Polinomios de Laguerre

• Las funciones de Laguerre se definen mediante funciones hipergeométricas confluentes y la transformación de Kummer como ^{[ 3 ].}$${\displaystyle L_{n}^{(\alpha )}(x):={n+\alpha \choose n}M(-n,\alpha +1,x).}$$dónde${\textstyle {n+\alpha \choose n}}$es un coeficiente binomial generalizado . Cuando n es un entero, la función se reduce a un polinomio de grado n . Tiene la expresión alternativa ^{[ 4 ].}$${\displaystyle L_{n}^{(\alpha )}(x)={\frac {(-1)^{n}}{n!}}U(-n,\alpha +1,x)}$$en términos de la función de segundo tipo de Kummer .
• La forma cerrada para estos polinomios de Laguerre generalizados de grado n es ^{[ 5 ].}$${\displaystyle L_{n}^{(\alpha )}(x)=\sum _{i=0}^{n}(-1)^{i}{n+\alpha \choose n-i}{\frac {x^{i}}{i!}}}$$derivado al aplicar el teorema de Leibniz para la diferenciación de un producto a la fórmula de Rodrigues.
• Los polinomios de Laguerre tienen una representación mediante operadores diferenciales , muy similar a la de los polinomios de Hermite, estrechamente relacionados. Es decir, sea ${\displaystyle D={\frac {d}{dx}}}$y considere el operador diferencial${\displaystyle M=xD^{2}+(\alpha +1)D}$. Entonces${\displaystyle \exp(-tM)x^{n}=(-1)^{n}t^{n}n!L_{n}^{(\alpha )}\left({\frac {x}{t}}\right)}$.
• Los primeros polinomios de Laguerre generalizados son:

• The coefficient of the leading term is (−1)ⁿ/n!;
• The constant term, which is the value at 0, is $${\displaystyle L_{n}^{(\alpha )}(0)={n+\alpha \choose n}={\frac {\Gamma (n+\alpha +1)}{n!\,\Gamma (\alpha +1)}};}$$
• The discriminant is^[6]$${\displaystyle \operatorname {Disc} \left(L_{n}^{(\alpha )}\right)=\prod _{j=1}^{n}j^{j-2n+2}(j+\alpha )^{j-1}}$$

Given the generating function specified above, the polynomials may be expressed in terms of a contour integral$${\displaystyle L_{n}^{(\alpha )}(x)={\frac {1}{2\pi i}}\oint _{C}{\frac {e^{-xt/(1-t)}}{(1-t)^{\alpha +1}\,t^{n+1}}}\;dt,}$$ where the contour circles the origin once in a counterclockwise direction without enclosing the essential singularity at 1

The addition formula for Laguerre polynomials:^[7]$${\displaystyle L_{n}^{(\alpha _{1}+\dots +\alpha _{r}+r-1)}\left(x_{1}+\dots +x_{r}\right)=\sum _{m_{1}+\dots +m_{r}=n}L_{m_{1}}^{(\alpha _{1})}\left(x_{1}\right)\cdots L_{m_{r}}^{(\alpha _{r})}\left(x_{r}\right).}$$Laguerre's polynomials satisfy the recurrence relations $${\displaystyle L_{n}^{(\alpha )}(x)=\sum _{i=0}^{n}L_{n-i}^{(\alpha +i)}(y){\frac {(y-x)^{i}}{i!}},}$$ in particular $${\displaystyle L_{n}^{(\alpha +1)}(x)=\sum _{i=0}^{n}L_{i}^{(\alpha )}(x)}$$ and $${\displaystyle L_{n}^{(\alpha )}(x)=\sum _{i=0}^{n}{\alpha -\beta +n-i-1 \choose n-i}L_{i}^{(\beta )}(x),}$$ or $${\displaystyle L_{n}^{(\alpha )}(x)=\sum _{i=0}^{n}{\alpha -\beta +n \choose n-i}L_{i}^{(\beta -i)}(x);}$$ moreover $${\displaystyle {\begin{aligned}L_{n}^{(\alpha )}(x)-\sum _{j=0}^{\Delta -1}{n+\alpha \choose n-j}(-1)^{j}{\frac {x^{j}}{j!}}&=(-1)^{\Delta }{\frac {x^{\Delta }}{(\Delta -1)!}}\sum _{i=0}^{n-\Delta }{\frac {n+\alpha \choose n-\Delta -i}{(n-i){n \choose i}}}L_{i}^{(\alpha +\Delta )}(x)\\[6pt]&=(-1)^{\Delta }{\frac {x^{\Delta }}{(\Delta -1)!}}\sum _{i=0}^{n-\Delta }{\frac {n+\alpha -i-1 \choose n-\Delta -i}{(n-i){n \choose i}}}L_{i}^{(n+\alpha +\Delta -i)}(x)\end{aligned}}}$$

They can be used to derive the four 3-point-rules $${\displaystyle {\begin{aligned}L_{n}^{(\alpha )}(x)&=L_{n}^{(\alpha +1)}(x)-L_{n-1}^{(\alpha +1)}(x)=\sum _{j=0}^{k}{k \choose j}(-1)^{j}L_{n-j}^{(\alpha +k)}(x),\\[10pt]nL_{n}^{(\alpha )}(x)&=(n+\alpha )L_{n-1}^{(\alpha )}(x)-xL_{n-1}^{(\alpha +1)}(x),\\[10pt]&{\text{or }}\\{\frac {x^{k}}{k!}}L_{n}^{(\alpha )}(x)&=\sum _{i=0}^{k}(-1)^{i}{n+i \choose i}{n+\alpha \choose k-i}L_{n+i}^{(\alpha -k)}(x),\\[10pt]nL_{n}^{(\alpha +1)}(x)&=(n-x)L_{n-1}^{(\alpha +1)}(x)+(n+\alpha )L_{n-1}^{(\alpha )}(x)\\[10pt]xL_{n}^{(\alpha +1)}(x)&=(n+\alpha )L_{n-1}^{(\alpha )}(x)-(n-x)L_{n}^{(\alpha )}(x);\end{aligned}}}$$

combined they give this additional, useful recurrence relations$${\displaystyle {\begin{aligned}L_{n}^{(\alpha )}(x)&=\left(2+{\frac {\alpha -1-x}{n}}\right)L_{n-1}^{(\alpha )}(x)-\left(1+{\frac {\alpha -1}{n}}\right)L_{n-2}^{(\alpha )}(x)\\[10pt]&={\frac {\alpha +1-x}{n}}L_{n-1}^{(\alpha +1)}(x)-{\frac {x}{n}}L_{n-2}^{(\alpha +2)}(x)\end{aligned}}}$$

Since ${\displaystyle L_{n}^{(\alpha )}(x)}$ is a monic polynomial of degree ${\displaystyle n}$ in ${\displaystyle \alpha }$, there is the partial fraction decomposition$${\displaystyle {\begin{aligned}{\frac {n!\,L_{n}^{(\alpha )}(x)}{(\alpha +1)_{n}}}&=1-\sum _{j=1}^{n}(-1)^{j}{\frac {j}{\alpha +j}}{n \choose j}L_{n}^{(-j)}(x)\\&=1-\sum _{j=1}^{n}{\frac {x^{j}}{\alpha +j}}\,\,{\frac {L_{n-j}^{(j)}(x)}{(j-1)!}}\\&=1-x\sum _{i=1}^{n}{\frac {L_{n-i}^{(-\alpha )}(x)L_{i-1}^{(\alpha +1)}(-x)}{\alpha +i}}.\end{aligned}}}$$ The second equality follows by the following identity, valid for integer i and n and immediate from the expression of ${\displaystyle L_{n}^{(\alpha )}(x)}$ in terms of Charlier polynomials: $${\displaystyle {\frac {(-x)^{i}}{i!}}L_{n}^{(i-n)}(x)={\frac {(-x)^{n}}{n!}}L_{i}^{(n-i)}(x).}$$ For the third equality apply the fourth and fifth identities of this section.

Differentiating the power series representation of a generalized Laguerre polynomial k times leads to $${\displaystyle {\frac {d^{k}}{dx^{k}}}L_{n}^{(\alpha )}(x)={\begin{cases}(-1)^{k}L_{n-k}^{(\alpha +k)}(x)&{\text{if }}k\leq n,\\0&{\text{otherwise.}}\end{cases}}}$$

This points to a special case (α = 0) of the formula above: for integer α = k the generalized polynomial may be written $${\displaystyle L_{n}^{(k)}(x)=(-1)^{k}{\frac {d^{k}L_{n+k}(x)}{dx^{k}}},}$$ the shift by k sometimes causing confusion with the usual parenthesis notation for a derivative.

Moreover, the following equation holds: $${\displaystyle {\frac {1}{k!}}{\frac {d^{k}}{dx^{k}}}x^{\alpha }L_{n}^{(\alpha )}(x)={n+\alpha \choose k}x^{\alpha -k}L_{n}^{(\alpha -k)}(x),}$$ which generalizes with Cauchy's formula to $${\displaystyle L_{n}^{(\alpha ')}(x)=(\alpha '-\alpha ){\alpha '+n \choose \alpha '-\alpha }\int _{0}^{x}{\frac {t^{\alpha }(x-t)^{\alpha '-\alpha -1}}{x^{\alpha '}}}L_{n}^{(\alpha )}(t)\,dt.}$$

The derivative with respect to the second variable α has the form,^[8]$${\displaystyle {\frac {d}{d\alpha }}L_{n}^{(\alpha )}(x)=\sum _{i=0}^{n-1}{\frac {L_{i}^{(\alpha )}(x)}{n-i}}.}$$ The generalized Laguerre polynomials obey the differential equation $${\displaystyle xL_{n}^{(\alpha )\prime \prime }(x)+(\alpha +1-x)L_{n}^{(\alpha )\prime }(x)+nL_{n}^{(\alpha )}(x)=0,}$$ which may be compared with the equation obeyed by the kth derivative of the ordinary Laguerre polynomial,

$${\displaystyle xL_{n}^{[k]\prime \prime }(x)+(k+1-x)L_{n}^{[k]\prime }(x)+(n-k)L_{n}^{[k]}(x)=0,}$$ where ${\displaystyle L_{n}^{[k]}(x)\equiv {\frac {d^{k}L_{n}(x)}{dx^{k}}}}$ for this equation only.

In Sturm–Liouville form the differential equation is

$${\displaystyle -\left(x^{\alpha +1}e^{-x}\cdot L_{n}^{(\alpha )}(x)^{\prime }\right)'=n\cdot x^{\alpha }e^{-x}\cdot L_{n}^{(\alpha )}(x),}$$

which shows that L^(α)_n is an eigenvector for the eigenvalue n.

Los polinomios de Laguerre generalizados son ortogonales sobre [ 0, ∞) con respecto a la medida con función de ponderación x ^α e ^{− x} : ^{[ 9 ]}

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

lo cual se deduce de

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

Si${\displaystyle \Gamma (x,\alpha +1,1)}$Si denotamos la distribución gamma , entonces la relación de ortogonalidad se puede escribir como

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

El polinomio núcleo simétrico asociado tiene las representaciones ( fórmula de Christoffel-Darboux )

$${\displaystyle {\begin{aligned}K_{n}^{(\alpha )}(x,y)&:={\frac {1}{\Gamma (\alpha +1)}}\sum _{i=0}^{n}{\frac {L_{i}^{(\alpha )}(x)L_{i}^{(\alpha )}(y)}{\alpha +i \choose i}}\\[4pt]&={\frac {1}{\Gamma (\alpha +1)}}{\frac {L_{n}^{(\alpha )}(x)L_{n+1}^{(\alpha )}(y)-L_{n+1}^{(\alpha )}(x)L_{n}^{(\alpha )}(y)}{{\frac {x-y}{n+1}}{n+\alpha \choose n}}}\\[4pt]&={\frac {1}{\Gamma (\alpha +1)}}\sum _{i=0}^{n}{\frac {x^{i}}{i!}}{\frac {L_{n-i}^{(\alpha +i)}(x)L_{n-i}^{(\alpha +i+1)}(y)}{{\alpha +n \choose n}{n \choose i}}};\end{aligned}}}$$

recursivamente

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

Además,

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

Las desigualdades de Turán se pueden derivar aquí, lo cual es $${\displaystyle L_{n}^{(\alpha )}(x)^{2}-L_{n-1}^{(\alpha )}(x)L_{n+1}^{(\alpha )}(x)=\sum _{k=0}^{n-1}{\frac {\alpha +n-1 \choose n-k}{n{n \choose k}}}L_{k}^{(\alpha -1)}(x)^{2}>0.}$$

La siguiente integral es necesaria en el tratamiento mecánico cuántico del átomo de hidrógeno ,

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

Sea una función con la siguiente expansión en serie (formal): $${\displaystyle f(x)=\sum _{i=0}^{\infty }f_{i}^{(\alpha )}L_{i}^{(\alpha )}(x).}$$

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

La serie converge en el espacio de Hilbert asociado L ² [0, ∞) si y solo si

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

${\displaystyle j_{\alpha ,m}}$es el${\displaystyle m}$-ésimo cero positivo de la función de Bessel${\displaystyle J_{\alpha }(x)}$.

${\displaystyle a_{m}}$es el${\displaystyle m}$-ésimo cero de la función de Airy${\displaystyle \operatorname {Ai} (x)}$, en orden descendente:${\displaystyle 0>a_{1}>a_{2}>\cdots }$.

${\displaystyle \nu =4n+2\alpha +2}$.

Si${\displaystyle \alpha >-1}$, entonces${\displaystyle L_{n}^{(\alpha )}}$tiene${\displaystyle n}$raíces reales. Por lo tanto, en esta sección asumimos${\displaystyle \alpha >-1}$por defecto.

${\displaystyle x_{1}<\dots <x_{n}}$son las verdaderas raíces de${\displaystyle L_{n}^{(\alpha )}}$.

Tenga en cuenta que${\displaystyle \left((-1)^{n-i}L_{n-i}^{(\alpha )}\right)_{i=0}^{n}}$es una cadena de Sturm .

Para${\displaystyle \alpha >-1}$, tenemos estos límites: ^{[ 16 ] [ 17 ] [ 6 ] [ 18 ]}

• ${\displaystyle x_{1}<{\frac {(\alpha +1)(\alpha +2)}{n+\alpha +1}}}$
• ${\displaystyle x_{1}<{\frac {(\alpha +1)(\alpha +3)}{2n+\alpha +1}}}$
• ${\displaystyle x_{1}<{\frac {(\alpha +1)(\alpha +2)(\alpha +4)(2n+\alpha +1)}{(\alpha +1)^{2}(\alpha +2)+n(5\alpha +11)(n+\alpha +1)}}}$
• ${\displaystyle x_{n}\leq 2n+\alpha -1+2{\sqrt {(n-2)(n+\alpha -1)}}}$cuando${\displaystyle n\geq 2}$
• ${\displaystyle x_{n}>4n+\alpha -16{\sqrt {2n}}}$
• ${\displaystyle x_{n}>3n-4}$
• ${\displaystyle x_{n}>2n+\alpha -1}$
• ${\displaystyle x_{n}>2n+\alpha -2+{\sqrt {n^{2}-2n+\alpha n+2}}}$
• $${\displaystyle {\begin{aligned}&(n+2)x_{1}&\geq \left(n-1-{\sqrt {n^{2}+(n+2)(\alpha +1)}}\right)^{2}-1\\&(n+2)x_{n}&\leq \left(n-1+{\sqrt {n^{2}+(n+2)(\alpha +1)}}\right)^{2}-1\end{aligned}}}$$
• $${\displaystyle {\begin{aligned}x_{1}&>{\frac {1}{2}}\nu -3-{\sqrt {1+4(n-1)(n+\alpha -1)}}\\x_{n}&<{\frac {1}{2}}\nu -3+{\sqrt {1+4(n-1)(n+\alpha -1)}}\end{aligned}}}$$

Para fijo${\displaystyle k=1,\dots ,n}$, ^{[ 16 ] [ 6 ] [ 17 ]}$${\displaystyle {\begin{aligned}\nu x_{k}&>j_{\alpha ,k}^{2}\\x_{k}&<{\frac {j_{\alpha ,k}^{2}}{\nu /2+{\sqrt {(\nu /2)^{2}-j_{\alpha ,k}^{2}}}}}\quad {\text{ if }}\nu /2>j_{\alpha ,k}\\x_{k}&<\left[\nu ^{1/2}+2^{-1/3}\nu ^{-1/6}a_{n-k+1}\right]^{2}\quad {\text{ if }}|\alpha |\geqslant 1/4\\x_{k}&<\nu +2^{\frac {2}{3}}a_{k}\nu ^{\frac {1}{3}}+2^{-{\frac {2}{3}}}a_{k}^{2}\nu ^{-{\frac {1}{3}}}\end{aligned}}}$$Para fijo${\displaystyle k}$, tenemos${\displaystyle \lim _{n\to \infty }\nu x_{k}=j_{\alpha ,k}^{2}}$, por lo tanto, la primera desigualdad es precisa.

Véase también. ^{[ 19 ]}

Los ceros satisfacen las relaciones de Stieltjes : ^{[ 20 ] [ 21 ]}$${\displaystyle {\begin{aligned}\sum _{1\leq j\leq n,i\neq j}{\frac {1}{x_{i}-x_{j}}}&={\frac {1}{2}}\left(1-{\frac {\alpha +1}{x_{i}}}\right)\\\sum _{1\leq j\leq n}{\frac {1}{x_{j}}}&={\frac {n}{\alpha +1}}\\\sum _{1\leq j\leq n,i\neq j}{\frac {1}{(x_{i}-x_{j})^{2}}}&=-{\frac {(\alpha +1)(\alpha +5)}{12x_{i}^{2}}}+{\frac {2n+\alpha +1}{6x_{i}}}-{\frac {1}{12}}\\\sum _{1\leq j\leq n,i\neq j}{\frac {1}{(x_{i}-x_{j})^{3}}}&=-{\frac {(\alpha +1)(\alpha +3)}{8x_{i}^{3}}}+{\frac {2n+\alpha +1}{8x_{i}^{2}}}\\\end{aligned}}}$$La primera relación puede interpretarse físicamente. Fijemos una partícula eléctrica en el origen con carga${\displaystyle +{\frac {\alpha +1}{2}}}$y producir un campo eléctrico constante de intensidad${\displaystyle -{\frac {1}{2}}}$Luego, coloque${\displaystyle n}$partículas eléctricas con carga${\displaystyle +1}$. La primera relación establece que los ceros de${\displaystyle L_{n}^{(\alpha )}}$son las posiciones de equilibrio de las partículas.

Dado que los ceros especifican el polinomio salvo por un factor de escala, esto proporciona una forma alternativa de caracterizar de manera única los polinomios de Laguerre.

Los ceros también satisfacen ^{[ 22 ]}$${\displaystyle \sum _{i=1}^{n}{\frac {1}{x-x_{i}}}=-\sum _{k=0}^{\infty }S_{k+1}x^{k},\quad S_{k}:=\sum _{i=1}^{n}x_{i}^{-k}}$$lo que permite el siguiente límite$${\displaystyle S_{m}^{-1/m}<x_{1}<S_{m}/S_{m+1},\quad m=1,2,\ldots }$$

Dejar${\displaystyle F_{n}(t):={\frac {1}{n}}\#\{i:x_{i}\leq t\}}$Sea la función de distribución acumulativa para las raíces, entonces tenemos la ley límite ^{[ 23 ].}$${\displaystyle \lim _{n\to \infty }F_{n}(4nt)={\frac {2}{\pi }}\int _{0}^{t}{\sqrt {\frac {1-s}{s}}}ds\quad \forall t\in (0,1]}$$lo cual puede interpretarse como la distribución límite del espectro del conjunto de Wishart .

Para fijo${\displaystyle \alpha >-1}$y fijo${\displaystyle k}$, como${\displaystyle n\to \infty }$, ^{[ 17 ]}$${\displaystyle {\begin{aligned}x_{n+1-k}=&\nu +2^{2/3}a_{k}\nu ^{1/3}+{\frac {1}{5}}2^{4/3}a_{k}^{2}\nu ^{-1/3}+\left({\frac {11}{35}}-\alpha ^{2}-{\frac {12}{175}}a_{k}^{3}\right)\nu ^{-1}\\&+\left({\frac {16}{1575}}a_{k}+{\frac {92}{7875}}a_{k}^{4}\right)2^{2/3}\nu ^{-5/3}-\left({\frac {15152}{3031875}}a_{k}^{5}+{\frac {1088}{121275}}a_{k}^{2}\right)2^{1/3}\nu ^{-7/3}+{\mathcal {O}}\left(\nu ^{-3}\right),\end{aligned}}}$$

Para${\displaystyle \alpha \in (-1,0)}$, ^{[ 22 ]}$${\displaystyle {\begin{aligned}x_{1}={\frac {\alpha +1}{n}}&+{\frac {n-1}{2}}\left({\frac {\alpha +1}{n}}\right)^{2}-{\frac {n^{2}+3n-4}{12}}\left({\frac {\alpha +1}{n}}\right)^{3}\\&+{\frac {7n^{3}+6n^{2}+23n-36}{144}}\left({\frac {\alpha +1}{n}}\right)^{4}\\&-{\frac {293n^{4}+210n^{3}+235n^{2}+990n-1728}{8640}}\left({\frac {\alpha +1}{n}}\right)^{5}+\cdots \end{aligned}}}$$