Laguerre polynomials

• Laguerre functions are defined by confluent hypergeometric functions and Kummer's transformation as[3]

${\displaystyle L_{n}^{(\alpha )}(x):={n+\alpha \choose n}M(-n,\alpha +1,x).}$

${\displaystyle {n+\alpha \choose n}}$ is a generalized binomial coefficient. When n is an integer the function reduces to a polynomial of degree n. It has the alternative expression[4]

${\displaystyle L_{n}^{(\alpha )}(x)={\frac {(-1)^{n}}{n!}}U(-n,\alpha +1,x)}$

in terms of Kummer's function of the second kind.

• The closed form for these generalized Laguerre polynomials of degree n is[5]

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

derived by applying Leibniz's theorem for differentiation of a product to Rodrigues' formula.

• The first few generalized Laguerre polynomials are:

${\displaystyle {\begin{aligned}L_{0}^{(\alpha )}(x)&=1\\L_{1}^{(\alpha )}(x)&=-x+\alpha +1\\L_{2}^{(\alpha )}(x)&={\frac {x^{2}}{2}}-(\alpha +2)x+{\frac {(\alpha +2)(\alpha +1)}{2}}\\L_{3}^{(\alpha )}(x)&={\frac {-x^{3}}{6}}+{\frac {(\alpha +3)x^{2}}{2}}-{\frac {(\alpha +2)(\alpha +3)x}{2}}+{\frac {(\alpha +1)(\alpha +2)(\alpha +3)}{6}}\end{aligned}}}$

• The coefficient of the leading term is (−1)ⁿ/n!;
• The constant term, which is the value at 0, is

${\displaystyle L_{n}^{(\alpha )}(0)={n+\alpha \choose n}={\frac {n^{\alpha }}{\Gamma (\alpha +1)}}+O\left(n^{\alpha -1}\right);}$

• If α is non-negative, then L_n^(α) has n real, strictly positive roots (notice that ${\displaystyle \left((-1)^{n-i}L_{n-i}^{(\alpha )}\right)_{i=0}^{n}}$ is a Sturm chain), which are all in the interval ${\displaystyle \left(0,n+\alpha +(n-1){\sqrt {n+\alpha }}\,\right].}$
• The polynomials' asymptotic behaviour for large n, but fixed α and x > 0, is given by[6][7]

${\displaystyle {\begin{aligned}&L_{n}^{(\alpha )}(x)={\frac {n^{{\frac {\alpha }{2}}-{\frac {1}{4}}}}{\sqrt {\pi }}}{\frac {e^{\frac {x}{2}}}{x^{{\frac {\alpha }{2}}+{\frac {1}{4}}}}}\sin \left(2{\sqrt {nx}}-{\frac {\pi }{2}}\left(\alpha -{\frac {1}{2}}\right)\right)+O\left(n^{{\frac {\alpha }{2}}-{\frac {3}{4}}}\right),\\[6pt]&L_{n}^{(\alpha )}(-x)={\frac {(n+1)^{{\frac {\alpha }{2}}-{\frac {1}{4}}}}{2{\sqrt {\pi }}}}{\frac {e^{-x/2}}{x^{{\frac {\alpha }{2}}+{\frac {1}{4}}}}}e^{2{\sqrt {x(n+1)}}}\cdot \left(1+O\left({\frac {1}{\sqrt {n+1}}}\right)\right),\end{aligned}}}$

and summarizing by

${\displaystyle {\frac {L_{n}^{(\alpha )}\left({\frac {x}{n}}\right)}{n^{\alpha }}}\approx e^{x/2n}\cdot {\frac {J_{\alpha }\left(2{\sqrt {x}}\right)}{{\sqrt {x}}^{\alpha }}},}$

where ${\displaystyle J_{\alpha }}$ is the Bessel function.

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:[8]

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

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

${\displaystyle {\begin{aligned}L_{n}^{(\alpha )}(x)&=L_{n}^{(\alpha +1)}(x)-L_{n-1}^{(\alpha +1)}(x)=\sum _{j=0}^{k}{k \choose j}L_{n-j}^{(\alpha +k)}(x),\\[10pt]nL_{n}^{(\alpha )}(x)&=(n+\alpha )L_{n-1}^{(\alpha )}(x)-xL_{n-1}^{(\alpha +1)}(x),\\[10pt]&{\text{or }}\\{\frac {x^{k}}{k!}}L_{n}^{(\alpha )}(x)&=\sum _{i=0}^{k}(-1)^{i}{n+i \choose i}{n+\alpha \choose k-i}L_{n+i}^{(\alpha -k)}(x),\\[10pt]nL_{n}^{(\alpha +1)}(x)&=(n-x)L_{n-1}^{(\alpha +1)}(x)+(n+\alpha )L_{n-1}^{(\alpha )}(x)\\[10pt]xL_{n}^{(\alpha +1)}(x)&=(n+\alpha )L_{n-1}^{(\alpha )}(x)-(n-x)L_{n}^{(\alpha )}(x);\end{aligned}}}$

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,[9]

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

This is evident from the contour integral representation below.

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

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

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

The generalized Laguerre polynomials are orthogonal over [0, ∞) with respect to the measure with weighting function x^α e^−x:[10]

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

which follows from

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

If ${\displaystyle \Gamma (x,\alpha +1,1)}$ denotes the Gamma distribution then the orthogonality relation can be written as

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

The associated, symmetric kernel polynomial has the representations (Christoffel–Darboux formula)

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

recursively

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

Moreover,

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

Turán's inequalities can be derived here, which is

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

The following integral is needed in the quantum mechanical treatment of the hydrogen atom,

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

Let a function have the (formal) series expansion

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

Then

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

The series converges in the associated Hilbert space L²[0, ∞) if and only if

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

Monomials are represented as

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

while binomials have the parametrization

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

This leads directly to

${\displaystyle e^{-\gamma x}=\sum _{i=0}^{\infty }{\frac {\gamma ^{i}}{(1+\gamma )^{i+\alpha +1}}}L_{i}^{(\alpha )}(x)\qquad {\text{convergent iff }}\Re (\gamma )>-{\tfrac {1}{2}}}$

for the exponential function. The incomplete gamma function has the representation

${\displaystyle \Gamma (\alpha ,x)=x^{\alpha }e^{-x}\sum _{i=0}^{\infty }{\frac {L_{i}^{(\alpha )}(x)}{1+i}}\qquad \left(\Re (\alpha )>-1,x>0\right).}$