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The Bessel function $J_{1/2}$ is a solution of the differential equation:
$$x^2 y '' + xy' +(x^2 - (1/2)^2)y = 0 $$
I am looking to find $J_{1/2}$ using a power series method, but letting $y = \sum_{n=0}^\infty a_n x^n$ gives $a_0 = a_1 = 0$ and $a_n = \frac{a_{n-2}}{n^2 - 1/4}$ which gives all coefficients equal to $0$, which is not desirable.
How do I find the correct power series expansion of $J_{1/2}?$
1 Answer
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If you instead want to use the differential equation, let $$ u(x)=\sqrt{x}y(x). $$ If I did the calculations correctly (you confirm), the differential equation transforms to $$ x^{3/2}(u''(x)+u(x))=0. $$ Please try to take it from here.
