Find the limits of the spherical bessel-function

Question

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.

Answer 1

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 $\sin(x)$.

Answer 2

$x \to 0$

For this case consider the relation

$$ j_l(x) = \sqrt{\frac{\pi}{2x}}J_{l + 1/2}(x) \tag{1} $$

with $J_\alpha(x)$ being the Bessel function of the first kind,

$$ J_\alpha(x) = \sum_{n=0}^{+\infty}\frac{(-1)^n}{n!\Gamma(n + \alpha + 1)} \left(\frac{x}{2} \right)^{\alpha + 2n} ~~\stackrel{x\to 0}{\approx}~~ \frac{1}{\Gamma(\alpha + 1)} \left(\frac{x}{2} \right)^\alpha \tag{2} $$

Putting together (1) and (2):

$$ j_l(x) \approx \frac{\pi^2}{2^{1/2}x^{1/2}}\frac{x^{l+1/2}}{2^{l+1/2}\Gamma(l + 1/2 + 1)} = \frac{\pi^{1/2}x^l}{2^{l+1}\Gamma(l +1/2 + 1)} \tag{3} $$

We can work out the value of $\Gamma$ as using the fact that $\Gamma(x + 1) = x \Gamma(x)$

\begin{eqnarray} \Gamma(l + 1/2 + 1) &=& (l + 1/2)\Gamma(l + 1/2) = (l + 1/2)(l - 1/2)\Gamma(l-1/2)\\ &=& (l + 1/2)(l-1/2)(l-3/2)\cdots (1/2)\Gamma(1/2) \\ &=& 2^{-(l+1)}(2l+1)(2l - 1)(2l - 3) \cdots (1) \Gamma(1/2)\\ &=& 2^{-(l + 1)}(2l + 1)!! \pi^{1/2} \tag{4} \end{eqnarray}

Replacing (4) in (3) we get the asymptotic behavior for small $x$

$$ \lim_{x\to 0} j_l(x) = \frac{x^l}{(2l + 1)!!} $$

$x\to \infty$

For this one consider the spherical Hankel functions

\begin{eqnarray} h_n^{(1)}(x) &=& j_n(x) + i y_n(x) \\ h_n^{(2)}(x) &=& j_n(x) - i y_n(x) \tag{5} \end{eqnarray}

with

$$ h_l^{(1)}(x) = (-i)^{l+1}\frac{e^{ix}}{x}\sum_{n=0}^{l}\frac{i^n}{n!(2x)^n}\frac{(l + n)!}{(l - n)!} ~~\stackrel{x\to \infty}{\approx}~~ -i\frac{e^{ix}}{x}(-i)^l = -i\frac{e^{i(x - l\pi/2)}}{x} $$

Using Eq. (5) you can conclude that

$$ \lim_{x\to \infty}j_l(x) = \frac{1}{x}\sin(x - l\pi/2) $$