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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 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 $h^n$.

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 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.

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 $e^{-x}$.

A mathematician associated with Rodrigues' formula and the Rodrigues rotation formula.

Assume first that $n\ge m$. Substituting Rodrigues' formula for $L_n$ givesHere integration by parts was performed $n$ times. Each boundary term vanishes: at infinity a polynomial times $e^{-x}$ tends to zero; at zero every derivative of $x^n{e}^{-x}$ of order less than $n$ vanishes. If $n>m$, the derivative $L_m^{(n)}$ is zero, proving orthogonality. The case $m>n$ follows by exchanging the two factors.

For $m=n$, the leading coefficient found above gives $L_n^{(n)}=(-1{)}^n$. ThereforeThe same conclusion includes $n=0$ directly. Thus the full Laguerre orthonormality relation is

There is a normalization conflict in the source: the printed Rodrigues' formula gives the standard Generalized Laguerre polynomial $P_n=L_n^{(m)}$ with $P_n(0)=\left(\genfrac{}{}{0ex}{}{n+m}{n}\right)$, which is generally not one. The norm requested here corresponds to that Rodrigues normalization. A solution normalized to one at zero is ${\hat{P}}_n=P_n/\left(\genfrac{}{}{0ex}{}{n+m}{n}\right)$.

Set $w=x^m{e}^{-x}$ and $p=x^{m+1}{e}^{-x}$. The equation becomes $(p{P}_n^{\prime }{)}^{\prime }+nw{P}_n=0$. Multiply the equations for $P_n,P_k$ by the opposite polynomial, subtract and integrate. The boundary term $p(P_n^{\prime }{P}_k-P_k^{\prime }{P}_n)$ vanishes at zero and infinity, soThis proves orthogonality for $k\ne n$ under either normalization.

For the squared norm, use the printed Rodrigues' formula and integrate by parts $n$ times:All boundary terms vanish: the required derivatives have a positive power of $x$ at zero and exponential decay at infinity. The leading coefficient is $(-1{)}^n/n!$, so $P_n^{(n)}=(-1{)}^n$. The Gamma function integral givesFor the source's alternative unit-at-zero normalization the correct answer instead isBoth cannot equal the printed norm when $m,n>0$. This distinction is the normalization of generalized Laguerre polynomials.

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.

Choose real monic orthogonal polynomials $p_j$ of degree $j$, $j=0,1,\dots$, for $w(x)=x^a{e}^{-x}$ on $[0,\infty )$, and letExtend each ${\varphi }_j$ by zero to negative $x$. The resulting functions form an orthonormal set in $L^2(\mathbb{R},dx)$. Define the orthogonal polynomial projection kernelThe formula on the right is for nonnegative arguments; the kernel is zero when either argument is negative. With ${\Phi }_{ij}={\varphi }_{j-1}(x_i)$, the kernel evaluation matrix is $\Phi {\Phi }^T$. By the determinant multiplication identity and the preceding monic basis change,Thus initially ${\hat{c}}_m=c_m{\prod }_j{h}_j$.

Use the symmetric labelled joint probability density on the full orthant. Expanding the two copies of $\det \Phi$ into permutations and integrating all variables, orthonormality makes a term vanish unless the two permutations agree. Each of the $m!$ surviving terms integrates to $1$. HenceIf the eigenvalues are instead listed in increasing order on a single ordered chamber, that chamber's density is $m!$ times this symmetric density. We keep the full-orthant convention throughout.

An explicit choice is $p_j(x)=(-1{)}^{j}j!L_j^{(a)}(x)$ in terms of the Generalized Laguerre polynomials. To verify the normalization independently, their Rodrigues' formula givesIntegrating by parts $j$ times proves orthogonality to every lower-degree polynomial. The boundary terms vanish at zero and infinity for $a\ge 0$. Taking the other factor to be $p_j$ and using $p_j^{(j)}=j!$ gives $h_j=j!\Gamma (a+j+1)$, where $\Gamma$ is the Gamma function. The Laguerre projection kernel is thereforefor $x,y\ge 0$.

Evaluate $R_n$ at $x=1$ using $(x^2-1{)}^n=(x-1{)}^n(x+1{)}^n$. In the Leibniz rule for the $n$th derivative, every term vanishes at $1$ except the term putting all $n$ derivatives on $(x-1{)}^n$. Therefore $R_n(1)=n!2^n$. Combining this with $P_n(1)=1$ gives the normalisation in the Rodrigues' formula:For $n=0$, the formula gives ${\alpha }_0=1$, as required. As an independent check, the leading term of $R_n$ is $(2n)!x^n/n!$, so the resulting leading coefficient of $P_n$ is $(2n)!/(2^n(n!{)}^2)$, consistent with the Legendre polynomial recurrence relation.

The displayed Rodrigues' formula gives the Legendre polynomialA four-node Gaussian rule is exact through degree $2\cdot 4-1=7$. Solving $q_4(x)=0$ first for $x^2$ gives the four nodesThe rule is not exact for every polynomial of degree eight: its quadrature value on $q_4^2$ is zero because every node is a root of $q_4$, whereas

Suppose $m<n$. The Rodrigues' formula and $n$ integrations by parts givebecause $p_m^{(n)}=0$. Every boundary term vanishes: the exponential controls infinity, while the remaining power of $x$ has positive exponent at zero. Symmetry of the inner product gives orthogonality whenever $m\ne n$.

For the norm, monicity gives $p_n^{(n)}=n!$, so the same calculation yields

Write $v=(x+1{)}^l(x-1{)}^l$. In the Leibniz rule for its $l$th derivative at one, any term with fewer than $l$ derivatives on $(x-1{)}^l$ vanishes. The sole surviving term leaves $(x+1{)}^l$ undifferentiated and differentiates the second factor $l$ times, givingThe normalization in Rodrigues' formula consequently givesThe same argument covers $l=0$, with both zeroth powers equal to one.

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 $L_n(0)=1$.

The coefficient formula from part (b) and the Leibniz rule yield Rodrigues' formula directly:The leading coefficient is $(-1{)}^n/n!$, so $p_n^{(n)}=(-1{)}^n$. Substitute this differentiated expression into one factor in the squared weighted norm and apply integration by parts $n$ times:Every endpoint term contains a derivative of $e^{-x}{x}^n$ of order less than $n$: it vanishes at zero, and its exponential decay beats every polynomial at infinity. This proves the normalization of the ordinary Laguerre polynomials for all $n$, including $n=0$ without any integrations by parts.

This instance of Rodrigues' formula constructs the Legendre polynomial for every nonnegative integer $l$. Put $v=(x^2-1{)}^l$; then $(x^2-1)v^{\prime }=2lxv$. Differentiating $l+1$ times by the Leibniz rule gives $(x^2-1)y^{\prime \prime }+2x{y}^{\prime }-l(l+1)y=0$ for $y=v^{(l)}$, which is the Legendre differential equation after reversing signs. The leading term of the normalized derivative is $(2l)!x^l/[2^l(l!{)}^2]$, proving its polynomial degree is exactly $l$. At $x=1$, the product $(x+1{)}^l(x-1{)}^l$ contributes only the term with all $l$ derivatives on the second factor, giving $v^{(l)}(1)=2^{l}l!$. Thus the normalization is $P_l(1)=1$.