Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 2 / 15D (Methods)

Substitute the power series into $(1-x^2)y^{\prime \prime }-2x{y}^{\prime }+\lambda y=0$. The coefficient of $x^k$ is $(k+2)(k+1)a_{k+2}+[\lambda -k(k+1)]a_k$, soStarting from $a_0,a_1$ determines the even and odd coefficient sequences separately. Unless a sequence terminates, the ratio of successive same-parity coefficients has absolute value tending to one, so its series converges for $|x|<1$ and solves the Legendre differential equation there.

With $\lambda =n(n+1)$, choose the parity matching $n$ and set the other starting coefficient to zero. All factors before index $n$ are nonzero, whereas the factor at $k=n$ vanishes. Thus the chosen series is a nonzero polynomial of exactly degree $n$. Its value at one cannot vanish: a zero of multiplicity $j\ge 1$ there would give a nonzero leading term $-2j^2(x-1{)}^{j-1}$, times its local leading coefficient, in the differential equation. We can therefore scale it to $P_n(1)=1$.

For $n=2$, $a_1=0$, $a_2=-3a_0$, and all later coefficients vanish. Normalization gives $a_0=-1/2$, yieldingFor even degree this scale is fixed by $a_0$ as in the question; for odd degree $a_0=0$ by parity and the scale is fixed by $a_1$.

The axis-regular separation of variables expansion for an axisymmetric harmonic potential isThe radial powers solve $r^2{R}^{\prime \prime }+2r{R}^{\prime }-n(n+1)R=0$, while the regular angular solutions are Legendre polynomials. For an exterior solution tending to zero at infinity, retain only the negative radial powers. The boundary polynomial decomposes as $1+x^2=\frac{4}{3}{P}_0(x)+\frac{2}{3}{P}_2(x)$. ThereforeIt has the required boundary value and decay. Uniqueness follows from the maximum principle for harmonic functions for a harmonic difference, using its zero boundary values and vanishing limit at infinity.