Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / 8

Insert the power series and equate the coefficient of $x^n$. In the Legendre differential equation this givesHence the recurrence relation isThe free coefficients $a_0,a_1$ generate the independent even and odd series, respectively. They converge at least for $|x|<1$, up to the nearest singular points of the equation.

For a nonnegative integer $\ell$, the chain of the same parity as $\ell$ reaches $n=\ell$, where the numerator of the recurrence relation vanishes. The next coefficient and every later coefficient in that chain are zero, producing a polynomial of degree $\ell$. Normalizing it at $x=1$ gives the Legendre polynomial $P_{\ell }$.

The opposite-parity chain never reaches $n=\ell$. Its other possible zero, $n=-\ell -1$, is negative and cannot occur, so a nonzero initial coefficient produces a nonterminating power series. It is linearly independent of $P_{\ell }$ and can be scaled to the Legendre function of the second kind $Q_{\ell }$.

The degree statement requires $\ell \ge 0$, as the examples in part (c) indicate. For a negative integer $\ell$, the same equation has the nonnegative degree $-\ell -1$, because $\ell (\ell +1)=(-\ell -1)(-\ell )$. A polynomial cannot literally have negative degree.

For $\ell =0$ the terminating even solution is constant; for $\ell =1$ the terminating odd solution is proportional to $x$. For $\ell =2$, the recurrence gives $a_2=-3a_0$; for $\ell =3$ it gives $a_3=-5a_1/3$. Imposing the Legendre polynomial normalization at $1$ yields

Let $W=P_{\ell }{Q}_{\ell }^{\prime }-P_{\ell }^{\prime }{Q}_{\ell }$. Differentiation cancels the cross terms, and the Legendre differential equation givesThus $[(1-x^2)W{]}^{\prime }=0$, sowith $A_{\ell }$ independent of $x$ on each interval avoiding the singular endpoints. This is the Abel identity. Linear independence forces $A_{\ell }\ne 0$, and its value depends on the chosen normalization of $Q_{\ell }$.

Since $P_0=1$, part (d) gives $Q_0^{\prime }=A_0/(1-x^2)$. A partial fraction decomposition and integration give, for $-1<x<1$,The condition at zero fixes $B=0$. Thus the full conclusion from the stated condition isThat condition alone does not fix the multiplicative constant. With the usual Legendre function of the second kind normalization $Q_0^{\prime }(0)=1$, one has $A_0=1$. On real exterior intervals replace the logarithm by $\log |(1+x)/(1-x)|$; complex continuation requires a choice of logarithm branch.