Series Expansion of Bessel Function of the First Kind/Negative Index

Theorem

Let $n \in \Z_{\ge 0}$ be a (strictly) non-negative integer.

Let $\map {J_n} x$ denote the Bessel function of the first kind of order $n$.

Then:

 $\displaystyle \map {J_{-n} } x$ $=$ $\displaystyle \dfrac {x^{-n} } {2^{-n} \, \map \Gamma {1 - n} } \paren {1 - \dfrac {x^2} {2 \paren {2 - 2 n} } + \dfrac {x^4} {2 \times 4 \paren {2 - 2 n} \paren {4 - 2 n} } - \cdots}$ $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {k! \, \map \Gamma {k + 1 - n} } \paren {\dfrac x 2}^{2 k - n}$

Proof

 $\text {(1)}: \quad$ $\displaystyle \map {J_n} x$ $=$ $\displaystyle \dfrac {x^n} {2^n \, \map \Gamma {n + 1} } \paren {1 - \dfrac {x^2} {2 \paren {2 n + 2} } + \dfrac {x^4} {2 \times 4 \paren {2 n + 2} \paren {2 n + 4} } - \cdots}$ $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {k! \, \map \Gamma {n + k + 1} } \paren {\dfrac x 2}^{n + 2 k}$

The result follows by substituting $-n$ for $n$ in $1$ and simplifying.

$\blacksquare$