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A mathematician associated with Laguerre polynomials and the Laguerre differential equation.

For the weight $e^{-x}$ on $[0,\infty )$, use the zeros of a degree-$n$ Laguerre polynomial as quadrature nodes. Divide a polynomial of degree at most $2n-1$ by that polynomial. The quotient term integrates to zero by orthogonality and vanishes at every node; interpolation of the remainder proves exactness. The two-node rule has nodes $2\pm \sqrt{2}$ and corresponding weights $(2\pm \sqrt{2})/4$, with the larger weight at the smaller node.

For polynomial solutions of the Laguerre differential equation, put $D_n=L_n^{\prime }-L_n$. The equation gives $D_n+t{D}_n^{\prime }=-(n+1)L_n$. Integrating $\int t{D}_n(L_m^{\prime }-L_m)e^{-t}$ by parts gives the displayed identity; the factor $t$ kills the endpoint at zero and exponential decay kills the endpoint at infinity. Its left side is symmetric in $n,m$, so subtracting the swapped identities proves orthogonality for distinct degrees. This provides a direct energy proof of orthogonality for the Laguerre polynomials without assuming their norm formula.

This equation has a regular singular point at zero and an irregular singular point at infinity. For a nonnegative integer $\nu =n$, the solution analytic at zero with value one is the Laguerre polynomial $L_n$. Its weighted self-adjoint form is $(x{e}^{-x}{y}^{\prime }{)}^{\prime }+\nu e^{-x}y=0$, connecting it to the Laguerre differential operator on polynomials.

On polynomials with weighted inner productthe operatoris self-adjoint because $e^{-x}Lf=(x{e}^{-x}{f}^{\prime }{)}^{\prime }$. Its eigenvalues on polynomials of degree at most $n$ are $0,-1,\dots ,-n$; its eigenvectors are scalar multiples of the Laguerre polynomials.

The ordinary Laguerre polynomials, normalized by $L_n(0)=1$, are orthonormal for the weighted inner product displayed above. Their coefficient formula is $L_n(x)={\sum }_{k=0}^n(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})x^k/k!$. To prove the relation, use Rodrigues' formula $e^{-x}{L}_n=(n!{)}^{-1}(x^n{e}^{-x}{)}^{(n)}$ and integrate by parts $n$ times against $L_m$. For $n>m$ its $n$th derivative vanishes. At zero every derivative of $x^n{e}^{-x}$ of order less than $n$ vanishes, and at infinity the exponential kills every polynomial boundary term. For $n=m$, $L_n^{(n)}=(-1{)}^n$, so the integral is $(n!{)}^{-1}{\int }_0^{\infty }{x}^n{e}^{-x}dx=1$. Symmetry covers $m>n$. Thus these are orthonormal without an extra normalization factor.

For the ordinary Laguerre polynomial $L_n(t)={\sum }_{j=0}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right)(-t{)}^j/j!$, apply the Laplace transform to each term, using $\mathcal{L}[t^j](p)=j!/p^{j+1}$ for $\mathrm{Re}p>0$. The binomial theorem givesThis proves the formula for every nonnegative integer $n$. In particular $L_2(t)=1-2t+t^2/2$ has transform $(p-1{)}^2/p^3$.

The Laguerre polynomial normalized by $L_n(0)=1$ has coefficients $L_n(x)={\sum }_{j=0}^n(-1{)}^j(\genfrac{}{}{0ex}{}{n}{j})x^j/j!$. By the Leibniz rule, this is equivalent to Rodrigues' formula, $e^{-x}{L}_n(x)=D^n(e^{-x}{x}^n)/n!$. For any polynomial $q$ of degree below $n$, $n$ applications of integration by parts give ${\int }_0^{\infty }{e}^{-x}{L}_{n}qdx=0$. All endpoint terms vanish: derivatives of $e^{-x}{x}^n$ of order below $n$ vanish at zero and decay at infinity. Since $L_n^{(n)}=(-1{)}^n$, the same calculation givesThus these orthogonal polynomials are already an orthonormal set for the weighted inner product with weight $e^{-x}$.

Substitute $y={\sum }_{k\ge 0}{a}_k{x}^k$ and compare the coefficient of $x^k$. The two derivative terms give $(k+1{)}^2{a}_{k+1}$, while the remaining terms give $(n-k)a_k$. ThusIterating the recurrence givesIndeed the recurrence first vanishes at $k=n$ and then remains zero. Consequently the Laguerre polynomial iswith exact degree $n$ and leading coefficient $(-1{)}^n/n!$. This is the regular power-series branch normalized at zero; the second branch at the repeated indicial root is not an additional analytic solution with this normalization.

For $m=0$ the two normalizations agree, and the Laguerre polynomials are orthonormal with weight $e^{-x}$. Take real $-1<h<1$. Their expansion coefficients for the requested function areSubstitute Rodrigues' formula and integrate by parts $n$ times. Differentiating $e^{-xh/(1-h)}$ transfers a factor $[h/(1-h){]}^n$ after the two sign factors cancel. ThereforeThe integral is $n!(1-h{)}^{n+1}$ by the Gamma function. Boundary terms vanish throughout this range.

The resulting generating function for Laguerre polynomials isOne can also verify convergence and equality directly: Rodrigues' formula gives $L_n(x)={\sum }_{j=0}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right)(-x{)}^j/j!$, and summing ${\sum }_{n\ge j}\left(\genfrac{}{}{0ex}{}{n}{j}\right)h^n=h^j/(1-h{)}^{j+1}$ reduces the double series to the exponential on the left. The absolute-value double series converges for fixed $x$ and $|h|<1$, justifying the interchange; the formula then also holds for complex $h$ in that disk.

Write $y={\sum }_{k\ge 0}{c}_k{x}^k$ with $c_0=1$. Substitution into the ordinary differential equation givesUnless the series terminates, its coefficient ratio is $O(1/k)$, so it converges for every finite $x$. A coefficient first vanishes exactly when $\lambda$ is a nonnegative integer $n$, and then $c_n=(-1{)}^n/n!\ne 0$, $c_{n+1}=0$. Thus $y$ is a polynomial precisely for $\lambda =n\in \{0,1,2,\dots \}$, in which caseis the normalized Laguerre polynomial.

Multiplying the differential equation by $e^{-x}$ gives $(x{e}^{-x}{y}_n^{\prime }{)}^{\prime }+n{e}^{-x}{y}_n=0$. Multiply by $y_m$, subtract the equation with the indices exchanged, and integrate over $[0,\infty )$. The boundary term $x{e}^{-x}(y_m{y}_n^{\prime }-y_n{y}_m^{\prime })$ vanishes both at zero and infinity. ThereforeThis is the weighted orthogonality of the Laguerre polynomials.

Their leading coefficients show $\mathrm{deg}(y_m{y}_n)=m+n$. Since $y_0,\dots ,y_{m+n}$ have distinct degrees, they form a basis of that polynomial space. Hence $y_m{y}_n={\sum }_{p=0}^{m+n}{a}_p{y}_p$, and comparison of leading coefficients gives $a_{m+n}=(m+n)!/(m!n!)\ne 0$. Likewise the degrees of $y_m$ and $y_m{y}_n$ for $0\le m<n$ run through $0,\dots ,n-1$ and $n,\dots ,2n-1$, respectively. A nontrivial linear combination cannot cancel its highest-degree term, so these $2n$ functions are linearly independent.

For $n=2$, this gives the basis $y_0,y_1,y_2,y_1{y}_2$ of the cubic polynomial space, establishing the specified expansion. Its weighted integral is $a_0$: $y_0=1$ integrates to one, $y_1$ and $y_2$ are orthogonal to $y_0$, and $y_1{y}_2$ integrates to zero. At the two roots ${\alpha }_1,{\alpha }_2$ of $y_2$, the expansion reduces to $f({\alpha }_i)=a_0+a_1{y}_1({\alpha }_i)$. Solving these two equations for $a_0$ givesFor explicit verification, the recurrence gives $y_1=1-x$, $y_2=1-2x+x^2/2$. Thus ${\alpha }_1=2-\sqrt{2}$, ${\alpha }_2=2+\sqrt{2}$ are distinct positive roots and $w_1=(2+\sqrt{2})/4$, $w_2=(2-\sqrt{2})/4$. This is the two-node Gauss-Laguerre quadrature rule, exact for every cubic.

A linear operator $\alpha$ on a real inner-product space is self-adjoint whenfor all $v,w$.

To prove the finite-dimensional spectral theorem, first note that the continuous quadratic function $⟨\alpha v,v⟩$ has a maximum on the unit sphere. The Lagrange multiplier equation at a maximizing vector $v$ gives $\alpha v=\lambda v$, so $\alpha$ has a real unit eigenvector. Its orthogonal complement is invariant becauseInduction on the dimension supplies an orthonormal eigenbasis of that complement and hence of $V$.

For $f\ne 0$,while symmetry and bilinearity are immediate, so the displayed formula defines an inner product on $P_n$. Sinceintegration by parts givesThe boundary term vanishes for polynomials, proving that $\alpha$ is self-adjoint.

On the monomial $x^k$,Thus the matrix of $\alpha$ in the monomial basis is triangular with diagonalThese are therefore its eigenvalues. For $n=2$, corresponding eigenvectors arefor eigenvalues $0,-1,-2$, respectively. They are the first Laguerre polynomials up to normalization, as described by the Laguerre differential operator on polynomials.

Write $\lambda =N$ with $N\in {\mathbb{Z}}_{\ge 0}$. ThenTake ${\gamma }_1$ to be a small positively oriented circle around $t=0$. For ${\gamma }_2$, start at $t=1$, travel to $-\infty$, and avoid the pole at zero by a fixed upper or lower detour. The endpoint factor vanishes at $t=1$, and the exponential controls $-\infty$.

The finite contour extracts the residue at zero:Only powers $z^0,\dots ,z^N$ occur, and the coefficient of $z^N$ is nonzero. Thus this solution is a polynomial of degree $N$ and, up to normalization, is the Laguerre polynomial $L_N(z)$:This also follows from the recurrence in part (i), because $a_{N+1}=0$ and every later coefficient vanishes.

Apply the Leibniz rule to the $n$th derivative in Rodrigues' formula. If $j$ derivatives fall on $e^{-x}$ and $n-j$ on $x^n$, the corresponding term is $\left(\genfrac{}{}{0ex}{}{n}{j}\right)(-1{)}^j{e}^{-x}(n!/j!)x^j$. Cancelling the prefactor givesThe leading coefficient $(-1{)}^n/n!$ is nonzero, so this is a polynomial of degree exactly $n$. In particular, the requested coefficients in $L_{10}$ areThese are the ordinary Laguerre polynomials, with normalization $L_n(0)=1$.

Write a second-order linear ordinary differential equation in monic form $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$. At a finite point $x_0$, an ordinary point has analytic $P,Q$. A singular point is a regular singular point if $(x-x_0)P$ and $(x-x_0{)}^{2}Q$ extend analytically there. A singular point failing this condition is an irregular singular point. The point at infinity is classified by the reciprocal coordinate $t=1/x$.

For the Laguerre differential equation, $P=1/x-1$ and $Q=\nu /x$. Thus $xP=1-x$ and $x^{2}Q=\nu x$ are analytic at zero, while $P$ itself is singular: zero is regular singular. Put $Y(t)=y(1/t)$. Since $y^{\prime }=-t^2{Y}^{\prime }$ and $y^{\prime \prime }=t^4{Y}^{\prime \prime }+2t^3{Y}^{\prime }$, the transformed equation isHere $t{P}_t=1+1/t$ is not analytic at zero, so infinity is an irregular singular point, for every real $\nu$.

For the Frobenius solution, the lowest power $x^{\sigma -1}$ has coefficient $a_0{\sigma }^2$. Hence the indicial equation is ${\sigma }^2=0$, and equating the remaining powers givesWith the repeated root $\sigma =0$, the Frobenius series for the Laguerre differential equation obeysIf the series does not terminate, its consecutive term ratio is asymptotic to $x/k$, so the ratio test gives infinite radius of convergence. It is an entire function, even though the differential equation has a singular coefficient at zero. The repeated indicial root yields only one analytic Frobenius branch; an independent solution generally has a logarithmic term there.

For $\nu =n\ge 0$, the factor $k-n$ stops the recurrence at degree $n$. With $a_0=1$ its coefficients are $a_k=(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})/k!$, giving the Laguerre polynomialsThe same termination argument includes $n=0$, as required for $L_0$ despite the earlier reference to positive integers.

For two real eigenfunctions, multiply the equation for $y_n$ by $y_m$ and the equation for $y_m$ by $y_n$, then subtract. The $q$ terms cancel. The resulting Lagrange identity for a Sturm-Liouville operator isIntegrating first over $[ϵ,R]$ and then taking the endpoint limits givesThe needed assumptions are that the improper integral exists and that the displayed boundary expression tends to zero at both endpoints. Bounded $y_m,y_n$ and their first derivatives are sufficient because $p\to 0$, but are stronger than necessary. Merely knowing that $p\to 0$ does not control arbitrarily growing eigenfunctions. With the vanishing boundary form and distinct eigenvalues, division by ${\lambda }_n-{\lambda }_m$ proves orthogonality of Sturm-Liouville eigenfunctions:For complex eigenfunctions in a real self-adjoint differential operator problem, use $\overline{y_m}$ instead, giving the corresponding weighted inner product.

Multiplying the particular equation by $e^{-x}$ puts it in Sturm-Liouville form:ConsequentlyIts polynomial eigenfunctions are the Laguerre polynomials. Products of these polynomials, with or without one derivative, grow only algebraically at infinity. The exponential in $p$ therefore makes the boundary form vanish there; its factor $x$ makes it vanish at zero. Their orthogonality is thus

For $\lambda =1$, substitution of $Ax+B$ gives $A+B=0$. For $\lambda =2$, substitution of $A{x}^2+Bx+C$ gives $4A+B=0$ and $B+2C=0$. Choosing $y_n(0)=1$, which fixes the otherwise arbitrary nonzero multiplicative constants, yieldsTheir product is $1-3x+\frac{5}{2}{x}^2-\frac{1}{2}{x}^3$. Using the Gamma function identity ${\int }_0^{\infty }{x}^k{e}^{-x}dx=k!$, also obtainable by repeated integration by parts, verifies the required orthogonality directly:

For real positive weight $w$, the weighted inner product isThe functions must have finite weighted norm; the conjugate is omitted for real functions. Multiplying the differential equation by $x{e}^{-x}$ givesThus the Sturm-Liouville form has $p=x{e}^{-x}$, $q=0$ and $w=e^{-x}$. With the positive-sign spectral convention defineThe existing Laguerre differential operator on polynomials uses the opposite operator $L=-\mathcal{A}$, with eigenvalues $-n$; keeping this sign convention explicit avoids confusing the two descriptions.

For admissible $f,g$, integration by parts givesThe boundary form must vanish for every pair in the self-adjoint domain. In the usual regular-at-zero Laguerre realization one imposes ${\lim }_{x\downarrow 0}x{e}^{-x}{y}^{\prime }=0$, excluding the logarithmic branch; at infinity one uses the weighted square-integrable domain with vanishing boundary form. A condition $y(\infty )=0$ would be inappropriate for polynomial eigenfunctions. For any pair of polynomials, the factor $x$ makes the boundary form zero at zero, while the exponential beats all polynomial growth at infinity. These functions also have finite weighted norms.

Self-adjointness yields $(n-m)⟨y_m,y_n{⟩}_w=0$, giving orthogonality of Sturm-Liouville eigenfunctions:For the specified polynomials, direct differentiation gives $\mathcal{A}u=u$ and $\mathcal{A}v=3v$. Hence their eigenvalues in the displayed convention are 1 and 3, respectively, or −1 and −3 for $L=-\mathcal{A}$.

To find the quadratic, put $h=a{x}^2+bx+c$. The equation $x{h}^{\prime \prime }+(1-x)h^{\prime }+2h=0$ reduces to $(4a+b)x+(b+2c)=0$. Thus $b=-4a$, $c=2a$. Since $h(1)=-a=1$,These are scalar multiples of the Laguerre polynomials of degrees one, three and two.

For $n=0$, the first integral is $x{e}^{-x}{y}^{\prime }=C$, so all solutions with the prescribed interior normalization areTwo independent choices areThey both equal one at $x_0$, and their Wronskian is $e^x/x\ne 0$. For sufficiently large $x$, the last unit interval contributes at least $e^{x-1}/x$, proving that $y_2\to \infty$. These are the zero-eigenvalue solutions of the Laguerre differential equation, with the second expressible through the exponential integral. It also has a logarithmic singularity at zero and is not an admissible polynomial or weighted square-integrable eigenfunction. An interior value alone does not enforce the endpoint conditions of the spectral problem.

Multiplication by a positive weight puts the differential equation into Sturm-Liouville form if $(xw{)}^{\prime }=(1-x)w$. Thus $x{w}^{\prime }=-xw$, and the constant factor can be chosen so thatThis fixes the otherwise arbitrary normalization of the weight consistently with the requested unit norms.

Write the polynomial as $p_n(x)={\sum }_{j=0}^n{c}_j{x}^j$. Comparing coefficients givesConsequently $c_j=(-1{)}^j(\genfrac{}{}{0ex}{}{n}{j})/j!$, identifying the solutions as Laguerre polynomials. In particular,Repeated integration by parts gives the moments ${\int }_0^{\infty }{e}^{-x}{x}^{j}dx=j!$. ThereforeFor later use, integration by parts in the weighted eigenvalue equations also givesThe endpoint expression vanishes because of the factor $x$ at zero and exponential decay at infinity. Hence distinct Laguerre polynomials are orthogonal, an instance of orthogonality of Sturm-Liouville eigenfunctions.