Zero term in Frobenius series in derivation of Bessel functions.

Question

When deriving Bessel functions by solving the Bessel equation \begin{equation} x^2y''+xy'+(x^2-n^2)y=0 \end{equation} using Frobenius method. In the resulting series \begin{equation} y = \sum\limits_{k=0}^{\infty} \frac{(-1)^k n!}{2^{2k} k! (k+n)!}a_0 x^{2k+n} \end{equation} why $a_0$ is chosen as $\frac{1}{2^n n!}$. The factor $n!$ in denominator can be intuitively understood to cancel the same arising in the numerator from recurrence realation. Powers of two are not that clear. Anyway more rigorous explanation is needed. Is it due to normalisation?

Answer 1

I'm not sure what kind of “rigorous explanation” that you're looking for, since it's just a definition. But anyway, I think the choice of $a_0$ is made simply to get a final formula that's convenient to write down. With those powers of two included in $a_0$, the whole thing becomes a power series in $x/2$: $$ y(x) = J_n(x) = \sum_{k=0}^{\infty} \frac{(-1)^k}{k! (n+k)!} \, \left( \frac{x}{2} \right)^{2k+n} . $$