= Solution
For $n\geq1$, the coefficient of $x^{n+\sigma}$ is
$$
\big[2(n+\sigma)(n+\sigma-1)-(n+\sigma)+1\big]a_n+a_{n-1}=0.
$$
Factoring the bracket gives the <recurrence relation>
$$
\boxed{a_n=-\frac{a_{n-1}}{[2(n+\sigma)-1](n+\sigma-1)}}.
$$
For $\sigma=1/2$ this is $a_n=-a_{n-1}/[n(2n-1)]$. Iteration and $(2n)!=2^n n!(2n-1)!!$ yield
$$
\boxed{a_n=\frac{(-1)^na_0}{n!(2n-1)!!}=\frac{(-2)^na_0}{(2n)!}\quad(\sigma=1/2)}.
$$
For $\sigma=1$, the recurrence is $a_n=-a_{n-1}/[n(2n+1)]$, giving
$$
\boxed{a_n=\frac{(-1)^na_0}{n!(2n+1)!!}=\frac{(-2)^na_0}{(2n+1)!}\quad(\sigma=1)}.
$$
Here the <double factorial> products are one at $n=0$, consistent with the chosen normalization. Recognizing the cosine and sine <power series>, for positive $x$ one may also write the two <Frobenius solutions> as
$$
y_{1/2}=a_0\sqrt{x}\cos\sqrt{2x},\qquad
y_1=a_0x\frac{\sin\sqrt{2x}}{\sqrt{2x}}.
$$
The constants multiplying the two independent solutions may be chosen separately.
Back to article page
