Past exam of the differential equations and applications course of the University of Oxford / 2011 / A / 3

After division by $x$, the ordinary differential equation has coefficients $P(x)=(m+1)/x-1$ and $Q(x)=n/x$. At zero, $xP\to m+1$ and $x^{2}Q\to 0$, making zero a regular singular point. The indicial equation isso the indicial exponents are $\boxed{0,-m}$, coincident when $m=0$.

At infinity put $t=1/x$ and $Y(t)=y(1/t)$. The transformed equation isThe coefficient of $Y^{\prime }$ already fails the regular-singularity test at $t=0$. Thus $\boxed{x=\infty \text{ is an irregular singular point}}$, including when $n=0$.

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.