= Solution
Termwise differentiation of the convergent <Frobenius series> and collection of powers in the <Bessel differential equation> gives the lowest coefficient $(\sigma^2-\nu^2)a_0$, the next coefficient $[(\sigma+1)^2-\nu^2]a_1$, and, for $n\geq2$, the coefficient $[(\sigma+n)^2-\nu^2]a_n+a_{n-2}$. Since $a_0\ne0$, the stated choices $\sigma=\pm\nu$, $a_1=0$ and the recurrence make all coefficients vanish.
For $\nu=1/2$, the two resulting even-coefficient series are
$$
\sigma=\tfrac12:\quad a_0x^{1/2}\left(1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots\right)=a_0\frac{\sin x}{\sqrt x},
$$
and
$$
\sigma=-\tfrac12:\quad a_0x^{-1/2}\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots\right)=a_0\frac{\cos x}{\sqrt x}.
$$
These are linearly independent on $x>0$ and give the <Bessel functions of order one half>, up to normalization. The specified condition $a_1=0$ is sufficient; in the negative-exponent branch at this particular order the next coefficient equation alone would allow $a_1$ to be free, adding a multiple of the positive-exponent solution.
For the <Legendre differential equation>, substitution gives the full conditions
$$
\boxed{\sigma(\sigma-1)=0,\qquad \sigma(\sigma+1)a_1=0,\qquad
a_{n+2}=\frac{(\sigma+n)(\sigma+n+1)-\lambda(\lambda+1)}{(\sigma+n+2)(\sigma+n+1)}a_n\quad(n\geq0).}
$$
Thus $\sigma=0$ permits arbitrary $a_1$, while $\sigma=1$ requires $a_1=0$. This <Frobenius recurrence for the Legendre equation> separates the even and odd series. The nearest singular points of the differential equation are $\pm1$, so the series about zero has radius at least one, and the generic nonterminating solutions have radius exactly one. Particular cancellations can give a larger radius.
Termination occurs when the degree reaches a nonnegative integer $N$ with $\lambda(\lambda+1)=N(N+1)$. For $\lambda\geq0$ this gives
$$
\boxed{\lambda=N\in\{0,1,2,\ldots\}.}
$$
Choose the even branch for even $N$ or the odd branch for odd $N$ to obtain a <Legendre polynomial>. The recurrence then terminates and gives a solution for all real $x$, with infinite radius of convergence. A nonzero entire solution at other $\lambda\geq0$ is impossible: each nonterminating parity has coefficient ratio tending to one in the variable $x^2$, and so radius one.
Back to article page
