Laguerre polynomials

Laguerre polynomials (named after Edmond Laguerre ) are special polynomials that form an orthogonal system of functions on the interval . They are the solutions of Laguera's differential equation . The Laguerre polynomials play an important role in theoretical physics , especially in quantum mechanics . ${\displaystyle [0,\infty ]}$

Differential equation and polynomials

Laguerre's differential equation

The Laguerean differential equation

${\displaystyle x\,y''(x)+(1-x)\,y'(x)+n\,y(x)=0}$,

is an ordinary second order linear differential equation for and ${\displaystyle x>0}$ ${\displaystyle n=0,1,2,\ldots }$

It is a special case of the Sturm-Liouville differential equation

${\displaystyle -\mathrm {e} ^{x}{\frac {\mathrm {d} }{\mathrm {d} x}}\left(x{\mathrm {e} }^{-x}{\frac {\mathrm {d} y}{\mathrm {d} x}}\right)=ny}$

First polynomials

The first five Laguerre polynomials

The first five Laguerre polynomials are

${\displaystyle {\begin{aligned}L_{0}(x)&=1\\L_{1}(x)&=-x+1\\L_{2}(x)&={\tfrac {1}{2}}(x^{2}-4x+2)\\L_{3}(x)&={\tfrac {1}{6}}(-x^{3}+9x^{2}-18x+6)\\L_{4}(x)&={\tfrac {1}{24}}(x^{4}-16x^{3}+72x^{2}-96x+24)\end{aligned}}}$

In physics, a definition is usually used according to which the Laguerre polynomials are larger by a factor . ${\displaystyle n!}$

properties

Recursion formulas

The Laguerre polynomial can be combined with the first two polynomials ${\displaystyle L_{n+1}(x)}$

${\displaystyle L_{0}(x)=1}$

${\displaystyle L_{1}(x)=1-x}$

Calculate using the following recursion formula

${\displaystyle (n+1)L_{n+1}(x)={\big (}(2n+1-x)L_{n}(x)-nL_{n-1}(x){\big )}.}$

The following recursion formulas also apply:

${\displaystyle L_{n}'(x)=L_{n-1}'(x)-L_{n-1}(x)}$,

${\displaystyle (x-n-1)L_{n}'(x)=-(n+1)L_{n+1}'(x)-(2n+2-x)L_{n}(x)+(n+1)L_{n+1}(x)}$,

${\displaystyle xL_{n}'(x)=nL_{n}(x)-nL_{n-1}(x)}$.

An explicit formula for the Laguerre polynomials is

${\displaystyle L_{n}(x)=\sum _{k=0}^{n}{\binom {n}{k}}{\frac {(-1)^{k}}{k!}}x^{k}}$.

example

The polynomial for is calculated. So ${\displaystyle L_{3}(x)}$ ${\displaystyle n=2}$

${\displaystyle L_{3}(x)={\tfrac {1}{3}}{\big (}(4+1-x)L_{2}(x)-2L_{1}(x){\big )}}$.

In order to obtain this polynomial it is necessary to determine the polynomial for . It turns out ${\displaystyle L_{2}(x)}$ ${\displaystyle n=1}$

${\displaystyle L_{2}(x)={\tfrac {1}{2}}{\big (}(2+1-x)L_{1}(x)-1L_{0}(x){\big )}={\tfrac {1}{2}}{\big (}(3-x)(1-x)-1{\big )}={\tfrac {1}{2}}(3-4x+x^{2}-1)={\tfrac {1}{2}}{\big (}2-4x+x^{2}{\big )}}$

So the polynomial is ${\displaystyle L_{3}(x)}$

${\displaystyle {\begin{aligned}L_{3}(x)&={\tfrac {1}{3}}{\big (}(4+1-x){\tfrac {1}{2}}(2-4x+x^{2})-2(1-x){\big )}={\tfrac {1}{6}}{\big (}(5-x)(2-4x+x^{2})-4+4x{\big )}\\&={\tfrac {1}{6}}(10-20x+5x^{2}-2x+4x^{2}-x^{3}-4+4x)={\tfrac {1}{6}}(6-18x+9x^{2}-x^{3}).\end{aligned}}}$

Rodrigues formula

The -th Laguerre polynomial can be represented with the Rodrigues formula as follows ${\displaystyle n}$

${\displaystyle L_{n}(x)={\frac {\mathrm {e} ^{x}}{n!}}{\frac {\mathrm {d} ^{n}}{\mathrm {d} x^{n}}}{\bigg (}x^{n}\mathrm {e} ^{-x}{\bigg )}}$

and

${\displaystyle L_{n}(x)={\frac {1}{n!}}{\bigg (}{\frac {\mathrm {d} }{\mathrm {d} x}}-1{\bigg )}^{n}x^{n}.}$

The Laguerre polynomial is calculated from the first equation with the product rule for higher derivatives and the identities , as well as according to ${\displaystyle \textstyle {\frac {\mathrm {d} ^{n}}{\mathrm {d} x^{n}}}{\big (}\mathrm {e} ^{-x}x^{n}{\big )}={\big (}\mathrm {e} ^{-x}x^{n}{\big )}^{(n)}}$ ${\displaystyle \left(\mathrm {e} ^{-x}\right)^{(k)}=(-1)^{k}\mathrm {e} ^{-x}}$ ${\displaystyle {\big (}x^{n}{\big )}^{(n-k)}={\tfrac {n!}{k!}}x^{k}}$

${\displaystyle {\begin{aligned}L_{n}(x)&={\frac {\mathrm {e} ^{x}}{n!}}{\frac {\mathrm {d} ^{n}}{\mathrm {d} x^{n}}}{\bigg (}\mathrm {e} ^{-x}x^{n}{\bigg )}={\frac {\mathrm {e} ^{x}}{n!}}{\bigg (}\mathrm {e} ^{-x}x^{n}{\bigg )}^{(n)}={\frac {\mathrm {e} ^{x}}{n!}}\sum _{k=0}^{n}{\binom {n}{k}}{\big (}\mathrm {e} ^{-x}{\big )}^{(k)}{\big (}x^{n}{\big )}^{(n-k)}\\\\&={\frac {\mathrm {e} ^{x}}{n!}}\sum _{k=0}^{n}{\binom {n}{k}}(-1)^{k}\mathrm {e} ^{-x}{\frac {n!}{k!}}x^{k}=\sum _{k=0}^{n}{\binom {n}{k}}{\frac {(-1)^{k}}{k!}}x^{k}.\end{aligned}}}$

The Laguerre polynomial with the binomial theorem and identity results from the second equation as follows ${\displaystyle {\big (}{\tfrac {\mathrm {d} }{\mathrm {d} x}}{\big )}^{(n-k)}x^{n}={\big (}x^{n}{\big )}^{(n-k)}={\tfrac {n!}{k!}}x^{k}}$

${\displaystyle {\begin{aligned}L_{n}(x)&={\frac {1}{n!}}{\bigg (}{\frac {\mathrm {d} }{\mathrm {d} x}}-1{\bigg )}^{n}x^{n}={\frac {1}{n!}}{\bigg (}-1+{\frac {\mathrm {d} }{\mathrm {d} x}}{\bigg )}^{n}x^{n}={\frac {1}{n!}}\sum _{k=0}^{n}{\binom {n}{k}}(-1)^{k}{\bigg (}{\frac {\mathrm {d} }{\mathrm {d} x}}{\bigg )}^{(n-k)}x^{n}\\\\&={\frac {1}{n!}}\sum _{k=0}^{n}{\binom {n}{k}}(-1)^{k}{\big (}x^{n}{\big )}^{(n-k)}={\frac {1}{n!}}\sum _{k=0}^{n}{\binom {n}{k}}(-1)^{k}{\frac {n!}{k!}}x^{k}=\sum _{k=0}^{n}{\binom {n}{k}}{\frac {(-1)^{k}}{k!}}x^{k}.\end{aligned}}}$

Orthogonal polynomials

Since the Laguerre polynomials are for and / or divergent, they do not form a Prehilbert space or a Hilbert space . Therefore a weight function is introduced which leaves the solution of the differential equation unchanged and which ensures that the Laguerre polynomials can be square-integrated . Under these conditions, the eigenfunctions form an orthonormal basis in the Hilbert space of square-integrable functions with the weight function . Hence applies ${\displaystyle n\to \infty }$ ${\displaystyle x\to \infty }$ ${\displaystyle L_{n}}$ ${\displaystyle L^{2}([0,\infty ],w(x)\mathrm {d} x)}$ ${\displaystyle w(x)=\mathrm {e} ^{-x}}$

${\displaystyle \langle L_{n},L_{m}\rangle =\int _{0}^{\infty }\mathrm {e} ^{-x}L_{n}(x)L_{m}(x)\mathrm {d} x=\delta _{nm}.}$

Here the Kronecker Delta means . ${\displaystyle \delta _{nm}}$

proof

Part 1: First of all, it is shown that the Laguerre polynomials with the weight are orthogonal, so for applies ${\displaystyle w(x)=\mathrm {e} ^{-x}}$ ${\displaystyle n\neq m}$ ${\displaystyle \langle L_{n},L_{m}\rangle =\int _{0}^{\infty }\mathrm {e} ^{-x}L_{n}(x)L_{m}(x)\mathrm {d} x=0.}$

With the Sturm-Liouville operator the following equations result for the Laguerre polynomials : ${\displaystyle \textstyle {\mathcal {L}}=-\mathrm {e} ^{x}{\frac {\mathrm {d} }{\mathrm {d} x}}\left(x{\mathrm {e} }^{-x}{\frac {\mathrm {d} }{\mathrm {d} x}}\right)}$ ${\displaystyle L_{n},L_{m}}$

(1) ${\displaystyle \quad {\mathcal {L}}L_{n}=-\mathrm {e} ^{x}{\frac {\mathrm {d} }{\mathrm {d} x}}\left(x{\mathrm {e} }^{-x}{\frac {\mathrm {d} L_{n}}{\mathrm {d} x}}\right)=nL_{n}}$

and

(2) . ${\displaystyle \quad {\mathcal {L}}L_{m}=-\mathrm {e} ^{x}{\frac {\mathrm {d} }{\mathrm {d} x}}\left(x{\mathrm {e} }^{-x}{\frac {\mathrm {d} L_{m}}{\mathrm {d} x}}\right)=mL_{m}}$

Equation (1) from the left with multiplied by equation (2) , which is also from the left and is multiplied, subtracted, so results in the two equations: ${\displaystyle L_{m}}$ ${\displaystyle L_{n}}$

(3) ${\displaystyle \quad L_{n}{\mathcal {L}}L_{m}-L_{m}{\mathcal {L}}L_{n}=-L_{n}\mathrm {e} ^{x}{\frac {\mathrm {d} }{\mathrm {d} x}}\left(x{\mathrm {e} }^{-x}{\frac {\mathrm {d} L_{m}}{\mathrm {d} x}}\right)+L_{m}\mathrm {e} ^{x}{\frac {\mathrm {d} }{\mathrm {d} x}}\left(x{\mathrm {e} }^{-x}{\frac {\mathrm {d} L_{n}}{\mathrm {d} x}}\right)}$

and

(4) . ${\displaystyle \quad L_{n}{\mathcal {L}}L_{m}-L_{m}{\mathcal {L}}L_{n}=(m-n)L_{m}L_{n}}$

First, equation (3) is summarized. With the product rule for derivatives , the term is not taken into account, the following presentations result ${\displaystyle \textstyle -\mathrm {e} ^{x}}$

${\displaystyle \textstyle L_{n}{\frac {\mathrm {d} }{\mathrm {d} x}}\left(x{\mathrm {e} }^{-x}{\frac {\mathrm {d} L_{m}}{\mathrm {d} x}}\right)={\frac {\mathrm {d} }{\mathrm {d} x}}\left(x{\mathrm {e} }^{-x}L_{n}{\frac {\mathrm {d} L_{m}}{\mathrm {d} x}}\right)-\left(x{\mathrm {e} }^{-x}{\frac {\mathrm {d} L_{m}}{\mathrm {d} x}}\right){\frac {\mathrm {d} L_{n}}{\mathrm {d} x}}}$

and

${\displaystyle \textstyle L_{m}{\frac {\mathrm {d} }{\mathrm {d} x}}\left(x{\mathrm {e} }^{-x}{\frac {\mathrm {d} L_{n}}{\mathrm {d} x}}\right)={\frac {\mathrm {d} }{\mathrm {d} x}}\left(x{\mathrm {e} }^{-x}L_{m}{\frac {\mathrm {d} L_{n}}{\mathrm {d} x}}\right)-\left(x{\mathrm {e} }^{-x}{\frac {\mathrm {d} L_{n}}{\mathrm {d} x}}\right){\frac {\mathrm {d} L_{m}}{\mathrm {d} x}}}$.

In this way it can be seen that the second term is the same in both derivatives and disappears when the difference is formed, i.e.:

(5) ${\displaystyle {\begin{aligned}\quad L_{n}{\mathcal {L}}L_{m}-L_{m}{\mathcal {L}}L_{n}&=-\mathrm {e} ^{x}{\frac {\mathrm {d} }{\mathrm {d} x}}\left(x{\mathrm {e} }^{-x}L_{n}{\frac {\mathrm {d} L_{m}}{\mathrm {d} x}}\right)+\mathrm {e} ^{x}{\frac {\mathrm {d} }{\mathrm {d} x}}\left(x{\mathrm {e} }^{-x}L_{m}{\frac {\mathrm {d} L_{n}}{\mathrm {d} x}}\right)\\\\&=-\mathrm {e} ^{x}{\frac {\mathrm {d} }{\mathrm {d} x}}{\bigg (}x{\mathrm {e} }^{-x}\left(L_{n}{\frac {\mathrm {d} L_{m}}{\mathrm {d} x}}-L_{m}{\frac {\mathrm {d} L_{n}}{\mathrm {d} x}}\right){\bigg )}\\\\&=-\mathrm {e} ^{x}{\frac {\mathrm {d} }{\mathrm {d} x}}{\bigg (}x{\mathrm {e} }^{-x}W(L_{n},L_{m}){\bigg )},\\\end{aligned}}}$

where means the Wronsky determinant of the functions . ${\displaystyle W(L_{n},L_{m})=\left|{\begin{smallmatrix}L_{n}&L_{m}\\L_{n}'&L_{m}'\end{smallmatrix}}\right|}$ ${\displaystyle L_{n},L_{m}}$

To calculate the Wronski determinant by means of the Abelian identity is the differential equation or considered, so that a removable singularity in arises. The coefficient matrix of the fundamental system is then and its trace is . Thus the Abelian identity is: ${\displaystyle \textstyle {\mathcal {L}}y=-\mathrm {e} ^{x}{\frac {\mathrm {d} }{\mathrm {d} x}}\left(x\mathrm {e} ^{-x}{\frac {\mathrm {d} }{\mathrm {d} x}}\right)y=-xy''-\mathrm {e} ^{x}{\big (}x\mathrm {e} ^{-x}{\big )}'y'=-xy''-{\big (}1-x{\big )}y'=0}$ ${\displaystyle \textstyle y''+{\frac {1-x}{x}}y'=0}$ ${\displaystyle x=0}$ ${\displaystyle \left({\begin{smallmatrix}0&1\\0&-{\tfrac {1-x}{x}}\end{smallmatrix}}\right)}$ ${\displaystyle \mathrm {Spur} {\Bigg (}\left({\begin{smallmatrix}0&1\\0&-{\tfrac {1-x}{x}}\end{smallmatrix}}\right){\Bigg )}=-{\frac {1-x}{x}}}$

${\displaystyle W(L_{n},L_{m})(x)=W(L_{n},L_{m})(0)\exp \left(\int _{0}^{x}{\bigg (}1-{\frac {1}{\xi }}{\bigg )}\mathrm {d} \xi \right)}$.

Since and are linearly independent, it is - on closer inspection - and the following result is obtained: ${\displaystyle L_{n}}$ ${\displaystyle L_{m}}$ ${\displaystyle W(L_{n},L_{m})(0)>0}$ ${\displaystyle W(L_{n},L_{m})(0)=1}$

${\displaystyle {\begin{aligned}W(L_{n},L_{m})(x)&=W(L_{n},L_{m})(0)\exp \left(\int _{0}^{x}{\bigg (}1-{\frac {1}{\xi }}{\bigg )}\mathrm {d} \xi \right)=W(L_{n},L_{m})(0)\exp {\Bigg (}{\bigg [}\xi -\ln \xi {\bigg ]}_{0}^{x}{\Bigg )}\\&=\lim _{\xi \to x}W(L_{n},L_{m})(0)\exp {\Big (}\xi -\ln \xi {\Big )}-\lim _{\xi \to 0}W(L_{n},L_{m})(0)\exp {\Big (}\xi -\ln \xi {\Big )}\\&=\lim _{\xi \to x}W(L_{n},L_{m})(0){\frac {\exp(\xi )}{\exp(\ln \xi )}}-\lim _{\xi \to 0}W(L_{n},L_{m})(0){\frac {\exp(\xi )}{\exp(\ln \xi )}}\\&=\lim _{\xi \to x}W(L_{n},L_{m})(0){\frac {\mathrm {e} ^{\xi }}{\xi }}+\lim _{\xi \to 0}W(L_{n},L_{m})(0){\frac {\mathrm {e} ^{\xi }}{\xi }}\\&=W(L_{n},L_{m})(0){\frac {\mathrm {e} ^{x}}{x}}+\lim _{\xi \to 0}W(L_{n},L_{m})(0){\frac {\mathrm {e} ^{\xi }}{\xi }}+C.\end{aligned}}}$

The constant of integration is chosen and equation (5) is multiplied by such that it follows: ${\displaystyle C=-\lim _{\xi \to 0}W(L_{n},L_{m})(0){\frac {\mathrm {e} ^{\xi }}{\xi }}}$ ${\displaystyle \mathrm {e} ^{-x}}$

${\displaystyle {\begin{aligned}\mathrm {e} ^{-x}{\big (}L_{n}{\mathcal {L}}L_{m}-L_{m}{\mathcal {L}}L_{n}{\big )}&=-{\frac {\mathrm {d} }{\mathrm {d} x}}{\bigg (}x{\mathrm {e} }^{-x}W(L_{n},L_{m})(0){\frac {\mathrm {e} ^{x}}{x}}{\bigg )}\\&=-{\frac {\mathrm {d} }{\mathrm {d} x}}{\bigg (}W(L_{n},L_{m})(0){\bigg )}\end{aligned}}}$

After transforming and separating the variables , the equation is now:

${\displaystyle -\mathrm {e} ^{-x}{\big (}L_{n}{\mathcal {L}}L_{m}-L_{m}{\mathcal {L}}L_{n}{\big )}\mathrm {d} x=\mathrm {d} {\bigg (}W(L_{n},L_{m})(0){\bigg )}}$

On both sides of the equation there are now one-dimensional Pfaffian forms and since is a constant function, the following applies . A suitable parameterization must be selected for the calculation of the remaining Pfaffian form . The integral is now: ${\displaystyle W(L_{n},L_{m})(0)}$ ${\displaystyle \mathrm {d} {\Big (}W(L_{n},L_{m})(0){\Big )}=0}$ ${\displaystyle \varphi (t)=t,\varphi (t_{0})=0,\varphi (t_{1})=\infty ,{\dot {\varphi }}(t)=1}$

${\displaystyle \int _{\varphi }\omega =\int _{0}^{\infty }\omega _{\varphi (t)}({\dot {\varphi }}(t))\,\mathrm {d} t=\int _{0}^{\infty }w{\Big (}L_{n}{\mathcal {L}}L_{m}-L_{m}{\mathcal {L}}L_{n}{\Big )}\mathrm {d} t=0}$.

Accordingly, the integral vanishes along the interval , so that using equation (4) : ${\displaystyle [0,\infty ]}$

${\displaystyle 0=(m-n)\int _{0}^{\infty }\mathrm {e} ^{-x}L_{m}L_{n}\mathrm {d} t}$

This condition can only be met if:

${\displaystyle \langle L_{n},L_{m}\rangle =\langle L_{m},L_{n}\rangle =0}$.

Part 2: In the following it is shown that the Laguerre polynomials are bounded with the weight , for is accordingly , or abbreviated . ${\displaystyle w(x)=\mathrm {e} ^{-x}}$ ${\displaystyle n=m}$ ${\displaystyle \langle L_{n},L_{m}\rangle =\int \mathrm {e} ^{-x}L_{n}(x)L_{m}(x)\mathrm {d} x=1}$ ${\displaystyle \langle L_{n},L_{n}\rangle =1}$

The series representation on the one hand and Rodrigues' formula on the other hand are used for the proof . The following applies: ${\displaystyle L_{n}(x)=\sum _{k=0}^{n}{\frac {(-1)^{k}}{k!}}x^{k}}$ ${\displaystyle L_{n}(x)={\frac {\mathrm {e} ^{x}}{n!}}{\frac {\mathrm {d} ^{n}}{\mathrm {d} x^{n}}}{\bigg (}\mathrm {e} ^{-x}x^{n}{\bigg )}}$

${\displaystyle \langle L_{n},L_{n}\rangle =\int _{0}^{\infty }\mathrm {e} ^{-x}\sum _{k=0}^{n}{\frac {(-1)^{k}}{k!}}x^{k}{\frac {\mathrm {e} ^{x}}{n!}}{\frac {\mathrm {d} ^{n}}{\mathrm {d} x^{n}}}{\bigg (}\mathrm {e} ^{-x}x^{n}{\bigg )}\mathrm {d} x=\sum _{k=0}^{n}{\frac {(-1)^{k}}{k!}}\int _{0}^{\infty }x^{k}{\frac {1}{n!}}{\frac {\mathrm {d} ^{n}}{\mathrm {d} x^{n}}}{\bigg (}\mathrm {e} ^{-x}x^{n}{\bigg )}\mathrm {d} x}$.

For with : ${\displaystyle n=0}$ ${\displaystyle \textstyle {\frac {\mathrm {d} ^{n=0}}{\mathrm {d} x^{n=0}}}{\big (}\mathrm {e} ^{-x}x^{0}{\big )}=\mathrm {e} ^{-x}x^{0}}$

${\displaystyle \langle L_{n},L_{n}\rangle =\int _{0}^{\infty }x^{0}{\bigg (}\mathrm {e} ^{-x}x^{0}{\bigg )}\mathrm {d} x=\int _{0}^{\infty }\mathrm {e} ^{-x}\mathrm {d} x=-{\bigg [}\mathrm {e} ^{-x}{\bigg ]}_{0}^{\infty }=1}$.

If the Laguerre polynomial is now decomposed, it follows: ${\displaystyle n>0}$

${\displaystyle \langle L_{n},L_{n}\rangle =\sum _{k=0}^{n-1}{\frac {(-1)^{k}}{k!}}\int _{0}^{\infty }x^{k}{\frac {1}{n!}}{\frac {\mathrm {d} ^{n}}{\mathrm {d} x^{n}}}{\bigg (}\mathrm {e} ^{-x}x^{n}{\bigg )}\mathrm {d} x+{\frac {(-1)^{n}}{n!}}\int _{0}^{\infty }x^{n}{\frac {1}{n!}}{\frac {\mathrm {d} ^{n}}{\mathrm {d} x^{n}}}{\bigg (}\mathrm {e} ^{-x}x^{n}{\bigg )}\mathrm {d} x.}$

This decomposition reduces the total degree of the polynomial by 1 and the following applies , as shown in part 1 . Only the second term remains, which is calculated with partial integration , i.e.: ${\displaystyle \langle L_{(n-1)},L_{n}\rangle =0}$

${\displaystyle {\begin{aligned}\langle L_{n},L_{n}\rangle &={\frac {(-1)^{n}}{n!}}\int _{0}^{\infty }x^{n}{\frac {1}{n!}}{\frac {\mathrm {d} ^{n}}{\mathrm {d} x^{n}}}{\bigg (}\mathrm {e} ^{-x}x^{n}{\bigg )}\mathrm {d} x\\&={\frac {(-1)^{n}}{n!}}{\bigg [}x^{n}{\frac {1}{n!}}{\frac {\mathrm {d} ^{(n-1)}}{\mathrm {d} x^{(n-1)}}}{\bigg (}\mathrm {e} ^{-x}x^{n}{\bigg )}{\bigg ]}_{0}^{\infty }-n{\frac {(-1)^{n}}{n!}}\int _{0}^{\infty }x^{(n-1)}{\frac {1}{n!}}{\frac {\mathrm {d} ^{(n-1)}}{\mathrm {d} x^{(n-1)}}}{\bigg (}\mathrm {e} ^{-x}x^{n}{\bigg )}\mathrm {d} x\end{aligned}}}$

The antiderivative was calculated using the product rule and it results in the limit value . The same result is obtained in the limit value . Since this result holds for all partial integrations, it follows: ${\displaystyle \textstyle \lim _{x\to 0}x^{n}{\frac {1}{n!}}{\frac {\mathrm {d} ^{(n-1)}}{\mathrm {d} x^{(n-1)}}}{\big (}\mathrm {e} ^{-x}x^{n}{\big )}=\sum _{k=0}^{n-1}\lim _{x\to 0}x^{n}{\frac {1}{n!}}{\binom {n}{k}}{\big (}\mathrm {e} ^{-x}{\big )}^{k}{\big (}x^{n}{\big )}^{(n-k)}=0}$ ${\displaystyle \textstyle \lim _{x\to \infty }}$ ${\displaystyle n}$

${\displaystyle {\begin{aligned}\langle L_{n},L_{n}\rangle &=(-1)^{1}n{\frac {(-1)^{n}}{n!}}\int _{0}^{\infty }x^{(n-1)}{\frac {1}{n!}}{\frac {\mathrm {d} ^{(n-1)}}{\mathrm {d} x^{(n-1)}}}{\bigg (}\mathrm {e} ^{-x}x^{n}{\bigg )}\mathrm {d} x\\&=(-1)^{2}n(n-1){\frac {(-1)^{n}}{n!}}\int _{0}^{\infty }x^{(n-2)}{\frac {1}{n!}}{\frac {\mathrm {d} ^{(n-2)}}{\mathrm {d} x^{(n-2)}}}{\bigg (}\mathrm {e} ^{-x}x^{n}{\bigg )}\mathrm {d} x\\&\;\;\vdots \\&=(-1)^{n}n!{\frac {(-1)^{n}}{n!}}\int _{0}^{\infty }x^{(n-n)}{\frac {1}{n!}}{\frac {\mathrm {d} ^{(n-n)}}{\mathrm {d} x^{(n-n)}}}{\bigg (}\mathrm {e} ^{-x}x^{n}{\bigg )}\mathrm {d} x\\&={\frac {(-1)^{2n}}{n!}}\int _{0}^{\infty }\mathrm {e} ^{-x}x^{n}\mathrm {d} x\\&={\frac {1}{n!}}\int _{0}^{\infty }\mathrm {e} ^{-x}x^{n}\mathrm {d} x\end{aligned}}}$

By means of further -fold partial integration or integration table follows and thus: ${\displaystyle n}$ ${\displaystyle \textstyle \int _{0}^{\infty }\mathrm {e} ^{-x}x^{n}\mathrm {d} x=n!}$

${\displaystyle \langle L_{n},L_{n}\rangle =1}$.

Of Part 1 and Part 2 gives:

${\displaystyle \langle L_{n},L_{m}\rangle =\int _{0}^{\infty }\mathrm {e} ^{-x}L_{n}(x)L_{m}(x)\mathrm {d} x=\delta _{nm}.}$

${\displaystyle \Box }$

Generating function

A generating function for the Laguerre polynomial is

${\displaystyle \sum _{n=0}^{\infty }L_{n}(x)\,t^{n}={\frac {1}{1-t}}e^{-{\frac {tx}{1-t}}}}$

Associated Laguerre polynomials

Some associated Laguerre polynomials

The assigned (generalized) Laguerre polynomials overhang with the ordinary Laguerre polynomials

${\displaystyle L_{n}^{k}(x)=(-1)^{k}\,{\frac {{\rm {d}}^{k}}{{\rm {d}}x^{k}}}\,L_{n+k}(x)\qquad k=0,1,\dotsc }$

together. Your Rodrigues formula is

${\displaystyle L_{n}^{k}(x)={\frac {\mathrm {e} ^{x}\,x^{-k}}{n!}}\,{\frac {{\rm {d}}^{n}}{{\rm {d}}x^{n}}}\,(\mathrm {e} ^{-x}\,x^{n+k}).}$

The assigned Laguerre polynomials satisfy the assigned Laguerre equation

${\displaystyle x\,y''(x)+(k+1-x)\,y'(x)+n\,y(x)=0,\qquad n=0,1,\dotsc }$

The first assigned Laguerre polynomials are:

${\displaystyle L_{0}^{k}(x)=1}$

${\displaystyle L_{1}^{k}(x)=-x+k+1}$

${\displaystyle L_{2}^{k}(x)={\frac {1}{2}}\,\left[x^{2}-2\,(k+2)\,x+(k+1)(k+2)\right]}$

${\displaystyle L_{3}^{k}(x)={\frac {1}{6}}\,\left[-x^{3}+3\,(k+3)\,x^{2}-3\,(k+2)\,(k+3)\,x+(k+1)\,(k+2)\,(k+3)\right]}$

The recursion formula

${\displaystyle (n+1)L_{n+1}^{k}(x)=(2n+1+k-x)L_{n}^{k}(x)-(n+k)L_{n-1}^{k}(x)}$

use.

The Sturm Liouville operator is

${\displaystyle {\mathcal {L}}=-\mathrm {e} ^{x}{\frac {\mathrm {d} }{\mathrm {d} x}}\left(x^{k+1}\mathrm {e} ^{-x}{\frac {\mathrm {d} }{\mathrm {d} x}}\right)}$

and with the weight function : ${\displaystyle \mathrm {e} ^{-x}}$

${\displaystyle \int \limits _{0}^{\infty }\mathrm {e} ^{-x}x^{k}L_{m}^{k}(x)L_{n}^{k}(x)\mathrm {d} x=0\qquad m\neq n}$

${\displaystyle \int \limits _{0}^{\infty }\mathrm {e} ^{-x}x^{k}\left(L_{n}^{k}(x)\right)^{2}\mathrm {d} x={\frac {\Gamma (n+k+1)}{n!}}\qquad n=0,1,\dotsc }$

Assigned Laguerre polynomials can be expressed as path integrals :

${\displaystyle L_{n}^{k}(x)={\frac {1}{2\pi i}}\oint \limits _{C}{\frac {\mathrm {e} ^{-{\frac {xt}{1-t}}}}{(1-t)^{k+1}\,t^{n+1}}}\;dt,}$

There is a path that circles the origin once in a counterclockwise direction and does not include the essential singularity at 1. ${\displaystyle C}$

Hydrogen atom

The Laguerre polynomials are used in quantum mechanics to solve the Schrödinger equation for the hydrogen atom or, in the general case, for a Coulomb potential. Using the assigned Laguerre polynomials, the radial component of the wave function can be written as

${\displaystyle R_{nl}(r)=D_{nl}\,\mathrm {e} ^{-\kappa \,r}\,(2\,\kappa \,r)^{l}\,L_{n-l-1}^{2\,l+1}(2\,\kappa \,r)}$

(Normalization constant , characteristic length , principal quantum number, orbital angular momentum quantum number ). The assigned Laguerre polynomials therefore play a decisive role here. The normalized total wave function is through ${\displaystyle D_{nl}}$ ${\displaystyle \kappa }$ ${\displaystyle n}$ ${\displaystyle l}$

${\displaystyle \Psi _{n,l,m}(r,\vartheta ,\varphi )={\sqrt {\frac {4\,(n-l-1)!}{(n+l)!\;n\,(na_{0}/Z)^{3}}}}\left[{\frac {2r}{na_{0}/Z}}\right]^{l}\exp {\left\{-{\frac {r}{na_{0}/Z}}\right\}}\;L_{n-l-1}^{2l+1}\left({\frac {2r}{na_{0}/Z}}\right)\;Y_{l,m}(\vartheta ,\varphi )}$

optionally, with the principal quantum number , the orbital angular momentum quantum number , the magnetic quantum number , the Bohr radius and the atomic number . The functions are the assigned Laguerre polynomials, the spherical surface functions . ${\displaystyle n}$ ${\displaystyle l}$ ${\displaystyle m}$ ${\displaystyle a_{0}}$ ${\displaystyle Z}$ ${\displaystyle L_{n}^{l}(r)}$ ${\displaystyle Y_{l,m}(\vartheta ,\varphi )}$

Web links

• Eric W. Weisstein : Laguerre Polynomial . In: MathWorld (English).
• Laguerre polynomials. ( Memento from February 29, 2016 in the Internet Archive ). At: ipf.uni-stuttgart.de.
• Laguerre's functions. At: stellarcom.org.
• Radial wave functions, Laguerre polynomials. At: physik.uni-ulm.de.

References and comments

1. ↑ Because of the linear parameterization, the differential can be selected as it is. ${\displaystyle \mathrm {d} t=\mathrm {d} x}$
2. ↑ In physics, the term standardized is usually used instead of restricted.
3. ↑ Harro Heuser : Ordinary differential equations , Vieweg + Teubner 2009 (6th edition), pages 352–354, ISBN 978-3-8348-0705-2