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.

At eigenvalue zero the Laguerre differential operator on polynomials gives $x{y}^{\prime \prime }+(1-x)y^{\prime }=0$, equivalently $(x{e}^{-x}{y}^{\prime }{)}^{\prime }=0$. Thus $y^{\prime }=D{e}^x/x$ and the displayed solutions follow from the exponential integral. For any $x_0>0$, all solutions with $y(x_0)=1$ can be written $y(x)=1+D{\int }_{x_0}^x{e}^{t}dt/t$. The choices $D=0$ and $D=1$ are linearly independent despite sharing that one prescribed value.

The second branch grows as $e^x/x$ at infinity and has a logarithmic singularity at zero. It is consequently excluded from the regular weighted square-integrable eigenproblem with weight $e^{-x}$. Only the constant zero mode remains in that self-adjoint realization. A normalization at one interior point alone is not enough to select a unique second-order solution.

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.

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.

The generating function holds for $|h|<1$ and fixed $x$. From Rodrigues' formula, $L_n(x)={\sum }_{j=0}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right)(-x{)}^j/j!$. Sum first over $n\ge j$, using ${\sum }_{n\ge j}\left(\genfrac{}{}{0ex}{}{n}{j}\right)h^n=h^j/(1-h{)}^{j+1}$, and then over $j$ to obtain the exponential function. The absolute-value double series is finite for $|h|<1$, justifying the rearrangement. Alternatively, the weighted inner product and the Laplace transform of a Laguerre polynomial extract the coefficient $h^n$.

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 $(p-1{)}^2/p^3$.

Set $v=x^n{e}^{-x}$. Direct differentiation gives $x{v}^{\prime }=(n-x)v$. Apply the Leibniz rule to differentiate $n+1$ times:Hence $x{v}^{(n+2)}+(x+1)v^{(n+1)}+(n+1)v^{(n)}=0$. By Rodrigues' formula, $v^{(n)}=n!e^{-x}{L}_n$, so the next two derivatives are $n!e^{-x}(L_n^{\prime }-L_n)$ and $n!e^{-x}(L_n^{\prime \prime }-2L_n^{\prime }+L_n)$. Substitution gives the displayed Laguerre differential operator on polynomials eigenvalue equation. This derivation also covers $n=0$.

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 $e^{-x}$.

The generalized Laguerre polynomial $L_n^{(\alpha )}$ is orthogonal on $[0,\infty )$ for the weight $x^{\alpha }{e}^{-x}$ when $\alpha >-1$.

For integers $m,n\ge 0$, the standard Rodrigues' formula $L_n^{(m)}=e^x{x}^{-m}{D}^n(x^{n+m}{e}^{-x})/n!$ has the endpoint value and squared norm displayed above. The lowest-power term after $n$ differentiations is $(n+m)!x^m/m!$, giving the endpoint value after division by $n!x^m$. For the norm, integrate by parts $n$ times against the same polynomial; its leading coefficient is $(-1{)}^n/n!$, so the final derivative is $(-1{)}^n$. All boundary terms vanish and the remaining Gamma function integral is $(n+m)!/n!$. Normalizing the value at zero to one divides the polynomial by $\left(\genfrac{}{}{0ex}{}{n+m}{n}\right)$ and its squared norm by the square of that factor. Only when $m=0$ or $n=0$ do these normalizations already coincide.