

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

运用以上式子可以得到以下四条关系式：

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

将它们组合在一起，就得到了最常用的递推关系式：

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

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

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

进一步可以得到部分分式分解：

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

将拉盖尔多项式对自变量x求导k次，得到：

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

进一步有：

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

运用柯西多重积分公式可以得到：

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

将拉盖尔多项式对参变量${\displaystyle \alpha }$求导，得到下面的有意思的结果：

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

广义拉盖尔多项式满足下面的微分方程：

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

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。

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

仅在此式中，${\displaystyle L_{n}^{(k)}(x)\equiv {\frac {dL_{n}(x)}{dx^{k}}}}$（后面这个符号又有了新的含义）。

于是，当${\displaystyle \alpha =0}$时，广义拉盖尔多项式可以用拉盖尔多项式的导数表示： ${\displaystyle L_{n}^{(k)}(x)=(-1)^{k}{\frac {dL_{n+k}(x)}{dx^{k}}}}$ 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x^α e ^−x正交：

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

这可由下式得到：

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

伴随对称核多项式可以用拉盖尔多项式表示为：

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

也有下面的递推关系：

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

进一步地，在伴L²[0, ∞)空间上，有：

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

在氢原子的量子力学处理中用到了下面的公式：

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

设一个函数具有以下的级数展开形式：

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

则展开式的系数由下式给出

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

这个级数在Lp空间${\displaystyle L^{2}[0,\infty )}$上收敛，当且仅当

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

一个相关的展开式为：

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

特别地

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

这可由下式得到：

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

还有，当${\displaystyle \operatorname {Re} {(2\alpha -\beta )}>-1}$时，

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

这个结果可以由下式导出，

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

幂函数可以展开为：

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

二项式可以展开为：

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

进一步可以得到：

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

更一般地

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

对于非负的整数${\displaystyle \beta }$，可以化简为：

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

当${\displaystyle \gamma =0}$时，可以化简为：

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

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

雅可比Theta 函数有下面的表示：

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

随意选定参量t，贝塞尔函数可以表示为： ${\displaystyle {\frac {J_{\alpha }(x)}{\left({\frac {x}{2}}\right)^{\alpha }}}={\frac {e^{-t}}{\Gamma (\alpha +1)}}\sum _{i=0}{\frac {L_{i}^{(\alpha )}\left({\frac {x^{2}}{4t}}\right)}{i+\alpha \choose i}}{\frac {t^{i}}{i!}};}$ Γ函数可以展开为：

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

低阶不完全伽玛函数可展开为：

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

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

还有：

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

于是，高阶不完全伽玛函数就是：

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

${\displaystyle _{2}F_{1}}$表示超几何函数。