I have to show following identity:
$$ j_l(x) \approx \frac{x^l}{(2l+1)!!}\quad \mathrm{for} \; x\rightarrow 0,\quad j_l(x)\approx \frac{1}{x} \sin\left(x-l\frac{\pi}{2} \right) \quad \mathrm{for} \; x\rightarrow \infty $$
I started by using the definition of the spherical bessel function
$$ j_l(x) = x^l \left( -\frac{1}{x} \frac{d}{dx} \right)^l \frac{\sin{x}}{x} $$
and expanding the $\sin x$ using its series representation:
$$ \sin x = \sum_{k=0}^{\infty} \frac{ (-1)^k x^{2k+1}}{ (2k+1)! } $$
$$ \Rightarrow j_l(x) = j_l(x) = x^l \left( -\frac{1}{x} \frac{d}{dx} \right)^l \frac{1}{x} \sum_{k=0}^{\infty} \frac{ (-1)^k x^{2k+1}}{ (2k+1)! }$$
I don't know how to continue from here. It would be great if someone could help me out.
Best Answer
For the zero limit note that you only need to keep the lowest order $x$ that leads to a non-zero value. ie where $2k=2l$. Note also that for each iteration of $(\frac{1}{xdx})$ you lose two powers of $x$. Thus: \begin{equation} (\frac{d}{xdx})^l x^m = x^{m-2l} m!!/(m-2l-2)!! \end{equation} given your series expansion what value of is the minimum necessary to ensure that your term in non-zero? What should the coefficient of that term be?
For $x$ large any variation in $\frac{1}{x}$ will be small thus to lowest order you should focus on taking derivatives of $\sin(x)$.